 So this is, okay, we go on with algebraic geometry. We were, last time we had talked about irreducible components and say the ideal of an algebraic set and finally we had started to talk about Hilbert's Nulsternsatz. So the theorem of zeros, Nulsternsatz. So which is, which tells us something about the relation between an algebraic set and the corresponding ideal. So first we have the weak version, which I stated the other time. So weak Nulsternsatz. Anyway, so as I said, Nulsternsatz means theorem of zeros. But for some reason it's always referred to by its German name. So this says that if I is an ideal in the polynomial ring, which is a strict ideal, so proper ideal. So an ideal which is not equal to the whole of the polynomial ring, then its zero set is non-empty. As I said, I will not prove this now. If you are interested, I will prove it later in the chapter on dimension. When it follows from the things that we do anyway. I will instead use it to prove the strong form of the Nulsternsatz and make some conclusions. So the way we usually want to use this theorem is in this kind of trivial reformulation, which is the following remark. Usually we will use this as follows. It's used in the following form. So if i is an ideal such that its zero set is empty, then it follows that one is contained in i. That's usually how one will use it. So sometimes you want to show that something is a polynomial. And then this ideal will somehow be the possible denominators. And then one of the denominators will be one. And so it's a polynomial or things like that. Anyway, this is how one usually uses it. And obviously this is equivalent. Because if an ideal contains one, it's equal to the whole of kx1 to xn. And then it's just a contraposition of the original statement. And one should also remark, we have assumed in the whole course that our ground field k is algebraically closed. Otherwise, this theorem is wrong. For instance, the most trivial example is if you take the ideal of x squared plus one in kx, in rx over the real numbers, then its zero set, if I call this i, zero set of i will be the set of all points a and k in r such that a squared is equal to minus one. And so this is the empty set. OK. And obviously this is not the whole of ix. This ideal doesn't contain the constants. So now we want to, so this is kind of the weak form. And the question is what's the precise relation between an ideal and its zero set. We had seen, so we know that if, say, j is an ideal, then we have that zero set of j is some fine algebraic set. And if we take its ideal, then this will contain j. Because just by definition, this ideal consists of all polynomials which vanish on the set. And these are among these polynomials. And conversely, if x is a subset, it's in a fine algebraic set, a n, a fine algebraic set, then we have that we can look at the zero set of the ideal of x. And we had seen that this is actually equal to x that we had seen. It's also easy. I mean, the obvious question is whether this is also true here, which means that we would have a bijection between ideals and zero sets. So is for all ideals. And so that, which would mean that we have a bijection between ideals and kx1 to xn and the fine algebraic sets. And the answer is no. And it's actually easy to see that this will not be the case because it comes from the fact that if you have a polynomial and the power of this polynomial, they have the same zero set. So for instance, we can just look in a1. We have that if we look at the zero set of the ideal generated by say x squared. This is obviously equal just to 0.0. But I mean the ideal of zero. So all the polynomials that vanish in one variable that vanish at zero are precisely the polynomials with zero constant term. So this is the ideal generated by x. So they are different. But the statement of the null stanzatz of the theorem of zeros is that in a suitable sense, this is precisely all that can go wrong. So that it is the difference between an ideal and the ideal of the zero set of the ideal is by being allowed to take powers of the elements. And so this we want to formalize by introducing this definition. So let I be an ideal in a ring R. So then I introduce the radical. And one can either write it like this, or I often will write it like square root. But it's obviously not the square root. So this radical of i is this which is the set of all elements in the ring such that some power of the element lies in i. So there exists an n bigger than zero such that r to the n is an element in i. It's an easy, maybe I don't do it. So exercise this radical of i is an ideal. If you just straightforwardly go through the definitions, we'll find that this satisfies the definition of an ideal. And so we call, so i is called a radical ideal if while one can say it in two words, so if it's equal to its own radical and you can easily see that this is equivalent to the fact that i is equal to the radical of some ideal. This comes from the fact that from the definition it easily follows that if you take the radical of the radical, this is equal to the radical. So now we see directly that this has something to do with this story here because we see that if x isn't a finite set, then its ideal is radical ideal. I mean to be a radical ideal, the reformulation is that if any power of an element lies