 So we had started talking about projective archbikes sets. So we'll review the notations and then we go on. So we had the projective space Pn, which was Cn plus 1 without 0, divided by the equivalence relation where we have that two vectors are called equivalent. If one is a multiple of the other by a nonzero constant, the quotient space is a projective space. And then we had seen that if we have a homogeneous polynomial, we can talk about point in Pn being 0 of that. So we say that f of the corresponding point here, the equivalence class is just done by the thing with square brackets. So if Pp equal to a0 to an, we say that f of P is equal to 0. If f of a0 to an of just this n plus 1 double of numbers is equal to 0, and this will be independent of the representative. And in the same way, now if we have a set, like in the affine case, if s is a set of homogeneous polynomials, we can talk about its zero set, the set of all points in Pn, where all these polynomials vanish. And then we had related, again, as in the affine case, these zero sets to ideals. So instead of taking a zero set of a set of homogeneous polynomials, we can take a so-called homogeneous ideal and take the zero set of that. And that, again, allows us to use a bit more algebra. So I don't recall the definition of homogeneous polynomial. One way is to say it's generated of homogeneous ideal. One way is to say that it's generated by homogeneous polynomials. So if i in kx0 to xn is a homogeneous ideal, then we can define the zero set of i to be a set of all Pn and Pn, such that f of P is equal to 0 for all homogeneous elements in i. Because until now, we have only said what it means to be 0 for homogeneous polynomial. And in the same way, we can associate a homogeneous ideal to a projective algebraic set. So I should say that sets of the form z of s for some set of homogeneous polynomials are called projective algebraic sets. So x in Pn is a projective algebraic set. And in fact, if it's any subset, but just say for in this case, then the ideal of x is defined to be the homogeneous ideal of x is defined to be the set of all the ideal generated by all the polynomials f in kx0 to xn, such that f of P is equal to 0 for all P and x. This is again also completely analogous to the defined case. But now we take the ideal generated by them because we have not set what it means to be 0 at f of P otherwise. I should say that later, we will sometimes need to look at the projective 0 set and the defined 0 set at the same time. So the 0 set of this ideal in Pn or the 0 set of this ideal in An plus 1. And then so we distinguish that. So in case of need, we call this zp of x of i. And the other one may be za of i, one for projective, one for affine. And this is the homogeneous ideal. So in case of need, we call it ih of x, the homogeneous ideal. We'll see that later. So after saying that, the story that we only know what it means to be 0 when the polynomial is homogeneous, I want to immediately contradict myself and say we can also define it if the polynomial is not homogeneous and also give the opposite answer to a query about this than I gave for the other time. But it's just one can remark, also an exercise. So if f in k at 0 to xn is not homogeneous, we can define the 0 set of f to be the set of all points P equal to P in. So P equals A0 to An in An such that f of A0 to An is equal to 0 for all representatives A0 to An of P. So in other words, for all, so it's 0. So if we fix any A0 to An here, it must be 0 for all lambda. Lambda A0 times lambda An must be 0. So we could say this is the 0 set of this polynomial. So the first thing that you can notice is, so quantum exercise. So if f equal to f, say, f0 plus 1 plus fd is the decomposition into homogeneous components, then the 0 set of f is the intersection from over all i, in this case, i from 0 to d of the 0 set of f i. So we see that, so in some sense, if f is not homogeneous, it doesn't just give us one equation, but many equations. And for instance, if f has a non-zero constant part, it means that the 0 set is always empty. Hope this is correct. But this is just a small thing. The actual exercise is another one. With this definition, so that 0 set of a polynomial is actually the set of all points where 0 for all representatives, we get, so if i in kx0 to xn is a homogeneous idea, then the 0 set of i can also be written, the 0 set of i, as we have defined it here, as defined before. This is also equal to the set of all p and pn, such that f of p is equal to 0, for all f in i. So if we take this strange definition for the 0 set, then actually, we don't have to restrict the homogeneous elements. And in the same way, if x in pn is a projective algebraic set, then it's ideal, as defined here. So the homogeneous idea is just equal to the set