## Galois Theory: Examples of Galois Groups

Armed with the fundamental theorem of Galois theory, the aspiring mathematician can now go and compute Galois groups of some easy extensions.

Example #1: K = Q, L = Q(SQRT(n)) for any squarefree integer n. The minimal polynomial of SQRT(n) is x^2 – n; L is obviously normal over K, so it’s Galois and the fundamental theorem applies. The degree of L over K is 2, so Gal(L/K) has order 2. There is only one group of size 2, Z/2Z. That group has only two subgroups, the improper subgroup consisting of the entire group and the trivial subgroup consisting of the identity element 0. The non-identity element of the Galois group is the automorphism sending a + bSQRT(n) to abSQRT(n).

Example #2: K = R, L = C. As with example #1, this is a quadratic extension with minimal polynomial x^2 + 1, so Gal(L/K) = Z/2Z.

Example #3: K = Q, L = Q(2^(1/3), w) where w is a primitive cube root of unity. This is simply the splitting field of x^3 – 2 over Q, since the three roots are 2^(1/3), 2^(1/3)w, and 2^(1/3)w^2. The degree of the extension must divide 6 = 3! since 3 is the degree of the polynomial it’s a splitting field of; it must also be divisible by 2 since [Q(w):Q] = 2 and by 3 since [Q(2^(1/3)):Q] = 3.

There are only two groups of size 6, Z/6Z and S(3), the group of permutations of three elements. If Gal(L/K) = Z/6Z, then every subgroup of Gal(L/K) is normal since it is abelian; therefore, every intermediate field is normal, contradicting the fact that Q(2^(1/3))/Q is not normal. So Gal(L/K) = S(3). The elements of S(3) can be written in disjoint cycle notation as {(1), (1 2), (1 3), (2 3), (1 2 3), (1 3 2)}; then the proper nontrivial subgroups are generated by (1 2), (1 3), (2 3), and (1 2 3).

Assigning 1 to 2^(1/3), 2 to 2^(1/3)w, and 3 to 2^(1/3)w^2, we get that (1 2) fixes Q(2^(1/3)w^2), (1 3) fixes Q(2^(1/3)w), (2 3) fixes Q(2^(1/3)), and (1 2 3) and (1 3 2) fix Q(w). The only one of the four that is normal is Q(w), corresponding to the fact that {(1), (1 2 3), (1 3 2)} is the only proper nontrivial normal subgroup S(3).

Example #4: K = Q, L = Q(2^(1/4), i) where i is the imaginary unit. This is the splitting field of x^4 – 2, which has degree divisible by 4 and dividing 8. To see that it has degree 8, it’s enough to prove that x^4 – 2 is irreducible over Q(i). But it has no root in Q(i) by taking norms, so it has no linear factor. If it has two quadratic factors, then they can be written as x^2 + ax + b and x^2 – ax – 2/b; equating x-coefficients gives a(2/b + b) = 0, i.e. a(2 + b^2) = 0, so a = 0 since x^2 + 2 has no root in Q(i). But then equating x^2-coefficients yields b = -(2/b), i.e. 2 + b^2 = 0, a contradiction.

There are five groups of size 8: Z/8Z, Z/4Z * Z/2Z, Z/2Z * Z/2Z * Z/2Z, Q(8), and D(4). D(n) is the group of symmetries of a regular polygon with n sizes, where D(3) = S(3); note that notations vary, and some people write what I call D(n) as D(2n). Q(8) = $\{\pm1, \pm i, \pm j, \pm k\}$ where $i^{2} = j^{2} = k^{2} = ijk = -1$; although it is not abelian, all of its subgroups are normal. But L/K has the non-normal intermediate field Q(2^(1/4)), so Gal(L/K) = D(4).

Example #5: L and K are finite fields (classified here), with K = F(p) and L = F(p^n). L is the splitting field for x^(p^n) – x over K, and K is a perfect field, so L/K is Galois. Hence, Gal(L/K) has size equal to [L:K] = n. In fact, it’s isomorphic to Z/nZ, generated by the Frobenius automorphism, which sends each a to a^p.

To see that the Frobenius automorphism f generates Gal(L/K), consider its fixed field. The fixed field contains K, obviously. In addition, a^p = a iff a is a root of the polynomial x^px; the p elements of K are then the roots of the polynomial, so L^{f} = K. Now, consider the subgroup of Gal(L/K) consisting of powers of f. Its fixed field is clearly K, so it must be the improper subgroup Gal(L/K); hence, f indeed generates Gal(L/K).

The group Z/nZ has one subgroup per divisor of n. The subgroup whose size is the divisor d is generated by n/d, so its fixed field is the set of all roots of x^(p^(n/d)) – x in L. But that’s just F(p^(n/d)), confirming that L/K has one intermediate field for each divisor of [L:K]. Note that this also solves the case of the extension L/K where K is any finite field; if K = F(p^m), then Gal(L/K) is generated by the largest subgroup of L/F(p) fixing K, i.e. by the automorphism that sends a to a^(p^m).