Math 203A - Solution 1 2 Problem 1. Show that the Zariski topology on A is not the product of the Zariski topologies 1 1 on A × A . Answer: Clearly, the diagonal 2 Z = {(x, y): x − y = 0} ⊂ A 2 is closed in the Zariski topology of A . We claim this Z is not closed in the product 1 1 2 topology of A × A . Assume otherwise. Then, the complement A \ Z is open in the 1 product topology. Therefore, there are nonempty open sets U, V in A such that 2 U × V ⊂ A \ Z. Now, the sets U and V have finite complements: 1 U = A \{p1, . . . , pn}, 1 V = A \{q1, . . . , qm}. We obtain a contradiction picking 1 z ∈ A \{p1, . . . , pn, q1, . . . , qm}. In this case, we clearly have 2 (z, z) ∈ U × V, while (z, z) 6∈ A \ Z.  Problem 2. A X is said to be Noetherian if it satisfies the ascending chain condition on open sets, i.e. any ascending chain of open sets eventually stabilizes. (i) Check that any subset Y ⊂ X of a Noetherian space is also Noetherian in the subspace topology. (ii) Check that a Noetherian space which is also Hausdorff must be a finite set of points. (iii) Show that a Noetherian space is quasi-compact i.e., show that any open cover of a Noe- therian space has a finite subcover. n (iv) Show that A is Noetherian in the Zariski topology. Conclude that any affine algebraic set is Noetherian.

Answer: (i) If {Ui}i≥1 is an ascending chain of open subsets of Y , there exists {Vi}i≥1 open subsets of X such that

Vi ∩ Y = Ui

for all i. Note that {Vi}i≥1 may not be ascending, but let i Wi = ∪k=1Vk.

Then {Wi}i≥1 is clearly ascending and i i i Wi ∩ Y = (∪k=1Vk) ∩ Y = ∪k=1(Vk ∩ Y ) = ∪k=1Uk = Ui.

Therefore {Wi}i≥1 is an ascending chain of open subsets of X. Since X is Noe-

therian, there exists n0 such that Wn = Wn0 for all n ≥ n0. Therefore,

Un = Wn ∩ Y = Wn0 ∩ Y = Un0

for all n ≥ n0, i.e. {Ui}i≥1 eventually stabilizes. This implies Y is Noetherian. 1 2

(ii) Since X is Noetherian and Hausdorff, every subset Y ⊂ X is also Noetherian and Hausdorff. By part (iii), Y is quasi-compact and Hausdorff, hence compact. In particular, X is compact. Since Y is compact, Y is closed. Since Y was ar- bitrary, X must have the discrete topology. In particular, X can be covered by point-sets {x}, for x ∈ X. Since X is compact, X must be finite. (iii) Order the open sets in X by inclusion. It is clear that any collection U of open sets in X must have a maximal element. Indeed, if this was false, we could inductively produce a strictly increasing chain of open sets, violating the fact that X is Noetherian. Now, consider an open cover [ X = Uα, α

and define the collection U consisting in finite union of the open sets Uα. The

collection U must have a maximal element Uα1 ∪ ... ∪ Uαn . We claim

X = Uα1 ∪ ... ∪ Uαn . Indeed, otherwise we could find

x ∈ X \ (Uα1 ∪ ... ∪ Uαn ) .

Now, x ∈ Uβ for some index β. Then,

Uβ ∪ Uα1 ∪ ... ∪ Uαn belongs to the collection U, and strictly contains the maximal element of U,

namely Uα1 ∪ ... ∪ Uαn . This is absurd, thus proving our claim. n (iv) It suffices to show that A satisfies the ascending chain condition on open sets. n If {Ui} be an ascending chain of open sets, let Yi be the complement of Ui in A . Then {Yi} is a descending chain of closed sets. Since taking ideal reverses inclu- sions, we conclude that I(Yi) is an ascending chain of ideals in k[X1,...,Xn]. This chain should eventually stabilize, so

I(Yn) = I(Yn+1) = ..., for n large enough. Applying Z to these equalities, we obtain

Yn = Yn+1 = ....

