[proofplan] View the $K$-vector space $L$ as a $K(\alpha)$-vector space with the restricted multiplication, and note that the multiplication-by-$\alpha$ map $m_\alpha: L \to L$ is $K(\alpha)$-linear. Choose a $K(\alpha)$-basis $e_1, \ldots, e_t$ of $L$ where $t = [L:K(\alpha)]$; this decomposes $L$ as a $K$-vector space into a direct sum $L = \bigoplus_{j=1}^t K(\alpha) \cdot e_j$, each summand $K(\alpha)$-invariant under $m_\alpha$. A block-diagonal argument shows the characteristic polynomial of $m_\alpha$ on $L$ equals the $t$-th power of its characteristic polynomial on the single summand $K(\alpha)$. On $K(\alpha)$, the matrix of $m_\alpha$ with respect to the $K$-basis $1, \alpha, \ldots, \alpha^{r-1}$ is the companion matrix of $p_\alpha$, whose characteristic polynomial is precisely $p_\alpha$. Combining, $\det(xI - m_\alpha) = p_\alpha^t = p_\alpha^{[L:K(\alpha)]}$. The norm and trace formulas then follow by reading off the constant term and the coefficient of $x^{n-1}$ of a product expansion of $p_\alpha^t = \prod_i (x - \alpha_i)^t$. [/proofplan]

[step:Set up the multiplication map $m_\alpha: L \to L$ and record its $K(\alpha)$-linearity]Let $n := [L:K]$ (a positive integer, since $L/K$ is a finite extension — the theorem is stated in the context where $L$ is a number field and $K$ a subfield, both finite-dimensional over $\mathbb{Q}$; in any case the statement presupposes $[L:K] < \infty$). The **multiplication-by-$\alpha$ map** is \begin{align*} m_\alpha: L &\to L \\ x &\mapsto \alpha x, \end{align*} which is well-defined because $L$ is a field (closed under multiplication). It is $K$-linear: for $x, y \in L$ and $c \in K$, \begin{align*} m_\alpha(x + y) &= \alpha(x + y) = \alpha x + \alpha y = m_\alpha(x) + m_\alpha(y), \\ m_\alpha(c x) &= \alpha \cdot (c x) = c \cdot (\alpha x) = c \cdot m_\alpha(x), \end{align*} using commutativity in the field $L$ and that $K \subseteq L$. Stronger: $m_\alpha$ is $K(\alpha)$-linear. For any $\lambda \in K(\alpha)$ and $x \in L$, \begin{align*} m_\alpha(\lambda x) = \alpha (\lambda x) = (\alpha \lambda) x = (\lambda \alpha) x = \lambda (\alpha x) = \lambda \cdot m_\alpha(x), \end{align*} using commutativity in $L$. The characteristic polynomial of $m_\alpha$ as a $K$-endomorphism of the $n$-dimensional $K$-vector space $L$ is the monic polynomial \begin{align*} \chi_{m_\alpha}(x) := \det(xI_n - M) \in K[x], \end{align*} where $M$ is the matrix of $m_\alpha$ with respect to any $K$-basis of $L$. This is independent of the chosen basis, because a change-of-basis matrix $P$ transforms $M \mapsto P^{-1} M P$, which does not affect $\det(xI - \cdot)$.[/step]

[guided]The multiplication-by-$\alpha$ map \begin{align*} m_\alpha: L \to L, \qquad x \mapsto \alpha x, \end{align*} is the central object of this proof. It is $K$-linear — as a direct check: $m_\alpha(x + y) = \alpha(x + y) = \alpha x + \alpha y = m_\alpha(x) + m_\alpha(y)$ by distributivity in the ring $L$, and $m_\alpha(c x) = \alpha(c x) = c(\alpha x) = c \cdot m_\alpha(x)$ for $c \in K$, using commutativity in $L$. A stronger observation: $m_\alpha$ is not just $K$-linear but $K(\alpha)$-linear. This is because commutativity in $L$ allows us to move any $\lambda \in K(\alpha) \subseteq L$ past $\alpha$: \begin{align*} m_\alpha(\lambda x) = \alpha \lambda x = \lambda \alpha x = \lambda \cdot m_\alpha(x). \end{align*} This $K(\alpha)$-linearity is what will let us decompose $m_\alpha$ into blocks, one per basis vector of $L$ over $K(\alpha)$. Define the characteristic polynomial of $m_\alpha$: \begin{align*} \chi_{m_\alpha}(x) := \det(xI_n - M) \in K[x], \end{align*} where $M \in K^{n \times n}$ is the matrix of $m_\alpha$ with respect to any chosen $K$-basis of $L$ (here $n = [L:K] = \dim_K L$). The polynomial $\chi_{m_\alpha}$ is independent of the basis: a change of $K$-basis via matrix $P \in GL_n(K)$ replaces $M$ by $P^{-1}MP$, and \begin{align*} \det(xI - P^{-1}MP) = \det(P^{-1}(xI - M)P) = \det(P^{-1})\det(xI - M)\det(P) = \det(xI - M). \end{align*} Hence $\chi_{m_\alpha}$ is an invariant of the operator $m_\alpha$ alone, not of the basis. Our ultimate goal is $\chi_{m_\alpha}(x) = p_\alpha(x)^{[L:K(\alpha)]}$ and the resulting norm/trace formulas.[/guided]

[step:Choose a $K(\alpha)$-basis of $L$ and decompose $L$ as a $K$-vector space]Let $t := [L:K(\alpha)]$ and $r := [K(\alpha):K] = \deg p_\alpha$ (the latter equality is the fundamental identity "$[K(\alpha):K] = \deg p_\alpha$" — verified below). By the [tower law](/pages/???), \begin{align*} n = [L:K] = [L:K(\alpha)] \cdot [K(\alpha):K] = t \cdot r. \end{align*} Choose a $K(\alpha)$-basis $e_1, \ldots, e_t$ of $L$: such a basis exists because $L$ is a finite-dimensional $K(\alpha)$-vector space of dimension $t$. For each $j \in \{1, \ldots, t\}$, the cyclic $K(\alpha)$-submodule \begin{align*} V_j := K(\alpha) \cdot e_j = \{\lambda e_j : \lambda \in K(\alpha)\} \subseteq L \end{align*} is a $K$-subspace of $L$ of dimension $r$ (it is isomorphic to $K(\alpha)$ as a $K$-vector space via $\lambda e_j \mapsto \lambda$, which is $K$-linear and bijective because $\{e_j\}$ is $K(\alpha)$-linearly independent). Since $\{e_1, \ldots, e_t\}$ is a $K(\alpha)$-basis of $L$, every $x \in L$ has a unique expression $x = \sum_{j=1}^t \lambda_j e_j$ with $\lambda_j \in K(\alpha)$. This is precisely the $K$-vector space direct sum decomposition \begin{align*} L = \bigoplus_{j=1}^t V_j \qquad \text{(as $K$-vector spaces)}. \tag{$\ast$} \end{align*} Each $V_j$ is $m_\alpha$-invariant: if $x = \lambda e_j \in V_j$ with $\lambda \in K(\alpha)$, then $m_\alpha(x) = \alpha \lambda e_j$, and $\alpha \lambda \in K(\alpha)$ since $K(\alpha)$ is closed under multiplication. Hence $m_\alpha(V_j) \subseteq V_j$.[/step]

[guided]Let $r := \deg p_\alpha$ and $t := [L:K(\alpha)]$. Note that $r = [K(\alpha):K]$: this is the standard fact that $K(\alpha) \cong K[x]/\langle p_\alpha \rangle$ (as $K$-algebras) via $\alpha \leftrightarrow x$, and the quotient $K[x]/\langle p_\alpha \rangle$ has $K$-basis $\{1, x, x^2, \ldots, x^{r-1}\}$, hence $K$-dimension $r$. (This is the [Minimal Polynomial Degree Equals Extension Degree](/theorems/???) principle.) Therefore, by the [tower law](/pages/???), \begin{align*} n = [L:K] = [L:K(\alpha)] \cdot [K(\alpha):K] = tr. \end{align*} Pick a $K(\alpha)$-basis $e_1, \ldots, e_t$ of $L$ (exists because $L$ is finite-dimensional over $K(\alpha)$, of dimension $t$). For each $j$, consider \begin{align*} V_j := K(\alpha) \cdot e_j = \{\lambda e_j : \lambda \in K(\alpha)\}. \end{align*} This is a $K$-subspace of $L$. Its $K$-dimension is $r$: the map $K(\alpha) \to V_j$, $\lambda \mapsto \lambda e_j$, is $K$-linear, surjective by definition of $V_j$, and injective because $\{e_j\}$ is linearly independent over $K(\alpha)$ (singleton subset of a basis), so $\lambda e_j = 0 \implies \lambda = 0$. Hence $V_j \cong K(\alpha)$ as $K$-vector spaces, and $\dim_K V_j = \dim_K K(\alpha) = r$. The direct sum decomposition $(\ast)$ follows from the fact that $\{e_1, \ldots, e_t\}$ is a $K(\alpha)$-basis of $L$: every $x \in L$ has a unique expression $x = \sum_j \lambda_j e_j$ with $\lambda_j \in K(\alpha)$, and the uniqueness gives the direct-sum structure. The total $K$-dimension of $\bigoplus_j V_j$ is $\sum_{j=1}^t r = tr = n$, matching $\dim_K L$, confirming that the decomposition exhausts $L$. Crucially, each $V_j$ is $m_\alpha$-invariant: for $\lambda e_j \in V_j$, we have $m_\alpha(\lambda e_j) = \alpha(\lambda e_j) = (\alpha \lambda) e_j$, and $\alpha \lambda \in K(\alpha)$ (because $\alpha \in K(\alpha)$ and $K(\alpha)$ is a subring of $L$, closed under multiplication). So $m_\alpha(\lambda e_j) \in V_j$. Invariance is the structural fact that enables block diagonalisation: because $m_\alpha$ preserves each summand, the matrix of $m_\alpha$ in a basis compatible with the decomposition is block-diagonal.[/guided]

[step:Express the matrix of $m_\alpha$ on $L$ as block-diagonal and reduce to the single block $K(\alpha) \cdot e_1$]Let $1, \alpha, \ldots, \alpha^{r-1}$ be the standard $K$-basis of $K(\alpha)$ (it is a $K$-basis because $K(\alpha) \cong K[x]/\langle p_\alpha \rangle$ and $\{1, x, \ldots, x^{r-1}\}$ is a $K$-basis of the quotient). For each $j \in \{1, \ldots, t\}$, the set \begin{align*} \mathcal{B}_j := \{e_j, \alpha e_j, \alpha^2 e_j, \ldots, \alpha^{r-1} e_j\} \end{align*} is a $K$-basis of $V_j$ (it is the image of $\{1, \alpha, \ldots, \alpha^{r-1}\}$ under the $K$-linear isomorphism $K(\alpha) \to V_j$, $\lambda \mapsto \lambda e_j$ from the previous step). Concatenating over $j$: \begin{align*} \mathcal{B} := \mathcal{B}_1 \sqcup \mathcal{B}_2 \sqcup \cdots \sqcup \mathcal{B}_t \end{align*} is a $K$-basis of $L$ (totalling $tr = n$ vectors, matching $\dim_K L$). This is an ordered basis: we list $\mathcal{B}_1$, then $\mathcal{B}_2$, etc. With respect to $\mathcal{B}$, the matrix $M \in K^{n \times n}$ of $m_\alpha$ is block-diagonal: \begin{align*} M = \begin{pmatrix} C & 0 & \cdots & 0 \\ 0 & C & \cdots & 0 \\ \vdots & & \ddots & \vdots \\ 0 & 0 & \cdots & C \end{pmatrix} \in K^{n \times n}, \tag{$\dagger$} \end{align*} where $C \in K^{r \times r}$ is the matrix of $m_\alpha$ restricted to $V_j$, expressed in the basis $\mathcal{B}_j$ — and this matrix is the **same** for every $j$. Indeed, for each $j$, the map $V_j \to K(\alpha)$ sending $\alpha^i e_j \mapsto \alpha^i$ is a $K$-linear isomorphism that intertwines $m_\alpha|_{V_j}$ with $m_\alpha|_{K(\alpha)}$ (both are "multiplication by $\alpha$"), so the matrix representations agree. The off-diagonal blocks are zero because each $V_j$ is $m_\alpha$-invariant (previous step): $m_\alpha(\alpha^i e_j) = \alpha^{i+1} e_j \in V_j$, never involving any $e_{j'}$ with $j' \neq j$. Taking determinants of $(xI_n - M)$ using the multiplicativity of the determinant on block-diagonal matrices: \begin{align*} \chi_{m_\alpha}(x) = \det(xI_n - M) = \prod_{j=1}^t \det(xI_r - C) = \det(xI_r - C)^t. \tag{$\ddagger$} \end{align*} So the characteristic polynomial of $m_\alpha$ on $L$ is the $t$-th power of the characteristic polynomial of $m_\alpha$ restricted to a single summand $K(\alpha) \cdot e_1 \cong K(\alpha)$.[/step]

