Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

For every ring $A$, the structural morphism of schemes $\pi_A : {\bf P}^n_{A} \to {\rm Spec}{A}$ is a closed map. The usual proof of this fact is not constructive : given equations of a closed subset $Z$ of ${\bf P}^n_{A}$, it doesn't produce equations for $\pi_A(Z)$.

In the case $A$ is a polynomial ring over an algebraically closed field $k$, this result is none other than the fundamental theorem of elimination theory : the image of a Zariski-closed subset of ${\bf P}^n(k) \times k^m$ under the second projection is a Zariski-closed subset of $k^m$. The first proofs of this theorem (Cayley, Kronecker, Sylvester) used resultants and thus were constructive.

In fact, the proof using elimination theory is universal in the following sense. Given integers $n,r \geq 1$, $d_1,\ldots,d_r \geq 1$, consider the universal homogenous polynomials $P_1,\ldots,P_r$ of degree $d_1,\ldots,d_r$ in the indeterminates $T_0,\ldots,T_n$, having coefficients in the polynomial ring $\widetilde{A} = \mathbf{Z}[Y_{i,\alpha} : 1 \leq i \leq r]$, where the indeterminates $Y_{i,\alpha}$ are the coefficients of $P_i$. Then there exists an explicit "resultant system" $R_1,\ldots,R_s \in \widetilde{A}$ such that $\pi_{\widetilde{A}}(V_+(P_1,\ldots,P_r))=V(R_1,\ldots,R_s)$. This means that specializations of $P_1,\ldots,P_r$ in some algebraically closed field $k$ have a common root in ${\bf P}^n(k)$ if and only if the corresponding specializations of $R_1,\ldots,R_s$ all vanish. Of course $s$ has to depend on $n,r,d_i$, but everything is explicit (at least from a theoretical point of view).

Now let $A$ be any ring and let $I=(f_1,\ldots,f_r)$ be an homogenous ideal of finite type of $A[T_0,\ldots,T_n]$. Then the resultant system above specialized at $f_1,\ldots,f_r$ provides explicit equations for $\pi_A(V_+(I))$ (this can be seen by studying the geometric fibers of $\pi_A$). In particular if $A$ is noetherian, then the map $\pi_A$ is closed, and we have a constructive proof for that. But in general, a closed subset of ${\bf P}^n_A$ need not be defined by finitely many equations. This raises the following questions :

  1. Is there a way to prove that the map $\pi_A$ is closed for every ring $A$, by some clever reduction to the noetherian case?

  2. If $Z$ is a closed subset of ${\bf P}^n_A$, given to us by infinitely many explicit equations $(f_i)_{i \in I}$, is there a way to produce explicit equations for $\pi_A(Z)$? In other words, is there a constructive proof of the fact that $\pi_A$ is closed?

  3. Regarding question 2, an obvious thing to do is to look at all finite subfamilies $(f_i)_{i \in J}$, where $J$ is a finite subset of $I$, and to consider the associated resultant systems. Are all these equations sufficient to define $\pi_A(Z)$?

EDIT. Will Sawin has proved that the answers to all these questions is yes. Following Daniel Litt's comment, we can also consider $\pi_A(Z)$ as a closed subscheme of $\operatorname{Spec} A$, namely the closed subscheme defined by the kernel of the morphism $A \to \mathcal{O}_Z(Z)$. Do the resultant systems generate this ideal of $A$?

share|improve this question
Do you want equations cutting out $\pi_A(Z)$ set-theoretically (which suffices for the claim of the title)? Or do you want generators for the ideal of the scheme-theoretic image $\pi_A(Z)$--namely, the kernel of $A\to \pi_*(\mcl{O}_Z)$? It seems to me that Will's answer meets the first requirement, but it's not obvious to me that it meets the second requirement... –  Daniel Litt Dec 19 '12 at 22:34
Dear Daniel, thanks for your comment. Indeed, this is something I also wanted to clarify. It's already not obvious to me in the noetherian case. Possibly it's enough to treat the case of the ring $\widetilde{A}$, but how would one go to prove that? –  François Brunault Dec 20 '12 at 8:02
I edited the post to incorporate Daniel's refined question. –  François Brunault Dec 20 '12 at 8:17

1 Answer 1

3, and thus 2 and 1: yes. By checking equality between the two sets at each point, we reduce to the case where the base is a point. But points are always Noetherian schemes, and the statement is obviously true for Noetherian schemes.

share|improve this answer
Dear Will, thanks for your answer. Your argument convices me! Following Daniel's comment, do you know if it also works scheme-theoretically? –  François Brunault Dec 20 '12 at 8:01
No, but I'll think about it. –  Will Sawin Dec 20 '12 at 18:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.