2. Automorphisms（自同构）
The main idea of Galois was to associate to any polynomial \(f\) a group of permutations of the roots of \(f\). In this section, we define and study this group and give some numerical information about it. Our description of this group is not the one originally given by Galois but an equivalent description given by Artin.
Let \(K\) be a field. A ring isomorphism from \(K\) to \(K\) is usually called an automorphism of \(K\). The group of all automorphisms of \(K\) will be denoted \(\mathrm{Aut}(K)\). Because we are interested in field extensions, we need to consider mappings of extensions. Let \(K\) and \(L\) be extension fields of \(F\). An \(F\)-homomorphism \(\tau : K \to L\) is a ring homomorphism such that \(\tau(a) = a\) for all \(a \in F\); that is, \(\tau|_F = \mathrm{id}\). If \(\tau\) is a bijection, then \(\tau\) is called an \(F\)-isomorphism. An \(F\)-isomorphism from a field \(K\) to itself is called an \(F\)-automorphism.
Let us point out some simple properties of \(F\)-homomorphisms. If \(\tau : K \to L\) is an \(F\)-homomorphism of extension fields of \(F\), then \(\tau\) is also a linear transformation of \(F\)-vector spaces, since \(\tau(\alpha a) = \tau(\alpha)\tau(a) = \alpha \tau(a)\) for \(\alpha \in F\) and \(a \in K\). Furthermore, \(\tau \ne 0\), so \(\tau\) is injective since \(K\) is a field. Also, if \([K : F] = [L : F] < \infty\), then \(\tau\) is automatically surjective by dimension counting. In particular, any \(F\)-homomorphism from \(K\) to itself is a bijection, provided that \([K : F] < \infty\).
Definition 2.1 Let \(K\) be a field extension of \(F\). The Galois group \(\mathrm{Gal}(K/F)\) is the set of all \(F\)-automorphisms of \(K\).
If \(K = F(X)\) is generated over \(F\) by a subset \(X\), we can determine the \(F\)-automorphisms of \(K\) in terms of their action on the generating set \(X\). For instance, if \(K\) is an extension of \(F\) that is generated by the roots of a polynomial \(f(x) \in F[x]\), the following two lemmas will allow us to interpret the Galois group \(\mathrm{Gal}(K/F)\) as a group of permutations of the roots of \(f\). This type of field extension obtained by adjoining to a base field roots of a polynomial is extremely important, and we will study it in Section 3. One use of these two lemmas will be to help calculate Galois groups, as shown in the examples below.
Lemma 2.2 Let \(K = F(X)\) be a field extension of \(F\) that is generated by a subset \(X\) of \(K\). If \(\sigma, \tau \in \mathrm{Gal}(K/F)\) with \(\sigma|_X = \tau|_X\), then \(\sigma = \tau\). Therefore, \(F\)-automorphisms of \(K\) are determined by their action on a generating set.
Proof. Let \(a \in K\). Then there is a finite subset \(\{\alpha_1, \ldots, \alpha_n\} \subseteq X\) with \(a \in F(\alpha_1, \ldots, \alpha_n)\). This means there are polynomials \(f, g \in F[x_1, \ldots, x_n]\) with \(a = f(\alpha_1, \ldots, \alpha_n)/g(\alpha_1, \ldots, \alpha_n)\); say
where each coefficient is in \(F\). Since \(\sigma\) and \(\tau\) preserve addition and multiplication, and fix elements of \(F\), we have
Thus, \(\sigma = \tau\), so \(F\)-automorphisms are determined by their action on generators. □
Lemma 2.3 Let \(\tau : K \to L\) be an \(F\)-homomorphism and let \(\alpha \in K\) be algebraic over \(F\). If \(f(z)\) is a polynomial over \(F\) with \(f(\alpha) = 0\), then \(f(\tau(\alpha)) = 0\). Therefore, \(\tau\) permutes the roots of \(\min(F,\alpha)\). Also, \(\min(F,\alpha) = \min(F,\tau(\alpha))\).
