 So now we came to the second chapter which is about functions and morphisms. So I've told you that somehow it's always when one, I mean usually in mathematics when one studies something, some spaces with a structure then one of the things that one always studies along with it is the maps between these spaces which are compatible with the structure. And so in our case these are the morphisms. These will be maps between a fine or quasi-protective or quasi-fine varieties which are compatible with the fact that these are varieties. And we have to make sense of this statement first to find out what we mean by that. And for that we first study the maps from varieties to just the ground field k which are compatible with the fact that we are looking at varieties. And so we first look at a very special case and look at some very special functions which are not as general as we will want them later. You know after all algebraic varieties are something which if we are in an N there are some zero sets of polynomials and so everything is somehow about polynomials. So certain kinds of functions that we can study are just the polynomials. And so the functions on a variety, on an affine variety, that we could study is just the restrictions of polynomials to that variety. That's certainly something we can look at. And this we will first study. And this is so-called co-ordinate ring. So first we look at the co-ordinate ring of an affine algebraic set. So actually we don't... So how's that? So we have seen if x, so x is in an affine algebraic set, so then we can consider the ideal of x. The ideal of x is Ix which is as you recall the set of all polynomials f in kx1 to xn such that say I could just say it like that, f restricted to x is equal to 0 for what? So for all points in x f will be 0. And the co-ordinate ring then, so often says the affine co-ordinate ring of x is just defined to be Ax equal to all the polynomials divided by the ideal of x. So this is a ring. And so this is a ring. And it is also, if you want, it is also a k-algebra. So that just means it's a ring which contains k as a subring. Because... So what can we say about it? The first thing is that we can also view it in a different way. That this can be identified with just the set of all restrictions of polynomials to x. So elements here, there's a ring isomorphism between this and the set of all restrictions of f to x. So these one could call polynomial functions. So definition. So we have x is still in a affine algebraic set. So a polynomial function on x is a function of the form from x to k such that f is equal to f restricted to x for f polynomial. OK, so just all the restrictions of polynomials to x. And you see that these can be, if you have such functions to k can always be added, multiplied, and so on by point-wise addition and multiplication. So this is a ring with point-wise addition and multiplication. So just, I mean, as usual so that f of f plus g of p is equal to f of p plus g of p and fg of p is equal to f of p times g of p for p. So as always, and now what you obviously have, I mean this is clear, is we have a ring homomorphism. So there's a ring homomorphism, say, from k x1 to xn to the set of polynomial functions on x, which just sends, obviously, a polynomial f to its restriction to x. And you can see that by definition this ring homomorphism is subjective and by definition of the ideal of x, the ideal of x is precisely the kernel of this map. So thus we get that we have a canonical isomorphism between. So thus we have an isomorphism of rings or k algebras if one wants that ax is isomorphic to the polynomial functions on x. I mean, by the obvious map, the class of a function f modulo, the ideal of x, is sent to f restricted to x. So in future we do not, in some sense, obviously these are the same things. So in future we will not distinguish between them. So we also write ax for the polynomial functions on x. So we can see that therefore this ring really parameterizes functions. So all possible functions from x to k, which are given as restrictions to polynomials. And we will write elements of ax either like that or like that, depending on what fits better into the argument. Okay, so this is this little statement. And we can, I want to just for, do I need this? For later usage, I want to make some tiny remark. Namely, so the zero set of a polynomial function is closed. So that means if x is in a finite algebraic set, f is an element in ax. Then if I look at the zero set of f, so this is the set of all points p and x, such that f of p is equal to zero, f, after all, is a function on x. So we can say what f of p is. Then this set is closed in x. And actually that's basically trivial because what does it mean? If f is in ax, means that we can write f equal to f restricted to x for some polynomial f in kx1 to xn. And what is this? Then z of f is by definition just a set of all points p and x, such that large f of p is equal to zero, which is nothing else than x intersected the zero set of f. And so it's closed in