The 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 \(F\).

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

Example

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 \(\sigma\).

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.

Lemma

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

Proof

Let \(|G| = n\). We must show that any set of \(n + 1\) elements \(\alpha_1, \ldots, \alpha_{n + 1}\) in \(E\) is linearly dependent over \(F\); that is, we need to find elements \(a_i \in F\), not all zero, such that \[\begin{aligned}\end{aligned}\]. Suppose that \(\sigma_1 = \identity, \sigma_2, \ldots, \sigma_n\) are the automorphisms in \(G\). The homogeneous system of linear equations \[\begin{aligned}\sigma_1( \alpha_1 ) x_1 + \sigma_1(\alpha_2) x_2 + \cdots + \sigma_1(\alpha_{n + 1} ) x_{n + 1} & = 0 \\ \sigma_2( \alpha_1 ) x_1 + \sigma_2(\alpha_2) x_2 + \cdots + \sigma_2(\alpha_{n + 1} ) x_{n + 1} & = 0 \\ & \vdots & \\ \sigma_n( \alpha_1 ) x_1 + \sigma_n(\alpha_2) x_2 + \cdots + \sigma_n(\alpha_{n + 1} ) x_{n + 1} & = 0\end{aligned}\] has more unknowns than equations. From linear algebra we know that this system has a nontrivial solution, say \(x_i = a_i\) for \(i = 1, 2, \ldots, n + 1\). Since \(\sigma_1\) is the identity, the first equation translates to \[\begin{aligned}\end{aligned}\]. The problem is that some of the \(a_i\)'s may be in \(E\) but not in \(F\). We must show that this is impossible.

Suppose that at least one of the \(a_i\)'s is in \(E\) but not in \(F\). 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 \(a_1 = 1\). 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 \(F\). Since \(F\) is the subfield of \(E\) that is fixed elementwise by \(G\), there exists a \(\sigma_i\) in \(G\) such that \(\sigma_i( a_2 ) \neq a_2\). Applying \(\sigma_i\) to each equation in the system, we end up with the same homogeneous system, since \(G\) is a group. Therefore, \(x_1 = \sigma_i(a_1) = 1\), \(x_2 = \sigma_i(a_2)\), \(\ldots\), \(x_{n + 1} = \sigma_i(a_{n+1} )\) 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, \[\begin{aligned}x_1 & = 1 -1 = 0 \\ x_2 & = a_2 - \sigma_i(a_2) \\ & \vdots & \\ x_{n + 1} & = a_{n + 1} - \sigma_i(a_{n + 1})\end{aligned}\] must be another solution of the system. This is a nontrivial solution because \(\sigma_i( a_2 ) \neq a_2\), 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 \(a_1, \ldots, a_{n + 1} \in F\).

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 \(E[x]\).

Theorem

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 \(E\).

Proof

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

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

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

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

Example

In we examined the automorphisms of \({\mathbb Q}( \sqrt{3}, \sqrt{5}\, )\) fixing \({\mathbb Q}\). 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.

Theorem: 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 \[\begin{aligned}\end{aligned}\].
\(F \subset K \subset L \subset E\) 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 \[\begin{aligned}\end{aligned}\].

Proof

(1) Suppose that \(G(E/K) = G(E/L) = G\). Both \(K\) and \(L\) are fixed fields of \(G\); 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 \(G\). Then \(F \subset K \subset E\); consequently, \(E\) is a normal extension of \(K\). Thus, \(G(E/K) = G\) and the map \(K \mapsto G(E/K)\) is a bijection.

(2) By Theorem , \(|G(E/K)| = [E:K]\); therefore, \[\begin{aligned}\end{aligned}\]. Thus, \([K:F] = [G(E/F):G(E/K)]\).

Statement (3) is illustrated in . 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 \(F\). If \(\sigma\) is in \(G(E/F)\) and \(\tau\) is in \(G(E/K)\), we need to show that \(\sigma^{-1} \tau \sigma\) is in \(G(E/K)\); that is, we need to show that \(\sigma^{-1} \tau \sigma( \alpha) = \alpha\) for all \(\alpha \in K\). Suppose that \(f(x)\) is the minimal polynomial of \(\alpha\) over \(F\). Then \(\sigma( \alpha )\) is also a root of \(f(x)\) lying in \(K\), since \(K\) is a normal extension of \(F\). Hence, \(\tau( \sigma( \alpha )) = \sigma( \alpha )\) or \(\sigma^{-1} \tau \sigma( \alpha) = \alpha\).

