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.

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