of all f in kx0 to xn, such that f of p is equal to 0 for all p and x. So if we change what we mean by the 0 set by saying it should be 0 for all representatives, then we can use the same definition as in the affine case. But it means something complicated to be 0 for all representatives, as you see here. So but I mean, I say this just now. We will not use this very much. We'll use it once. Otherwise, we usually restrict our attention to homogeneous polynomials and work with that, because it's just easier. OK, so now we can go ahead. So now, basically what we want to do now, I mean today, is to repeat everything that we did to the previous lectures for the affine case for the projective case. So with rather small differences, one can do the same things. And the proofs are very similar, so we can go over it rather fast. So the first thing that we had in the affine case is the Zariski topology. So we want, again, to say that the projective algebraic sets are the closed sets of a topology on pn. And that's very. So I just stated here as a proposition. So let's say that, so for instance, if x is contained in y, what? Into homogeneous components. So we have, yeah. So if I f right as a sum of its homogeneous component. So this is the part of degree 0. This is the part of degree 1, part of degree d. Then the 0 set of f defined in this way is the intersection of the 0 sets of the homogeneous. The degree is the number I put up here. So say if x and y in the n are projective algebraic sets, then we have the first thing that, say, the idea of x contains the idea of y. And the proof is in the same way, the same as before, because we just have less conditions here. I mean, if you look at the definition, this is completely obvious, so this is clear. Second, if, say, y, say x in pn is projective algebraic set, then if I take the 0 set of the deal of x, this is equal to x. And remember, we made some proof of that in the affine case. And you just repeat the proof. It was a formal argument, which we will use in the same way. In the same way, if I in kx0 to xn is a homogeneous ideal, if I then take the ideal of the 0 set of I, this will contain I. This is, again, obvious from the definition, because the ideal of the 0 set of I is all the polynomials which vanish, where all the polynomials in I vanish or all the homogeneous polynomials in I vanish. And obviously, this contains I. And do I really need this? OK. So and it's again clear that, so for instance, if s in kx0 to xn is a set of homogeneous polynomials, then the 0 set of s is equal to the 0 set of the ideal generated by s. And this is just the same proof as before. It's kind of clear that the ideal generated by s contains s. Therefore, the 0 set is possibly only smaller. But then you can see that if you write any element here as a combination of any element here as any combinations of elements here, you see that also the other inclusion is true. And then we come to the thing which gives us the risk topology. So these are the two things. So if I have a family, family or a set, say s alpha of sets of homogeneous polynomials, we find that if we take the 0 set of the union over all alpha of the 0 sets of s alpha, this is intersection over all alphas of the 0 set of s alpha. If you think of it, it's again completely obvious because we just, there is obviously, except that there's a misprint like this. It just means you put all these conditions at the same time. And this just means you take the section of the 0 sets where you have movies. And then the last one is that if s and t in k and 0 are sets of homogeneous polynomials, then the 0 set of if I write s times t, which is just a set of all products of one element in s and 1 in t, notice these will be also all homogeneous if all elements in s and all elements in t are homogeneous, is equal to the 0 set of s union 0 set of t. And this is, again, the same proof as before. If you just remember how we proved this, you can just see that an element in the product will be 0 at the given point if one of them is 0 there. And that's all that goes into the proof. So in particular, if you take the last two arbitrary intersections and finite unions of projective algebraic sets, are projective algebraic sets. Now, this is the content of 5 and 6. And we have already seen that the empty set and the whole of Pn are projective algebraic sets. So that means we have that the actions of the closed sets of a topology are fulfilled. So we have a topology whose closed sets, a topology on Pn whose closed sets are the projective algebraic sets. And this we recall the Zariski topology on Pn. And then every subset of Pn is given the induced topology. This is, again, the Zariski topology on that subset. So let me just write this down. Yes? Yeah, also that. No, so this, I didn't prove that