x. Okay, so this was kind of some kind of prelude about, before we really talk about the functions we want to study, we first look at a particular simple case of functions, which are just restrictions of polynomials. Now we want to define something which are regular functions. We will define them both for a fine varieties and for projective varieties. We'll only define them for varieties, so we will assume that our algebraic set is irreducible. So let's do it for on a fine varieties first. So these are the functions, these should be functions from our, well actually not a fine varieties, so far the fine varieties. So we have some open subset in a fine variety. We want to define what are the functions, so maps from this open subset. Okay, which are compatible with the structure of being a variety. Now in some senses everything is given by polynomials. One could think that this should be just again the restrictions of the polynomials to this quasi-fine variety. But if you think of it, you know, so if V subset X is a quasi-fine variety, so that X in AN is some sub-variety, then if we look at, we could define something which I for the moment call A of V, or maybe I'll just keep calling it A of V, which is the set of all restrictions f restricted to V, where f is a polynomial in K X1 to Xn. But, so again we have a map from K X1 to Xn to A of V, but what's the kernel? You know, we know that if you think of it, the ideal of V, so the set, you know, the kernel of, so we have a map again, K X1 to Xn to A of V, so objective by definition, just f maps to f restricted to V, and the kernel is I of V. But, you know, I of V is all the functions which vanish on V, but if they vanish on V, they also vanish on the closure of V, because the zero locus of a polynomial is closed. So this is the same as I of X. So we see that A of V actually is equal to A of X. So now we want that the functions that we consider somehow distinguish between the different spaces. So we see that this is not good if we just look at these, because we take an open subset and we don't, we get the same functions as before. They are not more or not less. So therefore we want to define it in a different way. Instead of taking polynomials, we take rational functions, so quotients of polynomials in such a way that there's no pole on V, so that they, you know, everywhere are defined. Okay, so let's do that. So a regular function on V should be a rational function, so quotient of two polynomials, which has no pole on V. And we have to see a bit more precisely what the correct definition is, but anyway, that's more or less what we want. And so now let's try to make an actual definition out of it. So V is an open subset, so I didn't say it, but V is supposed to be an open subset in X, okay? So the I of V is a set of all functions, of all polynomials which vanish on V. But if a, you know, the zero set of polynomial is closed, so if a polynomial vanishes on V, it also vanishes on the closure of V. And the closure of V is X. X is irreducible, so if I take an open subset of that, then, yeah? Okay. So I will still keep this notation A of V, although it's equal to A of X when I find it useful, okay? If I don't want to specify what the closure of my quasi-defined variety is, I just write A of V for this thing. Okay. So we have defined, so, but now I want to take a variety here. So V is irreducible. X is irreducible and V is irreducible. And I want to do this because we know, so if X is quasi-defined variety, then we have that the ideal of X is a prime ideal. In fact, it's equivalent, you know, that we had said. Now, this is equivalent to saying that the quotient of KX1 to Xn by this ideal is an integral domain. So this is equivalent to the fact that A of X is an integral domain. And if we have an integral domain, we can form the quotient field. So, and, you know, the quotient field only of a ring only exists if it is an integral domain. So that's why we restrict here attention to the case of varieties. There is, in fact, a way to do everything that I do here directly for the fine, quasi-fine algebraic sets by doing localization in the correct way. But somehow I kind of want to avoid that, but it's not particularly difficult, but I don't really need it. And so I don't do the general form of localization. I just look at integral domains in the quotient field, and then the localization of something is then always a suppering of the quotient field. Anyway, so you don't have to understand these words, so I just say them for who knows it. Now I will go ahead. So we have, so we can look at this quotient field. So again, assume we have, again, V in X quasi-fine variety, and then A V is equal to A X. So the quotient field cannot write any more. Q of A X, which is the same as the quotient field of A V, of say A X or A V is the field of rational functions, is called the field of rational functions on X, and denoted K X, or if I want it's also the field of rational functions on V, and then I denoted K V. And elements in this thing are called rational