1 Groebner Bases

Diane MaclaganHill 240 maclagan@mathOffice Hour M1-2Text: Eisenbud, recommended: Atiyah-MacdonaldOutline:Grobner BasesLocalizationAssociated PrimesIntegral DependenceBlow-ups/filtrationsFlatnessCompletionDimensionCM ringsGrobner Basics

Definition 1.1 (Affine Variety). Let S = k[x1, . . . , xn] and let I be an ideal ofS. Fact: S is Notherian so I = 〈f1, . . . , fk〉.

The affine variety V (I) = {v = (v1, . . . , vn) ∈ kn : f(v) = 0∀f ∈ I} = {v ∈kn : fi(v) = 0, 1 ≤ i ≤ k}.

For example, S = k[x, y] and I = 〈x − y〉 gives the line y = x and V (y2 −x3 + x) is an elliptic curve.

Definition 1.2 (Projective Space). Projective space, Pnk is the set of linesthrough the origin in kn+1. We write a point in Pn as v = (v0 : . . . : vn)and v ∼ v′ if v = λv′ for some λ ∈ k∗.

Definition 1.3 (Projective Variety). If we set S = k[x0, . . . , xn], and gradeS by deg(xi) = 1 for 0 ≤ i ≤ n, then a polynomial is homogeneous if everymonomial has the same degree. In fact, if f ∈ S is homogeneous of degree d,we can ask if f(~v) = 0 for ~v ∈ Pn, because f(λ~v) = λdf(~v).

A Projective Variety is V (I) = {~v ∈ Pn : f(~v) = 0∀f ∈ I homogeneous }

So, we ask the question, given I ⊂ k[x1, . . . , xn] = S, f ∈ S is f ∈ I? ie, ifI = 〈f1, . . . , fr〉, are there h1, . . . , hr ∈ S with f ∈

∑ri=1 hifi.

e.g., is x + 7 ∈ 〈x2 − 4x + 3, x2 + x − 2〉 ⊂ k[x]? Well, k[x] is a PID, soI = 〈f〉 for some f ∈ k[x], and x+ 7 ∈ I iff f |(x+ 7). So we use the EuclideanAlgorithm.

The Euclidean Algorithm gives x− 1 as the gcd, so I = (x− 1), so x+ 7 /∈ Ias x− 1 6 |x+ 7.

How about in several variables? Is x + 3y − 2z ∈ 〈x + y − z, y − z〉, ie, is(1, 3,−2) ∈ 〈(1, 1,−1), (0, 1,−1)〉? We can use Gaussian Elimination to say no.

Now, is xy2 − x in 〈xy + 1, y2 − 1〉? Naive attempts at division don’t work,but xy2 − x = x(y2 − 1), so it IS in the ideal.

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Definition 1.4 (Grobner Basis (Vague)). A Grobner Basis for an ideal I is agenerating set for which long division decides the ideal membership problem.

Definition 1.5 (Term Order). A term order is a total order on the monomialsin S = k[x1, . . . , xn] such that

1. 1 < xu := xu11 . . . xun

n for all u ∈ Nn \ {0}

2. xu < xv ⇒ xu+w < xv+w.

Examples include lexicographic ordering, that is, xu < xv if the first entryof v − u is positive, for example y7 < xz2 < x2.

Graded lex, xu < xv if deg(xu) < deg(xv) or deg(xu) = deg(xv) and xu <lexxv.

Reverse Lex (degree revlex), xu < xv if deg(xu) < deg(xv) or the last nonzeroelement of v − u is negative. eg, xz < y2 < z3.

Definition 1.6 (Lead Term, Initial Ideal). The leading term of a polynomialf ∈ I is the largest monomial appearing in it with respect to a term order, e.g.f = 3x2 − 7xy + 8z2 with lex gives in<(f) = x2.

The initial ideal in<(I) = 〈in<(f) : f ∈ I〉. WARNING: if I = 〈f1, . . . , fr〉,then in<(I) ⊇ 〈in(f1), . . . , in(fr)〉 but they are not, in general, equal.

eg, I = 〈xy + 1, y2 − 1〉, x + y ∈ I, y(xy + 1) − x(y2 − 1), so if < is lex,in(x+ y) = x ∈ in(I) 6= 〈xy, y2〉.

Definition 1.7 (Grobner Basis). A Grobner Basis for I ⊂ S is a generatingset G = {g1, . . . , gs} for I for which in(I) = 〈in(g1), . . . , in(gS)〉.

Point: We can define a division algorithm. Order the Grobner basis anddivide the polynomial by multiplying elements of the Grobner basis to cancelthe leading term of f if possible, otherwise pass to the next monomial, etc.

If G is a Grobner basis, then division by G will have remainder 0 if and onlyif the polynomial is in the ideal.

Facts: Every ideal in k[x1, . . . , xn] has a (finite) Grobner basis, and thereexists an algorithm called the Buchberger Algorithm to compute it.

Definition 1.8 (Division Algorithm). Input: f , {g1, . . . , gk}Output: Remainder on dividing f by {g1, . . . , gk}.Set f ′ = f , r = 0. While in(f ′) ∈ 〈in(g1), . . . , in(gk)〉, let j be the smallest

index for which in(f ′) = xuin(gj). Set f ′ = f ′ − lc(f ′)/lc(gj)xugjIf f ′ = 0, return r otherwise r = r+ lc(f ′)in(f ′) and f ′ = f ′ − lc(f ′)in(f ′),

and return to the while loop.

Note: this algorithm terminates because a term order has no infinite descend-ing chains. Also, this algorithm writes f =

∑higi + r for some polynomials

hi ∈ S with in(higi) ≤ in(f).

Proposition 1.1. If G = {g1, . . . , gs} is a Grobner basis for I then the remain-der on dividing f by G is 0 iff f ∈ I.

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Proof. If r = 0 then f ∈ I since f =∑higi.

Conversely, if f ∈ I, then in the while lop, f ′ ∈ I and we only leave it whenf ′ = 0, so r = 0.

Definition 1.9 (S-Pair). If f, g ∈ S their S-pair is S(f, g) = lcm(in(f),in(g))lc(f)in(f) f −

lcm(in(f),in(g))lc(g)in(g) g.

eg, if f = 3x2−7y2 and g = 8xy+z2, so S(f, g) = x2y3x2 (3x2−7y2)− x2y

8xy (8xy+z2) = − 7

3y3 − 1

8xz2.

The S is for syzygy.

Algorithm 1 (Buchberger). Input: {f1, . . . , fs} generating I, and a term order<.

Output: A Grobner basis for I wrt <.

1. Current={(fi, fj) : 1 ≤ i < j ≤ s}, G = {f1, . . . , fs}

2. While Current 6= ∅, do Pick (f, g) ∈Current, Current=Current\(f, g), r =remainderon dividing S(f, g) by G . If r 6= 0, then G = G∪{r} and Current=Current∪{(r, f) :f ∈ G }.

3. Output G

Corollary 1.2. If {g1, . . . , gs} ⊂ I and 〈in(g1), . . . , in(gs)〉 = in(I) then {g1, . . . , gs}generate I.

Definition 1.10 (Minimal Grobner Basis). A GB is minimal if each in(gi)is a minimal generator of in(I) and each minimal generator appears once in{in(gi)}.

Definition 1.11 (Reduced Grobner Basis). A Grobner basis is reduced if forall g ∈ G remainder on dividing g by G \ {g} is g.

Example: There is a unique reduced GB for each term order.

Proof. We must check that Buchberger’s Algorithm terminates and gives thecorrect answer.

At stage i of the algorithm, set Ii = 〈in(g) : g ∈ G 〉. Note that Ii+1 ) Ii.If the algorithm did not terminate, we would get an infinite ascending chainI1 ( I2 ( I3 ( . . . which would contradict the fact that S is Notherian.

If the output is not a Grobner basis, then there is f ∈ I, f =∑higi with

gi ∈ G , hi ∈ S with in(f) /∈ 〈in(g) : g ∈ G 〉. Write m = max in(higi), we willassume that m is minimal for such a counterexample. Let Im = {i : in(higi) =m}, since m ∈ 〈∈ (g) : g ∈ G 〉, we must have in(f) < m. Thus, |Im| ≥ 2.Second assumption, Im is minimal for such expressions.

Pick i, j ∈ I . Write S(gi, gj) =∑pjgk, pk ∈ S. Since in(gi), in(gj)|m,

lcm(in(gi), in(gj))|m, so there exit c ∈ k,m′ monomial such that in(cm′`igi) =in(higi) = in(cm′ljgj), so cm`igi = cm′(`jgj +

∑pkgk), where in(cm′pkgk) =

m′in(pkgk) < m.

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Replace higi + hjgj by (hi − cm′`i)gi which has initial term < m as does(hj + cm′`j)gj , so

∑cm′pkgk has initial term < m.

This gives either a set with smaller |Im| or smaller m, contradicting ourminimality assumption.

Applications of the Division Algorithm

1. Given I ⊂ k[x1, . . . , xn], compute I ∩ k[x2, . . . , xn] = I ′. V (I ′) ⊂ kn−1

is the closure of the projection of V (I) to kn−1. The algorithm is tocompute a lex GB for I with x1 largest and take the polynomials withoutx1 in them.

2. As V (I ∩ J) = V (I) ∪ V (J), we may want to compute I ∩ J . ComputeK = tI + (1− t)J ⊂ S[t], then compute K ∩ S.

3. I : J = 〈f |fg ∈ I for all g ∈ J〉. This is, geometrically, the closure ofV (I) \ V (J). I : J = ∩(I : fi) where J = {f1, . . . , fs}, and I : f iscomputed by computing I ∩ (f) and then dividing the generating set byf .

2 Hom, Tensor and Localization

Definition 2.1 (homR(M,N)). homR(M,N) is the set of R-module homo-morphisms from M to n, ie, ϕ : M → N is a group homomorphism withϕ(rm) = rϕ(m). It is, in fact, a group with (φ + ψ)(m) = φ(m) + ψ(m), andhas an R-module structure by (rφ)(m) = r(φ(m)).

This means homR(M,−) is a covariant functor from R-mod to R-mod. Themap on objects takes N to homR(M,N). If α : N → N ′ is an R-modulehomomorphism, then homR(M,α) : homR(M,N) → homR(M,N ′) by φ 7→α ◦φ. Similarly, homR(−, N) is a contravariant functor from R-mod to R-mod.

Proposition 2.1. hom is left exact. That is, if 0 → Aα→ B

β→ C then

0→ homR(M,A)homR(M,α)→ homR(M,B)

homR(M,β)→ homR(M,C).

WARNING: If 0→ A→ B → C → 0, we don’t expect 0→ homR(M,A)→homR(M,B)→ homR(M,C)→ 0 to be exact.

Example: 0 → Z x2→ Z → Z/2 → 0, apply hom(Z/2,−). This is where Extcomes from.

In general, 0 → A → B → C → 0 and if F is a functor, you don’t expect0 → F (A) → F (B) → F (C) → 0 to be exact. If f is left (right) exact we getderived functors from taking cohomology.

Tensor Product

Definition 2.2 (Tensor). If M,N are R-modules then M ⊗R N is the abeliangroup that is the quotient of the free abelian group on the symbols {m⊗n : m ∈M,n ∈ N} modulo the relations (am + bm′) ⊗ (cn + dn′) = acm ⊗ n + adm ⊗n′ + bcm′ ⊗ n+ bdm′ ⊗ n′. It is an R-module by r(m⊗ n) = rm⊗ n = m⊗ rn.

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Remember, not everything in M ⊗N is of the form m⊗ n.Example: What are the elements in kmn = km ⊗k kn of the form m ⊗ n?

The answer is the rank 1 matrices. These can be written as uvT where u ∈ kmand v ∈ kn.

Z/2⊗Z Z/3 = 0 since a⊗ b = 3a⊗ b = a⊗ 3b = a⊗ 0 = 0.Change of coefficients: k[x]⊗k R ' R[x] when R is a k-algebra.

Proposition 2.2. The tensor product satisfies the following universal property:if ϕ a bilinear map ϕ : M ×N → P that is bilinear, then there exists a uniqueϕ such that

P

M ×N M ⊗N.................................................................... ............

............................................................................................................................................................................ ............

ϕ

.............................................................................................................................

ϕ

Geometric InterpretationGiven X ⊆ kn, we can form I(X) = {f ∈ k[x1, . . . , xn] : f(x) = 0∀x ∈ X}.

Hilbert’s Nullstellensatz says that if k is algebraically closed, then I(V (I)) =√I.

Coordinate ring of a variety V is S/I(V ), and maps of varieties correspondto maps of rings in the other direction (see algebraic geometry for details).

If M is the coordinate ring of V , N is for U and R is for W . If f : V → Wand g : U → W , then the fiber product of V and U over W is F = {(u, v) ∈U × V : f(u) = g(v)}. The universal property it satisfies is that ϕ : Z → U × Vsuch that f ◦ π1 ◦ ϕ = g ◦ π2 ◦ ϕ thenZ

U × V

F

VU

W

................................................................................................................................ ............

.................................................................................................................................................................................................. ............

∃!

...................................................................................................................................................................................................................................................................................

....................................................................................................................................................................................................................................................................................................... ............

............................................................................................................................................

π1

................................................................................................................................ ............

π2...............................................................

...............

.................................................................. ............

................................................................................................................................ ............

f

............................................................................................................................................

g

Point: the map V →W gives a map R→M which makes M an R-moduleby rm = ϕ(r)m.

Claim: M⊗RN is the coordinate ring of the fiber product f , eg, k[x1, . . . , xn]⊗kk[y1, . . . , ym] ' k[x1, . . . , xn, y1, . . . , ym].

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Claim: −⊗RM is a right exact functor from R-mod to R-mod.If ϕ : M →M ′, then ϕ⊗N : M ⊗R N →M ′ ⊗R N comes from the bilinear

map ψ : M ×N →M ′ ⊗N by ψ(m,n) = ϕ(m)⊗R n.Now we check right exactness. Suppose A → B → C → 0 is exact. We

want to show that A ⊗R Nα⊗1→ B ⊗R N

β⊗1→ C ⊗R N → 0. To see that β ⊗ 1is surjective, note that if c ⊗ n ∈ C ⊗N then there is b ∈ B with β(b) = c, soβ ⊗ 1(b⊗ n) = c⊗ n.

To show that the other step is exact, we’ll show that C ⊗N ' B ⊗R N/α⊗1(A⊗R N). We’ll do this by checking that B ⊗N/A⊗N satisfies the universalproperty of C ⊗ N . Suppose that ϕ : C × N → P is a bilinear map withϕ(rc, n) = ϕ(c, rn) = rϕ(c, n). Define ψ : B ×N → P by ψ(b, n) = ϕ(β(b), n).Then ψ is bilinear. Thus, there exists a unique ψ : B ⊗R N → P an R-modulehomomorphism. We now show that α ⊗ 1(A⊗N) ⊆ ker ψ. ψ(α ⊗ 1(a⊗ n)) =ψ(α(a)⊗ n) = ϕ(β ◦ α(a), n) = ϕ(0, n) = 0.

Thus, we get a unique induced R-mod homomorphism ψ : B⊗RN/α⊗1(A⊗RN)→ P so by the universal property, B ⊗R N/α⊗ 1(A⊗R N) ' C ⊗R N .

Warning: Tensor is not always left exact! 0 → Z → Z → Z/2 → 0, tensorwith Z/2, and get Z⊗Z2 → Z⊗Z/2→ Z/2⊗Z/2→ 0 Specifically, Z⊗Z/2 'Z/2, but the map given by multiplication by 2 is the zero map, so it is notinjective.