the other time, but it's also true. It's also equal to, it's an exercise to prove that this is the same SS0 set of the sum. So all three are equal. But in the moment, I only need that, so I also only wrote that. But you're right. You can also take the sum. So this, if you want, you can do as an exercise. This is your set of the sum or the sum or the idea, whatever, we'll write sum. It's anyway the idea generated by the, you could say the idea generated by the union, which if you want, is the sum. I don't even know how I would want to write that. So maybe say, and if you want, you can call this. So you take the idea generated by the union of the alpha. This will give us that. And if you want, yeah, I think that's more correct. OK, so you want to talk about the Zariski. I wanted to just write down the definition of the Zariski topology so that we have it once in the blackboard, although it's obvious. So the Zariski topology on Pn is the topology whose close subsets are the projective algebraic sets. And as before, if x in Pn is any subset, we give it the induced topology. So the topology shows that such that a close subset of x will be intersection of x with a projective algebraic set. And in particular, we use this when x itself is a projective algebraic set. It automatically has topology. And this is also called, and this is a risky topology on x. So we have introduced this thing. And so we will most of the time talk about so-called quasi projective algebraic sets. These are open subsets of projective algebraic sets. So definition quasi projective is an open subset projective algebraic set. So these are sometimes the most general kind of algebraic set. So the most general things we want to study. OK. So the next thing that we had was the decomposition. So in the affine case, it was a decomposition into irreducible components. So if you have an affine algebraic set, it's in a unique way, a union of irreducible affine algebraic sets with a property that none is contained in another. And these are called the irreducible components. And so we want to see that the same solution here. And if you remember, this was done before by somehow using the fact that AN is a Neutering topological space. So that is, if you have a descending chain of algebraic sets, then it terminates. And this follows from the fact that kx1 to xn is Neutering as a ring. Because if you have an ascending chain of ideals, it becomes stationary. And so we want to see that, obviously, the same works here. And then we immediately get this decomposition into irreducible components. So remark, proposition. So we know that kx0 to xn is Neutering. This is, after all, the Hilbert-Basey's theorem. Whether we call the variables x0 to xn or x1 to xn is somehow of no relevance. So thus it follows that the same proof, as in the affine case, shows that Pn is a Neutering topological space. We have all the same repeated, but it's anyway a Neutering topological space. Maybe I can, namely, what was it? I mean, I will all the same, the same. So it first, x1, x2 is a chain of closed subsets of Pn. Then I can look at the corresponding chain of ideals. So x2 is a chain of ideals in kx0 to xn. And as this is a Neutering, it becomes stationary. It becomes stationary. So beyond a certain n or the i of xi are all the same. And as we know that xi is equal to the 0 set of i of xi, we know that if the ideals become stationary, then also the sets become stationary. Maybe this is like this. Also the original chain becomes stationary. So if at some point the ideals become always the same, then the 0 set becomes also the same. OK. So from this, it follows that Pn is a Neutering topological space. And we had stated the other time that if you have a Neutering topological space, then any subspace with the induced topology is also Neutering. So therefore, any subspace of Pn is a Neutering topological space. In particular, every quasi-projective variety is Neutering. So we have that. And then we had proven the existence and uniqueness of the decomposition of inter-reducible components in general for Neutering topological spaces. So it also holds here. So we have already proven that every Neutering topological space has decomposition into irreducible components. And so we have it. So in particular, we have, so it follows that quasi-projective varieties have a unique decomposition into irreducible components. So I mean the statement would be that if you have a quasi-projective variety, we can write it as a finite union of irreducible quasi-projective varieties. And if we require that none of these quasi-reducible quasi-projective varieties is contained in another one, then this decomposition is unique, is precisely as in defined case. So you just take the formulation that we had in defined case, and you replace a fine