Extension fields isomorphic to fields of matrices Suppose $K \subset L$ is a finite field extension of degree $m$. Is it true that there exists some natural $n$ such that $L$ is isomorphic to a subfield of $M^{n\times n}(K)$, the ring of $n\times n$ matrices with entries in $K$? What is the smallest dimension (as a function of $m$) for which an isomorphism exists?
The example I had in mind was the 2-dimensional extension $\mathbb{R} \subset \mathbb{C}$ with $\mathbb{C}$ being isomorphic to $\{ \begin{pmatrix}a&-b\\b&a\end{pmatrix} : a, b \in \mathbb{R}\}$.
 A: The answer to your first question is yes.  In particular, if $L/K$ has degree $m$ then $L$ is isomorphic to a subfield of $M_{m \times m}(K)$.  This can be seen via the (left) regular representation as follows.
Each $\alpha \in L$ gives rise to a (left) multiplication map $\lambda_\alpha : L \to L$, $x \mapsto \alpha x$ that is $K$-linear.  Choosing a basis of $L$ as a $K$-vector space allows us to write a matrix representation of $\lambda_\alpha$ with respect to this basis.  For instance, in your example $\mathbb{C}$ has basis $\{1, i\}$ as an $\mathbb{R}$-vector space.  Given $\alpha = a + ib \in \mathbb{C}$, then
\begin{align*}
\lambda_\alpha(1) &= \alpha = a + ib\\
\lambda_\alpha(i) &= \alpha \cdot i = (a + ib)i = -b + ia
\end{align*}
so the matrix representation of $\lambda_\alpha$ with respect to this basis is $\begin{pmatrix} a & -b\\ b & a \end{pmatrix}$.
Here's a partial answer to your second question.  Suppose $L/K$ has a primitive element, so $L = K(\beta)$ for some $\beta \in L$.  (This is true for instance if $L/K$ is a separable extension.)  Let $n$ be the minimal positive integer such that $L$ embeds into $M_{n \times n}(K)$; denote this embedding by $\iota : L \to M_{n \times n}(K)$.  One can show that $\beta$ satisfies the characteristic polynomial $f$ of $\iota(\beta)$, which has degree $n$.  Since $\beta$ generates the extension $L/K$ and $[L:K] = m$, then the minimal polynomial $g$ of $\beta$ over $K$ has degree $m$.  Since $\beta$ is also a root of $f$, then $g \mid f$, so $m \leq n$.  Thus $m=n$.
I don't know the answer to your second question in general.  Maybe someone with more background in representation theory will have an answer.
