Abstract Algebra

If $\{{\beta }_1,\dots ,{\beta }_s\}$ is a basis for $E∕F(\alpha )$, then $\{{\beta }_i{\alpha }^j\mid 1\le i\le s,0\le j\le d-1\}$ is a basis for $E∕F$. The matrix $A$ representing $m_{\alpha }$ with respect to this basis (with the lexicographical ordering on the pairs $(i,j)$) is block diagonal with $s$ blocks all equal to the matrix for $m_{\alpha }:F(\alpha )\to F(\alpha )$ for the ordered basis $\{1,\alpha ,\dots ,{\alpha }^{d-1}\}$. As we have an isomorphism of fields $F(\alpha )\cong \mathit{𝐹}[x]∕(f)$ fixing $F$ under which $\alpha$ is sent to the coset of $x$, the latter matrix is the companion matrix $A_f$.

By Lemma 8.10.10, we have $\mathrm{char}{m}_{\alpha }=f^s$. Lemma 8.7.21 tells us that

By expanding out the factorization of $f^s$ in $\overline{F}[x]$, we see that $N_{E∕F}\alpha$ and ${\mathrm{Tr}}_{E∕F}\alpha$ are as stated in this case. □

The distinct conjugates of $\alpha$ in a fixed algebraic closure $\overline{F}$ of $F$ are exactly the $\mathit{𝜏𝛼}$ for $\tau \in {\mathrm{Emb}}_F(F(\alpha ))$. These $\mathit{𝜏𝛼}$ are the distinct roots of the minimal polynomial of $\alpha$ over $F$, each occuring with multiplicity the degree ${[F(\alpha ):F]}_i$ of insparability of $F(\alpha )∕F$. Now, as in the proof of Lemma 6.10.17, each of these embeddings extends to ${[E:F(\alpha )]}_s$ distinct embeddings of $E$ into $\overline{F}$, and each extension $\sigma \in {\mathrm{Emb}}_F(E)$ of $\tau$ sends $\alpha$ to $\tau (\alpha )$. By Proposition 10.1.3, we have

and similarly for the trace. □

Since ${[K:F]}_i={[K:E]}_i{[E:F]}_i$, it suffices by Proposition 10.1.4 to show that

We extend each $\sigma$ to an automorphism $\tilde{\sigma }$ of $\overline{F}$ fixing $F$. We then have

For the trace map, we simply replace the products by sums.

Let $\sigma ,{\sigma }^{\prime }\in {\mathrm{Emb}}_F(E)$ and $\tau ,{\tau }^{\prime }\in {\mathrm{Emb}}_E(K)$, and suppose that

Since $\tilde{\sigma }\circ \tau {|}_E=\sigma {|}_E$, we have that $\sigma ={\sigma }^{\prime }$. Since $\tilde{\sigma }$ is an automorphism, we then apply its inverse to (10.1.2) to obtain $\tau ={\tau }^{\prime }$. As there are

terms of the product in (10.1.1), we have the result. □

Let $X$ be a set of linear characters $G\to E^{\times }$. Suppose by way of contradiction that $m\ge 1$ is minimal such that there $m$ distinct, linearly dependent elements of $G$. Choose $a_i\in E$ and ${\chi }_i\in X$ with $1\le i\le m$ for which

Also, let $h\in G$ be such that ${\chi }_1(h)\ne {\chi }_m(h)$. Set $b_i=a_i({\chi }_i(h)-{\chi }_m(h))$ for $1\le i\le m-1$. For any $g\in G$, we then have

Since $b_1\ne 0$ and ${\sum }_{i=1}^{m-1}{b}_i{\chi }_i$ has only $m-1$ terms, this contradicts the existence of $m$. □

Set $n=[E:F]$. Let $\beta \in E$, and note that

Next, suppose that $\alpha \in \ker N_{E∕F}$, and set

for $\gamma \in E$. The elements of $\mathrm{Gal}(E∕F)$, which is to say the powers of $\sigma$, are distinct $E$-valued characters on $E^{\times }$, and therefore they are $E$-linearly independent. Thus, there exists $\gamma \in E^{\times }$ such that $x_{\gamma }\ne 0$. We then note that

so $\alpha =\frac{\sigma (x_{\gamma }^{-1})}{x_{\gamma }^{-1}}$, finishing the proof. □

It suffices to show this in the case that the $v_i$ form a basis of $V$, since in this case, for any $w_1,\dots ,w_n\in V$ there exists a linear transformation $U:V\to V$ with $U(v_i)=w_i$ for all $i$. We then have

