
Section23.2The Fundamental Theorem

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 $F\text{.}$

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

\begin{equation*} \sigma_i(a \pm b) = \sigma_i(a) \pm \sigma_i(b) = a \pm b \end{equation*}

and

\begin{equation*} \sigma_i(a b) = \sigma_i(a) \sigma_i(b) = a b. \end{equation*}

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.

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

Example23.15

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

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,

\begin{equation*} |G| = [E: E_G] =[ E:F]. \end{equation*}

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

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 $|G| = n\text{.}$ 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

\begin{equation*} a_1 \alpha_1 + a_2 \alpha_2 + \cdots + a_{n + 1} \alpha_{n + 1} = 0. \end{equation*}

Suppose that $\sigma_1 = \identity, \sigma_2, \ldots, \sigma_n$ are the automorphisms in $G\text{.}$ The homogeneous system of linear equations

\begin{align*} \sigma_1( \alpha_1 ) x_1 + \sigma_1(\alpha_2) x_2 + \cdots + \sigma_1(\alpha_{n + 1} ) x_{n + 1} & = 0\\ \sigma_2( \alpha_1 ) x_1 + \sigma_2(\alpha_2) x_2 + \cdots + \sigma_2(\alpha_{n + 1} ) x_{n + 1} & = 0\\ & \vdots &\\ \sigma_n( \alpha_1 ) x_1 + \sigma_n(\alpha_2) x_2 + \cdots + \sigma_n(\alpha_{n + 1} ) x_{n + 1} & = 0 \end{align*}

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

\begin{equation*} a_1 \alpha_1 + a_2 \alpha_2 + \cdots + a_{n + 1} \alpha_{n + 1} = 0. \end{equation*}

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,

\begin{align*} x_1 & = 1 -1 = 0\\ x_2 & = a_2 - \sigma_i(a_2)\\ & \vdots &\\ x_{n + 1} & = a_{n + 1} - \sigma_i(a_{n + 1}) \end{align*}

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 $a_1, \ldots, a_{n + 1} \in F\text{.}$

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

(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 23.16, $E_{G(E/F)} = F\text{.}$ 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 23.5, 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 $\alpha\text{,}$ $f(x) = g(x)\text{.}$

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

\begin{equation*} [K : F ] \leq |G| \leq |G(K/F)| = [K:F]. \end{equation*}

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

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

Example23.20

In Example 23.4 we examined the automorphisms of ${\mathbb Q}( \sqrt{3}, \sqrt{5}\, )$ fixing ${\mathbb Q}\text{.}$ Figure 23.21 compares the lattice of field extensions of ${\mathbb Q}$ with the lattice of subgroups of $G( {\mathbb Q}( \sqrt{3}, \sqrt{5}\, ) /{\mathbb Q})\text{.}$ 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.

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

\begin{equation*} |G(E/F)| = [G(E/F):G(E/K)] \cdot |G(E/K)| = [E:F] = [E:K][K:F]. \end{equation*}

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

(3) Statement (3) is illustrated in Figure 23.23. 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 $\alpha \in K$

\begin{equation*} \tau( \sigma( \alpha ) ) = \sigma( \overline{\tau}( \alpha ) ) = \sigma( \alpha ); \end{equation*}

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 $F\text{,}$

\begin{equation*} G(K/F) \cong G(E/F) / G(E/K). \end{equation*}

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

\begin{equation*} \phi( \sigma \tau ) = (\sigma \tau)_K = \sigma_K \tau_K = \phi( \sigma) \phi( \tau ). \end{equation*}

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

\begin{equation*} |G(E/F)| / |G(E/K)| = [K:F] = |G(K/F)|. \end{equation*}

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

\begin{equation*} G(K/F) \cong G(E/F) / G( E/K ). \end{equation*}
Example23.24

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

Since $[ {\mathbb Q}( \sqrt[4]{2}\, ) : {\mathbb Q}] = 4$ and $i$ is not in ${\mathbb Q}( \sqrt[4]{2}\, )\text{,}$ it must be the case that $[ {\mathbb Q}( \sqrt[4]{2}, i ): {\mathbb Q}(\sqrt[4]{2}\, )] = 2\text{.}$ Hence, $[ {\mathbb Q}( \sqrt[4]{2}, i ):{\mathbb Q}] = 8\text{.}$ 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 ${\mathbb Q}\text{.}$ 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 $\sigma( i ) = i\text{,}$ and $\tau$ be the automorphism defined by complex conjugation; that is, $\tau(i ) = -i\text{.}$ 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 $\tau^2 = \identity\text{,}$ $\sigma^4 = \identity\text{,}$ and $\tau \sigma \tau = \sigma^{-1}$ are satisfied; hence, $G$ must be isomorphic to $D_4\text{.}$ The lattice of subgroups of $G$ is illustrated in Figure 23.25(b).

SubsectionHistorical Note

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.