RINGS AND FIELDS ‘07 - Auburn Universitywebhome.auburn.edu/~goetehp/algtex/gradalg2_07.pdf · RINGS AND FIELDS ‘07 Deﬂne ring, subrings, modules, and submodules. Example 1

RINGS AND FIELDS ‘07

RINGS AND FIELDS ‘07

Define ring, subrings, modules, and submodules.

Example 1. Rings:

(a) We have the well-known rings; Z, Q, R, and C.(b) Inside C we find rings as follows; select an element α that is

the root to some polynomial xn +an−1xn−1 +an−2xn−2 + · · ·+a0where the a′js are integers, and set R = Z[α] = {t ∈ C | t =bn−1αn−1 + bn−2αn−2 + · · ·+ b0, where each bi ∈ Z}.

(c) Given any ring R, we can construct a ring from the collectionof n× n matrices whose entries lie in R;

S = Matn(R),

where multiplication and addition are the standard matrix op-erations.

(d) The upper, 2×2 matrices over a commutative ring R with equaldiagonal entries forms a commutative subring of Mat2(R).

Example 2. Modules Over a Ring R:

(a) For an index set I, ΠIR and ⊕IR are modules over R.(b) An R-module F is called free on a subset X if for every map

f : X → A such that f(rx + sy) = rf(x) + sf(y) for x, y ∈X and r, s ∈ R, where A is an R-module, there is a uniquehomomorphism h : M → A such that h|X = f .

(c) An R-module M is called projective if, for every epimorphismf : A → B and homomorphism g : M → B, there is a homo-morphism

Theorem 1. An R-module M is free if and only if M ∼= ⊕IR.Theorem 2. An R-module M is projective if and only if M ⊕N = Ffor some free module F .

Let us use the projective and free modules as a foundation for ourcourse.

1

2 RINGS AND FIELDS ‘07

Theorem 3. Every R-module is free if and only if R is a division ring.

When is it true that every R-module is projective?

Artin-Wedderburn Theorem . Every R-module is projective if andonly if R is a finite direct product of matrix rings over division rings;i.e.,

R = Matn1(D1)×Matn2(D2)× · · · ×Matnk(Dk),for some natural numbers n1, n2, . . . , nk and division ringsD1, D2, . . . , Dk.

Example 3. Let H denote the (rational) Hamiltonian Quaternions;

H = {a + bi + cj + dk | a, b, c, d ∈ Q},where addition is component-wise and multiplication extends linearlyfrom i2 = j2 = k2 = −1. Then, the left ideal of R = Mat2(H) consist-ing of the matrices with second column zero is a projective (left) idealof R that is not free.

Example 4. Let R = Z[√−5] = {a + b√−5 | a, b ∈ Z and I = (3, 1 +√−5). We claim that I is projective but not free. First we will argue

that I 6= R. Notice 3R ⊆ I. Modulo 3R, the general element of I canbe represented as (1+

√−5) ·(a+b√−5) = (a−5b)+(a+b)√−5, whichis congruent to (a + b)[1 +

√−5] mod 3R. Therefore, I/3R ∼= Z/3Z,and since R/3R has order 9, I is properly contained in R.

If I were free, then I would be principal; say I = (a + b√−5). Since

I contains 3, 1+√−5, by computing norms, a2 +5b2 must divide 9 and

6. Hence a2 + 5b2 = 1 or 3. We rule out 1 since I 6= R. However,a2 + 5b2 = 3 is not possible. Therefore, I is not free.

Set

α =2

3(1 +

√−5), and β = −1.Note that α · I ⊆ R and β · I ⊆ R. Define the homomorphism θ :F = R ⊕ R → I by θ(r, s) = r3 + s(1 + √−5). The epimorphismθ is split by the map δ : I → F where δ(x) = (βx, αx). (Check:θ(δ(x)) = θ((βx, αx)) = 3βx + (1 +

√−5)αx = x(−3 + 236) = x.

Therefore, I is projective.

Example 5. The ring R = Z[√−5] of the last example has the property

that every nonzero nonunit can be expressed as a product of irreducibleelements (see homework), yet R is not a UFD (see the next example).

RINGS AND FIELDS ‘07 3

Example 6. Let d be a square-free integer 6= 1, and defineR = Z[

√d] provided d ≡ 2, 3 (mod 4),

or

R = Z[1 +

√d

2] when d ≡ 1(mod 4).

Then every ideal of R is projective. It is not known whether or notthere are infinitely many d′s for which R is a PID. However, for suchan R, R is a PID if and only if R is a UFD.


1. EXERCISE SET 1

In this homework set, R is an integral domain.A nonzero, nonunit r of R is said to be irreducible provided r = st

for s, t ∈ R can only happen when one of s or t is a unit. (Recall,u ∈ R is a unit if uv = 1 for some v ∈ R). A nonzero, nonunit r of Ris said to be prime if r | ab for a, b ∈ R, implies r | a or r | b.

R is said to have the ascending chain condition on principal ideals,if any increasing chain of ideals

(a1) ⊆ (a2) ⊆ (a3) ⊆ · · ·stablizes (i.e., there is an index n such that (am) = (an) for all m ≥ n).

1. Show that if R has the ascending chain condition on principalideals, then every nonzero nonunit r of R is a product of ir-reducible elements. Hint: To argue that r has an irreduciblefactor, either r is irreducible, or r = r1s1 with r1, s1 nonzerononunits. Note (r) is properly contained in (r1) in this case. Re-peat for r1 and use the hypothesis to eventually obtain r = s1t1with s1 irreducible. Now write r = r1v1 = r1r2v2 = . . . withr1, r2, . . . irreducible, and use the hypothesis to show that even-tually some vj is a unit.

2. Show that if R is a UFD then every irreducible element is primeand R has the ascending chain condition on principal ideals.

3. Show that if irreducible elements are prime in R and R has theascending chain condition on principal ideals, then R is a UFD.

4. Show that R is a UFD if and only if every nonzero nonunit is aproduct of primes.

5. Show any PID is a UFD.


2. Integral Domains

Theorem 4. Let R be an integral domain. Every submodule of ⊕nRfor any n is free if and only if R is a PID.

Proof. An ideal of R is a submodule of R. Hence if the condition isin force, R is necessarily a PID. Conversely, assume that R is a PID.We will show, by induction on n, that every submodule of ⊕nR is free.When n = 1, the nonzero submodules of R are ideals and are principal(hence isomorphic to R).

Inductively, let L be a nonzero submodule of F = ⊕nR. For πj : F →R equal to the projection map onto the jth coordinate, I = πj(L) 6= 0for some j. Since I = aR for some 0 6= a ∈ R, we can define a splittingmap f : aR → L by f(ar) = rx where x ∈ L is any element such thatπj(x) = a. By the standard argument,

L = Im f ⊕Ker πj|L ∼= aR⊕Ker πj|L.Note that Ker πj|L = {y ∈ L | y has a zero jth component} ≤ ⊕i6=jR,where the index i runs over 1, 2, . . . , j − 1, j + 1, . . . , n. By induction,Ker πj|L ∼= ⊕mR is free, and therefore, L ∼= ⊕m+1R. ¤

Given an ideal I 6= 0 of R, defineI−1 = {t ∈ Q | tIsubseteqR}.

The ideal I is said to be invertible, if

I · I−1 = R.Analogously, we have

Theorem 5. Let R be an integral domain. Every submodule of ⊕nRfor any n is projective if and only if every ideal of R is invertible (pro-jective).

In order to see this we need a few observations.

Bear Injective Test Lemma . The module U is injective if and onlyif U is injective relative to any sequence 0 → I → R where I is an idealof R.

Proof. Assume that U is injective with respect to any sequence 0 →I → R, and suppose that 0 → A → B is exact and f : A → U .Consider the set C of all 2-tuples (g, C) where A ≤ C ≤ B and g|A = f.There is a maximal element (g0, C0).

If C0 6= B let x ∈ B \ C0, and let I = {r ∈ R | rx ∈ C0}. SetC ′ = C0 + 〈x〉. If I = 0, then 〈x〉 ∩ C0 = 0 and so C ′ = C0 ⊕ 〈x〉 ≤ B;we then have the over-module C ′ of C0 and can define g′ : C ′ → U by


g′(c0, rx) = g(c0), contradicting the maximality of (g, C0). ThereforeI 6= 0.

By hypothesis, the map h : I → U given by h(r) = g(rx) canbe lifted to a map h′ : R → U . We then define g′ : C ′ → U byg′(c0 + rx) = g(c0) + h′(r). If c0 + rx = b0 + sx with c0, b0 ∈ C0 andr, s ∈ R, then (r − s)x = b0 − c0 ∈ C0 and so r − s ∈ I. But theng(b0 − c0) = g((r − s)x) = h(r − s) = h′(r − s) = h′(r) − h′(s) andso g(b0) − g(c0) = h′(r) − h′(s) implying g(b0) + h′(s) = g(c0) + h′(r)making g′ well-defined. It is easy to check that g′ is a homomorphismthat extends g which contradicts the maximality of (g, C0). Thus,B = C0 as desired. ¤Corollary 6. The quotient field Q of R is injective.

Proof. Let I be an ideal of R and 0 6= a ∈ I. If 0 6= f : I → Q,define h : R → Q to be multiplication by (1/a)f(a). If b ∈ I, thenh(b) = (1/a)f(a) · b = (1/a)f(ab) = (1/a)af(b) = f(b). By Baer’sInjective Test Lemma, Q is injective. ¤Corollary 7. If I, J are ideals of R, then Hom(I, J) is naturally iden-tifiable with {t ∈ Q | tI ⊆ J}.Proof. It is well-known that Hom(I, Q) can be identified with Q viamultiplication the elements of Q, and that Hom(I, J) is naturally asubmodule of Hom(I, Q). The assertion follows. ¤Lemma 8. Let I 6= 0 be an ideal of R. Then I is invertible if and onlyif I is projective.

Proof. Suppose II−1 + R. Then 1 = r1s1 + r2s2 + . . . + rnsn withrj ∈ I and sj ∈ I−1 for each j by the definition of II−1. The mapθ : F = ⊕nR → I given by θ(x1, x2, . . . , xn) = Σjrjxj is split by themap f : I → F given by f(a) = (as1, as2, . . . , asn) ∈ F . Therefore,I ⊕Ker θ ∼= F and I is projective.

Conversely, if I is projective, there is an epimorphism θ : F → Ithat is split by some map f : I → F . Let θ(ei) = xi ∈ I where ei is thestandard idempotent element of F having a 1 in the ith-component and0′s elsewhere. Write f = (f1, f2, . . . , fn) where fj : I → R. By the lastcorollary, regard fj ∈ Hom(I, R) = {t ∈ Q | tI ⊆ R} = I−1. It followsthat Σjxjfja = a for every a ∈ I, and consequently that Σjxjfj = 1.That is, I · I−1 = R.