So, assume that the $v_i$ form a basis of $V$, and let $A=(a_{\mathit{𝑖𝑗}})$ denote the matrix of $T$ with respect to the ordered basis $(v_1,\dots ,v_n)$ of $V$. We have $T{v}_i={\sum }_{k=1}^n{a}_{\mathit{𝑘𝑖}}{v}_k$ for each $i$, and therefore we have

As matrices, we then have

and the result follows as $\det T=\det A=\det A^T$. □

Note that

so the matrix $({\mathrm{Tr}}_{E∕F}({\beta }_i{\beta }_j))$ equals $Q^{T}Q$, where $Q\in M_n(E)$ satisfies $Q_{\mathit{𝑖𝑗}}={\sigma }_i({\beta }_j)$. □

We work by induction on $n\ge 1$, the case $n=1$ asserting the obvious fact that $\det Q(\alpha )=1$ for any $\alpha \in F$. To compute the determinant of $Q=Q({\alpha }_1,\dots ,{\alpha }_n)$, in order of descending $i\le n-1$ subtract ${\alpha }_1$ times the $i$th column of $Q$ from the $(i+1)$th column, which leaves the determinant unchanged. We then obtain

and the result now follows by induction. □

Let ${\sigma }_1,{\sigma }_2,\dots ,{\sigma }_n$ be the embeddings of $E$ in a fixed algebraic closure of $F$, and set ${\alpha }_i={\sigma }_i(\alpha )$. Then ${\sigma }_i({\alpha }^{j-1})={\alpha }_i^{j-1}$, so Lemmas 10.2.9 and 10.2.11 tell us that

The latter term is just $D(f)$. □

Let ${\alpha }_1,\dots ,{\alpha }_n$ be the conjugates of $\alpha$ in an algebraic closure $\overline{F}$ of $F$. Then

so we have

for each $i$, and the conjugates of $f^{\prime }(\alpha )$ in $\overline{F}$ are the $f^{\prime }({\alpha }_i)$. We then have

□

Since $L∕F$ is separable, there exists $\alpha \in L$ such that $L=F(\alpha )$. Then $(1,\alpha ,\dots ,{\alpha }^{n-1})$ is an ordered basis of $L∕F$, and there exists an invertible $F$-linear transformation $T:L\to L$ with $T({\alpha }^{i-1})={\beta }_i$ for $1\le i\le n$. By Lemma 10.2.6, we have that

It follows Proposition 10.2.12 that $D(1,\alpha ,\dots ,{\alpha }^{n-1})\ne 0$, so we have the result. □

The minimal polynomial $f$ of $\alpha$ divides $x^d-{\alpha }^d$ but not $x^m-{\alpha }^m$ for any $m$ dividing $d$. If ${\mu }_d$ has order $d$, then $f$ is separable as $x^d-{\alpha }^d$ is. If ${\mu }_d$ has order $m$ for some $m$ properly dividing $d$, then $f$ divides $x^d-{\alpha }^d={(x^m-{\alpha }^m)}^{d∕m}$ but not $x^m-{\alpha }^m$ so is inseparable. Note that $\mathrm{char}F$ does not divide $d$ if and only if ${\mu }_d$ has order $d$, so we suppose for the remainder of the proof that this holds.

Any field embedding $\sigma$ of $E$ in $\overline{F}$ fixing $F$ must send $\alpha$ to $\mathit{𝜁𝛼}$ for some $\zeta \in {\mu }_d$. If ${\mu }_d\subset E$, then every such element lies in $E$, so $E∕F$ is Galois. Conversely, if $E∕F$ is Galois, then since $f$ does not divide any $x^m-{\alpha }^m$ with $m$ properly dividing $d$ and ${\mu }_d$ has order $d$, it has a root of the form $\mathit{𝜁𝛼}$ with $\zeta \in {\mu }_d$ of order $d$. Then $\zeta =(\mathit{𝜁𝛼})\cdot {\alpha }^{-1}\in E$, so ${\mu }_d\subset E$.

