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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 $A$.


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. – Pierre-Yves Gaillard Jan 18 '12 at 14:47

You prematurely write "Then we must have that $\mathrm{det}(bI_n - (a_{ij})) = 0$".
At that stage you can only deduce (by multiplying by the adjoint of your matrix on the left) that all the $det\cdot c_i =0$.
However writing $1 = \alpha_1 c_1 + ... + \alpha_n c_n$ and multiplying by $det$ you do get

$$det=det\cdot 1= \alpha_1\cdot det\cdot c_1+...+\alpha_n\cdot det\cdot c_n=\alpha_1\cdot 0+...+\alpha_n\cdot 0=0$$

(This is a variation on the Cayley-Hamilton theorem, according to which the characteristic polynomial of a square matrix annihilates that matrix.)

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Doesn't it just come down to the fact that $c_i$ can't be $0$ for all $i$? – Pablo Fields Jan 18 '12 at 14:10
Dear @Pablo: no. Even if $c_{i_0}\neq 0$, you cannot deduce from $\det \cdot c_{i_0}=0$ that $\det=0$ because $B$ might have zero-divisors and $c_{i_0}$ might be one of them. – Georges Elencwajg Jan 18 '12 at 16:11

Another way to phrase it, slightly different to Georges's answer and comments, is as follows:

In the first paragraph of the proof, $B$ could be replaced by any f.g. $A$-module $M$, and $b$ could any endomorphism of that $A$-module. What we conclude is that every $\varphi \in End_A(M)$ is integral over $A$. In particular, if $M$ is in fact a $B$-module, then we conclude that the image of $B$ in $End_A(M)$ is integral over $A$.

The point of the second paragraph is to observe that (since $B$ is a ring with $1$), the natural map $B \to End_A(B)$ (given by $B$ acting on itself through multiplication) is injective, so that $B$ coincides with its image in $End_A(B)$. Only after making this additional observation can we conclude that $B$ is integral over $A$.

Just as something to think about, what you'll see is that the argument proves that if $B$ is an $A$-algebra which admits a faithful module which is f.g. over $A$, then $B$ is integral over $A$. On the other hand, if $B$ just admits a module that is f.g. over $A$, but not necessarily faithful, then we can't conclude that $B$ is integral over $A$. (See if you can find a counterexample.)

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