[guided]We upgrade the $K(\alpha)$-basis $\{e_1, \ldots, e_t\}$ of $L$ to a $K$-basis by refining each $e_j$ using the $K$-basis $\{1, \alpha, \ldots, \alpha^{r-1}\}$ of $K(\alpha)$. The set $\{1, \alpha, \ldots, \alpha^{r-1}\}$ is a $K$-basis of $K(\alpha)$: the isomorphism $K[x]/\langle p_\alpha \rangle \cong K(\alpha)$ (sending $x$ to $\alpha$) transfers the basis $\{1, x, \ldots, x^{r-1}\}$ of the quotient to $K(\alpha)$. For each $j$, define \begin{align*} \mathcal{B}_j := \{e_j, \alpha e_j, \alpha^2 e_j, \ldots, \alpha^{r-1} e_j\}. \end{align*} This is a $K$-basis of $V_j = K(\alpha) \cdot e_j$: applying the $K$-linear isomorphism $K(\alpha) \xrightarrow{\sim} V_j$, $\lambda \mapsto \lambda e_j$, to the $K$-basis $\{1, \alpha, \ldots, \alpha^{r-1}\}$ of $K(\alpha)$ gives a $K$-basis of $V_j$, which is exactly $\mathcal{B}_j$. Concatenating $\mathcal{B} := \mathcal{B}_1 \sqcup \cdots \sqcup \mathcal{B}_t$ produces a set of $tr = n$ vectors in $L$. Is $\mathcal{B}$ a $K$-basis of $L$? Yes: the $K$-direct sum $L = \bigoplus_j V_j$ from the previous step means $\mathcal{B}$ is a $K$-basis of $L$ iff each $\mathcal{B}_j$ is a $K$-basis of $V_j$, which we just verified. Now consider the matrix $M \in K^{n \times n}$ of $m_\alpha$ with respect to the ordered basis $\mathcal{B}$ (we fix the ordering: $e_1, \alpha e_1, \ldots, \alpha^{r-1} e_1, e_2, \alpha e_2, \ldots, \alpha^{r-1} e_t$). Since $m_\alpha$ preserves each $V_j$ (by invariance), applying $m_\alpha$ to any basis vector in $\mathcal{B}_j$ produces a $K$-linear combination of vectors in $\mathcal{B}_j$ alone — no "cross-block" contributions. The matrix $M$ is therefore block-diagonal with $t$ blocks of size $r$: \begin{align*} M = \operatorname{diag}(C_1, C_2, \ldots, C_t), \end{align*} where $C_j \in K^{r \times r}$ is the matrix of $m_\alpha|_{V_j}$ in the basis $\mathcal{B}_j$. Claim: $C_1 = C_2 = \cdots = C_t =: C$. Reason: for each $j$, the isomorphism $\psi_j: K(\alpha) \to V_j$, $\psi_j(\alpha^i) = \alpha^i e_j$, is $K$-linear and intertwines "multiplication by $\alpha$" — specifically, $m_\alpha \circ \psi_j = \psi_j \circ m_\alpha|_{K(\alpha)}$. In terms of matrices, the matrix of $m_\alpha|_{V_j}$ in $\mathcal{B}_j$ is the same as the matrix of $m_\alpha|_{K(\alpha)}$ in the basis $\{1, \alpha, \ldots, \alpha^{r-1}\}$ of $K(\alpha)$. This is independent of $j$. Denote the common block by $C$. Then $M = \operatorname{diag}(C, C, \ldots, C)$. Taking characteristic polynomials: for block-diagonal matrices, the determinant factors as the product of the determinants of the diagonal blocks (this is a standard determinant identity, following from Laplace expansion or the definition via permutations): \begin{align*} \chi_{m_\alpha}(x) = \det(xI_n - M) = \prod_{j=1}^t \det(xI_r - C) = \det(xI_r - C)^t. \end{align*} So we have reduced computing the characteristic polynomial of $m_\alpha$ on the $n$-dimensional $L$ to computing the characteristic polynomial of $m_\alpha$ on the $r$-dimensional $K(\alpha)$, namely $\det(xI_r - C)$.[/guided]

[step:Identify $C$ as the companion matrix of $p_\alpha$ and compute $\det(xI_r - C) = p_\alpha(x)$]We compute the matrix $C \in K^{r \times r}$ of $m_\alpha: K(\alpha) \to K(\alpha)$ in the $K$-basis $\{1, \alpha, \alpha^2, \ldots, \alpha^{r-1}\}$ of $K(\alpha)$. For $i \in \{0, 1, \ldots, r-2\}$, the action of $m_\alpha$ on the basis vector $\alpha^i$ is \begin{align*} m_\alpha(\alpha^i) = \alpha \cdot \alpha^i = \alpha^{i+1} \in K(\alpha), \end{align*} which is the next basis vector. For $i = r - 1$, we compute \begin{align*} m_\alpha(\alpha^{r-1}) = \alpha^r. \end{align*} Writing the minimal polynomial as $p_\alpha(x) = x^r + a_{r-1} x^{r-1} + \cdots + a_1 x + a_0$ with $a_i \in K$, the identity $p_\alpha(\alpha) = 0$ rearranges to \begin{align*} \alpha^r = -a_{r-1} \alpha^{r-1} - a_{r-2} \alpha^{r-2} - \cdots - a_1 \alpha - a_0. \end{align*} Reading off the coefficients: the matrix $C$ (with the convention that the $i$-th column of $C$ encodes $m_\alpha(\alpha^{i-1})$ in the basis) is \begin{align*} C = \begin{pmatrix} 0 & 0 & 0 & \cdots & 0 & -a_0 \\ 1 & 0 & 0 & \cdots & 0 & -a_1 \\ 0 & 1 & 0 & \cdots & 0 & -a_2 \\ 0 & 0 & 1 & \cdots & 0 & -a_3 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 1 & -a_{r-1} \end{pmatrix} \in K^{r \times r}. \tag{$\star$} \end{align*} This is the **companion matrix** of $p_\alpha$. We now compute $\chi_C(x) := \det(xI_r - C)$. [claim:Characteristic polynomial of the companion matrix equals the minimal polynomial] $\chi_C(x) = p_\alpha(x)$. [/claim] [proof] We compute $\det(xI_r - C)$ directly by Laplace expansion along the first row. Form the matrix \begin{align*} xI_r - C = \begin{pmatrix} x & 0 & 0 & \cdots & 0 & a_0 \\ -1 & x & 0 & \cdots & 0 & a_1 \\ 0 & -1 & x & \cdots & 0 & a_2 \\ 0 & 0 & -1 & \cdots & 0 & a_3 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & -1 & x + a_{r-1} \end{pmatrix}. \end{align*} We prove $\det(xI_r - C) = p_\alpha(x)$ by induction on $r$. **Base case $r = 1$:** $C = (-a_0) \in K^{1 \times 1}$ and $p_\alpha(x) = x + a_0$. Directly, $\det(xI_1 - C) = \det(x - (-a_0)) = x + a_0 = p_\alpha(x)$. **Inductive step:** Assume the claim for size $r - 1$. Expand $\det(xI_r - C)$ along the first row, which has two nonzero entries: $x$ in position $(1,1)$ and $a_0$ in position $(1, r)$. Using cofactor expansion: \begin{align*} \det(xI_r - C) = x \cdot \det(M_{11}) + (-1)^{1+r} a_0 \cdot \det(M_{1r}), \end{align*} where $M_{11}$ is the $(r-1) \times (r-1)$ submatrix obtained by deleting row $1$ and column $1$, and $M_{1r}$ is obtained by deleting row $1$ and column $r$. The submatrix $M_{11}$ is $xI_{r-1} - C'$, where $C' \in K^{(r-1) \times (r-1)}$ is the companion matrix of $p'(x) := x^{r-1} + a_{r-1} x^{r-2} + \cdots + a_2 x + a_1$. By the inductive hypothesis, $\det(M_{11}) = p'(x) = x^{r-1} + a_{r-1} x^{r-2} + \cdots + a_1$. The submatrix $M_{1r}$ is lower-triangular with $-1$'s on the diagonal: explicitly, \begin{align*} M_{1r} = \begin{pmatrix} -1 & x & 0 & \cdots & 0 \\ 0 & -1 & x & \cdots & 0 \\ \vdots & & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & -1 & x \\ 0 & 0 & \cdots & 0 & -1 \end{pmatrix} \in K^{(r-1) \times (r-1)}. \end{align*} The determinant of a triangular matrix is the product of its diagonal entries: $\det(M_{1r}) = (-1)^{r-1}$. Combining: \begin{align*} \det(xI_r - C) &= x \cdot (x^{r-1} + a_{r-1} x^{r-2} + \cdots + a_1) + (-1)^{1+r} a_0 \cdot (-1)^{r-1} \\ &= x^r + a_{r-1} x^{r-1} + \cdots + a_1 x + (-1)^{1+r}(-1)^{r-1} a_0. \end{align*} The final sign is $(-1)^{1+r+r-1} = (-1)^{2r} = 1$. Therefore \begin{align*} \det(xI_r - C) = x^r + a_{r-1} x^{r-1} + \cdots + a_1 x + a_0 = p_\alpha(x). \end{align*} [/proof][/step]

[guided]We compute the matrix $C$ of $m_\alpha$ restricted to $K(\alpha)$, in the $K$-basis $\{1, \alpha, \alpha^2, \ldots, \alpha^{r-1}\}$. For $i \in \{0, 1, \ldots, r-2\}$: \begin{align*} m_\alpha(\alpha^i) = \alpha \cdot \alpha^i = \alpha^{i+1}. \end{align*} This is already a single basis vector — the $(i+2)$-th basis vector, namely $\alpha^{i+1}$. So the $(i+1)$-th column of $C$ (with 1-indexing) is the standard basis vector $e_{i+2}$: a single $1$ in position $i+2$, zeros elsewhere. The interesting case is $i = r - 1$: \begin{align*} m_\alpha(\alpha^{r-1}) = \alpha^r. \end{align*} Now $\alpha^r$ is not itself one of the basis vectors $\{1, \alpha, \ldots, \alpha^{r-1}\}$ — it is one power higher. But $\alpha$ is a root of its minimal polynomial: \begin{align*} p_\alpha(\alpha) = \alpha^r + a_{r-1} \alpha^{r-1} + \cdots + a_1 \alpha + a_0 = 0, \end{align*} where