20 KIYOSHI IGUSA BRANDEIS UNIVERSITY

2. Projective varieties

For any field F , the standard definition of projective space Pn

(F ) is that it is the set ofone dimensional F -vector subspaces of Fn+1. For example, P1(R) is the set of lines throughthe origin in R2. These are specified by slope 0 ✓ < ⇡. So, P1(R) ⇠= R/⇡Z is a circle.However, there are many points missing. As an a�ne scheme, the circle implicitly alsocontains all points in the punctured disk (the “interior” of the circle) and a generic point.As a projective variety, the puncture will be filled in. (These are the two “circles at infinity”identified to one point.)

2.1. Lecture 7. Basic properties of projective space. In order to construct the schemetheoretic version of n-dimensional projective space P

n

(k) we need to take Pn

(⌦). This isdefined to be the space of all one dimensional ⌦ vector spaces in ⌦n+1. This is also thequotient space of ⌦n+1\0 modulo the identification x ⇠ ax for all a 6= 0 2 ⌦.

Note that the homogeneous coordinates Xi

are NOT functions Pn

(⌦)! ⌦. But homoge-neous polynomials define subsets of P

n

(⌦). First, a polynomial f [X] 2 k[X] is homogeneous

of degree d if for all � 2 ⌦ and x 2 ⌦n+1, f(�x) = �df(x). [Thanks to Mac for correctingan earlier version!] Given such a polynomial we can define

V (f) := {[x] 2 Pn

(⌦) | f(x) = 0}.If (x) ⇠ (�x) then f(�x) = �df(x) = 0. So, V (f) is well-defined.

Definition 2.1.1. If S = {f↵

} is a set of homogeneous polynomials f↵

2 k[X0, · · · , Xn

]then let

V (S) =\

f↵2SV (f

↵

) = {[x] 2 Pn

(⌦) | f↵

(x) = 0 8f↵

2 S}.

The Zariski topology or k-topology on Pn

(⌦) is defined by letting V (S) as above be theclosed subsets.

LetUi

= {[x0, · · · , xn] 2 Pn

(⌦) : xi

6= 0}.These are open subsets of P

n

(⌦) since they are the complements of V (Xi

). We claim thatUi

⇠= ⌦n for all i, but we took i = 0 for simplicity. Then we have a bijection:

'0 : ⌦n ⇠= U0

given by '0(x1, · · · , xn) = [1, x1, · · · , xn]. The inverse mapping is given by✓

x1x0

, · · · , xnx0

◆

[x0, · · · , xn] 2 U0

So, '0 is a bijection.

Proposition 2.1.2. '0 : ⌦n ⇠= U0 is a homeomorphism (using the Zariski topology on both

sets).

Proof. (1) As pointed out above, '0 is a bijection.(2) For any f(X) 2 k[X1, · · · , Xn

], let f̃(X) 2 k[X0, · · · , Xn

] be the unique homogeneouspolynomial of the same degree as f so that f̃(1, x1, · · · , xn) = f(x1, · · · , xn). The formulafor f̃ is as follows. Suppose that

f(X) =X

↵=(a1,··· ,an)

c↵

X↵ where X↵ = Xa11 Xa2

2 · · ·Xann

LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 21

Let d = max |↵| = deg f where |↵| =P

ai

. Then f̃(X) is given by

f̃(X) =X

↵

c↵

Xd�|↵|0 X↵

Then f̃(X) is homogeneous of degree d.(3) For any f(X) 2 k[X1, · · · , Xn

] we have

'0 (V (f) ✓ ⌦n) = V (f̃) \ U0

This is easy to check: (x1, · · · , xn) 2 V (f) i↵ f(x1, · · · , xn) = 0. But '0(x1, · · · , xn) =[1, x1, · · · , xn] which lies in V (f̃) \ U0 i↵ f̃(1, x1, · · · , xn) = 0. But f̃(1, x1, · · · , xn) =f(x1, · · · , xn). So, these are the same conditions and the two sets are equal as claimed.This proves that '0 is an open mapping (taking open sets to open sets) since it takes basicopen sets in ⌦n to basic open sets in U0.

