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Let $A$ be an integral domain which is finitely generated over a field $k$. Let $f \neq 0$ be a non-invertible element of $A$. Can one prove that there exists a prime ideal of $A$ containing $f$ without Axiom of Choice?

I came up with this question to solve this problem.

This is a related question.

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Dear Makoto, I see an impressive number of questions on this site , from many users, wondering whether this or that result can be proved without the axiom of choice. So I'm curious: why do you (and they) want to know that? Would a negative answer prevent you from using such a result? Is it a philosophical concern? Do you keep in mind the list of theorems requiring that axiom? Let me emphasize that I'm not criticizing you in the least. Since you seem to be a serious and dedicated person, I thought you might give me an articulate answer. But I welcome answers from other participants too. –  Georges Elencwajg Jul 11 '12 at 17:45
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@Georges: I don't know about other users, but I'm interested in these questions because 1) I want to know what parts of commutative algebra do and do not require choice so I know what can be made explicit in principle for doing actual computations and 2) I may at some point want to do commutative algebra in categories where the axiom of choice may be false. –  Qiaochu Yuan Jul 11 '12 at 17:50
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@GeorgesElencwajg I have asked some questions like this too. I did it purely out of curiosity. –  user23211 Jul 11 '12 at 17:50
    
@Qiaochu, this is exactly the type of explanation I was hoping for. Thank you. –  Georges Elencwajg Jul 11 '12 at 18:14
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Please let me know the reason for the downvotes. Unless you make it clear, it's hard to improve my question. –  Makoto Kato Jul 28 '12 at 5:23

3 Answers 3

up vote 3 down vote accepted

This answer builds on Qiaochu's and uses the same definition as Qiaochu, to wit: A ring $R$ is noetherian if, for any nonempty collection of ideals $\mathcal{I}$, there is some $I \in \mathcal{I}$ which is not properly contained in any $J \in \mathcal{I}$.

Theorem: If $R$ is noetherian, then $R[x]$ is noetherian.

This proof is basically taking the standard proof and rephrasing it to use Qiaochu's definition and be careful about choice.

I'm going to try to systematically use the following conventions: Ideals in $R[x]$ get capital letters; ideal in $R$ get overlined capital letters. Sets of ideals in $R[x]$ get calligraphic letters. I found that I could manage to write this without ever assigning a name to a set of ideals in $R$.

For any ideal $I \subseteq R[x]$, and any integer $j \geq 0$, define $$s_j(I) := \{ r \in R : \mbox{there is an element of $I$ of the form } r x^j + r_{j-1} x^{j-1} + \cdots +r_0 \}$$ Observe that $s_j(I)$ is an ideal and $s_j(I) \subseteq s_{j+1}(I)$.

Lemma: If $R$ is noetherian, and $I \subseteq R[x]$ is an ideal, then there is an index $j$ such that $s_k(I) = s_j(I)$ for $k \geq j$.

Proof: Let $s_j(I)$ be a maximal element in $\{ s_j(I) \}_{j \geq 0}$. Let $k \geq j$. Then we observed above that $s_k(I) \supseteq s_j(I)$. But, by the definition of a maximal element, we do not have $s_k(I) \supsetneq s_j(I)$, so $s_k(I) = s_j(I)$. $\square$.

We will denote the ideal $s_j(I)$ defined in the above lemma as $s_{\infty}(I)$. For $\mathcal{I}$ a collection of ideals in $R[x]$, we will write $s_j(\mathcal{I})$ or $s_{\infty}(\mathcal{I})$ for the result of applying $s_j$ or $s_{\infty}$ to each element of $\mathcal{I}$. So $s_j(\mathcal{I})$ is a set of ideals in $R$.

Let $\mathcal{I}$ be a collection of ideals in $R[x]$. Since $R$ is noetherian, there is a maximal element $\bar{J}$ in $s_{\infty}(\mathcal{I})$. Let $\mathcal{J}$ be the set of all $I \in \mathcal{I}$ with $s_{\infty}(I) = \bar{J}$.

Note that no element of $\mathcal{I} \setminus \mathcal{J}$ contains an element of $\mathcal{J}$, by the maximality of $\bar{J}$, so it is enough to show that $\mathcal{J}$ has a maximal element.