we write $p_\alpha(x) = x^r + a_{r-1} x^{r-1} + \cdots + a_1 x + a_0 \in K[x]$ (the coefficients $a_i$ lie in $K$ because $p_\alpha \in K[x]$). Rearranging: \begin{align*} \alpha^r = -a_0 - a_1 \alpha - a_2 \alpha^2 - \cdots - a_{r-1} \alpha^{r-1}. \end{align*} This expresses $\alpha^r$ as a $K$-linear combination of the basis vectors. So the $r$-th column of $C$ is $(-a_0, -a_1, \ldots, -a_{r-1})^\top$. Assembling: the matrix $C$ is the **companion matrix** of $p_\alpha$, displayed in $(\star)$. We claim $\det(xI_r - C) = p_\alpha(x)$. This is a classical fact about companion matrices, proved by induction on $r$ via Laplace expansion along the first row of $xI_r - C$: - The $(1,1)$ entry is $x$; its cofactor is the determinant of an $(r-1) \times (r-1)$ matrix, which — by induction — equals $x^{r-1} + a_{r-1} x^{r-2} + \cdots + a_1$ (the companion polynomial of the "shifted" $p_\alpha$). - The $(1, r)$ entry is $a_0$; its cofactor is $(-1)^{1+r}$ times the determinant of a lower-triangular matrix with $-1$'s on the diagonal, which equals $(-1)^{r-1}$. The signs combine to $(-1)^{1+r} \cdot (-1)^{r-1} = (-1)^{2r} = 1$, giving \begin{align*} \det(xI_r - C) = x \cdot (x^{r-1} + a_{r-1} x^{r-2} + \cdots + a_1) + a_0 = x^r + a_{r-1} x^{r-1} + \cdots + a_1 x + a_0 = p_\alpha(x). \end{align*} See the claim above for the full induction. The base case $r = 1$ is $C = (-a_0)$, $p_\alpha(x) = x + a_0$, and $\det(x - C) = \det(x + a_0) = x + a_0$, matching. The inductive step was sketched above. This is sometimes phrased: the minimal polynomial of an element $\alpha$ is equal to the characteristic polynomial of multiplication-by-$\alpha$ acting on $K(\alpha)$.[/guided]

[step:Combine to obtain $\det(xI - m_\alpha) = p_\alpha^{[L:K(\alpha)]}$] Combining $(\ddagger)$ with the claim in the previous step: \begin{align*} \chi_{m_\alpha}(x) = \det(xI_n - M) = \det(xI_r - C)^t = p_\alpha(x)^t = p_\alpha(x)^{[L:K(\alpha)]}. \end{align*} This is the first assertion of the theorem.[/step]

[guided]Assembling everything: - From step 3, the characteristic polynomial of $m_\alpha$ on $L$ factors as $\chi_{m_\alpha}(x) = \det(xI_r - C)^t$, where $C$ is the matrix of $m_\alpha$ on the single summand $K(\alpha)$ and $t = [L:K(\alpha)]$. - From step 4, $C$ is the companion matrix of $p_\alpha$ and $\det(xI_r - C) = p_\alpha(x)$. Substituting: \begin{align*} \chi_{m_\alpha}(x) = p_\alpha(x)^t = p_\alpha(x)^{[L:K(\alpha)]}. \end{align*} This is the main claim of the theorem. The formula has a clean structural interpretation: the characteristic polynomial of $m_\alpha$ is the minimal polynomial $p_\alpha$ raised to the "redundancy factor" $[L:K(\alpha)]$. The redundancy measures how much bigger $L$ is than the smallest field $K(\alpha)$ in which $m_\alpha$ already "uses up" all its algebraic content.[/guided]

[step:Derive the norm and trace formulas on $K(\alpha)$ from the constant and sub-leading coefficients of $p_\alpha$]Suppose $p_\alpha$ splits in some extension field $M \supseteq K$ as $p_\alpha(x) = \prod_{i=1}^r (x - \alpha_i)$, where $\alpha_1, \ldots, \alpha_r \in M$ are the roots (with multiplicity; for a number field in characteristic $0$ and a minimal polynomial, which is irreducible, the roots are in fact distinct). Expanding the product using [Vieta's formulas](/pages/???): \begin{align*} p_\alpha(x) = \prod_{i=1}^r (x - \alpha_i) = x^r - \left(\sum_i \alpha_i\right) x^{r-1} + \cdots + (-1)^r \prod_i \alpha_i. \tag{$\diamond$} \end{align*} So the coefficient of $x^{r-1}$ in $p_\alpha$ is $-\sum_i \alpha_i$, and