MATH 763 Algebraic Geometry Homework 2

Instructor: Dima Arinkin Author: Yifan Wei

September 26, 2019

Problem 1 Prove that a subspace of a noetherian topological space is noetherian in the induced topology. Proof Because in the subspace X the closed sets are exactly restrictions from the ambient space, a descending chain of closed sets in X takes the form · · · ⊃ Ci ∩ X ⊃ · · · , taking closure in the ambient space we get a descending chain which stablizes to, say Cn ∩ X, and we have for all i Ci ∩ X ⊂ Cn ∩ X. And the inequality can clearly be restricted to X, which means the chain in X stablizes.

Problem 2 Show that a topological space is noetherian if and only if all of its open subsets are quasi-compact. Proof If the space X is noetherian, then any open subset is noetherian hence quasi-compact. Conversely, if we have a descending chain of closed subsets C1 ⊃ C2 ⊃ · · · , then we have an open covering (indexed by n): ∩ ∪ ∪ X − Ci = X − Ci, n i≤n ∪ − ∩ − By quasi-compactness we have X Ci is equal to i≤n X Ci for some n, which means the descending chain stablizes to Cn.

Problem 3 Let S ⊂ k be an infinite subset (for instance, S = Z ⊂ k = C.) Show that the Zariski closure of the set

X = {(s, s2)|s ∈ S} ⊂ A2

1 is the parabola Y = {(a, a2)|a ∈ k} ⊂ A2.

Proof Consider the morphism ϕ : k[x, y] → k[t] sending x to t and y to t2, its kernel is precisely the vanishing ideal of the parabola Y . If f vanishes on X, then ϕf vanishes exactly on S, which is an infinite subset of k, therefore ϕf = 0 and f is in the kernel of ϕ. This shows I(X) = I(Y ), therefore the closure of X is Y .

Problem 4 Show that an algebraic set X is connected if and only if the algebra k[X] has no idempotents other than 0 and 1. (Recall that f is an idempotent if f 2 = f.)

Proof Since V (f(1 − f)) = V (f) ∪ V (1 − f), if f is a non-trivial idem- potent we would have a decomposation of X into two disjoint closed sets. Conversely if X = V (f) ∪ V (g) is a decomposation into non-empty disjoint closed sets. Then we have 0 = fg and 1 = af + bg for some a, b. Neither f nor g is a unit, so af is a non-trivial idempotent (af(1 − af) = afbg = 0).

Problem 5 Let

X = V (x2 − z2 + y, yz − y) ⊂ A3 and assume that char(k) =6 2. Show that X is a union of three irreducible components. Describe them and find their prime ideals.

Proof (Finding irreducible subsets is solving equations.) It’s easy to find three closed subsets of X, they correspond to these ideals:

(y, x + z), (y, x − z), (z − 1, x2 − y − 1).

For the convenience of notation, denote them as I1,I2 and I3. k[x, y, z]/I1 is isomorphic to k[x, z]/(x + z) (because k[x, z] → k[x, y, z]/I1 is surjective). The latter one is isomorphic to k[t] (via affine transformation), an entire ring (integral domain, I’m using the terminology of Serge Lang), hence I1 (as well as I2) is prime. For I3 it is also similar and we have k[x, y, z]/I3 isomorphic to coordinate ring of parabola, which is again entire, therefore all three ideals are prime.

2 What remains to show is that the union of the closed sets corresponding to these ieals is X, which is straightforward since points in X correspond to solutions to the equations x2 − z2 + y and yz − y, there are only three cases during the solution, which correspond to the three irreducible subsets. (It is simpler than showing the intersection of the three ideals equals (x2 − z2 + y, yz − y) is due to Nullstellensatz.)

Problem 6 Let f : X → Y be a regular map; consider the induced map f ∗ : k[Y ] → k[X]. Given an ideal I of k[Y ], consider the ideal k[X] · f ∗(I) generated by its image in k[X]. Describe the closed subset

V (k[X] · f ∗(I)) ⊂ X in terms of V (I) ⊂ Y. Claim: it is the preimage of V (I) under the map f.

Proof For the convenience of notation, denote k[X] · f ∗(I) as Ie, the ex- tension of ideal I. For any P in f −1(V (I)) any a in Ie, we have a(P ) = c · f ∗b(P ) = c·b(f(P )) = 0 (where b is in f ∗(I)), therefore f −1(V (I)) ⊂ V (Ie). Conversely, for any P in V (Ie) it must annihilates all functions in f ∗(I), which means f(P ) annihilates all functions in I, therefore f(P ) is in V (I). And we established the identity:

V (f ∗(I)) = f −1(V (I)) (notice that V (f ∗(I)) = V (Ie).)

This problem shows that (global) regular functions are continuous in Zariski topology. The converse may not be easy, in HW1 we have the cusp- idal cubic homeomorphic to A1, we could construct the inverse (x, y) 7→ x/y and (0, 0) 7→ 0. It is continuous since it’s the inverse of a homeomorphism, but it’s not polynomial in shape. (just came up with a better example, f : k → k mapping x to 1/x and 0 to 0, this is also a homeomorphism)

Problem 7 Let f : X → Y be a regular map; consider the induced map f ∗ : k[Y ] → k[X]. Given an ideal I of k[X], its preimage (f ∗)−1(I) is an ideal in k[Y ]. Describe the closed subset

V ((f ∗)−1(I)) ⊂ Y

3 in terms of V (I) ⊂ X. You may assume I is radical. Claim: it is the closure of f(V (I)).

Proof (Assuming I is radical.) For the convenience of notation, denote (f ∗)−1(I) as Ic, the contraction of I. Let f(P ) be a point in f(V (I)) and a an arbitrary function in V (Ic), then a(f(P )) = f ∗a(P ) = 0. Therefore f(V (I)) ⊂ V (Ic). Conversely, let a be a function which vanishes on f(V (I)), this means f ∗a vanishes on V (I), since I is radical this implies f ∗a ∈ I. So the closure of f(V (I)) is V (Ic).

Problem 8 Let f : X → Y be a regular map between algebraic sets. Suppose that the image of f is dense in the Zariski topology of Y . (Such maps f are called dominant.) Prove that if X is irreducible, then so is Y .

Proof Suppose Y can be written as a union of two closed subsets Y = −1 −1 Y1 ∪ Y2, then the preimages of them yield X = f (Y1) ∪ f (Y2). Since (global) regular maps are continuous from previous problem and the premise −1 that X is irreducible, we can assume that f (Y1) = X. Since f(X) is dense in Y and Y1 is closed we have Y1 = Y . Now let’s consider the algebraic approach. If a is in the kernel of f ∗, which means precisely a vanishes on f(X), since f(X) is dense, this shows that a vanishes everywhere hence a = 0, therefore f ∗ : k[Y ] → k[X] is an injection to an entire ring and k[Y ] must be entire as well.

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