Let $A$ be a commutative ring with $1$ and $\phi: A^n \to A^n$ be any injective $A$ linear map. Can I say $\phi $ is surjective ?

We know about the converse that surjectiveness implies injectivitness, but I don't know how to approach for this direction. I need some help. Thanks.

  • $\begingroup$ Interestingly you have the other implication when $A$ is noetherian, if $\phi$ is surjective then it must be injective. link $\endgroup$ – nowhere dense Apr 12 '18 at 16:38

No : for example the map $\mathbb{Z}\to \mathbb{Z}:n\mapsto 2n$ is injective and $\mathbb{Z}$-linear but not surjective.

  • $\begingroup$ If $A$ is a ring for which this does hold, will $A$ be self-injective? $\endgroup$ – Tobias Kildetoft Apr 11 '18 at 8:37

This is true if and only if every non-zero divisor of $A$ is invertible:

Let $n=1$ and $\phi$ be multiplication by a non-zero divisor $a \in A$. Then $\phi$ is injective, and surjective iff $a$ is invertible.

Conversely, let every non-zero divisor be invertible and $\phi : A^n \to A^n$ injective. Then $\det \phi$ is not a zero-divisor (see https://math.stackexchange.com/a/161544/43288) and thus invertible, so from $\phi \circ \phi^{ad} = \det \phi \cdot \operatorname{id}_{A^n}$ it follows that $\phi$ is invertible.


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