the constant term is $(-1)^r \prod_i \alpha_i$. By definition of norm and trace on the field $K(\alpha)$, where $m_\alpha: K(\alpha) \to K(\alpha)$ has characteristic polynomial $p_\alpha$: \begin{align*} N_{K(\alpha)/K}(\alpha) &:= \det(m_\alpha|_{K(\alpha)}) \in K, \\ \operatorname{tr}_{K(\alpha)/K}(\alpha) &:= \operatorname{tr}(m_\alpha|_{K(\alpha)}) \in K. \end{align*} For any $r \times r$ matrix with characteristic polynomial $x^r + c_{r-1} x^{r-1} + \cdots + c_1 x + c_0$: - The determinant equals $(-1)^r c_0$ (read off from $\chi(0) = c_0$ and $\det(-m_\alpha) = (-1)^r \det(m_\alpha)$; more directly, substituting $x = 0$ into $\det(xI - m_\alpha) = p_\alpha(x)$ gives $\det(-m_\alpha) = p_\alpha(0)$, so $(-1)^r \det(m_\alpha) = p_\alpha(0)$, i.e., $\det(m_\alpha) = (-1)^r p_\alpha(0)$). - The trace equals $-c_{r-1}$ (comparing coefficients of $x^{r-1}$ in $\det(xI - m_\alpha)$ and $x^r + c_{r-1} x^{r-1} + \cdots$; the $x^{r-1}$ coefficient of the characteristic polynomial of an $r \times r$ matrix is always $-\operatorname{tr}$ by direct expansion of the Leibniz formula). Applying these formulas to the characteristic polynomial $p_\alpha$ of $m_\alpha|_{K(\alpha)}$ with the expansion $(\diamond)$: \begin{align*} N_{K(\alpha)/K}(\alpha) &= (-1)^r p_\alpha(0) = (-1)^r \cdot (-1)^r \prod_i \alpha_i = \prod_i \alpha_i, \\ \operatorname{tr}_{K(\alpha)/K}(\alpha) &= -(-\sum_i \alpha_i) = \sum_i \alpha_i. \end{align*}[/step]

[guided]Suppose we have a splitting $p_\alpha(x) = \prod_{i=1}^r (x - \alpha_i)$ in some extension $M$ of $K$. Expanding the product by distribution gives the [Vieta formulas](/pages/???): \begin{align*} \prod_{i=1}^r (x - \alpha_i) = \sum_{k=0}^r (-1)^k e_k(\alpha_1, \ldots, \alpha_r) x^{r-k}, \end{align*} where $e_k$ is the $k$-th elementary symmetric polynomial. So \begin{align*} p_\alpha(x) = x^r - e_1(\alpha) x^{r-1} + e_2(\alpha) x^{r-2} - \cdots + (-1)^r e_r(\alpha), \end{align*} writing $e_k(\alpha) := e_k(\alpha_1, \ldots, \alpha_r)$ for brevity. Reading off specific coefficients: - Coefficient of $x^{r-1}$: $-e_1(\alpha) = -\sum_i \alpha_i$. - Constant term (coefficient of $x^0$): $(-1)^r e_r(\alpha) = (-1)^r \prod_i \alpha_i$. Now apply the definitions of norm and trace. For any $K$-linear endomorphism $T$ of a finite-dimensional $K$-vector space with characteristic polynomial \begin{align*} \chi_T(x) = x^N + c_{N-1} x^{N-1} + \cdots + c_1 x + c_0, \end{align*} the standard facts are: - $\det(T) = (-1)^N c_0$: substituting $x = 0$ into $\chi_T(x) = \det(xI - T)$ gives $c_0 = \chi_T(0) = \det(-T) = (-1)^N \det(T)$, so $\det(T) = (-1)^N c_0$. - $\operatorname{tr}(T) = -c_{N-1}$: expanding $\det(xI - T)$ by the Leibniz formula, the identity permutation contributes $\prod_i (x - T_{ii}) = x^N - (\sum_i T_{ii}) x^{N-1} + \cdots = x^N - \operatorname{tr}(T) x^{N-1} + \cdots$, and any other permutation contributes a term of $x$-degree at most $N - 2$; hence $c_{N-1} = -\operatorname{tr}(T)$. Apply these with $T = m_\alpha|_{K(\alpha)}$, $N = r$. The definitions of the field-theoretic norm and trace are \begin{align*} N_{K(\alpha)/K}(\alpha) &:= \det(m_\alpha|_{K(\alpha)}), \\ \operatorname{tr}_{K(\alpha)/K}(\alpha) &:= \operatorname{tr}(m_\alpha|_{K(\alpha)}). \end{align*} From the characteristic polynomial $p_\alpha$, we read: \begin{align*} c_0 &= (-1)^r \prod_i \alpha_i, \quad c_{r-1} = -\sum_i \alpha_i. \end{align*} Hence: \begin{align*} N_{K(\alpha)/K}(\alpha) &= (-1)^r \cdot c_0 = (-1)^r \cdot (-1)^r \prod_i \alpha_i = \prod_i \alpha_i, \\ \operatorname{tr}_{K(\alpha)/K}(\alpha) &= -c_{r-1} = -(-\sum_i \alpha_i) = \sum_i \alpha_i. \end{align*}[/guided]

