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Let $k$ be a field, $A$ a finitely generated local integral domain over $k$ of Krull dimension $1$, and $A'$ be the normalization of $A$ in the fraction field of $A$. Then is $A'/A$ a torsion $A$-module, i.e., $A' /A \cong \bigoplus A/g_i $ for some $g_i \in A$?

I want to use it to show that $\dim_k A'/A$ is finite and the equation $\dim_kA'/fA' = \dim_k A/fA$, where $f$ is a non-unit element of $A$.

I've shown that $A'$ is a finite $A$-module.

Thank you very much.

user26857
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k.j.
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  • It is torsion, but not necessarily of the form you suggest. What you say in the second paragraph is true.That is, if $M$ is any finitely generated torsion free $A$-module of rank 1, and $0\neq f\in A$, then the dimension formula holds for $M/fM$ (which might be zero if $f$ is a unit). – Mohan Dec 16 '18 at 20:41
  • @Mohan Thank you for your comment. Would you show me why $A'/A$ is torsion? – k.j. Dec 16 '18 at 20:45
  • Any element in $A'$ is of the form $a/b$ with $a,b\in A, b\neq 0$. So, its image in $A'/A$ is annihilated by $b$. This is true of every element of $A'/A$ and so it is torsion. – Mohan Dec 16 '18 at 20:49
  • @Mohan Thank you. I understand that $A'/A$ is annihilated. But if $A'/A$ can't be written as $\oplus A/g_i A$, I can't show that $\dim_k A'/A$ is finite. Could you show it? – k.j. Dec 16 '18 at 21:15
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    Since $A'$ and thus $A'/A$ is finitely generated, you can find a single $b\neq 0$ such that $b(A'/A)=0$. That means, $A'/A$ is a finitely generated module over $A/bA$. But $A/bA$ has finite dimension. – Mohan Dec 16 '18 at 22:59
  • @Mohan Thank you very much! – k.j. Dec 16 '18 at 23:14

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