# Krull-Akizuki theorem without Axiom of Choice

Motivation This question came from my efforts to solve this problem presented by Andre Weil in 1951.

We use the definitions in my answers to this question. Can we prove the following theorem without Axiom of Choice?

Theorem Let $A$ be a weakly Artinian integral domain. Let $K$ be the field of fractions of $A$. Let $L$ be a finite extension field of $K$. Let $B$ be a subring of L containing $A$. Then the following assertions hold.

(1) Every ideal of $B$ is finitely generated

(2) Every non-zero prime ideal of $B$ is maximal.

(3) $leng_A B/I$ is finite for every non-zero ideal $I$ of $B$.

• +1 because I don't understand why this question got 4 downvotes without explanation and with no obvious reason. – Rudy the Reindeer Jul 15 '12 at 9:47
• @Makoto Why did you delete your answer to this questions and three other questions? Was there serious errors? If not, could you please undelete them. – Bill Dubuque Jul 26 '12 at 14:27
• @BllDubuque You know the reason. Some people(perhaps even some moderators, too) complained in meta that answering my questions this way is not right in this site. Or something like that. – Makoto Kato Jul 26 '12 at 17:23

I borrowed the idea of the Bourbaki's proof of Krull-Akizuki theorem.

Lemma 1 Let A be a weakly Artinian integral domain. Let $M$ be a torsion $A$-module of finite type. Then $leng_A M$ is finite.

Proof: Let $x_1, ..., x_n$ be generating elements of $M$. There exists a non-zero element $f$ of $A$ such that $fx_i = 0$, $i = 1, ..., n$. Let $\psi:A^n \rightarrow M$ be the morphism defined by $\psi(e_i) = x_i$, $i = 1, ..., n$, where $e_1, ..., e_n$ is the canonical basis of $A^n$. Since $leng_A A^n/fA^n$ is finite and $\psi$ induces a surjective mophism $A^n/fA^n \rightarrow M$, $leng_A M$ is finite. QED

Lemma 2 Let A be a weakly Artinian integral domain. Let $K$ be the field of fractions of $A$. Let $M$ be a torsion-free $A$-module of finite type. Let $r = dim_K M \otimes_A K$. Let $f$ be a non-zero element of $A$. Then $leng_A M/fM \leq r(leng_A A/fA)$

Proof: There exists a $A$-submodule $L$ of $M$ such that $L$ is isomorphic to $A^r$ and $Q = M/L$ is a torsion module of finite type over $A$. Hence, by Lemma 1, $leng_A Q$ is finite. Let $n \geq 1$ be any integer. The kernel of $M/f^nM \rightarrow Q/f^nQ$ is $(L + f^nM)/f^nM$ which is isomorphic to $L/(f^nM \cap L)$. Since $f^nL \subset f^nM \cap L$, $leng_A M/f^nM \leq leng_A L/f^nL + leng_A Q/f^nQ \leq leng_A L/f^nL + leng_A Q$. Since $M$ is torsion-free, $f$ induces isomorphism $M/fM \rightarrow fM/f^2M$. Hence $leng_A M/f^nM = n(leng_A M/fM)$. Similarly $leng_A L/f^nL = n(leng_A L/fL)$. Hence $leng_A M/fM \leq leng_A L/fL + (1/n) leng_A Q$. Since $L$ is isomorphic to $A^r$, $leng_A L/fL = r(leng_A A/fA)$. Hence, by letting $n \rightarrow \infty$, $leng_A M/fM \leq r(Leng_A A/fA)$. QED

Lemma 3 Let A be a weakly Artinian integral domain. Let $K$ be the field of fractions of $A$. Let $M$ be a torsion-free $A$-module. Suppose $r = dim_K M \otimes_A K$ is finite. Let $f$ be a non-zero element of $A$. Then $leng_A M/fM \leq r(Leng_A A/fA)$

Proof: Let $(M_i)_I$ be the family of finitely generated $A$-submodules of $M$. $M/fM = \cup_i (M_i + fM)/fM =\cup_i M_i/(M_i \cap fM)$. Since $fM_i \subset M_i \cap fM$, $M_i/(M_i \cap fM)$ is isomorphic to a quotient of $M_i/fM_i$. Hence, by Lemma 2, $leng_A M_i/(M_i \cap fM) \leq r(leng_A A/fA)$. Hence, by By Lemma 4 of this, $leng_A M/fM \leq r(leng_A A/fA)$ QED

Lemma 4 Let A be a weakly Artinian integral domain. Let $K$ be the field of fractions of $A$. Let $L$ be a finite extension field of $K$. Let $B$ be a subring of $L$ containing $A$. Then $leng_A B/fB$ is finite for every non-zero element $f \in B$.

Proof: Since $L$ is a finite extension of $K$, $a_rf^r + ... + a_1f + a_0 = 0$, where $a_i \in A, a_0 \neq 0$. Then $a_0 \in fB$. Since $B \otimes_A K \subset L$, $dim_K B \otimes_A K \leq [L : K]$. Hence, by Lemma 3, $leng_A B/a_0B$ is finite. Hence $leng_A B/fB$ is finite. QED

Proof of the title theorem By Lemma 2 of my answer to this, $B$ is weakly Artinian. Hence, by this, we are done. QED

• Please wait to post your answer until it's complete. – Alex Becker Jul 16 '12 at 4:15
• This has been discussed in depth in meta, as you are well aware. In brief, it bumps your question an inordinate number of times. In the process it is not very helpful to other users at all, as it is really only useful to the person composing the lengthy proofs (i.e. you) and gives other users snippets they really can't do anything with. Some users, myself included, feel that in doing this you are forcing something inherently unfit for a Q-and-A site into the math.SE format, and that this is an abuse of the system and an inconvenience to the other users. – Alex Becker Jul 16 '12 at 4:32
• Downvoting for non-mathematical reasons is unjust. If you have a public complaint to me, please signal a flag or open a meta thread. – Makoto Kato Jul 16 '12 at 6:23
• @Peter Oh. My. God.. – Did Jul 16 '12 at 6:56
• How can he know when the answer is finished if he hasn't checked it? Please explain. – Asaf Karagila Jul 16 '12 at 12:46