Next, suppose that g 2 k[X0, · · · , Xn

] is homogeneous. Then

V (g(1, x1, · · · , xn)) = '�10 (V (g) \ U0)

This again is easy to prove. So, '0 is continuous. Since '0 is continuous and open, it is ahomeomorphism. ⇤

This proposition (⌦n ⇠= U0) is interpreted to mean that Pn

(⌦) is given by adding “pointsat 1” to ⌦n. To understand this, let

Hi

= {[x] 2 Pn

(⌦) |xi

= 0}Then P

n

(⌦) = Ui

`

Hi

and Ui

⇠= ⌦n, Hi

⇠= Pn�1(⌦). The second isomorphisms is given by

[x1, · · · , xn] 2 Pn�1(⌦)$ [x1, · · · , xi, 0, xi+1, · · · , xn] 2 H

i

The reason that H0 is considered to be the “set of points at 1” is the following calculationwhich only makes sense when ⌦ = C:

limt!1

[1, tx1, · · · , txn] = limt!1

1

t, x1, · · · , xn

�

= [0, x1, · · · , xn]

since 1t

! 0 as t!1. Thus

[0, x1, · · · , xn] = limt!1

'0(tx)

which corresponds to points in a straight line in ⌦n going to infinity.We will quickly verify that the theorems which hold in the a�ne case also hold for

projective space. The lemmas work as expected. But the theorem has a surprise that Ineed to explain.

Definition 2.1.3. A homogeneous ideal in k[X0, · · · , Xn

] is an ideal generated by homo-geneous polynomials.

Every polynomial is a sum of homogeneous polynomials: f =P

m

d=0 f[d] where f [d] is

called the homogeneous component of f of degree d. For example:

f = X3|{z}

f

[3]

+XY + Y Z| {z }

f

[2]

+ 5|{z}

f

[0]

Lemma 2.1.4. An ideal a is homogeneous i↵ it contains the homogeneous components f [d]

of each of its elements f .

22 KIYOSHI IGUSA BRANDEIS UNIVERSITY

Proof. (() This implication is clear since then a is generated by the homogeneous compo-nents of its elements.

()) Let a be a homogeneous ideal. Then it is generated by homogeneous polynomialsf↵

with degree d↵

. Any element h 2 a has the form

h =X

g↵

f↵

where g↵

are arbitrary polynomials. Then, the homogeneous components of h are:

h[d] =X

g[d�d↵]↵

f↵

where the sum is over those ↵ so that d↵

d. Since this is a linear combination of some ofthe f

↵

, h[d] 2 a for all d as claimed. ⇤Lemma 2.1.5. The radical of any homogeneous ideal is homogeneous.

Proof. Suppose that f 2 r(a). We need to show that each component f [d] of f lies in r(a).But f 2 r(a) implies fn 2 a. Write f as a sum of components:

f = f [d] + f [d�1] + · · ·Then

fn = (f [d])n + (lower terms) 2 a

By the previous lemma, the top degree homogenous component lies in a: (f [d])n 2 a. Butthis implies that f [d] 2 r(a). But then

f � f [d] = f [d�1] + (lower terms) 2 r(a)

By induction on d, f [i] 2 r(a) for all i. So, r(a) is a homogeneous ideal by the previouslemma. ⇤

Now we can follow all steps in the proof of Proposition 4 in [4].

Theorem 2.1.6. There is an order reversing bijection:

(

homogeneous radical ideal a ✓ k[X0, · · · , Xn

]

a 6= (X0, · · · , Xn

)

)

$ {closed subsets of Pn

(⌦)}

a �! V (a)

I(⌃) � ⌃

where I(⌃) is the ideal generated by all homogenous polynomials f which vanish on ⌃.

Proof. The reason that the homogeneous ideal (X0, · · · , Xn

) is excluded in the bijection isbecause there are two ideals mapping to the same set (the empty set) in P

n

(⌦), namely a =(X0, · · · , Xn

) and a = k[X]. However, I(;) = k[X]. So, the required equation I(V (a)) = adoes not hold for a = (X0, · · · , Xn

). Also the proof does not work for this ideal.It is clear that ⌃ = V (I(⌃)) for all closed sets ⌃. So, we need to show that a = I(V (a)).

But a ✓ I(V (a)) is clear. So, we just need to show that a ◆ I(V (a)).Let

V ⇤(a) = {x 2 ⌦n+1 | f(x) = 0 8f 2 a}Then we have

a = r(a) = I(V ⇤(a))

So, we just need to show that I(V (a)) ✓ I(V ⇤(a)) when V (a) is nonempty.There are three cases:

LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 23

(1) a = k[X], V ⇤(a) = ;, V (a) = ;(2) a = (X0, · · · , Xn

), V ⇤(a) = {0}, V (a) = ;(3) V (a) is nonempty and V ⇤(a) is the cone on V (a). I.e.,

(2.1) V ⇤(a) = {�x | [x] 2 V (a)}.

