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Consider a Field extension $K/k$ such that $K$ is $k$-finite dimensional. Is it true that $K$ can be written as a matrix algebra over $k$?

Motivating example: The complex numbers can be written as the $\mathbf R$-matrix algebra $\begin{pmatrix}\Re(z) & \Im(z)\\ -\Im(z) & \Re (z)\end{pmatrix}$.

Bubaya
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This is trivially true. Namely, $K$ comes with a $k$-vector space structure, so we can choose an isomorphism $K \cong k^{[K:k]}$ of $k$-vector spaces.

But now $K$ acts on itself freely by $K \times K \rightarrow K, (a,x) \mapsto a \cdot x$. Clearly this action is $k$-linear. Under the chosen isomorphism above, we get a free action of $K$ on $k^{[K:k]}$, but this just corresponds to an injection of $k$-algebras $K \hookrightarrow \mathrm{End}_k(k^{[K:k]}) \cong M_{[K:k]}(k)$. Note that this generally depends on quite some choices of basis vectors. However, trace and determinant (one also says norm) of an element in $K$ are well-defined and you might have come across those before.

Louis
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  • Indeed, this was too trivial; sorry. – Bubaya Jul 06 '18 at 08:51
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    That was not meant to be demeaning or saying your question was stupid. In fact not at all. It is just that I wanted to stress that the fact follows straightforwardly from definitions and no "deep" statements are needed. – Louis Jul 06 '18 at 08:53