functions in V, on V. So an element is called a rational function on V. So we want to use rational functions to define regular functions, I mean on open subsets. So first we define the local ring at a point. So let P be a point in V. The local ring of P of V at P is the set, say I write it O V P, maybe a bit larger, which is the set, what is it? So this should be all elements in this quotient field, which are in a suitable sense defined at P. So I can write like this, these are the sets of all H in K V such that there exists F and G in A V such that H can be written as F divided by G. You remember that the quotient field is given by equivalence classes of such pairs, and such an equivalence class is written as the quotient like that, such that G of P is different from zero. So these are, so what I mean by that, is the H can be written in many ways as a quotient of two functions you can always multiply with any element on A V on both sides, and if you choose one which is zero at P, if you multiply by some L, where L of P is equal to zero, then you have F L divided by G L, which, and then the statement is not fulfilled, but you have what is just said is that there is a way to write H as F divided by G such that G of P is non-zero. There is one such choice. Okay, in the equivalence class there is such an element. And as an abbreviation, so for simplicity, in future we write this and similar statements just, this is, we write this as O V P is equal to the set of all F divided by G in K V such that G of P is non-zero. So this is supposed to mean precisely this. Okay, just you, these are all elements in K V which can be written as F divided by G such that G of P is not zero, not that for every way of writing it. This is, okay, so this is the local ring, and you could say, okay, now maybe not, and then I say, so if U in V is an open subset, then we put say O V of U. So the regular functions on U are O V U, which is the set of, I can say it even more simply, this is the intersection over all P and U of O V P. Okay, you know, this by definition is a subset of, or suppering of, whatever, this is a subset of K V suppering, and I just take the intersection over all of these in K V. I can certainly do that. So in other words, I could also say it like this, this is the set of all H in K V such that for every P, there exists F and G such that H can be written as, for every P in U there exists F and G such that H can be written as F divided by G, such that G of P is non-zero. But it goes in that way, for every P there exists F and G. It's not claimed that, and it would be a different statement, that there exists F and G so that it works for every P. There's no reason to believe that that should be true. And it also isn't usually. Okay, so now these are called regular functions. So if they are called regular functions, they should be functions. A function is a map from U to K. So I have to see why this thing, which by itself is just abstractly defined as a, you know, as a suffering of K V, why this thing is actually a set of functions. So I have to see how, you know, I use this thing to map. And in some sense it's maybe obvious what you do if you have any P here, you send and you find such F and G at the P, which do this, you send P to F of P divided by G of P. Okay, so let's see whether that indeed works. So this is maybe remark. So we have a map, actually a ring homomorphism from O V of U to the functions from U to K. The functions are always into K, always a ring, but the point was this modification just in the, and how is this ring homomorphism given? So if we have an H in O V of U, we write, and we take a point P and U, we write H equal to F divided by G with G of P different from 0, and then we set H of P equal to F of P divided by G of P. As F and G are already elements in AV, so they are functions on V, and so I can just do that. And G of P is non-zero, so this makes sense. So thus we, this gives us a map like this. So we send H in O V of U is sent to something which we called also H from U to K. P is sent to H of P defined like this. So we have to be careful because we have to see that this makes sense. So is this map well defined? So after all, an element here is an element in this quotient field. So it's an equivalence class of pairs, so it must not depend on the representative. So if say H is equal to F divided by G, and it's also equal to F prime divided by G prime, where this G of P is different from 0, and also G prime of P is different from 0, well then the equivalence relation is that F times G prime is equal to F prime times G. And so it follows, now we can put P into this, these are already fine. So this is F of P, G prime of P is equal to F prime of P G of P. And then we can again divide, so now we are talking about numbers and we can look at the fraction of the numbers. So it follows that F of P divided by G of P is equal to F prime of P divided by G prime of P. So this is, and so the map is indeed well defined. So we have a well defined map like that, and what I also claim