 This lecture is part of an online commutative algebra course and will be about visualizing rings or how to draw pictures of them. So there are sort of three ways to draw a picture of a ring. We can draw a point of picture for each element of the ring, or we can draw a point for each basis element. If the ring is some vector space, or we can draw a point for each prime ideal. And this lecture, we're going to be discussing the first method where we draw a point for each element of the ring. So there are some examples of these we've had before. For example, if you've got the ring of real numbers, then of course we just draw it as a line with a point for each real number. If we've got the ring of complex numbers, we can draw it as a plane with points 0, 1, i, and so on. And if we've got the integers, we can think of it as being a subset of the reals of points 0, 1, 2, and so on. And what we're going to do is to find a few more examples like this. So the first example is the ring of Gaussian integers. And the ring of Gaussian integers is contained in the complex numbers, and so we can just draw it as a square lattice. So we have the point 0, 1, 2, and then we get i, i plus 1, and so on. So as an application of this, we will show that this ring is a unique factorization domain. So to do this, we'll just review Euclidean domains. So we recall Euclidean domain is an integral domain R with division with remainder algorithm. What this means is if we're given an A and B in R with B not equal 0, it's probably not a good idea to divide by 0, then A is equal to QB plus R, where this is the quotient and this is the remainder. And the key point is that R is smaller than B. Well, what do we mean by the absolute value of an element of a ring? Well, we have a map from the ring to some well-ordered set such as the non-negative integers, which just takes any element R to some element of the set denoted by its absolute value. So for the integers, we can just take the usual absolute value and so on. And now we recall that a Euclidean ring is a principal ideal domain, which is PID for short because it's a bit much to write out. And principal ideal domain has unique factorization into primes. So we'll just very quickly recall the proofs of these. So for instance, if we want to show that Euclide implies it's a principal ideal domain, what we do is we pick an ideal I in R and if I is not equal to the 0 ideal, we pick A in I with a minimal given that A is not 0. And then we can check that I is generated by the element A, so is principal. And if we take B in I, then we can write B is equal to A times Q plus R with R less than A, R in I, so R equals 0 because A was minimal. So all ideals in the Euclidean ring are principal. Also, we can show that principal ideal domains are unique factorization domains. And here we have the following key point. If P is irreducible, so this means P is not 0 or a unit. And if P equals AB, and A or B is a unit, then P is prime. So this means P is not 0 or a unit. And if P divides AB, then P divides A or P divides B. So vertical line here means the left-hand side divides the right-hand side. So sometimes primes are defined by this property, but usually in commut... Sorry, primes are often defined by this property here, but usually in commutative algebra. This property is referred to as being irreducible and being primus referred to as this property. For the integers, these two conditions are the same, so it doesn't matter. And the proof of this is, again, very easy. All you do is say, suppose P divides AB. If P does not divide A, then we look at the ideal generated by P and A, which must be the ideal generated by X for some element X, because it's a principal ideal domain. So X divides P, so X must either be... Since P is irreducible, X must either be a unit times P or must be a unit. But it can't be a unit times P because P does not divide A. So X is a unit, so PA is the ideal generation by one. So this means X times P plus Y times A equals one for some X, Y. And if we multiply by B, we find XPB plus YAB equals B. And this is divisible by P because P divides AB, so P divides B. So we've shown that if P does not divide A, it must divide B, so P is prime. And now from this, this easily implies unique factorization into primes. And I'll just give the key step of this. All we do is we say if P1, P2 and so on is equal to Q1, Q2 and so on, with all the PI and QI irreducibles, then P1 divides Q1, Q2 and so on. So P1 divides some QI by what we've just shown, that if an irreducible divides a product of numbers, it must divide one of them. So P1 must be equal to QI times a unit. And now we can remove P1 and QI from both factorizations and continue like that and we find the two factorizations must be the same, up to units and change of order, of course. So we have shown that Euclidean domains are unique factorization domains. Now what we want to do is to show that Z of I is Euclidean. So in particular, it's a unique factorization domain. And what we want to do is to show that if A and B are in Zi, B not equal 0, then we can find Q and R with this property here. Here we're going to take R to be the complex absolute value or, well, the complex absolute value isn't actually an integer, but it's given by the square root of M squared plus N squared if R is equal to M plus Ni, which is equal to the square root of an integer and square roots of integers are also well-ordered. So this is just as good as it being an integer. If you really want this to be an integer, you could just take it be the square of the usual absolute value. It doesn't really make a whole lot of difference. So we've got to find Q and R with this property. And now let's just divide by B. We've got to say A over B is equal to Q plus R over B. And now we want R over B to have absolute value less than one. In other words, R is less than the absolute value of B. So this number here is in Zi. And what this is saying, we want to show that any A over B is equal to something in Zi. That would be Q plus something with absolute value of less than one. So this would be the R over B. And so let's draw a picture of this. So here are the elements of Zi of I. And now let's draw all things that are within distance one of some element of Zi of I. But all that means is you draw a circle of radius one around each lattice point. And we notice that these unit circles, so the unit circles of radius one, I guess I would call them discs rather than circles because they're solid, cover the complex plane. I should say these are the open unit discs of radius one. So since these discs cover the plane, every complex number can be written as something in Zi plus something of absolute value less than one. So Zi of I is Euclidean. So we've got a very geometric picture of Zi of I being Euclidean. It's just saying that the plane is covered by open unit discs like this. Well, we can extend this argument. Suppose we look at Zi of root minus two and draw a picture of it. Well, here we sort of get rectangles. So here's nought one, here's root minus two and so on. And again, if we draw a unit circle around each point, then you can see that