with quasi-projective, and that's it. Actually, the statement was in general for Neutering topological spaces, so it follows anyway. OK. Oh. Yeah, well, I haven't yet, yeah, so algebraic sets. So it's actually, it's a debatable issue whether varieties are irreducible or not. That depends on the author. I mean, in some books, you find variety as a different word for algebraic set. And in some books, you find varieties always irreducible. So with me, a variety is always irreducible, except if I make a mistake. But if you look into a book, it sometimes is different. It depends basically on the country where the author comes from. So if you find a book in French, variety will usually not be irreducible. But if it's in English, it usually will be. So now, actually, this was precisely the point where I was now. So the definition goes now anticipated. But a quasi-projective variety is an irreducible quasi-projective algebraic set. Sometimes I just say variety instead of quasi-projective variety. There is a more general notion in algebraic geometry of abstract algebraic variety, which is really more general. But we are not going to deal with it in this course. So therefore, I don't have to distinguish from that. What else? So if we had identified, usually, identification from the beginning of the problem of projective algebraic sets, that an is viewed as a subset of pn, namely, an is supposed to be identified with this u0 in pn, a set of all points like that, where the first component is 1 or non-zero, which is the same, just by identifying a point a1 to an with 1, a1 to an. If we use this identification, we find that an is an open subset without the 0 set of the polynomial x0 is an open subset of pn. So an is a quasi-projective variety. And in fact, we get that any affine algebraic set is a quasi-projective algebraic set, just because it would be a closed subset of this open subset of pn. OK. So this was as much as we wanted to say about the Zariski topology. So the final thing we had in the section on the fine algebraic sets was the Nulsternsatz, the theorem of zeros, which gave us some kind of relation between the ideals in kx1 to xn and the fine algebraic sets, so that you can go backwards from taking the fine algebraic set. You can take its ideal, or if you have an ideal, you can take a 0 set. And these are not quite inverse projections, but you can precisely find out what the relation is. And this was the contents of the Nulsternsatz. Now we want to make a projective version of it. So this would be a fine cone and the projective Nulsternsatz. So we want to prove a projective version of the Nulsternsatz, which gives us a relation between, say, the projective algebraic sets in Pn as compared to the homogeneous ideals in kx0 to xn. So as before, we have these two maps in the two different directions. If you have an algebraic set, we can take its idea. And if we have a homogeneous idea, we can take its 0 set. And we have already seen that in the fine case, these are not precisely inverse to each other. But we can find out what the precise relation is. And now, however, the Nulsternsatz was a very difficult theorem. We certainly don't want to prove it once more. So what instead we want to do, we want to show that the statement for projective space follows from the statement for a fine space. And this we want to do by relating what happens on Pn to what happens in An plus 1 by taking the 0 set of an ideal in An plus 1 instead of Pn or taking the ideal of something. So after all, and how do we want to do that? We want to do this by looking at the fine cone of a projective algebraic set. These are all points in An plus 1, whose class lies in the projective algebraic set. So let's do this. So we want to prove this. I mean, I haven't said what the statement is, but whatever we want to prove, we want to prove by reducing to the affine case, N plus 1, making use of affine cones. So let me formalize this again by the definition. First, I have to say what a cone is and what the affine cone of a projective algebraic set is. So a non-empty affine algebraic set, say x in An plus 1 is called a cone. Well, what is a cone? A cone is a set which has a property that for every point it contains the whole line through the point. You are in a vector space. An plus 1 is just Kn plus 1. So the whole line through the origin is contained. So for any A0 to An, lambda times A0 to An will also be contained. So it's called a cone. If for all, say, p equal to A0 to An in x and all lambda in k, we have that lambda p by which I mean, obviously, the vector lambda A0 to lambda An is also in x. So that means, so if somehow here's the origin and we have that x contains this, then it also has to contain all the lines from this thing through the origin and also going on on the other side