is that this map is actually injective. So if we know what the function is associated, so the map P goes to this, P goes to H of P is, we know what H was as an element in this ring. So this map is injective. So the map which sends H in OV of U to the map to the thing which I called also H, you know later we want to identify them which sends, so the map which sends P to H of P from U to K. So this map is injective. So why is that? So we have to see, how is it? So we take another element H prime in OV of U such that H of P is equal to H prime of P for all P in U. We have to see that H is equal to H prime as elements in this ring. So as elements in the fraction field of OV, well, so let's see. So in other words this means that H, if I take H minus H prime of P, so I put L equal to H minus H prime. This is now as elements here. This is a ring homomorphism. So then this says then L of P is equal to 0 for all P in U. And we want to say that L is actually 0. Well, we write L to F divided by G where F and G are elements in whatever AX or AV. Maybe I write AX. And G is not 0. So this means G is not the 0 element in AX. Not that G is everywhere non-zero. But G is not the 0 element. But if G is non-zero, then this means that the 0 set of G is a close subset of X. So let W be equal to U minus the 0 set of G. This is a non-empty open subset U. And we know it's non-empty because if the thing is not 0, there will be some open subset where it's not 0. And this means that W is intersection of U with another open subset. If it's irreducible, it's easy. Being irreducible is equivalent to saying that the intersection of any non-empty open subsets is open. So this was maybe not stated explicitly. So if the X is irreducible and U and V are non-empty open subsets, then it follows that U intersected V is non-empty. This is actually a restatement of being irreducible. I think it's more or less, if you place the close subsets by their complements, then in the definition of irreducible, you have some statement about close subsets. And if you take their complements, which are open, you get this statement. So this is a non-empty open subset. And so thus for p in W, we have that 0, that if I take f of p divided by g of p, which after all is just L of p, but here I can just take this f and g because the g doesn't vanish there, is equal to f of p divided by g of p. So g of p is non-zero. So that means that f of p is 0 for all p and v in W. So that means that f lies in the ideal of W, which is the same as the ideal of X. Well, no, I mean, let me say it differently. We have seen that so f is an element in Ax. We know that the zero set of an element of Ax is closed. So that means f is 0 on a closed subset, which contains this open subset W. But the closure of W is the whole of X. So that means f is 0. So as the zero set of f is closed, it follows that f is equal to 0. And so therefore, by definition, L, which is f divided by g, is equal to 0. So this is the zero element in KX. So we see that indeed we have an injective ring homomorphism here. So we can therefore say that O v of U is embedded as a suffering of this. So if we want, we can identify O v of U with its image, which will be a certain ring of functions from U to K. And one can work out which ring it is. I mean, the map is obviously not subjective. So the result will be that O v of U, or rather its image in the set of functions from U to K, is the set of functions from U to K, which locally are given as quotients of polynomials in K. We are, after all, we have defined right in KX1 to Xn. So how does it look precisely? So the precise statement is the following, and I call it an exercise. It just is an exercise, just unraveling the definitions. So you have to kind of remember what AX was and then put this in the division there until you just put it together and you find, so this image. So I mean, I just say now the image of, and often we will just identify it with O v of U, is the set of functions H from U to K, such that, and now we, you know, every point has a neighborhood such that in that neighborhood it is a quotient of two polynomials, such that the denominator does not vanish there. So set of functions such that for all points P in U, there exists an open neighborhood W of P in U and functions and polynomials and G such that G is nowhere zero on W. So G of Q is zero for, is non-zero for all Q and W and H of Q is equal to F of Q divided by G of Q for all Q in W. So in some sense as I, so again, it is not claimed that it will not be true in general that we can write H everywhere as quotient of F divided by G so that G is nowhere zero on W. This will usually not be possible, but we can for each point find a neighborhood that in that neighborhood it can be written like that, okay. So, and then this thing obviously if you want to solve the exercise just given we have, you know, we have this class H in the quotient field in the field of rational functions. This can be written in several ways as restrictions of polynomials as elements of these quotients of