 So, yeah, so good afternoon, I'm sorry I'm a bit late. So, we were, it started to talk about morphisms, and so it first talked about regular functions, which were functions which were locally given as quotients of polynomials, and then we had defined morphisms in general as continuous maps, which are compatible with regular functions. So that means that if we have a regular function on an open subset, and we take the pullback of the regular map, that is the composition with the morphism, then this is a regular function on the inverse image of this open subset. And now we wanted to come to give some more concrete description of morphisms, because this definition is a little bit abstract. And so, I first wanted to talk about morphisms to subvarieties of the N, if you want to find varieties. So, and here the statement is that they are given by an n-tuple of regular functions. So, if theorem, so let x and y be varieties, and we assume that y is a subvariety of the N. So, a map phi from x to y is a morphism, if it is given by an n-tuple of regular functions, which are regular on the whole of x. So, if there exist, if and only if, there exist say f1 to fn, which are regular functions on the whole of x, such that we have p of p is equal to f1 of p until n of p for all p in x. So, we can then also write phi is just f1 to fn. Okay, so this is a characteristic characterization of morphisms. So, subvarieties of A and N. So, for instance, if x itself is an affine variety, close subvariety of some A and N, then these fi have to be elements of the co-ordinary, because we had this theorem. X of x is then equal to Ax. So, proof. So, we have, if and only if, we have to show both directions. So, first we take a morphism phi from x to y, a morphism. So, we somehow have to get these fi. And if you think of it, you know, they are just the, you know, if you compose phi with the coordinates on A and N, this will give us the fi. So, let's see. So, let y1 to yn be the restrictions of the coordinates on A and N to y. So, if I call the, I mean, if I call the coordinates on this A and N, if I call them, maybe also y1 to an, I can denote by the same letter, the restriction to y, which is now a regular function on the whole of y, because it's a restriction of polynomial. Yeah, so we have gotten some regular functions here. And, you know, note, I mean, this is duality, but I only want to keep it in mind. So, if we have a point q equals to A1 to An in y, we can write it. This, the i-th coordinate is just obtained by applying yi to q. It's just the i-th coordinate. So, now we put fi, we are supposed to produce these functions regularly. We define them to be the pullback of the yi. So, in other words, yi composed with p. And, as this is a morphism, these are regular functions on the whole of x, because the yi were regular functions on the whole of y. So, now I have to see that this does this kind of trivially. So, if we take a point p and x, then we can write phi of p equal to, so somehow there will be some points, p1 to bn, where the bi are just the yi of phi of p. This is just the i-th coordinate of p of p. And so, this is just equal to fi of p. So, I've set it up in such a way that indeed it is true. Okay. So, you just have to see that the components of the map are in a natural way the pullbacks of the coordinates. And therefore, then it's clear that these will be regular functions. So, now we have to do the other direction. So, conversely, we assume we are given phi in this form. So, let phi from equal to f1 to fn with fi in ox of x be a map. So, this is a map from x to y. And this notation means as before that phi of p is equal to f1 of p to fn of p. This is certainly a well-defined map. Just say what the coordinates of the map are. And we require that it goes from x to y. I mean, this doesn't know before, but we assume that. And so, now we have to see that this is indeed a morphism according to our definition. Because, you know, we have to find what a morphism is. You know, it's supposed to be continuous. And the pullback of regular functions should be regular. And so, let's see. So, to show phi is a morphism. So, let's see. So, first we have to show it's continuous. So, the inverse image of closed subsets is closed. So, let p subset x be closed. That is, we can write p equal to x intersected the zero set of some polynomials with g i some polynomial in k x1 to xn. Now, we know the closed subsets of this form. Because it's a risky topology. So, let g be one of the g i. Somehow, I want to describe the inverse image of p. So, we want to describe how the zero set of g composed with phi is. So, we can write g. Somehow, it's a polynomial. So, sum i1 to in, a i1 to in, x1 to the i1 until xn to the in. So, somehow it is a polynomial. So, what is now g composed with phi? Well, phi is given by component wise by taking the ith component as f i. So, if I compose it with, if I compose g with this, it means precisely that I have to replace the x i by the f i. So, this is just sum i1 to n. Also, you could also say it's g of f1, fn. And now, you see, this is a linear combination of products of the f i's with coefficients in k. And this f i itself is an element of o x of x. So, this whole expression is also an element of o of x, because this is a k algebra. Okay, what? Yeah, so, yeah, so I lost the check of, so the target is y. So, therefore, it is y here and this is y here. Okay. So, let's see. So, and so, therefore, if I take phi to the minus 1 of b, this is just a zero set of g1 composed with phi, gn, what is it? gm composed with phi. So, the common, and we know that the zero set, this is an element of o x of x, and we know a zero set of an element of o x of x is closed. And so, it's the intersection of these closed sets. So, we know z of g i phi is closed in x. So, this thing will be closed in x. So, therefore, the map is continuous, inverse image of a closed subset is closed. Then finally, we have to see that it's compatible with regular functions on open subsets. Well, so we have to write down any regular function and have to see what it is. So, the pre-image is set of all p in x, such that p of p is an element in p, which is, and p is equal