like this. And that's after what you think a cone looks like. A cone is something which looks like a cone. Maybe on the picture it doesn't really look like anything, but I think you can imagine. So now for a projective algebraic set, we can define a fine cone. So if x in An is a projective algebraic set, well, the fine cone is just the set of all points in An plus 1 whose equivalence class lies in x. So the set is, I call it C of x. So it's a fine cone is C of x, which is the set of all, say, A0 to An in An plus 1, such that if I take its class, this lies in x. And you know that the class of x just depends by this up to a non-zero multiple. So we know this thing will contain for every point the whole line through the origin except for the origin. That's not quite good enough, so we have to make sure that we also have the origin. And this would be a cone. So this is the fine cone. And we somehow see, in some sense, whether we talk about the projective algebraic set or it's a fine cone, it somehow in some sense is the same thing. I mean, we have the same information. We can get one from the other and vice versa. So well, that comes now. It turns out to be in a fine algebraic set. And I will precisely tell you now how come. OK? So this is a lemma. So let x be a non-empty fine algebraic set, projective algebraic set. Then there are two statements. The first thing is that if we take, so x is a projective algebraic set. So that means that x, so this may be x, we can write x is equal to the zero set. So I write now projective zero set. Of course, I projective space of some ideal i, some homogeneous ideal i. Then it follows that the cone of x is equal to just a usual zero set in a n plus 1 of i. In particular, it is in a fine algebraic set. And the second statement is also how the ideals are related. So if I'm in the same, maybe I, this is a general. So if the ideal, if I take the ideal of the cone of x. So this is the cone of x is now some subset of a n plus 1. I can take its ideal. This is equal to the homogeneous ideal of x. So if I have x is a projective variety, we have its homogeneous ideal. This is the same as the ideal of x. So you can see that from the algebraic point of view, if you relate to a projective variety, it's ideal and go backwards, somehow the cone is basically the same as the projective algebraic set. The ideal is precisely corresponds to it. But we are moving ourselves from the projective to the fine situation. So this I actually do not want to prove. This is also an exercise. It's actually quite easy. So it's a reformulation of this previous statement that we had that I said that if you take the zero set that if you take an ideal, if you have a homogeneous ideal and take a zero set. So then you can either define this as the common zero set of all homogeneous polynomials in that ideal, or you can define it as a zero set of all polynomials in the ideal with a slightly different definition of what a zero set is, namely if it holds for all representatives. And it's a reformulation of this statement. But I mean, it's not completely trivial reformulation remark. But it's easy to figure it out. So now, given this, we want to state and prove the Nulsternsatz, the theorem of zeros, projective Nulsternsatz. So well, it's supposed to be a thing that corresponds to the strong Nulsternsatz. There's only one very small difference. So let i in k x0 to xn be a homogeneous ideal. First statement is a bit different than the fine case. So the projective zero set of i is empty. You know that in the fine case, the zero set of an ideal is empty if and only if the ideal is the whole of k x0 to xn. Here, this is not quite the case. So if and only if i contains all homogeneous polynomials of degree n for some n. So if the ideal, it doesn't need to be that it contains all polynomials. It just is enough that for some given degree, it contains all homogeneous polynomials of that degree. So that doesn't look very similar to the fine case. But that's the only thing which is different. So if we look at an ideal whose zero set is not the empty set, we get the same statement of the strong Nulsternsatz as before. So if the projective zero set of i is not empty, then the ideal of the projective zero set of i is equal to the radical of i. And this means, again, that we have essentially one-one correspondence between the projective algebraic sets and the radical ideals in k x0 to xn, except that we have this tiny thing that the empty set is a little bit more complicated. So let's see the proof. And then we also see where this exception comes from. So we first prove one. So we put x to be the zero set in pn of this ideal. So maybe it starts here. So when does it happen that x is equal to the empty set? You have to remember the definition of the, I mean with respect