[step:Derive the norm and trace formulas on $L$ from $\chi_{m_\alpha} = p_\alpha^t$]The characteristic polynomial of $m_\alpha$ on $L$ is $p_\alpha(x)^t = \prod_{i=1}^r (x - \alpha_i)^t$, where $t = [L:K(\alpha)]$ and $n = rt = [L:K]$. Expanding $\prod_{i=1}^r (x - \alpha_i)^t$: this is a polynomial of degree $n = rt$ whose roots (in $M$, with multiplicity) are $\alpha_1, \ldots, \alpha_r$, each appearing with multiplicity $t$. By Vieta's formulas applied to this degree-$n$ polynomial: \begin{align*} \prod_{i=1}^r (x - \alpha_i)^t &= x^n - \left(\text{sum of roots with multiplicity}\right) x^{n-1} + \cdots + (-1)^n \left(\text{product of roots with multiplicity}\right). \end{align*} The "sum of roots with multiplicity" is $t \cdot \sum_i \alpha_i$ (each $\alpha_i$ contributing $t$ times). The "product of roots with multiplicity" is $\prod_i \alpha_i^t = (\prod_i \alpha_i)^t$. Applying the determinant/trace formulas from the previous step with $N = n$: \begin{align*} N_{L/K}(\alpha) &:= \det(m_\alpha) = (-1)^n c_0 = (-1)^n \cdot (-1)^n \left(\prod_i \alpha_i\right)^t = \left(\prod_i \alpha_i\right)^{[L:K(\alpha)]}, \\ \operatorname{tr}_{L/K}(\alpha) &:= \operatorname{tr}(m_\alpha) = -c_{n-1} = -\left(-t \sum_i \alpha_i\right) = [L:K(\alpha)] \sum_i \alpha_i. \end{align*} This completes the derivation of all four formulas in the theorem statement.[/step]

[guided]On $L$, the characteristic polynomial of $m_\alpha$ is $p_\alpha^t$ with $t = [L:K(\alpha)]$. Reading off the roots of $p_\alpha^t$ in $M$ (counted with multiplicity): \begin{align*} p_\alpha(x)^t = \prod_{i=1}^r (x - \alpha_i)^t. \end{align*} This polynomial has degree $rt = n = [L:K]$. Its roots (with multiplicity) are $\alpha_1, \ldots, \alpha_r$, each repeated $t$ times. Applying Vieta's formulas to this degree-$n$ polynomial: The coefficient of $x^{n-1}$ is negative of the sum of the roots counted with multiplicity: \begin{align*} -\left[\underbrace{(\alpha_1 + \alpha_1 + \cdots + \alpha_1)}_{t \text{ times}} + \cdots + \underbrace{(\alpha_r + \cdots + \alpha_r)}_{t \text{ times}}\right] = -t \sum_i \alpha_i. \end{align*} The constant term is $(-1)^n$ times the product of the roots counted with multiplicity: \begin{align*} (-1)^n \cdot \alpha_1^t \cdot \alpha_2^t \cdots \alpha_r^t = (-1)^n \left(\prod_i \alpha_i\right)^t. \end{align*} Apply the standard determinant-constant and trace-coefficient identities for an $n \times n$ matrix, this time with $N = n$: \begin{align*} N_{L/K}(\alpha) &= \det(m_\alpha) = (-1)^n c_0 = (-1)^n \cdot (-1)^n \left(\prod_i \alpha_i\right)^t = \left(\prod_i \alpha_i\right)^t, \\ \operatorname{tr}_{L/K}(\alpha) &= \operatorname{tr}(m_\alpha) = -c_{n-1} = t \sum_i \alpha_i. \end{align*} With $t = [L:K(\alpha)]$, these are the announced formulas: \begin{align*} N_{L/K}(\alpha) &= \left(\prod_i \alpha_i\right)^{[L:K(\alpha)]}, \\ \operatorname{tr}_{L/K}(\alpha) &= [L:K(\alpha)] \sum_i \alpha_i. \end{align*} Interpretation: the norm and trace over $L$ are obtained from the norm and trace over $K(\alpha)$ by raising the product to the power $t$ (for the norm, which is multiplicative) and multiplying the sum by $t$ (for the trace, which is additive). The factor $t$ records how many "copies" of the building block $K(\alpha)$ sit inside $L$, reflecting the multiplicative behaviour of $\det$ and the additive behaviour of $\operatorname{tr}$ on block-diagonal matrices with repeated diagonal blocks. This completes the proof of the theorem.[/guided]