Solvability by Radicals and Kummer Theory

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.

Multiplication operators compose as $m_{ab}=m_a\circ m_b$, because $(m_a\circ m_b)(x)=a(bx)=(ab)x$. Therefore [F1] and [L1] give $N_{K\mathrm{/}F}(ab)=\det (m_{ab})=\det (m_a)\det (m_b)=N_{K\mathrm{/}F}(a)N_{K\mathrm{/}F}(b)\mathrm{.}$

The operator identity $m_{a+b}=m_a+m_b$ and the scalar identity $m_{ca}=c{m}_a$ make trace additive and $F$-linear. Hence [F1] gives ${\mathrm{Tr}}_{K\mathrm{/}F}(a+b)={\mathrm{Tr}}_{K\mathrm{/}F}(a)+{\mathrm{Tr}}_{K\mathrm{/}F}(b),{\mathrm{Tr}}_{K\mathrm{/}F}(ca)=c{\mathrm{Tr}}_{K\mathrm{/}F}(a)\mathrm{.}$

Let $L\mathrm{/}K\mathrm{/}F$ be a finite tower, fix an algebraic closure $\mathrm{\Omega }\mathrm{/}F$, and write $i_{L\mathrm{/}F}$, $i_{L\mathrm{/}K}$, and $i_{K\mathrm{/}F}$ for the three inseparable degrees. For an $F$-embedding $\sigma :K\to \mathrm{\Omega }$, let ${\mathcal{E}}_{\sigma }$ be its restriction fibre in ${\mathrm{Hom}}_F(L,\mathrm{\Omega })$. Applying the embedding formulas [L2] to $L\mathrm{/}K$ after transporting scalars along $\sigma$ gives $\sigma (N_{L\mathrm{/}K}(a))={({\prod }_{\tau \in {\mathcal{E}}_{\sigma }}\tau (a))}^{i_{L\mathrm{/}K}},\sigma ({\mathrm{Tr}}_{L\mathrm{/}K}(a))=i_{L\mathrm{/}K}{\sum }_{\tau \in {\mathcal{E}}_{\sigma }}\tau (a)\mathrm{.}$ The fibres ${\mathcal{E}}_{\sigma }$ partition ${\mathrm{Hom}}_F(L,\mathrm{\Omega })$ by [L4].

By the ordinary tower law [L3] and separable-degree multiplicativity [L5], $i_{L\mathrm{/}F}=\frac{[L:F]}{[L:F{]}_s}=\frac{[L:K][K:F]}{[L:K{]}_s[K:F{]}_s}=i_{L\mathrm{/}K}{i}_{K\mathrm{/}F}\mathrm{.}$

Apply the outer embedding formulas [L2] for $K\mathrm{/}F$ to the two elements in step 1.3. Using the fibre partition and step 1.4 gives $N_{K\mathrm{/}F}(N_{L\mathrm{/}K}(a))={({\prod }_{\sigma }{\prod }_{\tau \in {\mathcal{E}}_{\sigma }}\tau (a{)}^{i_{L\mathrm{/}K}})}^{i_{K\mathrm{/}F}}={({\prod }_{\tau \in {\mathrm{Hom}}_F(L,\mathrm{\Omega })}\tau (a))}^{i_{L\mathrm{/}F}}=N_{L\mathrm{/}F}(a),$ and similarly ${\mathrm{Tr}}_{K\mathrm{/}F}({\mathrm{Tr}}_{L\mathrm{/}K}(a))=i_{K\mathrm{/}F}{i}_{L\mathrm{/}K}{\sum }_{\tau \in {\mathrm{Hom}}_F(L,\mathrm{\Omega })}\tau (a)={\mathrm{Tr}}_{L\mathrm{/}F}(a)\mathrm{.}$

Steps 1.1, 1.2, and 2.1 prove the three claims.

The two displayed identities are exactly the definitions of [F1], and [F2] identifies the determinant and trace on the right with the published operator notions. Since a field extension has positive degree, the zero-dimensional determinant convention never needs a separate case here.

This proves the stated dictionary identification.

For the forward implication from inseparability to degeneracy, suppose $K\mathrm{/}F$ is not separable. Then [L1] makes ${\mathrm{Tr}}_{K\mathrm{/}F}$ identically zero, so $T_{K\mathrm{/}F}(x,y)={\mathrm{Tr}}_{K\mathrm{/}F}(xy)=0$ for every $x,y\in K$. Thus every vector lies in both radicals, and the form is degenerate.

For the converse direction, suppose $K\mathrm{/}F$ is separable and let ${\sigma }_1,\dots ,{\sigma }_n\text{\hspace{0.17em}⁣}:K\to \mathrm{\Omega }$ be its distinct $F$-embeddings into an algebraic closure. Form the evaluation matrix $A=({\sigma }_i(e_j){)}_{i,j}$. By [L2], $A$ is invertible.

