##### Proposition23.13

Let \(\{\sigma_i : i \in I \}\) be a collection of automorphisms of a field \(F\). Then \begin{equation*}F_{\{\sigma_i \}} = \{ a \in F : \sigma_i(a) = a \text{ for all } \sigma_i \}\end{equation*} is a subfield of \(F\).

The goal of this section is to prove the Fundamental Theorem of Galois Theory. This theorem explains the connection between the subgroups of \(G(E/F)\) and the intermediate fields between \(E\) and \(F\).

Let \(\{\sigma_i : i \in I \}\) be a collection of automorphisms of a field \(F\). Then \begin{equation*}F_{\{\sigma_i \}} = \{ a \in F : \sigma_i(a) = a \text{ for all } \sigma_i \}\end{equation*} is a subfield of \(F\).

Let \(F\) be a field and let \(G\) be a subgroup of \(\aut(F)\). Then \begin{equation*}F_G = \{ \alpha \in F : \sigma( \alpha ) = \alpha \text{ for all } \sigma \in G \}\end{equation*} is a subfield of \(F\).

The subfield \(F_{ \{\sigma_i \} }\) of \(F\) is called the *fixed field* of \(\{ \sigma_i \}\). The field fixed for a subgroup \(G\) of \(\aut(F)\) will be denoted by \(F_G\).

Let \(\sigma : {\mathbb Q}(\sqrt{3}, \sqrt{5}\, ) \rightarrow {\mathbb Q}(\sqrt{3}, \sqrt{5}\, )\) be the automorphism that maps \(\sqrt{3}\) to \(-\sqrt{3}\). Then \({\mathbb Q}( \sqrt{5}\, )\) is the subfield of \({\mathbb Q}(\sqrt{3}, \sqrt{5}\, )\) left fixed by \(\sigma\).

Let \(E\) be a splitting field over \(F\) of a separable polynomial. Then \(E_{G(E/F)} = F\).

A large number of mathematicians first learned Galois theory from Emil Artin's monograph on the subject [1]. The very clever proof of the following lemma is due to Artin.

Let \(G\) be a finite group of automorphisms of \(E\) and let \(F = E_G\). Then \([E:F] \leq |G|\).

Let \(E\) be an algebraic extension of \(F\). If every irreducible polynomial in \(F[x]\) with a root in \(E\) has all of its roots in \(E\), then \(E\) is called a *normal extension* of \(F\); that is, every irreducible polynomial in \(F[x]\) containing a root in \(E\) is the product of linear factors in \(E[x]\).

Let \(E\) be a field extension of \(F\). Then the following statements are equivalent.

\(E\) is a finite, normal, separable extension of \(F\).

\(E\) is a splitting field over \(F\) of a separable polynomial.

\(F = E_G\) for some finite group \(G\) of automorphisms of \(E\).

Let \(K\) be a field extension of \(F\) such that \(F = K_G\) for some finite group of automorphisms \(G\) of \(K\). Then \(G = G(K/F)\).

Before we determine the exact correspondence between field extensionsand automorphisms of fields, let us return to a familiar example.

In Example 23.4 we examined the automorphisms of \({\mathbb Q}( \sqrt{3}, \sqrt{5}\, )\) fixing \({\mathbb Q}\). Figure 23.21 compares the lattice of field extensions of \({\mathbb Q}\) with the lattice of subgroups of \(G( {\mathbb Q}( \sqrt{3}, \sqrt{5}\, ) /{\mathbb Q})\). The Fundamental Theorem of Galois Theory tells us what the relationship is between the two lattices.

We are now ready to state and prove the Fundamental Theorem of Galois Theory.

Let \(F\) be a finite field or a field of characteristic zero. If \(E\) is a finite normal extension of \(F\) with Galois group \(G(E/F)\), then the following statements are true.

The map \(K \mapsto G(E/K)\) is a bijection of subfields \(K\) of \(E\) containing \(F\) with the subgroups of \(G(E/F)\).

If \(F \subset K \subset E\), then \begin{equation*}[E:K] = |G(E/K)| \text{ and } [K:F] = [G(E/F):G(E/K)].\end{equation*}

\(F \subset K \subset L \subset E\) if and only if \(\{ \identity \} \subset G(E/L) \subset G(E/K) \subset G(E/F)\).

