[proofplan] We reduce both identities to the [Trace and Norm via Minimal Polynomial](/theorems/1292) formula by analysing how the $n$ distinct $K$-homomorphisms $\varphi_i : L \hookrightarrow \bar{K}$ restrict to $K(\alpha)$. The restriction map $\varphi \mapsto \varphi|_{K(\alpha)}$ sends the $n = dr$ homomorphisms to the $d$ distinct $K$-homomorphisms $K(\alpha) \hookrightarrow \bar{K}$, and separability of $L/K(\alpha)$ guarantees that each fibre has exactly $r = [L : K(\alpha)]$ elements. By the [Roots-Homomorphisms Correspondence](/theorems/1256), the $d$ distinct restrictions biject with the $d$ roots of the minimal polynomial $P_\alpha$, so the sum $\sum_{i=1}^{n} \varphi_i(\alpha)$ collapses to $r$ times the sum of roots, which equals $\operatorname{Tr}_{L/K}(\alpha)$ by Vieta and the minimal polynomial formula. The norm follows by the same fibre-counting argument applied to products. [/proofplan] [step:Set up the degree decomposition and identify the restriction map] Let $\alpha \in L$. Define $d := [K(\alpha) : K] = \deg P_\alpha$ and $r := [L : K(\alpha)]$, so that $n = [L : K] = dr$ by the [Tower Law](/theorems/1248). Consider the restriction map \begin{align*} \operatorname{res}: \operatorname{Hom}_K(L, \bar{K}) &\to \operatorname{Hom}_K(K(\alpha), \bar{K}) \\ \varphi &\mapsto \varphi|_{K(\alpha)}. \end{align*} This map is well-defined because $K(\alpha) \subset L$, and any $K$-homomorphism $L \hookrightarrow \bar{K}$ restricts to a $K$-homomorphism $K(\alpha) \hookrightarrow \bar{K}$. [guided] The goal of the proof is to relate the sum $\sum_{i=1}^{n} \varphi_i(\alpha)$ and the product $\prod_{i=1}^{n} \varphi_i(\alpha)$ to the coefficients of the minimal polynomial $P_\alpha$, where the connection to the trace and norm is already established by [Trace and Norm via Minimal Polynomial](/theorems/1292). We begin by decomposing the degree. Since $K \subset K(\alpha) \subset L$, the [Tower Law](/theorems/1248) gives $[L : K] = [L : K(\alpha)] \cdot [K(\alpha) : K]$. We set $d := [K(\alpha) : K] = \deg P_\alpha$ (the degree of the minimal polynomial of $\alpha$ over $K$) and $r := [L : K(\alpha)]$, so $n = dr$. The key object is the restriction map $\operatorname{res}$. Every $K$-homomorphism $\varphi : L \hookrightarrow \bar{K}$ restricts to a map on the subfield $K(\alpha)$. This restriction is a $K$-homomorphism $K(\alpha) \hookrightarrow \bar{K}$ because $\varphi$ fixes $K$ pointwise and sends $K(\alpha)$ into $\bar{K}$. Injectivity of $\varphi$ guarantees injectivity of $\varphi|_{K(\alpha)}$, so $\varphi|_{K(\alpha)}$ is a $K$-embedding. Why study this restriction? Because the value $\varphi_i(\alpha)$ depends only on $\varphi_i|_{K(\alpha)}$, not on how $\varphi_i$ acts on the rest of $L$. If two homomorphisms $\varphi_i, \varphi_j$ agree on $K(\alpha)$, they send $\alpha$ to the same element of $\bar{K}$. So the sum $\sum \varphi_i(\alpha)$ can be reorganised by grouping homomorphisms according to their restriction. [/guided] [/step] [step:Count the fibres of the restriction map using separability] By the [Roots-Homomorphisms Correspondence](/theorems/1256) applied to the simple extension $K(\alpha)/K$ with target $\bar{K}$, the $K$-homomorphisms $K(\alpha) \hookrightarrow \bar{K}$ biject with the roots of $P_\alpha$ in $\bar{K}$. Since $L/K$ is separable, the element $\alpha \in L$ is separable over $K$, so $P_\alpha$ has $d = \deg P_\alpha$ distinct roots in $\bar{K}$. Therefore $|\operatorname{Hom}_K(K(\alpha), \bar{K})| = d$. We now determine the size of each fibre $\operatorname{res}^{-1}(\sigma)$ for $\sigma \in \operatorname{Hom}_K(K(\alpha), \bar{K})$. A homomorphism $\varphi \in \operatorname{Hom}_K(L, \bar{K})$ lies in $\operatorname{res}^{-1}(\sigma)$ if and only if $\varphi|_{K(\alpha)} = \sigma$, i.e., $\varphi$ is an extension of $\sigma$ to $L$. By the [Counting Homomorphisms](/theorems/1264) theorem applied to the extension $L/K(\alpha)$, the number of extensions of a fixed $K(\alpha)$-embedding to $L$ is at most $[L : K(\alpha)] = r$, with equality if and only if $L/K(\alpha)$ is separable. Since $L/K$ is separable and $K \subset K(\alpha) \subset L$, the extension $L/K(\alpha)$ is separable (every element of $L$ is separable over $K$, hence a fortiori separable over the larger field $K(\alpha)$). Therefore each fibre has exactly $r$ elements: \begin{align*} |\operatorname{res}^{-1}(\sigma)| = r \quad \text{for every } \sigma \in \operatorname{Hom}_K(K(\alpha), \bar{K}). \end{align*} As a consistency check, the total count is $d \cdot r = n = |\operatorname{Hom}_K(L, \bar{K})|$, which agrees with the [Counting Homomorphisms](/theorems/1264) theorem applied to the separable extension $L/K$. [guided] This step is the heart of the argument: we must show that the $n$ homomorphisms $\varphi_1, \ldots, \varphi_n$ partition into $d$ groups of $r$, one group for each root of $P_\alpha$. The [Roots-Homomorphisms Correspondence](/theorems/1256) states: for $\alpha$ algebraic over $K$ with minimal polynomial $P_\alpha$, and any extension $E/K$, the map $\sigma \mapsto \sigma(\alpha)$ is a bijection $\operatorname{Hom}_K(K(\alpha), E) \leftrightarrow \operatorname{Root}_{P_\alpha}(E)$. We apply this with $E = \bar{K}$. The hypothesis requires that $\alpha$ is algebraic over $K$, which holds since $L/K$ is finite (hence algebraic). The conclusion gives $|\operatorname{Hom}_K(K(\alpha), \bar{K})| = |\operatorname{Root}_{P_\alpha}(\bar{K})|$. Since $\bar{K}$ is algebraically closed, $P_\alpha$ splits completely in $\bar{K}$, and since $L/K$ is separable, the element $\alpha$ is separable over $K$, meaning $P_\alpha$ has no repeated roots. Therefore $|\operatorname{Root}_{P_\alpha}(\bar{K})| = \deg P_\alpha = d$. For the fibre count, fix $\sigma \in \operatorname{Hom}_K(K(\alpha), \bar{K})$. An extension of $\sigma$ to $L$ is a $K$-homomorphism $\varphi : L \to \bar{K}$ with $\varphi|_{K(\alpha)} = \sigma$. We can view $L$ as an extension of $K(\alpha)$, and $\sigma$ as giving $\bar{K}$ the structure of a $K(\alpha)$-extension (via $\sigma$). The [Counting Homomorphisms](/theorems/1264) theorem states that for a finite extension $L/K(\alpha)$, we have $|\operatorname{Hom}_{K(\alpha)}(L, \bar{K})| \le [L : K(\alpha)]$ with equality if and only if $L/K(\alpha)$ is separable. We verify separability of $L/K(\alpha)$. Since $L/K$ is separable, for any $\beta \in L$, the minimal polynomial $\operatorname{min}_K(\beta) \in K[t]$ has no repeated roots. The minimal polynomial $\operatorname{min}_{K(\alpha)}(\beta)$ divides $\operatorname{min}_K(\beta)$ in $K(\alpha)[t]$ (since $K \subset K(\alpha)$), so $\operatorname{min}_{K(\alpha)}(\beta)$ also has no repeated roots. Since $\beta \in L$ was arbitrary, $L/K(\alpha)$ is separable. Therefore each fibre has size $|\operatorname{res}^{-1}(\sigma)| = r$, and the $n$ homomorphisms partition into $d$ fibres