in the ideal, then the element itself lies there. So let's see. I mean that's again trivial but at least it's... So let's, we take a polynomial such that some power of it lies in the ideal of x. Well, so in other words, we have that f vanishes on the whole of x. So then for all points p and x, we have that f to the n of p is equal to 0. But this is just the same as f of p to the power n. And if a power... So we have, this is some number, its nth power is 0, so the number is 0. So as it follows, f of p is equal to 0. So we see that, and thus this means precisely that f lies in the ideal of x. So trivially by the definition, the ideal of a fine algebraic set is a radical ideal. And now we come to the formulation of a strong form of the Nullstandsatz, which as I said is, which says theorem, which says that, so let i be an ideal in kx1 to xn. Then if I take the ideal of the zero set of i, this is equal to the radical of i. So maybe you mind this notation that the i has two different meanings. So this i is just an ideal, and this is this process of taking the ideal of a zero set. I excuse myself, but the point is I would like to use the j in the proof, so I don't want to call it j. But anyway, I think you can distinguish what these two i's mean something different. Okay, so this is not so easy theorem. I mean actually the first part, the Nullstandsatz is more difficult. This is in some sense a consequence of the weak one. So we'll prove this as a consequence of the weak Nullstandsatz, but even this proof is not so straightforward. So how does it go? So before doing this, I, so in the proof, I will talk about the fraction field or the quotient field of an integral domain. So I briefly will introduce it because I talk about the field of rational functions in several variables. I mean it's something you I think already had in the algebra course, but you know I just briefly recall what you are supposed to know. So this is the quotient field of an integral domain. So R should be an integral domain. So the quotient field Q of R is a set of some equivalence classes of pairs. So is the set of equivalence classes and of pairs f, g, where f and g are in R and g is different from 0. Under the equivalence relation that f, g is equivalent to f prime, g prime. Well, if and only if f, g prime is equal to f prime, g in R. So these equivalence classes I write, so equivalence class of f, g is denoted f divided by g, which also comes somehow explains this thing because if the ring for instance would happen to be the integers, then these equivalence classes correspond precisely to the rational numbers. And you can do this however for any integral domain. And you have again multiplication, addition and multiplication are defined by the usual formulas for fractions as if you were in elementary school. So f divided by g plus f prime divided by g prime is f, g prime plus g, f prime if you want divided by g, g prime. And f divided by g times f prime divided by g prime is equal to f, f prime divided by g, g prime. So again, like you learn in elementary school, except that now we are in a general ring and not dealing with integers. The whole point of this is that this is all that happens. You can compute this if you had them. You can view r as a suppressing of qr. So first easy to see that this thing q of r is a field. And we have that r. So identifying an element r in r with the kind of fraction r is the one in q of r. So the map which sends such a thing to this is easy seen to be injective and you can just identify it if you want. Then we get that r is a suppressing of q of r. So this is all very standard. Obviously the inverse of f divided by g is g divided by f as in elementary school and so on. So in particular, we will only look at a very special case, namely that our ring is the polynomial ring in several variables. So in particular, the quotient field of the polynomial ring kx1 to xn is denoted usually like this with round brackets. It's called the field of rational functions in these variables. And you can see that these are just, so the elements are just of the form f divided by g with f and g polynomials and you compute with them in the obvious way. You just multiply them and add them and divide them really as fractions and this will be a field. So now after this thing we can, I will use this very briefly in a moment. Now we want to actually come to the proof of this null stands and we will use this thing. So we start with our idea. So we know that it's finitely generated. So we can as we write that, so proof, write i equal to f1 to fr for some polynomials fi, some polynomials in i. Okay. Now, I mean we know, we have seen that if we take the ideal of the zero set of i, this is a radical ideal and we also know it contains i. So it follows that the ideal of the zero set of i contains the radical of i. So we are supposed to prove the other conclusion. Well, and so basically now we just take an element here and we have to show it lies here. So let f be an element in the ideal of the zero set of i and we have to show there exists an n bigger than zero such that f to the n lies in i. So from the fact that it vanishes here we have to prove that. And that's actually, you know, if you think of it, that doesn't look so easy because, you know, for instance, it's not at all clear where you should get