¤The proof of the last theorem now follows easily from the proof at

the beginning this section in light of the last lemma.


3. Dedekind Domains

A domain such that every nonzero ideal is invertible is called aDedekind domain.

An R-submodule J of Q is called a fractional ideal of R if rJ ⊆ Rfor some 0 6= r ∈ R.

The domain R is said to be integrally closed if the only over-ring ofR inside Q that is fractional over R is R itself.

An ideal P of R is said to be prime, if ab ∈ P implies a ∈ P or b ∈ Pfor any a, b ∈ R. Equivalently, P is prime if R/P is an integral domain.

Recall that a UFD is a domain such that every nonzero nonunit is aproduct of primes.

Theorem 9. TFAE:

(a) R is Dedekind.(b) Every nonzero ideal is 11

2-generated (i.e., can be generated by

two nonzero elements with the first generator chosen arbitrar-ily).

(c) Every proper ideal of R is (uniquely) a product of prime idealsof R.

(d) R is integrally closed, each nonzero prime ideal is a maximalideal, and every ideal of R is finitely generated.

(e) R is Noetherian and for every prime ideal P of R, RP is is aPID.


4. A Fundamental Theorem

A module is free if it is isomorphic to a direct sum of copies of R.The number of copies of R is called the rank of the free module. Note⊕nR ∼= ⊕mR if and only if n = m. (One way to justify this is to notethat ⊕nR has at most n linearly independent elements).

If M is a module over an integral domain, the torsion submodule Tof M is

T = {x ∈ M | rx = 0 for some 0 6= r ∈ R}.A module whose torsion submodule is zero is said to be torsion-free.For example, the torsion submodule of M/T is zero, and so M/T istorsion-free.

We can rephrase Theorem 4.

Proposition 10. If R is an integral domain, then any finitely generatedtorsion-free module is isomorphic of a submodule of ⊕nR for some n.Proof. Let C be finitely generated with generating set {x1, x2, . . . , xn}of nonzero elements. Select x1; if x2 is independent of x1, select x2;otherwise reject x2. Having selected a subset S of {x1, x2, . . . , xm} inthis fashion, select xm+1 if S ∪{xm+1} is an independent set; otherwisereject xm+1. In this way we find a linearly independent subset U of{x1, x2, . . . , xn} such that for any xi /∈ U , U∪{xi} is linearly dependent.

Define F = ⊕x∈U〈x〉, a free module contained in C. Given any indexj, there is a nonzero ring element rj such that rjxj ∈ F (this is becauseU is a linearly independent set but U∪{xj} is linearly dependent. Thus,if c = Σjaixi ∈ C with ai ∈ R, multiply by r = Πjrj 6= 0 to obtainrc = Σjairxi ∈ F . Therefore, C ∼= rC ≤ F and C is isomorphic to asubmodule of the free module F (the isomorphism sends c 7→ rc whichis 1− 1 since C is torsion-free). ¤

A domain is said to be noetherian if each ideal of R is finitely gen-erated.

Theorem 11. The following are equivalent for an integral domain R:

(a) R is noetherian.(b) The finitely generated torsion-free modules are precisely the mod-

ules that are isomorphic to a submodule of ⊕nR for some n.(c) If M is a finitely generated module, then any submodule of M

is finitely generated.(d) If K is a finitely generated module and K1 ⊆ K2 ⊆ K3 ⊆ · · · are

submodules of K, then there is an index m such that Kn = Kmfor all n ≥ m.


Proof. (a) → (b). Clearly the submodules of ⊕nR are torsion-free.Induction on n shows that the submodules are finitely generated aswell. Conversely, the case n = 1 shows that

(b) → (c). Let N be a submodule of M . As M is finitely generated,there is a free module F = ⊕nR and an epimorphism φ : F → M → 0.By the Correspondence Theorem, φ−1(N) = K is a submodule of Fand by (b) is finitely generated. Therefore N = φ(φ−1(N)) is finitelygenerated.

(c) → (d). Set N = ∪nKn. Since K is a submodule of M , K isfinitely generated; generated by x1, x2, . . . , xm say. For each i, there isan index ni such that xi ∈ Kni . With m = max{n1, n2, . . . , xn}, eachxi ∈ Km and so N ⊆ Km ⊆ N .

(d) → (a). Given an ideal I of R, let x1 ∈ I and I1 = Rx1. If I 6= I1,there is an x2 ∈ I \ I1. Set I2 = Rx1 + Rx2. If I 6= I2 there is anx3 ∈ I \ I2. Set I3 = ΣRxi. Continuing in this manner, by (d), someIk must coincide with I. ¤

Furthermore we have a fundamental splitting result:

Corollary 12. If M is a finitely generated module over a PID, thenM = T ⊕N where T is the torsion submodule of M , and N is free, offinite rank.

Proof. The map M → M/T splits because M/T is finitely generated,torsion-free (hence free) by the last theorem. ¤

Finitely Generated Modules Over a PID . If R is a PID and Mis a finitely generated module, then M is a (finite) direct sum of cyclicmodules.

Proof. It remains to show that any finitely generated torsion mod-ule T is a direct sum of cyclic modules. Given a prime p ∈ R, letTp = {t ∈ T | pkt = 0 for some k ≥ 1}. The submodule Tp is called thep-primary submodule of T . We claim that T = ⊕p primeTp where theindex includes exactly one p from each associate class of a given prime.(Note that Tp = Tq when p and q are associated primes.) This reducesthe problem to finitely generated primary modules (i.e., a finitely gen-erated module K such that for some prime p, every element in K canbe annihilated by a power of p).

Let 0 6= t ∈ T . Then rt = 0 for some 0 6= r ∈ R. Express r as r =pe11 p

e22 · · · pemm with p1, p2, . . . , pm non-associate primes. Set rj = r/pejj

and note that the greatest common divisor of r1, r2, . . . , rm is 1. Thisis because, any prime dividing rj must divide rjp

ejj = r and some must

be one of the p′is; but pi does not divide ri.


In general, it is easy to see that the gcd of a1, a2, . . . , an is a whereaR = a1R + a2R + · · · + amR. Write b1r1 + b2r2 + · · · + bmrm = 1 forsome b1, b2, . . . , bm ∈ R. Then t = (b1r1t) + (b2r2t) + · · ·+ (bmrmt) andwith tj = bjrjt, p

ejj tj = 0 so tj ∈ Tpj . Thus t ∈ ΣjTpj .

If x ∈ Tp1 ∩ Σnj=2Tpj with p1, p2, . . . , pm non-associate primes, thenpf11 x = 0 and x = x2 + x3 + · · · + xn with pfjj xj = 0 for j ≥ 2.The elements pf11 and p

f22 · pf33 · · · pfnn are relatively prime (i.e., have

gcd 1) and so apf11 + bpf22 · pf33 · · · pfnn = 1 for some a, b ∈ R. Then

x = (apf11 + bpf22 · pf33 · · · pfnn )x = apf11 x + bpf22 · pf33 · · · pfnn x = 0.

From the next home set, we obtain that any finitely generated nonzeroprimary module M has a factorization

M = 〈x〉 ⊕K,where x 6= 0. Since K is either 0 or can be likewise factored

K = 〈x2〉 ⊕K2,we obtain progressive decompositions

M = 〈x〉 ⊕ 〈x2〉 ⊕ · · · 〈xj〉 ⊕Kj,which must stabilize since the chain of submodules 〈x〉 ≤ 〈x〉 ⊕ 〈x2〉 ≤〈x〉 ⊕ 〈x2〉 ⊕ · · · 〈xj〉 stabilizes (i.e., terminates with some Kj = 0).Therefore, M is a direct sum of cyclics and the proof is complete. ¤

Analogously, we have

Finitely Generated Modules Over a Dedekind Domain . If Ris Dedekind and M is a finitely generated module, then M = P ⊕ Cwhere P is a projective module and C is a finite direct sum of cyclicmodules.


5. EXERCISE SET 2

In this set R is a PID. A module M is said to be p-primary wherep is a prime element of R, if for every x ∈ M there exists a naturalnumber k such that pkx = 0.

1. Let M be a finitely generated p-primary module. Show thereexists a natural number n such that pnM = 0 (i.e., for everyx ∈ M , pnx = 0).

2. Let M be a finitely generated p-primary module. Argue thatchains of submodules K1 ⊆ K2 ⊆ · · · ⊆ Kn ⊆ · · · must stablize(i.e., there exists an m such that for all n ≥ m, Kn = Km).

3. Let M be a finitely generated p-primary module.(a) The order of an element 0 6= y ∈ M is the least positive

integer ` such that p`y = 0 (this is justified by Problem1). Use Problem 1 to show there is an element x ∈ M thathas maximal order k (i.e., pkx = 0, while pk−1x 6= 0 and if0 6= y ∈ M , then pky = 0).

(b) With the x in part (a), show there is a submodule K of Mthat is maximal with respect to 〈x〉 ∩K = 0. (Hint: UseProblem 2)

4. Let M be a finitely generated p-primary module. This problemshows that for x and K from Problem 3, M = 〈x〉 ⊕ K. LetX = 〈x〉.(a) Given r ∈ R a nonzero, nonunit, write r = pis such that

i ≥ 0 and p does not divide s (R is a UFD). Show thatry ∈ X ⊕ K if and only if piy ∈ X ⊕ K. Hint: For onedirection, pn, s relatively prime, where pny = 0, impliesapn + bs = 1 for some a, b ∈ R. Multiply by pi.

(b) To show that M = 〈x〉 ⊕K, it suffices to show that py ∈X⊕K implies y ∈ X⊕K, for every y ∈ M , using inductionand part (a).

(c) We will now assume that py ∈ X ⊕ K and show thaty ∈ X⊕K. Show there is an index j < k (where k is givenin Problem 1) such that pjpy = 0. (Take j = 0 if py = 0and otherwise compare the orders of y and x).

(d) Write py = rx + z with r ∈ R and z ∈ K. Using the jfrom part (c), conclude that p | r. Write r = ps.

(e) With K ′ = K + 〈y − sx〉, either K = K ′ or X ∩K ′ 6= 0.In the latter case, write 0 6= tx = z′ + a(y − sx) for somet, a ∈ R and z′ ∈ K. Argue that p does not divide a sinceotherwise, tx ∈ K.


(f) Continuing part (e), write up + va = 1, and conclude thaty = upy + vay ∈ X ⊕K.


6. EXERCISE SET 3

In this set, R is an integral domain.

1. Show that if F is a field, then R = F [x] is a PID. Hint: use thedegree function deg : F [x] → N and the division algorithm inF [x] to show that if f(x) 6= 0 is a polynomial of smallest degreein an ideal I 6= 0, then I = (f(x)).

