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How would I go about showing this?

Suppose $R$ is a commutative ring, $A \in R_n$, and the homomorphism $f: R^n \to R^n$ defined by $f(b) = Ab$ is surjective. Show $f$ is an isomorphism.

(Edit: $R_n$ is the ring of $n\times n$ matrices.)

I'm thinking that since $f$ is surjective on $R^n \to R^n$, for every $x \in R^n$, $x = Ab$, so that $f^{-1}(x) = A^{-1}x$. And since the inverse of any element in a ring is nonzero, $\ker(f) = 0$.

I'm not terribly confident with my answer though. Thanks!

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What do you mean by $A \in R_n$? – Isaac Solomon Jan 3 '13 at 0:32
$R_n$ is the ring of nxn matricies where $a_{i,j} \in R$ – randomafk Jan 3 '13 at 0:37… – user26857 Jan 3 '13 at 0:51
You have to prove that $A$ is invertible! – user26857 Jan 3 '13 at 1:27
up vote 2 down vote accepted

Denote by $e_i$ the column vector in $R^n$ that has $1$ on place $i$ and $0$ otherwise. Then there exists $b_i\in R^n$ such that $f(b_i)=e_i\Leftrightarrow Ab_i=e_i$. Take the matrix $B$ whose colums are the vectors $b_1,\dots,b_n$. Then $AB=I_n$, so $A$ is invertible and we are done.

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This is a great answer ! – user18119 Jan 3 '13 at 1:34

Apply the rank-nullity theorem, to get your answer.

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How do you define dimension over a commutative ring? – Chris Godsil Jan 3 '13 at 2:01
@kdd can you provide more details regarding your response? Regards – Amzoti Jan 3 '13 at 2:01

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