this n from. And so there must be some kind of trick and we somehow maybe also want to use the weak null stands, but it doesn't look very clear how one could do it. But the trick consists in using the weak null stands in one variable more. So we use weak null stands as in k x1 xn, t. It's not quite clear what this will do for us, but at least let's do that. And so for this we will have to produce an i, using i we will have to produce an ideal here whose zero set is empty. So let's write it down. J, so let J be now the ideal generated by these self-simple normials. And one more, we have here this f. So we take f times t minus one. So this is the ideal in k x1 xn, t. So we first have the ideal generated in this bigger ring by these polynomials which only depend on the first few and then we add one more polynomial which also depends on t. And so we take this ideal and we ask ourselves what is the zero set of J? So let's look at it. So we take a point, so let's say p, a be a point in an plus one. So where p is a point in an and a is a point in k. So then you know this is an n plus one pupil. And we ask ourselves what does it mean? So we want to know whether it can be that p, a is an element in the zero set of J. Well, when will this happen? So all of these polynomials must vanish. So these ones depend only on the first few coordinates. So they only give coordinates for p. So the first statement is that f1 of p until fr of p should be zero. But we know that the zero set of them is, well anyway, so we have that f1 of p is equal to fr of p equal to zero which in this itself is equivalent to the fact that p is an element in the zero set of i. And the second statement is that, well, f of p times a should be equal to y. But f lies in the ideal of the zero set of i. That is, if p lies in zero set of i then f of p is equal to zero. And so this cannot be true. So this implies that f of p is equal to zero and therefore this is impossible. So these two conditions can never be fulfilled at the same time. And so we deduce that indeed z of j is equal to the empty set. So and now I told you that we have this way how we usually use the weak null stanzatz. So this implies that j contains the element one. So by the weak null stanzatz we have that one is an element of j. Well, and we can write out what this means. No, just it's a linear combination with coefficients in kx1 and t of these elements. So that means we have thus we can write one is equal to say g zero times f t minus one plus sum i equals one to r g i f i. And this identity so where the g zero to g r are polynomials in x1 to xn and t. And this identity holds in kx1 to xn, t. So we have this identity. Now the obvious question that one can ask oneself here is why should this be helpful to us? We are interested in finding out something about i. Now we have made a different ideal in one more variable and we find that that contains one. So we have to have a way to get back to kx1 to xn. And how do we do this? We replace t by a polynomial. We express t in terms of something in x1 to xn. In fact, we replace t by one over f. But this only makes sense in the fractured field. So now we want to go back to kx1 to xn. But this is via the fraction field kx1 to xn. So we define a ring homomorphism say phi from this ring to the fraction field or the quotient field or the field of rational functions in x1 to xn which is done by taking any polynomial and sending it to this polynomial where we let the first two variables in the first variables we just take the same and for the last one we put one over f. Now this is clearly a homomorphism because if you take the sum of two polynomials you get here the sum of the rational functions. I mean it's clear what this notation means, I hope. And if you take the product and so on it will always be like this. But if the polynomial here contains some t you will get some power of f in the denominator. So you don't end up in the polynomial ring but in this fractured field. You have this homomorphism. So now you take this identity and want to see what happens to the identity by applying this homomorphism. So applying phi, so this may be a star or something. So apply phi on both sides to this identity star. We get one. Well, let's see. Before we do it we maybe have to look at it a little bit. So what happens to ft minus 1? So f is a polynomial in x1 to xn so nothing changes. It's sent to 1 over f. So we have f divided by f which is 1 minus 1. So this becomes 0. So this term is just not there. Then the fi again depend only on x1 to xn so they do not get changed. So what we find is that we get 1 is equal to sum i equals 1 to r. So phi of gi times fi. And this is an identity in this field of rational functions. Now we want to get rid of the denominator so that we get a statement in kx1 to xn. So we have to ask ourselves what are the denominators. So if you look at this, what have we done? So we have taken this polynomial. We have replaced t by 1 over f. So the only denominators that we can have are powers of f. Whenever we have some power of t, we get the same power of 1 over f. So these fi are anyway just polynomials in x1 to xn but it applies to this. So we have that there exist some numbers n i in non-negative integers such that phi of gi can be written as gi divided by f to the n i to the power n i where gi is