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Let $ R $ be commutative.

If $ \eta $ is a surjective endomorphism of $ R^{(n)} $ then $ \eta $ is bijective.

Proof:

Suppose $ (e_1, e_2, ..., e_n) $ is a base for $ R^{(n)} $, we denote $ \eta(e_1, e_2, ..., e_n) $ by $$ (f_1, f_2, ..., f_n)=(e_1, e_2, ..., e_n)A ,\ A\in M_n(R) $$

Since $ \eta $ is surjective, then there exists a matrix $ B\in M_n(R) $, such that $$ \eta((e_1, e_2, ..., e_n)B)=\eta(e_1, e_2, ..., e_n)B=(e_1, e_2, ..., e_n)AB=(e_1, e_2, ..., e_n) .$$
Hence $ AB=I_n $ and since $ R $ is commutative, we have $ A\in GL_n(R) $ and $ \eta $ is bijective.

If $ \eta $ is an injective endomorphism of $ R^{(n)} $, then $ \eta $ is not necessarily bijection.

Example:

$$ R=\mathbb Z, \quad \eta: \mathbb Z\to2\mathbb Z\subset \mathbb Z .$$

Obviously, $ \eta $ is injective, however, not a bijection.

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  • $\begingroup$ The proof relies on the fact that if $AB = I_n$, then $A \in \operatorname{GL}_n\left(R\right)$. This fact is true, but essentially a restatement of the claim of the question; often authors derive it from the latter. $\endgroup$ – darij grinberg Jan 20 at 9:37
  • $\begingroup$ What do you mean by rank-nullity? How would you measure the nullity? How would you compute the rank of a nonfree module over a general commutative ring? Generally speaking, rank would seem to be a local concept, and it could vary wildly from point to point. For example, take $F_2^{10}\times F_3^2$ over $F_2\times F_3 = \Bbb{Z}/6\Bbb{Z}$. $\endgroup$ – jgon Jan 20 at 16:35
  • $\begingroup$ @Philip: This doesn't seem to ask a question. Perhaps it would be clarified by framing it as asking for proof verification. This would be consistent with the responses you've gotten so far, and might avoid having your Question closed. $\endgroup$ – hardmath Jan 20 at 19:22

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