elements in AX and you can always, you know, for each point you can choose one such that denominator doesn't vanish at that point and if it doesn't vanish at the point it will also not vanish in some neighborhood of that point because the locals where it doesn't vanish is open and so if you write it down you see that this is the statement we want to show. Anyway, it's quite trivial, but it is just, we want to use in future both definitions so we also want to, sometimes we want to view or view of U as a set of functions like that and sometimes we want to view or view of U as, you know, a subring of the field of rational functions and it's okay because we have a canonical isomorphism between these two rings. One more thing I had called, we had called OVP, the local ring of V at P. So there's actually a concept in algebra of something being a local ring and so I want to tell you that the local ring of V at P is a local ring. Okay, so we have another thing, we can define something else. The maximal ideal, ideal at P is MP which is defined to be all the elements F in the local ring or maybe we call them H in the local ring at V at P such that H of P is zero. Remember that if it lies in local ring, we know what H of P is. We write H as F divided by G and it's F divided by G of P and it makes sense to say that H of P is zero. So we have the functions which are zero. And this is certainly, this is a maximal ideal in OVP. Namely, I mean we have, and obviously we have the evaluation map, homomorphism, whatever from this local ring to K which sends H to H of P. This is obviously a subjective ring homomorphism and obviously the kernel by definition is M of P. And so that means that MP is a maximal ideal because it's the kernel of a subjective ring homomorphism to a field. No, I mean ring is a maximal, I mean a subring. An ideal is a maximal ideal if it's the kernel of a map to a field or equivalently if the quotient by it is a field. Okay, and so this is a maximal ideal and we have something, it's however somewhat the situation is much more special. If we take, it's somehow the unique maximal ideal in this ring in a very strong sense. So if we have an element H in this local ring which does not lie in the maximal ideal, then this means then we can write H equal to F divided by G where F of P where first G of P is non-zero because that means that lies in the local ring. But you know the result H of P is F of P divided by G of P so we also have that F of P is non-zero. So therefore if we take 1 over H which is G divided by F this is also an element in the local ring. So thus it follows that H is a unit. So obviously a unit can never be an element of any non-trivial ideal. If you have an ideal which contains a unit it's the whole ring. So here we are in this special situation that we have a maximal ideal which contains all non-units. So it's as big as an ideal can possibly be. Yes? What? Yeah, I said it but I didn't write it. So we see that MP is a maximal ideal containing all non-units. And so this is the property of a local ring. So a ring is called local if it contains a maximal ideal. If you want unique but that obviously follows maximal ideal which consists of all non-units. So such that M such that R without M is equal to the units in R. So we have kind of an ideal which is really maximal in the most maximal sense. And so this is called a local ring and so in commutative algebra people often study local ring. But in some sense it actually is the other way around. So in algebraic geometry people have introduced these local rings and their main property is this one. And then people in commutative algebra study rings which look like the local ring of a variety and call them local rings. But anyway whatever it is the local ring of a variety is a local ring. Okay, so enough of that. Now we want to finally see. Now we want to study the regular functions on an affine variety. And in fact we want to study the regular functions which are regular everywhere on an affine variety. So according to our definition if X is in the affine variety then we have OX of X which is the functions on X which are regular on the whole of X. So we want to study those and the claim is that those are actually just the polynomial functions. So if we require that on an affine variety we have a function which has nowhere a pole which is everywhere defined then it actually is the restriction of polynomial. So fine variety, so functions which are regular everywhere are polynomial functions. So I should just to be sure we should recall that AX, so the coordinate ring of X can be viewed or be identified with a sub ring of the field of rational functions by just taking the rational functions where the denominator is 1. So it can be canonically identified with a set of all F divided by 1 such that F is an element in the X as a sub ring of the X. So this leads to the statement that if you have an integral domain it's always a sub ring of a field in this way. And now we want to come to this. So the statement is let X be an affine