Definition 2.3 (Flat Module). An R-module M is flat iff − ⊗R M is exact.That is, if P → P ′ is an injection, so is P ⊗M → P ′ ⊗M .

LocalizationMotivation: We put the Zariski Topology on kn, the closed sets are of the

form V (I) for some I. We ask: what are the rational functions defined every-where on kn \ V (f)? Well, they’re of the form p/f i where p ∈ S, i ≥ 0, that is,elements of S[f−1].

Definition 2.4 (Localization). Let U ⊂ R be a multiplicatively closed set(u, u′ ∈ U ⇒ uu′ ∈ U , 1 ∈ U).

For an R-module M , we set M [U−1] = {(m,u) : m ∈ M,u ∈ U}/ ∼ where(m,u) ∼ (m′, u′) iff ∃v ∈ U such that v(u′m − um′) = 0. We write (m,u) asm/u.

If M = R, then R[U−1] is a ring, with (r/u)(r′/u′) = rr′/uu′.

Example: R = M = Z, U = Z \ {0}, then R[U−1] = Q.Check: If M is an R-module, then M [U−1] is also an R-module by r(m,u) =

(rm, u) and (m,u) + (m′, u′) = (u′m+ um′, uu′).Example: If R is a domain, then U = R \ {0} gives R[U−1] is the field of

fractions, or quotient ring, of R. In general, let U = {nonzero divisors in R},then K(R) = R[U−1] is the total quotient ring of R.

Example: R = k[x], U = {xi : i ≥ 0}, then R[U−1] = k[x, x−1] is the ring ofLaurent Polynomials.

Warning: The localization of a nonzero module can be zero!Example: R = Z, M = Z/5. Let U = {5i : i ≥ 0}. M [U−1] = 0 as

(m,u) ∼ (0, 1) for all m ∈ Z/5 as 5(1m− u0) = 0.

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Proposition 2.3. Let U be a multiplicatively closed set of R and M an R-module, let ϕ : M →M [U−1] be an R-module homomorphism ϕ(m) = m/1.

Then ϕ(m) = 0 iff there is u ∈ U with um = 0, and if M is a finitelygenerated R-module, then M [U−1] is zero iff there is a u ∈ U that annihilatesM .

Proof. m/1 = 0/1 iff ∃u ∈ U with u(1m− u0) = um = 0.Suppose M is finitely generated by {m1, . . . ,ms}. If there exists u ∈ U

such that um = 0 for all m ∈ M , then m/u′ = 0/1 for all m/u′ ∈ M [U−1].Conversely, suppose that M [U−1] = 0, then mi/1 = 0 for all i, so there existsui such that uimi = 0 for each i, let u =

∏ui. Then uM = 0.

Notation: For any U ⊂ R, we’ll denote by R[U−1] the localization R[U−1]where U is the multiplicative closure of U .

The most important example is if P is a prime ideal of R and U = R \ P .Notation: R[(R \ P )−1] = RP , and M [(R \ P )−1]P .The residue class field κ(P ) = RP /PP , where PP is ϕ(P )RP .If R = Z, then Z0 = Q, κ(0) = Q. P = (p), so ZP = {a/b : p 6 |b}, and

PP = {a/b : p|a, p 6 |b}. Then κ(P ) = ZP /PP ' Z/P .RP is an example of a local ring.

Definition 2.5 (Local Ring). A ring R is local if it has a unique maximal ideal.

Proposition 2.4. Let ϕ : R→ R[U−1] be the map r 7→ r/1.For any ideal I ⊆ R[U−1], we have I = ϕ−1(I)R[U−1]. Thus that map

I 7→ ϕ−1(I) is an injection on the set of ideals in R[U−1] to the set of ideals ofR. This preserves inclusions and intersections, and takes primes to primes.

An ideal J is of the form ϕ−1(I) for some ideal I ⊂ R[U−1] iff J =ϕ−1(JR[U−1]) iff for u ∈ U , ur ∈ J ⇒ r ∈ J for r ∈ R. In particular,I 7→ ϕ−1(I) gives a bijection between the primes in R[U−1] and the primes ofR not meeting U .

Proof. I 7→ ϕ−1(I) gives an injection {primes of R[U−1]} → {primes of R}.Suppose that J is ϕ−1(I). Then ur ∈ J implies r ∈ J for u ∈ U , r ∈ R, soJ ∩ U = ∅, as otherwise u ∈ J ∩ U , then u1 ∈ J , so 1 ∈ J , so J = R.

Corollary 2.5. RP is a local ring.

Note: If ϕ : R→ S is a ring homomorphism with ϕ(u) a unit of S for all u ∈U , then if v(u′r−r′u) = 0, ϕ(v(u′r−ur′)) = 0, ϕ(v)(ϕ(u′)ϕ(r)−ϕ(u)ϕ(r′)) = 0.

Since ϕ(u) is a unit, ϕ(u′)ϕ(r)−ϕ(u)ϕ(r′) = 0 can be written as ϕ(r)ϕ(u)−1 =ϕ(r′)ϕ(u′)−1. So we can define ϕ : R[U−1]→ S by ϕ(r/u) = ϕ(r)ϕ(u)−1.

In fact, we have a universal property of localization: If ϕ : R → S is a ringhomomorphism with ϕ(u) a unit for all u ∈ U , then ∃!ϕ such that the followingdiagram commutes:

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R[U−1]

R S..................................................................................................................................................................... .........

................................................................................................................................

∃!ϕ

................................................................................................................. ............ϕ

If ϕ : M → N is an R-module homomorphism, then ϕ : M [U−1]→ N [U−1]defined by ϕ(m/n) = ϕ(m)/u is an R[U−1] homomorphism.

Check well defined: m/u = m′/u′ implies ∃v with v(mu′ − um′) = 0, sov(u′ϕ(m)− uϕ(m′)) = 0.

Lemma 2.6. The map of R-modules α : R[U−1]⊗RM →M [U−1] by α(r/u⊗m) = rm/u is an isomorphism.

Proof. We must first check that α is well defined. Define α : R[U−1] ×M →M [U−1] by (r/u,m) 7→ rm/u. This is bilinear and α(rs/u,m) = α(r/u, sm) =rsm/u, so we get a map on the tensor product.

Now define and inverse map β : M [U−1] → R[U−1] ⊗R M by m/u 7→1/u ⊗R m. This is well defined, as m′/u′ = m/u means that for some v ∈ U ,v(um′ − u′m) = 0, so β(m/u) = 1

u ⊗R m = vu′/vuu′ ⊗R m = 1vuu′ ⊗R vu

′m =β(m′/u′).

Check that it is an R-module homomorphism.Finally, check that β = α−1. β ◦ α(r/u ⊗ m) = β(rm/u) = 1/u ⊗ rm =

r/u⊗m.α ◦ β(m/u) = α(1/u⊗m) = m/u.

Lemma 2.7. R[U−1] is a flat R-module.

Proof. Suppose that 0→ Aα→ B → C → 0 is exact. It is enough to show that

A⊗R[U−1] α⊗1→ B⊗R[U−1] is injective. These are isomorphic, by the previouslemma, to A[U−1]→ B[U−1] with α(a/u) = α(a)/u is injective.

Suppose α(a/u) = 0 so there exists v ∈ U with vα(a/u) = 0, vα(a)/u) = 0in B[U−1], so there is v ∈ U with v′(vα(a) − u0) = 0, so v′v(α(a)) = 0, soα(v′va) = 0, α is injective, so v′va = 0, so a/u is 0 in A.

Exercise: M1,M2 ⊆ M are R-modules, show that (M1 ∩ M2)[U−1] =M1[U−1] ∩M2[U−1].

Hint: 0→M1 ∩M2 →M →M/M1 ⊕M/M2 is exact.Next: True Locally often implies True Globally

Lemma 2.8. Let R be a ring, M an R-module

1. If m ∈ M then m = 0 if and only if , /1 = 0 in each localization of M ata maximal prime m of R.

2. M = 0 iff Mm = 0 for each maximal ideal m of R.

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Proof. m/1 is zero in Mm if any only if ∃u /∈ m with um = 0. ie, the annihilatorof m in M , I = {r ∈ R : rm = 0} is not contained in m. So if m/1 = 0 in everylocalization of M at a maximal prime, then I is not contained in any maximalideal of R, which is a contradiction, so m = 0.

Corollary 2.9. Let α : M → N be an R-module homomorphism. Then α isinjective, surjective or isomorphism iff αm : Mm → Nm is injective, surjectiveor isomorphism for all maximal ideals m of R.

Proof. We have 0 → kerα → Mα→ N

s→ cokerα → 0. So 0 → ker(α)m →Mm → Nm → coker(α)m → 0 so if αm is injective for all m, then ker(α)m = 0for all m so kerα = 0, similarly for cokerα.

Next: (S is an R-algebra) S ⊗R hom(M,N)→ homS(S ⊗RM,S ⊗R N) bys⊗R ϕ 7→ s⊗ ϕ = s(1⊗ ϕ). Check that this is well-defined.

Define α : S×homR(M,N)→ homS(−,−) by α(ϕ) = s(1⊗ϕ) ∈ homS(S⊗RM,S⊗RN) is bilinear and respects the R-action, so α is well defined. We’ll seethat it is an isomorphism if S is a flat R-module and M is finitely presented.

Recall: M is finitely generated iff ∃α such that R2 α→ M → 0 is exact. Itfinitely presented if kerα is also finitely generated.

Fact: If R is Notherian then finitely generated implies finitely presented.A corollary of this is that if M is finitely presented, then homR(M,N) lo-

calizes, that is, homR[U−1](M [U−1], N [U−1]) ' homR(M,N)[U−1].

Lemma 2.10. If R is Notherian and M is finitely generated, then every sub-module of M is finitely generated. Thus, every finitely generated module isfinitely presented.

Proof. If M is f.g., then ∃ Rs ϕ→M → 0, and kerϕ is a submodule of the finitelygenerated R-module Rs, so the second sentence follows from the first.

Now suppose that N is a submodule of M , where M is gen by m1, . . . ,ms.The proof is by induction on s. If s = 1, then M ' Rm1, we let I = ker(R →M), so M ' R/I. Thus, ϕ−1(N) is an ideal in R, so ϕ−1(N) is finitely generatedby n1, . . . , nk, since R is Notherian, so ϕ(n1), . . . , ϕ(nk) generate N .

Now suppose that the lemma holds for all s′ < s. Consider N ⊆ M/Rm1

is generated by m2, . . . ,ms, so by induction N is generated by {g1, . . . , g` :gi ∈ N, 1 ≤ i ≤ `}. Also, N ∩ Rm1 is finitely generated by h1, . . . , hr. So forn ∈ N , n =

∑rigi and n −

∑rigi ∈ N ∩ Rm1, so n −

∑rigi =

∑rjhj so

g1, . . . , g`, h1, . . . , hr generate N .

This is NOT true if R is not Notherian. eg R = k[x1, x2, . . .], M = R,N = (x1, . . . , xn, . . .) is an infinitely generated submodule of M .

Next: Move towards localization of hom.

Proof. We first prove the proper when M = R. αR : S ⊗R homR(R,N) →homS(S ⊗R R,S ⊗R N) by s⊗ ϕ 7→ s(1⊗ ϕ).

S ⊗ ϕ(1)→ s((1⊗ ϕ)(1⊗ 1)) = s⊗ ϕ(1).

9

Similarly, if M = Rs, then αRs : S ⊗R Ns → (S ⊗ N)s, so finally, we takeM finitely presented.

As such, we have R` → Rs → M → 0. We write −′ for S ⊗ −, and so weapply this and get (R`)′ → (Rs)′ → M ′ → 0, we apply homR(−, N) and thenS ⊗−, and we get the correct diagram and apply the five lemma.

3 Primary Decomposition

Consider V (xy) ⊆ C2. Then V (xy) = V (x) ∪ V (y), so we can break it intoirreducible varieties.

Proposition 3.1. A variety V (I) ⊆ kn can be written uniquely as V1 ∪ . . .∪Vkwhere Vi are irreducible subvarieties and no Vi ⊂ Vj for i 6= j.

Proof. We first show that such a decomposition exists. Let S be the set ofall varieties V that do not have such a decomposition into irreducibles. Sincek[x1, . . . , xn] is Notherian, S has a minimal element, V (I). Since V (I) is in S ,it is not irreducible, so we can write V (I) = V1∪V2 for V1, V2 proper subvarieties.One of these must not have an irreducible decomposition, else V (I) would, butthis contradicts minimality of V (I)

For uniqueness, suppose that V (I) = V1 ∪ . . . ∪ Vk = V ′1 ∪ . . . ∪ V ′` . SoV ′1 = V ′1 ∩V (I) = (V ′1 ∩V1)∪ . . .∪ (V ′1 ∩Vk). V ′1 is irreducible, and each of theseis a subvariety of V ′1 , so as V ′1 is irreducible, we must have some Vi such thatV ′1 ∩Vi = V ′1 so V ′1 ⊆ Vi. Then Vi = (V ′1 ∩Vi)∪(V ′2 ∩Vi)∪ . . .∪(V ′` ∩Vi), so thereis a j such that Vi = V ′j ∩ Vi, so Vi ⊆ V ′j , so V ′1 ⊆ Vi ⊆ V ′j , so j = 1. ConsiderZ = V ′2 ∪ . . . V ′` = V1 ∪ . . . ∪ Vi ∪ . . . ∪ Vk = V (I) \ V ′1 , and then induction.

Note: If I =√I and V (I) is irreducible, I is prime.

Definition 3.1 (Associated Prime). Let R be a ring and M an R-module. Aprime P of R is associated to M if it is the annihilator of an element of M .We write AssR(M) for the set of all associated primes of M as an R-module.

Notation: If I ⊆ R is an ideal, write AssR(I) for AssR(R/I). We can getaway with this, because AssR(I) is rarely interesting, eg, if R is a domain, thenAssR(M) = {(0)} for M = I.

eg, if P is prime, AssR(P ) = {P}.eg, if V = V1 ∪ . . .∪Vk, is the irreducible decomposition of a variety V , then

I(Vi) will be the associated primes of I(V ).eg, R = Z, AssZ(n) = AssZ(Zn) =set of prime factors

Lemma 3.2. If R is Notherian then R[U−1] is Notherian.

Proof. Let I be an ideal in R. Then I = ϕ−1(I)R[U−1] where ϕ : R→ R[U−1]has ϕ(r) = r/1. Then ϕ−1(I) is an ideal of R, so finitely generated and, ϕ ofthose generators must generate I.

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This shows that ”Notherian” is ”better” than ”finitely generated over afield”, ie, quotient of a polynomial ring.

Lemma 3.3 (Prime Avoidance). Suppose that I1, . . . , In, J are ideals of R andJ ⊆ ∪nj=1Ij. If R contains an infinite field, or if all but two of the Ij are prime,then J is contained in one of the Ij. If R is Z-graded and J is generated byhomogeneous elements of deg > 0, and all the Ij are prime, then it is enoughto assume that all homogeneous elements of J are contained in ∪jIj.

Proof. If R contains an infinite field k, then J is a k-vector space, also eachJ ∩Ij is a k-vector subspace. If J 6⊆ Ij for all j, then J ∩Ij is a proper subspaceof J , but J = ∪nj=1J ∩ Ij , and we cannot write a vector space over an infinitefield as a finite union of proper subspaces. If one Ij ⊆ J , then quotient by itand repeat.

Now consider a general R, but all but two of the Ij are prime. The proof isby induction on n. If n = 1, then J ⊆ I1.