Proof. Let \(f(x) = a_0 + a_1 x + \cdots + a_n x^n\). Then
But, since each \(a_i \in F\), we have \(\tau(a_i) = a_i\). Thus, \(0 = \sum_i a_i \tau(\alpha)^i\), so \(f(\tau(\alpha)) = 0\). In particular, if \(p(x) = \min(F,\alpha)\), then \(p(\tau(\alpha)) = 0\), so \(\min(F,\tau(\alpha))\) divides \(p(x)\). Since \(p(x)\) is irreducible, \(\min(F,\tau(\alpha)) = p(x) = \min(F,\alpha)\). □
Corollary 2.4 If \([K : F] < \infty\), then \(|\mathrm{Gal}(K/F)| < \infty\).
Proof. We can write \(K = F(\alpha_1, \ldots, \alpha_n)\) for some \(\alpha_i \in K\). Any \(F\)-automorphism of \(K\) is determined by what it does to the \(\alpha_i\). By Lemma 2.3, there are only finitely many possibilities for the image of any \(\alpha_i\); hence, there are only finitely many automorphisms of \(K/F\). □
Example 2.5 Consider the extension \(\mathbb{C}/\mathbb{R}\). We claim that \(\mathrm{Gal}(\mathbb{C}/\mathbb{R}) = \{\mathrm{id}, \sigma\}\), where \(\sigma\) is complex conjugation. Both of these functions are \(\mathbb{R}\)-automorphisms of \(\mathbb{C}\), so they are contained in \(\mathrm{Gal}(\mathbb{C}/\mathbb{R})\). To see that there is no other automorphism of \(\mathbb{C}/\mathbb{R}\), note that an element of \(\mathrm{Gal}(\mathbb{C}/\mathbb{R})\) is determined by its action on \(i\), since \(\mathbb{C} = \mathbb{R}(i)\). Lemma 2.3 shows that if \(\tau \in \mathrm{Gal}(\mathbb{C}/\mathbb{R})\), then \(\tau(i)\) is a root of \(x^2 + 1\), so \(\tau(i)\) must be either \(i\) or \(-i\). Therefore, \(\tau = \mathrm{id}\) or \(\tau = \sigma\).
Example 2.6 The Galois group of \(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}\) is \(\{\mathrm{id}\}\). To see this, if \(\sigma\) is a \(\mathbb{Q}\)-automorphism of \(\mathbb{Q}(\sqrt[3]{2})\), then \(\sigma(\sqrt[3]{2})\) is a root of \(\min(\mathbb{Q}, \sqrt[3]{2}) = x^3 - 2\). If \(\omega = e^{2\pi i/3}\), then the roots of this polynomial are \(\sqrt[3]{2}\), \(\omega \sqrt[3]{2}\), and \(\omega^2 \sqrt[3]{2}\). The only root of \(x^3 - 2\) that lies in \(\mathbb{Q}(\sqrt[3]{2})\) is \(\sqrt[3]{2}\), since if another root lies in this field, then \(\omega \in \mathbb{Q}(\sqrt[3]{2})\), which is false since \([\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q}] = 3\) and \([\mathbb{Q}(\omega) : \mathbb{Q}] = 2\). Therefore, \(\sigma(\sqrt[3]{2}) = \sqrt[3]{2}\), and since \(\sigma\) is determined by its action on the generator \(\sqrt[3]{2}\), we see that \(\sigma = \mathrm{id}\).
Example 2.7 Let \(K = \mathbb{F}_2(t)\) be the rational function field in one variable over \(\mathbb{F}_2\), and let \(F = \mathbb{F}_2(t^2)\). Then \([K : F] = 2\). The element \(t\) satisfies the polynomial \(x^2 - t^2 \in F[x]\), which has only \(t\) as a root, since \(x^2 - t^2 = (x - t)^2\) in \(K[x]\). Consequently, if \(\sigma\) is an \(F\)-automorphism of \(K\), then \(\sigma(t) = t\), so \(\sigma = \mathrm{id}\). This proves that \(\mathrm{Gal}(K/F) = \{\mathrm{id}\}\).