to the cone. So remember that the cone of x was the set of all points in an plus one, such that its class lies in x union with a point zero. So when x is equal, x will be equal to the empty set if and only if the cone just consists of the point zero. Because that corresponds to the fact that there's no point in projective space that lies in x. This would correspond to the non-zero points in the cone. Well, we know that cx, after all, is equal to the zero set, the fine zero set of i. So the zero set in n plus 1 union zero. So the fact that the cone is equal to the point zero means that the fine zero set either was the point zero or it was the empty set. Thus x is equal to empty set if and only if z of i is equal to empty set or z of i is equal to zero. So now I can use the usual null tensor. So by the fine null tensor, we have that if I take the radical of i, this is either i itself. And therefore, the radical of i, if we are in the case of the empty set, will be the whole of kx1 to xn, x0 to xn. Or the radical ideal is the radical of i is the set of all polynomials which vanish at zero, which is just the ideal generated by x0 to xn. So in any case, the radical of i contains this. So this is equivalent to saying that the radical of i contains x0 to xn. So thus, if we have this, so for any, say, i from zero to n, there exists a number ji such that, say, xi to the ji lies in i. And so I can take, let n to be the, say, the sum of these ji's. Then we see, as all monomials, so if I take any monomial now in the xi, which has degree at least n, then the powers of one of the xi's must be at least ji. So any monomial of degree at least n lies in i in kx0 to xn in i. And so it follows that i contains all homogeneous polynomials of degree n for some n. But also, obviously, conversely, if it contains all homogeneous polynomials of degree n for some n, then its radical will contain x0 to xn, because it, in particular, contains x0 to some power and so on. So this is the first part. So this was this one little exception, which is also the more difficult part. And now the second part is quite easy. So let x to be, again, the 0 set of i. And we now know that this is non-empty. In the case it's non-empty, we want to prove something. So we just write down. So if I take the homogeneous idea, what is it, we have seen that this is the idea of the cone of x. And this is the same as the idea of the, say, a fine 0 set of i. So I take, in a n plus 1, I take the 0 set of i. And we know what x is a projective algebra. I wrote something else? Yeah, yeah, yeah. Well, OK. I mean, in principle, you see, I should have written the p here, you're right, because we have in the same two lines they're fine and the projective case. But if we're not in that situation, I'm free to write just z in both cases. But obviously, you're right that here I should write this because the next line I look also at the fine case. So and we know now by the usual null stanzatz that the ideal of the 0 set of an ideal is a radical by the null stanzatz. So you see the projective. So this is the projective null stanzatz or rather straightforward consequence of the fine one if one knows the story about cones. We have this slight complication with the empty set because it's just the case that if we have a, which comes from this relation between the cones and the projective algebraic sets because in order to talk about cones, we have to always add the 0 and this somehow messes up our argument here a little bit and therefore also the statement. But otherwise, it's precisely the same as in the projective case. Yes? OK. So actually, I maybe should have said it, I mean, so what we have said is something slightly different, which is equivalent. I had said as a remark and exercise, so you have to disprove it if you don't want to believe it, that if the 0 set of i is non-empty, we have that, let me see. So what was the statement? And then we have that the, so if x equal to the 0 set of i is non-empty, then we have that the cone of x, so this was the projective 0 set, is equal to the affine 0 set. This was one of the two statements I made. And then by definition, if x is empty, then the cone of x, just by definition, is equal to 0. If you, you can put this together and say this. I mean, it's maybe not the most intuitive way of saying the statement, but it's equivalent. I mean, OK, OK. So I want to just close this up by a few remarks. So one thing is that we see that for the statement 1, we got this exception, which came from this idea here. That if the 0 set is the empty set, we actually get that the radical of i will contain this. And so we say that the ideal x0 to xn is called the irrelevant ideal, so it's kind of a radical ideal which has the property that the 0 set is still empty, although it's not the whole