2. Show that Z[x] is not a PID. (Hint: Show that I = (2, x) is notprincipal.)

3. Let R be a UFD with quotient field Q, and let f(x) ∈ R[x].(a) Show that there is an element r ∈ R and a polynomial

g(x) ∈ R[x], such that f(x) = rg(x) and the gcd of thecoefficients of g(x) is 1. (r is called the content of f and wewrite c(f) = r).

(b) Show that if f(x) = g(x)h(x) with g(x), h(x) ∈ Q[x], thenthere exists polynomials g1(x), h1(x) ∈ R[x] each havingcontent 1, and elements 0 6= a, a1, b, b1 ∈ R such thata, a1 and b, b1 are coprime pairs, and ag(x) = a1g1(x) andbh(x) = b1h1(x). Thus, abf(x) = a1b1g1(x)h1(x).

(c) Reduce ab and a1b1 in part (b) as necessary to obtaindf(x) = eg1(x)h1(x) with the gcd(d, e) = 1. Show thatif c(f) = 1, then d and e are units.

(d) Conclude that f(x) ∈ R[x] is irreducible in R[x] if and onlyif f(x) is irreducible in Q[x].

4. Let R be a UFD. Using Problem 1 and Problem 3 (d), showthat R[x] is a UFD. Conclude that Z[x] is a UFD that is not aPID.

5. The Hilbert Basis Theorem asserts that if R is noetherian, thenso is R[x]. Assuming R is a noetherian domain, show that R[x]is noetherian. Hint: If J is an ideal in R[x], let In be the idealof elements that occur as coefficients of xn in some polynomialof degree n in J . Observe that I0 ⊆ I1 ⊆ I2 ⊆ · · · . Use this topick generators for J .

7. Primary Decompositions

In this section R is a commutative noetherian ring (it need not bea domain, but the issue never arises). Since we are not restrictingourselves to domains, the symbol Q is in play and will be used here todenote a special type of ideal.

Recall that an ideal P 6= R is called a prime ideal, if ab ∈ P impliesa ∈ P or b ∈ P . If one wishes, we can define the term I divides J whenI and J are ideals, to be the condition J ⊆ I. It is an easy exercise


to show that P 6= R is prime if and only if P divides I · J implies Pdivides I or P divides J , for ideals I and J .

We say that an ideal Q 6= R is primary, if for any a, b ∈ R, ab ∈ Qand a /∈ Q, implies bn ∈ Q for some natural number n. While primaryintegers are just the powers of primes, this is not the case for ideals ingeneral rings of our type.

The radical of an ideal I, rad(I), is defined to be

rad(I) = {r ∈ R | rn ∈ I for some n ≥ 1}.Proposition 13. (1) For any ideal I, there exists an n ≥ 1 such

that rad(I)n ⊆ I.(2) If Q is primary, then rad(Q) is prime.

Proof. (1) It is easy to see that rad(I) is an ideal, and since Risnoetherian, there exists a1, a2, . . . , am ∈ rad(I) such that rad(I) =〈a1, a2, . . . , am〉. For each j there exists an positive integer ki for whichakii ∈ I. Take n = k1 + k2 + · · · + km. Then, the general elementr1a1 +r2a2 + · · ·+rmam of rad(I) satisfies (r1a1 +r2a2 + · · ·+rmam)n =Σ n!

i1!i2!···im!(r1a1)i1(r2a2)

i2 · · · (rmam)im where the sum is over all non-negative integers i1, i2, . . . , im that sum to n (Multinomial Theorem).In any monomial (r1a1)

i1(r2a2)i2 · · · (rmam)im , one of the i′js must be

greater than kj and so that monomial lies in I. Therefore, rad(I)n ⊆ I.

(2) Let P = rad(Q) and suppose ab ∈ P with a /∈ P . By definition(ab)n ∈ Q for some n. Since a /∈ P , an /∈ Q and since Q is primary,(bn)m ∈ Q for some m. Then bnm ∈ Q and b ∈ P . ¤

A proper ideal I is called irreducible, if

I = J1 ∩ J3 =⇒ I = J1 or I = J2,for ideals J1, J2.

Proposition 14. Any irreducible ideal is primary.

Proof. Let I be irreducible and suppose that I is not primary. Thereexists a, b ∈ R, such that ab ∈ I, a /∈ I and bn /∈ I for every n ≥ 1.

Let In = {r ∈ R | rbn ∈ I}; rbn ∈ I implies rbn+1 ∈ I and soIn ⊆ In+1 for all n. Since R is noetherian, there is an index m suchthat In = Im for all n ≥ m. Let I ′ = {rbm + c | r ∈ R, c ∈ I}.

CLAIM: I ′ ∩ Im = I.If s ∈ Im∩I ′, then s = rbm+c and sbm ∈ I. Then rb2m = cbm−sbm ∈

I and so r ∈ I2m = Im. Therefore, s = rbm+c ∈ I. Clearly, I ⊆ I ′∩Im,and the claim is supported. But neither I ′ nor Im equal I, resulting ina contradiction. ¤


Primary Decompositions of Ideals Exist . Any proper ideal I ofR can be expressed as

I = I1 ∩ I2 ∩ · · · ∩ Im,with Ij primary ideals of R.

Proof. If there are ideals that do not possess primary decompositions,let I be maximal with respect to not having a primary decomposition.By the previous proposition, I is not irreducible, and so I = J1 ∩ J2with J1, J2 different (i.e., larger than) I. Hence J1, J2 have primarydecompositions, and therefore so does I; a contradiction. ¤

There are serious problems with the uniqueness issue. However, ageneral context under which there is uniqueness, is for noetherian do-mains such that every nonzero prime ideal is maximal, once the primarydecomposition has been adequately aligned.

Proposition 15. If I and J are primary ideals with radical P , thenI ∩ J is a primary ideal with radical P .

Hence, when considering a primary decomposition

I1 ⊕ I2 ⊕ · · · ⊕ In,we can select the distinct primes among rad(Ij), j = 1, 2, . . . , n; callthen P1, . . . , Pk, and set

Ji = ∩{Ij | rad(Ij) = Pi}.We then have I1⊕I2⊕· · ·⊕In = J1⊕J2⊕· · ·⊕Jk, with the J ′is primarywith distinct radicals. We now only consider primary decompositionsof the latter type, in seeking to show uniqueness.

Lemma 16. If I1, I2, . . . , Im are relatively prime (i.e., Ii + Ij = R forall i 6= j), then

I1 ∩ I2 ∩ · · · ∩ Im = I1I2 · · · Im.Proof. For two, I1 + I2 = R implies I1 ∩ I2 = (I1 ∩ I2)(I1 + I2) =(I1 ∩ I2)I1 + (I1 ∩ I2)I2 ⊆ I1I2. The other containment is obvious. Theinductive step is left as Exercise 4.4. ¤


Theorem 17. Let R be a noetherian domain such that every nonzeroprime ideal is maximal. If I1, I2, . . . , Im and J1, J2, . . . , Jn are collec-tions of primary ideals with distinct radicals, such that

I = I1 ∩ I2 ∩ · · · ∩ Im = J1 ∩ J2 ∩ · · · ∩ Jn,then n = m and after reindexing, Ij = Jj for all j.

Proof. Each primary ideal mentioned in the statement contains a powerof its radical by Proposition 13. Let Pi be the radical of Ii, so that

P kii ⊆ Ii for each i. Since P kii + P kjj = R for i 6= j (Exercise 4.2), itfollows that Ii+Ij = R for all i 6= j. Therefore, by the previous lemma,

I = I1I2 · · · Im = J1J2 · · · Jn.With k = max{k1, . . . , km}, P k1 · · ·P km ⊆ I. Likewise, if Qj is the

radical of Jj, then Q`1 · · ·Q`n ⊆ I. For the sake of later discussions, let

us assume that k ≥ ` (else, replace k by `). We now have, for any i,Qk1 · · ·Qkn ⊆ I ⊆ Ii ⊆ Pi = rad(Ii).

Since Pi is maximal (hence prime), each Pi contains and is thereforeequal to some Qj. Conversely, each Qj coincides with some Pi, andsince P1, P2, . . . , Pm are distinct and Q1, Q2, . . . , Qm are distinct, weconclude that n = m and after reindexing, Pi = Qi.

We will now show that I1 = J1; the proof for the others followsfrom the indexing being arbitrary. By Exercise 4.3, there exists a t ∈(P2 · · ·Pm)k \ P1. It follows that t ∈ I2 · · · Im and that ta ∈ I for everya ∈ I1, yet t /∈ I1 because t /∈ P1. Moreover, tj /∈ I1 for any j, becauseP1 is prime. Likewise, tr ∈ I for every r ∈ J1, yet t /∈ J1.

Suppose r ∈ J1, so that tr ∈ I. Since tr ∈ I1 and I1 is primary,either r ∈ I1 or tj ∈ I1 for some j. We’ve already observed that tj ∈ P1is impossible, so therefore, r ∈ I1. Similarly, I1 ⊆ J1, and the proof iscomplete. ¤


8. EXERCISE SET 4

1. Prove Proposition 15.

2. If I, J are comaximal ideals (i.e., I+J = R), show that Ik+J ` =R for any integers k, `.

3. Assume that I1, I2, . . . , Im are pairwise comaximal ideals of R.If k is a positive integer, show that

I1 + (I2 · · · Im)k = R.

4. Finish the proof of Lemma 16: If I1, I2, . . . , Im are pairwisecomaximal, then

I1 ∩ I2 ∩ · · · ∩ In = I1I2 · · · In.

5. Let R = Z[x]. Show that the ideal I = (x2, 2x) has the followingprimary decompositions with mutually distinct radicals:

I = (x) ∩ (x2, 2) = (x) ∩ (x2, 2 + x) = (x) ∩ (x2, 2x, 4).

6. Let I 6= R be an ideal that contains a power of a maximalideal. Then I is primary. Therefore, if R is a domain such thatevery nonzero prime ideal is maximal, the the primary idealsare precisely the ideals that contain a power of a maximal ideal.

7. Show that I = (x2, 2x) in R = Z[x] is not primary yet I contains(x)2. This shows the assumption in Exercise 4.6 that I containa power of a maximal ideal cannot be weakened to a power ofa prime.


9. Nakayama’s Lemma

In this section R is still a commutative ring.

Proposition 18. If I is an ideal in a commutative ring R, and I 6= R,then I is contained in a maximal ideal of R.

As a consequence, and element 0 6= r ∈ R is contained in somemaximal ideal if and only if r is not a unit.

The (Jacobson) radical of R is defined to be

J(R) = ∩{ M | M is a maximal ideal of R }.Nakayama’s Lemma . If K is a finitely generated module over acommutative ring R such that J(R)K = K, then K = 0.