For every [x] 2 V (a) and homogeneous f 2 I(V (a)) we have f(x) = 0. By (2.1) this impliesf(�x) = 0 for all �. So, f 2 I(V ⇤(a)). So, I(V (a)) ✓ I(V ⇤(a)) whenever (2.1) holds whichis Cases (1) and (3). So, the theorem holds in all cases but excluded Case (2). ⇤

We glossed over the general discussion of homogeneous prime ideals (in the next sub-section). I just said the following was clear using the obvious definitions: a closed subsetof P

n

(⌦) is irreducible if it is nonempty and not the union of two proper closed subsets,and, by a homogeneous prime ideal we mean an ideal having both of those properties.But we always need to exclude the case (X0, · · · , Xn

). (We include it in the definition of“homogeneous prime ideal” but exclude it in all theorems.)

Corollary 2.1.7. In the correspondence above, irreducible closed subsets correspond to

homogeneous prime ideals not equal to (X0, · · · , Xn

).

Proof. In Case 3 in the above proof, the homogeneous ideal s is prime i↵ V ⇤(a) is irreduciblei↵ V (a) is irreducible. Cases 1,2 are excluded in the definition of irreducible set. ⇤

Example 2.1.8. Let f = XW � Y Z. This is a homogeneous polynomial of degree 2 ink[X,Y, Z,W ]. Let ⌃ = V (XW � Y Z) ⇢ P3(⌦). This may be more familiar when we writeit this way:

⌃ =

⇢

x, yz, w

�

: det = 0

�

Since f cannot be factored as a product of polynomials of lower degree (necessarily linear),(XW � Y Z) is prime and ⌃ is irreducible. Last week we discussed regular functions onopen subsets of irreducible sets f : U ! ⌦. By definition these are functions which aregiven locally by rational functions f(x) = g(x)

h(x) . Mumford emphasized that, in some cases,the same fraction cannot always be used throughout the set U . Here is an example.

⌃X

=

⇢

x, yz, w

�

: x 6= 0

�

, ⌃Y

=

⇢

x, yz, w

�

: y 6= 0

�

Let U = ⌃X

[ ⌃Y

. Let f : U ! ⌦ be given by

f =

(

z

x

on ⌃X

w

y

on ⌃Y

on the intersection ⌃X

\ ⌃Y

we have that x, y are both nonzero and

z

x=

zw

xw=

zw

yz=

w

y

so f is well defined. (We forgot to check the case when w = 0. But the equation xw = yzimplies that z = 0 in that case and we get z

x

= 0 = w

y

.)This is an example of a regular function U ! ⌦ for which there is no single rational

functional expression which works on all of U .

24 KIYOSHI IGUSA BRANDEIS UNIVERSITY

2.2. Segre embedding. Mumford uses this example (in another book [5]) to introducethe following construction. (I changed the notation: elements of ⌦ should be lower caseletters. Capital letters are elements of the polynomial ring k[X0, · · · , Xn

].)

Definition 2.2.1. The Segre embedding

� : Pn

⇥ Pm

! Pn+m+nm

is given by

([xi

], [yj

]) 7! [zij

= xi

yj

] .

In other words, each of the n + 1 homogeneous variables Xi

is multiplied by each of them+ 1 variables Y

j

to get (n+ 1)(m+ 1) variables Zij

= Xi

Yj

:

⌦n+1 ⇥ ⌦m+1 ! ⌦(n+1)(m+1)

For example, when n = m = 1 we get:

P1(⌦)⇥ P1(⌦)! P3(⌦)

([s, t], [u, v]) 7! [ su|{z}

x

, sv|{z}

y

, tu|{z}

z

, tv|{z}

w

]

Theorem 2.2.2. The Segre embedding is a monomorphism with image

�(Pn

⇥ Pm

) = V (Zij

Zk`

� Zi`

Zkj

)

Proof from [5]. (We did the case n = m = 1 in class.) � is injective: Suppose �([x], [y]) =�([x0], [y0]). Then there is a � 6= 0 in ⌦ so that x0

i

y0j

= �xi

yj

for all i, j. By renumbering wemay assume x0, y0 are nonzero. Then x00y

00 = �x0y0 6= 0. So, we can rescale the coordinates

of x, y, x0, y0 and assume that x0 = y0 = x00 = y00 = 1. But then � = 1. So, for every i,x0i

y00 = xi

y0 making x0i

= xi

. Similarly y0j

= yj

for all j. So, ([x], [y]) = ([x0], [y0]).� is surjective: Suppose [z] 2 V (Z

ij

Zk`

� Zi`

Zkj

) ✓ Pn+m+nm

(⌦). Then, one of thecoordinates is nonzero. Renumbering and rescaling we can assume that z00 = 1. Then letxi

= zi0 and y

j

= z0j . Then

zij

= zij

z00 = zi0z0j = x

i

yj

for all i, j and [z] = �([x], [y]). So, � is a bijection. ⇤

Lecture 8

Corollary 2.2.3. The image of the Segre embedding is irreducible.

