 OK. So this is the second lecture of algebraic geometry. So we introduced the fine algebraic sets. So these are subsets 0 set of s in a k, where s is some set of polynomials, the mimecate n in n variables. And z of s just means it's a set of all points p in a n, such that f of p is equal to 0 for all f in s. We had seen that these fine algebraic sets also can be just obtained as 0 sets of ideals. So of course, the 0 set of s is equal to the 0 set of the ideal generated by s. And we had also introduced the ideal of a fine algebraic set. So if x subset, we have the ideal of x, which is just a set of all polynomials f in k x1 to xn, such that f restricted to x is the 0 polynomial. And say, if we assume that x is in a fine algebraic set, we can consider this. So we have a way how to associate to an ideal. It's 0 set, which is a fine algebraic set. And we have also the possibility of associating to a fine algebraic set an ideal. And this ideal is somehow associated in a natural canonical way to x. So you can somehow hope that the properties of the ideal, I mean the algebraic properties of the ideal of x, are closely related to the geometric properties of x. Let us see that this is the case. And then what else do we have? We had the Hilbert-Base's theorem, which used in the form that we have that k x1 to xn is Neutering. So that means, in particular, one way of formulating is that every ideal in this ring is finitely generated. And as a corollary, we would get that every fine algebraic set is a 0 set of finitely many polynomials. So this is just clear. So we know that x is 0 set of some ideal. And this ideal is finitely generated because every ideal in k x1 to xn is finitely generated. So we can write i because the ideal is generated by some elements f1 to fr for some r. And then we have that x is 0 set of the ideal generated by the fi, which is the same as 0 set of the fi. So in our ritual definition, we allow 0 sets of arbitrary sets of polynomials. But in the end, one only needs finitely many, so it somehow feels a bit more comfortable. So now, who's that? Now I want to give an application of a geometric application in some sense of this Hilbert-Basey's theorem. I already hinted at it. So the statement is that every fine algebraic set is a finite union of a fine algebraic set which cannot be further decomposed. So we'll have to see precisely what I mean by that. So there is some kind of way how you can really say that in a fine algebraic set is a union of finitely many pieces. And these pieces will be called the irreducible components. So irreducible components. So you know, for instance, that if in A2, if you take the 0 set of x times y, where x and y are the coordinates, then we know this is the 0 set of z of x, union is 0 set of y. If you look at the real points, then it would look this one would be the union of two lines, the horizontal and the vertical line. And you can evidently see that this thing in some sense has two pieces. And somehow we want to see that it's always like that. And what one cannot see here immediately, but which turns out to be true, is that these pieces cannot be further decomposed. One cannot write it as a union of a fine algebraic sets. I mean, at least not a finitely many. So in order to make sense of this statement and prove it, we want to formulate it in a topological way. So we have this risky topology. In terms of this, we can say what we mean by something being irreducible. And then we want to show this statement about the irreducible components. So let's see. So formulate it more generally as a topological statement. Then later, such statements can be applied in slightly larger generality. So we say definition. A topological space x is called reducible if we can write it as a union of two closed subsets, none of which is equal to x. So if x is equal to x1 union x2, where x1 and x2 are closed subsets, and x1 and x2 are strictly contained in x. So to say it, so for instance here, this is evidently the case. This zxy is the union of z of x and z of y, and both of them are closed subsets. And none of them is equal to this union. And then what we really are interested in is irreducible, which is the opposite. So something is called irreducible if it's not reducible. So explicitly it says that, I mean, because we use it most like in that way. So if x is irreducible, if x is the union of x1 union x2 with xi in x closed subsets, then it follows that x is equal to x1, or x is equal to x2. So that is what it means to be irreducible. OK, that's just the definition. So maybe I should say that these topological statements, we will apply to find algebraic varieties with the Zariski topology. And if you look at it in the usual sense, if you look at subsets of Rn with the usual topology, these kind of concepts make essentially no sense. I mean, if for instance in a Hausdorff topological space, the only irreducible subsets are the points. So it's not such an exciting concept if you use this in the things you usually study. But the Zariski topology is very strange. This concept does make sense. So one thing is that, for instance, if x is irreducible and u subsets x an open subset, then it follows. So and u is non-empty. So any non-empty open subset of an irreducible space, then the closure of u, then u is dense in x. So if you have an irreducible topological space, then any non-empty open