Proof. Suppose r ∈ J(R). If 1 − r is not a unit of R, then by thelast proposition, 1 − r ∈ M for some maximal ideal M . But then1 = 1− r + r ∈ M , a contradiction. Thus, 1− r is a unit.

Let x1, x2, . . . , xn be a minimal generating set for K. By hypothesis,we can write

x1 = Σjrjxjwith rj ∈ J(R) for all j. Therefore, x1 = (1−r1)−1Σnj=2rjxj, contradict-ing the minimality of the generating set, unless n = 1 and x1 = 0. ¤

A ring R is called quasi-local, if R has a unique maximal ideal M .In this case J(R) = M .

Kaplansky . If R is a quasi-local ring, then any projective R-moduleis free.

Proof. We will only include the proof that finitely generated projectivemodules are free. Let W be a finitely generated projective module, andlet x1, x2, . . . , xn be a minimal generating set for W . There is a freemodule F of rank n and a natural epimorphism f : F → W given byf(ei) = xi where e1, e2, . . . , en is the standard basis for F .

Let K = Ker f . Since W is projective, there is a monomorphismg : W → F such that F = K ⊕ Im g. Observe that K ⊆ MF ,because if f(r1, r2, . . . , rn) = Σjrjxj = 0, and r1 (say) is not in M ,then r1 is a unit, allowing a reduction in the size of the generatingset; a contradiction. Thus, K ⊆ MF and since K = π(F ) whereπ : F → K is the coordinate projection map, it follows that MK = K.By Nakayama’s Lemma, K = 0. ¤


10. Subrings of Quadratic Number Fields

Let us look to the classical examples of Dedekind to illustrate someof the discussions thus far. Call an element α ∈ C integral over Z ifthere is a monic polynomial 0 6= f(x) ∈ Z[x] that has α as a root. Iff(x) has degree n, then

R = Z[α] = {b0 + b1α + · · ·+ bn−1αn−1 | bj ∈ Z},is an integral domain that is generated by 1, α, α2, . . . , αn−1 over Z. Itturns out, α is integral if and only if α belongs to a subring R of Cthat is finitely generated as a Z-module.

We say that α is algebraic (over Q) if α is a root to some polynomial0 6= f(x) ∈ Z[x]. Integral elements are algebraic but not conversely.The number 1√

2is not integral, being a root to the irreducible polyno-

mial 2x2 − 1, which is not monic.Given an algebraic element α, there is an irreducible polynomial in

Z[x] having alpha as a root. Denote this polynomial by irrZ(α). It iseasy to check that a complex number α is integral if and only if it isalgebraic and irrZ(α) is monic.

The elements of the form n√

m where m,n are integers and n > 0 areintegral. For our examples here we will consider the case n = 2 and mis square-free. The field

Q[√

m] = {a + b√m | a, b ∈ Q}is called a quadratic number field.

A subring R of a quadratic number field Q[√

m] is said to be full, ifR contains at least one nonrational number. In this case, R is calledintegrally closed if each α ∈ Q[√m] that is integral actually belongs inR.

Example 7. A full subring R of Q[√

m] that is finitely generated overZ is integrally closed if and only if

R = Z[√

m] when m ≡ 2 or 3 (mod 4), andR = Z[1+

√m

2] when m ≡ 1 (mod 4).

It then turns out that a full subring S of Q[√

m] is integrally closedif and only if

S = ∩P∈PRPwhere R is one of the rings of the previous example, and P is a setof maximal ideals of R. By the theory of localizations, the ideals ofS are IS where I is an ideal of R. Hence S is noetherian if R isnoetherian. Since R is finitely generated as a Z-module, any ideal is


finitely generated as a Z-module and therefore as an R-module. HenceR and S are noetherian.

If P is a nonzero prime ideal of R, and 0 6= β ∈ P , then nβ−1 ∈ Rfor some n (since Q[

√m] is a field), and therefore nβ−1β = n ∈ P .

Consequently, since R is an image of Z ⊕ Z, R/P is an image of Z ⊕Z/n(Z⊕Z) ∼= Z/nZ⊕Z/nZ. Therefore R/P is a finite integral domain,implying that R/P is a field. I.e., P is maximal.

To recap, an integrally closed full subring of Q[√

m] is

(i) integrally closed,(ii) noetherian, and(iii) every nonzero prime ideal is maximal.

Such rings are called Dedekind domains, after the person to first for-mally study these rings, and the initiator of formal commutative ringtheory.

Theorem 19. A full subring R of Q[√

m] is a Dedekind domain ifand only if every ideal I is a product prime ideals. Furthermore, theseconditions are equivalent to the property that for any ideal I 6= 0 of R,I−1 = {t ∈ Q[√m] | tI ⊆ R} satisfies I−1I = R.

The proof of this is within the scope of our course, but as time isof the essence, will be omitted. Class Number Theory and AlgebraicNumber Theory study the field Q[

√m] by considering the group of

isomorphism classes of ideals of R (as in Example 7) where [I] · [J ] =[IJ ]; here, [I]−1 = [I−1].

Corollary 20. Let R be a full subring of Q[√

m]. Then every primaryideal is a power of a prime ideal if and only if R is Dedekind.

Proof. Since R is noetherian such that nonzero prime ideals are maxi-mal, any proper ideal I of R can be expressed as

I = I1I2 · · · Im,where I1, I2, . . . , Im are primary ideals with distinct radicals. If foreach I, every Ij is a power of a prime, then I is a product of primeideals and R is Dedekind by the theorem. Conversely, if R is Dedekind,then a primary ideal J must also be a product of prime ideals. Since Jcontains a power of its radical P = rad(J), P k ⊆ J = P1P2 · · ·P` ⊆ Pj,implies P = Pi for any i. I.e., J is a power of a prime ideal. ¤Corollary 21. Let R be a full subring of Q[

√m]. Then R is a UFD

if and only if R is a PID. In this case, R is Dedekind.

Proof. Assume that R is a UFD and let r be a prime in R. Then (r) isa prime ideal, and because R is a full subring of Q[

√m], (r) is maximal.


Conversely, if M is maximal then M = (r) for some prime r (that is,if p1p2 · · · pn ∈ M , then some pj ∈ M and therefore M = (pj)).

Because the maximal ideals of R are of the form (p) for some primep, if b ∈ R is not a multiple of p, then

Rb + Rp = R.

Therefore, sb + tp = 1 for some s, t ∈ R.If I is a primary ideal, then (pn) ⊆ I ⊆ (p) for some prime p. Let

k be the smallest index such that (pk) ⊆ I, so that pk−1 /∈ I. Supposethere exists a ∈ I \ (pk). Since R is a UFD we can write a = pjb withp, b relatively prime. Note j < k since a /∈ (pk).

Since b, p are coprime, there exists s, t ∈ R such thatsb + tp = 1.

Thus,

pk−1 = spk−1b + tpk = spk−1−jpjb + tpk ∈ I,contrary to the choice of k. Therefore, I = (pk) and every primaryideal is a power of a prime (maximal) ideal. ¤

There are only finitely many known PID’s of the form of Example7. Here are some interesting heavy-duty algebraic number theoreticresults: Let R be as in Example 7:

(a) There are infinitely many integral primes p such that pR isprime.

(b) There are infinitely many integral primes p such that pR = P1P2for distinct maximal ideals P1, P2.

(c) There are only finitely many p′s for which pR = P 2; these arethe prime divisors of m with the possible exception of p = 2.

Some research problems suggested by the topics that came up in thisregard:

1. Which full subrings S of Q[√

m] have every ideal generatedby an integer? This pertains to a problem of P. Hill which Ianswered in the 1990’s.

2. If R ⊆ S ⊆ Q with R a domain, when is it the case that idealsof S are extendable from R; i.e., every ideal J of S is of theform IS where I is an ideal of R (i.e., I = J ∩ R)? I answeredthis to a degree in the late 90’s.


3. If R is noetherian, when is every overring of R inside Q of theform ∩P∈PRP ? These turn out to be the Dedekind domains.

4. If R is a noetherian domains such that every nonzero prime idealis maximal, under what conditions does the Krull-Schmidt The-orem hold relative to finitely generated, torsion-free modules?I.e., when is it the case that whenever

A1 ⊕ A2 ⊕ · · · ⊕ An ∼= B1 ⊕B2 ⊕ · · · ⊕Bn,with Ai, Bj indecomposable, torsion-free finitely generated R-modules, then

m = n and after reindexing Ai ∼= Bi ∀ i?


11. Semi-Simple Rings

Let R be a ring (not necessarily commutative). In this section, wewill consider right R-modules and and right ideals. A module refersto a right R-module. The case for left modules can be made indepen-dently and analogously. A module is called simple if is had no propersubmodules. A module M is called semi-simple if it is a sum of simplesubmodules;

M = ΣiNi,

where each Ni is simple.

Schur’s Lemma . If I and J are simple modules and f ∈ Hom(I, J),then either f = 0 or f is an isomorphism. In particular, EndR(I) ={f | f : I → I is a module homomorphism } is a division ring.Proof. Let 0 6= f : I → J . Since Ker f, Im f are submodules of Iand J respectively, and since I, J are simple,Im f = I and Ker f = 0.In particular, any g : I → I is an automorphism. Thus, g−1 : I → Iis a well-defined homomorphism, implying that EndR(I) is a divisionring. ¤

The Jacobson Radical of R, J(R), is defined to be

J(R) = ∩{M | M is a maximal right ideal}.The ring R is called (Jacobson) semi-simple if J(R) = 0. R is called

right artinian if R has the descending chain condition on right ideals:i.e., if

I1 ⊇ I2 ⊇ · · · ,are right ideals, then for some index m, In = Im for all n ≥ m.


Chinese Remainder Theorem . If P1, P2, . . . , Pk be parwise comax-imal ideals of R, then

R/ ∩j Pj ∼= R/P1 ×R/P2 × · · · ×R/Pkas rings.

Proof. Define a map θ : R → R/P1 ⊕ R/P2 ⊕ · · · ⊕ R/Pk by θ(r) =(r + P1, r + P2, . . . , r + Pk). Clearly Ker θ = ∩jPj so it remains toshow that θ is an epimorphism.

We have P1+P2 = R, implying that P1P3+P2P3 = P3, from which itfollows that P1 +P2P3 = P1 +P1P3 +P2P3 = P1 +P3 = R. Continuingin this way, P1 + Πj≥2Pj = R. Now let P̂j = Πi6=jPi (with the Pi’sappearing with increasing index moving from left to right). We can

induct on k to show that P̂1 + P̂2 + · · · + P̂k = R. The case k = 2 isobvious. Inductively, for n ≥ 3, let

Qj = Πn−1i6=j Pi.

