[proofplan] We reduce to a purely inseparable simple extension. Since $L/K$ is not separable, there exists $\alpha \in L$ whose minimal polynomial $P_\alpha \in K[t]$ is inseparable. Writing $P_\alpha(t) = Q(t^{p^k})$ for some separable irreducible $Q \in K[t]$ and maximal $k \ge 1$, we set $F := K(\alpha)$ and consider the tower $K \subset K(\alpha^{p^k}) \subset F$. The extension $F/K(\alpha^{p^k})$ is purely inseparable of degree $p^k$, and we show $\operatorname{Tr}_{F/K(\alpha^{p^k})} \equiv 0$ by observing that every element of $F$ has a minimal polynomial of the form $t^{p^m} - c$ over $K(\alpha^{p^k})$ (whose $t^{p^m - 1}$ coefficient is zero) and that $\operatorname{Tr}_{F/K(\alpha^{p^k})}(1) = p^k = 0$ in characteristic $p$. The [Transitivity of Trace and Norm](/theorems/1293) applied twice -- first to $K \subset K(\alpha^{p^k}) \subset F$, then to $K \subset F \subset L$ -- yields $\operatorname{Tr}_{L/K} \equiv 0$. [/proofplan] [step:Locate an inseparable element and decompose its minimal polynomial] Since $L/K$ is a finite extension that is not separable, there exists $\alpha \in L$ such that the minimal polynomial $P_\alpha \in K[t]$ of $\alpha$ over $K$ is inseparable (i.e., has a repeated root in an algebraic closure $\bar{K}$). Set $F := K(\alpha)$ and $d := [F:K] = \deg P_\alpha$. Since $P_\alpha$ is irreducible over $K$ and inseparable, the characteristic of $K$ is a prime $p > 0$, and $P_\alpha$ has zero formal derivative: $P_\alpha'(t) = 0$ in $K[t]$. In characteristic $p$, $P_\alpha'(t) = 0$ if and only if $P_\alpha$ is a polynomial in $t^p$. Extracting the maximal power of $p$, there exist $k \ge 1$ and an irreducible polynomial $Q \in K[t]$ with $Q'(t) \neq 0$ (i.e., $Q$ is separable) such that \begin{align*} P_\alpha(t) = Q(t^{p^k}). \end{align*} Write $e := \deg Q$, so $d = e \cdot p^k$. The element $\alpha^{p^k}$ is a root of $Q$ (since $Q(\alpha^{p^k}) = P_\alpha(\alpha) = 0$), and since $Q$ is irreducible over $K$, it is the minimal polynomial of $\alpha^{p^k}$ over $K$. Therefore $[K(\alpha^{p^k}) : K] = e$. [guided] Why must such an inseparable element exist? An extension $L/K$ is separable if and only if every $\alpha \in L$ has a separable minimal polynomial over $K$. Since $L/K$ is assumed not separable, the negation provides an $\alpha \in L$ whose minimal polynomial $P_\alpha \in K[t]$ is inseparable. The structure of inseparable irreducible polynomials in characteristic $p$ is determined by the formal derivative criterion. An irreducible polynomial $f(t) \in K[t]$ of degree $\ge 1$ is separable if and only if $f'(t) \neq 0$. When $f'(t) = 0$, every coefficient $a_j$ with $p \nmid j$ must satisfy $j a_j = 0$ in $K$. Since $\operatorname{char}(K) = p$ and $\gcd(j, p) = 1$ implies $j \neq 0$ in $K$, the condition $j a_j = 0$ forces $a_j = 0$. Therefore only coefficients $a_j$ with $p \mid j$ survive, meaning $f(t) = g(t^p)$ for some $g \in K[t]$. Iterating, we extract the maximal power $p^k$ such that $f(t) = Q(t^{p^k})$ with $Q'(t) \neq 0$. The irreducibility of $Q$ follows from the irreducibility of $P_\alpha$: if $Q = Q_1 