subset is dense. And this is easy because we can certainly write x equal to x minus u, which is a closed subset because u is open, union, the closure of u, which is also a closed subset. And as x is irreducible, it follows that x is equal to x minus u or x is equal to the closure of u. But u was non-empty. So this is obviously not true. This is not true. So it follows that x is equal to the closure of u and u is dense in x. And in a similar way, you can show it's an easy exercise. So under these assumptions, also u is irreducible. So if x is irreducible and u is a non-empty open subset, then u is also irreducible. And it's an easy exercise in the definitions to see that. So now we give some examples in the examples. So first, as I said, a point P in an is irreducible. Because if you write it as a union of two closed subsets, then one of them must be the point. And then we also see that the zero set of x, y in a2 is reducible because it's a union of z of x and z of y. And you can easily see that none of these two is equal to x to the zero set of these two. Take the point, whatever, 1, 0. It will lie in one. If you take the point 0, 1, it lies in the other. And so this is reducible. Now we want to show the statement that I made about irreducible components. We want to show that a fine algebraic sets can be written as a union of finitely many irreducible fine algebraic sets, where I call in the fine algebraic set irreducible as it's irreducible as a topological space. And again, to make it more generally applicable, we will prove a slightly more general statement in a topological context. So we first define something which will be true for the fine algebraic sets. And then we prove that the statement about decomposition into irreducible components holds for each such thing that we define. So this is the following statement. Definition, a topological space is called Neuterium if the following holds for if every descending chain, so x contains x1, contains x2. So an infinite chain like this of closed subsets becomes stationary. So that means from some point onwards, they are all equal for some n. So this is obviously very similar to the definition of a Neuterium ring. So there we also have some chain condition. But for a Neuterium ring, we had the ascending chain condition. So if you have a chain of ideals where they become larger and larger, then they will become stationary. And here we have it for closed subsets, but they become smaller and smaller. But if you think of it, this should somehow correspond. And we'll see that because for a fine algebraic sets, if we have an inclusion i1 is contained in i2, then the zero set of i2 is contained in i1. We have the inclusion gets reversed so that in ascending chain will become a descending chain. So its definition is precisely made in such a way to match, via this kind of duality, the Neuterium ring for rings. OK, so I make some mark. So first, any subspace of a Neuterium topological space topological is Neuterium. So we have a subspace, we have a topological space. We will call the subspace y of a Neuterium topological space x is Neuterium. So subspace mean is a subset with induced topology, as usual. So this is almost obvious. Let's see. So assume we have a descending chain. Take a descending chain in y. So y contains y1, contains y2, a descending chain of closed subsets. Well, the subspace topology means precisely that I can take these closed subsets as the intersection of y with closed subsets of x. So then we have that for all i, yi can be written as x intersected xi, where xi in x is a closed subset. So now one could be tempted to say that the xi form a descending chain in x, but this is actually not true. But with the tiny modification, they do. So we put xi prime to be the intersection for all j smaller equal to i. So one smaller equal to j smaller equal to the given i. So for all j from 1 to i of the xi. Then it is clear that if I take xi prime intersected y, this is the same as x intersected. It's the same as yi. Because you know that the intersections of the xi with y form a descending chain. So if we first intersect them in this way, we get the same intersection with y. And now these xi prime form a descending chain in x. So then x descending chain of closed subsets of x. So it becomes stationary. So for some n, we have that xn prime is equal to xn plus 1 prime. Well, and as the yi are just given as intersection of y with this thing, with these, so then they also become stationary. Thus also this chain becomes stationary. So these are some very easy observations. So I also need a second statement to now finally come back to our fine algebraic sets. So the claim is that an is in a terren topological space. And thus by 1, every a fine algebraic set, and in fact any subset of an with induced topology is in a terren. So we only have to prove this statement. The second follows by 1 because we have the statement that any subspace of a terren topological space is in a terren. So how is that? Well, and this is precisely the story with the chains of closed subsets corresponding to chains in the other direction of ideals. So let say x contains x1, contains x2 be a chain. So x is an, after all, of closed subsets in an. Well, then I can look at the ideals of these closed subsets. So the ideal of x1 will be contained in the ideal of x2 is an ascending chain of