Norm and trace from embeddings, with the inseparable exponent in the norm formula

Suppose first that $K\mathrm{/}F$ is separable. Choose an $F$-basis $u_1,\dots ,u_n$ of $K$, list the embeddings as ${\sigma }_1,\dots ,{\sigma }_n$, and form the evaluation matrix $A=({\sigma }_i(u_j){)}_{i,j}$. By the same argument used in Artin's fixed-field lower bound, [L1] makes $A$ invertible.

For general $K\mathrm{/}F$, let $K_s$ be the separable closure of $F$ in $K$. By [L3], the extension $K\mathrm{/}{K}_s$ is purely inseparable of degree $[K:F{]}_i$, and the $F$-embeddings of $K$ are exactly the extensions of the embeddings of $K_s$, one extension for each embedding because [L4] gives uniqueness over $K_s$. Thus $\mathrm{\Sigma }$ may be identified with ${\mathrm{Hom}}_F(K_s,\mathrm{\Omega })$, and its cardinality is $[K:F{]}_s$.

Let $M$ be the matrix of $m_a$ in the basis $u_1,\dots ,u_n$. Because ${\sigma }_i(a{u}_j)={\sigma }_i(a){\sigma }_i(u_j)$ for every $i,j$, one has $AM=\mathrm{diag}({\sigma }_1(a),\dots ,{\sigma }_n(a))A\mathrm{.}$ Hence $M=A^{-1}\mathrm{diag}({\sigma }_1(a),\dots ,{\sigma }_n(a))A,$ so determinant and trace of [F1] give $N_{K\mathrm{/}F}(a)={\prod }_i{\sigma }_i(a),{\mathrm{Tr}}_{K\mathrm{/}F}(a)={\sum }_i{\sigma }_i(a)\mathrm{.}$

Choose an $F$-basis $v_1,\dots ,v_s$ of $K_s$ and a $K_s$-basis $b_1,\dots ,b_i$ of $K$, where $i=[K:F{]}_i$. Writing $a{b}_r={\sum }_{q=1}^i{c}_{qr}{b}_q(c_{qr}\in K_s),$ the matrix of $m_a$ on the product basis $(b_q{v}_j)$ is the block matrix $C^{\mathrm{♯}}=(M(c_{qr}){)}_{q,r}$, where $M(c_{qr})$ is the matrix of multiplication by $c_{qr}$ on $K_s\mathrm{/}F$. Conjugating each block by the separable evaluation matrix of step 1.1 for $K_s\mathrm{/}F$ turns $C^{\mathrm{♯}}$ into a block-diagonal matrix with diagonal blocks $\sigma (C)$ as $\sigma$ runs through $\mathrm{\Sigma }$. Therefore $N_{K\mathrm{/}F}(a)={\prod }_{\sigma \in \mathrm{\Sigma }}\det (\sigma (C)),{\mathrm{Tr}}_{K\mathrm{/}F}(a)={\sum }_{\sigma \in \mathrm{\Sigma }}\mathrm{tr}(\sigma (C))\mathrm{.}$

Over $K_s$, the extension $K\mathrm{/}{K}_s$ is purely inseparable of degree $i$, so every conjugate of $a$ over $K_s$ equals $a$. Accordingly the characteristic polynomial of the $K_s$-linear operator represented by $C$ is $(X-a{)}^i$, and therefore $\det (C)=a^i,\mathrm{tr}(C)=ia\mathrm{.}$ Applying each $\sigma \in \mathrm{\Sigma }$ in step 3.1 gives $\det (\sigma (C))=\sigma (a{)}^i,\mathrm{tr}(\sigma (C))=i\sigma (a)\mathrm{.}$

Substituting step 4.1 into step 3.1 yields $N_{K\mathrm{/}F}(a)={\prod }_{\sigma \in \mathrm{\Sigma }}\sigma (a{)}^i={({\prod }_{\sigma \in \mathrm{\Sigma }}\sigma (a))}^i,$ and ${\mathrm{Tr}}_{K\mathrm{/}F}(a)={\sum }_{\sigma \in \mathrm{\Sigma }}i\sigma (a)=i{\sum }_{\sigma \in \mathrm{\Sigma }}\sigma (a)\mathrm{.}$ This is the stated formula, with the separable case already proved in step 2.1.

If $\mathrm{char}F=p>0$ and $i>1$, then [L2] makes $i$ a positive power of $p$, so $i\cdot 1_F=0$. The trace formula of step 5.1 is then identically zero.