Example 2.8 Let \(F = \mathbb{F}_2\). The polynomial \(1 + x + x^2\) is irreducible over \(F\), since it has no roots in \(F\). In fact, this is the only irreducible quadratic over \(F\); the three other quadratics factor over \(F\). Let \(K = F[x]/(1 + x + x^2)\), a field that we can view as an extension field of \(F\); see Example 1.6 for details on this construction. To simplify notation, we write \(M = (1 + x + x^2)\). Every element of \(K\) can be written in the form \(a + bx + M\) by the division algorithm. Let us write \(\alpha = x + M\). The subfield \(\{a + M : a \in F\}\) of \(K\) is isomorphic to \(F\). By identifying \(F\) with this subfield of \(K\), we can write every element of \(K\) in the form \(a + b\alpha\) with \(a, b \in F\). Then \(K = F(\alpha)\), so any \(F\)-automorphism of \(K\) is determined by its action on \(\alpha\). By Lemma 2.3, if \(\sigma\) is an \(F\)-automorphism of \(K\), then \(\sigma(\alpha)\) is a root of \(1 + x + x^2\). By factoring \(1 + x + x^2\) as \((x - \alpha)(x - \beta)\) and expanding, we see that the other root of \(1 + x + x^2\) is \(\alpha + 1\). Therefore, the only possibility for \(\sigma(\alpha)\) is \(\alpha\) or \(\alpha + 1\), so \(\mathrm{Gal}(K/F)\) has at most two elements. To see that \(\mathrm{Gal}(K/F)\) has exactly two elements, we need to check that there is indeed an automorphism \(\sigma\) with \(\sigma(\alpha) = \alpha + 1\). If \(\sigma\) does exist, then \(\sigma(a + b\alpha) = a + b(\alpha + 1) = (a + b) + b\alpha\). We leave it as an exercise (Problem 7) to show that the function \(\sigma : K \to K\) defined by \(\sigma(a + b\alpha) = (a + b) + b\alpha\) is an \(F\)-automorphism of \(K\). Therefore, \(\mathrm{Gal}(K/F) = \{\mathrm{id}, \sigma\}\).
The idea of Galois theory is to be able to go back and forth from field extensions to groups. We have now seen how to take a field extension \(K/F\) and associate a group, \(\mathrm{Gal}(K/F)\). More generally, if \(L\) is a field with \(F \subseteq L \subseteq K\), we can associate a group \(\mathrm{Gal}(K/L)\). This is a subgroup of \(\mathrm{Gal}(K/F)\), as we will see in the lemma below. Conversely, given a subgroup of \(\mathrm{Gal}(K/F)\) we can associate a subfield of \(K\) containing \(F\). Actually, we can do this for an arbitrary subset of \(\mathrm{Aut}(K)\). Let \(S\) be a subset of \(\mathrm{Aut}(K)\), and set
It is not hard to see that \(\mathcal{F}(S)\) is a subfield of \(K\), called the fixed field of \(S\). A field \(L\) with \(F \subseteq L \subseteq K\) is called an intermediate field of the extension \(K/F\). Therefore, if \(S \subseteq \mathrm{Gal}(K/F)\), then \(\mathcal{F}(S)\) is an intermediate field of \(K/F\).
The following lemma gives some simple properties of Galois groups and fixed fields.
Lemma 2.9 Let \(K\) be a field.
- If \(L_1 \subseteq L_2\) are subfields of \(K\), then \(\mathrm{Gal}(K/L_2) \subseteq \mathrm{Gal}(K/L_1)\).
- If \(L\) is a subfield of \(K\), then \(L \subseteq \mathcal{F}(\mathrm{Gal}(K/L))\).
- If \(S_1 \subseteq S_2\) are subsets of \(\mathrm{Aut}(K)\), then \(\mathcal{F}(S_2) \subseteq \mathcal{F}(S_1)\).