ideals in kx1, xn. So we know it becomes stationary. Thus we have that same, it becomes stationary. It means i xn is equal to i of xn plus 1. And but note that we have seen that for a close, for a fine algebraic set, the zero set of the ideal is equal to the set itself. So thus also xn, which is the zero set of i of xn, is equal to xn plus 1. So this proves that an is in a terren topological space. We have a chain, our given chain of ideals of closed subsets becomes stationary. And so whatever we have said about a new terren topological spaces applies to a fine algebraic sets. And also to open subsets of a fine algebraic sets, whatever we want. Now we want to make a statement about these irreducible components. This is theorem and also definition. So we prove that something exists and then give it a name. So let x be a terren topological space. Then there are two statements. First, x is a union of finitely many irreducible topological spaces. So I can let x is equal to x1 union. Maybe irreducible, I don't think I want like this, irreducible closed subsets. And the second statement is that in a suitable sense, this decomposition is unique. There's some stupid thing you can do. You can always add to this decomposition some closed subset, which is contained in one of the xi's. You can add as many additional components as you want by just taking something very small, which fits into one of them. But if you don't do such a stupid thing, so if you don't allow some of these components to be contained in another one, then this decomposition is unique. So if we require that xi is not contained in xj for i different from j, then this decomposition is unique up to reordering. So let's see. So I mean, you can. So we want to see. I mean, the most important thing is the first part. We want to see that we can write x as a union of finitely many irreducible closed subsets. So let's start with the proof. So we first show, in some sense, the existence. So basically 1. So for the existence of a decomposition like this, also if you want to have this property, it's enough to show the existence of any such decomposition without this property. Let me say it more precisely. So it's enough to show the existence of a decomposition. x is equal to x1 union xr with xi irreducible closed subsets. So for the existence part of 2, which means there exists such a decomposition where the xi are not contained in the xj, is trivial. If some xi is contained in another xj, we just throw it away. And the union of all the xi's is still x. Because after all, the thing we threw away was contained in another one. So by removing the xi with xi contained in xj for i different from j, we get a decomposition as in 2. So x equal to the union. And xi is not contained in xj for i different from j. So this is kind of clear. Now we want to prove just that we have a decomposition into finitely many irreducible closed subsets. And this we do by an indirect argument. We assume such a thing does not exist and we get to a contradiction. So assume x does not have a decomposition into finitely many irreducible closed subsets. So we make this assumption. We will want to come to a contradiction. So one thing, if x doesn't have such a decomposition into finitely many irreducible closed subsets, certainly x is not irreducible. Because otherwise we could take x equal to x as the decomposition. So in particular, x is reducible. So we can write x is union of x1 and y1, where x1 and y1 are closed subsets. And it can also not be that both x1 and y1 have a decomposition into finitely many irreducible closed subsets. Because otherwise we just take the finitely many for this plus the finitely many for this. And we still have finitely many for x. So one of the two does not have a decomposition into finitely many closed irreducible closed subsets. So one of the two and whose x1 and y1 is just names that we give. So say y1, say x1, does not have a decomposition into finitely many irreducible closed subsets. But if you look at it now, this is precisely the assumption we made about x. The only assumption we made about x is that it does not. So I wrote something else. I said x does not have a decomposition. And I wrote it does. But obviously the assumption was that. Otherwise, how do we bring it to a contradiction? So we assume that x does not have a decomposition into finitely many irreducible closed subsets. Then in particular, x is irreducible. And now we find that one of the two does not have a decomposition into finitely many irreducible closed subsets. This is precisely the assumption we made about x. So whatever we have deduced about x, we also deduce of x1. So we can again write x1 as a union of two things, one of which does not have a decomposition and so on. So we just repeat the argument with it. So this is the same assumption, same statement as the assumption about x. So we get, we can repeat the argument. So repeat the argument. Then we get that x1 can be written as x2 union y2, where x2 does not have a decomposition into finitely many irreducible closed subsets. So we get, we can again apply to this. So we get inductively that in descending chain, x contains x1, x2, and so on. And if you look at it here, I mean, I haven't been careful in writing that x is the union of x1 and y1. But part of the story is it's the union of two such things, none of which is equal to x. We