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Let $A$ be a commutative ring with $1$. It is a standard result that every set of $n$ generators of the free $A$-module $A^{n}$ is actually a basis. The proof uses tensor products.

I was reading a "proof" of this argument:

Let $\{m_{1},..,m_{t}\}$ be a set of $n$-generators of $A^{n}$. Define the map $f: A^{n} \rightarrow A^{n}$ by $e_{i} \mapsto m_{i}$ where $e_{i}$ is the canonical basis of $A^{n}$.

Let $K=ker(f)$ then we have the following exact sequence:

$0 \rightarrow K \rightarrow A^{n} \rightarrow A^{n} \rightarrow 0$

Then the sequence splits so $A^{n} \cong A^{n} \oplus K$. Therefore $K=0$.

Can you please explain why the relation $A^{n} \cong A^{n} \oplus K$ implies $K=0$? how do we know that $K$ is free?

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1 Answer 1

up vote 5 down vote accepted

What we know a priori is that $K$ is finitely generated projective. The identity $K = 0$ can be checked locally, so we are reduced to the case in which $A$ is local. But a finitely generated projective module over a local ring is free, and by tensoring to the residue field $k = A/\mathfrak{m}$ one obtains a contradiction.

(Actually any projective module over a local ring is free, but this is much harder to prove for infinitely generated modules.)

By the way, this question is very similar to this previous one, for which several answers were given. In particular I recorded an answer there which proves something more general: every finitely generated module over a commutative ring is Hopfian.

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