The idea is to use the ⌦-topology. The ⌦-closed subsets of Pn

(⌦) are

V (S) = {[x] 2 Pn

(⌦) : f(x) = 0 8f 2 S ✓ ⌦[X]d

}

where ⌦[X]d

denotes the set of degree d homogeneous polynomials in variables X withcoe�cients in ⌦. Since ⌦ = ⌦ we are in the realm of standard algebraic geometry. Sincek ⇢ ⌦, there are more ⌦-closed set than k-closed sets. I.e., every k-closed set is ⌦-closed.The following is clear.

Lemma 2.2.4. Every k-closed set which is ⌦-irreducible is k-irreducible.

Lemma 2.2.5. For every [c] 2 Pm

(k), the mapping �c

: Pn

(⌦) ! Pn+m+nm

(⌦) given by

�c

([x]) = �([x], [c]) is continuous in the k-topology.

LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 25

Proof. Every k-closed set in Pn+m+nm

(⌦) is an intersection of sets

V (f) = {[z] 2 Pn+m+nm

(⌦) : f(z) = 0}

for homogeneous f 2 k[Z]. The inverse image of this in Pn

(⌦) is

��1c

(V (f)) = {[x] 2 Pn

(⌦) : f(xi

cj

) = 0}

Since f has coe�cients in k and cj

2 k, this is a k-closed set in Pn

(⌦). ⇤

We need to go to the ⌦-topology to get the statement that �c

: Pn

(⌦) ! Pn+m+nm

(⌦)is continuous for all [c] 2 P

m

(⌦).

Lemma 2.2.6. Pn

(⌦) is irreducible in the ⌦-topology.

Proof. Pn

(⌦) = V (0) and 0 is a homogeneous prime ideal in ⌦[X]. I forgot one subtlety:We didn’t prove the correspondence between prime ideals in ⌦[X] and ⌦-irreducible subsetsof P

n

(⌦). But, we can go to V ⇤(0) = ⌦n+1. This is ⌦-irreducible since ⌦ is infinite: If n = 0this is clear since proper closed subsets of ⌦ are finite. Suppose it is true for n � 1. Thenfor any two proper closed subsets Y, Z, let y, z 2 ⌦n so that y ⇥ ⌦ 6⇢ Y and z ⇥ ⌦ 6⇢ Z.Then there are only finitely many points in y ⇥ ⌦ \ Y and z ⇥ ⌦ \ Z. Let w 2 ⌦ so that(y, w) /2 Y, (z, w) /2 Z. Then Y \⌦n⇥w and Z\⌦n⇥w are proper closed subsets of ⌦n⇥wwhose union is the whole. But, by induction on n, ⌦n⇥w is irreducible. Contradiction. ⇤

We can now complete Mumford’s proof [5] that the image ⌃ of the Segre embedding� : P

n

(⌦)⇥Pm

(⌦)! Pn+m+nm

(⌦) is irreducible in the ⌦-topology and therefore irreduciblein the k-topology.

Proof of Corollary 2.2.3. Suppose not. Then ⌃ is the union of two proper closed nonemptysubsets. The complements of these closed subsets are disjoint nonempty open sets U1, U2 ✓⌃. We would like to say that the inverse images of these are open subsets of P

n

(⌦)⇥Pm

(⌦).However, we do not (yet) know the topology on such a product space. (It is NOT the producttopology!) But we can use the continuity in each variable.

