Abstract Algebra: Theory and Applications

Let ${\sigma }_i(a)=a$ and ${\sigma }_i(b)=b\text{.}$ Then

and

If $a\ne 0\text{,}$ then ${\sigma }_i(a^{-1})=[{\sigma }_i(a){]}^{-1}=a^{-1}\text{.}$ Finally, ${\sigma }_i(0)=0$ and ${\sigma }_i(1)=1$ since ${\sigma }_i$ is an automorphism.

Let $G=G(E/F)\text{.}$ Clearly, $F\subset E_G\subset E\text{.}$ Also, $E$ must be a splitting field of $E_G$ and $G(E/F)=G(E/E_G)\text{.}$ By Theorem 23.7,

Therefore, $[E_G:F]=1\text{.}$ Consequently, $E_G=F\text{.}$

Let $|G|=n\text{.}$ We must show that any set of $n+1$ elements ${\alpha }_1,\dots ,{\alpha }_{n+1}$ in $E$ is linearly dependent over $F\text{;}$ that is, we need to find elements $a_i\in F\text{,}$ not all zero, such that

Suppose that ${\sigma }_1=\mathrm{id},{\sigma }_2,\dots ,{\sigma }_n$ are the automorphisms in $G\text{.}$ The homogeneous system of linear equations

has more unknowns than equations. From linear algebra we know that this system has a nontrivial solution, say $x_i=a_i$ for $i=1,2,\dots ,n+1\text{.}$ Since ${\sigma }_1$ is the identity, the first equation translates to

The problem is that some of the $a_i$’s may be in $E$ but not in $F\text{.}$ We must show that this is impossible.

Suppose that at least one of the $a_i$’s is in $E$ but not in $F\text{.}$ By rearranging the ${\alpha }_i$’s we may assume that $a_1$ is nonzero. Since any nonzero multiple of a solution is also a solution, we can also assume that $a_1=1\text{.}$ Of all possible solutions fitting this description, we choose the one with the smallest number of nonzero terms. Again, by rearranging ${\alpha }_2,\dots ,{\alpha }_{n+1}$ if necessary, we can assume that $a_2$ is in $E$ but not in $F\text{.}$ Since $F$ is the subfield of $E$ that is fixed elementwise by $G\text{,}$ there exists a ${\sigma }_i$ in $G$ such that ${\sigma }_i(a_2)\ne a_2\text{.}$ Applying ${\sigma }_i$ to each equation in the system, we end up with the same homogeneous system, since $G$ is a group. Therefore, $x_1={\sigma }_i(a_1)=1\text{,}$ $x_2={\sigma }_i(a_2)\text{,}$ $\dots \text{,}$ $x_{n+1}={\sigma }_i(a_{n+1})$ is also a solution of the original system. We know that a linear combination of two solutions of a homogeneous system is also a solution; consequently,

must be another solution of the system. This is a nontrivial solution because ${\sigma }_i(a_2)\ne a_2\text{,}$ and has fewer nonzero entries than our original solution. This is a contradiction, since the number of nonzero solutions to our original solution was assumed to be minimal. We can therefore conclude that $a_1,\dots ,a_{n+1}\in F\text{.}$

Since $F=K_G\text{,}$ $G$ is a subgroup of $G(K/F)\text{.}$ Hence,

It follows that $G=G(K/F)\text{,}$ since they must have the same order.

(2) By Theorem Theorem 23.7, $|G(E/K)|=[E:K]\text{;}$ therefore,

Thus, $[K:F]=[G(E/F):G(E/K)]\text{.}$

Conversely, let $G(E/K)$ be a normal subgroup of $G(E/F)\text{.}$ We need to show that $F=K_{G(K/F)}\text{.}$ Let $\tau \in G(E/K)\text{.}$ For all $\sigma \in G(E/F)$ there exists a $\bar{\tau }\in G(E/K)$ such that $\tau \sigma =\sigma \bar{\tau }\text{.}$ Consequently, for all $\alpha \in K$

hence, $\sigma (\alpha )$ must be in the fixed field of $G(E/K)\text{.}$ Let $\bar{\sigma }$ be the restriction of $\sigma$ to $K\text{.}$ Then $\bar{\sigma }$ is an automorphism of $K$ fixing $F\text{,}$ since $\sigma (\alpha )\in K$ for all $\alpha \in K\text{;}$ hence, $\bar{\sigma }\in G(K/F)\text{.}$ Next, we will show that the fixed field of $G(K/F)$ is $F\text{.}$ Let $\beta$ be an element in $K$ that is fixed by all automorphisms in $G(K/F)\text{.}$ In particular, $\bar{\sigma }(\beta )=\beta$ for all $\sigma \in G(E/F)\text{.}$ Therefore, $\beta$ belongs to the fixed field $F$ of $G(E/F)\text{.}$

Finally, we must show that when $K$ is a normal extension of $F\text{,}$

For $\sigma \in G(E/F)\text{,}$ let ${\sigma }_K$ be the automorphism of $K$ obtained by restricting $\sigma$ to $K\text{.}$ Since $K$ is a normal extension, the argument in the preceding paragraph shows that ${\sigma }_K\in G(K/F)\text{.}$ Consequently, we have a map $\varphi :G(E/F)\to G(K/F)$ defined by $\sigma \mapsto {\sigma }_K\text{.}$ This map is a group homomorphism since

The kernel of $\varphi$ is $G(E/K)\text{.}$ By (2),

Hence, the image of $\varphi$ is $G(K/F)$ and $\varphi$ is onto. Applying the First Isomorphism Theorem, we have