Elements of field extensions as matrices (D&F 13.2.19)

Let \(K\) be an extension of \(F\) of degree \(n\).

(a) For any \(\alpha \in K\) prove that \(\alpha\) acting by left multiplication on \(K\) is an \(F\)-linear transformation of \(K\).
(b) Prove that \(K\) is isomorphic to a subfield of the ring \(M_n(F)\), so \(M_n(F)\) contains an isomorphic copy of every extension of \(F\) of degree \(\le n\).

(a) Trivial.
(b) Each element \(\alpha \in K\) can be associated with a unique matrix \(M_\alpha \in M_n(F)\) as follows: let \(k_1, \cdots, k_n\) be a basis for \(K\) over \(F\). For each \(k_i\), we can write \(k_i \alpha = m_{i1}k_1 + \cdots m_{in}k_n\), where \(m_{ij} \in F\). Then let \((M_\alpha)_{ij} = m_{ij}\). Now we show that this is an isomorphism.

Let \(\alpha, \beta \in K\), \(m_{ij} = (M_\alpha)_{ij}\) and \(n_{ij} = (M_\beta)_{ij}\). Then \(k_i(\alpha+\beta) = \sum_{j} m_{ij}k_j + \sum_{j} n_{ij}k_j = \sum_{j} (m_{ij}+n_{ij})k_j = \sum_{j} (M_\alpha + M_\beta)_{ij}k_j\), so the map is additive.

Now for simplicity's sake assume that \(k_1 = 1\), so \(\beta = \sum_{j} n_{1j}k_j\) (hopefully it works). Thus \(k_i(\alpha\beta) = (\sum_{j} m_{ij}k_j)(\sum_{j} n_{1j}k_j) = \sum_{a,b} m_{ia}k_a(n_{1b}k_b) = \sum_{a,b} m_{ia}n_{ab}k_b\) \(= \sum_b (\sum_{a} m_{ia}n_{ab}) k_b = \sum_{b} (M_{\alpha}M_{\beta})_{ib}k_b\).