MHB Is the Endomorphism Ring of a Finite Dimensional Vector Space Dedekind Finite?

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Here is this week's POTW:

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Prove that if $V$ is a finite dimensional vector space over field $\Bbb k$, then the endomorphism ring $\text{End}_{\Bbb k}(V)$ is Dedekind finite.

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This week's problem was correctly solved by Opalg. Here is his solution:

Spoiler

Choosing a (finite) basis for $V$, we can associate a matrix, and hence a determinant, with each element of $\mathrm{End}_{\mathbb k}(V).$

Suppose that $S,T \in \mathrm{End}_{\mathbb k}(V)$, and $ST = \mathrm{id}$. Then $\det(S)\det(T) = 1$ and so $\det(S) \ne0$. Therefore $S$ is invertible. It follows that $T = (S^{-1}S)T = S^{-1}(ST) = S^{-1}$. Hence $TS = S^{-1}S = \mathrm{id}$.

Here is also my solution, which uses a non-determinantal argument:

Spoiler

Let $n = \mathrm{dim}_{\Bbb k}(V)$. The ring $\rm{End}_{\Bbb k}(V)$ is identified with the matrix ring $M_n(\Bbb k)$ via the isomorphism $\Lambda : M_n(\Bbb k)\to \rm{End}_{\Bbb k}(V)$, $A\xrightarrow{\Lambda} (x \mapsto Ax)$. So it suffices to show $M_n(\Bbb k)$ is Dedekind finite. Let $A, B\in M_n(\Bbb k)$ such that $AB = I$. Then $BA = I$ if and only if $A$ is invertible. By way of contradiction, suppose $A$ is not invertible. Then there is a sequence $E_1,\ldots, E_s$ of elementary matrices such that $E_1\cdots E_s A$ has a row of zeros. Then $E_1\cdots E_s = E_1\cdots E_sAB$ has a row of zeros. This contradicts invertibility of $E_1\cdots E_s$. Thus $BA = I$ and consequently $M_n(\Bbb k)$ is Dedekind finite.