 I will stop by discussing several examples. The first example is the problem of Pythagorean triangles. As everybody knows, right-angled triangles satisfy Pythagoras' theorem, where the sum of the squares of the sides is the square of the hypotenuse. So we have the usual three, four, five triangles, the simplest example. So this leads to the following problem. How can we classify all Pythagorean triangles? So there are several ways of doing this. I'm going to discuss an algebraic way and a geometrical way. So first of all, let's do the algebraic solution. So we want to solve x squared plus y squared equals z squared with x, y, and z integers. So here's the algebraics. We may as well assume that x, y, and z are co-prime because if any two have a common factor, we can just divide out by it. If we look at the equation mod four, we know that x squared, y squared, and z squared are all congruent to zero or one, mod four. And since they can't all be even, we see that z must be odd and one of x, y must be even and the other must be odd. So we may as well take y to be odd without any loss of generality and x to be even. Then we can rearrange the equation as y squared equals z minus x times z plus x. And we notice that these two numbers are co-prime. Since x and z are co-prime, the only possible common factors, these two numbers would be two, but since z is odd and x is even, they can't have a common factor of two. So they're co-prime. Now, if we've got two numbers that are co-prime whose product is a square, then both the numbers must be squares, at least opt to sign. So by changing the sign of this, if necessary, we can assume that this is a square, say r squared, and this is a square, say s squared. Well, then we can solve for z and x. We find z is equal to r squared plus s squared over two and x is equal to s squared minus r squared over two and y is equal to rs. Here, r and s will be odd. And co-prime. And for example, you can read off some solution. So if r and s is one, three, or one, five, or three, five, then we find the solutions, three squared plus four squared equals five squared, or five squared plus 12 squared equals 13 squared, or eight squared plus 15 squared equals 17 squared, and so on. So that's a perfectly good solution, but it's not completely satisfactory. It's slightly messy because we've had to assume, we've had to pick one of these numbers to be odd and we have r and s must both be odd and so on. So it's a slightly messy. So that's the algebraic solution. Now we're going to discuss a geometric solution in the style of algebraic geometry. For this, we want to solve x squared plus y squared equals z squared in integers. Well, integers don't work terribly well geometrically. So let's instead put big x equals x over z and big y equals y over z. And then we find x squared plus y squared is equal to one, x, y, rational. And I guess you all recognize this equation. It's just the equation of a circle. So what we're trying to do is to find rational points on a circle where a rational point is, of course, one whose coordinates are rational. Now, since we've made this geometric, we can draw a picture of what is going on and it just looks like this. So here we have a unit circle and we're trying to find rational points on it. And there's a very simple way of doing this. What we do is we pick this point here and here we have some point with rational coordinates x, y. And we just draw the line through x, y and this special point minus one, zero. And it intersects the y-axis in a point zero t. I'm using t instead of y because I've already used y for that. So if you look at this for about very briefly, you notice that t is just the slope of the line. So the slope is equal to t. And we can calculate t. We see that t is just y over x plus one. And conversely, so if we're given x and y, we can construct a rational number t. And conversely, given t, we can find x and y and these will be got by solving some quadratic equation and we know one of the roots of the equation is rational. So the other root must also be rational. Anyway, we can do this explicitly as follows. So we know y is equal to t times x plus one and substituting in, we find t squared x plus one squared plus x squared equals one. Rearranging it slightly, we find x plus one times t squared plus one x plus t squared minus one equals zero. And this factors and of course, x plus one is just this point here, which we're not really all that interested in. And this root here corresponds to this point here. So we find x equals one minus t squared over one plus t squared y equals two t over one plus t squared. And obviously, big x and big y, both rational with t is rational. So what we have here is a correspondence between points on the circle. They almost correspond to points on the y-axis. Well, they don't quite correspond because there's one slight problem. If we chose this point here, then we would kind of be trying to find the intersection of a vertical line with the y-axis, which doesn't really work. So this is the points on the circle which are not point minus one zero. So we almost have a one-to-one correspondence between the circle and the y-axis, a straight line. In algebraic geometry, this is called a birational correspondence or a birational equivalence. So a birational equivalence roughly means one that is an equivalence except on subsets of code I mentioned at least one. If you're doing something like studying smooth manifolds or differential geometry, then this notion of birational equivalence is kind of stupid because if you take two reasonable smooth manifolds of the same dimension, then, and you're allowed, and you can throw away a co-dimension one subset of both of them in order to make them both just a union of both. So this concept is uninteresting in the smooth manifolds and there's a characteristic feature of algebraic geometry. This correspondence between the line and the circle is also used in calculus. Here it's called the Weierstrich substitution. The Weierstrich substitution is one for use used for evaluating integrals of some function of sine theta and cosine theta. So what I'm going to do is I'm going to take this point X, Y, if this angle is theta, then X is equal to cosine of theta and Y is equal to sine of theta. And T can be worked out because if you remember Euclidean geometry, you can see this angle here is theta over two. So T is tan of theta over two. And now you may recognize the Weierstrich substitution. You remember if you've got an integral involving cosine theta and sine theta, you put T equals tan theta over two and turn cosine theta and sine theta into these rational functions of T. So any integral of