\(K\) is a normal extension of \(F\) if and only if \(G(E/K)\) is a normal subgroup of \(G( E/F)\). In this case \begin{equation*}G(K/F) \cong G(E/F) / G( E/K ).\end{equation*}

In this example we will illustrate the Fundamental Theorem of Galois Theory by determining the lattice of subgroups of the Galois group of \(f(x) = x^4 - 2\). We will compare this lattice to the lattice of field extensions of \({\mathbb Q}\) that are contained in the splitting field of \(x^4-2\). The splitting field of \(f(x)\) is \({\mathbb Q}( \sqrt[4]{2}, i )\). To see this, notice that \(f(x)\) factors as \((x^2 + \sqrt{2}\, )(x^2 - \sqrt{2}\, )\); hence, the roots of \(f(x)\) are \(\pm \sqrt[4]{2}\) and \(\pm \sqrt[4]{2}\, i\). We first adjoin the root \(\sqrt[4]{2}\) to \({\mathbb Q}\) and then adjoin the root \(i\) of \(x^2 + 1\) to \({\mathbb Q}(\sqrt[4]{2}\, )\). The splitting field of \(f(x)\) is then \({\mathbb Q}(\sqrt[4]{2}\, )(i) = {\mathbb Q}( \sqrt[4]{2}, i )\).

Since \([ {\mathbb Q}( \sqrt[4]{2}\, ) : {\mathbb Q}] = 4\) and \(i\) is not in \({\mathbb Q}( \sqrt[4]{2}\, )\), it must be the case that \([ {\mathbb Q}( \sqrt[4]{2}, i ): {\mathbb Q}(\sqrt[4]{2}\, )] = 2\). Hence, \([ {\mathbb Q}( \sqrt[4]{2}, i ):{\mathbb Q}] = 8\). The set \begin{equation*}\{ 1, \sqrt[4]{2}, (\sqrt[4]{2}\, )^2, (\sqrt[4]{2}\, )^3, i, i \sqrt[4]{2}, i (\sqrt[4]{2}\, )^2, i(\sqrt[4]{2}\, )^3 \}\end{equation*} is a basis of \({\mathbb Q}( \sqrt[4]{2}, i )\) over \({\mathbb Q}\). The lattice of field extensions of \({\mathbb Q}\) contained in \({\mathbb Q}( \sqrt[4]{2}, i)\) is illustrated in Figure 23.25(a).

The Galois group \(G\) of \(f(x)\) must be of order 8. Let \(\sigma\) be the automorphism defined by \(\sigma( \sqrt[4]{2}\, ) = i \sqrt[4]{2}\) and \(\sigma( i ) = i\), and \(\tau\) be the automorphism defined by complex conjugation; that is, \(\tau(i ) = -i\). Then \(G\) has an element of order 4 and an element of order 2. It is easy to verify by direct computation that the elements of \(G\) are \(\{ \identity, \sigma, \sigma^2, \sigma^3, \tau, \sigma \tau, \sigma^2 \tau, \sigma^3 \tau \}\) and that the relations \(\tau^2 = \identity\), \(\sigma^4 = \identity\), and \(\tau \sigma \tau = \sigma^{-1}\) are satisfied; hence, \(G\) must be isomorphic to \(D_4\). The lattice of subgroups of \(G\) is illustrated in Figure 23.25(b).

Solutions for the cubic and quartic equations were discovered in the 1500s. Attempts to find solutions for the quintic equations puzzled some of history's best mathematicians. In 1798, P. Ruffini submitted a paper that claimed no such solution could be found; however, the paper was not well received. In 1826, Niels Henrik Abel (1802–1829) finally offered the first correct proof that quintics are not always solvable by radicals.

Abel inspired the work of Évariste Galois. Born in 1811, Galois began to display extraordinary mathematical talent at the age of 14. He applied for entrance to the École Polytechnique several times; however, he had great difficulty meeting the formal entrance requirements, and the examiners failed to recognize his mathematical genius. He was finally accepted at the École Normale in 1829.

Galois worked to develop a theory of solvability for polynomials. In 1829, at the age of 17, Galois presented two papers on the solution of algebraic equations to the Académie des Sciences de Paris. These papers were sent to Cauchy, who subsequently lost them. A third paper was submitted to Fourier, who died before he could read the paper. Another paper was presented, but was not published until 1846.

Galois' democratic sympathies led him into the Revolution of 1830. He was expelled from school and sent to prison for his part in the turmoil. After his release in 1832, he was drawn into a duel possibly over a love affair. Certain that he would be killed, he spent the evening before his death outlining his work and his basic ideas for research in a long letter to his friend Chevalier. He was indeed dead the next day, at the age of 20.