Q_2$ with $\deg Q_i \ge 1$, then $P_\alpha(t) = Q_1(t^{p^k}) Q_2(t^{p^k})$ is a nontrivial factorisation of $P_\alpha$ in $K[t]$, contradicting irreducibility. Setting $F := K(\alpha)$ gives an intermediate field $K \subset F \subset L$ with $[F:K] = d = ep^k$. The strategy is to show $\operatorname{Tr}_{F/K} \equiv 0$ and then invoke transitivity to kill $\operatorname{Tr}_{L/K}$. [/guided] [/step] [step:Identify $F/K(\alpha^{p^k})$ as a purely inseparable extension of degree $p^k$] Consider the tower $K \subset K(\alpha^{p^k}) \subset F = K(\alpha)$. We have $[K(\alpha^{p^k}):K] = e$ (established in the previous step) and $[F:K] = ep^k$, so \begin{align*} [F : K(\alpha^{p^k})] = \frac{[F:K]}{[K(\alpha^{p^k}):K]} = \frac{ep^k}{e} = p^k. \end{align*} The element $\alpha$ satisfies the polynomial $t^{p^k} - \alpha^{p^k} \in K(\alpha^{p^k})[t]$. In characteristic $p$, this polynomial factors as \begin{align*} t^{p^k} - \alpha^{p^k} = (t - \alpha)^{p^k}, \end{align*} so its only root is $\alpha$ with multiplicity $p^k$. Since $[F:K(\alpha^{p^k})] = p^k = \deg(t^{p^k} - \alpha^{p^k})$, this polynomial is the minimal polynomial of $\alpha$ over $K(\alpha^{p^k})$. The extension $F/K(\alpha^{p^k})$ is therefore purely inseparable of degree $p^k$, with $K(\alpha^{p^k})$-basis $\{1, \alpha, \alpha^2, \ldots, \alpha^{p^k - 1}\}$. [guided] Why does $(t - \alpha)^{p^k} = t^{p^k} - \alpha^{p^k}$ hold? This is the Frobenius identity in characteristic $p$: for any elements $a, b$ in a commutative ring of characteristic $p$, $(a - b)^p = a^p - b^p$ (all binomial coefficients $\binom{p}{j}$ for $1 \le j \le p-1$ are divisible by $p$ and hence vanish). Iterating $k$ times gives $(a - b)^{p^k} = a^{p^k} - b^{p^k}$. Could $t^{p^k} - \alpha^{p^k}$ fail to be the minimal polynomial of $\alpha$ over $K(\alpha^{p^k})$? If the minimal polynomial had degree $d' < p^k$, then $[K(\alpha^{p^k})(\alpha) : K(\alpha^{p^k})] = d' < p^k$, giving $[F:K] = [F:K(\alpha^{p^k})] \cdot [K(\alpha^{p^k}):K] = d' \cdot e < p^k \cdot e = d$, contradicting $[F:K] = d$. So $t^{p^k} - \alpha^{p^k}$ is indeed the minimal polynomial. [/guided] [/step] [step:Show $\operatorname{Tr}_{F/K(\alpha^{p^k})} \equiv 0$ using the vanishing of the sub-leading coefficient] We show that $\operatorname{Tr}_{F/K(\alpha^{p^k})}(\alpha^j) = 0$ for every $0 \le j \le p^k - 1$. Since these elements form a $K(\alpha^{p^k})$-basis of $F$, $K(\alpha^{p^k})$-linearity of the trace will then force $\operatorname{Tr}_{F/K(\alpha^{p^k})} \equiv 0$. **Case $j = 0$.** The [Trace and Norm via Minimal Polynomial](/theorems/1292) gives $\operatorname{Tr}_{F/K(\alpha^{p^k})}(1) = [F : K(\alpha^{p^k})] = p^k$, viewed as an element of $K(\alpha^{p^k})$. Since $\operatorname{char}(K) = p$, we have $p^k \cdot 1 = 0$ in $K(\alpha^{p^k})$. Therefore $\operatorname{Tr}_{F/K(\alpha^{p^k})}(1) = 0$. **Case $1 \le j \le p^k - 1$.