Solutions to Exercises 1 and 2

Math 8205
Spring 1998

Solutions to Exercises 1 and 2

Last updated Friday, May 15, 1998.

Solution to Exercise 1a
Solution to Exercise 1b
Solution to Exercise 1c
Solution to Exercise 1d

Solution to Exercise 2a
Solution to Exercise 2b
Solution to Exercise 2c

Solution to Exercise 1a Let J = I(X) $\cap$ K[x₁,..., x_n_-1]. Clearly, every f $\in$ J vanishes at every point of $\pi$ (X) and hence, by continuity, at every point of Y. Therefore, J $\subseteq$ I(Y). Conversely if f = f(x₁,..., x_n_-1) $\in$ I(Y), then f, considered as an element of K[x₁,..., x_n], vanishes at every point of X. Therefore f $\in$ J. We conclude that I(Y) $\subseteq$ J.
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Solution to Exercise 1b
First, a correction: We should have said that f $\in$ JI(X) is of minimal x_n-degree subject to the condition that f $\notin$ (g₁..., g_r), the extension of J = I $\pi$ K[x₁,..., x_n_-1] to K[x₁,..., x_n].

To see the solution, consider F $\in$ I(X), and suppose that F $\notin$ (g₁..., g_r). In K[x₁,..., x_n] [1/h_d], the highest degree coefficient of x_n in f is invertible, so that we can do long division:

F = f·g(x₁,..., x_n, 1/h_d) + r(x₁,..., x_n, 1/h_d). By clearing denominators, we get: h_d^s r(x₁,..., x_n, 1/h_d) $\in$ I(X), and of lower x_n-degree, and hence in (g₁..., g_r). This shows that a general element of I(X) is already in the ideal (f,g₁..., g_r).
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Solution to Exercise 1c There are two cases to consider. The notation is as in part b.

If I(X) = (g₁..., g_r), then X = $\pi$ (X) $\times$ A¹.
If I(X) = (g₁..., g_r, f), where f is defined as in part b., then every point p $\in$ U $\cap$ Y has at least 1, and at most d preimages. Specifically, if p = (a₁,..., a_n_-1), then we solve the equation f(a₁,..., a_n_-1, t) = 0 to find the last coordinate, namely a_n.

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Solution to Exercise 1d Since U is locally closed in Aⁿ, we have U = $\Ubar$ $\cap$ V, where V is open in Aⁿ. We set X = $\Ubar$ and Z = X - U. We define Y $\subseteq$ A^n-1 to be the closure of $\pi$ (X). By the result of part c., $\pi$ (X) contains a dense open subset U₀ $\subseteq$ Y. As in previous parts of this exercise, there are two cases to consider. In either case, the conclusion will follow if it is true that the restriction of $\pi$ , considered as a regular map $\pi$ : Z -> Y is not dominant. Indeed, this statement implies that $\pi$ (Z) $\subseteq$ W, where W is a proper closed subset of Y. It follows from this that

U₀ $\cap$ (Y - W) $\subseteq$ $\pi$ (U). Now, we discuss the two cases separately:

Case 1: X = $\pi$ (X) $\times$ A¹. By the discussion above, we only need to understand what happens when $\pi$ : Z -> Y is dominant. In this situation, Y is the closure of $\pi$ (Z). Applying the result of part c to Z, we see that $\pi$ : Z -> Y must be generically finite to one. We take S $\subseteq$ $\pi$ (X) to be the set of points p such that $\pi$ ^-1 (p) $\subseteq$ Z. It follows from what we have already proved that S is a proper subset of Y. To see that S is a closed subset, we view the elements of I(Z) as polynomials in x_n with coefficients in K[x₁,..., x_n_-1]. Then S is defined by the vanishing of the coefficient polynomials, so that it is closed. Then, Y - S is contained in $\pi$ (U), which completes this case of the proof.
Case 2: $\pi$ : X -> Y is generically finite to one. We claim that $\pi$ : Z -> Y is not dominant in this case. There are at least 3 different ways to check this fact. Since the method that you were "supposed to use" is not really my favorite, I'll just sketch it briefly, and then proceed with the others.

Method 1 (for proving that $\pi$ : Z -> Y is not dominant in case 2.) Whether or not this method is what you were "supposed to do", it does involve applying the results of previous parts of this exercise.

We assume, to the contrary, that $\pi$ : Z -> Y is dominant, and show that this leads to a contradiction. It is given that $\pi$ : X -> Y and $\pi$ : Z -> Y are both generically finite to one, and are both dominant. So, we can apply the result of part b. to the ideals I(X) and I(Z). We obtain generating sets:

I(X) = (f,g₁..., g_r). and I(Z) = (f',g₁..., g_r). Because I(Z) $\supseteq$ I(X), we have deg_xn (f') $\leq$ deg_x
n (f). After adjoining the reciprocals of the leading coefficients, we can apply the division algorithm to obtain: f = qf' + r,

in K[x₁,..., x_n], where deg _xn (r) < deg _xn (f). Since f has minimal positive xn-degree, we conclude that:

r $\in$ I(X) $\cap$ K[x₁,..., x_n_-1] = I(Z) $\cap$ K[x₁,..., x_n_-1]. This, however, implies that f $\in$ I(Z), which is a contradiction. We conclude that $\pi$ : Z -> Y is not dominant.

Method 2 (for proving that $\pi$ : Z -> Y is not dominant in case 2.) The method described here is a more standard technique in commutative algebra. We will show that if Z₀ is an irreducible component of Z which is contained in an irreducible component X₀ of X, then $\pi$ (Z₀) is not dense in $\pi$ (X₀). Thus, we may assume that X and Z are irreducible.