- If \(S\) is a subset of \(\mathrm{Aut}(K)\), then \(S \subseteq \mathrm{Gal}(K/\mathcal{F}(S))\).
- If \(L = \mathcal{F}(S)\) for some \(S \subseteq \mathrm{Aut}(K)\), then \(L = \mathcal{F}(\mathrm{Gal}(K/L))\).
- If \(H = \mathrm{Gal}(K/L)\) for some subfield \(L\) of \(K\), then \(H = \mathrm{Gal}(K/\mathcal{F}(H))\).
Proof. The first four parts are simple consequences of the definitions. We leave the proofs of parts 2, 3, and 4 to the reader and prove part 1 for the sake of illustration. If \(\sigma \in \mathrm{Gal}(K/L_2)\), then \(\sigma(a) = a\) for all \(a \in L_2\). Thus, \(\sigma(a) = a\) for all \(a \in L_1\), as \(L_1 \subseteq L_2\), so \(\sigma \in \mathrm{Gal}(K/L_1)\).
To prove part 5, suppose that \(L = \mathcal{F}(S)\) for some subset \(S\) of \(\mathrm{Aut}(K)\). Then \(S \subseteq \mathrm{Gal}(K/L)\), so \(\mathcal{F}(\mathrm{Gal}(K/L)) \subseteq \mathcal{F}(S) = L\). But \(L \subseteq \mathcal{F}(\mathrm{Gal}(K/L))\), so \(L = \mathcal{F}(\mathrm{Gal}(K/L))\). For part 6, if \(H = \mathrm{Gal}(K/L)\) for some subfield \(L\) of \(K\), then \(L \subseteq \mathcal{F}(\mathrm{Gal}(K/L))\), so
However, by part 4 we have \(H \subseteq \mathrm{Gal}(K/\mathcal{F}(H))\), so \(H = \mathrm{Gal}(K/\mathcal{F}(H))\). □
Corollary 2.10 If \(K\) is a field extension of \(F\), then there is 1-1 inclusion reversing correspondence between the set of subgroups of \(\mathrm{Gal}(K/F)\) of the form \(\mathrm{Gal}(K/L)\) for some subfield \(L\) of \(K\) containing \(F\) and the set of subfields of \(K\) that contain \(F\) of the form \(\mathcal{F}(S)\) for some subset \(S\) of \(\mathrm{Aut}(K)\). This correspondence is given by \(L \mapsto \mathrm{Gal}(K/L)\), and its inverse is given by \(H \mapsto \mathcal{F}(H)\).
Proof. This follows immediately from the lemma. If \(\mathcal{G}\) and \(\mathcal{F}\) are respectively the set of groups and fields in question, then the map that sends a subfield \(L\) of \(K\) to the subgroup \(\mathrm{Gal}(K/L)\) of \(\mathrm{Aut}(K)\) sends \(\mathcal{F}\) to \(\mathcal{G}\). This map is injective and surjective by part 5 of the lemma. Its inverse is given by sending \(H\) to \(\mathcal{F}(H)\) by part 6. □
If \(K/F\) is a finite extension, under what circumstances does the association \(L \mapsto \mathrm{Gal}(K/L)\) give an inclusion reversing correspondence between the set of all subfields of \(K\) containing \(F\) and the set of all subgroups of \(\mathrm{Gal}(K/F)\)? A necessary condition from part 5 is that \(F = \mathcal{F}(\mathrm{Gal}(K/F))\). We shall see in Section 5 that this is actually a sufficient condition.
The next three results aim at getting more precise numerical information on \(|\mathrm{Gal}(K/F)|\) for a finite extension \(K/F\). We first need a definition.
Definition 2.11 If \(G\) is a group and if \(K\) is a field, then a character is a group homomorphism from \(G\) to \(K^*\).