variety then OX of X is equal to AX. Okay, so let's see. This is actually not, if you think of it this is not so, doesn't look so very trivial because it a little bit contradicts some of the things that I warned you about before. You know an element in OX of X is something which I can write so that for every point in X I can find two polynomials or can find two regular functions so that I can write it as a quotient of these two regular functions so that the denominator is not zero at that point. But for different points I might have to use different denominators, different such representations. But the claim is here that I can find one which is nowhere zero in fact I can take it to be a constant. And so one has to see how that would be the case. So let's look at the proof. So obviously we have that O of X of X that AX is contained in OX of X. An element in AX can be viewed as F divided by one and this is certainly a quotient of two polynomials so that the denominator does not vanish anywhere. So that's fine. So we have to show the other conclusion. So well let's see. So we take an element H in OX of X. So then we go by the definition. So if we fix any P in X then it follows the axis polynomials say I call them F of P and G of P. So we assume here maybe that X is a close sub-variety of AN also to fix notation in K X1 to XN such that well that the class of this thing which is an element in AX. So if I take F divided P divided by G of P this is an element in OX of X a quotient of two elements in AX that this is equal to H but also that G of P of P is non-zero. So we can write it as a quotient of the classes of polynomials which just means the same as a quotient of elements in AX such that at P it is non-zero. And we can find such representation for every P and we want to somehow find a representation which works for all P at the same time. In fact the denominator should not be 0. So we can also say it like this maybe I forget about this P. So equivalently for all P in X there exists a polynomial G which he was called GP but anyway we don't really need it such that if I take H times the class of G this is an element in A of X. Because you know if I take so maybe I do write here GP in this notation before you know it's just if I take H and multiplied by GP I get FP. Okay, so we have this wonderful statement. Well, so we use this to write down some ideal which we want to show that actually is the whole ring namely let J be the set of all polynomials G such that if I take H times the class of G this is an element in AX. Okay, so J is an ideal because if I take H by G1 I get some element in AX if I take H times G2 I get an element in A of X. So if I take what where no N in I hope very strongly I said in but I equal so this is an ideal this is kind of obvious know H times G1 is an element A in AX and H times G2 is an element B in AX and H times G1 times G2 plus G2 is A plus B and so on with the product with it. Okay, so this is an ideal and furthermore and J contains the ideal of X because if I have an element in the ideal of X and this class is zero and so I get zero and that's certainly in A of X. Okay, so this is an ideal which contains the ideal of X and we can look at something else. What is the zero set? What is the zero set of J? So let's say we look at zero set of J intersected with X. What is it? This is set of all points such that all points in X such that if I have an element here with this property then this element has to vanish at that point. Okay, but we have seen here that is precisely not the case. We know that for every point in X that exists a polynomial such that if I multiply with the class of the polynomial we are here and G of P is non-zero. So we see that this is the empty set. Okay, on the other hand we have that J contains I of X so that means so J contains I of X this means that zero set of J is contained in X. No? So if then the intersection, you know, so that means the intersection of C of J with X is just C of J. So it follows that Z of J is the empty set. And now we should remember, you know, the whole point of introducing this strange ideal is that now we are in a situation to use the null stanzatz. Now remember we had the statement of the null stanzatz that if we have an ideal with the property that it's zero set is the empty set then this ideal contains the element one. Okay, so by the null stanzatz we have that one is an element in J. And so what this means is that if I take, you know, what is J is a set of all polynomials that if I multiply by its class I get into AX. So we have that H times one which is just H is an element in AX. So but you can see that in some sense, I mean the proof is not now so very complicated. But in some sense you see it's a very deep result because we actually have to use this null stanzatz which requires a very long proof. Because so to go from this fact that you locally have this everywhere to the fact that you actually can choose the denominator to be one actually requires some reasonably big gun. Okay, so this was the