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This question is motivated by this and this. Can the following proposition be proved without Axiom of Choice?

Proposition: Let $k$ be a field. Let $A$ and $B$ be commutative algebras without zero-divisors which are finitely generated over $k$. Suppose that $A$ is a subring of $B$ and $B$ is integral over $A$. Let $P$ be a prime ideal of $A$. Then there exists a prime ideal $Q$ of $B$ such that $P = A \cap Q$.

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Makoto Kato
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    Your $A$ and $B$ (and hence everything in sight) are then Noetherian, right? Why isn't that enough? – Dylan Moreland Jul 07 '12 at 21:26
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    @Dylan: Even so, there are many definitions of "noetherian" and their equivalence depends on AC. E.g., ACC and every ideal is finitely generated depends on some choice, and of course the characterization in terms of nonempty families having maximal elements. So one needs to specify which definition of "noetherian" one is using if you are also working without AC. – Arturo Magidin Jul 07 '12 at 21:31
  • @Arturo That's a good point. Really the only time I come in contact with Choice is when an author remarks, “We used Zorn's lemma but if $A$ were Noetherian, and most of our $A$ are, then we would not have had to.” My guess is that one [and not this one :)] would have to go back to the proof of the Hilbert basis theorem and see whether it can be tweaked appropriately; I just wanted to bring it up. – Dylan Moreland Jul 07 '12 at 21:40
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    @DylanMoreland: The "strongest" definition of noetherian (in the sense that it implies the other two even in the absence of AC) is "if $\mathcal{C}$ is a nonempty family of submodules, then $\mathcal{C}$ has a maximal element." That definition essentially says "Zorn's Lemma holds" (one proves it from ACC using Zorn's Lemma), so I can see how one would not need to invoke Zorn's Lemma if you assume it holds. It implies ACC (just take the set of submodules in your chain) and finite generation of submodules (take the collection of finitely generated submodules of the given module), even without AC – Arturo Magidin Jul 07 '12 at 21:47
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    This is a very good reflection of Tarski's characterization of finiteness, vs. T-finite sets. Tarski proved that $A$ is finite if and only if for every non-empty ${\cal A\subseteq P}(A)$ there is a maximal element; and $A$ is T-finite if and only if every chain in ${\cal P}(A)$ has a maximal element. ZF only proves finite implies T-finite and needs some choice for the other direction. These things are very common in mathematics, actually, where choice principles are reflected like this, but I digress (I was off-topic to begin with!)... – Asaf Karagila Jul 07 '12 at 21:53
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    I don't think that you can prove it without AC. And to be honest, I think it is a little bit misguided to develop commutative algebra without AC. – Martin Brandenburg Jul 07 '12 at 22:39
  • You have a lot of these questions about choice. Is there a particular reason? – Asaf Karagila Jul 08 '12 at 08:10
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    @AsafKaragila "You have a lot of these questions about choice. Is there a particular reason?" I prefer not using AC. I'd like to use it only when it's absolutely necessary. – Makoto Kato Jul 08 '12 at 20:03
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    But what difference does it make, if you do use it once you can use it always. This is mathematics, not World War II and AC is not a nuclear weapon. If you assumed AC for the existence of a basis for a vector space, there is absolutely no harm in use AC for proving a theorem about modules. If you were intentionally limiting yourself to models without AC to begin with, that would be a whole other story. However from your comment it seems that you still accept AC as necessary sometimes and does use it in those cases, so you have to assume it is true in the universe... – Asaf Karagila Jul 08 '12 at 20:05
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    @Asaf: Dear Asaf, There are plenty of reasons to be mindful of whether or not AC is being used. Naturally if we talk about all vectors spaces (arbitrary, and of arbitrarily large infinite dimension) then we will need AC to say things. But one expects that in more finitistic contexts AC would not be needed. As one example (that is close to my own interests and concerns), it is natural to ask whether AC is used in the proof of Fermat's Last Theorem. Certainly various forms of the result being asked about in this question are used, and so it is natural to wonder if AC is used. Regards, – Matt E Jul 09 '12 at 01:47
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    @Matt: Naturally there is no need for Zorn's lemma when dealing with tame objects. However the insistence of not using any choice to do mathematics outside the context of set theory and AC research seems unreasonable to me (and I spend my days in models where choice is negated very strongly). Even the axiom of determinacy gives you some choice. Most of ordinary mathematics, and often finitary mathematics, would need to assume roughly dependent/countable choice, or be reformulated in a very precise manner to rid of this need. There is no real escape from that, except into [ultra]finitism. – Asaf Karagila Jul 09 '12 at 04:57
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    @Asaf: Dear Asaf, Thanks for this thoughtful reply. Although I freely use any form of choice in my arguments, I am probably some combination of ultra-finitist and ultra-platonist at heart (!), which is why I find these discussions interesting. Best wishes, – Matt E Jul 09 '12 at 14:59
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    @AsafKaragila As Matt E wrote, there are several reasons to mind this type of questions. One reason: Generally speaking I think it's not meaningless to prove a theorem under fewer conditions. Another reason: A constructive proof can be useful when you solve a concrete problem. Of course you may need a lot of works to get a computer algorithm, but at least you can be sure that you have a more constructive proof at hand than the one using AC. – Makoto Kato Jul 09 '12 at 19:07