a polynomial in kx1 to xn. And so now if we have that, we can clear the denominators. We can multiply with whatever the maximum of the f to the maximum of the n i. Then there will be no denominators here until we get the power of f which actually will turn out to prove the theorem. So let n be the maximum of the n i. So from i equals 1 to r. So we multiply by f to the n. Then we get the identity f to the n is equal to what is gi? So sum i equals 1 to r gi times f to the power n minus n i which is some non-negative number times fi. But now what do we see? So this is a linear combination of the fi's. So the fi generates the ideal i, large i. So it means that this thing is an element of i. And we have proven what we wanted to prove. We have found some power of f which lies in i. But it comes about in this very roundabout way as the maximal denominator we need in this expression. So the fact that before we couldn't have a guess for what power of n one should need to take is reflected in the complicateness in the proof in such a way that there's no way how you can find out. So anyway, so this proves this theorem. And you can see it's a bit tricky. Now the moment we will not really have very many applications of it. So we now find that we have indeed this bijection between, say, a fine algebraic set in an and the radical ideals in kx1 to xn. So we have a bijection in both directions. So in the one direction one goes with the ideal. The other one with the zero set. And these are two bijections in both directions which are inverse to each other. So precise correspondence between radical ideals and a fine algebraic sets. Okay, so I want to give some small corollary. I had, we had proven as a statement that and a fine algebraic set is irreducible if and only if its ideal is a prime ideal. That sounds like a nice statement, but in practice it's not, wouldn't be so nice because it's very difficult. It's not so easy to find out what the ideal of an affine algebraic set is. Usually you are given the affine algebraic set as a zero set of some, some polynomials. And so it's not so obvious. But in this case as a corollary you can get a more direct statement. So the first is so if say I in kx1 to xn is a prime ideal then its zero set is irreducible. And the second statement which is in some sense a corollary is that if f in kx1 to xn is an irreducible polynomial so it cannot be written as a product of two polynomials unless one of them is constant then its zero set is also irreducible. So the other statement was that we had before is that if the ideal of the zero set is a prime ideal then it's irreducible. So that's not a part of quite the same thing but it actually is almost trivially so because from the definition it follows that prime ideals are radical because what does it say? To be a prime ideal means that if you have f and g in the ring such that f times g lies in the ideal then it follows f lies in the ideal or g lies in the ideal. And now if you have instead a power of an element so you have an element in the ring so that the power lies in the ideal then it should follow that already the element lies there this is a special case where these f and g can be taken as one f and the other one some power of f. And then you can make induction. So thus if i is a prime ideal then it is a radical ideal so if I take the ideal of the zero set of i this is just i. And so the statement and this is still a prime ideal and therefore it's irreducible. And we know so if that is a prime ideal so if I write so if I write x equal to z of i we have that i of x is prime so thus x is irreducible by what we had seen before. And the second part is a special case if one so for two if f is irreducible so it is a fact which was maybe mentioned maybe also proven but I don't know in the algebra course that k x1 to xn is a unique factorization domain I will not recall what that means but one of the things that follow is that every irreducible element is a prime element so that means if f in k x1 to xn is irreducible then it follows that the ideal generated by f is a prime ideal so this is something which holds in unique factorization domain well and so and so thus if this irreducible then this is a prime ideal and thus the zero set of f which is the same as the zero set of the ideal generated by f is irreducible so that was for the moment so this is the as much as I want to say about the nulstern that's for the moment so now I want to come to a different topic which is projective varieties and projective algebraic sets so you know we here do everything over some algebraic very closed field but for the intuition it's in some sense best to think about the complex numbers and so if you for instance take some a fine algebraic set which is does not happen just to be a finite set of points but it has some is somehow bigger over the complex numbers you will find that it always somehow goes to infinity so it will always be non-compact and this somehow I mean somehow it's nice to also look at compact things I mean they have a they behave nicer in many ways and in some sense the more interesting and more useful things to study I mean also from the viewpoint of topology or whatever anyway so therefore we want to somehow you know do something like compactifying these fine varieties so we want to look at varieties which