This clearly implies that the chain {Ui} stabilizes.  3 Problem 3. Let A be the 3-dimensional affine space with coordinates x, y, z. Find the ideals of the following algebraic sets: (i) The union of the (x, y)-plane with the z-axis. 1 3 2 3 (ii) The image of the map A → A given by t → (t, t , t ). Answer: (i) The ideal of the (x, y)-plane is (z). The ideal of the z-axis is (x, y). By Problem 6, the ideal we’re looking for is (z) ∩ (x, y) = (xz, yz). 3

2 3 3 1 3 2 (ii) Let I be the ideal of the set (t, t , t ) in A , as t ∈ A . Observe (z − x , y − x ) is clearly contained in I. We claim that I = (z − x3, y − x2). We need to show every f ∈ I is in the ideal (z − x3, y − x2). Indeed, let us consider the image of f in the quotient A = k[x, y, z]/(z − x3, y − x2). We would like to show that the image of f is 0 in A. Now, in the ring A the relations z = x3, y = x2 are satisfied. Thus f(x, y, z) = f(x, x2, x3) in A. Consider the g(x) = f(x, x2, x3). Since f(t, t3, t5) = 0, we see g(t) = 0 for all t. Therefore g ≡ 0 as a polynomial in x. (This uses that the ground field is infinite). Therefore, 3 2 f(x, y, z) = 0 in A = C[x, y, z]/(z − x , y − x ), which is what we wanted.  n m n Problem 4. Let f : A → A be a polynomial map i.e. f(p) = (f1(p), . . . , fm(p)) for p ∈ A , where f1, . . . , fm are in n variables. Are the following true or false: m n (1) The image f(X) ⊂ A of an affine algebraic set X ⊂ A is an affine algebraic set. −1 n m (2) The inverse image f (X) ⊂ A of an affine algebraic set X ⊂ A is an affine algebraic set. n n+m (3) If X ⊂ A is an affine algebraic set, then the graph Γ = {(x, f(x)) : x ∈ X} ⊂ A is an affine algebraic set. Answer: (i) False. Consider the following counterexample which works over infi- nite ground fields. Consider the hyperbola 2 X = {(x, y): xy − 1 = 0} ⊂ A . Clearly, X is an algebraic set. Let 2 1 f : A → A , f(x, y) = x 1 be the projection onto the first axis. The image of f in A is {x : x 6= 0}. This 1 is not an algebraic set because the only algebraic subsets of A are empty set, 1 finitely many points or A . (ii) True. Suppose m X = Z(F1,...,Fs) ⊆ A , where F1,...,Fs are polynomials in m variables. Define

Gi = Fi(f1(X1,...,Xn), . . . , fm(X1,...,Xn)), 1 ≤ i ≤ s. These are clearly polynomials in n variables. In short, we set

Gi = Fi ◦ f, 1 ≤ i ≤ s We claim −1 n f (X) = Z(G1,...,Gs) ⊆ A , which clearly exihibits f −1(X) as an algebraic set. This claim follows from the remarks −1 p ∈ f (X) ⇐⇒ f(p) ∈ Z(F1,...,Fs) ⇐⇒ F1(f(p)) = F2(f(p)) = ... = Fs(f(p)) = 0 4

⇐⇒ G1(p) = ... = Gs(p) ⇐⇒ p ∈ Z(G1,...,Gs). (iii) True. Assume that X is defined by the vanishing of the polynomials F1,...,Fs in the variables X1,...,Xn. Set

G1 = Y1 − f1(X1,...,Xn),...,Gm = Ym − fm(X1,...,Xn).

We regard the F and G’s as polynomials in the n+m variables X1,...,Xn,Y1,...,Ym. We claim that the graph Γ is defined by the n+m Γ = Z(F1,...,Fs,G1,...,Gs) ⊆ A . Indeed,

(x, y) ∈ Z(F1,...,Fs,G1,...,Gs) ⇐⇒ F1(x) = ... = Fs(x) = 0, y1 = f1(x), . . . , ym = fm(x) ⇐⇒ x ∈ X, y = f(x) ⇐⇒ (x, y) ∈ Γ.  3 Problem 5. Let X be the union of the three coordinate axes in A . Determine generators for the ideal I(X). Show that I(X) cannot be generated by fewer than 3 elements. Answer: We claim that I(X) = (xy, yz, zx). Clearly, xy, yz, zx vanish on X. Conversely, if a polynomial X i j k f = aikjx y z (i,j,k) vanishes on the x-axis, then f(x, 0, 0) = 0, hence

