Algebraic Geometry, PS 5

Bryden Cais^{∗}

November 13, 2002

1 Lecture 10

1. (a) Let Y = V (Z_{1}^{2} − Z_{2}^{3}) and X = 𝔸^{1}. Define f : X → Y via f(x) = (x^{3}, x^{2}). Then clearly, O(X) = k[Z_{2}] and O(Y) = k[Z_{1}, Z_{2}]/(Z_{1}^{2} − Z_{2}^{3}) and f^{∗} : O(Y) → O(X) is given by f^{∗}(z_{1}) = Z_{2}^{3}, f^{∗}(z_{2}) = Z_{2}^{2}, where z_{i} is the image of Z_{i} under the natural homomorphism k[Z_{1}, Z_{2}] → O(Y). Obviously, f^{∗} is injective. Moreover, observe that

O(X) = f^{∗}(O(Y))[Z_{1}]

and that Z_{1} is a root of the monic polynomial

Z^{2} − f^{∗}(z_{1}) = 0.

It follows that O(X) is integral over f^{∗}(O(Y)) and hence that f is finite.

(b) Let X = Y = 𝔸^{2}, and suppose f : X → Y is given by f(x, y) = (xy, y). By Proposition 3 of Lecture 10, we know that a finite map is surjective. But observe that f is not surjective: the fibre over any point (a, 0) with a ∈ 𝔸^{1} \ {0} is clearly empty. It follows that f is not finite. Indeed, we have O(X) = O(Y) = k[Z_{1}, Z_{2}] and f^{∗} : O(Y) → O(X) is given by f^{∗}(Z_{1}) = Z_{1}Z_{2}, f^{∗}(Z_{2}) = Z_{2}. Observe that

O(X) = f^{∗}(O(Y))[Z_{1}]

and that Z_{1} is a root of the degree one equation

f^{∗}(Z_{2})Z − f^{∗}(Z_{1}) = 0,

which is not monic.

∗in collaboration with Eric, Chris, Cornelia, and Jesse

2. Let f : X → Y be a finite map. We show that the image of any closed subset Z ⊂ X under f is closed. First, it is enough to show this for irreducible closed subsets of X since every closed subset is a finite union of irreducible closed subsets. So assume that Z is irreducible. If we show that f : Z → \overline{f(Z)} is a finite map then we are done, since by Proposition 3 any finite map is surjective (i.e. we will have shown that f(Z) = \overline{f(Z)} so that f(Z) is closed). Let I be the prime ideal such that

O(Z) = O(X)/I

and let I^{c} = (f^{∗})^{−1}(I). We claim that O(Y)/I^{c} = O(\overline{f(Z)}). Observe that p ∈ O(Y) vanishes on \overline{f(Z)} if and only if p vanishes on O(f(Z)), if and only if p ∘ f = f^{∗}(p) ∈ O(X) vanishes on Z, that is, if and only if f^{∗}(p) ∈ I. It follows that we have the following commutative diagram:

\begin{matrix}O(Y) & ⟶^{f^{∗}} & O(X) \\ ↓ & & ↓ \\ O(\overline{f(Z)}) & ⟶_{f^{∗}} & O(Z) \end{matrix},

where the vertical maps are the obvious quotient maps. Since O(X) is integral over f^{∗}(O(Y)), it follows that O(Z) is integral over f^{∗}(O(\overline{f(Z)})) and hence that f^{∗} : Z → \overline{f(Z)} is finite.

3. Let f : X → Y and g : X′ → Y′ be two finite regular maps. Then f, g are surjective and O(X), O(X′) are finitely generated modules over f^{∗}(O(Y)), g^{∗}(O(Y′)), respectively. We showed on a previous problem set that f × g is regular. Since f, g are surjective, so is f × g : X × X′ → Y × Y′. Moreover, we have seen on a previous homework that

O(X × X′) = O(X) ⊗ O(X′), O(Y × Y′) = O(Y) ⊗ O(Y′).

Suppose that O(X) is generated by a_{1}, …, a_{m} over f^{∗}(O(Y)) and that O(X′) is generated by b_{1}, …, b_{n} over g^{∗}(O(Y′)). Then we see that O(X) ⊗ O(X′) is generated by {a_{i} ⊗ b_{j}} for 1 ≤ i ≤ m and 1 ≤ j ≤ n over f^{∗}(O(Y)) ⊗ g^{∗}(O(Y′)) = f^{∗} × g^{∗}(O(Y) ⊗ O(Y′)), and hence that O(X) ⊗ O(X′) is integral over f^{∗} × g^{∗}(O(Y) ⊗ O(Y′)). It follows that f × g is a finite regular map.