Choose an ideal $K \in \mathcal{J}$. (Making one choice does not use AC.) Let $m$ be an index such that $s_m(K) = \bar{J}$. Let $\mathcal{K}$ be the collection of ideals $I \in \mathcal{J}$ for which $s_m(I) = \bar{J}$. Note that no element of $\mathcal{J} \setminus \mathcal{K}$ can properly contain an element of $\mathcal{K}$, so it is enough to show that $\mathcal{K}$ has a maximal element.

We now make finitely many dependent choices. Choose a maximal element $\bar{J}^{m-1}$ in $s_{m-1}(\mathcal{K})$; let $\mathcal{K}_{m-1}$ be the set of $I \in \mathcal{K}$ with $s_{m-1}(I)=\bar{J}^{m-1}$; it is enough to show that $\mathcal{K}_{m-1}$ has a maximal element.

Choose a maximal element $\bar{J}^{m-2}$ in $s_{m-2}(\mathcal{K}_{m-1})$; let $\mathcal{K}_{m-2}$ be the set of $I \in \mathcal{K}^{m-1}$ with $s_{m-2}(I)=\bar{J}^{m-2}$; it is enough to show that $\mathcal{K}_{m-2}$ has a maximal element.

Continue in this manner to construct $\mathcal{K}_{m-3}$, $\mathcal{K}_{m-4}$, ..., $\mathcal{K}_0$. Since we are only making finitely many choices, we don't need AC; see my answer here.

At the end, we have a nonempty collection $\mathcal{K}_0$ of ideals such that, for any $I$ and $J \in \mathcal{K}_0$, and any $j \geq 0$, we have $s_j(I)= s_j(J)$. I claim that any element of $\mathcal{K}_0$ is maximal.

Let $I$ and $J \in \mathcal{K}_0$ and suppose that $I \supseteq J$. I will prove that $I=J$. This shows that every element of $\mathcal{K}_0$ is maximal.

Let $I_{\leq d}$ be the set of polynomials in $I$ of degree $\leq d$. I will show by induction on $d$ that $I_{\leq d} = J_{\leq d}$. The base case is $d=-1$, where both sides are $\{ 0 \}$. Since $I \supseteq J$, I just need to show that $I_{\leq d} \subseteq J_{\leq d}$.

Let $f \in I_{\leq d}$ and let the leading term of $f$ be $r x^d$. Then $r \in s_d(I) = s_d(J)$ so there is some $g \in J_{\leq d}$ with leading term $r$. Since $I \supseteq J$, we have $g \in I$ and hence $f-g \in I$. Since $\deg(f-g) < d$, by the induction hypothesis, we have $f-g \in J$. So $f = (f-g)+g \in J$. QED

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Great! I was about to start doing this but you seem to have beaten me to it. I'm pretty satisfied now. –  Qiaochu Yuan Jul 11 '12 at 19:03
    
@DavidSpeyer Perhaps I'm slow. May I ask why $\mathcal{L}^{m-1}$ is non-empty? –  Makoto Kato Jul 16 '12 at 21:24
    
$\mathcal{L}^{m-1}$ is the set of values of $I^{m-1}$ for $I \in \mathcal{K}$, so it is nonempty because $\mathcal{K}$ is nonempty. Did you mean to ask why $\mathcal{K}_{m-1}$ is nonempty? Because it is constructed to contain that element of $\mathcal{K}$ which gave raise to the element $J^{m-1}$ in $\mathcal{L}^{m-1}$. I might need better notation here. –  David Speyer Jul 16 '12 at 21:37
    
@DavidSpeyer I thought $I^{m-1}$ might be empty for $I \in \mathcal{K}$. No? –  Makoto Kato Jul 16 '12 at 21:50
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$I^{m-1}$ is an ideal. It certainly isn't empty, because it contains $0$. But we also don't care whether the ideal is empty, we care whether $\mathcal{L}^{m-1}$, which is a set of ideals, is empty. You are convincing me I need better notation, though... –  David Speyer Jul 16 '12 at 22:14

This problem reduces to proving Hilbert's basis theorem without choice, which I don't currently know how to do, but which I believe can be done. Let me explain the reduction. Below "ring" means "commutative unital ring."

Definition: A ring $R$ is Noetherian if any non-empty collection of ideals of $R$ has a maximal element.

