## Every degree n extension of a field F is embedded in the set of nxn matrices over F

Let $F$ be a field, and let $K$ be an extension of $F$ of finite degree.

1. Fix $\alpha \in K$. Prove that the mapping ‘multiplication by $\alpha$‘ is an $F$-linear transformation on $K$. (In fact an automorphism for $\alpha \neq 0$.)
2. Deduce that $K$ is isomorphically embedded in $\mathsf{Mat}_n(F)$.

Let $\varphi_\alpha(x) = \alpha x$. Certainly then we have $\varphi_\alpha(x+ry) = \alpha(x+ry) = \alpha x + r \alpha y$ $= \varphi_\alpha(x) + r \varphi_\alpha(y)$ for all $x,y \in K$ and $r \in F$; so $\varphi_\alpha$ is an $F$-linear transformation. If $\alpha \neq 0$, then evidently $\varphi_{\alpha^{-1}} \circ \varphi_\alpha = \varphi_\alpha \circ \varphi_{\alpha^{-1}} = 1$.

Fix a basis for $K$ over $F$; this yields a ring homomorphism $\Psi : K \rightarrow \mathsf{Mat}_n(F)$ which takes $\alpha$ and returns the matrix of $\varphi_\alpha$ with respect to the chosen basis. Suppose $\alpha \in \mathsf{ker}\ \Psi$; then $\varphi_\alpha(x) = 0$ for all $x \in K$, and thus $\alpha = 0$. So $\Psi$ is injective as desired.