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
x^{2}=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 Kautomorphisms
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 x^{3}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 x^{2}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 e^{r} 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 e^{r}.)
Proof: I claim that if f is a Kautomorphism
of L over K then then f(a) is a root of
m. To see this, suppose that
m=x^{n}+b_{n1}x^{n1}+...+b_{0},
for some b_{i} in K.
Thus we have 0=a^{n}+b_{n1}a^{n1}+...+b_{0}.
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}+b_{n1}f(a)^{n1}+...+b_{0}.
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 Kautomorphism of K(a) that maps a to b.
Firstly, the evaluation map ev_{a}: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
can_{a}:K[x]/m(x)K[x]>K(a).
Likewise we have the isomorphism
can_{b}:K[x]/m(x)K[x]>K(b).
Composing can_{b} and can_{a}^{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 Kautomorphism of K(a) that maps a to b.
Finally, any Kautomorphism of K(a) is obviously determined
once we know where it maps a, so we are done.