By setting \(G = K^*\), we see that \(F\)-automorphisms of \(K\) can be viewed as characters from \(G\) to \(K^*\). The next lemma will lead to a bound on \(|\mathrm{Gal}(K/F)|\).
Lemma 2.12 (Dedekind's Lemma) Let \(\tau_1, \ldots, \tau_n\) be distinct characters from \(G\) to \(K^*\). Then the \(\tau_i\) are linearly independent over \(K\); that is, if \(\sum_i c_i \tau_i(g) = 0\) for all \(g \in G\), where the \(c_i \in K\), then all \(c_i = 0\).
Proof. Suppose that the lemma is false. Choose \(k\) minimal (relabeling the \(\tau_i\) if necessary) so that there are \(c_i \in K\) with \(\sum_i c_i \tau_i(g) = 0\) for all \(g \in G\). Then all \(c_i \ne 0\). Since \(\tau_1 \ne \tau_2\), there is an \(h \in G\) with \(\tau_1(h) \ne \tau_2(h)\). We have \(\sum_{i=1}^k (c_i \tau_1(h)) \tau_i(g) = 0\) and
for all \(g\). Subtracting gives \(\sum_{i=1}^k (c_i (\tau_1(h) - \tau_i(h))) \tau_i(g) = 0\) for all \(g\). This is an expression involving \(k - 1\) of the \(\tau_i\) with not all of the coefficients zero. This contradicts the minimality of \(k\), so the lemma is proved. □
There is a vector space interpretation of Dedekind's lemma. If \(V\) is the set of all functions from \(G\) to \(K\), then \(V\) is a \(K\)-vector space under usual function addition and scalar multiplication, and Dedekind's lemma can be viewed as showing that the set of characters from \(G\) to \(K^*\) forms a linearly independent set in \(V\).
Proposition 2.13 If \(K\) is a finite field extension of \(F\), then \(|\mathrm{Gal}(K/F)| \le [K : F]\).
Proof. The group \(\mathrm{Gal}(K/F)\) is finite by Corollary 2.4. Let \(\mathrm{Gal}(K/F) = \{\tau_1, \ldots, \tau_n\}\), and suppose that \([K : F] < n\). Let \(\alpha_1, \ldots, \alpha_m\) be a basis for \(K\) as an \(F\)-vector space. The matrix
over \(K\) has \(\mathrm{rank}(A) \le m < n\), so the rows of \(A\) are linearly dependent over \(K\). Thus, there are \(c_i \in K\), not all zero, such that \(\sum_i c_i \tau_i(\alpha_j) = 0\) for all \(j\). If we set \(G = K^*\), then for \(g \in G\) there are \(a_j \in F\) with \(g = \sum_j a_j \alpha_j\). Thus,
All the \(c_i\) are then 0 by Dedekind's lemma. This contradiction proves that \(\mathrm{Gal}(K/F) \le [K : F]\). □
The following question arises naturally from this proposition: For which field extensions \(K/F\) does \(|\mathrm{Gal}(K/F)| = [K : F]\)? The inequality in the proposition above may be strict, as shown in Examples 2.6 and 2.7.
The next proposition determines when \(|\mathrm{Gal}(K/F)| = [K : F]\), provided that the group \(\mathrm{Gal}(K/F)\) is finite.
Proposition 2.14 Let \(G\) be a finite group of automorphisms of \(K\) with \(F = \mathcal{F}(G)\). Then \(|G| = [K : F]\), and so \(G = \mathrm{Gal}(K/F)\).