Since � : Pn

⇥ Pm

! ⌃ is onto by definition, we have disjoint nonempty subsets��1(U1),��1(U2). Let ([a], [b]) 2 ��1(U1) and ([c], [d]) 2 ��1(U2). This means that

[a] 2 ��1b

(U1), [c] 2 ��1d

(U2)

So, we have two nonempty subsets ��1b

(U1), ��1d

(U2) of Pn

(⌦) which are open in the ⌦-topology. Since P

n

(⌦) is ⌦-irreducible, any two open sets meet. So, there is [e] in the inter-section. So, ([e], [b]) 2 ��1(U1) and ([e], [d]) 2 ��1(U2). This implies that ��1

e

(U1),��1e

(U2)are nonempty ⌦-open subsets of P

m

(⌦). Since Pm

(⌦) is ⌦-irreducible, they meet at somepoint ([e], [f ]). This means �([e], [f ]) lies in U1 \ U2 which is a contradiction. ⇤

Another proof of the same thing can be given using graded rings and graded ideals.

2.3. Graded rings and graded ideals.

Definition 2.3.1. A graded ring is a ring R which is decomposed as an infinite directsum: R = R0 � R1 � R2 � · · · so that R

n

Rm

✓ Rn+m

. Elements in Rd

are said to behomogeneous of degree d. The condition R

n

Rm

✓ Rn+m

means deg(fg) = deg f + deg g iff, g are homogeneous. The equation R =

L

Rd

means that every element of R is equal toa sum of homogeneous components which all lie in R.

26 KIYOSHI IGUSA BRANDEIS UNIVERSITY

The basic example is R = k[X1, · · · , Xn

].A graded ring can be constructed from its pieces as follows. Let R0 be a ring and let R

n

be R0-modules. Let µn,m

: Rn

⇥Rm

! Rn+m

be an R0-bilinear mapping which is

(1) commutative: µn,m

(a, b) = µm,n

(b, a) for all a 2 Rn

, b 2 Rm

.(2) associative: µ(a, µ(b, c)) = µ(µ(a, b), c) for all a, b, c in R

n

, Rm

, Rk

.(3) µ : R0 ⇥R

n

! Rn

gives the action of R0 on Rn

.

Then R =L

Rn

is a graded ring. A graded ideal can be constructed using the same idea.

Definition 2.3.2. A graded ideal in a graded ring R =L

Rd

is a sequence of R0 submodulesIn

✓ Rn

so that

In

Rm

✓ In+m

.

Proposition 2.3.3. Graded ideals are the same as homogeneous ideal in R =L

Rn

.

Proof. A graded ideal I =L

In

is clearly an ideal. It is homogeneous since it is generatedby elements of

S

In

. (Keep in mind that there is one element 0 which is in all In

and all Rn

but, otherwise, the sets Rn

are disjoint.) Conversely, any homogeneous ideal a in a gradedring R is a graded ideal since a contains I

n

= a \Rn

and a =L

In

. ⇤

The grading makes it easier to talk about projective algebraic sets:

V (I) = {[x] 2 Pn

(⌦) : f(x) = 0 8f 2 Id

8d}

Definition 2.3.4. Let R =L

Rn

, S =L

Sm

be graded k-algebras. Then a graded k-homomorphism of degree d is a homomorphism of k-algebras ' : R! S so that '(R

n

) ✓ Sdn

for all n. (Note that all rings are Z-algebras.)

Proposition 2.3.5. (a) The kernel of any graded homomorphism ' : R ! S of graded

rings is a graded ideal in R: ker' =L

(ker'n

: Rn

! Sdn

)(b) Conversely, any graded ideal I ✓ R is the kernel of a degree 1 graded homomorphism:

' : R ⇣ S =L

Rn

/In

given in degree n by the quotient map

Rn

⇣ Sn

=R

n

In

.

Going back to our example, one way to show that an ideal is prime is to show that it isthe kernel of a graded homomorphism. In our case, I = (XW �XY ) is the kernel of the(degree 2) graded homomorphism:

' : k[X,Y, Z,W ] ! k[S, T, U, V ]

'(X) = SU

'(Y ) = SV

'(Z) = TU

'(W ) = TV

It is clear that XW � Y Z 2 ker'. To show that ker' = I = (XW � Y Z), we shouldcompare:

ker'd

: Rd

! S2d

and

Id

= (XW � Y Z)Rd�2

LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 27

Lemma 2.3.6. If R = k[X0, · · · , Xn

] then R0 = k and Rd

a vector space over k of dimen-

sion

�

n+d

n

�

with basis given by the monomials

X↵ = Xa00 · · ·Xan

n

where |↵| :=P

ai

= d and all ai

� 0.