If n = 2, J ⊆ I1 ∪ I2, if J 6⊂ I1, J 6⊂ I2 we can find x1 ∈ J ∩ I1 \ I2 andx2 ∈ J ∩ I2 \ I1. But then x1 +x2 ∈ J1, so x1 +x2 ∈ I1∪ I2, but x1 +x2 /∈ I1, I2,a contradiction.

Suppose n > 2. J ⊆ ∪Ij , J 6⊂ Ij . So again we take xi ∈ J∩Ii\∪i 6=jIj . Afterreordering, we may assume that I1 is prime. Consider f = x1 +

∏nj=2 xj ∈ J .

x1 ∈ I1 \∪nj=2Ij , the product is in ∩nj=2Ij \ I1, since I1 is prime. So f ∈ ∪Ii butf /∈ Ij for any j, a contradiction.

Finally, assume R is graded. The Proof is almost the same. In the n = 2case, we need to consider xk1 +x`2 for some k, ` to make xk1 +x`2 homogeneous.

Proposition 3.4. Let R be a ring and M an R-module. If I is maximal amongall ideals of R that are annihilators of elements of M then I is prime andso belongs to AssR(M). Thus, if R is Notherian, AssRM is nonempty and∪P∈AssR MP = 0 ∪ {zero divisors of M}. (that is, r ∈ R nonzero such that∃m 6= 0 with rm = 0)

Proof. Let P be such an ideal maximal with respect to the property of an-nihilating an element of M . Let rs ∈ P for r, s ∈ R. Let m ∈ M haveP = AnnR(m). Then if sm = 0, we have s ∈ P . Otherwise (rs)m = r(sm) = 0so r ∈ AnnR(sm). But also if p ∈ P , psm = s(pm) = 0, so r + P ⊆ AnnR(sm).Since sm 6= 0, AnnR(sm) 6= R, so AnnR(sm) = P . So r ∈ P . This showsthat P is prime. We now check that ∪P∈AssR(M)P = 0 ∪ {zero divs}. If0 6= p ∈ P ∈ AssR(M), then ∃m ∈ M with pm = 0, so p is a zero divisoron M . This shows ⊆. Conversely, if r 6= 0 is a zero divisor, then ∃m ∈ Mwith rm = 0, so r ∈ AnnRM . Let P be an ideal containing AnnRM that ismaximal with respect to annihilating some element of M . Then P is prime soP ∈ AssRM , so r ∈ ∪P .

Corollary 3.5. Suppose that M is a module over a Notherian ring R.

1. If m ∈ M , then m = 0 iff m/1 = 0 ∈ MP for all maximal associatedprimes of M .

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2. If K is a submodule of M , then K = 0 iff KP = 0 for all maximalP ∈ AssM .

3. If ϕ : M → N is an R-module homomorphism, then ϕ is an injective iffϕp : Mp → Np is an injection for each P ∈ AssM .

Proof. If 0 6= m ∈ M , then ∃P ∈ AssM containing Annm. Then some P ∩Annm = {0}, we get m/1 6= 0 in MP . If AnnM = 0, then any P works. Wetake P maximal to get the result. Part 1 implies 2 and 2 implies 3.

Q: How can we find ALL associated primes?

Lemma 3.6. If R is a Notherian ring and M is a finitely generated R-module,then M has a filtration 0 = M0 ( M1 ( . . . ( Mn = M with Mi+1/Mi ' R/Pifor some prime Pi of R.

Proof. If M 6= 0 then there is an associated prime P1 ∈ Ann(m1) for m1 ∈ M .Set M1 = Rm1 ' R/P1. If M1 6= M , consider M/M1. This has an associatedprime P2 = Ann(m2), set M2 = M1 +Rm2. By construction, M2/M1 ' R/P2.Continue in this fashion, this must terminate with some Mi = M , else we wouldhave an infinite ascending chain of submodules of M . This is impossible as Mis finitely generated over a Notherian ring.

Lemma 3.7. M is an R-module.

1. If M = M ′ ⊕M ′′, then AssR(M) = AssR(M ′) ∪AssR(M ′′)

2. If 0 → M ′i→ M

p→ M ′′ → 0 is a s.e.s. of R-modules, then AssR(M ′) ⊆AssR(M) ⊆ AssR(M ′) ∪AssR(M ′′).

Proof. 1. Follows from 2.

2. Suppose m ∈ M ′ with AnnR(m) = P prime. Then AnnR(i(m)) = P , soO ∈ AssR(M).

Now suppose P ∈ AssR(M) \R (M ′). P = AnnR(m) for m ∈ M . SoRm ' R/P . Now for all r ∈ R with rm 6= 0, we have AnnR(rm) = P ,since s ∈ P ⇒ srm = 0 and if srm = 0 then sr ∈ P , and r /∈ P (sincerm 6= 0) so s ∈ P . This means that Rm ∩ i(M ′) = {0}.We now claim that AnnR(p(m)) = P since Rp(m) = p(Rm) ' Rm =R/P , so P ∈ AssR(M ′′).

Corollary 3.8. If R is Notherian, M finitely generated, and 0 = M0 ( M1 (. . . ( Mn = M with Mi/Mi−1 ' R/Pi, then AssR(M) ⊆ {P1, . . . , Pn}, soAssR(M) is a finite nonempty set.

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Proof. When n = 1, M = R/P1, which has AssR(R/P1) = {P1}.For n > 1, 0 → M1 → M → M/M1 → 0 is ses, both ends have filtrations,

the first 0 ( M1, the second 0 ( M2/M1 ( M3/M1 ( . . . ( M/M1. Byinduction, AssR(M1) ⊆ {P1}, and AssR(M/M1) ⊆ {P2, . . . , Pn}, so AssR(M) ⊆{P1, . . . , Pn}.

eg, k[x, y]/(x2y, xy2) has 0 ( Ry2/(x2y, xy2) ( R{x2, y2}/(x2y, xy2) (R{xy, x2, y2}/(xy2, x2y) ( M .

Now Ry2/(x2y, xy2) = R/(x) and R{x2, y2}/(xy2, x2y)/Ry2/(xy2, x2y) =Rx2/(x2y, xy2) ' R/(x) andM3/M2 ' R/(x, y). SetM4 = R{x, y2}/(xy2, x2y),M4/M3 ' R/(x, y). Set M5 = R{x, y}/(x2y, xy2), M5/M4 ' R/(x, y), andM = R{1}/(x2y, xy2) and M/M5 ' R/(x, y). So we now have a filtration0 ( M1 ( M2 ( M3 ( M4 ( M5 ( M , so P1 = (x), P2 = (y), P3 = P4 = P5 =(x, y).

Warning: There is not always a filtration with all Pi associated! e.g. R =k[x, y, z, w], I = (x, y) ∩ (z, w) = (xz, xw, yz, yw).

Claim: AssR(I) = {(x, y), (z, w)}. Claim 2: There is no filtration 0 ( M1 (M with M1 ' R/(x, y), M/M1 ' R/(z, w).

Theorem 3.9. R Notherian, M is finitely generated.

1. Associated primes commutes with localization, ie AssR[U−1](M [U−1]) ={PR[U−1]|P ∈ AssR(M) with P ∩ U = ∅}

2. AssR(M) contains all primes minimal over AnnR(M).

Proof. If P ∈ AssR(M), then there exists m ∈ M with P = AnnR(m). SoRm ' R/P , and we get an inclusion R/P → M . As localization is exact,(R/P )[U−1] → M [U−1] is an inclusion, so if P ∩ U = ∅, PR[U−1] is prime inR[U−1], then PR[U−1] ∈ AssR[U−1](M [U−1]).

Conversely, suppose that Q ∈ AssR[U−1](M [U−1]), then Q = PR[U−1] forsome prime P of R with P ∩U = ∅. Since R is Notherian, we know that R[U−1]is, so PR[U−1] is finitely generated. Thus, R[U−1]/PR[U−1] is finitely pre-sented. Thus, homR[U−1](R[U−1]/PR[U−1],M [U−1]) ' homR(R/P,M)[U−1],thus the inclusion ϕ : R[U−1]/Q→M [U−1] must be f/u for some g ∈ homR(R/P,M)since ϕ is injective, so is f , and so R/P →M is an injection, so P ∈ AssR(M).

For part 2, we consider the RP module MP with P a minimal prime overAnnR(m). AssRP

(MP ) 6= ∅ and if Q ∈ AssRP(MP ), then P ′P = Q ( PP , but

if P ′P ( PP then P ′ 6⊇ AnnR(M), so there is r ∈ AnnR(M) with r /∈ P ′ sothat r/1 /∈ P ′P and r/1 ∈ AnnRP

(MP ). Thus, AnnRP(MP ) 6( P ′P , so P ′P is not

associated.This means that AssRP

(MP ) = {PP }, so P ∈ AssR(M).

Fact: If I is radical then I = ∩P for P a prime containing I or for P primesminimal wrt containing I.

Lemma 3.10. If R is commutative, U is multiplicatively closed and I is max-imal among ideals not meeting U , then I is prime.

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Proof. Suppose that fg ∈ I. If f, g /∈ I, then I+(f)∩U 6= ∅, I+(g)∩U 6= ∅, sothere exists i, i′ ∈ I, r, r′ ∈ R with i+ rf, i′ + r′g ∈ U so (i+ rf)(r′ + r′g) ∈ U ,is equal to ii′ + ir′g + rf ′i+ rr′fg ∈ I, contradicting that I ∩ U = ∅.

So either f ∈ I or ∈ I, so I is prime.

Corollary 3.11. I ⊆ R is an ideal, then√I = ∩P over primes containing I.

In particular, the intersection of all primes is√

0, the set of nilpotents.

Proof.√I ⊆ ∩P is straightforward.

Suppose that f ∈ ∩P \√I. U = {f i : i ≥ 0}, then

√I ∩ U = ∅. Let J be

an ideal containing√I max wrt J ∩ U = ∅, then J is prime which contains I,

so f ∈ J , contradiction, so ∩P =√I.

Corollary 3.12. If I =√I, then I = ∩P , P minimal over I, and AssR(I) =

{P |P is minimal over I}.

Next time, I = ∩Qi,√Qi ∈ AssR(I).

As√I = ∩P , P prime and I ⊆ P . So ∩P over all primes are the nilpotent

elements, and ∩P∈AssP =√

AnnRM .First, IOUs.

1. If R is f.g. over a field, R[U−1] doesn’t have to be. k[x](x), for example.Suppose that k[x](x) as a k-algebra by f1/g1, . . . , fr/gr. Assume k is al-gebraically closed, look at the factors of the gi. Only finitely many x− αshow up, but k is infinite, so these cannot generate everything.

2. We looked at = (x, y) ∩ (z, w) = ∩P . Suppose that P1 ∩ P2 = P3 ∩ P4 ∩. . . ∩ Pn irredundant, all Pi necessary. So P1 ∩ P2 ⊆ P3, so P1P2 ⊆ P3, soP1 ⊆ P3 or P2 ⊆ P3, and so by irredundancy, P3 = P1 and P4 = P2.

Lemma 3.13. If I =√I, then Ass(I) = {P minimal over I}.

This is because I = ∩P over the primes minimal over I.In general, we replace primes by ideals with only one associated prime.

Ass(R/P ) = {P}.

Definition 3.2 (P -primary). Let R be a Notherian ring and M a finitely gen-erated R-module.

A submodule N ⊆M is P -primary if AssR(M/N) = {P}.

Proposition 3.14. Let P be a prime ideal in R. Then TFAE

1. AssR(M) = {P}

2. P is minimal over AnnRM and every element not in P is a non zerodivisor on M .

3. A power of P annihilates M and every element not in P is a non zerodivisor.

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Proof. 1⇒ 2 : If P is not minimal over AnnRM , then there exists P ′ minimalover AnnRM with AnnRM ⊆ P ′ ( P . So P ′ ∈ AssRM . Also, every zerodivisor is in some associated prime, so they all lie in P .

2⇒ 3 : Since elements outside of P are non zero divisors, we get M →MP

injective, so if a power of PP annihilates MP that power of P annihilates M .But PP is minimal over AnnRP

MP and every prime is contained in PP . So ∩Qover Q ∈ RP prime, minimal over AnnRP

MP , is equal to PP =√

AnnRPMP .

Thus, ∃k such that P kP ⊆ AnnRPMP .

3 ⇒ 1 : Since P k ⊆ AnnRM ⊆ P , and P must be contained in any primecontaining AnnRM , so P is minimal over AnnP M , so P ∈ AssRM . But also,every associated prime is contained in P , from the nonzero divisor assumption,so P is the only associated prime.

Corollary 3.15. Let I be an ideal in R. TFAE

1. I is P -primary

2. I contains a power of P and for all r, s with rs ∈ I and r /∈ I, we haves ∈ P

3.√I = P and for all r, s ∈ R with rs ∈ I, either r ∈ I or there is a k such

that sk ∈ I.

Proof. 1 and 2 are equivalent because they are 1 and 3 of the prop.2 ⇒ 3 : If rs ∈ I, r /∈ I, then s ∈ P , so ∃k with sk ∈ I. Also, P k ⊆ I for

some k, P ⊆√I. If f ∈

√I \ P , then ∃` such that g = f ` ∈ I \ P . But now

g · 1 ∈ I, 1 /∈ I so g ∈ P , contradiction. Thus,√I ⊆ P .

3 ⇒ 1 : Since√I = P , we know P k ⊆ I for some k, and if rs ∈ I, r /∈ I,

then ∃` such that s` ∈ I, so s ∈√I = P .

Theorem 3.16. A proper submodule M ′ of M is the intersection of primarysubmodules.

Definition 3.3 (Irreducible Submodule). A submodule N of M is irreducible ifit cannot be written as the intersection N = N1∩N2 with N ( N1 and N ( N2.

Lemma 3.17. If M ′ is a proper submodule of M , then we can write M ′ =∩ki=1Mi where each Mi is irreducible.

Proof. R Notherian and M finitely generated implies that M is Notherian.So if the lemma is false, there exists a submodule N maximal with respect

to not having an irreducible decomposition. In particular, N is not irreducible,so we write N = N1 ∩N2, with N ( N1 and N ( N2. But N1 = ∩ki=1Mi andN2 = ∩`j=k+1Mj , so N = ∩`i=1Mi, contradiction.

Lemma 3.18. If N ⊆M is irreducible, then N is primary.

Proof. Suppose P 6= Q ∈ AssR(M/N). Then R/P ' Rm1 for some m1 ∈M/N ,and R/Q ' Rm2 for some m2 ∈M/N .

Rm1 ∩Rm2 = {0}, so (N +m1) ∩ (N +m2) = N .

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Theorem 3.19. Let M ′ be a proper submodule of M and write M ′ = ∩Mi

where Mi is Pi-primary. Then

1. Every associated prime of M/M ′ occurs among the Pi.

2. If the decomposition is irredundant, then the Pi are precisely the associatedprimes.

3. If the intersection is minimal, then each associated prime occurs exactlyonce, and if P is a minimal associated prime, then Mi is the P -primarycomponent of M ′.

Proof. We first note that if M ′ = ∩Mi with AssR(M/Mi) = {Pi}, then 0 =∩(Mi/M

′) ⊆ M/M ′ with AssR((M/M ′)/(Mi/M′)) = {Pi}. ie, we can replace

M by M/M ′ and M ′ by 0, so we will assume that M ′ = 0.

