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Let \(E\) be a field extension over \(F\) of degree \(n\). Let
\[\mathcal{L}(E/F) = \{T:E\to E\mid T \text{ is a } F \text{-linear transformation} \},\]which is a ring. Especially, if \(\alpha\) is a basis of \(E\), then \(T\to [T]_\alpha\) defines a ring isomorphism from \(\mathcal{L}(E/F)\) to \(M_n(F)\).
For all \(a \in E\), define
\[L_a(x) = ax,\]which is an element of \(\mathcal{L}(E/F)\).
\(F\)-linear means that it is a linear transformation over \(E\) where we view \(E\) as a vector space over \(F\).
換句話說，不用理他。
The map \(\rho: E\to \mathcal{L}(E/F)\) defined by \(\rho(a) = [L_a]_\alpha\) is an injective ring homomorphism from \(E\) to \(M_n(F)\).
homomorphism
\[L_{a+b}(x) = (a+b)x = ax + bx = L_a(x) + L_b(x), \\ L_{ab}(x) = (ab)x = a(bx) = L_a(L_b(x)) = L_aL_b(x).\]one-to-one
For all \(a\in \text{ker}(\rho)\),
\[\rho(a) = 0 = L_a(x) = ax, \forall x \in E.\]which implies that \(a=0\) and \(\rho\) is injective. ◼
\(E\) is a field and has no zero divisors!
Every field extension \(E\) over \(F\) of degree \(n\) is isomorphic to some subring of \(M_n(F)\).
於是所有 field extension 的計算可以用矩陣完成！
Let \(\alpha\) be a zero of \(x^3+x+1\). Find \((\alpha^2-\alpha+1)^{-1}\).
Solution
Let \(\beta = \{1, \alpha, \alpha^2\}\), we have
\[\begin{align*} [L_1]_\beta &= \begin{pmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{pmatrix} \\ [L_\alpha]_\beta &= \begin{pmatrix} 0&0&-1 \\ 1&0&-1 \\ 0&1&0 \end{pmatrix} \\ [L_{\alpha^2}]_\beta &= \begin{pmatrix} 0&-1&0 \\ 0&-1&-1 \\ 1&0&-1 \end{pmatrix} \end{align*}\]and then
\[[L_{\alpha^2-\alpha+1}]_\beta^{-1} = \big([L_{\alpha^2}]_\beta-[L_\alpha]_\beta+[L_1]_\beta \big)^{-1} = \begin{pmatrix} 1&-1&1 \\ -1&0&0 \\ 1&-1&0 \end{pmatrix}^{-1} = \begin{pmatrix} 0& \\ 0&\cdots \\ 1& \end{pmatrix}.\]Since \((\alpha^2-\alpha+1)^{-1}\) is be a linear combination of the vectors in \(\beta\), \([L_{\alpha^2-\alpha+1}]_\beta^{-1}\) must be a linear combination of \([L_{\alpha^i}]_\beta\) as well. Thus, from the first column of \([L_{\alpha^2-\alpha+1}]_\beta^{-1}\), which is \((0, 0, 1)^T\), we can easily identify that \([L_{\alpha^2-\alpha+1}]_\beta^{-1} = [L_{\alpha^2}]_\beta\). That is to say,
\[(\alpha^2-\alpha+1)^{-1} = \alpha^2. ◼\]首先，\([L_{\alpha^i}]_\beta\) 是怎麼算出來的？\([L_1]_\beta\) 就只是 \(n\times n\) 單位矩陣。重點來了：因為我們定義 \(L_a(x) = ax\)，所以 \(L_\alpha(x) = \alpha x\)；也就是說，\([L_{\alpha^{i+1}}]_\beta\) 就僅僅是將 \([L_{\alpha^i}]_\beta\) 中所有 column 往左移一單位，並在最右的空缺填上經由 \(\text{Irr}(\alpha, F)\) 算出的係數！
\([L_{\alpha^i}]_\beta\) 的第一 column 對應 \(1\in \beta\)、第二 column 對應 \(\alpha\in \beta\)、第三對應 \(\alpha^2 \in \beta\)。
以 \([L_\alpha]_\beta\) 為例：前兩 column 可以輕鬆得出（\(L_\alpha(1) = \alpha, L_\alpha(\alpha) = \alpha^2\)），而最後一 column 則是，
\[L_\alpha(\alpha^2) = \alpha^3 = -1-\alpha = (-1, -1, 0)^T，\]並且
\[L_{\alpha^2}(\alpha) = L_\alpha(\alpha^2), \\ L_{\alpha^2}(\alpha^2) = \alpha^4 = \alpha(-1-\alpha) = (0, -1, -1)^T。\]利用上述這些等式，就能看出第一段所說的性質為何成立了！
此外，計算反矩陣時，利用 Cramer’s Rule，算第一 column 就夠了，因為 \(n\) 個矩陣的第一 column 組合起來正是單位矩陣。
Note
上述方法只適用於 simple extension，也就是當 basis 為 cyclic 的時候。如果是 \(\Bbb{Q}(\sqrt{2}, \sqrt{3})\)、 basis 為 \(\beta = \{1, \sqrt{2}, \sqrt{3}, \sqrt{6} \}\)，我們就只能一個一個算出 \([L_1]_\beta, [L_\sqrt{2}]_\beta\cdots\) 了！例如
\[[L_\sqrt{2}]_\beta = \begin{pmatrix} 0 & 2 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 1 & 0 \end{pmatrix},\]因為左乘 \(\sqrt{2}\) 後，
\[\begin{align*} 1 &\to \sqrt{2} \\ \sqrt{2} &\to 2 \\ \sqrt{3} &\to \sqrt{6} \\ \sqrt{6} &\to 2\sqrt{3}. \end{align*}\]