have in particular also that x1 is different from x and y1 is different from x. Because being reducible means that you can write it as a union of closed subsets of two closed subsets, none of which is equal to x. And the same applies here. So this means here x1 is not equal to x, x2 is not equal to x1, and so on. So we get a descending chain where at each step it gets strictly smaller. So that means the chain is not stationary. And this is a contradiction to x be in the terrain. So we can conclude that we were wrong in the beginning. And x does indeed have a decomposition into finitely many irreducible closed subsets. So we get a statement of one. OK. So now the second is sometimes. So you see, that's somehow how one usually proves results using this notoriety that you somehow make such a chain and bring things to contradiction. It always leads to some indirect argument which is very non-constructive. I mean, it actually gives you no idea how you would find these components. It just is an existence result. But this is what this assumption will give you. So let's prove two, I mean, or rather the uniqueness. So we assume now we have a decomposition into x1 union xr such that no xi is contained in another xj. Now we have seen that we can get that. And actually we assume we have two such decomposition. ys, the two compositions as in two. So x is the union of the xi. It's also union of the yj. These are all irreducible closed subsets. And none of the xi is contained in another xj. And none of the yi is contained in another yj. And we want to show that these are the same things. r is equal to s. And the yi are permutation of the xj. So we can certainly write xi is the intersection of xi with x, which is the same as the intersection as union. i equals 1 to s of xi intersected j equals 1 to s yj. Because after all, the union of these is just x. And so this is certainly the case. But now xi is irreducible. And these are closed subsets. So it follows that if xi is equal to the union of these finitely, many it must be equal to one of them. It follows that xi is equal to xi intersected yj for sum j. But what does that mean? It means that xi is contained in yj. So we find that each xi is contained in yj. In the same, I mean, each xi is contained in sum yj. In the same way, we can exchange the role of the x and y here. So similarly, we have that yj is contained in xk for some k. So we have this chain that xi is contained in yj is contained in xk. But then obviously, xi is contained in xk. And we have excluded that. The only possibility how this can be the case is that xi is equal to xk. And actually, the k is equal to i. So it is only possible if i is equal to k. But in this case, you see that yj is squeezed here in between. It contains this. And it's contained in the same. So they are equal. And yj is equal to xi. So we have that each xi is equal to sum yj. In the same way, you can exchange the word each yj is equal to sum xi. And this just means that the yj's are just the permutation of the xi's. We have here r different things. We have here s different things. But each of them is equal to one of these. And each of them is equal to one of these. So the number of things is the same. And they are just permutations of each other. So r is equal to s. And the xi's, or yj's, are a permutation, the xi's. So this was quite, OK. So here, this is actually quite simple. So we just use this condition that we put here. If one is contained in the other, they must be equal. Then just uses this together with this condition of irreducibility here. And one gets a diminutive. So the reason why we introduce these irreducible algebraic sets is that it's, from some point onwards, we will mostly just consider those. So in the future, but not immediately in the future, but most of the time, we will mostly consider only irreducible algebraic sets. And that's OK, because we know that if we have a general defined algebraic set, it's just a finite union of these. So if we know all the pieces, we know the thing. It's not such a thing. And so we consider it important enough that we give it a name. So the definition and the fine variety is an irreducible defined algebraic set. And I might see in a moment. And now one thing I had mentioned when I talked about the ideal of an defined algebraic set that because the ideal is kind of associated in a canonical way to the defined algebraic set, one should have that the algebraic properties of the ideal should somehow reflect some geometric properties of the defined algebraic set. And the first instance we have here, an defined algebraic set is irreducible if and only if its ideal is a prime ideal. This is a proposition. x in a defined algebraic set. Then x is irreducible if and only if. The ideal of x is a prime ideal. So we have some, at least, we have a somewhat interesting concept in commutative algebra, which is related to this geometric property of being irreducible. And this is actually quite simple. If you think of it, so we have these two directions. I would want to prove that. So we assume, say, x is irreducible. And then we have to show it's prime ideal. So recall the definition of a prime ideal, so an ideal. So that means ix is an ideal. And in addition, if a product f times g, where f and g are elements of kx1 to xn, so prime ideal in kx1 to xn. If f and g are two polynomials that take their