2 Answers2

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First we show

Lemma If $A\to B$ is a finite homomorphism of rings with $A$ local and $B\ne 0$, then the maximal ideal $P$ of $A$ is the pre-image of a maximal ideal $Q$ of $B$.

Proof. By Nakayama's lemma, $PB\ne B$. The quotient $B/PB$ is a finite $k$-algebra (where $k$ is the field $A/P$) and is no zero. The set of proper ideals of $B/PB$ is non-empty and has an element of maximal $k$-vector space dimension. The latter is then a maximal ideal, hence equal to $Q/PB$ for some maximal ideal $Q$ of $B$ containing $P$. The pre-image $P'$ of $Q$ is maximal because $A/P'$ is contained in $B/Q$ and the later is finite over $A/P'$. So $P'=P$.

Now we prove your proposition. As $B$ is a finitely generated $A$-algebra, $B$ integral over $A$ implies that $B$ is finite over $A$. Hence $A_P\to A_P\otimes_A B$ is finite with $A_P\otimes_A B\ne 0$. By the above lemma, $PA_P$ is the pre-image of a maximal ideal of $A_P\otimes_A B$. The existence of $Q$ as desired follows from standard arguments on localizations.

This been said, I agree with the second part of Martin Brandenburg's comment.

  • Have you proved that Nakayama's lemma does not require choice? – Asaf Karagila Jul 07 '12 at 23:02
  • @AsafKaragila: the standard proof doesn't involve AC (take a minimal system of generators, no need of AC here, and find a contradiction). –  Jul 07 '12 at 23:06
  • Yeah, I keep forgetting the proof of all those algebra theorems - even though I'd seen them not too long ago, some not for the first time too - and for a good reason. I just wanted to make sure that you are not missing anything. – Asaf Karagila Jul 07 '12 at 23:08
  • Some proof(for example the one using determinant) of Nakayama use AC. – Makoto Kato Jul 07 '12 at 23:13
  • @AsafKaragila: your question is welcome. I also had some doubt while writing... –  Jul 07 '12 at 23:15
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    @MakotoKato: I don't see why using determinant requires AC. Anyway, the point is there exists one proof without AC. –  Jul 07 '12 at 23:17
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    The proof using determinant shows that there exists an elemet $r$ in $A$ such that $r \equiv 1$ (mod $P$) and $rB = 0$. Hence $B$ = 0, because if $r$ is not invertible, $r \in P$ by Zorn's lemma. – Makoto Kato Jul 07 '12 at 23:26
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    I agree with Makoto. The last step uses AC. See also my summary at MO: http://mathoverflow.net/questions/41836/elementary-proof-of-nakayamas-lemma With ZF it is consistent that in some rings there are no prime ideals at all ... – Martin Brandenburg Jul 07 '12 at 23:40
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    I see. So we have to be careful on the definition of local rings. One (strong) form is the maximal ideal contains all non-invertible elements. The localization $A_P$ at a prime ideal is consistent with this definition. –  Jul 08 '12 at 08:46
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Definition 1 Let $A$ be a commutative ring. Suppose $A$ has a unique maximal ideal. We say $A$ is a local ring in the usual sense.