1. So 0 = ∩Mi, soM → ⊕M/Mi is an injection. So AssR(M) ⊆ AssR(⊕M/Mi) =∪AssR(M/Mi) = {P1, . . . , Pn}.

2. If the decomposition is irredundant, then ∩i 6=jMi 6= 0 for any j. So∩i6=jMi = ∩i 6=jMi/(∩i 6=j(Mi ∩Mj)) ' (∩i 6=jMi + Mj)/Mj ⊆ M/Mj . SoAssR(∩i 6=jMi + Mj) ⊆ AssR(M/Mj) = {Pj}, so AssR(∩i 6=jMi + Mj) ={Pj}, so Oj ∈ AssR(M).

3. We first note that if N1, N2 ⊆ M with Ass(M/Ni) = {P} for i = 1, 2,then M/N1 ∩N2 →M/N1⊕M/N2 is an inclusion, so Ass(M/N1 ∩N2) ⊆Ass(M/N1 ⊕M/N2) = {P}, so Ass(M/N1 ∩N2) = {P}. Thus if Pi = Pjwe can replace Mi,Mj by Mi ∩Mj to get a primary decomposition withfewer terms. So each Pi shows up at most once. But minimal impliesirredundant, so each Pi shows up exactly once.

Now suppose that Pi is minimal over AnnRM . We want to show thatMi = ker(M →MPi). Consider the diagram

M

M/Mi

MPi

(M/Mi)Pi.......................................................... ............β

......................................................................α

...............................................................δ

................................................... ............γ

We have Mi = kerβ. So show that Mi = kerα, it suffices to check thatδ, γ are injections. δ is because Ass(M/Mi) = {Pi}. Since ∩Mj = 0,φ : M → ⊕M/Mj localizes to φPi

: MPi→ ⊕(M/Mj)Pi

. γ is the ithcomponent of φPi . To see that γ is injective, it suffices to note that(M/Mj)Pi = 0. If it weren’t zero, we would have AssRPi

((M/Mj)Pi) ={QRPi

|Q ∈ AssR(M/Mj) with Q ∩ (R \ P ) = ∅} = {QRPi|Q = Pj and

Q ⊆ Pi}, which is empty as Pi is minimal.

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eg, I = (x2y4, x3y3, x4y2) = (x2) ∩ (y2) ∩ (x4, xy4, y7, x3y3) = (x2) ∩ (y2) ∩(x10, x4y, x3y3, y4)

Proposition 3.20. Let A be a finitely generated torsion free abelian group andlet R = ⊕a∈ARa and M be a graded R-module. If P = AnnRM for anym ∈ M and P is prime, then P is homogeneous and P is the annihilator of ahomogeneous element.

Proof. A ' Zk for some k. Choose the isomorphism and w ∈ Rk sufficientlygeneral and set u ≺ u′ for u, u′ ∈ Rk if w ·u < w ·u′. (sufficiently general meansthat this is a total order).

Note that if u < u′ then u+ v < u′+ v. If P is not homogeneous, then thereexists f ∈ P and a ∈ A with fa /∈ P where f =

∑a∈A fa. We can assume in

fact that no fa ∈ P . Write m =∑ma. Then 0 = fm = f1m1 + HOT where

deg f1 = min{ai|fai6= 0} and degm1 = {ai|mai

6= 0}. So f1m1 = 0. So ifm = m1 done. Otherwise, this is the base case of an induction on the numberof nonzero components, b.

Suppose that it is true for small b. Write f1m =∑ma 6=m1

f1ma, this hasfewer terms, so P ⊆ I = AnnR f1m. If P = I, then P is homogeneous byinduction. Otherwise, there is g ∈ I \ P with gf1m = 0 so gf1 ∈ P so f1 ∈ Pas required.

Lemma 3.21. If M is a noetherian R-module and M ′ ⊆ M with M ′ = ∩Mi

with Mi Pi-primary is minimal, let U be a multiplicatively closed set of R. ThenM ′[U−1] = ∩Mi[U−1] over the submodules with Pi∩U = ∅ is a minimal primarydecomposition of M ′[U−1] as an R[U−1]-module.

Proof. M ′[U−1] = ∩Mi[U−1], since localization commutes with intersection. IfPi ∩ U 6= ∅, then Mi[U−1] = M [U−1] since (M/Mi)[U−1] has no associatedprimes, it must be the zero module.

AsM ′[U−1] = ∩Mi[U−1] with P∩U = ∅, and AssR[U−1](M [U−1]/M ′[U−1]) ={PR[U−1]| where P ∈ AssR(M/M ′) and P ∩ U = ∅}. So since ∩Mi was mini-mal, each Pi was in AssR(M/M ′) and showed up exactly once, so in this newintersection.

4 Integral Dependence

Cayley-Hamilton TheoremIn linear algebra, pA(X) = det(A− xI)Cayley-Hamilton says pA(A) = 0.Slightly more abstractly, if V is an n-dimensional vector space over k, and

ϕ : V → V is a linear map, then there exists a0, . . . , an−1 ∈ k such thatϕn +

∑n−1i=0 aiϕ

i = 0.

Theorem 4.1. Let R be a ring and M a finitely generated R-module that hasa generating set with n elements. Let ϕ : M → M be an R-module homomor-phism. If ϕ(M) = IM for an ideal I ⊆ R, then there exists a monic polynomial

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p(x) = xn + p1xn−1 + . . .+ pn with pj ∈ Ij for each j such that p(ϕ) = 0 as an

endomorphism of M .

Proof. Let m1, . . . ,mn be generators for M . Write ϕ(mj) =∑ni=1 aijmi. Let

A be the matrix A = (aij) and m = (m1, . . . ,mn)T ∈ Mn. Regard M as anR[x]-module where x ·m = ϕ(m).

Then (Ix − A)m = 0 for all m ∈ M , where Ix : M → M where if m =∑rimi, Ix(m) =

∑riϕ(mi), then Ix(mi) = ϕ(mi), and Ami =

∑nj=1 ajimj .

Let A′ be the matrix of cofactors of Ix − A, recall from linear algebra that ifA = (aij) then the cofactor matrix B is (bij) with bij = (−1)i+j det(Aij).

Then A′(Ix−A) = det(Ix−A)I, that is, det(Ix−A)m = 0 for all m ∈M ,so det(Ix−A) ∈ R[x] and is in AnnR[x]M .

Let p(x) = det(Ix − A). p(x) has degree n and p(ϕ) = 0 and if p(X) =xn +

∑n−1i=0 pix

n−i then pi ∈ Ii.

More linear algebra: If ϕ : V → V is surjective, then V is injective, if V is afinite dimensional vector space.

Corollary 4.2. Let R be a ring and let M be a finitely generated R-module.

1. If α : M → M is a surjective R-module homomorphism, then α is anisomorphism.

2. If M ' Rn then every set of n elements that generates M forms a freebasis, in particular, the rank of M is well-defined.

Proof. 1. Regard M as a module over R[t] with t acting as α, so tm = α(m).Set I = (t) ⊆ R[t], then IM = M since α is surjective. Apply Cayley-Hamilton to the identity homomorphism, 1 : M → M so there existsxn + p1x

n−1 + . . . + pn with pi ∈ Ii and (1 + p11 + . . . + pn1)m = 0, so(1 + p1 + . . .+ pn)m = 0, write this as 1− tq(t), so there exists q(t) ∈ R[t]with (1−tq(t))m = 0 for all m ∈M . So (1−q(α)α)m = 0, so q(α)◦α = 1.Thus α is injective.

2. A set of n generators for M corresponds to a surjection β : Rn → M .Since M is free of rank n, there exists γ : M → Rn, then β ◦ γ : M →Mis surjective, and thus it is an isomorphism. So β = (βγ) ◦ γ−1 is anisomorphism, so that the given generators form a free basis.

To finish we check that rank is well defined. 1) suppose that Rm ' Rn form < n, we extend our generating set of size m to one of size n by addingn − m zeros. Then part 1 says that this is a free basis, but it containszero, so it is a contradiction.

A second proof is that we let P be a maximal ideal of R, then R/P⊗RM '(R/P )m ' (R/P )n

Note: This is not true for injections! eg α : Z→ Z by α(x) = 2x.Integral Dependence

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Definition 4.1 (Integral over R). Let S be an R-algebra and let p(x) be apolynomial in R[x]. We say that s ∈ S satisfies p if p(s) = 0. The element s iscalled integral over R if it satisfies some monic polynomial.

The equation p(s) = 0 is the equation of integral dependence for s over R.If every element of S is integral over R, we say S is integral over R.

eg S = Q(√

2) with R = Z.√

2 is integral over Z, as it satisfies x2 − 2 = 0.In fact, a+ b

√2 for any a, b ∈ Z is integral over Z. (x− a)2 = 2b2.

Claim: These are all the elements integral over Z. eg, K = Q(√

5), thenx = 1+

√5

2 is integral over Z, as it satisfies (2x−1)2 = 5 which is 4x2−4x−4 = 0,so x2 − x− 1 = 0.

Definition 4.2 (Integral Closure). The collection of all elements of S integralover R is called the integral closure, or normalization, of R in S.

Definition 4.3 (Number Field). A finite field extension of Q is called a numberfield. The integral closure of Z in a number field is called the ring of integers inthat number field.

Definition 4.4 (Normal). If R is a domain, then it’s normalization is its inte-gral closure in its field of fractions. If R is equal to its normalization, then wesay that it is normal.

eg, Z is normal, though proving this really proves the following:

Lemma 4.3. Any UFD is normal.

Proof. Consider r/s with r, s relatively prime. If (r/s)n+p1(r/s)n−1+. . .+pn =0, then rn + p1sp

n−1 + . . . + pnsn = 0, which contradicts the relatively prime

assumption.

e.g. k[x] is normal.

Theorem 4.4. Let R be a ring and let J ⊆ R[x] be an ideal. Let S = R[x]/Jand let s be the image of x in s.

1. S is generated by ≤ n elements as an R-module iff J contains a monicpolynomial of deg ≤ n. In this case, S is generated by {1, s, . . . , sn−1}.In particular, S is a finitely generated R-module iff J contains a monicpolynomial.

2. S is a finitely generated free R-module iff J can be generated by a monicpolynomial. Then S has a basis of the form {1, s, . . . , sn−1}.

eg Z[x]/(2x+ 1) is not finitely generated.

Proof. 1. The powers of x generated R[x] as an R-module, and so generateS as well. So if J contains a monic polynomial, p, of degree n, then ford ≥ n, we can write sd in terms of smaller powers of s using sd−np(s) = 0,so {1, . . . , sn−1} generate S. Conversely, suppose that S is generated by

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≤ n elements as an R-module. Multiplication by s is an endomorphism ofS, so Cayley-Hamilton says that there exists a monic p of deg ≤ n withp(s) = 0, so p(x) ∈ J .

2. Suppose that J is generated by a monic polynomial p of degree n. Thenfrom a, {1, . . . , sn−1} generates S. if these do not form a free basis, thenthere are ai ∈ R with

∑ais

i = 0. Thus,∑aix

i ∈ J of degree n− 1. Thiscontradicts the fact that J is generated by p which has degree n. Thus, Sis a free R-module with free basis {1, . . . , sn−1}. Conversely, we supposethat S is a free R-module of rank n. Then S is finitely generated, so by1, J contains a monic p of degree n. Suppose there is q ∈ J \ (p). Usethe division algorithm to write q = ap+ r for a, r ∈ R[x] with deg r < n.Then r ∈ J , but if r 6= 0, then this would contradict {1, . . . sn−1} being afree basis. So r = 0 and q ∈ (p), thus (p) = J .

Definition 4.5 (Finite). An R-algebra S is finite over R if S is a finitelygenerated R-module.

Corollary 4.5. An R-algebra S is finite over R iff S is generated as an R-algebra by finitely many integral elements.

Proof. If S is finite over R and s ∈ S, multiplication by s is an endomorphismof s. So Cayley-Hamilton shows that there exists monic p with coefficients in Rand p(s) = 0, so s is integral over R¿

Thus, since S is a finitely generated R-algebra, it is generated by finitelymany integral elements.

Conversely, suppose that S is generated by t integral elements as an R-algebra.

If t = 1, then S ' R[x]/J for some J , and by the theorem, J contains amonic polynomial, so S is finite over R by the theorem.

We may assume that t > 1 and that the result is true for t = 1. Let S′ be thesubalgebra of S generated by the first t−1 generators of S. Then, by induction,S′ is finite over R, so S′ is generated, as an R-modules, by {s1, . . . , s`}. Theextra generator s for S is integral over R, so it is integral over S′, so S is finiteover S′. So S is generated as an S′-module by t1, . . . , tm, but then {sitj : 1 ≤i ≤ `, 1 ≤ j ≤ m} generates S as an R-module. So S is finite over R.

Corollary 4.6. If S is an R-algebra and s ∈ S, then s is integral over R iffthere exists an S-module N and a f.g. R-submodule M ⊆ N nor annihilated byany nonzero element of S such that sM ⊆M .

In particular, S is integral iff R[s] is a finitely generated R-module.

Proof. If s is integral over R, take N = S, M = R[s] ⊂ S is a finitely generatedR-module and not annihilated by anything in S, since s′1 = s′ 6= 0.

Conversely, if ∃M ⊆ N with these properties, then multiplication by s is anR-module homomorphism. So by Cayley-Hamilton, there exists a monic p withcoefficients in R such that p(s)M = 0, then p(s) = 0 in S, so S is integral overR.

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Theorem 4.7. Let R be a ring and S be an R-algebra.. The set of all elementsof S integral over R is a subalgebra of S.

In particular, if S is generated by elements integral over R, then S is integralover R.

Proof. Let S′ be the set of elements of S integral over R. We need to showthat if s, s′ ∈ S′ then ss′ and s + s′ are in S′. Let M = R[s], M ′ = R[s′],which are finitely generated R-module. Let MM ′ = R{fg, f ∈ M, g ∈ M ′},then MM ′ is a finitely generated R-module. ss′MM ′ = (sM)(s′M ′) ⊆ MM ′,and (s + s′)MM ′ = (sM)M ′ + M(s′M ′), so by the corollary, since MM ′ ⊆ Sis a finitely generated submodule not annihilated by any element of S, since1 ∈MM ′, so ss′ and s+ s′ are integral over R.

Corollary 4.8 (To Cayley-Hamilton). If M is a finitely generated R-moduleand I is an ideal of R such that IM = M then ∃r ∈ I such that (1− r)M = 0.

Proof. By CH, we get p(x) ∈ R[x] with xn+p1xn−1 + . . .+pn with pj ∈ IJ such

that p(id)M = 0, that is, (1+p1+. . .+pn) idM = 0, so set r = −(p1+. . .+pn) ∈I. So (1− r)M = 0.

Definition 4.6 (Jacobson Radical). The Jacobson radical of R is the intersec-tion of all maximal ideals.

eg, R = Z, then the Jacobson Radical is (0). If R is local, then the JacobsonRadical is the unique maximal ideal.

Corollary 4.9 (Nakayama’s Lemma). Let I be an ideal contained in the Ja-cobson radical of R and let M be a finitely generated R-module.

1. If IM = M then M = 0.

2. If m1, . . . ,mn ∈ M have images in M/IM that generate M/IM as anR-module, then m1, . . . ,mn generated M as an R-module.

Proof. 1. By the corollary, ∃r ∈ I such that (1 − r)M = 0, since (1 − r) isnot in any maximal ideal (as r is in all of them), we have 1 − r is a unitof R, so M = 0.

2. Suppose that m1, . . . , mn generate M/IM . Let N = M/(∑Rmi). Then

N/IN = M/(IM +∑Rmi) = M/M = 0, so IN = N , so N = 0.

Mostly, we use this in the case (R,P ) is local, then M/PM is an R/P -module, and R/P is a field, so it is a vector space.