Proof. There are�

n+d

n

�

monomials of degree d in n+1 variables. This is obvious from some

examples: Take n = 2, d = 7. The statement is that there are�

n+d

n

�

=�92

�

= 36 monomialsin three letters of degree 7. To see this consider

X3Y 2Z2 = XXX|Y Y |ZZ

We put partitions between the letters. There are d letters and n partitions. The monomialis completely determined by the position of the partitions. For example:

| • • • •| • • •corresponds to Y 4Z3 (no X’s): You are only allowed to put X

i

between the ith and i+1stpartitions. Since you choose n out of n+d placed to put your partitions, you get

�

n+d

n

�

. ⇤We say that X↵ has multidegree ↵ = (a0, · · · , an). In the case at hand we have:

dimRd

=

✓

d+ 3

3

◆

=d+ 1

6(d+ 3)(d+ 2)

dim Id

= dimRd�2 =

✓

d+ 1

3

◆

=d+ 1

6d(d+ 1)

and the di↵erence is

dimRd

/Id

=d+ 1

6(d2 + 5d+ 6� d2 � d) = (d+ 1)2.

On the other hand, '(Rd

) is spanned by monomials SaT bU cV e where a + b = d = c + e.There are (d + 1)2 such monomials. So, by linear algebra, I

d

= ker'd

. So, I = ker' is aprime ideal. (Maybe not the most e�cient way to prove this. But, when you get used toit, you can do this calculation very fast.)

A similar calculation works for the general case

k[Z]! k[X,Y ]

'(Zij

) = Xi

Yj

Example 2.3.7. Here is an example in [4]. The twisted cubic is V (I) ✓ P3(⌦) where I isthe kernel of the degree 3 graded homomorphism

' : k[X,Y, Z,W ] ! k[S, T ]'(X) = S3

'(Y ) = S2T

'(Z) = ST 2

'(W ) = T 3

Since ' is graded, the kernel I is a graded ideal in k[X,Y, Z,W ]. The claim is:

I = (XZ � Y 2, Y W � Z2, XW � Y Z)

Certainly, these lie in ker'. It is also easy to see that

'd

: Rd

! S3d

28 KIYOSHI IGUSA BRANDEIS UNIVERSITY

is surjective. Furthermore, any two monomials in the inverse image of SaT 3d�a di↵er by aelement of I

d

. So, I = ker'. (Proof: We can assume a 3d � a. Then for any monomialXiY jZkW ` in '�1(SaT 3d�a), we have k + ` � 0. So, we can exchange all X terms usingthe relations XZ ⌘ Y 2, XW ⌘ Y Z modulo I

d

. Then we have i = 0. Next we can eliminateeither Y or W in the expression using YW = Z2. We are left with either Y ⇤Z⇤ or Z⇤W ⇤

which is unique.)The graded homomorphism ' : k[X,Y, Z,W ]! k[S, T ] gives a morphism

f'

: P1(⌦)! P3(⌦)

defined byf'

([s, t]) = [s3, s2t, st2, t3]

This is well-defined since, if we replace [s, t] with the equivalent [s0, t0] = [�s,�t] then

f'

([s0, t0]) = [�3s3,�3s2t,�3st2,�3t3]

Claim: Let ⌃ denote the image of this mapping. Then

⌃ = V (I)

where I = (XZ � Y 2, Y W � Z2, XW � Y Z).Proof: It is clear that ⌃ ✓ V (I) since each generator of I is a homogeneous polynomial

which is zero on the image of f'

. Conversely, suppose that [x, y, z, w] 2 V (I). Thenxz = y2, yw = z2, xw = yz. Also, either x 6= 0 or w 6= 0 (otherwise x = y = z = w = 0which is not allowed). Suppose w 6= 0. Then we may assume w = 1. Then, y = z2 andx = yz = z3. So,

[x, y, z, 1] = [z3, z2, z, 1] = f'

([z, 1])

is in the image of f'

.

Lecture 8 ends here. We skipped the next subsection because it is routine.

2.4. Irreducible algebraic sets in Pn

. We need to verify that the arguments used in thea�ne case carry over to the homogeneous case.

Lemma 2.4.1. A homogeneous ideal a in k[X] is prime i↵ it satisfies the following. For

any two homogenous polynomials f, g, if fg 2 a then either f 2 a or g 2 a.

Remark 2.4.2. This can be rephased (and generalized) in two equivalent ways:(1) A graded ringR =

L

Rn

is a domain i↵ there are no nonzero elements f 2 Rn

, g 2 Rm

so that fg = 0.(2) A graded ideal I =

L

Id

in a graded ring R =L

Rd

is prime i↵ for all f 2 Rn

\In

,g 2 R

m

\Im

, fg 2 Rn+m

\In+m

.

