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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 $B$.

EDIT Why worry about the axiom of choice?

  • 4
    +1 because I don't understand why this question got 4 downvotes without explanation and with no obvious reason.2012-07-15
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    @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.2012-07-26
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    @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.2012-07-26

1 Answers 1

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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

  • 8
    *Please* wait to post your answer until it's complete.2012-07-16
  • 0
    @AlexBecker May I ask the reason?2012-07-16
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    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.2012-07-16
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    @AlexBecker Yes, we discussed well and as far as I know it's conclusion is that there's no harm to anyone. If you disagree, please discuss it there. Regards.2012-07-16
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    Downvoting for non-mathematical reasons is unjust. If you have a *public* complaint to me, please signal a flag or open a meta thread.2012-07-16
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    @MakotoKato I side with you on that.2012-07-16
  • 2
    @Peter [Oh. My. God.](http://www.imdb.com/title/tt0078714/quotes?qt=qt0419913).2012-07-16
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    @did I think the downvoting is not called for, that's all.2012-07-16
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    @PeterTamaroff Thanks.2012-07-16
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    @JasperLoy You don't have to check my incomplete answer every time I edit. I recommend you to read my answer after I finished editing.2012-07-16
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    How can he know when the answer is finished if he hasn't checked it? Please explain.2012-07-16
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    @AsafKaragila Easy. Check it after a few days later.2012-07-16