Application: If (R,P ) is local, then PM = M ⇒M = 0. ie, if 0 = M/PM ,then M = 0. ie, if M/PM = R/P ⊗RM = 0, then M = 0.

Corollary 4.10. If M and N are f.g. R-modules and M ⊗R N = 0 thenAnnRM + AnnRN = R, in particular, if R is local, then M = 0 or N = 0.

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Proof. We first prove the local case. Suppose M ⊗R N = 0 but M 6= 0. ThenNakayama says that M/PM 6= 0. Now M/PM is an R/P -vector space, sothere exists a surjection M/PM → R/P and thus there exists a surjectionM → R/P . So 0 = M ⊗R N → R/P ⊗R N is surjective, since ⊗ is right exact,so R/P ⊗R N = 0, so N/PM = 0, thus PN = N so N = 0.

Now suppose that R is general. M ⊗R N = 0 but AnnRM + AnnRN 6=R. Then there exists prime ideal P with P ⊇ AnnRM + AnnRN . ThenMP ⊗RP

NP = 0, so WLOG, MP = 0, so M = 0 since AnnRM ⊆ P . But thenAnnRM = R, contradiction.

Recall that we shows that integral closure is an R-subalgebra of S. Next:Integral closure commutes with localization.

Proposition 4.11. Let R ⊆ S be rings and let U be a multiplicatively closedsubset of R. If S′ is the integral closure of R in S, then S′[U−1] is the integralclosure of R[U−1] in S[U−1].

Proof. Any element of S integral over R is integral over R[U−1], so S′ is integralover R[U−1] and thus, S′[U−1] is integral over R[U−1]. So we just need to showthat if s/u ∈ S[U−1] is integral over R[U−1], then there exists u′ ∈ U withsu′ ∈ S′ (then s/u = su′/uu′ ∈ S′[U−1]).

If (s/u)n + r1/u1(s/u)n−1 + . . . + rn/un = 0, multiply by (uu1 . . . un)n toget (su1 . . . un)n+ . . .+ rn(uu1 . . . un)n = 0, so su1 . . . un is integral over R, andu1 . . . un ∈ U .

Warning: If R is a Notherian domain, then the integral closure of R in itsquotient field is not necessarily Notherian.

It is Notherian if the integral closure is a finitely generated R-algebra (⇒ R-module). Also Notherian if R is a finitely generated domain containing a fieldor the integers (Nother)

eg, R = k[x1, . . . , xn]/I, the normalization is of the form k[y1, . . . , ym]/J .Next: relationship between the primes in the integral closure of R and the

primes in R.

Proposition 4.12 (Lying Over and Going Up). Suppose R ⊆ S is an integralextension of rings, given a prime P ⊆ R¡ then there exists Q ⊆ S prime withQ ∩R = P . (Lying Over)

Also, Q may be chosen to contain any ideal Q1 ⊆ S with Q1∩R ⊆ P . (GoingUp)

eg, R = Z, S = Z[√

2] and P = (7).

Proof. Factor out Q1 and R∩Q1, to see that we just need to find a prime Q inS with R ∩Q = P . Let U = R \ P , so R[U−1] = RP . If we show ∃Q′ ∈ S[U−1]with Q′ ∩ RP = Pp, then since Q′ = Q[U−1] for some prime Q of S, we wouldhave Q ∩R = P , so we can assume that R is local with maximal ideal P .

Then, if PS 6= S, any maximal ideal Q of S containing PS will have P ⊆Q ∩R, so P = Q ∩R. So we just need PS 6= S.

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If PS = S, then 1 =∑`i=1 sipi, and let S′ be the R-algebra generated by

{s1, . . . , s`}, then 1 ∈ PS′ so PS′ = S′ and S′ is a f.g. R-module, since itis a finitely generated R-algebra over R, so by Nakayama, S′ = 0, which is acontradiction, so PS 6= S.

Proposition 4.13. Let R ⊆ S be domains, if K(S) is algebraic over K(R),then every nonzero or S intersects R nontrivially.

If R ⊆ S is an integral extension of domains, then S is a field iff R is afield. Equivalently, if S is an integral R-algebra, P a prime of S, then P is amaximal ideal of S iff P ∩R is a maximal ideal of R.

Proof. For the first statement, if sufficies to show it for a principal ideal bS of S.If b ∈ S, then there exist ai ∈ K(R) with

∑ni=0 aib

i = 0. Clearing denominatorsand dividing by a power of b if necessary, we get

∑mi=0 a

′ibi = 0 with a′0 6= 0,

a′i ∈ R. Then a′0 ∈ bS ∩ R, a′0 6= 0, so the ideal generated by b intersects Rnontrivially.

If R ⊆ S is an integral extension of domains, then K(S) is alg over K(R),if s/u ∈ K(S), then ∃ai ∈ R with

∑ais

i = 0, so is alg over R (IOU) and∑bju

j = 0⇒∑mi=1 bju

j−m

Suppose R is a field. Let P be a maximal ideal in S. Then P ∩R 6= {0}, soP ∩R = R, this contains 1, so P = S.

Suppose instead that S is a field. Let P be a prime of R. Then by LyingOver, there is a prime Q of S with Q ∩ R = P . But Q must equal (0), soP = (0) ∩R = (0). So the only prime in R is (0), and R is a field.

Finally, take S/P and R/(P ∩R). The statement follows from the field state-ment once we check that S/P is integral over R/(P ∩R). This is because ”inte-gral dependence persists mod P”, ie, if xn+

∑aix

i = 0, then xn+∑n−1i=0 aix

i = 0in S/P , and ai = ai ∈ R/(P ∩R).

Corollary 4.14 (Incompatibility). Suppose R ⊆ S is an integral extensionof rings. Two distinct primes of S having the same intersection with R areincomparable, ie, neither is contained in the other.

Without integrality, we have, for example, k ⊆ k[x, y], (0) = k ∩ (x) =k ∩ (x, y), but (x) ⊆ (x, y).

Proof. If Q ⊆ Q1 ⊆ S with Q ∩ R = Q1 ∩ R = P ⊆ R, factor out P ⊆ R, andQ ⊆ S. Then we get 0 in R/P equals 0 ⊂ Q1/Q ⊂ S/Q. So we are in the casewhere R′ and S′ are domains. S′ is still integral over R′, so K(S′) is alg overK(R′), so if Q1 6= Q = 0, Q1∩R 6= (0), contradiction. So Q1 = Q = 0 in S.

5 Blowup Algebra

Geometric MotivationBl0 C2 ”the blowup of C2 at 0”. We want to be able to take a curve and

separate out the strands going through the origin, we’ll cut out the origin and

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glue in a P1. Algebraically, we replace C[x, y] by C[x, y, u, v]/(xv − yu), so wereplace C2 by V (xv − yu) ⊆ C2 × P2.

If (x, y) 6= (0, 0), then WLOG, x 6= 0, so if xv−yu = 0, then v = y/xu. So ify 6= 0, (u : v) = (1 : x/y) = (x : y). If y = 0, then (u : v) = (1 : 0) = (x : y). Soif (x, y) 6= (0, 0), there is a unique (u : v) with (x, y)× (u : v) ∈ V (xv − yu). If(x, y) = (0, 0), then there are no conditions, so any (0, 0)× (u : v) ∈ V (xv−yu).

Ie, consider the map π : Bl0 C2 → C2 by (x, y) × (u : v) → (x, y), this mapis 1-1 away from the origin, and π−1(0, 0) = P1, the ”exceptional divisor”.

Now C[x, y, u, v]/(xv − yu) ' C[x, y, xt, yt] ⊆ C[x, y, t], deg t = 1,deg x =deg y = 0.

Definition 5.1 (Blow-Up Algebras). If R is a ring and I is an ideal, then theblow-up algebra of I in R is the R-algebra BI(R) = R⊕I⊕I2⊕. . . ' R[It] ⊆ R[t].

eg, B(x,y)C[x, y] is the coordinate ring of Bl0 C2.Set deg(r) = 0 for r ∈ R and deg(t) = 1. Then BIR is a Z-graded ring.The ideal I ⊆ BI(R) is homogeneous, thus the quotient BI(R)/I is graded.

Definition 5.2 (Associated Graded Ring). grI R = R[It]/IR[It] is the assci-ated graded ring of R.

[Can also do for any diltration R ) I1 ) I2 ) . . . with IjIk ⊆ Ij+k.]For C[x, y, tx, ty]/(x, y) ' C[x, y, u, v]/(xv−yu, x, y) ' C[u, v] = gr(x,y) C[x, y],

the coordinate ring of P1. We can also do this for modules:

Definition 5.3. Let R be a ring, M an R-module, and I an ideal of R, thenI : M = M0 ⊃M1 ⊃M2 ⊃ . . . is a filtration of R-modules. It is an I-filtrationif IMi ⊆ Mi+1 for all i > 0 and it is I-stable if ∃n > 0 such that ∀i ≥ 0,IMn+i = Mn+i+1 = Ii+1Mn.

OWED RESULT: If R ⊂ S is an integral extension of domains, then K(S)is algebraic over K(R). Suppose s1/s2 ∈ K(S). Then there exist ai, bj with∑mi=0 ais

ii = 0, am = 1 and

∑nj=0 bjs

j2 = 0 with bn = 1 and b0 6= 0. Then∑n

j=0 bj/b0sj−n2 = 0 =

∑nj=0 bj/b0(1/s2)n−j , this shows that 1/s2 is integral

over K(R), so s1/s2 = s1 ∗ 1/s2 is integral over K(R).

Definition 5.4. If I is an I-filtration, then grIM = M/M1 ⊕M1/M2 ⊕ . . .and BlI M = M ⊕M1 ⊕ . . ..

Note: grI R is an R/I-algebra. If I is finitely generated, then grI R is afinitely generated R/I-algebra. If I is a maximal ideal, then grI R is a finitelygenerated algebra over a field.

Also grI M is a graded grI R module.

Proposition 5.1. Let I be an ideal in R and let M be a finitely generated R-module. If I : M = M0 ⊃M1 ⊃ . . . is an I-stable filtration by finitely generatedR-submodules of M , then grI M is a finitely generated module over grI R.

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Proof. Suppose IM1 = Mi+1 for i ≥ n. Then (I/I2)(Mi/Mi+1) = Mi+1/Mi+2

for i ≥ n. So the union of any set of generateors for Mi/Mi+1 0 ≤ i ≤ ngenerates grI M as a grI R-module.

Since the Mi are finitely generated R-modules, so are the Mi/Mi+1, so weget a finite set of generators.

Proposition 5.2. Let R be a ring, I ⊆ R an ideal, M a finitely generatedR-module with I-filtration I = M0 ⊇M1 ⊇ . . . by finitely generated Mi.

Then the filtration is I-stable iff the BIR-module BIM is finitely generated.

Proof. If BIM is finitely generated, then its genertaors appear in the first nsteps for some n, so BIM is gneerated by M0, . . . ,Mn. So Mn⊕. . . is generated,as a BIR-module, by Mn. This means that Mn+1 = IiMn for i ≥ 0 so I isI-stable.

Conversely, if I is I-stable, then ∃n such that IiMn = Mn+i for all i ≥ 0,so a generating set for M0, . . . ,Mn generated BIM .

Lemma 5.3 (Artin-Rees). Let R be a Notherian ring, I ⊆ R an ideal, andlet M ′ ⊂ M be finitely generated R-modules. IF M = M0 ⊃ M1 ⊃ . . . isw anI-stable filtration, then the induced filtration M ′0 = M ′ ⊂ M ′1 = M1 ∩M ′ ⊃ . . .is also I-stable, so ∃n such that M ′ ∩Mn+i = Ii(M ′ ∩Mn).

Note: IF R is Notherian, then so is BIR, since I is finitely generated. SoBIR is a finitely generated R-algebra, so is Notherian.

Proof. Let I ′ = M ′ = M ′0 ⊃ M ′1 ⊃ . . .. BI ′M′ is naturally a BIR submod-

ule of BIM . If I is stable, then BIM is a finitely generated BIR-module,so all submodules are finitely generated and, in particular, BI ′M

′ is finitelygenerated, so I ′ is I-stable.

Q: Can we have 0 6= I5 = I20 in a nice ring? If so, actually have Ij = I5 forall j ≥ 5. So we’d get ∩j≥1I

j = I5.

Corollary 5.4 (Krull Intersection Theorem). Let I ⊂ R be an ideal in aNotherian ring. If M is a finitely generated R-module, then there is an r ∈ Isuch that (1 − r)(∩∞j=1I

jM) = 0. If R is a domain or a local ring, and I is aproper ideal, then ∩∞j=1I

j = 0.

Proof. For any P , ∩j≥1IjM = (∩j≥1I

jM) ∩ Ip+1M , now Artin-Rees appliedto ∩IjM ⊂ M for the I-stable filtration Mj = IjM says that ∃p such thatI(∩j≥0I

jM ∩ IpM) = (∩j≥0IjM) ∩ Ip+1M = ∩j≥0I

jM . ie, ∃p such thatI(∩j≥0I

jM) = ∩j≥0(IjM). So ∃r ∈ I such that (1− r) ∩j≥0 IjM = 0.

6 Flatness

A Flat Family: A ”family” of varieties is one that varies with parameters.Eg: V (x2 − a2) for a ∈ k is {a,−a}.Eg: V (x) ∪ V (y − ax) = V (x(y − ax)), two lines through the origin.

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A way to think about V (x2 − a2) is as a union of two lines projected downto a line.

A family, then, is a map of varieties π : Y → X with the fibers π−1(x) ⊂ Yfor x ∈ X.

Corresponding map of rings R → S in the other direction. IE, k[a] →k[x, a]/(x2 − a2). So a family over Spec(R) is an R-algebra S.

Definition 6.1 (Fiber). The fiber over any prime P ⊂ R is K(R/P )⊗R S

eg (a− 7) ⊆ k[a] has fiber k[a]/(a− 7)⊗k[a] k[a, x]/(x2− a2) ' k[a, x]/(x2−a2, a− 7) ' k[x]/(x2 − 49) is the ring of {7,−7}.

eg, Z/2Z is a Z-algebra, so Spec(Z2) → Spec(Z) is a family. The fiber over7 is Z7 ⊗Z Z2 ' 0. So the fiber over (2) is Z2 ⊗Z Z2 and is trivial elsewhere.

We replace locally trivial is flat. A family is nice if it is flat, that is, S is aflat R-module.

Recall that if 0 → A → B → C → 0 is a ses of R-mods, then A ⊗M →B ⊗M → C ⊗M → 0 is exact for any R-module M . If 0→ A⊗M → B ⊗Mis exact, then M is called a flat R-module.

eg R = k[x] and S = k[x, y]/(x− y). Check: S is flat (since S ' k[x] = R asan R-module).

Free R-modules are always flat.eg: S = k[x, t]/(tx− 1), R = k[t]. Then S ' k[t, t−1], and R[U−1] is flat for

all multiplicative sets U ⊂ R.eg: R = k[a, b], S = R[a, b, x, y]/(ax+ by), then S is an R-module but is not

flat!Important Example: ”Grobner Degeneration”. If S = k[x1, . . . , xn], and

I ⊂ S is an ideal. Let w ∈ Rn and consider ≤w where xu < xv if w · u < w · vor (other condition)

Grobner theory studies the ideal ∈w (I). So we define, given f ∈ S, f =∑cux

u, set f = f(xi/tw1 , . . . , xn/twn)tb ∈ S[t] where b = maxcu 6=0 w · u. ie,

f =∑cux

utb−w·u.So if S = k[x, y], w = (2, 3) and f = x2+y3, then f = x3+y2, if f = x3+3xy

then F = x3 + 3xyt.Define It ⊂ S[t] to be It = 〈f |f ∈ I〉, then It|t=1 = I and It|t=0 =∈w (I).