Proof. Prime ideals certainly have this property. Conversely, suppose that a is homogeneousand has this property. Let f, g be arbitrary polynomials so that fg 2 a. Let f [n], g[m] bethe highest degree terms in f, g. Then f [n]g[m] 2 a since it is the highest degree term infg. So, either f [n] or g[m] lies in a, say f [n] 2 a. Then (f � f [n])g 2 a. By induction, eitherf � f [n] 2 a (which implies f 2 a) or g 2 a. So, a is prime. ⇤

By a similar argument we have the following.

Lemma 2.4.3. A homogeneous ideal a in k[X] is primary i↵ it satisfies the following. For

any two homogenous polynomials f, g, if fg 2 a then either f 2 r(a) or g 2 a.

Write this in terms of graded ideals in any graded ring.

LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 29

Proof. We may assume that none of the homogeneous components of g lie in a. Thenf [n]g[m] 2 a implies that f [n] 2 r(a). Let k � 1 be minimal so that (f [n])kg 2 a. Then

(f � f [n])(f [n])k�1g = (f [n])k�1fg � (f [n])kg 2 a

and (f [n])k�1g /2 a. So, f � f [n] 2 r(a) (which implies f 2 r(a) by induction on the numberof homogeneous components of f). ⇤

Using these lemmas, the proof in the a�ne case carries over to show the following.

Proposition 2.4.4. Every homogeneous radical ideal in any graded Noetherian ring is the

intersection of finitely many homogeneous prime ideals.

Theorem 2.4.5. Irreducible closed subsets of Pn

(⌦) correspond to homogeneous prime

ideal in k[X] not equal to (X0, · · · , Xn

). Furthermore, every closed subset of Pn

(⌦) can be

expressed uniquely as a finite union of irreducible subsets.

Note: the uniqueness is a topological property. The proof of Proposition 1.4.2 works inany topological space.

Lecture 9.

2.5. Projective schemes. The purpose of this section is to construct the projective ana-logue of Spec(R) and prove the analogue of Theorem 1.9.3. We continue using the colorcode: red for schemes, blue for varieties.

Definition 2.5.1. Let R =L

Rd

be a graded ring. Then Proj(R) is defined to be thefollowing topological space (together with a sheaf which will be constructed later). As aset Proj(R) is the set of homogeneous prime ideals in R which are not equal to the idealL

d>0Rd

. In the special case R = k[X0, · · · , Xn

], this is the ideal (X0, X1, · · · , Xn

) whichwas excluded before. The topology on Proj(R) is given as follows. For any f 2 R, let

V (f) = {P 2 Proj(R) : f 2 P}These sets are defined to be closed. General closed sets are intersections of these sets:

V (S) =\

f2SV (f) = {P 2 Proj(R) : S ✓ P}

In the case R = k[X] we have a bijection:

irreducible closed subsets of Pn

(⌦)$ P 2 Proj(k[X])

Theorem 2.5.2. There is a continuous epimorphism

⇡ : Pn

(⌦) ⇣ Proj(k[X])

given by ⇡([x]) = P where P 2 Proj(k[X]) corresponds to the irreducible set [x].

Proof. We first show ⇡ is surjective.Let P be a homogeneous prime ideal in R = k[X0, · · · , Xn

], P 6= (X0, · · · , Xn

). Then

P 6= (X0, · · · , Xn

) prime) V ⇤(P ) 6= {0} ✓ ⌦n+1 irreducible

We know that V ⇤(P ) has a generic point y 2 ⌦n+1. And y 6= 0 since, otherwise, V ⇤(P ) =0 = {0}, the excluded case. So, y gives a point [y] 2 P

n

(⌦). Let ⌃ = V (P ).Claim: [y] = ⌃. (Thus ⇡([y]) = P .)Pf: [y] = ⌃0 = V (I) where I is some graded ideal. But y 2 V ⇤(I)

closed

✓ V ⇤(P ) impliesV ⇤(I) = V ⇤(P ) implies I = P . So, ⌃0 = ⌃.

30 KIYOSHI IGUSA BRANDEIS UNIVERSITY

Next we show ⇡ is continuous.Take a closed subset V (f) ⇢ Proj(R). Then

⇡�1(V (f)) = {[y] 2 Pn

(⌦) : f 2 I([y])} = V (f).