We assume that $\pi$ : Z -> Y is dominant, and show that this leads to a contradiction. Since I(Z) $\supset$ I(X) and the inclusion is proper, we can consider I₀:= I(Z)/I(Z) as a nonzero prime ideal in A(X). Since $\pi$ : Z -> Y is dominant, I₀A(X) = (0). Now, A(X) is generated, as a ring extension of A(Y), by the additive coset of x_n. We know that x_n satisfies a polynomial relation with coefficients in A(Y). Let a $\in$ A(Y) be the highest coefficient. After adjoining 1/a, we may assume that x_n is integral. Specifically, x_n is integral over A(Y)[1/a]. Since x_n generates the ring extension, we conclude that A(X)[1/a] is an integral extension of A(Y)[1/a], and hence finitely generated as a module over A(Y)[1/a]. It follows that A(X)[1/a] is finitely generated as a module over A(Y)[1/a]. Therefore, every element of A(X)[1/a] is integral over A(Y)[1/a]. Now, let g be a nonzero element of I₀, and consider a relation of integral dependence, of minimal degree:
g^m + c₁g^m-¹ + . . . + c_m-₁g + c_m = 0.

Then, we must have c_m $\neq$ 0; otherwise we could cancel a factor and obtain a relation of lower degree. Now, we can re-write the relation of integral dependence as follows:

g(g^m-¹ + c₁g^m-2+ . . . + c_m-₁ ) = - c_m .

This implies that c_m $\in$ I(Z) $\cap$ K[x₁,..., x_n_-1] = (0), which is a contradiction. We conclude that $\pi$ : Z -> Y is not dominant.

Method 3 (for proving that $\pi$ : Z -> Y is not dominant in case 2.) In this case, we'll consider the dimensions of X and Z, and the dimensions of the fibers of the restrictions of $\pi$ to X and Z respectively. (The "legality" of this approach will be discussed below.) As in the previous method, we just need to consider the case where X is irreducible. (There seems to be no compelling reason to assume that Z is irreducible, however.) Since Z is a proper closed subset of X, we have dim(Z) < dim(X). Because we are in "case 2", the restriction of $\pi$ is a generically finite-to-one map $\pi$ : X -> Y. Therefore dim(Y) = dim(X). On the other hand, we will certainly have dim( $\pi$ (Z)) $\leq$ dim(Z) < dim(X). (Literally we should speak about the dimension of the closure of $\pi$ (Z) . . . ) Since dim( $\pi$ (Z)) < dim(Y), we conclude that $\pi$ : Z -> Y is certainly not dominant!!

Remark. Since we didn't study dimension until somewhat after this exercise was assigned, there could be some question about whether it as really appropriate to use that method in this situation. To be absolutely sure that we haven't engaged in circular reasoning we would have to review our proofs of the results about dimension to see that no version of this result is used here. I haven't actually done that, but it certainly is known to be possible to prove those results without referring to the result of Exercise 1d. For example, see Chapter I, sections 7 and 8 of Mumford's "red notes".

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Solution to Exercise 2a As in the hint, we proceed by induction on n - m. The case n - m = 1 is immediate from Exercise 1a. The most effective way to handle the inductive step is to consider $\pi$ as a composition Aⁿ -> A^m+1 -> A^m. Let $\pi$ ₁: Aⁿ -> A^m+1 and $\pi$ ₂: A^m+1 -> A^m be the two "intermediate" projections. We define Z to be the closure of $\pi$ ₁(X) in A^m+1, and we recall that Y is the closure of $\pi$ (X) in A^m. By the inductive hypothesis, $\pi$ ₁(X) contains a dense open subset U $\subseteq$ Z. Using standard facts from topology, we check that $\pi$ (X), $\pi$ ₂(Z) , and $\pi$ ₂(U) all have the same closure in A^m+1. By the result of Exercise 1d, $\pi$ ₂(U) contains a dense open subset of its closure. Hence $\pi$ (X) contains a dense open subset of Y.

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Solution to Exercise 2b We may assume that X $\subset$ A^m. Let $\Gamma$ = $\Gamma$ be the graph of $\varphi$ . Then, $\Gamma$ is a closed subset of A^m $\times$ Aⁿ = A^m+n, and $\varphi$ (X) = $\pi$ ( $\Gamma$ ), where $\pi$ : A^m+n -> Aⁿ is the linear projection. Therefore, the result follows from Exercise 2a.

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Solution to Exercise 2c Let Y be the closure of $\varphi$ (X) in Pⁿ. Let {U₀, ..., U_n} be the standard affine open covering of Pⁿ. Without loss of generality, we may assume that U₀ $\cap$ Y is a dense open subset of Y. Let V be an affine open subset of $\varphi$ ^-1(U₀)which is dense in X. Let $\psi$ : V -> Aⁿ be the restriction of $\varphi$ , where Aⁿ is identified with U₀. By the result of Exercise 2b., $\varphi$ (V) = $\psi$ (V) contains a dense open subset of its closure in Aⁿ, and hence a dense open subset of its closure in Pⁿ. By standard facts from topology, $\varphi$ ( $\Vbar$ ) is contained in the closure of $\varphi$ (V). Since V $\subseteq$ X $\subseteq$ $\Vbar$ , it follows that $\varphi$ (X) contains a dense open subset of its closure, as claimed.

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Math 8205 Spring 1998

Solutions to Exercises 1 and 2

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Math 8205
Spring 1998