4
$\begingroup$

I would like to show that in a map of affine rings (i.e. finitely generated $k$-domains), the preimage of a maximal ideal is maximal. I know a more general statement is true (we just need a map of $k$-algebras where the target is finitely generated), and I know how to prove it, but the proofs that I know are not super obvious to think about geometrically.

Is there a geometric proof of the statement above (and maybe even the more general one)?

For reference as to what I'm looking for, if I assume $k$ is algebraically closed, here is a proof I came up with that I think works and is fairly okay to think about geometrically:

Proof: Let $f:A \to B$ be a map between f.g. $k$-domains. Since the category of affine varieties is equivalent to the (opposite) category of f.g. $k$-domains under our assumption, we may write our map as $f:A(X) \to A(Y)$ for $X,Y$ affine varieties and consider the induced map $\phi:Y \to X$. In particular, we can describe $f$ as sending $g$ to $g \circ \phi$ (identifying $A(-)$ with $\mathcal O(-)$). Now, let $\mathfrak m$ be a maximal ideal of $A(Y)$. It corresponds to some point $p \in Y$. Then, we have $$f^{-1}(\mathfrak m) = \{h \in A(X) \mid h \circ \phi \in \mathfrak m\} = \{h \in A(X) \mid h(\phi(p))=0\}$$ which is the maximal ideal corresponding to $\phi(p)$.

Thinking about it geometrically, the algebraic statement is essentially equivalent to the fact that image of a point is a point. So my question is, can we do something similar (perhaps salvage the above proof, assuming it's correct) to show the situation when $k$ need not be algebraically closed? Thanks in advance!

Edit: Also if possible, an elementary proof would be nice (e.g. nothing beyond the first chapter of Hartshorne). This is for a talk to an audience without much knowledge of Algebraic Geometry (which is why I'm looking for a geometrically intuitive proof) so I don't want to use things like affine schemes if avoidable. Thanks again!

$\endgroup$
7
  • $\begingroup$ I'm not sure your proof works: I think establishing that the category of "classical $k$-varieties" is dual to the category of finite type $k$-domains requires the result you are trying to prove! $\endgroup$ Commented Apr 12 at 20:12
  • $\begingroup$ You might be right, I'd have to take a closer look at the proof, though in the exposition of e.g. Hartshorne it didn't seem to be used, at least at first glance (e.g. see Proposition I.3.5). But I imagine at the very least, it probably follows as an immediate corollary of some lemma needed to prove the equivalence. In any case, this isn't too much of a problem for my purposes, since I just want to provide some intuition in the general case, though if it is used, then I guess a truly rigorous geometric proof would be even nicer! $\endgroup$ Commented Apr 12 at 20:24
  • $\begingroup$ Ok I took a closer look at the proof in Hartshorne it seems that the result I want to prove wasn't used (at least as far as I could see) to prove the equivalence. Please correct me if I'm incorrect. Thanks! $\endgroup$ Commented Apr 12 at 21:23
  • $\begingroup$ Establishing the classical ideal-variety correspondence doesn't require this result, I believe. (The idea is that after selecting generators of $A$ and $B$, the possible images of the generators are precisely the possible components of morphisms of varieties. Then the ring morphism corresponds to the pullback of functions. No invocation here.) On how to generalize the proof... when $k$ is not algebraically closed, there are more maximal ideals than "classical" points. So you need to add more points in and pass to affine schemes. Once you do this, I believe the proof holds with no changes. $\endgroup$ Commented Apr 12 at 21:23
  • $\begingroup$ More precisely, when $k$ is not algebraically closed, the key idea (that you can interpret $\mathfrak{m}$ as being a point) falls apart in the classical setting. $\endgroup$ Commented Apr 12 at 21:25

1 Answer 1

Reset to default
3
$\begingroup$

Let $k$ be an algebraically closed field. We never particularly need it but it clarifies a geometric discussion if you insist on avoiding the language of schemes. I will present a scheme-theoretic picture anyway, since I find it to be very lucid.

Let $f:X\to Y$ be the associated map of affine schemes. Restrict $f$ along the closed point $\ast\hookrightarrow X$ and take the integral subscheme $Z\subset Y$ which is the closure of $f(\ast)$: we have a morphism $f':\ast\to Z$ of varieties with dense image.

Your question is really just the question: "when is there a dominant morphism of varieties $\ast\to Z$"? To which the answer is, only when $Z$ itself is a point (thus, $f(\ast)$ was already a closed point). This is intuitively absolutely clear; more rigorously, it is clear by thinking that dominant morphisms should have a certain behaviour on dimension.