of $r$ elements each. This is the combinatorial fact that drives the remaining computation. [/guided] [/step] [step:Compute $\sum_{i=1}^{n} \varphi_i(\alpha)$ by grouping over fibres to obtain the trace formula] Let $\alpha_1, \ldots, \alpha_d \in \bar{K}$ be the $d$ distinct roots of $P_\alpha$. By the [Roots-Homomorphisms Correspondence](/theorems/1256), the $d$ distinct $K$-embeddings $\sigma_1, \ldots, \sigma_d \in \operatorname{Hom}_K(K(\alpha), \bar{K})$ satisfy $\sigma_j(\alpha) = \alpha_j$ for $j = 1, \ldots, d$. For each $\varphi_i$, the value $\varphi_i(\alpha)$ depends only on $\operatorname{res}(\varphi_i) = \varphi_i|_{K(\alpha)}$. Grouping the sum $\sum_{i=1}^{n} \varphi_i(\alpha)$ according to the fibres of $\operatorname{res}$: \begin{align*} \sum_{i=1}^{n} \varphi_i(\alpha) &= \sum_{j=1}^{d} \sum_{\substack{\varphi \in \operatorname{res}^{-1}(\sigma_j)}} \varphi(\alpha) = \sum_{j=1}^{d} \sum_{\substack{\varphi \in \operatorname{res}^{-1}(\sigma_j)}} \sigma_j(\alpha) = \sum_{j=1}^{d} r \cdot \alpha_j = r \sum_{j=1}^{d} \alpha_j. \end{align*} The second equality holds because every $\varphi$ in the fibre $\operatorname{res}^{-1}(\sigma_j)$ satisfies $\varphi(\alpha) = \varphi|_{K(\alpha)}(\alpha) = \sigma_j(\alpha) = \alpha_j$, and the third equality uses $|\operatorname{res}^{-1}(\sigma_j)| = r$. By Vieta's formulas applied to $P_\alpha = t^d + a_{d-1}t^{d-1} + \cdots + a_0$, the sum of roots satisfies $\sum_{j=1}^{d} \alpha_j = -a_{d-1}$. Therefore: \begin{align*} \sum_{i=1}^{n} \varphi_i(\alpha) = r \cdot (-a_{d-1}) = -r a_{d-1}. \end{align*} By [Trace and Norm via Minimal Polynomial](/theorems/1292), $\operatorname{Tr}_{L/K}(\alpha) = -r a_{d-1}$, where $r = [L : K(\alpha)]$ and $a_{d-1}$ is the coefficient of $t^{d-1}$ in $P_\alpha$. Comparing, we obtain \begin{align*} \operatorname{Tr}_{L/K}(\alpha) = \sum_{i=1}^{n} \varphi_i(\alpha). \end{align*} [guided] The idea is to reorganise the sum over all $n$ homomorphisms into a sum over the $d$ distinct roots of $P_\alpha$, using the fibre structure established in the previous step. Each $\varphi_i \in \operatorname{Hom}_K(L, \bar{K})$ restricts to some $\sigma_j \in \operatorname{Hom}_K(K(\alpha), \bar{K})$, and the value $\varphi_i(\alpha)$ depends only on this restriction. Precisely, $\varphi_i(\alpha) = \varphi_i|_{K(\alpha)}(\alpha) = \sigma_j(\alpha) = \alpha_j$, where $\alpha_j$ is the root of $P_\alpha$ corresponding to $\sigma_j$ under the [Roots-Homomorphisms Correspondence](/theorems/1256). This means all $r$ homomorphisms in a given fibre contribute the same value $\alpha_j$ to the sum. Partitioning $\{1, \ldots, n\}$ into the $d$ fibres (each of size $r$): \begin{align*} \sum_{i=1}^{n} \varphi_i(\alpha) &= \sum_{j=1}^{d} \sum_{\substack{\varphi \in \operatorname{res}^{-1}(\sigma_j)}} \varphi(\alpha) = \sum_{j=1}^{d} r \cdot \alpha_j = r \sum_{j=1}^{d} \alpha_j. \end{align*} Now we connect $\sum_{j=1}^{d} \alpha_j$ to the coefficients of $P_\alpha$. Write $P_\alpha = t^d + a_{d-1}t^{d-1} + \cdots + a_0 \in K[t]$. Since $\alpha_1, \ldots, \alpha_d$ are the $d$ roots of $P_\alpha$ in $\bar{K}$ (all distinct by separability), the factorisation of $P_\alpha$ in $\bar{K}[t]$ is \begin{align*} P_\alpha = \prod_{j=1}^{d}(t - \alpha_j). \end{align*} Expanding and comparing the coefficient of $t^{d-1}$: the coefficient on the right-hand side is $-(\alpha_1 + \cdots + \alpha_d)$. On the left-hand side it is $a_{d-1}$. Therefore $\sum_{j=1}^{d} \alpha_j = -a_{d-1}$ (Vieta's formula for the sum of roots). Substituting: \begin{align*} \sum_{i=1}^{n} \varphi_i(\alpha) = r(-a_{d-1}) = -ra_{d-1}. \end{align*} The [Trace and Norm via Minimal Polynomial](/theorems/1292) theorem states that for a finite extension $L/K$, the trace satisfies $\operatorname{Tr}_{L/K}(\alpha) = -ra_{d-1}$ where $r = [L : K(\alpha)]$ and $a_{d-1}$ is the sub-leading coefficient of $P_\alpha$. The two expressions coincide, giving $\operatorname{Tr}_{L/K}(\alpha) = \sum_{i=1}^{n} \varphi_i(\alpha)$. [/guided] [/step] [step:Compute $\prod_{i=1}^{n} \varphi_i(\alpha)$ by the same fibre-grouping to obtain the norm formula] Applying the same fibre decomposition to the product: \begin{align*} \prod_{i=1}^{n} \varphi_i(\alpha) &= \prod_{j=1}^{d} \prod_{\substack{\varphi \in \operatorname{res}^{-1}(\sigma_j)}} \varphi(\alpha) = \prod_{j=1}^{d} \alpha_j^r = \left(\prod_{j=1}^{d} \alpha_j\right)^r. \end{align*} The second equality uses $\varphi(\alpha) = \alpha_j$ for all $\varphi \in \operatorname{res}^{-1}(\sigma_j)$ and $|\operatorname{res}^{-1}(\sigma_j)| = r$. By Vieta's formulas, the product of roots of $P_\alpha = t^d + a_{d-1}t^{d-1} + \cdots + a_0$ satisfies $\prod_{j=1}^{d} \alpha_j = (-1)^d a_0$. Therefore: \begin{align*} \prod_{i=1}^{n} \varphi_i(\alpha) = \left((-1)^d a_0\right)^r = (-1)^{dr} a_0^r. \end{align*} By [Trace and Norm via Minimal Polynomial](/theorems/1292), $\operatorname{N}_{L/K}(\alpha) = (-1)^{dr} a_0^r$. Comparing: \begin{align*} \operatorname{N}_{L/K}(\alpha) = \prod_{i=1}^{n} \varphi_i(\alpha). \end{align*} [guided] The argument is structurally identical to the trace computation, replacing sums with products. We partition the product $\prod_{i=1}^{n} \varphi_i(\alpha)$ according to the fibres of $\operatorname{res}$. Within each fibre $\operatorname{res}^{-1}(\sigma_j)$, every homomorphism sends $\alpha$ to $\alpha_j$, and the fibre has $r$ elements, so the contribution is $\alpha_j^r$. Taking the product over all $d$ fibres: \begin{align*} \prod_{i=1}^{n} \varphi_i(\alpha) = \prod_{j=1}^{d} \alpha_j^r = \left(\prod_{j=1}^{d} \alpha_j\right)^r. \end{align*} For the Vieta step, we need the product of all roots of $P_\alpha = t^d + a_{d-1}t^{d-1} + \cdots + a_1 t + a_0$. From the factorisation $P_\alpha = \prod_{j=1}^{d}(t - \alpha_j)$, we set $t = 0$: \begin{align*} a_0 = P_\alpha(0) = \prod_{j=1}^{d}(0 - \alpha_j) = (-1)^d \prod_{j=1}^{d} \alpha_j. \end{align*} Solving for the product: $\prod_{j=1}^{d} \alpha_j = (-1)^d a_0$. Raising to the $r$-th power: \begin{align*} \left(\prod_{j=1}^{d} \alpha_j\right)^r = (-1)^{dr} a_0^r. \end{align*} The [Trace and Norm via Minimal Polynomial](/theorems/1292) theorem gives $\operatorname{N}_{L/K}(\alpha) = (-1)^{dr} a_0^r$ (via the determinant of the companion matrix of $P_\alpha$ repeated $r$ times). This matches $\prod_{i=1}^{n} \varphi_i(\alpha)$, completing the proof. Note the structural parallel: the trace formula uses the additive Vieta relation ($\sum \alpha_j = -a_{d-1}$) with the additive fibre-counting ($r$ copies summed), while the norm formula uses the multiplicative Vieta relation ($\prod \alpha_j = (-1)^d a_0$) with multiplicative fibre-counting ($r$-th power). This reflects the fact that trace is defined via the trace of a linear map (an additive invariant) while norm is defined via the determinant (a multiplicative invariant). [/guided] [/step]