Abstract Algebra: Theory and Applications, Remixed!

Proof.

Let \(\sigma_i(a) = a\) and \(\sigma_i(b)=b\text{.}\) Then

and

If \(a \neq 0\text{,}\) 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.

Proof.

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 9.15,

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

Proof.

Let \(|G| = n\text{.}\) We must show that any set of \(n + 1\) elements \(\alpha_1, \ldots, \alpha_{n + 1}\) 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 \(\sigma_1 = \identity, \sigma_2, \ldots, \sigma_n\) 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 \(i = 1, 2, \ldots, n + 1\text{.}\) 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 \(\alpha_2, \ldots, \alpha_{n + 1}\) 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 \(\sigma_i( a_2 ) \neq a_2\text{.}\) Applying \(\sigma_i\) to each equation in the system, we end up with the same homogeneous system, since \(G\) is a group. Therefore, \(x_1 = \sigma_i(a_1) = 1\text{,}\) \(x_2 = \sigma_i(a_2)\text{,}\) \(\ldots\text{,}\) \(x_{n + 1} = \sigma_i(a_{n+1} )\) 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 \(\sigma_i( a_2 ) \neq a_2\text{,}\) 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 \(a_1, \ldots, a_{n + 1} \in F\text{.}\)

Proof.

(1) \(\Rightarrow\) (2). Let \(E\) be a finite, normal, separable extension of \(F\text{.}\) By the Primitive Element Theorem, we can find an \(\alpha\) in \(E\) such that \(E = F(\alpha)\text{.}\) Let \(f(x)\) be the minimal polynomial of \(\alpha\) over \(F\text{.}\) The field \(E\) must contain all of the roots of \(f(x)\) since it is a normal extension \(F\text{;}\) hence, \(E\) is a splitting field for \(f(x)\text{.}\)

(2) \(\Rightarrow\) (3). Let \(E\) be the splitting field over \(F\) of a separable polynomial. By Proposition 9.21, \(E_{G(E/F)} = F\text{.}\) Since \(| G(E/F)| = [E:F]\text{,}\) this is a finite group.

(3) \(\Rightarrow\) (1). Let \(F = E_G\) for some finite group of automorphisms \(G\) of \(E\text{.}\) Since \([E:F] \leq |G|\text{,}\) \(E\) is a finite extension of \(F\text{.}\) To show that \(E\) is a finite, normal extension of \(F\text{,}\) let \(f(x) \in F[x]\) be an irreducible monic polynomial that has a root \(\alpha\) in \(E\text{.}\) We must show that \(f(x)\) is the product of distinct linear factors in \(E[x]\text{.}\) By Proposition 9.13, automorphisms in \(G\) permute the roots of \(f(x)\) lying in \(E\text{.}\) Hence, if we let \(G\) act on \(\alpha\text{,}\) we can obtain distinct roots \(\alpha_1 = \alpha, \alpha_2, \ldots, \alpha_n\) in \(E\text{.}\) Let \(g(x) = \prod_{i = 1}^{n} (x -\alpha_i)\text{.}\) Then \(g(x)\) is separable over \(F\) and \(g( \alpha ) = 0\text{.}\) Any automorphism \(\sigma\) in \(G\) permutes the factors of \(g(x)\) since it permutes these roots; hence, when \(\sigma\) acts on \(g(x)\text{,}\) it must fix the coefficients of \(g(x)\text{.}\) Therefore, the coefficients of \(g(x)\) must be in \(F\text{.}\) Since \(\deg g(x) \leq \deg f(x)\) and \(f(x)\) is the minimal polynomial of \(\alpha\text{,}\) \(f(x) = g(x)\text{.}\)

Proof.

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.

Proof.

(1) Suppose that \(G(E/K) = G(E/L) = G\text{.}\) Both \(K\) and \(L\) are fixed fields of \(G\text{;}\) hence, \(K=L\) and the map defined by \(K \mapsto G(E/K)\) is one-to-one. To show that the map is onto, let \(G\) be a subgroup of \(G(E/F)\) and \(K\) be the field fixed by \(G\text{.}\) Then \(F \subset K \subset E\text{;}\) consequently, \(E\) is a normal extension of \(K\text{.}\) Thus, \(G(E/K) = G\) and the map \(K \mapsto G(E/K)\) is a bijection.

(2) By Theorem Theorem 9.15, \(|G(E/K)| = [E:K]\text{;}\) therefore,

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

Statement (3) is illustrated in Figure 9.29. We leave the proof of this property as an exercise.

(4) This part takes a little more work. Let \(K\) be a normal extension of \(F\text{.}\) If \(\sigma\) is in \(G(E/F)\) and \(\tau\) is in \(G(E/K)\text{,}\) we need to show that \(\sigma^{-1} \tau \sigma\) is in \(G(E/K)\text{;}\) that is, we need to show that \(\sigma^{-1} \tau \sigma( \alpha) = \alpha\) for all \(\alpha \in K\text{.}\) Suppose that \(f(x)\) is the minimal polynomial of \(\alpha\) over \(F\text{.}\) Then \(\sigma( \alpha )\) is also a root of \(f(x)\) lying in \(K\text{,}\) since \(K\) is a normal extension of \(F\text{.}\) Hence, \(\tau( \sigma( \alpha )) = \sigma( \alpha )\) or \(\sigma^{-1} \tau \sigma( \alpha) = \alpha\text{.}\)

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 \(\overline{\tau} \in G(E/K)\) such that \(\tau \sigma = \sigma \overline{\tau}\text{.}\) Consequently, for all \(\alpha \in K\)

hence, \(\sigma( \alpha )\) must be in the fixed field of \(G(E/K)\text{.}\) Let \(\overline{\sigma}\) be the restriction of \(\sigma\) to \(K\text{.}\) Then \(\overline{\sigma}\) is an automorphism of \(K\) fixing \(F\text{,}\) since \(\sigma( \alpha ) \in K\) for all \(\alpha \in K\text{;}\) hence, \(\overline{\sigma} \in G(K/F)\text{.}\) 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, \(\overline{\sigma}(\beta) = \beta\) 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 \(\phi:G(E/F) \rightarrow G(K/F)\) defined by \(\sigma \mapsto \sigma_K\text{.}\) This map is a group homomorphism since

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

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