(This definition is equivalent to the usual definitions in the presence of dependent choice, but in the absence of dependent choice I believe it is known that this definition is stronger. In any case it implies the other definitions.)

Fields are obviously Noetherian. Note that Noetherian rings contain maximal ideals by definition.

Proposition: Let $R$ be a Noetherian ring and $f : R \to S$ a surjective ring homomorphism. Then $S$ is Noetherian.

Proof. Let $I_i$ be a non-empty collection of ideals in $S$. Then $f^{-1}(I_i)$ is a non-empty collection of ideals in $R$ which by assumption has a maximal element $I_j$. Since $f$ is surjective, $I_j$ is also maximal in $S$.

Assumption: If $R$ is Noetherian, then $R[x]$ is Noetherian.

From this assumption it follows that any finitely-generated ring over a Noetherian ring is Noetherian, hence any ideal in such a ring is contained in a maximal ideal (in particular a prime ideal).


Let me explain the problem with the standard proof. This proof (as found in e.g. Atiyah-MacDonald) shows that if every ideal of $R$ is finitely-generated, then every ideal of $R[x]$ is finitely-generated (and I am not sure this works without dependent choice either). As far as I know, this condition is not equivalent to Noetherian (as defined above) in ZF. What we can prove in ZF is the following.

Proposition: If $R$ is Noetherian, then every ideal of $R$ is finitely-generated.

Proof. Let $I$ be an ideal of $R$. Then the collection of finitely-generated ideals contained in $I$ has a maximal element $J = (r_1, ... r_n)$. If $I$ is not all of $J$, then there exists $r_{n+1} \in I$ which is not in $J$, but then $(r_1, ... r_{n+1})$ is a finitely-generated ideal containing $J$, which contradicts the minimality of $J$.

However, as far as I can tell, we cannot prove the converse without dependent choice.

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Beat me to it! I think the proof of Hilbert basis theorem which I know, arguing about ideals of leading coefficients, does not use choice (explicitly...). –  Andrew Jul 11 '12 at 18:03
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@Andrew: see my edit. –  Qiaochu Yuan Jul 11 '12 at 18:13
    
Once the basic tools have been established, constructing a counterexample to the equivalence between the definitions of a Noetherian ring is not a hard task (what is harder is the proof that DC is really needed and not something weaker like countable choice). Your belief that the definitions are not equivalent is indeed true. –  Asaf Karagila Jul 11 '12 at 18:16
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@QiaochuYuan If you directly look at the standard proof without using the word finitely generated anywhere, I think you can directly show $R$ noetherian implies $R[x]$ noetherian; see my answer. –  David Speyer Jul 11 '12 at 18:58
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For future reference, Wilfrid Hodges wrote an excellent paper called Six impossible rings [J. Algebra 31 (1974)] where he examines the three Noetherian conditions and the three Artinian conditions. Using six pathological rings, he concludes that no implications between these six conditions other than the obvious ones are provable in ZF. –  François G. Dorais Jul 17 '12 at 0:34

To simplify the notations, we assume the charcteristic of $k$ is $0$. The positive characteristic case can be proved similarly.

By Noether normalization lemma(this can be proved without AC), there exist algebraically independent elements $x_1, ..., x_n$ in $A$ such that $A$ is a finitely generated module over the polynomial ring $A' = k[x_1,..., x_n]$. Let $K$ and $K'$ be the fields of fractions of $A$ and $A'$ respectively. Let $L$ be the smallest Galois extension of $K'$ containing $K$. Let $G$ be the Galois group of $L/K'$. Let $B$ be the integral closure of $A$ in $L$. It is well known that $B$ is a finite $A'$-module. Let $g = \prod_{\sigma \in G} \sigma(f)$. Since $A'$ is integrally closed, $g \in A'$. Since $g$ is non-invertible, there exists an irreducible polynomial $h$ which divides $g$. Then $P = hA'$ is a prime ideal of $A'$ containing $g$. By this, there exists a prime ideal $Q$ of $B$ lying over $P$. Since $g \in Q$, there exists $\sigma \in G$ such that $\sigma(f) \in Q$. Then $f \in \sigma^{-1}(Q)$. Hence $f \in A \cap \sigma^{-1}(Q)$ QED

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