product, if that lies in ix, then it follows that f is in ix or g is in ix. This is the definition of a prime ideal, once it's an ideal. So let x be irreducible. Well, and then we take f and g, some polynomials, such that f times g lies in the ideal of x. So then our task is to prove that one of the two already lies in the ideal of x. Well, that's very simple. So the ideal of x is a set of all polynomials which vanish on x. So it means that if I take f times g vanishes on the whole of x, or saying it in a different way, we have that x is contained in the zero set of f times g. And so in particular, we can write x to be in similar fashion as here, to be x intersected the zero set of f. And recall that this is just the zero set of f union zero set of g. And from this one can already see that these concepts should be related. So therefore, we can write x as the union of x intersected the zero set of f, union x intersected the zero set of g. Because union of c of f and g of g contains x. So x is union is equal to that. So these are two closed subsets of x. So two closed subsets of x. And x is irreducible, so one of them must be equal to x. So it follows x is equal to x intersected z of f, or x is equal to x intersected z of g. But what does it mean? If x is equal to x intersected z of f, it means that x is contained in the zero set of f, which means that f lies in the ideal of x. And this was precisely what we had to see. So this is the one direction. And now the other direction is not much different, except that we decide to do it indirect. So we assume that x is reducible, so not irreducible. And we have to show that ix is not a prime ideal. So assume x reducible, show ix is not a prime ideal. Well, so again, we just spell out the definitions. And then we cannot avoid proving the result. So we assume it's reducible. So x is equal to x1 union x2, where the xi are closed subsets of x, and none of them is equal to x. So we have the zero set of the ideal of x1. We know that this is equal to x1. And this is contained in x, which is the zero set of the ideal of x. And it's not only contained, it's strictly contained. So we have this and same for x2. So we have that as we have the strict inclusion like this, we must have the strict inclusion the other around between the ideals. Because if the ideals were equal, then the zero sets could not be different. So it follows that i of x1 contains strictly i of x. And in the same way, obviously, ix2 constrain strictly i of x. So we can find an element f in i of x1, which does not lie here. And we can find in g here, which does not lie here. So let f in i of x1 on i of x, in g in i of x2 without i of x. Well, so here we have found two elements which do not lie in i of x, but the claim is obviously that the product lies in i of x. And so it's not a prime ideal. So obviously, if I take f times g, so f vanishes on x1. g vanishes on x2. So f times g vanishes on x1 union x2. So z vanishes just as a union of the zero sets. And this is equal to x. And as the ideal of x is just the zero set is just the set of all functions which vanishes on x. It means that f of times g is an element in the ideal of x. So we have indeed found two elements, f and g, which do not lie in the ideal of x that the product does. So i of x is not a prime ideal. OK, and so this proves this statement. So in the moment, this doesn't help us so very much to decide whether a given fine algebraic set is irreducible. So one thing we can see as an example is that An is irreducible. That was maybe not so clear before, but now it's clear because we know that An, what is the ideal of An? So what are all the polynomials which vanish on the whole of k to the n? I mean, a polynomial vanishes on the whole of k to the n precisely when it's 0. And we know that kx1 to xn doesn't have zero revives or anything, so 0 is a prime ideal, which is a prime ideal. So we see that An is indeed irreducible. I should maybe say that this fact that if x is irreducible, then its ideal is a prime ideal is actually the main reason why we will want later to restrict attention to irreducible algebraic sets to a fine varieties. Because it is very useful for us that the ideal is a prime ideal. The main reason is the following. So this is a remark. So if x is in a finite algebraic set, so we will consider the following, the so-called a fine coordinate ring, A of x. So I assume that x is maybe contained in kx1 in An. So Ax will be just defined to be all polynomials divided by the ideal of x. And we will use this ring very much because we can view this as somehow functions on x. And we use this to make maps between such things. So these are somehow our functions on x. And so we will make a lot of use of this. And the point is that if x is irreducible, if and only if ix is a prime ideal, if and only if this Ax is an integral domain. And it's very useful for us that this thing is an integral domain because it's much easier to work with integral domains than with rings which have zero devices. In particular, as we will want to consider the so-called the field of fractions of A of x, which doesn't even exist if it's not an integral domain. OK. Yeah, well, these are polynomial functions somehow. But I mean, we will come back to that. So you find, if you think of it, that these are all the functions from x to k which can be obtained as restrictions of