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Let $A$ be a left Noetherian ring. How do I show that every element $a\in A$ which is left invertible is actually two-sided invertible?

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up vote 7 down vote accepted

Recall that we get an isomorphism $A^\text{op} \to \operatorname{End}_A(A)$ by sending $a$ to the endomorphism $x \mapsto xa$. Here $A^\text{op}$ is the opposite ring. If $a$ is left invertible then the corresponding endomorphism $f$ is surjective, and if we can show that $f$ is injective then $f$ is invertible in $\operatorname{End}_A(A)$, whence $a$ is invertible in both $A^\text{op}$ and $A$.

It isn't any harder to prove a more general statement: If an endomorphism of a Noetherian module is surjective, then it is an isomorphism.

Here are some ideas for this. If $g$ is such an endomorphism then the increasing sequence of submodules $\{\operatorname{Ker}(g^n)\}$ must stabilize. Use this and the fact that each $g^n$ is surjective to show that the kernel is trivial.

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