So S[t]/It is a k[t]-module defining a family. Check that, for all a 6= 0, the fiberover t = a is isomorphic to S/I.

Lemma 6.1. S[t]/It is a free, and thus flat, k[t]-module.

Recall: B = {xu|xu /∈∈w (I)} is a basis for S/I as a k-vector space.

Proof. We claim that B is a k[t] basis for S[t]/It. That is, S[t]/It ' ⊕k[t]xu

over xu ∈ B.The key point is that if G = {g1, . . . , gr} is a Grobner basis for I, then

{g1, . . . , gr} generate It and gi =∈w (gi) + t(other stuff). Given f ∈ S[t],dividing by {gi} gives a polynomial

∑pu(t)xu.

”linear independence” as before. If f =∑pu(t)xu = 0 in S[t]/It, then

f = tk(∑p′u(t)xu) are not all divisible by G so equals

∑q(x, t)gi.

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Summary: S[t]/It is a free and thus flat R-module, so V (It) ⊆ kn × k is aflat family.

We say this gives a Grobner Degeneration from V (I) to V (∈ (I)).Check: S[t]/It ⊗k[t] k[t]/(t) ' S/ ∈w (I) and S[t]/It ⊗k[t] k[t]/(t− a) ' S/I.eg: I = (xy− y2) ⊂ k[x, y], w = (2, 1), so It = (xy− t2y2), we get the union

of the x axis and a line of slope 1/t2, and ∈w (I) = (xy).eg: I = (x2 + y2 − 4) ⊆ C[x, y], then It = (x2 + t2y2 − 4t2) ⊆ C[x, y, t],

∈≤ (T ) = (x2).Let R = k[a, b] and S = k[a, b, x, y]/(ax + by), and S is not flat. Look at

(a, b)⊗ S → R⊗ S.TorFlatness says that 0 → A → B → C → 0 exact implies 0 → A ⊗M →

B ⊗M → C ⊗M → 0, but we get all but the first anyway.If M is not flat, we would like to define Tor1(M,C)→ ker(A⊗M → B⊗M).’General Homological Algebra’If F is a right exact functor from R-modules to R-modules, then given

0→ A→ B → C → 0 we get . . .→ L1FB → L1FC → FA→ FB → FC → 0.

Definition 6.2. Let P : . . . → P2 → P1 → P0 → M → 0 be a projectiveresolution of an R-module M (ie P is exact with each Pi projective).

If R = k[x1, . . . , xn] and M = R/I ten we can construct a free resolutionusing Grobner Bases by Schreyer’s Algorithm.

Definition 6.3. Given a projective resolution P , define FP to be . . .→ FP2 →FP1 → FP0 → FM → 0. This is still a chain complex, so we take homology,and LFi = kerFϕi/ ImFϕi+1.

For us: Tori(M,N) = LFi(N) where F is−⊗RM . So to compute Tori(M,N),we compute a projective resolution of N , tensor with M , and take the ith ho-mology.

Basic Facts:

1. It doesn’t matter what resolution we take.

2. Tori(M,N) = Tori(N,M).

3. Tor0(M,N) = M ⊗R N .

4. If M is projective, then Tori(M,N) = 0 for all i > 1.

Examples: If x ∈ R is a nonzero divisor then 0 → Rx→ R → R/(x) → 0.

Claim: This is a free resolution ofR/(x). Thus Tori(R/(x),M) = R/(x)⊗RM =M/xM if i = 0, is ker(M x→ M) = (0 : mx) = {m ∈ M : xm = 0} if i = 1 and0 else.

eg. R = k[x, y] and M = N = k[x, y]/(x, y) ' k. What is TorRi (k, k). Then

0←M ← R(x,y)← R2 (y,−x)t

← R← 0 is a free resolution.

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So then Tori(k, k) = k if i = 0, k2 is i = 1 and k if i = 2, 0 else. In general,if R = k[x1, . . . , xn] and M = R/(x1, . . . , xn), then Tori(M,N) = kβi for someβi, and we call the βi the Betti numbers.

A long exact sequence: If 0→ A→ B → C → 0 then we get Tori(A,M)→Tori(B,M) → Tori(C,M) → Tori−1(A,M) → . . . → Tor1(C,M) → A ⊗M →B ⊗M → C ⊗M → 0.

More facts about Tor: If S is a flat R-algebra, then S ⊗R TorRi (M,N) =TorSi (S ⊗RM,S ⊗R N).

If we have a short exact sequence 0→ A→ B → C → 0 we get a long exactsequence → TorRi (A,M) → TorRi (B,M) → TorRi (C,M) → TorRi−1(A,M) →. . .→ Tor1(B,M)→ Tor1(C,M)→ A⊗M → B ⊗M → C ⊗M → 0.

Proposition 6.2. Let R be a ring and M an R-module. If I is an ideal of Rthen the multiplicative map I⊗RM →M is an injection iff TorRi (R/I,M)→ 0.

The module M is flat iff this condition is satisfied for all finitely generatedI.

Proof. Consider the ses 0→ I → R→ R/I → 0. From this we get a long exactsequence Tor1(R,M)→ Tor(R/I,M)→ I ⊗M → R⊗M = M .

As Tor1(R,M) = 0, I⊗M →M is injective iff Tor1(R/I,M) = ker(I⊗M →M) is zero.

Recall that M is flat iff M ⊗N ′ →M ⊗N is an inclusion for all inclusionsN ′ ⊆ N .

First we assume this for all N ′ finitely generated I and N = R. We mustshow that this implies the general condition for N ′ ⊆ N . First, let I be a generalideal of R and x ∈ I⊗M . Then x =

∑si=1 ri⊗mi. Let I ′ be the ideal generated

by < r1, . . . , rs >. Then x ∈ I ′ ⊗M →M is an inclusion, so x 6= 0 in M .Now consider N ′ ⊆ N . BY the same argument we may assume that N is

finitely generated. ie, x ∈ N ′⊗M , then x =∑ni⊗mi, let N be the submodule

generated by the ni and any necessary relations. We can thus assume that Nis finitely generated. So we can find a filtration N ′ = N0 ⊂ N1 ⊂ . . . ⊂ Nr = Nwith each Ni/Ni−1 cyclic.

It suffices to show that Ni ⊗M → Ni+1 ⊗M is an inclusion, so we mayassume that N/N ′ ' Rx ' R/I. Now from 0→ N ′ → N → N/N ′ → 0 we getTor1(N/N ′,M)→ N ′ ⊗M → N ⊗M . Tor1(N/N ′,M) = Tor1(R/I,M) = 0 byhypothesis. So N ′ ⊗M → N ⊗M is an inclusion for arbitrary N ′ ⊆ N , so Mis flat.

The point was that I ⊗M →M being an inclusion for all finitely generatedI implies that N ′ ⊗M → N ⊗M is an inclusion for all N ′ ⊆ N .

So to check flatness, it is enough to check finitely generated ideals.

Corollary 6.3. Let k be a field and R = k[t]/t2, and M an R-module. Then Mis flat iff multiplication by t from M to tM induces an isomorphism M/tM →tM .

Proof. The only nonzero ideal in R is (t). So M is flat iff (t) ⊗M → M isan injection. As (t) ' R/(t) as an R-module by t ↔ 1, we have (t) ⊗M '

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R/(t)⊗M 'M/tM , so the map M/tM → tM by m 7→ tm is the compositionR/t⊗M → t⊗M →M . So it is injective.

Corollary 6.4. If a ∈ R is a nzd in R and M is a flat R-module, then a is anzd on M .

If R is a PID, then the converse is true: M is flat iff M is torsion free.

Proof. Let a ∈ R be a nzd and M flat. I = Ra ' R by 1 7→ a. So we haveR ⊗M → I ⊗M → R ⊗M by 1 ⊗m → a ⊗m → a ⊗m, so m 7→ am. Simcethe map m 7→ am is injective, a is a nzd on M .

Suppose that R is a PID and M is torsion free, so no element of R annihilatesan element of M . Then for any a 6= 0 in R, Ra⊗M →M is an injection, since0⊗M →M is an injection as well, this means that I ⊗M →M is an inclusionfor all finitely generated I, thus M is flat.

Definition 6.4 (Rees Algebra). The Rees Algebra of R with respect to I,R[R, I] = R[t, t−1I] ⊆ R[t, t−1]. It is

∑∞n=−∞ Int−n with In = R for n ≤ 0.

If R is a k-algebra (k a field) we’ll see that R[R, I] is a flat k[t]-algebra.Facts: R[R, I]/iR[R, I] = grI R = R/I ⊕ I/I2 ⊕ I2/I3 ⊕ . . ..If a 6= 0 in R, then R[R, I]/(t− a)R[R, I] ' R.So R[R, I] is a family over [t] with fiber over t = 0 grI R and fiber over

everything else R.

Lemma 6.5. If R is a k-algebra, then S = R[R, I] is flat over k[t].If ∩∞d=1I

d = 0 then every element of the form 1− ts with s ∈ S is a nzd onR[R, I].

Proof. Since k[t] is a PID, it suffices to observe that S is torsion free as k[t]-module. This is immediate from the fact that S = R[t, It−1] ⊆ R[t, t−1].

For the second statement, suppose first that p(1− ts) = 0 ∈ S. This meansthat p ∈ (t), p = qt. t is a nzd on S, so q(1 − ts) = 0 in S, so q ∈ t, so p ∈ t2.Continue to get that p ∈ tnS for all n.

Now p =∑ji=−j pit

j . Since p ∈ tnS for all n, we must have pi ∈ Im for allm.

What to take away: Flatness is a niceness property, and flat families preservea lot of properties (ie, dimension)

7 Completions

The basic idea is that the open sets in the Zariski topology are too big, so welook for smaller neighborhoods.

eq if R = k[x1, . . . , xn] and m = (x1, . . . , xn), then Rm = {f/g : g(0) 6= 0}.We replace this by R = k[[x1, . . . , xn]] formal power series, and we get a naturalmap Rm → R.

One advantage is that we get a version of the inverse function theorem.

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Definition 7.1 (Inverse Limit). If {Gi|i ∈ N} is a sequence of abelian groupswith homomorphisms ϕi : Gi → Gi−1. Then lim←−Gi = {g ∈

∏∞i=1Gi|ϕi(gi) =

gi−1} is the inverse limit, which is an abelian group under coordinatewise addi-tion.

We will be interested in the case where we start with a ring R and a filtrationm1 ⊃ . . .mn ⊂ . . . of ideals and set Gi = R/mi. Write R = lim←−R/mi.

R is a ring by coordinate multiplication. Most important case is mi = mi

for some ideal m ⊂ R. Notation is Rm. eg R = k[x], m = (x), then Rm =lim←− k[x]/xi.

Claim: Rm = k[[x]].

Proof. ϕ : k[[x]]→ Rm, a→ (b1, b2, . . .) by∑aix

i 7→ bi =∑i−1n=0 anx

n.This is a well-defined homomorphism, so we just need to check that it is iso.

For the inverse map, given b = (b1, . . .) ∈ Rm, each bi has a representation of theform

∑i−1j=0 aijx

j and if k < ` then akj = a`j for j < k. Define ψ : Rm → k[[x]]by b 7→

∑∞j=0 aijx

j .

Definition 7.2 (Complete with respect to m). There is a natural map, R→ Rm

by r 7→ (r, r, r, . . .), if this is an isomrophism, then R is complete.

Theorem 7.1 (Cohen Structure Theorem). If R is a complete local ring con-taining a field, then R = k[[x1, . . . , xn]]/I for some I.

eg, the p-adics, p ∈ Z, then Zp = lim←−Z/pn, with ϕ : Z/pn → Z/pn−1 bya 7→ a.

Then we can write elements of Zp as∑∞i=0 aip

i for 0 ≤ ai < p with additionis ”add with carrying”

In Z2, 1 + 2 + 4 + 8 + 16 + . . . is b = (1, 3, 7, . . .) = (−1,−1,−1, . . .), and1 = (1, 1, 1, . . .).

If we have m1 ⊃ m2 ⊃ . . . we get an ideal m1 = {g = (g1, . . .)|gj = 0∀j ≤i} ⊆ R. When mi = mi, then m1 = m.

Lemma 7.2. When m is a maximal ideal in R, then Rm is a local ring. Forany filtration, we have R/mi = R/mi.

Proof. If g ∈ R/mi, then map g + mi 7→ gi + mi is the ”projection homomor-phism”

If ϕ(g+mi) = 0 then gi ∈ mi, so gj ∈ mj for j < i. So g = (0, 0, 0, . . . , 0,−, . . .)+mi, so g ∈ mi, so g = 0.

It is surjective, since R→ R→ R/mi by r 7→ (r, r, r, r, . . .)→ r + mi.Thus R/mi ' R/mi. If m is a maximal ideal in R, then R/m is maximal.

Thus, R/m is a field, and so m is a maximal ideal of R. We now show that ifg ∈ R \ m, then g is a unit.

Note that each R/mi is a local ring with maximal ideal m. So if gi ∈ R/mi\m,then gi is a unit in R/mi. If g = (g1, g2, . . .) ∈ Rm \ m, then g1 6= 0, so eachgi /∈ mR/mi.

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Since gi = gj mod m, then g−1i = g−1

j mod m, so set g−1 = (g−11 , . . .), and

note that gg−1 = (1, 1, 1, . . .).

eg k[[x]] is local with maximal ideal (x). Zp is local with maximal ideal (p).Note: If m is a maximal ideal, R/mi ' (R/mi)m = Rm/mmi.So we get the same Rm if we localize at m first.So if m is maximal, we say that Rm is a complete local ring.Note: If R is complete with respect to m, then ∩∞j=0m

j = 0.

Definition 7.3 (Convergence). A sequence a1, a2, . . . ∈ R converges to an ele-ment a ∈ R if ∀n∃in such that a−aj ∈ mn for j ≥ in, ie ain = (a1, a2, . . . , an, other).

A sequence is Cauchy if ∀n∃in such that ai − aj ∈ mn for i, j ≥ in.A sequence converges iff it is Cauchy.

This is the usual notion of convergence in the m-adic topology which has abase of open sets {a+ mi|a ∈ R, i ≥ 1}.

Cauchy implies Convergence: we set a = lim ai, ie, if {ai} is Cauchy, ai =(ai1 , ai2 , . . .), set a = (b1, b2, . . .) with bn = ajn for any j > in.

This is well defined, since if j, k ≥ in, aj − ak ∈ mn so ajn = akn. Checka ∈ R: If j > in, then bn = ajn and bn−1 = ajn−1 so bn = bn−1 mod mn−1.

Check: {ai} converges to a. Given n, ∃in such that ∀i, j ≥ in, ai− aj ∈ mn.By construction, for such an i, ai−a ∈ mn, so ∀i ≥ in, ai−a ∈ mn, so {ai} → a.

From analysis: we know that if ai → a and bi → b, then ai + bi → a+ b andaibi → ab.

Application:

Proposition 7.3. If R is complete with respect to m, then U = {1− a|a ∈ m}are units in R.

Proof. If a ∈ m, set ai =∑ij=0 a

j = 1 + a+ a2 + . . .. Then the sequence {ai} isCauchy. If i, j > n− 1, then ai − aj ∈ mn. So it converges to some b ∈ R. And(1− a)ai converges to (1− a)b.

But (1− a)ai = 1− ai+1, so it converges to 1.

We saw this with a = 2 in Z2, there 1 + 2 + 4 + 8 + . . ., so the limit of theCauchy sequence

∑ij=0 2i is −1.

