display | more...
The Galois group is the object studied in Galois theory. It was invented by Evariste Galois well before the concept of a group had been properly formulated. For him, a group was a collection of permutations. Galois was interested in solving polynomial equations in one variable and the Galois group tells us about the symmetries of the roots of such equations. (The permutations Galois looked at are permutations of the roots.) Before I give any precise definitions we can at least get a flavour by thinking about the equation x2=2. This has solutions sqrt(2) and -sqrt(2) and the symmetry here is given by flipping over sqrt(2) and -sqrt(2). In this case the Galois group is cyclic of order 2.

Start with a field extension L of K. The Galois group of L over K is the the set of all K-automorphisms of L, made into a group under the binary operation of composition of functions. It is denoted by Gal(L/K).

Let's work out the Galois group of something substantial. Suppose then that a is algebraic over K. This just means that a is a zero of a nonconstant polynomial with coefficients in K. We are going to compute Gal(K(a)/K). Suppose that m(x) is the minimal polynomial of a over K.

Proposition Gal(K(a)/K) is in bijection to the zeroes of m(x) which lie in K(a). Under this bijection the zero b of m(x) corresponds to an automorphism that maps a to b.

Before we look at the proof, note that we can use it to work out a few examples quite easily.

• Let a be the real cubed root of 2. Then by the above proposition the Gal(Q(a)/Q)={1} is trivial. This is because the elements of Q(a) are all real numbers but the other two roots of the minimal polynomial x3-2 both have a nonzero imaginary part.
• Gal(Q(sqrt(2))/Q) is cyclic of order two. This is because Q(sqrt(2)) contains the two roots of the minimal polynomial x2-2.
• Finally, something quite tasty, let Q(e) be the nth cyclotomic field extension of Q. i.e e is a primitive nth complex root of unity. In this case Gal(Q(e)/Q) is isomorphic to the multiplicative group of the ring of integers modulo n. This follows from the above with a little bit of thinking. (Briefly, this time the minimal polynomial is the nth cyclotomic polynomial and the roots of this are er for r coprime to n. The unit of the integers modulo n which is the residue class of r corresponds to the automorphism that maps e to er.)

Proof: I claim that if f is a K-automorphism of L over K then then f(a) is a root of m. To see this, suppose that m=xn+bn-1xn-1+...+b0, for some bi in K. Thus we have 0=an+bn-1an-1+...+b0. Apply f to both sides and use that f is a ring homomorphism and fixes elements of K. Then we get 0=f(a)n+bn-1f(a)n-1+...+b0. In other words f(a) is a zero of m(x).

Next, suppose that b is a zero of m(x) that happens to lie inside K(a). I am going to show that there exists a K-automorphism of K(a) that maps a to b. Firstly, the evaluation map eva:K[x]-->K(a) is a surjection. Furthermore one checks (essentially from the defintion) that the kernel of this map is the principal ideal generated by m(x). By the first isomorphism theorem it follows that there is an isomorphism cana:K[x]/m(x)K[x]-->K(a). Likewise we have the isomorphism canb:K[x]/m(x)K[x]-->K(b). Composing canb and cana-1 we get an isomorphism of rings K(x)-->K(b). It is not hard to see that this isomorphism fixes the elements of K and furthermore, an argument based on counting vector space dimensions shows that K(a)=K(b). We have thus established that there is a K-automorphism of K(a) that maps a to b.

Finally, any K-automorphism of K(a) is obviously determined once we know where it maps a, so we are done.

Log in or register to write something here or to contact authors.