By induction Q1 + Q2 + · · · + Qn−1 = R. Note P̂j = QjPn if j < n.Thus Q1Pn + Q2Pn + · · ·+ Qn−1Pn = Pn, and therefore

P̂1 + P̂2 + · · ·+ P̂n−1 + P̂n = Pn + P̂n = R,as claimed.

Write 1 = x1 + x2 + · · · + xk where xj ∈ P̂j. Let r1, r2, . . . , rk ∈ Rbe given, and take r = r1x1 + r2x2 + · · · rkxk. Modulo Pj,

r ≡ rjxj ≡ (rjx1 + · · ·+ rjxn) ≡ rj.Therefore, the map from R → R/P1 ⊕ R/P2 ⊕ · · · ⊕ R/Pk is an epi-morphism which is clearly an R-module map.

Note 12 = 1 = Σi,jxixj which maps to (x21+P1, x

22+P2, . . . , x

2k+Pk) =

(x1+P1, x2+P2, . . . , xk+Pk). Multiplying 1 = x1+x2+· · ·+xk throughby r and s respectively, we obtain

r = rx1 + rx2 + · · ·+ rxkand

s = sx1 + sx2 + · · ·+ sxk.So, rs maps to (rsx21+P1, rsx

22+P2, . . . , rsx

2k+Pk) = (rsx1+P1, rsx2+

P2, . . . , rsxk + Pk). The former term (rsx21 + P1, rsx

22 + P2, . . . , rsx

2k +

Pk) is equal to (rx1 + P1, rx2 + P2, . . . , rxk + Pk) · (rsx1 + P1, rsx2 +P2, . . . , sxk + Pk)

¤


Artin-Wedderburn Theorem . The following are equivalent for aring R:

1. R is semi-simple as a module.2. R is semi-simple artinian as a ring.3. There exists natural numbers n1, n2, . . . , nk and division rings

D1, D2, . . . , Dk such that

R ∼= Matn1(D1)×Matn2(D2)× · · · ×Matnk(Dk).Proof. (1) → (3): If R is semi-simple as a right module, write R = ΣjIj,and in particular, 1 = xj1 +xj2 + · · ·+xjm with xji ∈ Iji . If r ∈ R, thenr = xj1r+xj2r+· · ·+xjmr. Furthermore, r = rak1+· · ·+rakm ∈ Σmi=1Ikifor any r ∈ R, and after relabeling Iki as Ii, we obtain

R = I1 + I2 + · · ·+ Im,a finite sum.

Reorder the I ′js so that I1, I2, . . . , Ik are pairwise non-isomorphic andif j ≥ k, then Ij ∼= Ii for some i ≤ k. We have that

R ∼= In11 ⊕ In22 ⊕ · · · ⊕ Inkk ,with HomR(Ii, Ij) = 0 for all i 6= j by Schur’s Lemma.

We now have the isomorphisms

R ∼= EndR(R) ∼= HomR(In11 ⊕ In22 ⊕ · · · ⊕ Inkk , In11 ⊕ In22 ⊕ · · · ⊕ Inkk )∼= HomR(In11 , In11 )×HomR(In22 , In21 )× · · · ×HomR(Inkk , Inkk ),

and in turn this last ring is isomorphic to

Matn1(D1)×Matn2(D2)× · · · ×Matnk(Dk).The first isomorphism is called the left regular representation of R

and sends r ∈ R to λr ∈ EndR(R) where λr(a) = ra (note, this isa right module map). If f ∈ EndR(R), then f(a) = f(1)a and sof = λf(1). The rest of the details concerning this isomorphism are easyto check.

The second isomorphism is an application of the obvious fact that ifA ∼= B as modules, then EndR(A) ∼= EndR(B) as rings. Specifically, ifθ : A → B is a right module isomorphism, then the map δ 7→ θδθ−1 isan isomorphism from EndR(A) onto EndR(B).

The third isomorphism employs the principal that if Hom(A,B) = 0,then

Hom(⊕ni=1Ai,⊕ni=1Bi) =

f11 f12 · · · f1nf21 f22 · · · f2n. . . . . . . . . . . . . . . . . .fn1 fn2 · · · fnn

,


where fij : Ai → Bj. Such a matrix sends (a1, a2, . . . , an) ∈ ⊕jAj to(Σjf1j(aj), Σjf2j(aj), . . . , Σjfnj(aj)) ∈ ⊕jBj (via left multiplication bythe matrix on (a1, a2, . . . , an)).

The final isomorphism follows from Schur’s Lemma and propertiesof Hom. Let I be a simple right ideal of R, and D = EndR(I). Then

EndR(Im) ∼= Matm(D),

as rings. To examine this we will distinguish between different copiesof I; let Xi be the copy of I generating the i

th component. Then α ∈EndR(I

m) can be identified with the m×m matrix whose i, jth entryis πjα|Xi where πj : ⊕`X` → Xj is projection onto the jth coordinate.Conversely, a matrix A in Matm(D) represents an endomorphism of⊕iXi via left multiplication by A on the m-tuples inside X1 ⊕ X2 ⊕· · · ⊕ Xm. In this way we identify elements of EndR(Im) with thecorresponding elements of Matm(D).

(3) → (2): Follows from Exercise 5.5.(3) → (1): Follows from Exercise 5.4.(2) → (3): According to the Chinese Remainder Theorem, it is

sufficient to find comaximal two-sided ideals P1, P2, . . . , Pn whose in-tersection is zero such that R/Pi is a matrix ring over a division ringfor each i.

Let P be the right annihilator of a simple right R-module A. Then

A(rP ) ⊆ (Ar)P = 0,implying that P is a two-sided ideal. Certainly P 6= R. To the end ofshowing that R/P is a matrix ring over a division ring, we embed R/Pinto EndD(A) where D = EndR(A) (is a division ring), via right mul-tiplication (i.e., r ∈ R maps to right multiplication by r in EndD(A)).

We will establish a lemma below which will aide us. Assuming thisfor now, suppose A is not finite dimensional over D; say x1, x2, . . . areindependent over D. Set Ij equal to the right annihilator of{x1, x2, . . . , xj}. Then

I1 ⊇ I2 ⊇ · · · ,is a chain of right ideals of R. By the lemma below, for each j, thereexists r ∈ Ij such that xj+1r 6= 0, implying Ij 6= Ij+1. This contradictsR being right artinian. Hence A is finite dimensional. By Exercise 5.1,EndD(A) is the matrix ring Matn(D) where n = dimD A.

We have the embedding of R/P into EndD(A). It remains to showthat this map is onto. Let x1, x2, . . . , xn be a basis for A over D and lety1, y2, . . . , yn ∈ A. Set Bj equal to the D-subspace of A generated by{x1, x2, . . . , xn} \ {xj} and let Ij = annR (Bj)R. By the lemma below,xjIj 6= 0 and since A is right simple over R, xjIj = A for all j.


Write yj = xjtj for tj ∈ Ij and setr = t1 + t2 = · · ·+ tn.

Thenxjr = xjt1 + xjt2 · · ·+ xjtn = xjtj = yj,

since xjri = 0 when i 6= j. This shows that the map R/P → EndD(A)is onto.

Now, as R/P is a matrix ring over a division ring, P is the intersec-tion of finitely many maximal right ideals of R. Therefore ∩{ P | P isthe right annihilator of a simple right R-module } ⊆ J(R) = 0. But,R is artinian, so there are finitely many distinct ideals P1, P2, . . . , Pmthat are annihilators of simple right R-modules, such that

P1 ∩ P2 ∩ · · · ∩ Pm = 0.The rings R/Pi are simple by Exercise 5.2 and what we have shownabove, so each Pi is a maximal two-sided ideal. By the Chinese Re-mainder Theorem,

R = R/ ∩i Pi ∼= R/P1 ×R/P2 × · · · ×R/Pm,and R is a finite product of matrix rings over division rings. ¤Lemma 22. Let R be right artinian, and A a simple R-module withright annihilator P . If B is a finite dimensional subspace of A over thedivision ring D = EndR(A), and I = annR BR, then aI = 0 impliesa ∈ B.Proof. The proof is by induction on dimD B = n. If n = 0, thenR = annR BR, so aR = 0 implies a = 0. Let B be an n-dimensionalsubspace of A, and let B1 be a subspace of B of dimension n − 1.By the induction hypothesis, with I1 = annR (B1)R, aI1 = 0 impliesa ∈ B1 for any a ∈ A. We must show that aI = 0 implies a ∈ B whereI = annR BR. Equivalently, cI 6= 0 for any c ∈ A \B.

Let b ∈ B \ B1, and let a ∈ A \ B. We must show that aI 6= 0. Ifbr1 = 0 yet ar1 6= 0 for some r1 ∈ I1, then r1 ∈ I, yet 0 6= ar1 ∈ aI; wehave finished. Suppose, on the other hand, that br1 = 0 implies ar1 = 0for all r1 ∈ I1. In this case, br1 = br2 if and only if b(r1 − r2) = 0 ifand only if a(r1 − r2) = 0 if and only if ar1 = ar2. Therefore, the mapθ : bI1 → aI1 given by θ(br1) = ar1 is a well-defined R-linear map.

By induction, bI1, aI1 6= 0, and because A is simple, bI1 = aI1 = A.Thus, the map θ ∈ D = EndR(A). Note, θ(br1)−ar1 = 0 = (θ(b)−a)r1for all r1 ∈ I1. By induction again, θ(b) − a ∈ B1 and consequently,a = a − θ(b) + θ(b) ∈ B since B is a D-linear space. This contradictsa /∈ B, and therefore there exists r1 ∈ I1 such that br1 = 0 whilear1 6= 0; that is, aI 6= 0 as desired. ¤


Corollary 23. R is simple artinian if and only if R ∼= Matn(D) forsome n and some matrix ring D.

Proof. Assume that R is simple and artinian. As in the proof of thetheorem, if P is the right annihilator of a simple right R-module, thenR/P ∼= Matn(D) for some n and some matrix ring D. But R is simple,so P = 0. The converse is Exercise 5.2. ¤Corollary 24. The following are equivalent:

(1) R is semi-simple as a right module.(2) R is semi-simple as a left module.(3) Every right R-module is projective.(4) Every left R-module is projective.(5) Every right R-module is injective.(6) Every left R-module is injective.

Proof. For pedagogical reasons, the proof of this theorem will be carriedout for the finitely generated case; the general case holds with very littleadaptation.