Corollary 7.4. R is a local ring with maximal ideal P , then R[[x1, . . . , xn]] islocal with maximal ideal P + (x1, . . . , xn).

Proof. R[[x1, . . . , xn]] = R[x1, . . . , xn](x1,...,xn). If f ∈ P + (x1, . . . , xn), thenf has constant term f0 /∈ P , so f0 is a unit in R. So f−1

0 f = 1 + g forg ∈ (x1, . . . , xn).

So 1 + g is a unit, thus f is a unit as well.

Definition 7.4. Let R = m0 ⊃ m1 ⊃ . . . be a filtration of ideals, and let grRbe the associated graded ring.

Given f ∈ R let m = max{j|f ∈ mj}. Define in(f) = f + mm+1 ∈mj/mj+1 ∈ grR.

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Proposition 7.5. Suppose that R is a ring complete with respect to m1 ⊃ m2 ⊃. . .. Suppose I ⊂ R is an ideal, a1, . . . , as ∈ I. If in(a1), . . . , in(as) generatein(I) = (in(f)|f ∈ I) ⊂ grR. Then {a1, . . . , as} generate I.

Proof. Let I ′ = (a1, . . . , as). We may assume that none of the ai are 0. Choosed >> 0 so that ai /∈ md for all i.

Given f ∈ I with in(f) of degree e, we can write in(f) =∑sj=1Gjin(aj) with

Gj ∈ grmR homogeneous of degree deg(in(f)) − deg(in(aj)). Take g1, . . . , gswith in(gj) = Gj . Then f −

∑sj=1 gjaj ∈ me+1.

Repeat: we can get f ′ ∈ I ′ with f − f ′ ∈ md+1. Now keep repeating, notingthat now deg(Gj) ≥ e− d > 0.

gj ∈ me−d, so get f −∑j g

(0)j aj −

∑j g

(1)j aj − . . . −

∑j g

(n)j aj with g

(i)j ∈

me−d+i.fn ∈ me+n+1, so the series

∑∞i=0 g

(i)j converges, and we will call the limit

hj . Now look at f −∑hjaj . This is 0, since ∩mi = 0, and so f ∈ I ′, thus the

ai generate I.

Theorem 7.6. Let R be a Notherian ring and let m be an ideal of R.

1. Rm is Notherian.

2. mn = mnRm, so grm R = grmR.

Proof. Write gr R for the associated graded ring of R with respect to mi. Thensince R/mi ' R/mi, we have gr R = grmR. Since R is Notherian, so is R/m.

The ring grmR is finitely generated as an R/m algebra (since m is a finitelygenerated ideal). So grmR ' R/m[x1, . . . , x`]/J for some ideal `, J , thus grmRis Notherian by the Hilbert Basis Theorem.

So gr R is Notherian. So for any ideal I ⊆ R, the ideal in(I) ⊆ gr R is finitelygenerated by the initial forms of a1, . . . , as, so a1, . . . , as generate I, and so Ris Notherian.

To show that mn = mnRm it suffices to show that both have the sameinitial ideals in gr R. (This uses Notherian, as we need ideals in gr R to befinitely generated to apply the prop). But both initial ideals consist of all termsof degree ≥ n.

Theorem 7.7. Let R be a Notherian ring. Let I is an ideal of R.

1. If M is a finitely generate R-module, then the natural map R ⊗R M →lim←−M/IjM = M is an isomorphism.

2. R is a flat R-module.

Lemma 7.8 (Hensel’s Lemma). Let R be a ring that is complete with respectto an ideal m. Let f(x) ∈ R[x]. If a is an approximate root of f in the sensethat f(a) ≡ 0 mod f ′(a)2m then there is a root b of f near a in the sense thatf(b) = 0 and b ≡ a mod f ′(a)m.

If f ′(a) is a nzd on R, then b is unique.

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Most often used when f ′(a) is a unit.ApplicationIs 8 a square in Z7?Take c ∈ Zp. Write c = pnb where p 6 |b. (ie b =

∑∞i=0 bip

i, b0 6= 0). Thenc is a square iff n is even and b is a square. So we’ll assume c =

∑∞i=0 cip

i hasc0 6= 0.

If c = d2, d =∑∞i=0 dip

i, then c0 = d20 mod p.

Consider f(x) = x2 − c. (assume p > 2). Then f(d0) = d20 − c ∈ (p).

f ′(x) = 2x, so f ′(d0) = 2d0 6= 0, so is a unit in Zp. So Hensel’s lemma says thatthere is a root of f , so c is a square.

So as 8 = 1 + 7, and 1 is a square mod 7, 8 must be square.Idea: use Newton’s method to construct a1, a2, a3, . . . by ai+1 = ai−f(ai)/f ′(ai).

Claim: This is a convergent sequence with limit b and f(b) = 0. Recall Tay-lor’s Theorem that f(x + y) = f(x) + f ′(x)y + h(y)y2 for some polynomialh(y) ∈ R[x][y].

Fact 1: If f ′(a1) is a unit, so is f ′(ai) for all i (assume that f(ai) ∈ m).bi = f ′(ai+1) − ai = −f(ai)/f ′(ai) = f ′(ai) + f ′′(ai)bi + hai(bi)b2i is a unit

plus something in m. As such, f(ai + 1) is a unit.Fact 2: If f(a1) ∈ mi, f(ai+1) ∈ m2i. f(ai+1) = f(ai + bi) = f(ai) +

f ′(ai)bi + hai(b1)b2i = ha(b1)b2i ∈ m2i.f(ai+1) ∈ m2i implies that bi+1 ∈ m2i, so ai+1−ai ∈ m2i, so {ai} is Cauchy.

Let b be the limit. Argue that f(b) = 0.Question: What about better approximation? ie, can we replace m by

m2, . . . ,mn?Yes, and the same proof/statement works, because if R is complete wrt m,

it is complete wrt mn.

Proposition 7.9 (Universal Property of Inverse Limit). If {Gi} is a set ofabelian groups with ϕi : Gi → Gi−1 and H is an abelian group with ψi : H → Gisuch that ϕiψi = ψi−1, then there exists a unique map ψ : H → lim←−Gi such thateverything commutes.

Note: If R is a ring and S is and R-algebra that is complete with respect ton and f1, . . . , fn ∈ n, then ∃! R-algebra homomorphism ϕ : R[[x1, . . . , xn]]→ Ssending xi to fi for each i. Theorem 7.16.

8 Dimension

Definition 8.1 (Krull Dimension). The Krull dimension is the supremum ofthe lengths of ascending chains of distinct prime ideals.

The motivation is: The dimension of a vector space is the length of thelongest (or any maximal) chain of subspaces.

eg. Let k be a field. Then dim k = 0.dim k[x] = 1, as 0 ⊆ (x) and if 0 ⊆ (p) ⊆ (q) then (q) = (p).This actually proves the following:

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Lemma 8.1. If R is a PID, then dimR = 1.

In affine algebraic geometry, the dimension of a variety V (I) ⊆ An is thelength of the longest chain of subvarieties V (I) = Vr ⊃ Vr−1 ⊇ . . . ⊇ V0, whichis dim k[x1, . . . , xn]/I.

For rings of the form R = k[x1, . . . , xn]/I every maximal chain of primes hasthe same length (a ring with this property is called Catenary)

Properties of Dimension

1. dimR = supP dimRP over all P prime in R.

2. Nilpotents don’t affect dimension. ie, if I is a nilpotent ideal of R (soIk = 0 for some k), then dimR = dimR/I.

3. Dimension is preserved by maps with finite fibers. If R ⊆ S are rings suchthat S if a finitely generated R-module, then dimR = dimS.

4. Calibration: if k is a field, then dim k[x1, . . . , xn] = n.

5. If R is an affine domain over a field R ' k[x1, . . . , xn]/I for I prime, thendimR = tr degk R.

6. If R is Notherian local with maximal ideal m, then dimR is the minimumn such that ∃n elements f1, . . . , fn ∈ m not in any prime other than m.

7. If R is an N-graded ring R0 = k generated in degree 1, then there existsa polynomial P such that P (n) = dimk Rn for n >> 0. Then dimR =1 + degP . This polynomial is called the Hilbert Polynomial.

eg, R = k[x], deg(x) = 1, then P (n) = 1 for all n, so dimR = 1. Thisgives an algorithm to compute dimension for R = k[x1, . . . , xn]/I.

Fact: Flat implies that the fibers have the same hilbert polynomialBy convension: if I ⊆ R is an ideal, the dimension of I is the dimension of

the ring R/I.If M is an R-module, then the dimension of M is the dimension of AnnM

(which is dimR/AnnM). Think about this in the same way as we do primarydecomposition, so we just ignore the dimension of I as an R-module.

If I is prime, then the codimension of I is the dimension of RI , that is, thelength of the longest chain of primes contained in I. As dim I = dimR/I andcodim I = dimRI , dimR ≥ dim I + codim I.

For general I, codim I = min of codimP where P is a prime containing I.If M is an R-module, then codimM = codim AnnM .Today: 0-dimensional Notherian ringsIf R has dimension zero, then all primes are maximal.If R is a zero-dimensional domain, then 0 is a maximal ideal and so R is a

field.Recall: A ring R is Artinian iff it satisfies the descending chain condition on

ideals.

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Definition 8.2 (Composition Series). If M is an R-module, a compositionseries for M is M = M0 ⊇ M1 ⊇ . . . ⊇ Mn = 0 with Mi/Mi+1 a nonzerosimple module (has no nontrivial submodules)

The length of M is the smallest length of a composition series for M or ∞if M has no finite composition series.

eg: the length of Z as a Z-module if ∞.eg: the length of R = k[x]/x2 as an R-module is R ⊃ (x) ⊇ 0, so 2.

Proposition 8.2. Let R be a ring and let M be an R-module. M has a finitecomposition series iff M is Artinian and Notherian.

In tis case, any filtration of submodules of M has length at most n and canbe refined to a composition series.

Proof. Suppose that M is Artinian and Notherian. Then by the ACC, M hasa maximal proper submodule M1, which has a maximal proper submodule M2,etc. This gives a descending chain of proper submodules M ⊇ M1 ⊇ M2 ⊇ . . .where each Mi/Mi+1 is simple. As M is Artinian, this chain must be finite, sosome Mn = 0.

Suppose now that M has a finite composition series M = M0 ⊇M1 ⊇ . . . ⊇Mn = 0. We first show that if M ′ ⊆M is a proper submodule, then the lengthof M ′ is less than the length of M .

Indeed, consider M ′i = M ′ ∩Mi. Then M ′ = M ′0 ⊇ M ′1 ⊇ . . . ⊇ M ′n = 0.And M ′i/M

′i+1 = (M ′ ∩Mi)/(M ′ ∩Mi+1) = M ′i +Mi+1/Mi+1 ⊆Mi/Mi+1. So

M ′i/M′i+1 is either Mi/Mi+1 and is simple, or it is zero. There must be at least

one i for which M ′i/M′i+1 = 0, because otherwise we would get M ′ ⊇ Mi for

all i by descending induction, Mn = 0 ⊆ M ′, so Mi = M ′i + Mi+1 ⊆ M ′ byinduction. Then, M ′ ⊃M0, which is a contradiction.

This gives a filtration of M ′ = M ′0 ⊇ M ′1 ⊇ . . . ⊇ M ′n = 0 where we leaveout any repeated factor to get length < n. This is a composition series becausesuccessive quotients are simple. Thus, if we start with a composition series forM of minimal length, then M ′ has smaller length.

Now suppose that M = N0 ⊇ N1 ⊇ . . . ⊇ Nk is a chain of submodules of M .By assumption, M has a composition series of length n. We will show that k ≤n. When n = 0, M = 0, so k = 0. Assume now that for m < n a compositionseries of lenght m implies that all filtrations of submodules have length ≤ m.Now N1 is a proper submodule of M , and so has length < length(M) ≤ n. Sothis means that k − 1 < n, so k ≤ n.

Thus, every chain of submodules is finite, and so in particular every ascend-ing chain and every descending chain is, so M is Artinian and Notherian.

Corollary 8.3. If M has length n, then every composition series of M haslength n.

Corollary 8.4. If R is of finite length as an R-module, then R is Artinian andNotherian.

Theorem 8.5. TFAE

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1. R is Notherian and all primes are maximal.

2. R is of finite length as an R-module.

3. R is Artinian.

Proof. 1 ⇒ 2: Let R be Notherian with all primes maximal. Suppose that Ris not of finite length as an R-module. Let I be an ideal maximal with respectto the property that R/I is not of finite length as an R-module. We claimthat I is prime. If not, we can find a, b ∈ R \ I with ab ∈ I. Then consider0 → R/(I : a) → R/I → R/(I, a) → 0. Both (I : a) and (I, a) are strictlylarger ideals, so R/(I : a) and R/(I, a) have finite length as R-modules. Wecan now get a finite composition series for R/I from the ones for R/(I : a) andR/(I, a), because length is additive in short exact sequences. This contradictsour assumption on I, so I is prime. As I is prime, it is maximal, so R/I is afield. Which has finite length. This contradiction means that I does not exist,so R has finite length as an R-module.

2⇒ 3: follows from the previous proposition.3 ⇒ 1: Suppose that R is Artinian. We’ll first show that 0 is a prime of

finitely many maximal ideals. Since R is Artinian, we may choose from all idealsin R that are products of finitely many maximal ones a minimal one J . We wantto show that J = 0. For each maximal ideals M , we must have MJ = J , soJ ⊂M , so J is contained in the intersection of all maximal ideals. J2 = J

If J is nonzero, we can find an ideal I minimal wrt not annihilating J . Since(IJ)J = IJ2 = IJ 6= 0, IJ ⊆ I. We must have IJ = I by minimality of I.Also, ∃f ∈ I with fJ 6= 0, so I = (f). Since IJ = I, ∃g ∈ J with fg = f , sof(g − 1) = 0, but g − 1 is in no maximal ideal, so it is a unit. Thus f = 0, andso I = 0 and J = 0.

The above is LEMMA: If R has finite length as an R-module, then 0 =M1 . . .Mt where Mi are maximal.

Consider the R-module Vs = M1 . . .Ms/M1 . . .Ms+1 for 0 ≤ s < t. Thisis an R/Ms+1-module, and so a vector space. Subspaces of Vs correspond toideals of R containing M1 . . .Ms+1 and contained in M1 . . .Ms. Since R isArtinian, Vs must be finite dimensional, so has a finite composition series. Wenow glue together the composition series for R/M1,M1/M1M2, . . . to get a finitecomposition series for R.

So R has finite length as an R-modules, and is thus Notherian. Now let Pbe a prime in R. Then 0 = M1 . . .Mt ⊆ P , so at least one Mi ⊆ P . But sinceMi is maximal, Mi = P . Thus we have only a finite number of primes in R,each of which is maximal.

Corollary 8.6. If R is Notherian and dimension 0, then R is Artinian andthere are only finitely many primes. Also, (0) is the product of powers of theseprimes.

Corollary 8.7. Let V = V (I) ⊆ kn be a variety over k = k. TFAE

1. V is a finite set.

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2. k[x1, . . . , xn]/√I is a finite dimensional k-vector sapce whose dimension

is |V |.

3. k[x1, . . . , xn]/√I is Artinian.

Proposition 8.8. Let R be Notherian and M a finitely generated R-module.TFAE

1. Some finite produt of maximal ideals∏si=1 Pi annihilates M .

2. All primes that contain the annihilator of M are maximal.

3. R/Ann(M) is Artinian.

Proof. Suppose that∏Pi annihilates M . Let P ⊃ Ann(M). Then P ⊇

∏Pi.