4. Let X ⊂ 𝔸^{2} be given by the equation T_{2}^{2}T_{1} −T_{2} = 0 and let Y = 𝔸^{1}. It is not hard to see that X consists of two irreducible components: a copy of 𝔸^{1} given by (T_{1}, 0) and the hyperbola T_{1}T_{2} = 1. Define f : X → Y by

f(a, b) = a.

It is clear that f is a surjective regular map (since, in particular, the point a of K has preimage containing (a, 0) ∈ X(K); regularity follows from the fact that f is given by polynomials). Moreover, the fibres of f are finite: the preimage of a point a consists of the points (a, 0), (a, 1/a) if a ≠ 0 and of the single point (0, 0) if a = 0. However, f is not a finite map. If it were, it would take closed sets to closed sets by problem 2. However, the algebraic subset of X given by T_{1}T_{2} = 1 is obviously a closed set in X, but its image in 𝔸^{1} is not closed since its complement is a single point, which is clearly closed.

5. (a) Let A be an integral domain and let Q(A) denote the field fractions of A. Let Ā be the integral closure of A in Q(A). We claim that Ā is normal. First, it is obvious that Q(A) = Q(Ā). Now suppose that x ∈ Q(A) is integral over Ā. Then Ā(x) is an integral extension of Ā. Since Ā is integral over A, it follows that Ā(x) is integral over A and hence that x is integral over A. Thus x ∈ Ā by definition.

(b) Let A = k[Z_{1}, Z_{2}]/(Z_{1}^{2} − Z_{2}^{2}(Z_{2} + 1)). We claim that Q(A) = k(z_{1}/z_{2}), where z_{1}, z_{2} are the images of Z_{1}, Z_{2} in the quotient. Certainly we have Q(A) = k(z_{1}, z_{2}). But (z_{1}/z_{2})^{2} − 1 = z_{2} and (z_{1}/z_{2})^{3} − (z_{1}/z_{2}) = z_{1} so that k(z_{1}, z_{2}) ⊂ k(z_{1}/z_{2}). On the other hand, we certainly have k(z_{1}/z_{2}) ⊂ Q(A) = k(z_{1}, z_{2}). Our claim follows. Observe, however, that z_{1}/z_{2} is integral over A: it is obviously a root of X^{2} − 1 − z_{2}. Thus, k[z_{1}/z_{2}] ⊂ Ā. But since k[z_{1}/z_{2}] ≃ k[T] by the obvious isomorphism, (and k[T] is a PID) it follows that k[z_{1}/z_{2}] is integrally closed in k(z_{1}/z_{2}) = Q(A) and by our above remarks, A ⊂ k[z_{1}/z_{2}]. It follows that Ā = k[z_{1}/z_{2}] ≃ k[T], as required.

(c) Let us show that A = k[Z_{1}, Z_{2}, Z_{3}]/(Z_{1}Z_{2} − Z_{3}^{2}) is normal. We have an isomorphism of A with k[T_{1}^{2}, T_{2}^{2}, T_{1}T_{2}] given by

Z_{1} ↦ T_{1}^{2}
Z_{2} ↦ T_{2}^{2}
Z_{3} ↦ T_{1}T_{2}.

(Surjectivity is obvious; injectivity just as trivial). But this shows that A is isomorphic to a direct summand of k[T_{1}, T_{2}] (recalling that we can write k[T_{1}, T_{2}] = ⨁_{i=0}^{∞} S_{i}, where S_{i} consists of all homogenous polynomials of degree i, we see that k[T_{1}, T_{2}] = k[T_{1}^{2}, T_{2}^{2}, T_{1}T_{2}] ⊕ ⨁_{r=1}^{∞} S_{2i−1}). Since k[T_{1}, T_{2}] is normal, it follows from commutative algebra that any direct summand of a normal ring is normal, and hence that A is normal.

6. Let B = k[Z_{1}, Z_{2}]/(Z_{1}Z_{2}^{2} +Z_{2} +1), and let A = k[z_{2} +z_{1}z_{2}], where z_{i} is the image of Z_{i} under the natural homomorphism k[Z_{1}, Z_{2}] → B. Then we claim that B is finite over A. Indeed, first notice that A[z_{2}] contains z_{2} (trivially) and z_{1} since we have

(z_{1}z_{2})^{2} + z_{1}z_{2} = −z_{1}

and z_{1}z_{2} ∈ A[z_{2}], clearly. It follows that A[z_{2}] = B. On the other hand, z_{2} is a root of the monic polynomial

X^{2} − (z_{2} + z_{1}z_{2} + 1)X − 1 = 0

whose coefficients are clearly in A, so that B is integral (and hence finite) over A. Finally, we have an isomorphism

A → k[T]

given by

z_{2} + z_{1}z_{2} ↦ T.

2 Lecture 11

1. (a) Let X be the two point space {a, b} with a open and b closed. This is obviously a topological space. Moreover, a is dense in X since it is open. Further, it is clear that X, b are closed and irreducible. However, dim a = 0 while dim X = 1 since we have the chain b ⊂ X.