The matrix of $T_{K\mathrm{/}F}$ in the basis $e_1,\dots ,e_n$ is $B=({\mathrm{Tr}}_{K\mathrm{/}F}(e_i{e}_j){)}_{i,j}$. Because $K\mathrm{/}F$ is separable, [L1] gives ${\mathrm{Tr}}_{K\mathrm{/}F}(e_i{e}_j)={\sum }_{r=1}^n{\sigma }_r(e_i){\sigma }_r(e_j),$ so $B=A^{\mathsf{T}}A\mathrm{.}$

Since $A$ is invertible, the matrix $B=A^{\mathsf{T}}A$ is invertible. Therefore [F2] makes the trace form nondegenerate.

Steps 1.1 and 3.1 prove the equivalence.

By [L1], the trace form of $K\mathrm{/}F$ is nondegenerate. If the trace map itself were zero, then ${\mathrm{Tr}}_{K\mathrm{/}F}(xy)=0$ for every $x,y\in K$, so the vector $1\in K$ would lie in the radical, contradicting nondegeneracy. Therefore the trace map is not the zero linear functional.

The image of an $F$-linear map $K\to F$ is an $F$-subspace of the one-dimensional vector space $F$. A nonzero subspace of $F$ is all of $F$, so the trace map is surjective.

For the forward direction from 2 to 1, suppose $b=\beta \mathrm{/}\sigma (\beta )$ for some $\beta \in K^{\times }$. Since the embeddings of $K\mathrm{/}F$ are $1,\sigma ,\dots ,{\sigma }^{n-1}$, [L1] gives $N_{K\mathrm{/}F}(b)={\prod }_{i=0}^{n-1}\frac{{\sigma }^i(\beta )}{{\sigma }^{i+1}(\beta )}=1,$ the numerator and denominator cancelling cyclically because ${\sigma }^n=1$.

For the converse, assume $N_{K\mathrm{/}F}(b)=1$. For $0\le i<n$, set $c_i:={\prod }_{j=0}^{i-1}{\sigma }^j(b),$ with the empty product $c_0=1$. Then the distinct automorphisms $1,\sigma ,\dots ,{\sigma }^{n-1}$ restrict to distinct characters $K^{\times }\to K^{\times }$, so [L2] implies that the $K$-linear operator $T(x):={\sum }_{i=0}^{n-1}{c}_i{\sigma }^i(x)$ is not identically zero. Choose $x\in K$ with $\beta :=T(x)\mathrm{\ne }0$.

The coefficients satisfy $c_{i+1}=c_i{\sigma }^i(b)$ for $0\le i<n-1$, and $c_n=N_{K\mathrm{/}F}(b)=1$. Therefore $\sigma (c_i)=b^{-1}{c}_{i+1}(0\le i<n-1),\sigma (c_{n-1})=b^{-1}\mathrm{.}$ Applying $\sigma$ to $\beta =T(x)$ and re-indexing the sum gives $\sigma (\beta )=b^{-1}\beta \mathrm{.}$ Hence $b=\beta \mathrm{/}\sigma (\beta )$.

Steps 1.1 and 2.1 prove the equivalence.

For the forward direction from 2 to 1, suppose $b=\alpha -\sigma (\alpha )$. Summing the conjugates gives ${\mathrm{Tr}}_{K\mathrm{/}F}(b)={\sum }_{i=0}^{n-1}{\sigma }^i(\alpha -\sigma (\alpha ))={\sum }_{i=0}^{n-1}({\sigma }^i(\alpha )-{\sigma }^{i+1}(\alpha ))=0,$ again by telescoping and ${\sigma }^n=1$.

For the converse, assume ${\mathrm{Tr}}_{K\mathrm{/}F}(b)=0$. By [F1] and [L1], choose $c\in K$ with ${\mathrm{Tr}}_{K\mathrm{/}F}(c)=1$. Define $\alpha :={\sum }_{i=0}^{n-1}({\sum }_{j=0}^{i-1}{\sigma }^j(b)){\sigma }^i(c),$ where the inner sum is $0$ for $i=0$.

Put $s_i={\sum }_{j=0}^{i-1}{\sigma }^j(b)$, so $s_0=0$, $s_n={\mathrm{Tr}}_{K\mathrm{/}F}(b)=0$, and $s_i=b+\sigma (s_{i-1})$ for $1\le i\le n-1$, while $0=s_n=b+\sigma (s_{n-1})$. Applying $\sigma$ to the coefficients as well gives $\sigma (\alpha )={\sum }_{i=0}^{n-1}\sigma (s_i){\sigma }^{i+1}(c)=\sigma (s_{n-1})c+{\sum }_{i=1}^{n-1}\sigma (s_{i-1}){\sigma }^i(c)\mathrm{.}$ Therefore every coefficient of $\alpha -\sigma (\alpha )$ is $b$, so $\alpha -\sigma (\alpha )=b{\sum }_{i=0}^{n-1}{\sigma }^i(c)=b{\mathrm{Tr}}_{K\mathrm{/}F}(c)=b\mathrm{.}$

Steps 1.1 and 2.1 prove the equivalence.

