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**/2**Z**. 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* + *b*SQRT(*n*) to *a* – *b*SQRT(*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**/2**Z**.

*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**/6**Z** and S(3), the group of permutations of three elements. If Gal(L/K) = **Z**/6**Z**, 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**/8**Z**, **Z**/4**Z** * **Z**/2**Z**, **Z**/2**Z** * **Z**/2**Z** * **Z**/2**Z**, 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(2*n*). Q(8) = where ; 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**/*n***Z**, 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*^*p* – *x*; 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**/*n***Z** 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*).

Nice sum-up of some of the common Galois extensions. This is definitely helpful for someone who understands most of the theory but is looking for some extra examples. thanks!

I admire your piece of work, thanks for all the useful blog posts.