So there exists i with Pi ⊆ P , and so Pi = P , as Pi is maximal.R is Notherian, so R/Ann(M) is, and every prime in R/Ann(M) is maximal,

so by the theorem, R/Ann(M) is Artinian.IfR/Ann(M) is Artinian and Notherian, then 0/Ann(M) =

∏(Pi/Ann(M)),

so∏Pi annihilate M .

Recall: If P ⊆ R is prime, codimP = dimRP . (if (R,m) is local, codimm =dimR)

For general I, codim I = min codimP for P ⊃ I.Goal: R Notherian.

Theorem 8.9. If x1, . . . , xc ∈ P and P is minimal among primes containingx1, . . . , xc, then codimP ≤ c.

This is a generalization of

Theorem 8.10 (Principle Ideal Theorem). If x ∈ R and P is minimal amongprimes of R containing x, then codimP ≤ 1.

Proof. We will show that every prime Q properly contained in P has codim 0.Since P is minimal over x, all primes in RP /x are maximal (since PP /x is

the only one) and RP /x is Notherian, so RP /x is Artinian. Also, x /∈ Q.Look at the chain QQ + (x), Q2

Q + (x) + . . . in RQ. It must stabilize, soQnQ + (x) = Qn+1

Q + (x). So for any f ∈ QnQ, we can write it as f = ax+ g withg ∈ Qn+1

Q and a ∈ RQ.So ax ∈ QnQ, and thus a ∈ QnQ, since x is not.Thus, QnQ = xQnQ+Qn+1

Q , so the finitely generatedRQ/Qn+1Q -moduleQnQ/Q

n+1Q

satisfies QnQ/Qn+1Q = xQnQ/Q

n+1Q .

The Idea WOULD HAVE BEEN appeal to Nakayama to get that QnQ =Qn+1Q , and thus, QnQ = 0 and thus codimQ = 0. Look in book, and attempt to

fill in details.

Theorem 8.11. If x1, . . . , xc ∈ R and P is minimal over primes containingx1, . . . , xc, then codimP ≤ c.

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Proof. Localize at P to assume that R has a unique maximal ideal. Then sinceall primes in R/(x1, . . . , xc) are maximal, as P is the only one, we know thatAnn(R/(x1, . . . , xc)) ⊃ P k for some k. Choose P1 prime with P1 ( P with noprimes between them. As R is Notherian, this exists if codimP > 0.

We will show that P1 is minimal over an ideal generated by c− 1 elements,which will, by induction that codimP1 ≤ c − 1, and so since P1 was arbitrary,codimP ≤ c.

Since P is minimal over (x1, . . . , xc), there must be an xi, say x1, withx1 /∈ P1. Thus P is minimal over (P1, x1), and so ∃s with P s ⊆ (P1, x1).

So for 2 ≤ j ≤ c, we can write xsj = ajx1 + yj with yj ∈ P1. Now we claimthat P1 is minimal over (y2, . . . , yc). Indeed, P is minimal over (x1, y2, . . . , yc).

So codimP/(y2, . . . , yc) in R/(y2, . . . , yc) is ≤ 1 by PIT. So P1/(y2, . . . , yc)has codim 0, and so P1 is minimal over (y2, . . . , yc).

Corollary 8.12. (x1, . . . , xn) ⊂ k[x1, . . . , xn] has codim = n.

Corollary 8.13. Prime ideals in a Notherian ring satisfy the DCC with thelength of a chain of ideals descending from a prime bounded by the number ofgenerators of P

Corollary 8.14 (Converse to PIT). Any prime P of codimP = c is minimalover an ideal generated by c elements.

eg, R = k[a, b, c, d] and I = (ad− bc, ac− b2, bd− c2). dimR/I = 2.

Corollary 8.15. Any prime P of codim c is minimal over an ideal generatedby c elements. (actually of codim c)

Proof. The proof is by induction on c.If c = 0, then P is minimal over 0, which has codim 0.Now suppose that the corollary is true for smaller c and choose P1 prime,

maximal proper in P of codim c− 1.Then, by induction, P1 is minimal over (x1, . . . , xc−1) of codim c− 1. Each

prime Qi minimal over (x1, . . . , xc−1) has codimension c− 1.So P ( Qi for any Qi minimal over (x1, . . . , xc−1), so by prime avoidance,

P 6⊆ ∪Qi. So we can find xc ∈ P \ ∪Qi. Then we must have P minimalover (x1, . . . , xc), as if P ′ ( P contains it, then there exists P ′′ minimal over(x1, . . . , xc−1) which has codim c− 1, so codimP > c, which is impossible.

Also, if Q is minimal over (x1, . . . , xc), then ∃P ′ with (x1, . . . , xc−1) ⊆P ′ ( Q and codim P ≥ c − 1. So codimQ ≥ c. By PIT codimQ = c, socodim(x1, . . . , xc) = c.

Corollary 8.16. Let (R,m) be a local ring. Then dimR = min{d|∃x1, . . . , xd ∈m with mn ⊂ (x1, . . . , xd) for n >> 0}.

Proof. If mn ⊂ (x1, . . . , xd) ⊂ m, then m is minimal over (x1, . . . , xd). Socodimm = dimR ≤ d by the PIT.

Conversely, it d = dimR = codimm, then by the converse to the PIT,∃x1, . . . , xd with m minimal over (x1, . . . , xd). But R/(x1, . . . , xd) has only the

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one prime ideal, we have that all elements ofm are nilpotent module (x1, . . . , xd).So ∃k such that mk ⊂ (x1, . . . , xd), because the ring is Notherian. (since ifm = (y1, . . . , ys), yNi = 0 for all i, then msN = 0)

Definition 8.3 (System of Parameters). A system of parameters for m ⊂R with (R,m) local, is a sequence x1, . . . , xd, d = dimR such that mn ⊂(x1, . . . , xd) for n >> 0.

Parameters are a ”local coordinate system” (up to finite ambiguity). eg,k[x, y](x,y), parameters (x2 − y, y), so (up to the square) you are picking ahorizontal line and a parabola to determine a point (determines 2, but that’sfinite ambiguity).

Recall: A family of varieties is U π→ B. The fiber over b is ”π−1(b)” Alge-braically, this is a map R→ S with fiber over P ⊆ R being S ⊗R/P ' S/PS.

We expect dimU ≤ dimB+dimπ−1(b) for each b. Recall the blowup, π−1(0)has dimension 1 and dimB = dimU = 2 for the plane at the origin.

Theorem 8.17. If (R,m) → (S, n) is a map of local rings, then dimS ≤dimR+ dimS/mS.

Proof. Write d = dimR, e = dimS/mS. Then there exist x1, . . . , xd, s withms ⊆ (x1, . . . , xd) and y1, . . . , ye, t such that nt ⊆ (y1, . . . , ye) +mS.

Then, nst ⊆ ((y1, . . . , ye)+mS)s ⊆ msS+(y1, . . . , ye) ⊆ (x1, . . . , xd, y1, . . . , ye)S,so n is minimal over na ideal generated by d+e elements, so codimn = dimS ≤d+ e = dimR+ ∼ S/mS.

Theorem 8.18. If (R,m) → (S, n) is a map of local rings and S is flat overR, then dimS = dimR+ dimS/mS.

We’ll need the following:

Lemma 8.19. If N is a flat R-module and R → T is any ring map, thenN ⊗R T is flat over T .

Lemma 8.20. Suppose ϕ : R → S is a map of rings with S flat over R. IfP ⊃ P ′ are primes of R, and Q is a prime of S with ϕ−1(Q) = P , then ∃Q′ ofS contained in Q with ϕ−1(Q′) = P ′.

Corollary 8.21. If (R,m) is a local ring, then dim Rm = dimR.

Proof. Rm is flat over R and R/mR ' R/m is a field, so dim R/mR = 0.

Theorem 8.22. 1. If k is a field, then dim k[x1, . . . , xn] = n.

2. In general, dimR[x] = dimR+ 1

3. If P is a prime of R, then ∃ a prime Q of R[x] with Q ∩ R = P , andfor a maximal such ideal, dimR[x]Q = 1 + dimRP , so codimR[x]Q =1 + codimR P .

Proof. 1. Follows from 2 by induction on n.

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2. If P1 ⊂ . . . ⊂ Pd of R, we get P1R[x] ⊂ P2R[x] ⊂ . . . ⊂ PdR[x] ⊂PdR[x] + (x). (using PR[x] ∩ R = P and if P is prime then PR[x] isprime). So dimR[x] ≥ dimR+ 1.

Conversely, if Q is a maximal ideal of R[x], then Q is maximal amongprimes meeting Q ∩ R, so by part 3, we get codimQ = 1 + codimQ ∩ R,so dimR[x] ≤ 1 + dimR.

3. We first prove the case where R is a field and P = 0. Then Q∩R = 0 forany proper prime of R[x], so we take Q to be any maximal ideal of R[x].Then codimQ = 1 = 1 + dimR.

In general, PR[x] is a prime in R[x] with PR[x] ∩ R = P . Localize at Pto assume that R is local and P is maximal. Let Q be a maximal ideal ofR[x] containing P . Then Q ∩ R = P . So all the remains to be shown isthat codimQ = 1 + codimP .

If P0 ⊂ . . . ⊂ Pd = P is a chain of primes in P , then P0R[x] ⊂ . . . PdR[x]is a chain in R[x]. Look at Q/PR[x] ⊆ R[x]/PR[x] ' (R/P )[x], thencodimQ/PR[x] = 1. So codimQ ≥ d+ 1.

Then codimQ ≤ dimRP+dimR[x]Q/PR[x]Q = dimRp+1 = codimP+1.

Therefore, dim k[x1, . . . , xn] = n.

Definition 8.4 (Regular Local Ring). A ring (R,m) of dimension d is a regularlocal ring if m can be generated by d elements.

If R is a regular local ring, then Nakayama says that dimR/mm/m2 = d.

Thus, every generating set for m has size d.A generating set of size d is called a regular sequence of parameters.

Definition 8.5 (Regular Sequence). In general, a regular sequence is a se-quence x1, . . . , xd such that (x1, . . . , xd) is proper and xi+1 is a nonzero-divisoron R/(x1, . . . , xi) for all i.

Definition 8.6 (Cohen-Macaulay). A ring R is Cohen-Macaulay if there is aregular sequence of length dimR.

Goal: degPR + 1 = dimR.Let R = ⊕i≥0Ri be a Notherian graded ring with R0 a field and R is gen-

erated as an R0-algebra by R1. This is sometimes called a standard-gradedalgebra.

This means R = k[x1, . . . , xn]/I for some n and homogeneous I.Why? Because R is Notherian, R1 is a finite dimensional R0-vector space.

Then k[x1, . . . , xn] surjects onto R, so take I to be the kernel, which is homo-geneous as this is a graded homomorphism of rings.

Definition 8.7 (Hilbert Function). The Hilbert function of R is HR(t) =dimk Rt.

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In homework, we showed that there exists a polynomial PR(t) which is equalto the Hilbert Function for sufficiently large t.

If M is a graded R-module, then HM (t) = dimkMt. Again, it eventuallywill agree with a polynomial.

Lemma 8.23. If 0 → M ′ → M → M ′′ → 0 is a graded (ie, degree 0) shortexact sequence of modules, then HM (t) = HM ′(t) +HM ′′(t).

Proof. The ses is the direct sum of ses 0 → M ′d → Md → M ′′d → 0 of k-vectorspaces, so additivity follows.

Lemma 8.24. If x is a nonzero divisor on R, homogeneous of degree 1, thenHR(t) = HR(t − 1) + HR/x(t) so PR/x(t) = PR(t) − PR(t − 1) for all t, sodegPR = degPR/x + 1.

Proof. Consider the ses 0 → R[−1] x→ R → R/x → 0 where R[−1] is R withgraded R[−1]n = Rn−1. In general, R[a]b = Ra+b.

So HR(t) = HR[−1](t) +HR/x(t) = HR(t− 1) +HR/x(t).

Fact: If R is graded etcetera, then dimR = max{codimQ|Q is a homoge-neous prime}.

Proposition 8.25. If deg(x) > 0 and x is a nonzerodivisor on R then dimR/x =dimR− 1.

Proof. Since x is a nzd, x is not contained in any associated prime, so in par-ticular, x does not lie in any minimal prime.

If P0 ⊂ . . . ⊂ Pd is a chain of primes in R/x, that is, a chain of primes in Rcontaining x, then ∃ prime P ( P0 so dimR > dimR/x.

It remains to show that dimR/x ≥ dimR − 1. Since R is graded anddimR < ∞, there is a homogeneous prime Q with codimQ = dimR. Supposethat dimR/x = e < dimR− 1 with d = dimR.

We know that x ∈ Q since (Q, x) is proper (since Q, x are homogeneous) socontained in a maximal ideal. But Q is maximal. This menas that RQ/x hasdim ≤ e. So there are x1, . . . , xe ∈ Q qwith QnQ ⊆ (x1, . . . , xe) + (x) for n >> 0.So QnQ ⊆ (x1, . . . , xe, x), and so codimQ = dimRQ ≤ e+ 1, so d ≤ e+ 1.

Theorem 8.26. dimR = degPR + 1

Proof. The proof is by induction on dimension. If dimR = 0, then R is Artinian.So finite length ` as an R-module, so every filtration has length ≤ `. If

R`+1 6= 0 R ⊂ R>0 ⊂ R>1 ⊂ . . . ⊂ R>` is a filtration of length ` + 1. IfR>j = R>j+1 then Rj+1 = 0, which implies that Rn = 0 for n ≥ j + 1.

So this means that Rn = 0 for n >> 0, so PR(t) = 0 which has degree −1by convention.

We now assume that dimR > 0 and that the theorem is true for smallerdimension. We first reduce to the case that m = R>0 is not an associatedprime. Since the zero divisors on R are the union of the associated primes, thiswill let us find a homogeneous nonzerodivisor (of degree 1)

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Let J = (0 :R m∞) = {f ∈ R|∃k with fmk = 0} = f ∈ R|∃k with fg = 0for all g ∈ mk}.

Let R′ = R/J . First note that m is not associated to R′, since if (0 : x) = m,then x ∈ J . Also note that PR = PR′ because Jt = 0 for t >> 0, becauseJ = (f1, . . . , fs) for some s, where we can take fi homogeneous.

Then there is a k such that fimk = 0 for all i, then Jt = 0 for t >>k + max dim fi. So Rt = R′t for t >> 0 so PR = PR′ .

Also, dimR = dimR′, so since if P0 ⊂ . . . ⊂ Pd is a chain of length d = dimRin R, then since dimR > 0, P0 6= m, so we can find x ∈ m \ P0 homogeneous,and then if f ∈ J , fxk = 0 for k >> 0 so fxk ∈ P0, so f ∈ P0. Thus J ⊆ P0,so dimR′ = dimR.

Thus, we assume that m is not an associated prime of R. So we can findx ∈ R1 a nonzerodivisor on R. Then dimR/x = degPR/x+1, by induction. Andso, dimR = dimR/x+1 and degPR = degPR/x+1, so dimR = degPR+1.

Fact: R graded, etc, then dimR = max codim of homogeneous Q. Why?Follows from dim k[x1, . . . , xn]/I = tr degk R and all maximal ideals have thesame codimension. This itself follows from Nother Normalization.

Theorem 8.27 (Nother Normalization). If R = k[x1, . . . , xn]/J then there existy1, . . . , yd ∈ R such that k[y1, . . . , yd] ⊆ R and R is finite over S. (can take yito be homogeneous).

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