Conversely, let \(G(E/K)\) be a normal subgroup of \(G(E/F)\). We need to show that \(F = K_{G(K/F)}\). Let \(\tau \in G(E/K)\). For all \(\sigma \in G(E/F)\) there exists a \(\overline{\tau} \in G(E/K)\) such that \(\tau \sigma = \sigma \overline{\tau}\). Consequently, for all \(\alpha \in K\) \[\begin{aligned}\end{aligned}\] hence, \(\sigma( \alpha )\) must be in the fixed field of \(G(E/K)\). Let \(\overline{\sigma}\) be the restriction of \(\sigma\) to \(K\). Then \(\overline{\sigma}\) is an automorphism of \(K\) fixing \(F\), since \(\sigma( \alpha ) \in K\) for all \(\alpha \in K\); hence, \(\overline{\sigma} \in G(K/F)\). Next, we will show that the fixed field of \(G(K/F)\) is \(F\). Let \(\beta\) be an element in \(K\) that is fixed by all automorphisms in \(G(K/F)\). In particular, \(\overline{\sigma}(\beta) = \beta\) for all \(\sigma \in G(E/F)\). Therefore, \(\beta\) belongs to the fixed field \(F\) of \(G(E/F)\).

Finally, we must show that when \(K\) is a normal extension of \(F\), \[\begin{aligned}\end{aligned}\]. For \(\sigma \in G(E/F)\), let \(\sigma_K\) be the automorphism of \(K\) obtained by restricting \(\sigma\) to \(K\). Since \(K\) is a normal extension, the argument in the preceding paragraph shows that \(\sigma_K \in G( K/F)\). Consequently, we have a map \(\phi:G(E/F) \rightarrow G(K/F)\) defined by \(\sigma \mapsto \sigma_K\). This map is a group homomorphism since \[\begin{aligned}\end{aligned}\]. The kernel of \(\phi\) is \(G(E/K)\). By (2), \[\begin{aligned}\end{aligned}\]. Hence, the image of \(\phi\) is \(G(K/F)\) and \(\phi\) is onto. Applying the First Isomorphism Theorem, we have \[\begin{aligned}\end{aligned}\].

Example

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 \[\begin{aligned}\end{aligned}\] 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 (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 (b).

Historical 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 (18021829) 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.

Symbols used here

\prod_{k=1}^{n} a_k

product
Multiply a_k for k = 1 up to n.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

x \in A,\ A \subseteq B

element of, subset
x belongs to A; every element of A is in B.

\neg,\ \wedge,\ \vee,\ \Rightarrow,\ \Leftrightarrow

not, and, or, implies, iff
Logical connectives.

\sigma,\ s,\ \sigma^2

standard deviation, sample s.d., variance
Typical distance from the mean; its square.

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

i

imaginary unit
i² = −1.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\neq

not equal
The two sides are different.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

\blacksquare\ \text{or}\ \square

end of proof (halmos)
Marks the point where the statement has been established.

a \equiv b \pmod n

congruent modulo n
n divides a − b; a and b have the same remainder.

a \mid b,\ \gcd(a,b)

divides, greatest common divisor
b is a multiple of a; the largest number dividing both.

(G, \cdot),\ e,\ g^{-1}

group, identity, inverse
A set with an operation; the do-nothing element; the element that undoes g.

G \cong H,\ G / N

isomorphic, quotient group
Same structure; the group of cosets of a normal subgroup N.

Questions people ask

What is a group, in plain words?

A set with one operation that is associative, has an identity, and lets every element be undone. Symmetries of any object form a group — that is where the idea came from.

What is the difference between a ring and a field?

A ring has addition and multiplication that behave like the integers (you cannot always divide); a field is a ring where every non-zero element has a reciprocal, like the rationals or the reals.

The Fundamental Theorem