(b) Let X = ℤ^{+} with the discrete topology and let Y = ℤ^{+} with the following topology: the

closed sets are the sets

C_{N} = {n ∈ ℤ^{+} : n ≤ N}.

By part c), Y is infinite dimensional while X is clearly one dimensional since the only closed irreducible subsets are single points. Moreover, the identity map id : X → Y is obviously surjective and continuous, since every finite subset of X is closed.

(c) Let Y be as in part b). We claim that C_{n} is irreducible for all n. For we have the trivial

property

C_{i} ∪ C_{j} = C_{max(i,j)},

from which it follows that any union of closed sets that is equal to C_{n} contains C_{n} as one of the elements of the union. Certainly Y is noetherian: if Z_{0} ⊃ Z_{1} ⊃ … is any descending chain of closed irreducible subsets then either every Z_{i} = Y, in which case there is nothing to prove, or there is some i such that Z_{i} = C_{m} for some m. Since C_{m} has dimension m (by considering the chain C_{m} ⊃ C_{m−1} ⊃ … ⊃ C_{1} it follows that the chain {Z_{i}} must stabilize. However, Y is infinite dimensional since it contains closed subsets of dimension m for any positive integer m (namely C_{m}).

2. Let X be a closed irreducible subset of ℙ^{n}(K) or 𝔸^{n}(K) of codimension 1. Since X is not all of ℙ^{n}(K) (resp. 𝔸^{n}(K)) since dim X = n − 1, there is some nonzero polynomial f vanishing on X. Moreover, since X is irreducible, there is some irreducible factor g of f vanishing on X. Let Y = V (g). Then dim Y = n − 1 by Theorem 4 of Lecture 11. Moreover, by construction, we have X ⊂ Y. But X, Y are irreducible and have the same dimension. It follows that the inclusion map X → Y is finite and hence surjective; that is, X = Y. Thus, X is a hypersurface.

4. Let X be an irreducible algebraic set of dimension n − 1. Then we have seen that I(X) is a prime ideal of Krull dimension n − 1, that is, there exists a chain of n prime ideals

I(X) = ℘_{0} ⊊ ℘_{1} ⊊ … ⊊ ℘_{n}.

To any closed irreducible subset Z of X we can associate the prime ideal 𝔮 = I(Z) ⊋ I(X). Then as a consequence of the well-known Jordan Hölder Theorem, there exists a chain of prime ideals of length n + 1 containing 𝔮, say

I(X) = 𝔮_{0} ⊊ 𝔮_{1} ⊊ … ⊊ 𝔮_{n}

with 𝔮_{i} = 𝔮 for some i. Then

V (𝔮_{n}) ⊊ V (𝔮_{n−1}) ⊊ … ⊊ X

is a chain of strictly decreasing closed irreducible subsets of X with Z as a member of the chain. Define

codim(Y, X) = max{k : ∃a chain of closed irreducible subsets Y = Z_{0} ⊊ … ⊊ Z_{k} = X}.

Suppose that dim Y = d − 1. Then we have a chain of closed irreducible subsets

Y_{0} ⊊ Y_{1} ⊊ … ⊊ Y_{d} = Y.

This gives a chain of maximal length = k + d + 1 (by the definition of codimension and dimension) given by

Y_{0} ⊊ Y_{1} ⊊ … ⊊ Y_{d} ⊊ Z_{1} ⊊ … ⊊ Z_{k} = X.

It follows from the above that k + d = dim X, that is,

dim Y + codim(Y, X) = dim X

7. Recall that a point of the veronese curve v_{r}(ℙ^{1}) ⊂ ℙ^{n}(K) is given by

(X_{0}^{r}, X_{0}^{r−1}X_{1}, X_{0}^{r−2}X_{1}^{2}, …, X_{1}^{r})

for (X_{0}, X_{1}) ∈ ℙ^{1}. Let

a_{0}T_{0} + a_{1}T_{1} + … + a_{n}T_{n}

be a hyperplane in ℙ^{n}. Then the number of points of intersection of v_{r}(ℙ^{1}) with this hyperplane is clearly the number of solutions to

a_{0}X_{0}^{r} + a_{1}X_{0}^{r−1}X_{1} + … + a_{n}X_{1}^{r} = 0.

Observe that if X_{1} = 0 we are forced to take X_{0} = 0 (since we may assume that a_{0} ≠ 0, as we shall see momentarily), and this is not a point of projective space. We may therefore dehomogenize to obtain a polynomial p of degree r in one variable. Now if the coefficients a_{i} belong to the dense, zariski open subset of the space of coefficients given by the nonvanishing of the discriminant of the a_{i}, then the polynomial p has r distinct roots. The degree is therefore r, since this dense open set nontrivially intersects the dense open set as in problem 6.