** The element $\alpha^j$ satisfies the polynomial \begin{align*} t^{p^k} - \alpha^{jp^k} = (t - \alpha^j)^{p^k} \in K(\alpha^{p^k})[t], \end{align*} since $\alpha^{jp^k} = (\alpha^{p^k})^j \in K(\alpha^{p^k})$. The minimal polynomial of $\alpha^j$ over $K(\alpha^{p^k})$ divides $(t - \alpha^j)^{p^k}$, so it has the form $(t - \alpha^j)^{p^m} = t^{p^m} - \alpha^{jp^m}$ for some $1 \le m \le k$ (it cannot have degree $1$, since $\alpha^j \notin K(\alpha^{p^k})$ for $1 \le j \le p^k - 1$ with $p \nmid j$; when $p \mid j$, write $j = p^s j'$ with $\gcd(j', p) = 1$ and $1 \le s < k$, so the minimal polynomial has degree $p^{k-s} \ge p$). In all cases, the minimal polynomial of $\alpha^j$ over $K(\alpha^{p^k})$ is $t^{p^m} - c$ for some $c \in K(\alpha^{p^k})$ and some $m \ge 1$. This polynomial has the form \begin{align*} t^{p^m} + 0 \cdot t^{p^m - 1} + \cdots + 0 \cdot t + (-c), \end{align*} where every coefficient between the leading term and the constant term is zero (since $p^m \ge p \ge 2$, there are no terms of degree $1, 2, \ldots, p^m - 1$ except possibly $t^0$). In particular, the coefficient of $t^{p^m - 1}$ is $0$. Set $r_j := [F : K(\alpha^{p^k})(\alpha^j)]$. By the [Trace and Norm via Minimal Polynomial](/theorems/1292): \begin{align*} \operatorname{Tr}_{F/K(\alpha^{p^k})}(\alpha^j) = -r_j \cdot 0 = 0. \end{align*} Since $\{1, \alpha, \ldots, \alpha^{p^k-1}\}$ is a $K(\alpha^{p^k})$-basis for $F$ and $\operatorname{Tr}_{F/K(\alpha^{p^k})}$ is $K(\alpha^{p^k})$-linear, every element $\beta \in F$ satisfies \begin{align*} \operatorname{Tr}_{F/K(\alpha^{p^k})}(\beta) = \sum_{j=0}^{p^k - 1} c_j \operatorname{Tr}_{F/K(\alpha^{p^k})}(\alpha^j) = 0, \end{align*} where $\beta = \sum_{j=0}^{p^k-1} c_j \alpha^j$ with $c_j \in K(\alpha^{p^k})$. Therefore $\operatorname{Tr}_{F/K(\alpha^{p^k})} \equiv 0$. [guided] This is the heart of the proof. The extension $F/K(\alpha^{p^k})$ is purely inseparable: it is generated by $\alpha$, whose only conjugate over $K(\alpha^{p^k})$ is itself (with multiplicity $p^k$). The companion matrix of $t^{p^k} - \alpha^{p^k}$ with respect to the basis $\{1, \alpha, \ldots, \alpha^{p^k - 1}\}$ is the $p^k \times p^k$ matrix \begin{align*} C = \begin{pmatrix} 0 & 0 & \cdots & 0 & \alpha^{p^k} \\ 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & \cdots & 0 & 0 \\ \vdots & & \ddots & & \vdots \\ 0 & 0 & \cdots & 1 & 0 \end{pmatrix}, \end{align*} which has all diagonal entries equal to $0$, so $\operatorname{Tr}(C) = 0$. This confirms $\operatorname{Tr}_{F/K(\alpha^{p^k})}(\alpha) = 0$ directly from the matrix representation. For the other basis elements $\alpha^j$, we cannot simply look at powers of $C$ (which would give the matrix of multiplication by $\alpha^j$), because the trace of $C^j$ could in principle be nonzero. Instead, we use the [Trace and Norm via Minimal Polynomial](/theorems/1292) formula, which reduces the computation to the sub-leading coefficient of the minimal polynomial of $\alpha^j$ over $K(\alpha^{p^k})$. The crucial structural fact is that every element of $F$ is purely inseparable over $K(\alpha^{p^k})$: if $\beta \in F$ satisfies $\beta^{p^k} \in K(\alpha^{p^k})$, then its minimal polynomial over $K(\alpha^{p^k})$ divides $(t - \beta)^{p^k} = t^{p^k} - \beta^{p^k}$, so it has the form $t^{p^m} - \beta^{p^m}$ for some $m \ge 0$. When $m \ge 1$ (i.e., $\beta \notin K(\alpha^{p^k})$), the polynomial $t^{p^m} - \beta^{p^m}$ is a pure $p^m$-th power minus a constant, with no intermediate terms. The sub-leading coefficient is $0$. When $m = 0$ (i.e., $\beta \in K(\alpha^{p^k})$), we use $K(\alpha^{p^k})$-linearity: $\operatorname{Tr}_{F/K(\alpha^{p^k})}(\beta) = \beta \cdot \operatorname{Tr}_{F/K(\alpha^{p^k})}(1) = \beta \cdot p^k = 0$ in characteristic $p$. Either way, the trace vanishes. The two mechanisms -- vanishing sub-leading coefficient for $m \ge 1$ and vanishing characteristic for $m = 0$ -- conspire to kill the trace on every element. [/guided] [/step] [step:Deduce $\operatorname{Tr}_{F/K} \equiv 0$ by transitivity through $K(\alpha^{p^k})$] We apply the [Transitivity of Trace and Norm](/theorems/1293) to the tower $K \subset K(\alpha^{p^k}) \subset F$. For every $\beta \in F$: \begin{align*} \operatorname{Tr}_{F/K}(\beta) = \operatorname{Tr}_{K(\alpha^{p^k})/K}\!\bigl(\operatorname{Tr}_{F/K(\alpha^{p^k})}(\beta)\bigr) = \operatorname{Tr}_{K(\alpha^{p^k})/K}(0) = 0. \end{align*} The first equality is the transitivity formula. The second uses $\operatorname{Tr}_{F/K(\alpha^{p^k})} \equiv 0$, established in the previous step. The third uses $K$-linearity of $\operatorname{Tr}_{K(\alpha^{p^k})/K}$. Therefore $\operatorname{Tr}_{F/K} \equiv 0$. [/step] [step:Conclude $\operatorname{Tr}_{L/K} \equiv 0$ by transitivity through $F$] We apply the [Transitivity of Trace and Norm](/theorems/1293) to the tower $K \subset F \subset L$. For every $\beta \in L$: \begin{align*} \operatorname{Tr}_{L/K}(\beta) = \operatorname{Tr}_{F/K}\!\bigl(\operatorname{Tr}_{L/F}(\beta)\bigr). \end{align*} The element $\operatorname{Tr}_{L/F}(\beta) \in F$, and since $\operatorname{Tr}_{F/K}$ vanishes identically on $F$: \begin{align*} \operatorname{Tr}_{L/K}(\beta) = \operatorname{Tr}_{F/K}\!\bigl(\operatorname{Tr}_{L/F}(\beta)\bigr) = 0. \end{align*} Since $\beta \in L$ was arbitrary, $\operatorname{Tr}_{L/K} \equiv 0$. [guided] The transitivity formula $\operatorname{Tr}_{L/K} = \operatorname{Tr}_{F/K} \circ \operatorname{Tr}_{L/F}$ reduces the problem to showing that the "lower trace" $\operatorname{Tr}_{F/K}$ is the zero map. We do not need to understand $\operatorname{Tr}_{L/F}$ at all -- it maps $L$ into $F$, and $\operatorname{Tr}_{F/K}$ kills everything in $F$, so the composition is zero regardless. This is the reason for the reduction strategy: rather than analyzing the full multiplication-by-$\beta$ matrix for arbitrary $\beta \in L$ (which would require understanding the $[L:K] \times [L:K]$ matrix), we factor the trace through the much smaller extension $F/K$ where inseparability is directly visible. The hypothesis that $L/K$ is not separable is used only once, at the very beginning: to produce the inseparable element $\alpha \in L$. From that point on, the argument is purely structural, depending on the characteristic-$p$ form of inseparable minimal polynomials (all of the form $t^{p^m} - c$ with no sub-leading term) and two applications of the transitivity of trace. [/guided] [/step]