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This set of questions refers to the proof of the Krull-Akizuki theorem given in Matsumura's Commutative Ring Theory, pages 84-85. For those who don't have the text, i will provide the details.

The Krull-Akizuki theorem reads as follows: Let $A$ be a one-dimensional Noetherian integral domain with field of fractions $K$, let $L$ be a finite algebraic extension of $K$, and $B$ a ring with $A \subset B \subset L$; then $B$ is a Noetherian ring of dimension at most 1, and if $J$ is a non-zero ideal of $B$ then $B/J$ is an $A$-module of finite length.

The proof starts by proving the following Lemma: Let $A$ and $K$ be as in the theorem, and let $M$ be a torsion-free $A$-module of finite rank $r$. Then for nonzero $a \in A$ we have $l(M/aM) \le r \cdot l(A/aA)$, where $l(\cdot)$ is the length function.

Proof of Lemma: First assume that $M$ is finitely generated. Choose elements $\xi_1,\cdots,\xi_r \in M$ linearly independent over $A$ and set $E=\sum A\xi_i$; then for any $\eta \in M$ there exists $t \in A$ with $t\neq0$ such that $t \eta \in E$...

...If we set $C=M/E$ then from the assumption on $M$ we see that $C$ is also finitely generated, so that $t C=0$ for suitable nonzero $t \in A$...we can find $C=C_0 \supset C_1 \supset \cdots \supset C_m=0$, such that $C_i/C_{i+1} \cong A/p_i$ with $p_i \in \operatorname{Spec}(A)$. Now $t \in p_i$, and since $A$ is one-dimensional each $p_i$ is maximal, so that $l(C)=m < \infty$...

Question 1: What exactly is the role of $t$ in showing that $l(C) < \infty$? It seems to me that since $A$ Noetherian and $C$ is finitely generated by construction, we obtain directly the chain $C=C_0 \supset \cdots \supset C_m=0$, with $C_i/C_{i+1} \cong A/p_i, p_i \in \operatorname{Spec}(A)$ (this is essentially theorem 6.4). From this chain and the fact that each $A/p_i$ is a field, we get that the length of $C$ is finite. What am i missing?

Question 2:Further down in the proof of the theorem, the proof shows the existence of an element $a_0 \in J \cap A$. How do we deduce from this that $l(B/J) \le l(B/a_0B)$? If $B,J$ are finite, then i can see that $l(B/J)=l(B)-l(J) \le l(B) - l(a_0B)=l(B/a_0B)$. But we don't know a priori this finiteness condition.

luxerhia
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Manos
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1 Answers1

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Q1. Think about extreme case $C=\mathbb{Z}^2$, $A=\mathbb{Z}$. In this case, you won't have such finite sequence of modules. The existence of such $t$ means $C$ is a torsion module, and being a torsion module ensures the existence of such finite sequence.

Q2. Finiteness of $B$, $J$ cannot be assumed. I think this is just because $a_0 B \subset J$.

Sungjin Kim
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  • I don't have Matsumura's book, so I don't know about Theorem 6.4, but this torsion thing seems to be the only difference between my example and the module $C$ in the proof. – Sungjin Kim May 23 '13 at 23:08
  • it took me a while, but i now see that your argument in $Q1$ is not correct. In fact, for your "counterexample" $C=\mathbb{Z}^2,A=\mathbb{Z}$ we have the sequence $\mathbb{Z}^2 \supset (1,0)\mathbb{Z} \supset 0$. Now each quotient in the above sequence is isomorphic to $\mathbb{Z} \cong \mathbb{Z} / 0$ and $0 \in Spec(\mathbb{Z})$. Could you please reconsider your answer? – Manos Jun 07 '13 at 18:09
  • Another requirement of the sequence $C_i$, is that the quotient should be $A/p_i$ and each $p_i$ is maximal($0$ is not a maximal ideal in $\mathbb{Z}$). For my example such sequence is not obtained. – Sungjin Kim Jun 07 '13 at 21:16