these unit circles still cover the plane. So unit circles, so unit discs still cover the plane. So Zi root minus two is also a unique factorization domain. These are the open ones. Well, now let's move on to Zi root minus three and see what happens. So if we take Zi root minus three, then we get these points one, zero, one, root minus three, one plus root minus three. And now something a little bit different happens because if you draw the... Sorry, that should have gone through there. If you draw the open discs around each point, then there's a point here that they don't quite cover. So this point here is the point one plus root minus three over two. And it is distance one from all these points, zero, one, root minus three, one plus root minus three. So the previous argument fails and it's not clear that this ring is Euclidean. In fact, it's not Euclidean. So Zi root minus three is not Euclidean. It's not Euclidean for any absolute value. I mean, we've shown it's not Euclidean for one particular absolute value, but maybe there's another clever thing we could define that made it Euclidean. Well, that fails because it's not Euclidean. It's not a principal ideal domain and it's not a unique factorization domain. If you want to see it's not a unique factorization domain, you just notice that two times two is equal to one plus root minus three times one minus root minus three and this gives two different factorizations of four and these don't differ by units or anything else. And we can also see directly that it's not a principal ideal domain. Well, let's see an example of a non-principle ideal. Well, a non-principle ideal is the ideal generated by two and one plus root minus three. And to see this as non-principle, the easiest way is to draw a picture of it. So let's draw the ring and now you notice the ring is sort of rectangular. So the black dots is just the principal ideal one, and all things will form M plus N root minus three. So here's naught one, here's root minus three. And now I want to draw the ideal generated by two and one plus root minus three. And if you draw this ideal, you get all these points here and you can easily check this as an ideal. And now these points here, if you join them up, they form a nice triangular lattice. You see like this. And cover the plane with equilateral triangles. So you can think of this as being a triangular lattice with equilateral triangles. On the other hand, if you look at the principal ideal, it's a sort of rectangular lattice. You know, if I join the top points up like this, I'm getting a lot of rectangles. Any other principal ideal has to be of the form A times R and is also a rectangular lattice. That's because if we multiply by some complex number, that corresponds to rotating by the argument of the complex number and rescaling by the absolute value of the complex number. So if you take this lattice and multiply it by A, you'll get some sort of rescaled and rotated rectangular lattice. You might get a lattice looking something like that, but it will always kind of look rectangular and it will never look like a bunch of equilateral triangles. So this ideal is non-principle because it's sort of a different shape from the principal ideal in some sense. So you can visibly see this ideal is non-principle just by drawing a picture of it. However, we can modify this result because we can replace the ring Z root minus 3 by the ring Z1 plus root minus 3 over 2. And this is still a ring. If we call this element omega, then we know omega squared plus omega plus 1 is equal to naught. So this ring is just everything of the form M plus N omega and because omega squared is a linear combination of these, this is indeed a ring. Now if we draw a picture of this ring, we now get a triangular lattice and now if we draw unit circles around each of these points, these obviously cover the plane. So unit discs are covering the plane. So Z1 plus root minus 3 over 2 is Euclidean and a unique factorization domain and so on. So this gives a method of giving geometric proofs that various rings are Euclidean and therefore unique factorization domains. I should comment that Euclidean domains are actually rather rare even among unique factorization domains. For example, you may or may not remember that the ring of polynomials is a unique factorization domain if R is a unique factorization domain. So in particular, if you've got a field, the field of polynomials in several variables is a unique factorization domain, but it's not a principal ideal domain. For example, the ideal generation by X, Y is not principal. So Euclidean domains are only very special sorts of unique factorization domains. You can also ask, are all principal ideal domains Euclidean domains? Well, the answer is most principal ideal domains in practice are Euclidean. So this includes things like Z ring of polynomials over a field, discrete valuation rings that we'll talk about later, and the Gaussian integers. In fact, it's quite difficult to find an example of a principal ideal domain that isn't Euclidean. The simplest example I know of is the ring Z one plus root minus 19 over two. So this is a principal ideal domain, but it's not Euclidean. If you want to see it's a principal ideal domain, you go to an algebraic number theory course. If you want to see it's not Euclidean, you can show this as follows. So if R is Euclidean, what we do is we let A be the smallest element with A not equal to zero, A not equal to a unit. Then every element of R over A is represented by zero or a unit by the division with remainder algorithm. However, the ring Z one plus root minus 19 over two has only two units, plus or minus one. So the quotient by this minimal element must have at most three elements. However, you can easily check that every quotient of the ring by some, this ring by some element is either the zero ring with no elements or has at least four elements. The quotient by two has exactly four elements and it's got no quotient by three elements. So this ring can't be Euclidean no matter what sort of absolute value you put on it. I mean, I'm not saying you, I'm saying it's, you might find some absolute value other than the obvious one where you take the absolute value for a complex number and even for some other absolute value this isn't going to be Euclidean. So this method of drawing pictures of a ring by drawing a point for each element works fine for rings whose additive group can be embedded in a vector space. So this applies to rings of integers of algebraic number fields and is very useful in algebraic number theory. For example, you can use it to prove Minkowski's theorem that the ideal class group of the ring of integers of an algebraic number field is finite. However, there are apart from rings of integers of algebraic number fields that aren't really very many rings which can be embedded in finite dimension Euclidean spaces. So this method of drawing pictures of rings is very powerful when it works but doesn't work for most rings. So next lecture we will discuss another method of drawing pictures of rings where you draw a picture of the basis of the ring over some field.