Proof. By the previous proposition, \(|G| \le [K : F]\) since \(G \subseteq \mathrm{Gal}(K/F)\). Suppose that \(|G| < [K : F]\). Let \(n = |G|\), and take \(\alpha_1, \ldots, \alpha_{n+1} \in K\) linearly independent over \(F\). If \(G = \{\tau_1, \ldots, \tau_n\}\), let \(A\) be the matrix
Then the columns of \(A\) are linearly dependent over \(K\). Choose \(k\) minimal so that the first \(k\) columns of \(A\) are linearly dependent over \(K\) (relabeling if necessary). Thus, there are \(c_i \in K\) not all zero with \(\sum_{i=1}^k c_i \tau_j(\alpha_i) = 0\) for all \(j\). Minimality of \(k\) shows all \(c_i \ne 0\). Thus, by dividing we may assume that \(c_1 = 1\). If each \(c_i \in F\), then \(0 = \tau_j(\sum_{i=1}^k c_i \alpha_i)\) for each \(j\), so \(\sum_{i=1}^k c_i \alpha_i = 0\). This is false by the independence of the \(\alpha_i\) over \(F\). Take \(\sigma \in G\). Since \(\sigma\) permutes the elements of \(G\), we get \(\sum_{i=1}^k \sigma(c_i) \tau_j(\alpha_i) = 0\) for all \(j\). Subtracting this from the original equation and recalling that \(c_1 = 1\) gives \(\sum_{i=2}^k (c_i - \sigma(c_i)) \tau_j(\alpha_i) = 0\) for all \(j\). Minimality of \(k\) shows that \(c_i - \sigma(c_i) = 0\) for each \(i\). Since this is true for all \(\sigma \in G\), we get all \(c_i \in \mathcal{F}(G) = F\). But we have seen that this leads to a contradiction. Thus \(|G| = [K : F]\). In particular, \(G = \mathrm{Gal}(K/F)\), since \(G \subseteq \mathrm{Gal}(K/F)\) and \(|G| = [K : F] \ge |\mathrm{Gal}(K/F)|\). □
The field extensions described in Proposition 2.14 are those of particular interest to us, as they were to Galois in his work on the solvability of polynomials.
Definition 2.15 Let \(K\) be an algebraic extension of \(F\). Then \(K\) is Galois over \(F\) if \(F = \mathcal{F}(\mathrm{Gal}(K/F))\).
If \([K : F] < \infty\), then Proposition 2.14 gives us a numerical criterion for when \(K/F\) is Galois.
Corollary 2.16 Let \(K\) be a finite extension of \(F\). Then \(K/F\) is Galois if and only if \(|\mathrm{Gal}(K/F)| = [K : F]\).
Proof. If \(K/F\) is a Galois extension, then \(F = \mathcal{F}(\mathrm{Gal}(K/F))\), so by Proposition 2.14, \(|\mathrm{Gal}(K/F)| = [K : F]\). Conversely, if \(|\mathrm{Gal}(K/F)| = [K : F]\), let \(L = \mathcal{F}(\mathrm{Gal}(K/F))\). Then \(\mathrm{Gal}(K/L) = \mathrm{Gal}(K/F)\) by Proposition 2.14, and so \(|\mathrm{Gal}(K/F)| = [K : L] \le [K : F]\). Since \(|\mathrm{Gal}(K/F)| = [K : F]\), this forces \([K : L] = [K : F]\), so \(L = F\). □
The previous corollary gives us a numerical criterion for when a finite extension is Galois. However, to use it we need to know the Galois group of the extension. This group is not always easy to determine. For extensions of \(F\) of the form \(F(\alpha)\), we have a simpler criterion to determine when \(F(\alpha)/F\) is Galois.
Corollary 2.17 Let \(K\) be a field extension of \(F\), and let \(\alpha \in K\) be algebraic over \(F\). Then \(|\mathrm{Gal}(F(\alpha)/F)|\) is equal to the number of distinct roots of \(\min(F,\alpha)\) in \(F(\alpha)\). Therefore, \(F(\alpha)\) is Galois over \(F\) if and only if \(\min(F,\alpha)\) has \(n\) distinct roots in \(F(\alpha)\), where \(n = \deg(\min(F,\alpha))\).