Definition 2 Let $A$ be a commutative ring. Let $P$ be an ideal of $A$. Suppose that every element of $A - P$ is invertible. Then we say $A$ is a strictly local ring. Clearly $P$ is the unique maximal ideal of $A$. Hence $A$ is a local ring in the usual sense.

Note 1 It can be proved by using AC that a local ring in the usual sense is a strictly local ring. However, we are not supposed to use AC here.

Lemma 1 Let $A$ be commutative ring. Let $P$ be a prime ideal of $A$. Then $A_P$ is a strictly local ring.

Proof: Clear.

Definition 3 Let $A$ be a domain. Let $P$ be a prime ideal of $A$. Let $k$ be the field of fractions of $A/P$. Let $K$ be a field which contains $A$ as a subring. Let $x$ be an element of $K$. Let $x'$ be an element of some field containing $k$. We say $x'$ is a specialzation of $x$ over $P$ and we write $x \rightarrow x'$ over $P$, if there exists a homomorophism $\psi:A[x] \rightarrow (A/P)[x']$ extending the canonical homomorphism $A \rightarrow A/P$ and $\psi(x) = x'$. This is equivalent to that $\tilde{f}(x') = 0$ whenever $f(X) \in A[X]$ such that $f(x) = 0$,where $\tilde{f}(X) \in (A/P)[X]$ is the reduction of $f(X)$ mod $P$.

Lemma 2 Let $A$ be a strictly local domain. Let $\mathfrak{m}$ be the unique maximal ideal of $A$. Let $k = A/\mathfrak{m}$. Let $K$ be a field which contains A. Let $x \in K$. Suppose $x$ has no specializaion in any finite extension field of $k$ over $\mathfrak{m}$. Then $x$ is non-zero and $1/x \rightarrow 0$ over $\mathfrak{m}$.

Proof: Let $P$ = {$f \in A[X]$; $f(x) = 0$}. $P$ is an ideal of $A[X]$. Let $P'$ be the ideal of $k[X]$ generated by the set {$f(X)$ (mod $\mathfrak{m}$); $f(X) \in P$}. We claim that $P' = k[X]$. Suppose otherwise. Then $P'$ is generated by a polynomial $h(X) \in k[X]$, where $h(X)$ is not a non-zero constant. Hence $h(X)$ has a root $x'$ in a finite extension of k. Then $x \rightarrow x'$ over $\frak{m}$ This is a contradiction. Hence there exists $f(X) \in P$ such that $f(X)$ (mod $\mathfrak{m}$) is a non-zero constant. Let $f(X) = a_mX^m + ... + a_0$. Then $a_0 \in A - \mathfrak{m}$, $a_i \in \mathfrak{m}$ for $i > 0$. We assume that $m$ is minimal among the degrees of such polynomials. Let $y = 1/x$. Let $g(Y) = a_0Y^m + ... + a_m$ be a polynomial in $A[Y]$. Then $g(y) = 0$. Let $h(Y)$ be any polynomial in $A[Y]$ such that $h(y) = 0$. Since $A$ is a strict local ring, $a_0$ is invertible. Hence $Z^{m-1}h(Y) = g(Y)q(Y) + r(Y)$, where $q(Y), r(Y) \in A[Y]$ and deg $r \leq m - 1$. Substituting $Y$ by $y$ we get $r(y) = 0$. Taking the reductions mod $\mathfrak{m}$ of the both sides, we get $Y^{m-1}\tilde{h}(Y) = \tilde{a_0}Y^m\tilde{q}(Y) + \tilde{r}(Y)$. Hence $\tilde{h}(0) = \tilde{r}(Y)$. Since $\tilde{r}(Y)$ cannot be a non-zero constant by the minimality of $m$, $\tilde{h}(0) = 0$. Hence $z \rightarrow 0$ over $\mathfrak{m}$. QED