(1) → (3): Every module H has a projective resolution:⊕nR → H,

where n is the size of a generating set for H. Write R = I1⊕I2⊕· · ·⊕Imfor simple (right or left) ideals I1, I2, . . . , Im of R (Artin-WedderburnTheorem). Since the image of a simple module is simple,

H = ΣjXj,

where Xj is isomorphic to one of the Ii’s.We can easily find a sub-collection of the Xj’s indexed by X1, X2, . . . , Xk

(after reindexing) so that H = ⊕jXj. Since each Xj is isomorphic tosome Ii, we can write H ∼= I`11 ⊕ I`22 ⊕ · · · ⊕ I`mm . With F = ⊕`R,` = max{`1, `2, . . . , `m},

F ∼= H ⊕K,where K = Ik11 ⊕ Ik22 ⊕ · · · ⊕ Ikmm with kj = `− `j, for all j.

(3) ↔ (5): Recall that a module A (respectively C) is injective(respectively projective) if and only if every short exact sequence

0 → A → B → C → 0,splits. Hence, every module is injective if and only if every module isprojective.

(5) → (1): Let soc(R) (read, the socle of R) be the submodule ofR generated by all of the simple right ideals of R. We need to showthat soc(R) = R. By (5), R = soc(R) ⊕ X since soc(R) is injective.


Suppose 0 6= r ∈ X. Let M be a right submodule of X maximal withrespect to r /∈ M . Then every submodule of X/M must contain r+M .It follows that the submodule C of X/M generated by r+M is simple.

By (5), the simple submodule C of X splits out of X, so that X =C ⊕ Y for some module Y . Thus, soc(R)⊕C is contained in R and isa sum of simple modules, contrary to the maximality of soc(R) as thesum of all simple modules. Therefore X = 0 and R = soc(R).

The equivalence of (2), (4) and (6) is handled analogously. The equiv-alence of (1) and (2) is a consequence of the Artin-Wedderburn Theo-rem. ¤


12. EXERCISE SET 5

1. Let V be the vector space of dimension n as a left module overthe division ring D; where V consists of all n-tuples with entriesin D.(a) Viewing V as a collection of row vectors, show that S =

Matn(D) is equal to EndD(V ) (which operates via rightmultiplication).

(b) Show that V is a simple S-module.

2. A ring R is called simple if R has no proper 2-sided ideals.Show that R = Matn(D) where D is a division ring is simpleand artinian on either side (you can pick a side and do thatcase).

3. Let R = R1 × R2 × · · · × Rm as rings. Given a side, show thatR is artinian if and only if each Rj is artinian.

4. Let R = R1×R2×· · ·×Rm as rings. Show that R is semi-simpleas a right (left) R-module if and only if each Rj is semi-simpleas a right (left) Rj-module.

5. Show that a matrix ring over a division ring is semi-simpleartinian (on either side). Conclude that a product of matrixrings over division rings is semi-simple artinian (on either side).

6. Let A1, A2 be simple right R-modules with endomorphism ringsDi = EndR(Ai). Show A1 ∼= A2 as R-modules if and only ifD1 ∼= D2 as rings. This in turn is equivalent to one of Di beingisomorphic to a subring of the other.


13. Field Theory

Throughout this subject matter, F is a field with subfield K. We willinvestigate the relationship between K and a subfield E of F containingK. We assume that the notation K ≤ E acknowledges the relationshipthat K is a subfield of E.

For fields K ≤ E, E is a vector space over K; the dimension of Eover K, is also called the degree of E over K, and is denoted by

[E : K] = dimK E.

Theorem 25. Let K ≤ E ≤ F be fields. Then[F : K] = [F : E] · [E : K].

Proof. Let A = { αi | i ∈ I} be a basis for E as a vector space overK, and B = { βj | j ∈ J} be a basis for F as a vector space over E.We claim that the elements αiβj for i ∈ I and j ∈ J form a basis forF over K.

SupposeΣi,jai,jαiβj = 0,

for ai,j ∈ K (almost all zero) and i ∈ I, j ∈ J . Since the sum is finite,we obtain

Σj(Σiai,jαi)βj = 0,

which implies Σiai,jαi = 0 for each j since Σiai,jαi ∈ E and B is a basisfor F over E. But then, ai,j = 0 for all i, j because A is a basis for Eover K. Therefore αiβj = αi′βj′ if and only if i = i

′ and j = j′, and

{αiβj | i ∈ I, j ∈ J },is a K-independent set of cardinality |I| · |J |.

If δ ∈ F , then δ = Σjbjβj for some bj ∈ E (almost all zero) since Bis a basis for F over E. For each j, write

bj = Σiai,jαi,

with ai,j ∈ K (almost all zero). Thenδ = Σj(Σiai,jαi)βj = Σi,jai,jαiβj.

ThereforeAB = { αβ | α ∈ A, β ∈ B },

is a basis for F over K of cardinality |A| · |B|. ¤


Example 8. The element√

2 + ι ∈ C is algebraic of degree 4 over Q.Proof. Let α =

√2+ι. Then (α−√2)2 = −1, and so α2+2+1 = 2√2α.

Thus, α is a root to (x2 + 3)2− 8x2, and so the degree of α is less thanor equal to 4. However, the degree of α coincides with the dimensionof Q[α] over Q.

Observe,√

2 = α−1(α2 + 3)/2 ∈ Q[α], and consequently ι ∈ Q[α] aswell. We have Q ≤ Q[√2] ≤ Q[α], and so

[Q[α] : Q] = [Q[α] : Q[√

2]][Q[√

2] : Q].

Since [Q[α] : Q[√

2]] is not 1 (ι /∈ Q[√2]), but is less than or equal to2 (α is a root to x2 − 2√2x + 3), we conclude [Q[α] : Q[√2] = 2, andα has degree 4. ¤

There are possible distinctions between subfields. For example, Q[ 3√

2]contains a root to x3 − 2 while Q[ 3√2,√3ι] contains all roots.Kronecker’s Theorem . Let f(x) ∈ K[x] be irreducible.

(1) There exists a field extension E of K that contains a root α tof(x).

(2) If K ′ is a field isomorphic to K under σ : K → K ′, and E ′ isa field extension of K ′ containing a root α′ of σf (where σf isthe polynomial formed by applying σ to the coefficients of f),then there exists an isomorphism of fields

K[α] ∼= K ′[α′]with k ∈ K 7→ σ(k) and α 7→ α′.

Proof. (1): Let E = K[x]/(f), a field containing K canonically. Theroot to g(y) is α = x + (g). Hence E = K[α] contains a root to f(y).

(2): The extension of σ to a map from K[x] → K ′[x] which sendsg(x) ∈ K[x] to σg, the polynomial in K ′[x] obtained by applying σ toeach of the coefficients of g(x) is an isomorphism of rings;

K[x] ∼= K ′[x].Hence,

K[α] ∼= K ′[α′],by the construction in (1) (necessarily σf is irreducible).

¤

Corollary 26. Given any polynomial f(x) ∈ K[x], there is a fieldextension E of K such that f(x) splits into a product of linear factors.Moreover, E = K[α1, α2, . . . , αm] where αj are the roots of f(x) in E.


The field E in this corollary is called a splitting field for f(x) overK. It is an easy proof by induction on the degree of f(x) that splittingfields are unique up to isomorphism.

The Galois group of F over K is

G = AutK(F );

that is, G is the group (under composition of maps) of K-linear au-tomorphisms σ : F → F . If σ ∈ G, then σ(k) = kσ(1) = k for allk ∈ K (i.e., each element of K is fixed under σ). Conversely, if eachelement of K is fixed under a field automorphism σ : F → F , thenσ(k · a) = σ(k)σ(a) = kσ(a), so σ is K-linear.Corollary 27. If σ ∈ G and α ∈ F is a root to f(x) ∈ K[x], thenσ(α) is also a root to f(x).

Proof. The map σ induces an automorphism of F [x]. On the one hand,σf = f since σ fixes K point-wise, and on the other, f(x) = (x−α)g(x)for some g(x) ∈ K[x] and so f = σf = (x− σ(α))σg. ¤

As a consequence, the more roots to a given f(x), F possesses, themore automorphisms F has. Note that if θ : F → F is an field au-tomorphism and F contains Q, then nθ(m/n) = f(m) = m and soθ(m/n) = m/n. I.e., θ fixes Q automatically.

Example 9. The extension field Q[ 3√

2] of Q contains a single rootof x3 − 2 while Q[ 3√2,√3ι] contains all roots e2kπι/3 3√2, k = 0, 1, 2 tox3 − 2. The Galois group of Q[ 3√2] over Q is the trivial group, whilethere is an automorphism σ : Q[ 3

√2,√

3ι] → Q[ 3√2,√3ι] sending 3√2 toe2πι/3 3

√2.

To see this, by the last theorem, there is an isomorphism

θ : Q[ 3√

2] ∼= Q[e2πι/3 3√

2],

sending 3√

2] 7→ e2πι/3 3√2. By the theorem again, this isomorphismextends to an automorphism of

Q[ 3√

2,√

3ι] =

(Q[ 3√

2])[√

3ι] = (Q[e2πι/3 3√

2])[√

3ι]

extending θ and sending√

3ι 7→ √3ι.Example 10. In the last example we constructed an automorphism ofF = Q[ 3

√2,√

3ι] sending 3√

2 to e2πι/3 3√

2 and sending√

3ι to√

3ι. Like-wise, there are 5 other automorphisms of F ; calling the one mentionedabove, σ1, we have:


σ0 :3√

2 7→ 3√2 √3ι 7→ √3ισ1 :

3√

2 7→ e2πι/3 3√2 √3ι 7→ √3ισ2 :

3√

2 7→ e2πι/3 3√2 √3ι 7→ −√3ισ3 :

3√

2 7→ e4πι/3 3√2 √3ι 7→ √3ισ4 :

3√

2 7→ e4πι/3 3√2 √3ι 7→ −√3ισ5 :

3√

2 7→ 3√2 √3ι 7→ −√3ιNote that F is the splitting field of x3 − 2 and is of degree 6 over Q,

and there are 6 distinct automorphisms of F .The splitting field of xp − 1 has degree p. The splitting field in this

case is Q[ρ] where ρ = e2πι/p. Since xp−1 is irreducible (undergraduateexercise; (x + 1)p− 1 is irreducible by Eisenstein’s Criterion) with ρ asa root, Q[ρ] has degree p over Q). So the obvious conjecture that thesplitting field of an irreducible polynomial of degree n has degree n! isincorrect.

The Fundamental Theorem of Galois is a detailed description of thecorrespondence between subgroups of G and subfields of F containingK.

Given a subgroup H of G,

H ′ = { t ∈ F | θ(t) = t ∀ θ ∈ H }.Given a subfield E of F containing K,

E ′ = AutE(F ) = {θ ∈ G | θ(t) = t ∀ t ∈ E }.In Hungerford, the field F is called a Galois extension of K, provided

G′ = K.