Proof. If \(\tau \in \mathrm{Gal}(F(\alpha)/F)\), we have seen that \(\tau(\alpha)\) is a root of \(\min(F,\alpha)\). Moreover, if \(\sigma, \tau \in \mathrm{Gal}(F(\alpha)/F)\) with \(\sigma \ne \tau\), then \(\sigma(\alpha) \ne \tau(\alpha)\), since \(F\)-automorphisms on \(F(\alpha)\) are determined by their action on \(\alpha\). Therefore, \(|\mathrm{Gal}(F(\alpha)/F)| \le n\). Conversely, let \(b\) be a root in \(F(\alpha)\) of \(\min(F,\alpha)\). Define \(\tau : F(\alpha) \to F(\alpha)\) by \(\tau(f(\alpha)) = f(b)\) for any \(f(z) \in F[z]\). This map is well defined precisely because \(b\) is a root of \(\min(F,\alpha)\). It is straightforward to show that \(\tau\) is an \(F\)-automorphism, and \(\tau(\alpha) = b\) by the definition of \(\tau\). Thus, \(|\mathrm{Gal}(F(\alpha)/F)|\) is equal to the number of distinct roots of \(\min(F,\alpha)\) in \(F(\alpha)\). Since \([F(\alpha) : F] = \deg(\min(F,\alpha))\), we see that \(F(\alpha)\) is Galois over \(F\) if and only if \(\min(F,\alpha)\) has \(n\) distinct roots in \(F(\alpha)\). □
There are two ways that a field extension \(F(\alpha)/F\) can fail to be Galois. First, if \(p(x) = \min(F,\alpha)\), then \(p\) could fail to have all its roots in \(F(\alpha)\). Second, \(p(x)\) could have repeated roots. The next two sections will address these concerns. We finish this section with a number of examples of extensions for which we determine whether or not they are Galois. Here and elsewhere in this book, we use the idea of the characteristic of a field (or a ring with identity). For the reader unfamiliar with this notion, the characteristic \(\mathrm{char}(F)\) of a field \(F\) is the order of the multiplicative identity \(1\) as an element of the additive group \((F, +)\), provided that this order is finite, or else \(\mathrm{char}(F) = 0\) if this order is infinite. Note that the characteristic of a field is either \(0\) or is a prime number. More information on the characteristic of a ring can be found in Appendix A or in the last six problems in the previous section.
Example 2.18 The extension \(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}\) is not Galois, for we have seen that \([\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q}] = 3\) but \(|\mathrm{Gal}(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q})| = 1\). The polynomial \(x^3 - 2\) has three distinct roots, but only one of them lies in \(\mathbb{Q}(\sqrt[3]{2})\).
Example 2.19 Let \(k\) be a field of characteristic \(p > 0\), and let \(k(t)\) be the rational function field in one variable over \(k\). Consider the field extension \(k(t)/k(t^p)\). Then \(t\) satisfies the polynomial \(x^p - t^p \in k(t^p)[x]\). However, over \(k(t)\) this polynomial factors as \(x^p - t^p = (x - t)^p\). Thus, the minimal polynomial of \(t\) over \(k(t^p)\) has only one root; consequently, \(\mathrm{Gal}(k(t)/k(t^p)) = \{\mathrm{id}\}\). Thus, \(k(t)/k(t^p)\) is not Galois.
The previous two examples illustrate the two ways a field extension of the form \(F(\alpha)/F\) can fail to be Galois. The remaining examples are examples of extensions that are Galois.
Example 2.20 Let \(F\) be a field of characteristic not 2, and let \(a \in F\) be an element that is not the square of any element in \(F\). Let \(K = F[x]/(x^2 - a)\), a field since \(x^2 - a\) is irreducible over \(F\). We view \(F\) as a subfield of \(K\) by identifying \(F\) with the subfield \(\{\alpha + (x^2 - a) : \alpha \in F\}\) of \(K\). Under this identification, each coset is uniquely expressible in the form \(\alpha + \beta x + (x^2 - a)\) and, hence, is an \(F\)-linear combination of \(1 + (x^2 - a)\) and \(x + (x^2 - a)\). Thus, \(1\) and \(u = x + (x^2 - a)\) form a basis for \(K\) as an \(F\)-vector space, so \([K : F] = 2\). If \(\sigma\) is defined by
then \(\sigma\) is an automorphism of \(K\) since \(u\) and \(-u\) are roots in \(K\) of \(z^2 - a\). Thus, \(\mathrm{id}, \sigma \in \mathrm{Gal}(K/F)\), so \(|\mathrm{Gal}(K/F)| = 2 = [K : F]\). Consequently, \(K/F\) is a Galois extension.
