Intereting Posts

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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.

- Can one avoid AC in the proof that in Noetherian rings there is a maximal element for each set?
- There's non-Aleph transfinite cardinals without the axiom of choice?
- Cardinality of set of well-orderable subsets of a non-well-orderable set
- What's an example of a vector space that doesn't have a basis if we don't accept Choice?
- Without the Axiom of Choice, does every infinite field contain a countably infinite subfield?
- Why is the axiom of choice separated from the other axioms?

(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?

- Can infinitely many primes lie over a prime?
- Proof of Steinitz Theorem
- Showing an ideal is prime in polynomial ring
- Dimension of the total ring of fractions of a reduced ring.
- Compactness of $\operatorname{Spec}(A)$
- What's the motivation of the definition of primary ideals?
- Is the localization of a PID a PID?
- Checking the maximality of a principal ideal in $R$
- Definition of a finitely generated $k$ - algebra
- Irreducible polynomials and affine variety

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**

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