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Let $X$ and $Y$ be affine varieties such that the coordinate ring $A(Y)$ is a subring of $A(X)$. Let $$\pi:X\to Y$$ be the morphism induced by the inclusion $A(Y)\subseteq A(X)$. I need to show that $\pi$ is surjective if it satisfies the following condition:

If $J=(g_1,\ldots,g_n)\subseteq A(Y)$ is an ideal such that $J\cdot A(X)=A(X)$, then $J=A(Y)$.

I have a lot of trouble understanding how this condition can imply that $\pi$ is surjective. I don't know where to start. I start with "let $P\in Y$", but then what ideal $J$ do we take?

user323124
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  • You can take $J$ to be the maximal ideal of $A(Y)$ corresponding to $P$. Personally, I always just identify the maximal ideals of $A(Y)$ with the points of $Y$. – D_S Mar 16 '16 at 04:39

3 Answers3

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Let $k$ be the field you're working over. If $A$ is the coordinate ring of an affine variety $Y$, then $Y$ can be identified with the set of maximal ideals of $A$. Let $X$ be the set of maximal ideals of a ring $B$, and assume $A$ is a subring of $B$. The inclusion map $A \rightarrow B$ corresponds to the morphism of varieties $X \rightarrow Y$ given by $\mathfrak M \mapsto \mathfrak M \cap A$. (Since $A, B$ are finitely generated $k$-algebras, $\mathfrak M \cap A$ is actually a maximal ideal of $A$.)

Your problem now becomes the following: suppose $A, B$ are rings with $A$ a subring of $B$. Suppose that whenever $J$ is a proper ideal of $A$, $JB$ is a proper ideal of $B$. Show that every maximal ideal of $A$ is equal to $\mathfrak M \cap B$ for some maximal ideal $\mathfrak M$ of $B$.

Here's how you can start the proof: let $\mathfrak m$ be any maximal ideal of $A$. By hypothesis, $\mathfrak m B$ is a proper ideal of $B$. So it must be contained in at least one maximal ideal $\mathfrak M$ of $B$. Now what?

D_S
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    @GeorgesElencwajg We have ${\frak m}\subseteq{\frak M}\cap A$, where ${\frak M}\cap A$ is an ideal in $A$. Since ${\frak m}$ is maximal it suffices to show that ${\frak M}\cap A\neq A$. But if ${\frak M}\cap A=A$ then $1\in{\frak M}$ (since $A$ is a subring of $B$), so ${\frak M}=B$ contradicting that ${\frak M}$ is maximal. Thus, ${\frak m}={\frak M}\cap A$. – user323124 Mar 16 '16 at 12:30
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    @D_S: yes, of course! I was thinking of the case where the morphism $A\to B$ is not assumed injective, since I believe that the result is still true in that case. But of course you were perfectly entitled to use the hypothesis given by the OP! Sorry for my unfair criticism (now deleted) and +1 for your answer! – Georges Elencwajg Mar 16 '16 at 15:04
  • @Georges Your comments are always appreciated. I am still pretty bad at algebraic geometry, so it's good for me to scrutinize my proofs. – D_S Mar 16 '16 at 15:07
  • Dear D_S: you are too modest and you have a definitely laudable approach to criticism, including shamefully unfair criticism :-) Bravo! – Georges Elencwajg Mar 16 '16 at 15:15
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The key point is that given a subvariety $Z\subset Y$ with ideal $I(Z)=J\subset A(Y)$, its inverse image is the variety $\pi^{-1}(Z)=V(J.A(X))$ determined by the extension ideal $J^e=J.A(X)$.
[Atiyah-Macdonald, Chapter 1, Exercise 21 ii)]
In particular, taking $Z=\{y\}$, a single point of $Y$, we have $J=I(Z)\neq A(Y)$ so that your hypothesis (displayed in yellow) implies by contraposition that $J. A(X)\neq A(X)$.
The weak Nullstellensatz [Fulton, Chapter 1, §7, page 10] then implies that $\pi^{-1}(Z)=V(J. A(X))\neq \emptyset$, so that indeed $y\in\pi(X)$ and $\pi$ is surjective, just as you wished.

Edit (optional !)
Notice that the assumption that $A(Y)$ be a subring of $A(X)$ was not used: the proof works if one only assumes that one starts with an arbitrary morphism of rings $u:A(Y)\to A(X)$ with the highlighted property on ideals $J\subset A(Y)$.
However, since $\pi=u^\ast$ has now been proved to be surjective, we can actually deduce that indeed $u$ was injective all along, thanks to Atiyah-Macdonald's Chapter 1, Exercise 21 v).

Georges Elencwajg
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$\pi$ is induced by a morphism $f:A(Y)\rightarrow A(X)$. The condition implies that if $I$ is an ideal distinct of $A(Y)$, $f^{-1}(I)$ is distinct of $A(X)$. In particular, if $I$ is a prime, $A(X)/f^{-1}(I)\rightarrow A(Y)/I$ is an injective map, thus $A(X)/f^{-1}(I)$ is a non trivial principal domain. Thus $f^{-1}(I)\in Spec(A(X))$ and $\pi(f^{-1}(I))=I$.

Tsemo Aristide
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  • I don't get it. The last sentence "Thus $f^{-1}(I)\in A(Y)$ and $\pi(f^{-1}(I))=I$" makes no sense to me. Could you elaborate? – user323124 Mar 15 '16 at 18:38
  • This means that $f^{-1}(I)$ is a prime ideal since $A(X)/f^{-1}(I)$ is a principal domain. Thus $f^{-1}(I)\in Spec(A(X))$. – Tsemo Aristide Mar 15 '16 at 18:40
  • Thanks for your answer. Could you explain a bit more what you are saying. I am new to this subject. Why does $f^{-1}(I)\in{\rm Spec}(A(X))$ implies that $\pi(f^{-1}(I))=I$? And why does that imply that $\pi$ is surjective? – user323124 Mar 15 '16 at 18:49
  • By definition if $f:A\rightarrow B$ $\pi:Spec(B)\rightarrow Spec(A)$ is defined by $\pi(P)=f^{-1}(P)$. The argument above shows that for every prime $I$, $f^{-1}(I)$ is a prime, so by definition, $\pi(f^{-1}(I))=I$. – Tsemo Aristide Mar 15 '16 at 18:53
  • $Spec(A)$ is the space of prime ideal and $f^{-1}(I)$ is a prime ideal. – Tsemo Aristide Mar 15 '16 at 18:55
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    Dear Tsemo, I don't understand your proof at all. In particular, your notation doesn't make sense: If $I\subset A(Y)$ you can't take $ f^{-1}(I)$. Moreover you don't use that $f$ is injective and I have no idea why you claim that $A(X)/f^{-1}(I)$ is a principal ideal domain: principal ideal domains have nothing to do in this question. Finally you are mixing up $\pi$ and $f$ in your formula $\pi(f^{-1}(I))=I$. – Georges Elencwajg Mar 16 '16 at 08:23
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    @GeorgesElencwajg Thanks for your comment! I felt bad because I didn't understand anything about that proof. Now I understand why... – user323124 Mar 16 '16 at 08:31
  • Just for information: despite my comment I didn't downvote this answer,since I think that pointing out a possible flaw in an answer is more useful than downvoting. – Georges Elencwajg Mar 16 '16 at 15:06