The extension \(K = F(\alpha)\) is generated by an element \(\alpha\) with \(\alpha^2 = a\). We will often write \(F(\sqrt{a})\) for this extension. The notation \(\sqrt{a}\) is somewhat ambiguous, since for an arbitrary field \(F\) there is no way to distinguish between different square roots, although this will not cause us any problems.
Example 2.21 The extension \(\mathbb{Q}(\sqrt[3]{2}, \omega)/\mathbb{Q}\) is Galois, where \(\omega = e^{2\pi i/3}\). In fact, the field \(\mathbb{Q}(\sqrt[3]{2}, \omega)\) is the field generated over \(\mathbb{Q}\) by the three roots \(\sqrt[3]{2}\), \(\omega \sqrt[3]{2}\), and \(\omega^2 \sqrt[3]{2}\), of \(x^3 - 2\), and since \(\omega\) satisfies \(x^2 + x + 1\) over \(\mathbb{Q}\) and \(\omega\) is not in \(\mathbb{Q}(\sqrt[3]{2})\), we see that \([\mathbb{Q}(\sqrt[3]{2}, \omega) : \mathbb{Q}] = 6\). It can be shown (see Problem 3) that the six functions
extend to distinct automorphisms of \(\mathbb{Q}(\sqrt[3]{2}, \omega)/\mathbb{Q}\). Thus,
and so \(\mathbb{Q}(\omega, \sqrt[3]{2})/\mathbb{Q}\) is Galois.
One reason we did not do the calculation that shows that we do get six automorphisms from these formulas is that this calculation is long and not particularly informative. Another reason is that later on we will see easier ways to determine when an extension is Galois. Knowing ahead of time that \(\mathbb{Q}(\sqrt[3]{2}, \omega)/\mathbb{Q}\) is Galois and that the degree of this extension is six tells us that we have six \(\mathbb{Q}\)-automorphisms of \(\mathbb{Q}(\sqrt[3]{2}, \omega)\). There are only six possibilities for the images of \(\sqrt[3]{2}\) and \(\omega\) under an automorphism, and so all six must occur.
Example 2.22 This example shows us that any finite group can occur as the Galois group of a Galois extension. We will use this example a number of times in later sections. Let \(k\) be a field and let \(K = k(x_1, x_2, \ldots, x_n)\) be the field of rational functions in \(n\) variables over \(k\). For each permutation \(\sigma \in S_n\), define \(\sigma(x_i) = x_{\sigma(i)}\). Then \(\sigma\) has a natural extension to an automorphism of \(K\) by defining
The straightforward but somewhat messy calculation that this does define a field automorphism on \(K\) is left to Problem 5. We can then view \(S_n \subseteq \mathrm{Aut}(K)\). Let \(F = \mathcal{F}(S_n)\). By Proposition 2.14, \(K/F\) is a Galois extension with \(\mathrm{Gal}(K/F) = S_n\). The field \(F\) is called the field of symmetric functions in the \(x_i\). The reason for this name is that if \(f(x_1, \ldots, x_n)/g(x_1, \ldots, x_n) \in F\), then
for all \(\sigma \in S_n\). Let
The polynomial \(s_i\) is called the \(i\)th elementary symmetric function. We see that each \(s_i \in F\), so \(k(s_1, \ldots, s_n) \subseteq F\). Note that
From this fact, we shall see in Section 3 that \(F = k(s_1, \ldots, s_n)\). This means that every symmetric function in the \(x_i\) is a rational function in the elementary symmetric functions.