have some kind of which at least if we would look at complex numbers with the standard topology would be compact now and so this is done by adding some points at infinity so a fine variety is always you know they are not bounded somehow but you somehow add something at infinity to get some projective varieties and this will have many nice properties I mean one of so the simplest projective variety or whatever would be projective space or something and there you can you have this thing that if you are in the usual space if you have two lines then you know usually they will intersect but sometimes they are parallel and they don't but in projective space what you call lines there they will always intersect if they don't intersect they are parallel they intersect at infinity and so somehow looking at projective against instead of the fine varieties will also kind of remove very many case distinctions and things like that they are just kind of nice in many ways so anyway after regardless of that I mean they you know in some sense it turns out that projective varieties are more important to study than the fine varieties and fine varieties you know you would more think of them by thinking at you know how locally the projective variety looks like so to look at a piece of it but the total thing you want to study would be rather a projective variety and so we want to now introduce them so I first tell you what the projective space is and then projective varieties will be 0 sets of polynomials in projective space where we even have to worry what we mean by that so first projective space and projective algebraic sets so on so this is a definition on KN so KN plus 1 without the origin so the origin in KN plus 1 I have an equivalence relation so I say that an N plus 1 tuple A0 to AN is equivalent to B0 to BN if and only if they differ by multiplying all of them by the same non-zero constant there exists a lambda in K without 0 such that A0 to AN is equal to lambda B0 until lambda BN so just the whole vector is obtained by multiplying by the same constant and I so the quotient so maybe write first A0 for N for the equivalence class with a square bracket and the quotient I call projective space so n dimensional projecting so projective N space or n dimensional projective space is the quotient quotient set divided by this equivalence relation so an element in projective space is just an N plus 1 tuple of elements in K or a vector in KN plus 1 up to multiplying by non-zero constant so which is equivalent to saying it is a line so in a one dimensional subspace of KN plus 1 because such a line is determined by a non-zero vector on the line up to multiplication by a non-zero constant so you can also view projective space as a space of lines in KN plus 1 lines means linear subspaces so lines to the origin so now we want to somehow show that this is somehow some relation also with a fine space so we want to somehow find that there are N plus 1 subsets which look a little bit like AN and we use one of them to identify AN as a subset of this so definition so let we take UI for I is an element from 0N we define this to be the set of all A0 to AN in PN we denote this as PN projective space in PN such that AI is non-zero and now I define a bijection of this to AN so I take phi I from UI to AN which sends a point to AN to a point here in AN so I take E0 divided by AI and then it goes on A1 divided by AN at some point I arrive at AI divided by AI and this I leave out I want only N coordinates and then it goes on until AN divided by AI so this will be something in AN and I am allowed to make this quotient because I have assumed that AI is non-zero so this is certainly a well defined map and I claim it's a bijection because I can write down the inverse so say UI from AN to UI which sends where I write like this I want to have to write down an N tuple here I do this by writing an N plus 1 tuple of which I leave one out so I take say B0 and Bi is not there until BN we send this to B0 and in the ith position I put 1 I mean the ith position is not here but you know just where and BN and you can check that these two maps are well defined and inverse to each other I mean I don't know whether you might have already seen something like this in different geometry or something if you define many folds in a projective space as a many fold by giving local charts to it and this would somehow be such a local would be such a set of local charts anyway we have this it's very easy to check that this is a bijection with this property so usually we fix we concentrate on the and and we want to we want to view AN as a subset of PN by identifying the point say A1 to AN in AN with its image under this A1 AN so I mean later we will actually do this I mean obviously they are not precisely the same thing but in some sense you can see that precisely the same information is contained it's just a notation issue whether you write this or whether you add the 1 in the beginning and put the brackets around it contains the same information so but I mean strictly speaking the statement would be so okay I don't say that but given this so with this identification we have PN is equal to the union of AN and H infinity where H infinity defined to be well whatever in PN does not lie in the image of this U0 namely this would be the set of