Without the language of schemes, this has reduced to the standard commutative algebra proof. Let us try again: say $\phi:A\to B$ is a morphism of f.g. $k$-algebras, $\mathfrak{m}\subset B$ maximal. Present $A$ by generators $X_1,\cdots,X_n$, $B$ by generators $Y_1,\cdots,Y_m$, and say $\mathfrak{m}\sim(\beta_1,\cdots,\beta_m)$ is a classical point.

$Z:=\mathbb{V}(\phi^{-1}(\mathfrak{m}))\subseteq\mathbb{A}^n$ defines a classical algebraic variety (irreducible). We have to show $Z$ is a single point. We know $u(X_\bullet)\to u(\phi(X_\bullet)(\beta))$ determines an injective linear functional $\mathscr{K}(Z)\to k$ (this is obvious with no theory) so it follows that $\mathscr{K}(Z)=k$ is one-dimensional, in particular algebraic; there are no transcendental elements, and $Z$ is $0$-dimensional i.e. a point $(\alpha_1,\cdots,\alpha_n)$.

It follows that $\phi(\beta):=\alpha$ is well-defined, but also by the vanishing locus/ideal correspondence that $\phi^{-1}(\mathfrak{m})=(X_1-\alpha_1,\cdots,X_n-\alpha_n)$ a maximal ideal.

For this proof, we have to presuppose the dimension theory of classical varieties, but this sits at a lower, more accessible level than the equivalence of categories you mention (which contains the result you want immediately). I strongly suspect we cannot offer a proof which both (1) does not do any commutative algebra at all and (2) avoids the language of schemes. Schemes were invented, in part, to have exactly this kind of flexibility.

If, in the above, you want a general $\mathfrak{m}$, then we take the map $\mathscr{K}(Z)\to\kappa$, $u\mapsto\overline{u(\phi(X_\bullet))}$ where $\kappa:=B/\mathfrak{m}$ the residue field and the overline denotes a residue class in the quotient. This is again an injective linear functional, so if $\dim Z>0$ we must have that $\kappa/k$ is a transcendental field extension. However, $\kappa$ is a finitely generated $k$-algebra so this would contradict the Zariski lemma. We have now been forced to use some commutative algebra.

A more elementary but yet more algebraic explanation: assuming the Zariski lemma, the above injective functional shows that $\mathscr{K}(Z)=\operatorname{Frac}(A/\phi^{-1}(\mathfrak{m}))$ is also a $k$-vector space of finite dimension. Then, $A/\phi^{-1}(\mathfrak{m})$ is too: multiplication by a nonzero element of this domain defines a $k$-linear injection on this vector space, forced by finitude to then be a surjection and a bijection, so we learn that $A/\phi^{-1}(\mathfrak{m})=\mathcal{O}(Z)$ is already a field - $\phi^{-1}(\mathfrak{m})$ is a maximal ideal.

$\endgroup$
12
  • $\begingroup$ Thanks a lot for your answer! Would you mind clarifying a few points? It seems to me you are using algebraic closure (though you claim we don't need it) e.g. when you say $\mathfrak m \sim (\beta_1, \dots, \beta_m)$. Without algebraic closure, I don't see how your geometric argument works (or how it's different from my initial argument) which is what I'm after. $\endgroup$ Commented Apr 12 at 23:03
  • $\begingroup$ I take a classical point because if $k$ is not algebraically closed and $\mathfrak{m}$ is not a classical point, we cannot pass to classical affine schemes (and I got the feeling that you wanted to do that). You could nevertheless follow the thread of the argument and replace 'evaluate at $\beta$' with 'take the residue class in $B/\mathfrak{m}$' and apply Noether's normalisation lemma, say, to obtain what we want, but this is now doing the commutative algebra explicitly -- we are no longer able to be slick and avoid any algebra. $\endgroup$ Commented Apr 12 at 23:17
  • $\begingroup$ There is also not the same correspondence between classical vanishing loci and ideals in the non-algebraically closed case, so we cannot faithfully geometrically interpret the a priori non-maximal ideal $\phi^{-1}(\mathfrak{m})$ as an embedded affine subset of $k^n$. $\endgroup$ Commented Apr 12 at 23:19
  • 1
    $\begingroup$ Ah okay thanks for the clarification, sorry I was misunderstanding your argument. I guess one still needs to know only points admit dominant morphisms from points (which I suppose follows by Zariski's lemma, so it's not that different from the algebraic proof). In any case, this is good enough for my purposes of a talk and certainly more geometric, so I will accept your answer. Thanks again for all your help! $\endgroup$ Commented Apr 13 at 1:14
  • 1
    $\begingroup$ @respect_the_cone Very welcome! There's also the dimension argument (yet more intuitive, 'no algebra'): if $\ast\to Z$ is a dominant map of varieties, $0=\dim(\ast)\ge\dim(Z)$ forces $Z$ to be a point $\endgroup$ Commented Apr 13 at 3:57

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.