statement. So we somehow see that in some sense that's also maybe what we wanted. We wanted the irregular functions somehow generalize the polynomial functions on a fine varieties. So on a fine varieties we would want to get back the polynomial functions. We wanted to get more functions on smaller sets but we wanted this to be just this. Okay, so this was it for the, so these are the regular functions on projective algebraic sets. Now on a fine algebraic set, now we want to go to projective varieties. And here we somehow can see that this thing of having to take these quotients of polynomials is much more evident. Because if we take a polynomial and take a restriction of a polynomial to projective variety, this is not a function. Because if you multiply the coordinates, an element in the projective space is an n plus one tuple of elements in K up to a multiplying by a constant. If I multiply by a constant and let's put it into a polynomial I get something different. So polynomials never define functions on projective or quasi-projective varieties. So we have to anyway use something else than polynomials. But what will turn out is that if we take quotients of homogeneous polynomials of the same degree, then they define functions. Okay, so if x is a projective variety or quasi-projective variety, x in pn, then elements of polynomials in K x0 to xn will not define functions. I mean just nowhere or no open set of x, so x to K. But we can take quotients of homogeneous polynomials of the same degree, quotients f divided by g of homogeneous polynomials of the same degree. So we will want to do this in a moment. So we, and we can again instead of looking at the polynomials, we can look at elements in K x0 to xn divided by the homogeneous ideal of x. So definition, so we look at the generalization of the coordinate ring. So let x subset pn be a projective algebraic set, so it also works for projective algebraic sets. The homogeneous coordinate ring of x, so I denoted by, is by sx. So this is defined to be just all the polynomials divided by the homogeneous ideal of x. So this is something that we can define, but we have to remember that elements of sx do not define functions on x. They only would if polynomials would define functions on x, but they don't. But we can again look at the quotient field of this thing, and inside these quotient fields we look just at the elements which are quotients of homogeneous polynomials of the same degree. Okay, but maybe I can first say, so again s of x is an integral domain. So we can form, we have q of x, q of s of x is its quotient field. And maybe I state again what I just said. So we have that elements of sx do not define functions x to k. But h is equal to f divided by g, where f and g are homogeneous polynomials. So either homogeneous polynomials or classes of the same degree, say t. So what then? Then if I take any point, if g of p is non-zero, so p is equal to a0 to an in pn. So if g of p is non-zero, I can put h of p to be f of p divided by g of p. And notice that if I write it out, if I take another representative of p, we have that f of a0 to an divided by g of a0 to an on the one side. We can also look at another representative, so f of lambda a0 to lambda an by g of lambda a0 to lambda an. Well, I know that if it's homogeneous of degree d, this is equal to lambda to the d times f of this is equal to lambda to the d like this. So I can write this is equal to this lambda to the d divided by lambda to the d. And so we are equal. So this cancels. So we see that the, so h of p is independent of the representative, it's well defined. This is well defined. Okay, so time is already up. So, yeah, maybe it is, let me see. Yeah, maybe it's not good to seriously go into this now. So I will next time do the proper, I mean finish the definition correctly. And then, so it's kind of clear that we do the same thing. Our regular functions will be on an open subset will be, so will be elements in the quotient field of Sx, which however can be written as quotients of homogeneous polynomials of the same degree, such that denominator doesn't vanish at the given point p, and then for all p in open set u. And then with these things we can work in precisely the same way as in the fine case. And so we will develop this and then we'll see that this actually allows us to say that this is, well, anyway, then we will use this to define morphisms. So morphisms will be continuous maps between such varieties which are compatible with these regular functions. And we will find that this tells us that, for instance, this map from an into pn, which we had used to identify an with an open subset of pn, actually turns out to be a morphism. And it really embeds an is an open subset of pn. And anyway, and then so that after that somehow it gets unified again. So there's no diff, we really have to look only at quasi-projective varieties because whatever we have done in the case of an is a special case of what we do for pn. Well, okay, maybe that's as much as we hope to do next time.