The idea is that we have a polynomial $p(x)$, for example $x^2-2$. Galois theory is concerned with symmetries in the solutions of $p(x)=0$. For example, if $p(x)=x^2-2$ then the solutions are $\pm \sqrt{2}$. A symmetry of the roots is a way of swapping the solutions around in a way which "doesn't matter" in some sense. So, $\sqrt{2}$ and $-\sqrt{2}$ are "the same" because any polynomial expression involving $\sqrt{2}$ will be the same if we replace $\sqrt{2}$ by $-\sqrt{2}$. For example, we know that $\sqrt{2}{}^2+\sqrt{2}+1=3+\sqrt{2}$. Or ${\alpha }^2+\alpha +1=3+\alpha$ when $\alpha =\sqrt{2}$. However, the same equation is true when $\alpha =-\sqrt{2}$, and this will be true for any expression involving only adding and multiplying $\sqrt{2}$.

One of the central results of Galois theory is that there are polynomials ( $x^5+x+1$ is one I think) whose solutions cannot be written in terms of only rational numbers ( $p/q$ for $p$, $q$ integers) addition, multiplication, division and taking things to the power $1/n$ for $n$ an integer. In fact, it turns out that any polynomial whose degree (the highest power of $x$) is less than or equal to 4 can be solved using only these operations, but there are polynomials of degree 5 and higher which cannot. The way this is proved (using Galois theory) is to prove a result about the collection of symmetries among the roots of a polynomial given that the roots are built up using only the special operations above. (It turns out that the collection of symmetries must form what is called a soluble group. More on this in a minute.) Then you find a polynomial for which the symmetries of the roots does not have this special property, so you know that the roots couldn't be built up from the special operations.

Let's be a bit more precise about that now.

A group $G$ is a collection of objects with an operation " $.$" satisfying the following rules (axioms):

1. For any two elements $x$ and $y$ in the group $G$ we also have $x.y$ is in the group $G$.
2. There is an element (usually written 1 or $e$, but sometimes 0) called the identity in $G$ such that for any $x$ in the group $G$ we have $1.x=x=x.1$.
3. For any elements $x$, $y$, $z$ in $G$ we have $(x.y).z=x.(y.z)$ (so it doesn't matter what order we do the calculations in). This property is called associativity, it means we can write $x.y.z$ unambiguously (otherwise it would not be clear what we meant by $x.y.z$, would it be $x.(y.z)$ or $(x.y).z$?)
4. Every element $x$ in $G$ has a unique inverse $y$ (sometimes written $-x$ or $x^{-1}$) so that $xy=yx=1$.

However, the integers are not a group with multiplication, because the identity on the integers with multiplication is 1, and there is no integer $n$ with $2n=1$.

Important finite groups are things like $C_p$ which is the cyclic group of order $p$, this is the set of elements 1, $x$, $x^2$, ldots, $x^{p-1}$ with the relation that $x^p=1$. So, for example, in $C_5$ we have that $x^2.x^4=x^6=x^5.x^1=x$. You can tell this is a group because the inverse of $x^n$ is $x^{p-n}$. Another important example of a finite group is $S_n$, the symmetric group on $n$ elements. Suppose we rearrange the numbers 1,..., $n$. For example, we could rearrange 1,2,3 to 2,3,1. The collection of all of these rearrangements forms a group. The operation is "do the second one, then the first". So, if we write $\sigma$ for the rearrangement 1,2,3 goes to 2,3,1 and $\tau$ for the rearrangement 1,2,3 goes to 3,2,1 then the rearrangement $\sigma .\tau$ does the following: it rearranges 1,2,3 to 3,2,1 (that's $\tau$) then it rearranges this to 2,1,3 (because $\sigma$ takes the first element (which is 3) to the last element, takes the middle element (which is 2) to the first element, and takes the last element (which is 1) to the middle element). So the group $S_n$ is the collection of rearrangements of 1,2,3,..., $n$.

At this point, you may want to check you've followed so far. See if you can prove that $S_n$ is a group and that it has $n!$ elements.

A field $F$ is a bit like a group, but we have two operations, usually written " $.$" and " $+$". $F$ is a field if $F$ has elements 0 and 1 such that $F$ with the operation " $+$" is a group, the set $F$ without the element 0 is a group with the operation " $.$" and we have relations like $(x+y).z=x.z+y.z$.

A good example of a field is the real numbers or the rational numbers. (Check the axioms.)

A field extension of a field $F$ is a field $K$ containing $F$. For example, the real numbers are a field extension of the rational numbers, because the reals form a field and every rational is also a real number.