all A0 well whatever I can write this is 0 to N in PN such that A0 is equal to 0 this would be called the hyperplane at infinity so when I write this notation strictly speaking what it really means is that PN is equal to U0 of AN union H infinity but I kind of use this U0 you know as it's a bijection and it even I mean not so much happens but I kind of can pretend that this is really just that I identify the image with what I started it so anyway but if we have this picture we can really see that PN is obtained from AN by adding something namely this so it is you know AN plus something added at infinity yeah in some that is well it is somehow in some sense it is PN minus 1 I mean we don't yet know what it means to be isomorphic but it is in some sense it will be isomorphic to PN minus 1 it's very close to PN minus 1 because it's just you know if you have 0 and then A1 to AN so that you know A1 to AN so in some sense this will be this is essentially PN minus 1 okay so yeah you could see it is okay now that we have the projective space we want to talk about projective algebraic sets so we want to define projective algebraic sets as 0 sets of polynomials in this projective space and we immediately run into some small problem that it doesn't completely make sense so we want to define projective algebraic sets as 0 sets of polynomials so of sets of polynomials in so so if we are projective algebraic sets in PN this will be of polynomials in k x0 to xn and now there is a problem namely it is not true that the polynomial in k x0 to xn defines a function on PN so but f in k x0 to xn does not define a function on PN no if we have no obvious it is just not true actually will never be true that I mean unless f is constant or something that f of A0 to AN is always different from f lambda A0 lambda PN we don't have that is a function so it's not really clear how we could talk about so we cannot say what f of P is for a point P in PN because it depends on the representative but it still sometimes makes sense about a 0 of a function namely if it's 0 for all representatives and this makes sense if f is homogeneous can still say whether P in PN is a 0 of f or not and so why is that well this is the you know so I hope you remember what it means for polynomial to be homogeneous so you have that it's a linear combination of monomials so some product of the XIs with some powers and it's homogeneous if the degree of all monomials is the same and that would then be called the degree of this polynomial so this is the following remark if f is homogeneous is homogeneous of degree D then we have what if we take f of lambda a 0 until lambda a n then this will be equal to lambda to the d times f of a 0 a n and this is actually kind of obvious because if you take any monomial of this polynomial it has degree D so it's the product of d of this XI counted with that power so if you put it into you get d of these lambda AIs so out of each factor you get out of each of these you get one factor lambda so you get lambda to the d comes out of every monomial we have that so in particular so thus whether f of a 0 to a n is equal to 0 depends only on the class of a 0 on the point in projective space because lambda is a non-zero if you multiply by non-zero constant then the question whether or not it is 0 is not changed so with this we can define what we mean by a point in projective space to be a 0 of a homogeneous polynomial it is a 0 if for any representative or equivalently for all representatives it is a 0 so let G be a homogeneous polynomial so a point p equal to a 0 a n is a 0 of G and we will write this just we will just write this as G of p is equal to 0 although I mean in general G of p doesn't make sense but the question whether G of p is 0 makes sense if well G of a 0 to a n is equal to 0 for one and equivalently all representatives of this point so if we have we can p is the equivalence class of some n plus 1 tuple and we just take you know a representative and put into the polynomial if we get 0 we say it's this and it doesn't depend on which representative we take because of this statement now we can talk about 0 sets of polynomials so let S in K X 0 X n is set of polynomials set of homogeneous polynomials so then the 0 set of S is defined to be so Z of S the set of all points p in p n such that f of p is equal to 0 for all f in S as before in the affine case with this definition of what f of p is equal to 0 means so a subset of p n of the form C of S is called a projective algebraic set so now we have defined what projective algebraic set is so we only had this tiny difficulty with saying what the 0 of a polynomial is but for homogeneous polynomial sets okay and now we want to also I mean you want to basically repeat what we did for the affine algebraic sets so for instance you want to show that every projective algebraic set is a 0 set of an ideal but one has to be a little bit more careful because we have somehow have been forced to talk only about homogeneous polynomial so somehow that will also play a role here and so we will have to deal with so-called homogeneous ideals before that maybe I can give a couple of trivial examples which anyway use so for instance we have that the empty set is a 0 set of 1 and p n p n is say the 0 set of the empty set of polynomials so we again have the most