But V (f) ✓ Pn

(⌦) is a closed subset by definition. So, ⇡ is continuous. ⇤Corollary 2.5.3. The continuous epimorphism ⇡ : P

n

(⌦) ⇣ Proj(R) is also an open

mapping. Therefore, Proj(R) has the quotient topology.

Proof. Every open subset of Y = Pn

is a union of basic open sets Yf

where f 2 k[X] ishomogeneous. But Y

f

is the complement of V (f). And ⇡�1(V (f)) = V (f) implies that⇡(Y

f

) is the complement of V (f) in Proj(R). So, ⇡ is an open mapping as claimed. ⇤Example 2.5.4. Let k = R and consider P1(⌦) ! Proj(R[X,Y ]). Then P1(R) = R [1is a circle and P1(⌦) = ⌦[1. The elements of P1(⌦) are [1, x] for x 2 ⌦ and 1 = [0, x] =[0, y] 8x, y.

P1(⌦) has three kinds of points:

(1) P1(R) is the set of closed point x = {x}, x 2 R and 1.(2) P1(C) is the set of x = [y] so that x is finite (consisting of [y] and [y], the complex

conjugate of [y].(3) P1(⌦)\P1(C) is the set of generic points. This comes from dimension theory. The

only homogeneous ideal in R[X,Y ] which is not maximal is 0.

Consider Proj(R[X,Y ]) as the quotient space of P1(⌦) under the identification [x] ⇠ [y]if ⇡([x]) = ⇡([y]). This gives the following.

P1(C) = C [ 1 is a sphere. The equator is P1(R). The identification is z ⇠ z. Thisidentifies the top hemisphere with the bottom hemisphere. All other points are generic andtherefore are identified to one point in Proj. So, Proj(R[X,Y ]) is a 2-dimensional diskwith one generic point which we view as being “smeared” over the entire disk (like peanutbutter on a cracker).

Spec(R) = [ (generic point)

We did another example. The key point was:Given ' : R! S a graded homomorphism of degree d, there is an induced mapping

'⇤ : Proj(S)! Proj(R)

sending P to Q = '�1(P ). This mapping is continuous since

('⇤)�1(V (f)) = {P : f 2 '�1(P )} = V ('(f))

since f 2 '�1(P ), '(f) 2 P .

2.6. Projective space as union of a�ne spaces. Recall that Pn

(⌦) =S

Ui

whereUi

= {xi

6= 0}. Then⌦n ⇠= U

i

(x1, · · · , xn) 7! [x1, · · · , xi, 1, xi+1, · · · , xn]Projective space (together with its structure sheaf) is the basic example of a variety whichis not a�ne but rather made up of a�ne varieties pasted together. We need to study this so

LECTURES ON ALGEBRAIC GEOMETRY MATH 202A 31

that we can generalize the construction later. The a�ne pieces Ui

are pasted together alongtheir intersection. This pasting is given by isomorphisms of open subsets of irreducible sets.We want to see how U

i

, Uj

are pasted together. By symmetry we can assume i = 0, j = 1.We look at U0 \ U1 and see where it goes to under the equivalence ⌦n ⇠= U

i

.

x0 6= 0 x1 6= 0

U0 U1

U0 \ U1

x0, x1 6= 0

⇠= ⇠=

x y

V0 V1

⌦n ⌦n

x1 6= 0 x1 6= 0

The image of U0\U1 under the equivalences Ui

⇠= ⌦n are shaded. The image under bothmaps is the set V0 = V1 = {x1 6= 0}. The mapping : V0 ! V1 is the pasting map:

: (x1, · · · , xn) 7! [1, x1, · · · , xn]/x1 7!✓

1

x1,x2x1

, · · · , xnx1

◆

This is a regular function with 2 = id. In the drawing, (x) = y.Since V0, V1 are given by X1 6= 0, the ring of regular functions on V0 and V1 is

�V0 = �V1 = k[X0, X1, X2, · · · , Xn

]/(X0X1 � 1) = R

Then '

: �V1⇠= �V0 is given by '

(X0) = X1,'

(X1) = X0 and '

(Xi

) = X0Xi

fori � 2. This is an automorphism of R as k-algebra with ('

)2 = id.In my notes I verified that the ring isomorphism '

induces an isomorphism

('

)⇤ : Spec(�V0)! Spec(�V1)

which is compatible with the homeomorphism in the sense that ('

)⇤ sends the primeideal ⇡(x) to ⇡(y) if y = (x). But I don’t think I got to that in class. We will do thisagain later (verify that the pasting together of a�ne varieties is compatible with the pastingtogether of corresponding a�ne schemes).