A less obvious example of a field (the important example for Galois theory) is $Q[\sqrt{2}]$. This is the set of all numbers which can be written $a+b\sqrt{2}$ for $a$ and $b$ rational numbers. It is not immediately obvious that this is a field, because we do not know, for example, if $1/(a+b\sqrt{2})$ can be written $c+d\sqrt{2}$. However, you can always do this. If $x=1/(a+b\sqrt{2})$ then (multiplying the top and bottom by $a-b\sqrt{2}$) $x=(a-b\sqrt{2})/((a+b\sqrt{2})(a-b\sqrt{2}))$. And $(a+b\sqrt{2})(a-b\sqrt{2})=a^2-2b^2=p$ say. So we have that that $x=a/p-(b/p)\sqrt{2}$. So $Q[\sqrt{2}]$ really is a field, and it is a field extension of $Q$ (since every rational number $a$ can be written as $a+0\sqrt{2}$).

More generally, if $\alpha$ is a real number with the property that $p(\alpha )=0$ for some polynomial $p(x)$, then $Q[\alpha ]$ is a field. We can think of $Q[\alpha ]$ in two ways: (1) as the set of elements $a_0+a_1\alpha +\dots +a_{n-1}{\alpha }^{n-1}$ where each $a_i$ is a rational number and $n$ is the smallest integer such that there is a polynomial $p(x)$ of degree $n$ with $p(\alpha )=0$. (2) as the smallest field extension of $Q$ containing $\alpha$.

Here's where the Galois theory bit starts. Given a polynomial $p(x)$ we have what is called the ßplitting field" of $p(x)$ which is the smallest field extension of $Q$ which contains all the roots of $p(x)$. So, if $p(x)=x^2-2$ then the splitting field of $p(x)$ is $Q[\sqrt{2}]$ (it contains all the roots of $p$ and if it had any less elements it either wouldn't contain all the roots or wouldn't be a field).

So where do the symmetries come in? Here's where the idea of a field automorphism comes in. Let's use $Q[\sqrt{2}]$ as an example. If we define a function $f:Q[\sqrt{2}]\to Q[\sqrt{2}]$ by taking $f(a+b\sqrt{2})=a-b\sqrt{2}$ then we find that $f$ is what is called a field automorphism. A field automorphism $f$ has to be an invertible function (which the $f$ above clearly is) such that $f(x+y)=f(x)+f(y)$, $f(ax)=f(a)f(x)$, $f(1/x)=1/f(x)$. You can check that for the function $f$ above it really does satisfy all the conditions.

More specifically, if we have a field extension $K$ of a field $F$, then a $K$-automorphism of $F$ is an automorphism $f$ of $K$ with the additional property that $f(x)=x$ for all $x$ in $F$. This is the precise way of defining the symmetry of the roots that I talked about above. It turns out that for $Q[\sqrt{2}]$ the function $f$ I defined above is the only $Q$-automorphism other than the obvious $g(x)=x$.

So for the polynomial $p(x)=x^2-2$ we have the following:

(a) The splitting field of $p(x)$ is $Q[\sqrt{2}]$
(b) The $Q$-automorphisms, which we can think of as the symmetries of the roots, of $p(x)$ are $f(a+b\sqrt{2})=a-b\sqrt{2}$ and $g(x)=x$.

Now, if we have a field $F$ which is a field extension of $Q$ and a collection $G$ of $Q$-automorphisms of $F$. This collection $G$ is a group (with the operation defined by $.$; if $f$ and $g$ are in $G$ then $f.g$ is an automorphism defined by $(f.g)(x)=f(g(x))$ - check that this really is a group). It is called the Galois group of the field extension $F$ over $Q$. If $F$ is the splitting field of a polynomial $p(x)$ then $G$ is called the Galois group of the polynomial $p(x)$.

So, taking the polynomial $p(x)=x^2-2$, we have $G=\{f,g\}$ where $f(a+b\sqrt{2})=a-b\sqrt{2}$ and $g(x)=x$. Here, $g$ is the identity element of the group, and we have that $f.f=g$, because $(f.f)(a+b\sqrt{2})=f(f(a+b\sqrt{2})=f(a-b\sqrt{2})=a+b\sqrt{2}=g(a+b\sqrt{2})$. So, the group $G$ is the same as $C_2$, the cyclic group of order 2, or $S_2$, the symmetric group of order 2, because we have a single element $f$ with $f^2=f.f=1$ the identity on the group.

I think this is about as far as I'll go for the moment, because to go any further would be too complicated. I'll sketch the rest of the proof of existence of polynomials which cannot be solved with $+$, $-$, $\times$, $/$ and $n$th root operations on $Q$.