Example 11. The extension field Q[ 3√

2] of Q has Galois group { 1F }and { 1F }′ = F 6= Q, so Q[ 3

√2] is not a Galois extension of Q. We

will show shortly that Q[ 3√

2,√

3ι] is Galois over Q.

Fundamental Theorem of Galois . Let F be a finite dimensionalGalois extension of K, and let G be the Galois group of F over K.

(i) There is a bijective, order-reversing correspondence between thesubfields of F containing K and the subgroups of G given by

E 7→ AutE(F ) and H ≤ G 7→ H ′ ≤ F.(ii) Under this correspondence, an intermediate field K ≤ E ≤ F

corresponds to a normal subgroup of G if and only if E is Galoisover K; and [E : K] is the index of AutE(F ) in G.


14. EXERCISE SET 6

1. Let α ∈ F . Show that K[α] is a field if it has finite dimensionover K.

2. Let α, β ∈ F be algebraic over K having degrees n and mrespectively. Show that

[K[α, β] : K] ≤ m · n,and equality holds if n and m are relatively prime.

3. Let f(x), g(x) ∈ K[x] and let F be an extension field of K.Show that the gcd of f and g is 1 in F [x] if and onlt if the gcdof f and g is 1 in K[x].

4. A irreducible polynomial f(x) ∈ K[x] is called separable iff(x) = a(x−α1)(x−α2) · · · (x−αk) with a ∈ K and α1, α2, . . . , αkdistinct, in E for any splitting field E for f(x). Show that anirreducible polynomial f(x) ∈ K[x] is separable if and only thegcd of f and its derivative f ′ is 1 in K[x].

5. Recall that a finite field is a field extension of Zp for some pand consequently must have order pn for some n. Show that Fis a finite field if and only if F is the splitting field of xp

n − xfor some prime p and natural number n. Consequently, twofinite fields F1, F2 are isomorphic if and only if |F1| = |F2|.(Hint: ⇒ If |F | = pn, note that F ∗ = F \ {0} is a finite abeliangroup of order pn−1. Furthermore, f = xp

n−x has distinct roots(consider the gcd of f with its derivative f ′). ⇐ In the splittingfield of xp

n − x, the roots form a field; hence the splitting fieldis the collection of the roots.)

6. Let E, E ′ be subfields of F containing K and suppose σ : E ∼=E ′ is an isomorphism fixing K. Assume that F is the splittingfield of a separable polynomial f(x) ∈ K[x]. Using Kronecker’sTheorem and induction on [F : E], argue that there are[F : E] automorphisms σ̃ ∈ G such that σ̃|E = σ.

7. Diminishing Returns: Let G be the Galois group of F overK. Show that (AutH′(F ))

′ = H ′ for every H ≤ G and thatAutE(F ) = Aut(AutE(F ))′(F ) for every intermediate field K ≤E ≤ F .


15. The Fundamental Theorem of Galois

Before embarking on a proof of the Main Theorem of Galois, we willexamine the hypothesis of that result. Let us recall the FundamentalTheorem of Linear Algebra:

If M is an m× n matrix over a field, thenrank M + nullity M = n.

To prove this, row reduce M . The number of columns in the row-reduced form that do not contain a leading nonzero entry of a row isthe nullity of M (i.e., the dimension of the nullspace of A), and thenumber of columns containing a leading nonzero entry of a row is therank of M (rank M = dimension of the column space of M = dimensionof the row space of M).

Assume

[F : K] < ∞throughout, and let

G = Gal(F/K) = AutK(F ).

Remember that an automorphism σ : F → F fixes K if and only if σis K-linear.

A Lemma of Dedekind . If σ1, σ2, . . . , σm ∈ G are distinct, thenthey are linearly independent.

Proof. The proof will be by induction on m; clearly the case m = 1 issettled in the affirmative. Suppose, for the sake of induction, that

(1) a1σ1 + a2σ2 + · · ·+ amσm = 0,with not all a′is equal to zero. By induction, we assume that all a

′is are

nonzero. Multiplying through by a−1m we may assume that am = 1.Since σ1 6= σn, there exists t ∈ F such that σ1(t) 6= σm(t). Evaluating

(1) at ts for s ∈ F arbitrary, and multiplying both sides by σm(t)−1,we obtain

σm(t)−1[a1σ1(ts) + a2σ2(ts) + · · ·+ σm(ts)] =

a1σm(t)−1σ1(t)σ1(s) + · · ·+ a1σm(t)−1σm−1(t)σm−1(s) + σm(s) = 0.

Evaluating (1) at s and subtracting this last equation, we have

a1[1− σm(t)−1σ1(t)]σ1(s) + a2[1− σm(t)−1σ2(t)]σ2(s)+· · ·+ am−1[1− σm(t)−1σ1(t)]σm−1(s) = 0,


for every s ∈ F . That is,a1[1− σm(t)−1σ1(t)]σ1 + · · ·+ am−1[1− σm(t)−1σ1(t)]σm−1 = 0.

This contradicts the inductive hypothesis since, in particular, a1[1 −σm(t)

−1σ1(t)] 6= 0. ¤

Lemma 28. If H ≤ G, then[F : H ′] ≥ |H|.

Proof. Let H consist of σ1, σ2, . . . , σm, and let u1, u2, . . . , uk be a basisfor F over H ′. Suppose to the contrary that m > k, and consider thesystem of linear equations

σ1(u1)x1 + σ2(u1)x2 + · · ·+ σm(u1)xm = 0σ1(u2)x1 + σ2(u2)x2 + · · ·+ σm(u2)xm = 0

· · · · · · · · · · · · · · ·σ1(uk)x1 + σ2(uk)x2 + · · ·+ σm(uk)xm = 0

over F . We know there exists a nontrivial solution (x1, x2, . . . , xm).For any β ∈ F , write β = Σjbjuj for bj ∈ H ′. Multiplying the ith

equation by bi and adding we obtain

σ1(β)x1 + σ2(β)x2 + · · ·+ σm(β)xm = 0contradicting Dedekind’s Lemma. ¤

Theorem 29. If H ≤ G, then[F : H ′] = |H|.

Proof. It remains to show that

[F : H ′] ≤ |H|.Let H consist of σ1, σ2, . . . , σm, and suppose that F contains m +1 elements u1, u2, . . . , um+1 which are linearly independent over H

′.Consider the system of linear equations

σ1(u1)x1 + σ1(u2)x2 + · · ·+ σ1(um+1)xm+1 = 0· · · · · · · · · · · · · · ·

σm(u1)x1 + σm(u2)x2 + · · ·+ σm(um+1)xm+1 = 0Let j be the minimal number of nonzero components in a nontriv-

ial solution to the system. Choose a nontrivial solution with j termsand reorder the u′js (if necessary) so that this solution is of the form


(a1, a2, . . . , aj, 0, . . . , 0) such that a1, . . . , aj are nonzero, and aj = 1.Note that j > 1. Because any σi is H

′-linear, not all of the a′iscan belong to H ′. Assume (by reordering the u′is) that a1 /∈ H ′; so,σk(a1) 6= a1 for some k.

Applying σk to the `th-row in the system

(1) σ`(u1)a1 + σ`(u2)a2 + · · ·+ σ`(uj)aj = 0we obtain

σkσ`(u1)σk(a1) + σkσ`(u2)σk(a2) + · · ·+ σkσ`(uj)σk(aj) = 0.But σkH = H, and so for every i there exists an ` such that σkσ` = σi.

Subtract the equation

σi(u1)σk(a1) + σi(u2)σk(a2) + · · ·+ σi(uj)σk(aj) = 0.from the equation pointed to by (1) but corresponding to i to obtain anew system whose ith row is

σi(u1)[a1 − σk(a1)] + · · ·+ σi(uj)[aj − σk(aj)] =σi(u1)[a1 − σk(a1)] + · · ·+ σi(uj−1)[aj−1 − σk(aj−1)] = 0,

since aj = 1. But a1 − σk(a1) 6= 0, so we have found a solution to theoriginal system with fewer than j nonzero terms; a contradiction. ¤

The hypothesis of the Fundamental Theorem of Galois poses a re-striction on the field extension F (unlike the previous results of thissection). F is called a Galois extension of K if F satisfies any (henceall) of the conditions of the next result.

Galois Extension Theorem . The following are equivalent:

(1) [F : K] = |G|.(2) G′ = K.(3) Every monic irreducible polynomial in K[x] that has a root in

F , is separable and splits in F [x].(4) F is the splitting field of a separable polynomial in K[x].

Proof. (1) → (2) By the last theorem,[F : G′] = |G|.

Since

|G| = [F : K] = [F : G′][G′ : K] = |G|[G′ : K],[G′ : K] must be 1.


(2) → (3) By the previous theorem, [F : G′] = |G| = [F : K]. Letp(x) ∈ K[x] be monic, irreducible, with a root α in F , and consider

g(x) = Πj(x− αi),where α1, . . . , αk are the distinct members of the set {σ(α) | σ ∈ G}.If we apply any σ ∈ G to the coefficients of g(x) we find that σ fixesg(x), and therefore g(x) ∈ G′[x] = K[x]. Since the gcd of g(x) andp(x) in F [x] is not 1, the gcd of g(x) and p(x) in K[x] cannot be 1.Therefore, p divides g implying that p is separable.

(3) → (4) If α ∈ F , then α is algebraic and is the root of a separable(irreducible) polynomial p(x). If the splitting field E of p(x) inside F ,differs from F , let β ∈ F \ E; β is a root to a separable polynomialp2(x) ∈ K[x] ≤ E[x], and the splitting field of p(x)p2(x) inside Fproperly contains E. Since [F : K] < ∞, we will obtain F as asplitting field of a separable polynomial in K[x] after a finite numberof steps.

(4) → (1) By Exercise 6.6, there are [F : K] automorphisms whichextend the identity map 1K : K → K. I.e., [F : K] = |G|.

¤Let Sub G denote the set of subgroups of G and Int F/K the (lattice

of) intermediate subfields of F containing K. Both sets Sub G andInt G have containment as the partial order.

Fundamental Theorem of Galois . Let F be a Galois extension ofK. Then:

(i) The map Sub G → Int F/K given by H 7→ H ′ is an order-reversing bijection with inverse E 7→ AutE(F ).

(ii) AutH′(F ) = H and E = AutE(F )′.

(iii) [E : K] = [G : AutE(F )] and [G : H] = [H′ : K].

(iv) E is Galois over K if and only if AutE(F ) is normal in G.

Proof. If H ′1 = H′2, then it is easy to see that (H1H2)

′ = H ′1 = H′2.

Theorem 29 reveals

[F : (H1H2)′] = |H1H2| = [F : H ′1] = |H1|.