polynomials to x. OK. And these are the kind of natural functions that one might want to consider. But we will not, in the moment, do this. But this is the main reason why we later will restrict our attention to irreducible algebraic sets and call them varieties. Yeah, I want to very briefly also touch some other subject which we will not give us. So just that you have seen it when we finally do it. So I want to just say two words about dimension. So there's a concept of dimension for a fine algebraic set and which surprisingly turns out to work rather nicely and similarly to what one would expect. It's part of the fact that the definition will look rather strange to you. And so again, we introduce the dimension more generally for so-called, for neutral anthropological spaces. So we have a, but you think of a fine algebraic set. And the idea, the basic idea is very simple. So see if x is irreducible and y is also irreducible and y is a close subset in x, then somehow y is smaller than x. And we would really feel that y is given by some close condition in x. It should really have a smaller dimension. So then we want the dimension of y is strictly smaller than the dimension of x. And somehow if we cannot fit another irreducible close subset between y and x, we would want that the difference of dimension is precisely y. This somehow should define the dimension. So one can actually turn this into a definition. As I said, we are not going to work with this now. We will introduce it once. And then in the second half of this course, we will actually work with it and show that one actually can prove something about it. But in the moment, we don't have the tools. So let x be a, say, non-empty irreducible topological space. The dimension, which we call dmx of x, is the largest n so that we have some chain of close subsets. So it's the largest n is an ascending chain, empty set, x0 strictly contained in x1, strictly contained in x2, and so on. And finally, we come to xn, which is equal to x, of irreducible close subsets. And if x is not irreducible, if just in the terrarium space, not irreducible, we take the maximum of the dimensions of the irreducible components. So if x is in the terrarium topological space, then the dimension of x equal to the maximum of the dimensions of the irreducible components of x. Now this, I think, reminds me of the fact that I did not, or that I at least don't recall, but I don't think I properly did it. When we have had this theorem and definition, I somehow forgot the definition part. So we had the statement that if, so I just make this as a side remark to have it once completely on the blackboard. So we had x in a terrarium topological space. Then we had that x can be written as x1 union xr, finite decomposition with the xi irreducible close subsets. And this was unique if we assumed that xi is not contained in xj for i different from j. And then under these assumptions, so when we have this thing, this unique decomposition, the xi are called the irreducible components of x. So now this definition here makes sense if we have an irreducible topological space. The dimension is the length of the longest chain of the largest n for the length of a chain of irreducible close subsets. And if x is not irreducible, we just do this for each irreducible component of x. And we take the maximum of the dimensions. And that's the dimension of the whole thing. So this definition in the moment is not very practical because we don't have any idea how to find out something about all close subsets of an affine algebraic set. So there are very few things that we can say. We can say that, first, the points p in n have dimension 0 because the chain will only consist of the point itself. And then that's it. As it also should be, a point has dimension 0. And it's also comforting that a1 has dimension 1. And this we actually proved the other time without seeing it. Namely, we had seen that the only irreducible close subsets of a1 are the points and a1 itself. Because we have seen that the close subsets are the empty set, the finite sets, and a1. And so if you have a finite set of points, it's not irreducible, unless it's just one point. So we only have this. So the maximum length of a chain is the empty set contained in a point, contained in the whole of a1. So the dimension is 1. And something which we cannot see, which obviously is necessary for the whole thing to make any kind of sense, is that in the moment we cannot prove, but we will later, but it's true that an has dimension n. If you look at this definition, this is not so clear. But it's evident that the dimension of n is at least n. So it's easy. It's bigger equal to n. I mean, if, for instance, you are in two-dimensional space, what kind of chain can you take? You can start with a point. Then you take a line through the point. And then you take the whole thing. So this is a chain of length 2. You can do the same thing in an. You take a point. You take a one-dimensional linear subspace. Then a two-dimensional linear subspace. And each of these are fine algebraic sets. I mean, you can see that they are irreducible. Maybe that is not so evident. But it is quite evident. But anyway, as we are not proving it, so let me just write it down. So we have an ascending chain. So 0.0 is contained, say, in the 0 set of the variables x2 to xn. This is contained