of kx1, so it somehow doesn't do anything. And so we have this corollary. We have a 1, 1 correspondence to a bijection, inverse bijections between the homogeneous radical ideals. So if i in kx0 to xn is an ideal, which is not the irrelevant ideal, different. We call it relevant. It's kind of stupid, but anyway. So we take the homogeneous, relevant radical ideals in kx0 to xn, in 1, 1 correspondence by these two things. We can take this projective 0 set, and we get a projective algebraic set in pn. And if you have a projective algebraic set, we can go back and take its homogeneous ideal. And this gives us bijections between the homogeneous radical ideals in x1, x0 to xn, and the projective algebraic sets in pn. The only thing is in order to have a 3D bijection, we don't miss out one ideal. We have to throw away the irrelevant ideal, which comes from this statement here. And as a corollary, we get the same thing that we had in the fine case that a projective algebraic set is irreducible if it's a 0 set of a homogeneous prime ideal. So a projective algebraic set x in pn is irreducible if x is a 0 set of i, where i in kx0 to xn is a homogeneous prime ideal. So it's a homogeneous ideal, which is also a prime ideal. And then as a straightforward consequence, if f is a homogeneous irreducible polynomial and irreducible, then we have that its 0 set is irreducible. So the second statement follows in the same way from the first as before. It's just that if f is irreducible, then the ideal generated by f is a prime ideal. And then we apply the first part. So to prove it, we only have to prove the second. And we proved a similar thing in the fine case. So the proof is analogous, so only slightly complicated by the fact that we have to somehow talk about homogeneous polynomials. So we somehow have to, we have something about that something. If something is a product of polynomials, then if a product of polynomials lies in i, then one of them must lie in i. But in order to do this with a homogeneous case, we somehow have to make the polynomials homogeneous. And so this, you need some tiny extra argument. But it's not a big deal. So first, assume x is irreducible, so it's reducible. So then we have to show that the ideal of this is not prime. We'll just see. Yeah. So it appears, I'm pretty sure that this statement is also true, but this is not what I'm claiming. So I will maybe stick to what I claim, so that I don't have to, you can see whether you can prove this. No, no, but I think I will stick to it. I will just prove it as is. So then, no, but I think I have to, no, no, but yeah, yeah, that's what I have in my notes. I think what's written here is also true, but as I haven't worked out the proof, I will stick to what I have. So reducible if and only if is a prime idea. I mean, this is a somewhat less useful statement, but that is what I have proven. Not quite sure. I think in that case, it's not so evident, well, I think to be on the safe side, I will want, yeah, anyway, let's see. So assume x is reducible. Then it follows that we can write x equal to x1 union x2 for closed subsets x1 strictly contained in x and x2 strictly contained in x. So we have two closed subsets whose union is x, but none of them is equal to x. Well, it follows then obviously that if I take the cone of x, this is the cone of x1 union the cone of x2, no? Because the way the cone is defined, this is evident. And these are both closed subsets of the cone of x. And none of them is equal to the cone of x, because it's not true here. So these are closed subsets with a c of x1 and c of x2 strictly contained in c of x. So it follows that c of x is reducible. So it follows that the homogeneous idea of x, which is equal to the idea of the cone of x, is not a prime idea. So that was quite simple. Now we want to go the other direction. So we assume the ideal of x is not prime. So I just write i for it. So that means there exists an element f and g in i such that in kx1.xn such that f times g is in i, and f and g are not in i. So normally you would say, I mean you would want now to work with f and g, take the intersection of x with the zero set of f and the zero set of g, and then we would conclude in the same way as in the fine case. The only problem is that we want to talk about, I mean, zero sets of homogeneous polynomials. So we would need that these are homogeneous polynomials. But there's no reason why they should be, because that's not part of being a prime idea that you can do that. So we want to somehow replace f and g by homogeneous polynomials. This is by looking at homogeneous components. So let i and j be some positive, non-negative integers minimal such that the homogeneous component of f of degree i is not in i. And the homogeneous component of degree j of g is also not in