ai00 = 0, for i ≥ 0. Similarly, ai00 = a0j0 = a00k = 0, for all i, j, k ≥ 0. This means that f ∈ (xy, yz, zx). Now, I(X) cannot be generated by one polynomial f since then f will have to divide xy, yz, zx which have no common factors. Assume that I(X) is generated by two elements (f, g). Since f ∈ (xy, yz, zx), we see that f = xyP + yzQ + zxR for some polynomials P, Q, R. Letting f˜ be the degree 2 component of f, and p, q, r be the free terms in P, Q, R, we obtain that f˜ = pxy + qyz + rzx only contains the monomials xy, yz, zx. The similar remark applies to the degree 2 piece g˜ of the polynomial g. Furthermore, there must be polynomials A1,B1 such that

yz = A1f + B1g.

Look at the degree 2 terms of the above equality. Letting a1, b1 be the free terms in A1,B1, we obtain yz = a1f˜+ b1g.˜ 5

Similarly,

xz = a2f˜+ b2g,˜

xy = a3f˜+ b3g.˜ To reach a contradiction, consider the V the three dimensional vector space of poly- nomials of degree 2 which contain the monomials xy, yz, zx only. We observed that f,˜ g˜ ∈ V. Moreover, the last three equations above show that f˜ and g˜ span the three dimensional vector space V . This is a contradiction showing that I(X) cannot be generated by fewer than 3 elements.  n Problem 6. Let X1, X2 be affine algebraic sets in A . Show that (i) I(X1 ∪ X2) = I(X1) ∩ I(X2), p (ii) I(X1 ∩ X2) = I(X1) + I(X2). Show by an example that taking radicals in (ii) is necessary.

Answer: (i) If f ∈ I(X1 ∪ X2), f vanishes on X1 ∪ X2. In particular, f vanishes on X1. Thus, f ∈ I(X1). Similarly, we have f ∈ I(X2). Then f ∈ I(X1) ∩ I(X2). We proved that I(X1 ∪ X2) ⊆ I(X1) ∩ I(X2).

On the other hand, if f ∈ I(X1) ∩ I(X2), then f ∈ I(X1) and f ∈ I(X2). There- fore, f vanishes on both X1 and X2, and then also on X1 ∪ X2. This implies that f ∈ I(X1 ∪ X2). Hence

I(X1) ∩ I(X2) ⊆ I(X1 ∪ X2).

(ii) We assume the base field is algebraically closed. Because X1, X2 be affine alge- braic sets, we can assume

X1 = Z(a),X2 = Z(b).

Furthermore, we may assume a and√ b are radical. Otherwise, if for instance a is not radical, we can replace a by a. This doesn’t change the algebraic sets in question as √ X1 = Z(a) = Z( a). By Hilbert’s Nullstellensatz, p p q√ √ √ I(X1) + I(X2) = I(Z(a)) + I(Z(b)) = a + b = a + b

and √ I(X1 ∩ X2) = I(Z(a) ∩ Z(b)) = I(Z(a + b)) = a + b . Therefore, p I(X1 ∩ X2) = I(X1) + I(X2).

Note: In the above, we made use of the equality Z(a) ∩ Z(b) = Z(a + b). 6

This can be argued as follows. If x ∈ Z(a) ∩ Z(b), then f(x) = g(x) = 0 for all f ∈ a and g ∈ b. Thus x ∈ Z(a + b) proving that Z(a) ∩ Z(b) ⊆ Z(a + b). Conversely, if x ∈ Z(a + b), then (f + g)(x) = 0 for every f ∈ a and g ∈ b. In particular, picking g = 0, we get f(x) = 0 for every f ∈ a, so x ∈ Z(a). Similarly x ∈ Z(b). Therefore Z(a + b) ⊆ Z(a) ∩ Z(b).

2 Counterexample: Consider the following two algebraic subsets of A : 2 X1 = (y − x = 0),X2 = (y = 0). Graphically, these can be represented by a parabola and a line tangent to it. It is clear that X1 ∩ X2 = {0}, so I(X1 ∩ X2) = (x, y). On the other hand, 2 2 2 I(X1) + I(X2) = (y − x ) + (y) = (y − x , y) = (y, x ) . Therefore I(X1 ∩ X2) 6= I(X1) + I(X2).