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I'm going through a proof of the statement:


Let $A$ and $B$ be commutative rings. If $A \subseteq B$ and $B$ is a finitely generated $A$-module, then all $b \in B$ are integral over $A$.


Proof:

Let $\{c_1, ... , c_n\} \subseteq B$ be a set of generators for $B$ as an $A$-module, i.e $B = \sum_{i=1}^n Ac_i$. Let $b \in B$ and write $bc_i = \sum_{j=1}^n a_{ij}c_j $ with $a_{ij} \in A$, which says that $(bI_n - (a_{ij}))c_j = 0 $ for $ 1 \leq j \leq n$. Then we must have that $\mathrm{det}(bI_n - (a_{ij})) = 0 $. This is a monic polynomial in $b$ of degree $n$.

Why are we not done here? The proof goes on to say:

Write $1 = \alpha_1 c_1 + ... + \alpha_n c_n$, with the $\alpha_i \in A$. Then $\mathrm{det}(bI_n - (a_{ij})) = \alpha_1 (\mathrm{det}...) c_1 + \alpha_2 (\mathrm{det}...) c_2 + ... + \alpha_n (\mathrm{det}...) c_n = 0$. Hence every $b \in B$ is integral over $A$.

I understand what is being done here on a technical level, but I don't understand why it's being done. I'd appreciate a hint/explanation. Thanks

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    Dear Pablo: What's behind this argument are, I think, the following facts. (a) Each $A$-module $M$ is a quotient of a free module $F_M$. (b) $M$ is finitely generated iff (a) holds for some $F_M$ of finite rank. (c) Each $A$-linear map $\phi:M\to N$ admits a lift (obvious definition) $\Phi:F_M\to F_N$ (obvious notation). - I don't know if this is the kind of things you're looking for, but if it is, and if you want more details, I'd be happy to try to spell them out.2012-01-18

2 Answers 2