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Let $A$ be a nonzero commutative ring with $1$, and $n$ a positive integer. If $b_1,\ldots,b_n\in A^n$ generate $A^n$ as an $A$-module, is it true that $b_1,\ldots,b_n$ are linearly independent over $A$ (hence a basis)? I suspect this is true (It is true if $n=1$), but I can't seem to prove it.

I guess an equivalent way of asking this question is: if a linear map $A^n\to A^n$ is surjective, is it necessarily injective?

Any help is welcome!

ashpool
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    If you make the assumption that $A$ is an integral domain, then you can use the quotient field to prove your claim (extend it to an linear map $\operatorname{Quot(A)}\rightarrow \operatorname{Quot(A)}$). Just commutative will not be enough, the trivial ring gives a counterexample. – Severin Schraven Jul 22 '17 at 22:49
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    Your equivalent statement is a particular case of Corollary 0.2 in my A constructive proof of Orzech's theorem. But there are also easier proofs if you just want to get your claim. For example (hint): Encode your linear map as an $n \times n$-matrix. The surjectivity shows that this matrix has a left inverse. Hence, its determinant is invertible. Hence, the matrix itself is invertible (because $M \operatorname{adj} M = \det M \cdot I_n$ for any $n \times n$-matrix $M$), so that the map is bijective. – darij grinberg Jul 22 '17 at 22:52
  • Thanks for the hint. I just thought of the same answer! :) – ashpool Jul 22 '17 at 23:00
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    @SeverinSchraven: In Commutative Algebra, the convention is that $1\ne 0$, usually. – Bernard Jul 23 '17 at 00:17
  • @Bernard I'm aware of that convention. I just wanted to emphasize the importance of the convention in this case. – Severin Schraven Jul 23 '17 at 10:02
  • Eisenbud's commutative algebra book has a proof of this using Cayley-Hamilton theorem if you are interested in that approach. – user7090 Jul 26 '17 at 20:34

2 Answers2

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Yes. Let $B$ be the $n\times n$ matrix consisting of $b_1 \cdots b_n$ as its columns, and $e_1,\ldots,e_n\in A^n$ the standard basis. Since $b_1,\ldots,b_n$ generate $A^n$, there exist $c_{ij}\in A$ such that $e_i=c_{i1}b_1+\cdots+c_{in}b_n$ for each $i$. Let $C$ be the $n\times n$ matrix whose $ij$-th entry is $c_{ji}$. Then $BC=I$, where $I$ is the $n\times n$ identity matrix. This implies that $\det B\in A$ is a unit, hence the linear map $B:A^n\to A^n$ is injective.

ashpool
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  • Note that this is in general not equivalent to the first question and that one requires for this that the ring is unitary. – Severin Schraven Jul 22 '17 at 23:03
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    @Severin Schraven: Thanks for pointing that out. I edited the question so that the ring is nonzero and has 1. – ashpool Jul 23 '17 at 00:39
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A proof of the equivalent way :

If $\;u\colon A^n\longrightarrow A^n$ is surjective, $u$ is injective.

Indeed, consider the short exact sequence: $$0\longrightarrow K \overset{i}\longrightarrow A^n\overset{u}\longrightarrow A^n\longrightarrow 0,$$ where $i$ is the canonical injection of $K=\ker u$. This is a split exact sequence, since the last term is free, so $i$ has a linear retraction, which is surjective. In other words, $K$ is a finitely generated submodule of $A^n$.

Tensoring with each of the residual fields $\kappa(\mathfrak m)=A_\mathfrak m/\mathfrak mA_\mathfrak m$, $\;u\otimes_A\kappa(\mathfrak m)\colon\kappa(\mathfrak m)^n\longrightarrow \kappa(\mathfrak m)^n$ is an automorphism, hence the image of $K\otimes_A\kappa(\mathfrak m)$ in $\kappa(\mathfrak m)^n$ is $0$.

We conclude that $K_\mathfrak m=\{0\}$ for each maximal ideal, whence $K=\{0\}$ since $K$ is finitely generated.

Bernard
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