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Let $A$ be a commutative ring with one, and $A^n, A^m$ be free modules over $A$. Let $\varphi:A^m\rightarrow A^n$. I want to show that if $\varphi$ is injective, then $m\leq n$. Similarly, if $\varphi$ is surjective, then $m\geq n$.

My attempt: Let $\varphi:A^m \rightarrow A^n$ be $A$-linear. Let $x_1,...,x_m$ and $y_1,...,y_n$ be bases of $A^n$ and $A^m$, respectively. By the universal property of free modules, $\varphi$ is induced by a function $\tilde{\varphi}:\{x_1,...,x_m\}\rightarrow \{y_1,...,y_m\}$. We now know that there exists an injective map between two sets $X$ and $Y$ if and only if $\mid X\mid \leq \mid Y \mid$. A similar argument delivers the second statement.

Is my solution correct? If not, where does it fail?

bliipbluup
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1 Answers1

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As I have stated in the comments, this statement fails to hold without further assumptions on $A$. Since your proof doesn't use these assumptions, it must be wrong somewhere.

Concretely the problem with your proof is the following: the universal property allows us to say that $\varphi : A^m \to A^n$ is induced by a function $\tilde \varphi:\{x_1,\dots,x_m\} \to A^n$. However, we cannot state a priori that the target of $\tilde \varphi$ is some fixed basis $\{y_1,\dots,y_n\}$ of $A^n$.

Ben Grossmann
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  • So, one would have to say that $\tilde{\varphi}$ is onto some finite subset ${\varphi(x_1),...,\varphi(x_m)}\subset A^n$, then show that any subset of order $\geq n$ is linearly independent? – bliipbluup Dec 17 '19 at 10:14
  • First of all, your comment is wrong by any interpretation: any subset of order $> n$ would be linearly dependent, and we can certainly construct linearly dependent subsets that are as small as you like. What you should say is that if $\varphi$ is injective, then ${\varphi(x_1),...,\varphi(x_m)}\subset A^n$ must be linearly independent, which means that the set (of $m$ elements) contains at most $n$ elements. – Ben Grossmann Dec 17 '19 at 10:21
  • again made a couple of typos. Thanks for the clarification! – bliipbluup Dec 17 '19 at 10:36