For the forward direction, assume $K\mathrm{/}F$ is cyclic and choose a generator $\sigma$ of $\mathrm{Gal}(K\mathrm{/}F)$. The distinct powers $1,\sigma ,\dots ,{\sigma }^{n-1}$ are distinct characters $K^{\times }\to K^{\times }$, so [L1] makes the resolvent operator $R_{\sigma ,\zeta }$ of [F1] nonzero. Choose $\gamma \in K$ with $\alpha :=R_{\sigma ,\zeta }(\gamma )\mathrm{\ne }0$.

For the converse direction, assume $K=F(\alpha )$, ${\alpha }^n=a\in F$, and $x^n-a$ is irreducible. If $\alpha =0$, then $a=0$ and the irreducible polynomial is $x^n$, which forces $n=1$. Hence $K=F$, so the extension is cyclic. Assume from now on that $\alpha \mathrm{\ne }0$. Then $[K:F]=n$, and [L2] shows every root of $x^n-a$ is ${\zeta }^i\alpha$ with ${\zeta }^i\in F$. Because [L3] makes the $n$ roots of $x^n-1$ distinct, these roots ${\zeta }^i\alpha$ are distinct as well. Hence all roots of $x^n-a$ already lie in $K$, and the polynomial is separable, so $K$ is its splitting field and [L4] makes $K\mathrm{/}F$ Galois.

In the nonzero case of step 1.2, the root $\zeta \alpha$ of the irreducible polynomial $x^n-a$ determines, by [L5], an $F$-embedding $K=F(\alpha )\to K$ sending $\alpha$ to $\zeta \alpha$, hence an automorphism $\tau$ of $K\mathrm{/}F$. Its powers send $\alpha$ to ${\zeta }^i\alpha$, so $\tau$ has order $n$. Because $[K:F]=n$ and a finite Galois extension has at most $[K:F]$ automorphisms, $\mathrm{Gal}(K\mathrm{/}F)=⟨\tau ⟩$ is cyclic of order $n$.

Because $\sigma (\alpha )={\sum }_{i=0}^{n-1}{\zeta }^{-i}{\sigma }^{i+1}(\gamma )=\zeta \alpha ,$ the element ${\alpha }^n$ is fixed by $\sigma$ and hence by the whole cyclic Galois group, so ${\alpha }^n\in F$. If $0<d<n$ and ${\alpha }^d\in F$, then ${\alpha }^d=\sigma ({\alpha }^d)={\zeta }^d{\alpha }^d,$ so ${\zeta }^d=1$, contradicting primitivity of $\zeta$. Therefore no smaller positive power of $\alpha$ lies in $F$. The minimal polynomial of $\alpha$ divides $x^n-{\alpha }^n$ and has degree $n=[K:F]$, so it is exactly $x^n-{\alpha }^n$ and $K=F(\alpha )$.

Steps 2.2 and 2.1 prove the equivalence.

For the forward direction, assume $K\mathrm{/}F$ is cyclic and choose a generator $\sigma$ of its Galois group. Since $[K:F]=p$ and $\mathrm{char}F=p$, one has ${\mathrm{Tr}}_{K\mathrm{/}F}(-1)=-p=0.$ By [L1], choose $\alpha \in K$ with $-1=\alpha -\sigma (\alpha ),$ so $\sigma (\alpha )=\alpha +1$.

For the converse direction, assume ${\alpha }^p-\alpha =a\in F$ and $x^p-x-a$ is irreducible. For each $i\in {\mathbb{F}}_p$ one has $(\alpha +i{)}^p-(\alpha +i)={\alpha }^p-\alpha =a,$ so the roots of the polynomial are exactly $\alpha +i$ for $i\in {\mathbb{F}}_p$. Thus all roots lie in $F(\alpha )$, the derivative is $-1\mathrm{\ne }0$, and [L2] makes $F(\alpha )\mathrm{/}F$ Galois. Because the polynomial is irreducible of degree $p$, this extension has degree $p$.

In characteristic $p$, one has $(\alpha +1{)}^p-(\alpha +1)={\alpha }^p-\alpha ,$ so the element $a:={\alpha }^p-\alpha$ is fixed by $\sigma$ and therefore lies in $F$. Also $\alpha \mathrm{\notin }F$ because step 1.1 gives $\sigma (\alpha )\mathrm{\ne }\alpha$. Since $[K:F]=p$ is prime and $F\subseteq F(\alpha )\subseteq K,$ one must have $F(\alpha )=K$. The polynomial $x^p-x-a\in F[x]$ has root $\alpha$ and degree $p=[F(\alpha ):F]$, so it is the minimal polynomial of $\alpha$ over $F$ and is irreducible.

The rule $\tau (\alpha )=\alpha +1$ permutes the root set and fixes $F$, so it extends to an $F$-automorphism of $F(\alpha )$. Its $p$-th power fixes $\alpha$ and each smaller positive power moves $\alpha$, so $\tau$ has order $p$. A degree-$p$ finite Galois extension has at most $p$ automorphisms, hence exactly the cyclic group generated by $\tau$. Therefore $F(\alpha )\mathrm{/}F$ is cyclic of degree $p$.

Steps 2.1 and 2.2 prove the equivalence, and the displayed root set in step 1.2 proves the final sentence.