Finally, if $E∕F$ is Galois and ${\mu }_d\subset F$, then the map ${\chi }_a$ as defined in the statement is bijective by what we have already said, and it satisfies

for $\sigma ,\tau \in \mathrm{Gal}(E∕F)$, noting that $\sigma$ fixes ${\mu }_d$. □

Let $\zeta$ be a primitive $n$th root of unity in $F$. Note that $N_{E∕F}(\zeta )={\zeta }^n=1$, so Hilbert’s Theorem 90 tells us that there exists $\alpha \in E$ and a generator $\sigma$ of $\mathrm{Gal}(E∕F)$ with $\frac{\sigma (\alpha )}{\alpha }=\zeta$. Note that

so setting $a=-N_{E∕F}(-\alpha )$, we have ${\alpha }^n=a$. Since $\alpha$ has $n$ distinct conjugates in $E$, we have that $E=F(\alpha )$. □

Since ${\mu }_n\subset F$, Proposition 10.3.3 tells us that the map taking $a\in \mathrm{\Delta }$ to its Kummer cocycle ${\chi }_a$ yields an injection

This gives rise to the bimultiplicative Kummer pairing $⟨,⟩$, and it implies that any $a\in \mathrm{\Delta }∕F^{\times n}$ of order $d$ dividing $n$ pairs with some element of $\mathrm{Gal}(E∕F)$ to a $d$th root of unity.

We claim that $\psi$ is surjective. Let $\chi :\mathrm{Gal}(E∕F)\to {\mu }_n$ be a homomorphism, and let $H=\ker \chi$, which by the fundamental theorem of Galois theory corresponds to some cyclic extension $K∕F$ of degree dividing $n$. By Proposition 10.3.5, we have that $K=F(\alpha )$ for some $\alpha$ with $a={\alpha }^n\in \mathrm{\Delta }$, and then $\chi ={\chi }_a^k$ for some $k\ge 1$. That is, $\chi ={\chi }_b$ with $b=a^k\in \mathrm{\Delta }$, so $\psi (b)=\chi$. Since $\mathrm{\Delta }∕F^{\times n}$ is therefore finite of degree $[E:F]$, we have that the map

induced by the pairing is an isomorphism as well, and thus the Kummer pairing is perfect. □

If $K∕F$ is solvable by radicals, then there exists a field $K^{\prime }$ containing $F$ that is a radical extension of $F$. Suppose that $K^{\prime }=K_s^{\prime }$ where $K_0^{\prime }=F$ and $K_{i+1}^{\prime }=K_i^{\prime }(\sqrt[n_i]{{\alpha }_i})$ for some $n_i\ge 1$ with $n_i\nmid \mathrm{char}F$ and ${\alpha }_i\in K_i^{\prime }$ for $0\le i<s$. In fact, we may redefine the fields $K_i^{\prime }$ by setting $K_1^{\prime }=F({\mu }_m)$ where $m={\prod }_{i=1}^s{n}_i$ by replacing $K_j^{\prime }$ by $K_{j-1}^{\prime }({\mu }_m)$ for $1\le j\le s$ and defining it as such for $j=s+1$. Then each $K_{i+1}^{\prime }∕K_i^{\prime }$ for $1\le i\le s$ is a cyclic extension of exponent dividing $m$. Now set $K_i=K_i^{\prime }\cap K({\mu }_m)$ for each $1\le i\le s$, and note that $K_{i+1}∕K_i$ is then also cyclic of degree dividing $m$, while $K_1∕K_0$ is abelian. Thus the groups $\mathrm{Gal}(K∕K_i)$ for $1\le i\le s+1$ form a subnormal series in $\mathrm{Gal}(K∕F)$ with abelian composition facts, and $\mathrm{Gal}(K∕F)$ is solvable.

Conversely, if $\mathrm{Gal}(K∕F)$ is solvable, then we have intermediate fields $K_i$ with $K_0=F$, $K_s=K$, and $K_i\subset K_{i+1}$ such that $K_{i+1}∕F$ is Galois and $K_{i+1}∕K_i$ is cyclic of degree dividing $n=[K:F]$. Then $K_{i+1}({\mu }_n)∕K_i({\mu }_n)$ is also cyclic of degree dividing $n$, so there exists ${\alpha }_i\in K_i({\mu }_n)$ such that $K_{i+1}({\mu }_n)=K_i({\mu }_n)(\sqrt[n]{{\alpha }_i})$. Then $K({\mu }_n)∕F({\mu }_n)$ is a radical extension, and since $F({\mu }_n)∕F$ is given by adjoining an $n$th root of unity, $K({\mu }_n)∕F$ is also radical. As $K$ is contained in $K({\mu }_n)$, we conclude that $K∕F$ is solvable by radicals. □

We know that $\mathrm{Gal}(K∕F)$ is isomorphic to a subgroup of $S_n$ for $n$ equal to the degree of the polynomial defining $K$, and $S_n$ is solvable for $n\le 4$, so $\mathrm{Gal}(K∕F)$ is solvable as well. □

Note that we have a canonical isomorphism $K[x]{\otimes }_{K}M\underset{}{\overset{\sim }{\to }}M[x]$ that gives rise to the first map in the composition