First, you define a cyclotomic field extension to be a field extension of $F$ where you take an element $x$ in $F$ and add the $n$th root. So, $Q[\sqrt{2}]$ is a cyclotomic field extension of $Q$.

Second, you define a radical field extension $K$ of a field $F$ to be a field extension which you can get to only using cyclotomic field extensions. So, $Q[\sqrt{1+\sqrt{2}}]$ is a radical field extension because you can start with $Q$, add $\sqrt{2}$ to form $Q[\sqrt{2}]$. Now, $1+\sqrt{2}$ is in $Q[\sqrt{2}]$, so taking the square root of this you get $Q[\sqrt{1+\sqrt{2}}]$. If the polynomial $p(x)$ has roots which can be described using only $+$, $-$, $\times$, $/$ and $n$th root, then the splitting field $F$ of $p(x)$ is a radical field extension of $Q$ (can you see why?).

Third, you prove that the Galois group of any radical field extension is ßoluble". This is the hardest part by a long, long way. In fact, I'm not even going to attempt to explain what a soluble group is here, because it would take too long.

Fourth, you prove that the group $S_5$ (the symmetric group on 5 elements) is not soluble.

Fifth, you find a polynomial $p(x)$ whose Galois group is $S_5$. The splitting field of this polynomial cannot be a radical field extension (because all radical field extensions have soluble Galois group), so the roots of $p(x)$ cannot be built up from $+$, $-$, $\times$, $/$ and $n$th root.

That's pretty long, and very hard, don't worry if you can't follow it because Galois theory is usually covered in the last year of university maths.

Suppose we want to find the Galois group of $p(x)=a{x}^2+bx+c$, with $a$, $b$, $c$ rational numbers as always. We know what the solutions of this are, they are $x=(-b\pm \sqrt{b^2-4ac})/(2a)$. How do we find the splitting field of $p(x)$?

Well, we know that $(-b\pm \sqrt{b^2-4ac})/(2a)$ are in the splitting field, and we also know that every rational number is in the splitting field $F$, and that we can add, multiply, divide (and so forth) elements in a field and remain in the field. So, adding $b/(2a)$ and multiplying by $(2a)$ we get that $\sqrt{b^2-4ac}$ is in the splitting field, and that any field extension of $Q$ containing $\sqrt{b^2-4ac}$ contains all the roots of $p(x)$.

If $b^2-4ac=r^2$ for some rational number $r$, then the splitting field $F=Q$, the rational numbers, because the solutions can be written with rational numbers only. Clearly in this case, the only $Q$- automorphism of $Q$ is the identity $f(x)=x$. So the Galois group is the 0 group with only an identity.

If $b^2-4ac$ cannot be written as $r^2$ for some rational $r$ then $b^2-4ac$ is called square-free, and we know that the splitting field $F=Q[\sqrt{b^2-4ac}]$. We can define a $Q$-automorphism $f(x+y\sqrt{b^2-4\mathrm{ac}})=x-y\sqrt{b^2-4\mathrm{ac}}$, which satisfies $f.f=1$. And these are the only $Q$-automorphisms of $F$. So in this case we have that the Galois group is $C_2$, the cyclic group of order 2.

So, summing up, if $b^2-4ac$ is the square of a rational number the Galois group is $C_1=0$, and if $b^2-4ac$ is not the square of a rational number, the Galois group is $C_2$.

Now we can easily find the Galois group of any quadratic polynomial. For example, $p(x)=x^2+x+1$ has $b^2-4ac=-3$ which is not a square so this polynomial has Galois group $C_2$.

We can do a similar thing for cubics and quartics, but it's a lot harder, so I won't go into it here...

It's quite difficult to set questions at this level because the interesting questions require sophisticated theory to answer. However, here are a couple:

You might like to try and prove that a $Q$-automorphism of the splitting field $F$ of a polynomial $p(x)$ permutes the roots (i.e. if $p(\alpha )=0$ and $\sigma$ is a $Q$-automorphism of $F$, and $\beta =\sigma (\alpha )$ then $p(\beta )=0$, and no two roots are sent, by $\sigma$, to the same root). You could also show, by giving an example, that not any permutation (rearrangement) of the roots of a polynomial comes from a $Q$-automorphism of $F$. However, you can also show that if you take a $Q$-automorphism of $F$, define a permutation $\sigma$ of the roots using the $Q$-automorphism, then the permutation alone gives you enough information to reconstruct the $Q$-automorphism. In other words, if you have a $Q$-automorphism it is enough to know where it sends the roots of $p(x)$ to know what it does to any element in the splitting field of $p(x)$. (If you know about group theory, you can deduce that if $p(x)$ is degree $n$, then the Galois group of $p(x)$ is a subgroup of $S_n$.)