Thus, H1 = H1H2. Similarly H2 = H1H2 = H1. Note; the hypothesiswas not needed for this part. We next argue that AutE1(F ) = AutE2(F )implies E1 = E2. Now, E1E2 (consists of finite sums of products a1a2,ai ∈ Ei) is finite dimensional over K, hence is a subfield of F containingboth E1, E2. It readily checks that AutE1E2(F ) = AutEi(F ), so we mayassume that E1 ⊆ E2, in order to show that E1 = E2.


If there exists an α ∈ E2 \E1, then α is a root to an irreducible andseparable polynomial p(x) ∈ K[x] which splits in F (Galois ExtensionTheorem). Let g(x) be an irreducible factor of p(x) in E1[x] that hasα as a root. Then g(x) is separable (it divides p(x)) and has degreeat least 2, so there exists another root α′ to g(x) in F . By Exercise6.6, there exists σ ∈ G such that σ fixes E1 and σ(α) = α′. Thenσ ∈ AutE1(F ) but is not in AutE2(F ). Thus, E1 = E2.

To finish (i) we need to establish (ii). But (ii) follows from Exercise6.7; we know H ′ = (AutH′(F ))′ and so by what we have shown above,H = AutH′(F ). Similarly, E = (AutE(F ))

′. We next verify the claimsabout the indices:

By Theorem 29, [F : H ′] = |H| and by Lagrange’s Theorem andTheorem 25, [H ′ : K] = [G : H]. Putting in E = H ′ we obtain[E : K] = [G : AutE(F )] since H = AutE(F ) from (ii).

If E is Galois over K, then E is the splitting field of a separablepolynomial g(x) in K[x]. If σ ∈ G, then σ must permute the rootsof g(x) in some fashion, and so σ(E) ⊆ E. Therefore σ−1δσ(a) =σ−1σ(a) = a for every a ∈ E and σ−1δσ ∈ AutE(F ), for δ ∈ AutE(F ).Conversely, suppose p(x) ∈ K[x] is irreducible with a root α ∈ E (wealready know that p is separable since F is Galois), yet another rootα′ of p is in F \ E. Then, there exists σ ∈ G such that σ(α) = α′ andσ fixes E by Exercise 6.6. For H = AutE(F ), H1 = σHσ

−1 fixes α′.Since

[F : H ′1] = |H1| = |H| = [F : H ′] = [F : E],it is impossible for H ′1 to contain E. Hence H1 6= H. I.e., if E is notGalois, then AutE(F ) is not normal in G. ¤


16. The Galois Group, Finite Fields and Simple Extensions

Every finite dimensional extension F of K is

F = K[α1, α2, . . . , αk],

for some elements α1, α2, . . . , αk, algebraic over K. It turns out, in thecase of a Galois extension, that

F = K[α]

for a single algebraic element α. This is also the case in the contextQ ⊆ K.Lemma 30. Any finite subgroup of F ∗ = F \ 0 is cyclic.Proof. Observe that an abelian group C of order n is cyclic if and onlyif for each divisor d of n, C has at most 1 cyclic subgroup of order d. Tosee this, if C = 〈c〉 is cyclic (written multiplicatively), then 〈cn/d〉 is theonly cyclic subgroup of order d. Conversely, write C = C1×C2×· · ·×Cjwith Ci primary. Then, C is cyclic if and only if the C

′is correspond to

distinct primes. But, if Ci and C` are p-primary for a given prime p,then C has two cyclic subgroups of order p contrary to the assumption.

Let C be a subgroup of order n of the multiplicative group F ∗ andlet d divide n. Any cyclic subgroup B of F ∗ of order d consists of theroots of xd − 1 in F . Therefore, there is at most one such subgroup Band so C can have at most one cyclic subgroup of order d. Thus, C iscyclic. ¤

Corollary 31. If F is a finite field, then F ∗ is cyclic.

Primitive Element Theorem . If F is a finite dimensional Galoisextension of K, then

F = K[α],

for some algebraic element α.

Proof. By the lemma, it suffices to assume that F is infinite. It sufficesto argue that for α, β ∈ F , that K[α, β] = K[δ] for some δ ∈ F .By the Galois Correspondence Theorem, there are only finitely manyintermediate subfields of F containing K. On the other hand, thereare infinitely many distinct elements of the form α + aβ for 0 6= a ∈ F .Choose two such elements; α + aβ, α + bβ with a 6= b, such that

K[α + aβ] = K[α + bβ].

Evidently, α, β ∈ K[α+aβ], and so δ = α+aβ satisfies our purpose. ¤


If F is the splitting field of a separable polynomial f(x) ∈ K[x] ofdegree n, then clearly

|G| = [F : K] ≤ n!,since any σ ∈ G must permute the roots of f(x) and is determined byhow σ permutes the roots of f(x). That is,

G ≤ Sn,the symmetric group on n letters.

Theorem 32. If F is the splitting field of an irreducible polynomialf(x) of degree p (a prime) over Q, and f(x) has exactly two non-realroots, then

G ∼= Sp.Proof. Since |G| = [F : Q] is divisible by p, G has an element σof order p by Cauchy’s Theorem. We now argue that σ is a cycle oflength p: Express σ as a product of disjoint cycles δ1 · · · δi; the orderof σ, p, is the least common multiple of the lengths of the δ′js; henceone of the δ′s (hence σ) is a cycle of length p.

Let E be the subfield of F generated by all of the real roots of fand let α and ᾱ be the two complex roots. Since F is the splittingfield of f over E, the irreducible polynomial for α over E must havedegree 2 (i.e., the irreducible factor of f in E[x] having α as a rootmust be of degree 2 since the only roots that do not split out of f overE are α and ᾱ). There is an automorphism of F fixing E but sendingα 7→ ᾱ. In summary, under the identification of G with Sp, G containsa transposition and a cycle of length p. Since Sp is generated by anysuch pair, G = Sp. ¤

Definition . Given a field extension K of Q, a polynomial f(x) ∈ K issaid to be solvable by radicals if the roots of f(x) can be expressed usingalgebraic combinations and radical combinations of the coefficients off(x).

Example 12. For f(x) = x3+ax+b, the roots of f(x) are given below:

x = A + B = −A + B2

+A−B

2

√−3 = −A + B2

− A−B2

√−3,

where A = 3√− b

2+

√b2

4+ a

3

27, and B = 3

√− b

2−

√b2

4+ a

3

27. The

general cubic x3 + ux2 + vx + w can be transformed into a cubic in the


above form by setting

a =1

3(3v − u2), and b = 1

27(2u3 − 9uv + 27w).

Example 13. Given the general quartic equation x4+ax3+bx2+cx+d,the roots can be found as follows:

Let y be any root to y3 − by2 + (ac− 4d)y − a2d + 4bd− c2; and

R =

√a2

4− b + y.

If R 6= 0, take

D =

√3a2

4−R2 − 2b + 4ab− 8c− a

3

4R

and

E =

√3a2

4−R2 − 2b− 4ab− 8c− a

3

4R.

If R = 0, take

D =

√3a2

4−R2 − 2b + 2

√y2 − 4d

and

E =

√3a2

4−R2 − 2b−+2

√y2 − 4d.

The four roots of the original quartic are

x = −a4

+R

2± D

2and − a

4− R

2± E

2.

Recall that a finite group G is solvable if there exists a chain

G0 = 〈1〉 < G1 < · · · < Gn = Gsuch that Gi is normal in Gi+1, and Gi+1/Gi is cyclic for all i.


Big Theorem of Galois . Let K be a subfield of C, and let f(x) ∈K[x]. Then, f(x) is solvable by radicals if and only if the Galois groupof the splitting field of f(x) over K is a solvable group.

Example 14. The polynomial x5 − 4x2 + 2 ∈ Q[x] is irreducible byEisenstein’s Criterion, and has exactly 2 non-real roots. Hence, theGalois group of x5− 4x2 + 2 over Q is S5; a group that is not solvable.Therefore, the polynomial is not solvable by radicals.

Example 15. The polynomial f(x) = xp− 1 splits into (x− 1)(xp−1 +· · ·+ 1). For any prime p, g(x) = xp−1 + · · ·+ x + 1 is irreducible. Tosee this observe that f(x + 1) = xg(x + 1) is equal to

xp +

(p

1

)xp−1 +

(p

2

)xp−2 + · · ·+

(p

p− 1

)x + 1− 1.

So g(x + 1) = xp−1 +(

p1

)xp−2 +

(p2

)xp−3 + · · ·+

(p

p−1).

The coefficient of xj, j < p− 1, is(

pp−j

)and is divisible by p, while

the constant term, p, is not divisible by p2. So g(x + 1), hence g, isirreducible. Therefore, the splitting field F = Q[ρ] of f(x) (and g) hasdegree p− 1 where ρ = e2πι/p. The Galois group of f (and g) is abelianand is isomorphic to the cyclic group Z∗(p). The isomorphism sendsσ ∈ G to j ∈ Z∗(p) where σ(ρ) = ρj.

Theorem 33. Let K be a subfield of C. The splitting field of xn − 1over K has an abelian Galois group of order at most ϕ(n) (where ϕ isthe Euler phi-function). If K = Q, then G has order ϕ(n).

Proof. The splitting field F is obtained as F = K[ρ] where ρ = e2πι/n.If σ ∈ G (the Galois group), then σ(ρ) = ρj. Evidently, ρj must be aprimitive nth root of unity, and so j must be relatively prime to n. I.e.,the map from G into Z∗(n) has image in Zu(n), the group of units in Z∗(n).This establishes the first part.

Setgm(x) = Πδ(x− δ),

where the product is indexed over all primitive mth roots of unity.Evidently

(xn − 1) = Πd|ngd(x).We will prove by induction that gd(x) ∈ Z[x].


For the sake of induction, set f(x) = Π{(x−δ) | δ is a primitive dth−root of unity, for d|n, d < n} = Πd|n,d


17. EXERCISE SET 7

1. In the Galois extension F = Q[ 3√

2,√

3ι] over K = Q, findall intermediate subfields. Acknowledge the subfields that areGalois over K

2. Do the same as in 1. but for x4 − 5 over Q.3. Determine the Galois group of (the splitting field for) (x3 −

2)(x2 − 3) over Q.4. Assume that F is Galois over K. Show that [H1 : H2] =

[H ′2 : H′1] for any subgroups H2 ≤ H1 of G, and [E1 : E2] =

[AutE2(F ) : AutE1(F )] for any intermediate fields E2 ≤ E1between F and K.

5. Let G be the Galois group of the polynomial x11 − 1 over therational field Q. Determine this well-known group. Show all ofthe necessary work.

6. Determine the Galois group of the polynomial x10 − 1 over therational field Q. Show all of the necessary work.

7. Determine the Galois group of x5−6x+3 over the rational fieldQ. Show all of the necessary work.

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