trivial sets then we also can the points also find algebraic sets so if p equal to a 0 to a n is a point in p n then we have that where p can be written as the 0 set of what so we say for instance we take x 1 minus a 1 x 0 minus a 0 x 1 a 2 x 0 minus a 0 x 2 and so on a n x 0 minus a 0 x n so if I'm not mistaken you find that this precisely will determine this equivalence class I mean so precisely that you know that up to multiplying by a non-zero constant the coordinates have to be these anyway you can check that so now we want to come to the story with the ideals so if we have a polynomial we can always write it as a sum of homogenous polynomials so the polynomial f in k x 0 x n can be written uniquely as f equal to say f 0 plus 1 plus whatever f d where d is the degree of the polynomial where the fi homogenous of degree i so we just take all the monomials of degree i together with their coefficients and we take the sum of those and put them together so these fi are called homogenous components of f now an ideal is called homogenous if with every polynomial it contains all its homogenous all its homogenous components so an ideal i is called homogenous, so homogenous ideal if for every f in i also all homogenous components fi i in i so this is like the strange definition but there's a you can formulate it in another way namely have the following proposition some more remark an ideal is homogenous if it's generated by homogenous polynomials if and only if it is generated by homogenous polynomials so if and only if there's a set of generators so that all the generators are homogenous i don't know whether i can also prove it although it's not very exciting but so we assume we assume that i is homogenous and so we take some system of generators well, so then for each of the generators we can take the homogenous components if we take all the homogenous components of all the generators they will certainly generate because we can write the generators as a combination of them so then the f i for i for i and i are a set of homogenous generators because after all f alpha is the sum of the f alpha i so whatever you can express in terms of the f alpha you can do in terms of the f and the other direction is ever so slightly more complicated so let i be generated by homogenous polynomials generated where these are the i are homogenous okay so I have to show that for every f and i I can write it as a we have that all the homogenous components of f also in i so we take an f in i so then thus we can write f is equal to sum over i a finite sum and a i g i because after all the g i are generators with a i sum elements in k x0 to xn not necessarily homogenous and now but notice but g i is homogenous so the homogenous part of a certain degree of this product is just this g i multiplied by the corresponding homogenous part of a i so thus the homogenous part of a i g i of degree d is just g i times the homogenous part of degree sorry the homogenous part of degree d minus the degree of g i you get something of degree d by taking here something of degree d minus the degree of this that's the way how you get the polynomials the sum of all the monomials of a i which have the degree that adds up to d well but if that is so then we know how to get the homogenous part of degree d of f we just take this part so thus it follows that fd is the sum of all i of a i to the d minus the degree so the homogenous part of degree d minus g i times g i and so the g i so it is a linear combination of the generator so it follows that this thing is indeed an element in i so I'm essentially up but I can at least write down the definition we were in for and we come back at the next time so we want to say what the zero set of an ideal is so according to our and we want to say this for homogenous idea but we cannot say it's the zero set of all the elements in the idea because we have only talked about zero sets of homogenous polynomial so we only talk it's the zero set of all the homogenous elements in the idea definition that i in k x0 to xn in ideal so so in some sense projective but we will not write it zero set of i is denoted again z of i which is the set of all points p in pn such that f of p is equal to 0 for all homogenous elements I do I want an ideal or I want a homogenous I want a homogenous idea for all homogenous elements f in i so the zero set of i is not the zero set of all the common zero sets of all elements in i because we have only said what it means to be zero for homogenous polynomial and conversely we can talk about the ideal of a projective algebra set this is just the ideal which is generated by all homogenous polynomials which vanish on that set or just for a subset x in pn the homogenous but again we will not say ideal of x is i of x which is defined to be a set of all f in k x0 to xn such that f is homogenous and f vanishes on the whole of x f of p is equal to 0 for all p in x and by definition this is a homogenous ideal so we have this definition now next time we will work a little bit with it so we have seen we have kind of made the same I mean the analogous definitions to the affine case we want to then prove that most of the results we proved in the affine case also hold in a suitable way in the projective case and then ok that we will do the next time so thank you and see you next week