Note 2 The idea of the proof of Lemma 2 is borrowed from Weil's Foundations of algebraic geometry. According to him, the idea is due to Chevalley. See also Note 3 below.

Lemma 3 Let $A$ be a domain. Let $P$ be a prime ideal of $A$. Let $k$ be the field of fractions of $A/P$. There exists a unique homomorphism $A_P \rightarrow k$ extending the canonical homomorphism $A \rightarrow A/P$.

Proof: Clear.

Lemma 4 Let $A$ be a domain. Let $P$ be a prime ideal of $A$. Let $k$ be the field of fractions of $A/P$. Let $K$ be a field which contains A as a subring. Let $x \in K$. Suppose $x$ has no specializaion in any finite extension of $k$ over $P$. Then $x$ is non-zero and $1/x \rightarrow 0$ over $P$.

Proof: Let $B = A_P$. By lemma 1, $B$ is a strictly local ring. Suppose $x \rightarrow x'$ over $PA_P$. By Lemma 3, $x \rightarrow x'$ over $P$. Hence $x$ has no specializaion in any finite extension of $k$ over $PA_P$. By Lemma 2, $1/x \rightarrow 0$ over $PA_P$. Hence, by Lemma 3, $1/x \rightarrow 0$ over $P$. QED

Note 3 Lemma 4 is a generalization of the one given in Weil's Foundations. He proved it when $A$ is a finitely generated domain over a field. Our Lemma 4 treats not only a case where $A$ and $A/P$ have equal characteristics but also a case of unequal ones.

Note 4 As van der Waerden and Weil showed, Lemma 4 has vast applications in algebraic geometry. For example, Hilbert Nullstellensatz can be proved by using it.

Lemma 5 Let $A$ be a domain. Let $P$ be a prime ideal of $A$. Let $k$ be the field of fractions of $A/P$. Let $K$ be a field which contains $A$ as a subring. Let $x \in K$. Suppose $x$ is integral over $A$. Then there exist a finite extension $k'$ of $k$ and $x' \in k'$ such that $x \rightarrow x'$ over $P$.

Proof: Suppose there exists no such x'. By Lemma 4, there exists a homomorophism $\psi:A[1/x] \rightarrow k$ such that $\psi(1/x) = 0$ extending the canonical homomorphism $A \rightarrow A/P$. Let $y = 1/x$. Since $x$ is integral over $A$, $x^n + a_1x^{n-1} + ... + a_0 = 0$. Hence $1 + a_1y + ... + a_0y^n = 0$. Applying $\psi$, we get $1 = 0$. A contradiction. QED

Proposition Let $B$ be a domain. Let $A$ be a subring of $B$. Suppose $B$ is finitely generated as an $A$-module. Let $P$ be a prime ideal of $A$. Let $k$ be the field of fractions of $A/P$. Then there exist a finite extension $k'$ of $k$ and a homomorophism $\psi:B \rightarrow k'$ extending the canonical homomorphism $A \rightarrow A/P$.

Proof: This follows Immediately from Lemma 5.

Corollary Let $B$ be a domain. Let $A$ be a subring of $B$. Suppose $B$ is finitely generated as an $A$-module. Let $P$ be a prime ideal of $A$. Then there exists a prime ideal $Q$ of $B$ such that $P = A \cap Q$.

Proof: This follows Immediately from the proposition.

Makoto Kato
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