sine theta and cosine theta can be turned into an integral of some rational function of T. In calculus textbooks, this is obviously presented with this really mysterious puzzling substitution that comes from nowhere at all. But from algebraic geometry, it's clear where this substitution comes from. It's simply the easiest, my rational equivalence between a straight line and a circle. So what you're doing is you're turning integrals over a circle into integrals over a straight line in the simplest possible way. Anyway, we should just have a quick example. So what this says is if you choose a rational number T, you get a solution to, you get a Pythagorean triangle. So let's try T equals a half. Then we see X is equal to three-fifths. Y is equal to four-fifths. So we get the solution three squared plus four squared equals five squared. And similarly for other rational values of T. Well, treating this as a geometrical problem rather than algebraic problem gives you additional insight into the solutions. In particular, circle forms a group. In fact, you can identify it with just the group of rotations of the plane fixing the origin. And we can work out the group operation fairly easily. For instance, suppose you've got two X1, Y1 and X2, Y2, and you wish to multiply them by the group operation. Well, this is just X1, X2 minus Y1, Y2 to Y2 plus X2, Y1. Where does this formula come from? Well, it's just the well-known formula for the cosine and sine of some of two angles. So we can think of this as being cosine theta one, sine theta one. When I say it's well-known, I mean, it used to be well-known. Of course, nobody memorizes these things anymore. So this is cosine theta two, sine theta two. And this is just cosine theta one, cosine theta two, minus sine theta one, sine theta two. And I'm not only getting the board of writing this out, but I've run out of room, so I'll leave you to fill in this one. So of course, this is just the well-known formula for cosine theta one plus theta two. So here we have the addition of angles, which is the group operation of the circle corresponds to this operation on the circle in rectangular coordinates. This is one of the simplest examples of something called an algebraic group. So informally, an algebraic group is a group whose group operation is given by polynomials. And so here the polynomials of that and that. There's another way of thinking of algebraic groups or for that matter algebraic varieties. You can think of this as being a functor from rings to groups. I explain what this means. Suppose we take any commutative ring R, pretty much all rings in algebraic geometry are commutative by default. Then we can map this to the set of pairs x, y in R squared with x squared plus y squared equals one. And now we can define the group operation by this formula here. Well, we should of course also have an identity of a group or the identity we can just take to be the point one zero and the inverse x, y inverses, of course, just going to be x and y and y and y and y. And you can easily check that if R is any commutative ring whatsoever, then the set of solutions of this equation forms a group under these operations. For example, if we call this algebraic group G, we can ask what is G of the complex numbers. So if you put the real numbers in here, we just get the unit circle. So if we put the real numbers in here, we just get the unit circle. So what would the analog of the circle be for the complex numbers? Well, it's the set of points x squared plus y squared equals one with x, y complex numbers. So let's see if we can figure out what that is. Well, that's easy enough because of the complex numbers, this factorizes as x plus i, y times x minus i, y. And we can set x squared plus i, y minus i, y. So x plus i, y equal to z, in which case this is equal to z bar. Next, not z bar, it's equal to z minus one. So all we have, one of the set of complex numbers that have an inverse. So this is the set of non-zero complex numbers z. So the circle group, you can say it's real points are just the usual circle of radius one say, and it's complex points turn out to be, you can identify them with the group of non-zero complex numbers. So to summarize this example, we have seen several different ways of thinking about a circle. So this is ways to view the circle. So first of all, we can just think of it as being a subset of the real plane. Secondly, we can think of it as being a polynomial x squared plus y squared minus one. And of course, if we've got any collection of polynomials in several variables, we can think of them as being a subset of some vector space. So the set of zeros of such collection of polynomials is often called an algebraic set. Thirdly, we can think of the ideal generated by x squared plus y squared minus one in the ring of polynomials in two variables. So this ideal is of course just the ideal of all functions that vanish on the circle. So we can try going from ideals to subsets of space. So we just, if we've given an ideal, we can take the set of all common zeros. And conversely, if we've got some subset of the plane, then we can look at the ideal of all polynomials vanishing on it. We'll be discussing this in much more detail in later lectures. Well, if we've got an ideal, we can take a quotient might, so we can look at the ring r x y modulo, this ideal. This is called the coordinate ring of the circle. And we can think of this as being the collection of all functions on the circle. Well, you may think it's not all functions on the circle, but if you're an algebraic geometry, you can use all functions on the circle, because the only functions we're going to allow are going to be polynomials. So this is the set of all polynomial functions on the circle, which are the only functions, algebraic geometers are really comfortable with. You can also think of this as being a smooth manifold. So one thing you can do an algebraic geometry is pretend that things are smooth manifolds and start applying methods of differential geometry to this. This doesn't always work because in the first place, we want to work over fields like finite fields where smooth manifolds don't make a whole lot of sense. And secondly, as we will see fairly soon, you can also have solutions of equations that aren't smooth. Number six, we can think of this as being a group or rather an algebraic group. So algebraic group theory is quite a big area in algebraic geometry. And another way of looking at it is as a functor from rings to groups or sets, if we don't want to think of it as being group. So this is a very powerful method of studying algebraic sets introduced mainly by growth and dick. Thank you.