i. We have that f and g are both not in i. So if all the homogeneous components were in i, then f and g also would be in i. So there must be some smallest degree for which the homogeneous component is not in i. So this certainly exists. Now, so subtracting homogeneous components of f from f and of g, so of lower degree, from f and g, we can assume that f starts in degree i and g starts in degree j. Why is that? We know that all the homogeneous components of f of lower degree than i actually lie in the ideal i. So if I subtract them, I don't change the fact that f does not lie in i. As this thing itself, if I have the sum of two elements in i, then if I take something which is not in i and I subtract from it something which is in i, then afterwards it's also not in i because i is an idiot in the same here. So we find that if we subtract the lower degree homogeneous components from f and g, what we get afterwards is still not in i. So we can replace f by what we get by subtracting it, then f starts in degree i and same g in degree j. So but now we have that f i times g j is the homogeneous component of minimal degree in f times g of f times g. If we multiply things, the degree will add. And the lowest possible degree that we can get is if we take the lowest degree part here and the lowest degree part here. If we take one of them higher degree, we get the higher degree. So the part of degree i plus j of f times g is this product. But now one has to remember that f times g is an element in i. And i is a homogeneous polynomial, a homogeneous ideal. So then by the definition of homogeneous ideal, an ideal is homogeneous, if an element lies in the ideal, then all its homogeneous components lie in the ideal. So therefore also this lies in the ideal. And now we can argue in the same way as in the defined case. So we let x1 to be the zero set of i union, say, f i. So we take the ideal i and we add to it this one thing. And x2, the zero set. So in other words, I could write it as z of i, an intersected zero set of f i. x2 is the zero set of i, intersected zero set of g j. Well, we know that as f i is not in the ideal of x, it follows that this is strictly contained in x. But on the other hand, we know that if I take their union, then this corresponds to taking the product here. We have that the product of these two elements is f i g j, which lies in i. So it's the same as the zero set of i. It follows that n is the zero set. So x1 union x2 is equal to x. So thus x is reduced. So now from this, one can actually deduce what I wrote on the blackboard as part one. Let me that if i is a homogeneous ideal, so this would be in some sense part two. So this should be, so if we have in the look at the theorem of this proposition, we have this which I call part two. This should in some sense be part three. And the thing which I first wrote on the blackboard would be part two. So if i in k x0 to xn is a homogeneous idea with a zero set of i is empty, is non-empty, then it follows that this zero set of i is irreducible. And this part two follows in the same way as in the fine case. Because in the fine case, we had seen that because we have this correspondence, prime ideals are radical. So that if we have this homogeneous prime ideal with this property, then this is actually the ideal of its zero set. And therefore part one applies. And then part two follows from, part three follows from part two directly in the same way as in the fine case. Because if you have an irreducible homogeneous polynomial, it generates a homogeneous prime ideal. And therefore part two applies. OK, so my time is already up. I was a bit, I was a bit, I was a bit, I was a bit slower today. But in some sense that's maybe just as well because this was, this finished the first part of this course, which was, you know, introduction of the final project of varieties. And the next time we will start to talk about morphisms. So I mean, the idea is that when you study something, say vector spaces, many folds and whatever. So one of the main things you use to study it is the, the maps between these things which are compatible with the structure. So linear maps between vector spaces or continuous maps between topological spaces. And so, you know, and now we have to define what are the maps between varieties which are compatible with the structure of being a variety. And these maps will call morphisms. You know, a little, not so much fantasy. And somehow very important in the definition is to say what are the maps from a variety just to K to the field. These we would call the functions, the regular functions. We'll actually find out that the way to define the morphisms is to say something about what they do to the functions and therefore we first have to look at the functions. Okay, we start this next time.