in and always strictly 0 set of x3 to xn and so on. 0 set of xn contained in an. And this is a chain of length n. It's not completely evident that these are all irreducible. We will see this later soon anyway. But if you believe that for the moment, we have an ascending chain of length n of irreducible closed subsets. And so we get the dimension of n is at least n. And later, we will prove really something about the theory of dimensions. We'll see that there's actually a lot that one can say. But for this, we need to develop quite a few tools. So now we just go. So this was just some kind of preview for what we want to do. So now we come to another big theorem, which is the Hilbert-Nulsteinsatz. So let me, this is another section kind of. So this is again after the well-known German mathematician David Hilbert. And Nulsteinsatz is a German word, which is how everybody calls it in spite of just because that's how Hilbert called it. And this means theorem of zeros. But if you look in a book, it's called like that, not like that, even if it's an English book. I mean Nulsteinsatz means zero. Nulsteinsatz means the place where the zero is. And Zatz means theorem. So it's all easy. Now, so this is about the relations of ideals and their zero sets. So remember, if you have an ideal in the polynomial ring, we can associate to it, we have its zero set, which is in a fine algebraic set. On the other hand, if x is in a fine algebraic set, we can associate to it the ideal of x, which is an ideal. So if you want, you can say we have two maps between the ideals and the fine algebraic sets in both directions. So we have here, say, which we do want it, the fine algebraic sets in the N. And here, we have the ideals in kx1 to xn. And we have two maps. We can associate to a fine algebraic set its ideal. And we have a map in the other direction, which associates to an ideal its zero set. Now, in the best of all worlds, you could hope that, for instance, these are inverse bijections. So that we somehow talking about the fine algebraic sets is the same as talking about ideals. Just you can go backwards and forwards. We have to figure out whether that's true. And it actually turns out to be not quite true. So are these inverse to each other? If not, what is the precise relation? So we had already seen, this was an exercise we know, that if we take the zero, so if x is in a fine algebraic set, and we take the zero set of the ideal of x, this is equal to x. More generally, if x is any subset of an, then the zero set of the ideal of x is the closure of x. So we know, at least, that if we start from here, we go here, and here we get the identity, but it turns out, we'll see it's not true, that in the other way, we get the identity, that the ideal of the zero set of an ideal j, the j ideal, is equal to j. It is clear again, whether it's again obvious, but clearly we have that the ideal of the zero set of j contains j. This is just by definition, because what do we have here? We have the zero set of j is a set of all points, where all point on what? We will see, I write it a little bit larger, we will see. This is not true. In one direction, it's true. The ideal of the zero set of an ideal is that ideal. And I also, even in the other direction, it's almost, there's some part of it, too. If we take, so it is not, this will not be true, that this is equal, but we will see that it's almost, that at least this is true, this we see immediately, and I was just explaining it. So the zero set of j is a set of all points, where all elements of j vanish. And the ideal of the zero set is a set of all polynomials, which vanish on the set. As the elements of j vanish on the set, j belongs to it. So it's just by pure elementary logic, this is true. But as I said, the other inclusion will generally not be true, and the answer will be a little bit more interesting. But first, before one does that, our time is almost up. Let me see. But we can ask a simpler question first. First, we ask a simpler question. Namely, when is it true that the zero set of an ideal is the empty set, or when it's not the empty set? So we know that the zero set of the whole polynomial ring is the empty set, because in particular, this polynomial contains the constant polynomial 1, which vanishes nowhere. And the first theorem, which is the weak form of the Nolsternsatz, says that this is the only ideal for which the zero set is the empty set. So let i, strictly contain in x1, px1, xn, your proper ideal. So if I have an ideal which is not equal to the whole of the polynomial ring, then the zero set of i is non-empty. So this is a surprisingly difficult theorem. It's so difficult that I don't want to prove it now. So it's not outrageously difficult, but the proof is very long. I mean, it would take me maybe, I don't know, two lectures or something to prove it. So in the notes, there is actually a proof, but it's in the chapter about dimension. So if you have time and you are interested, I will prove it then, because then we have developed the tools to do this. If now we have to do it from scratch, it would take a very long time. So we, for the moment, assume this result. And then we will prove as a consequence the strong form of the nullstellensatz, which tells you about the precise relation between ideals and their zero sets. And OK, and then we will do next time. OK, thanks.