to z of g1 of gn. So, this is equal to the set of all p in x, such that if I take g i composed with phi to zero, okay, maybe with a bracket. So, now we want to come to, we want to show it's compatible with regular functions on open subsets. So, we take, so let h be a regular function on an open subset u of y, and say we write w, w to be the inverse image of u. So, we have to show that h composed with phi is regular on this. Now, we had this before, that we are always allowed, when we want to check that something is regular, we are always allowed to replace an open subset by a smaller open neighborhood of any point in this open subset. So, we can always make this open subset smaller. But, you know, we might then need several open subsets for one here, but for every point, we have to be able to find an open neighborhood for which whatever we want to prove is true. And we always use this in the way, in the way that we can assume that our regular function is a quotient of polynomials. No? Because we know that locally a regular function is a quotient of polynomials. And so we will do this again. So, replacing u by a smaller neighborhood of any given point of any, say, p in u, we can assume that, say, h of q is equal to f of q divided by g of q for all q in u, where f and g are some polynomials, and g has no zero on u. And now, what happens if we take h composed with phi? h composed with phi is f composed with phi divided by g composed with phi. That is, according to what we have seen here, we just take f and replace the variables by f1 to fn and the same with g. And so we have just seen that if we take a polynomial and we put into it regular functions like this, we get regular functions. So, this is a regular function and this is a regular function, a regular on the whole of x. And g of f1 to fn has no zero. We're actually not on the whole of x. We have a, I mean, wherever we were. So, we were on u, o, x, w. And we had seen that if we are in u, then g of a point p in u is not zero. And the points of w all map into u. So, it means that g of f1 to fn has no zero on w. So, by the result, we proved the other time, we know that if we have a quotient of two regular functions such that the denominator has no zero on the corresponding open set, then this quotient is also regular. And so this shows that this is indeed amorphous. And so, anyway, you see that in order to get this relation, we just have to see that the coordinates themselves on the target in the n are regular functions and that we can define the regular functions which give us these, which define as the map in terms of the coordinates just as a pullback of the coordinates and then we don't just have to check the definitions. So, in particular, we get what we had before in special case. So, the regular functions on the variety x are the same as the morphisms for a1 because this is the special case of having one variable. It just says that the coordinate of the map is the only one, must be a regular function. And I also mentioned this before that so if I have two affine varieties or two close varieties of, sub varieties of a, a, n and a, m then the morphisms between them are precisely the polynomial maps. So, this is because we have just proven that if we have a morphism from anything to y, it's given by an n, in this case an m tuple of regular functions, which are regular on the whole of x. And now, I'm going to give you a simple example of what we can do. So, we have a regular on the whole of x and now we know that if x itself is a close sub variety of a, n then the functions which are regular on the whole of x are precisely the elements of a, x of the coordinate ring and a polynomial map is one which is given by an n tuple of elements in the polynomial ring which is the same as given by an n tuple of polynomials. So, it's, this is straightforward. So, we can also have a kind of more abstract algebraic description of that to say that somehow talking about morphisms of sub varieties of a, n and a, m is the same as talking about homomorphisms of k algebras. So, so this is, this is, you know, rather, I mean we are not going to use it so very much but it's rather important when one goes to more advanced topics in algebraic geometry that somehow this somehow leads to the generalizations that one makes there when one talks about schemes and so on. So, but here we just have the statement so let x and y be varieties and assume y in a, n is a close sub variety. There's a bijection. So, we can look at the morphisms from x to y is in bijection to the k-algebra homomorphisms from in this case from E y to O x of x. And the map is the obvious one to morphism phi we associate the pullback by phi. Not just the composition with this we know that if phi is a morphism then the pullback by it is a homomorphism of k algebras like that that we had seen. And the claim is that if we know this k-algebra homomorphism we also know the morphism. And all k-algebra homomorphisms come about in this way. So, I will not give a complete proof of this. It's not very difficult. I will just, I will construct the inverse map and then one has to check that is indeed the inverse. That somehow the most difficult part you have to see if you have such a homomorphism how from that do you get the morphism back. But in some sense it's very similar to what we already did. So proof. So first so I mean I can just set if phi from x to y is a morphism then we know phi star from A y to O x of x is a k-algebra homomorphism. This is clear. And now we want to construct the inverse map. Define inverse map. So we have to start with the k-algebra homomorphism. So let phi from A y to A x of x be a k-algebra homomorphism. We want to construct a kind of phi of which it was the pullback, a small phi. So how do we do this? Now we do in some sense what we already did because we know that somehow it's kind of clear what the pullback of the coordinates I mean that the coordinates play a role and they are pullbacks. So let say y1, yn in A y be the coordinate function. So in other words we have the coordinates if you want x1 to xn on this, well actually it should be m because we are in A m. We have the coordinates here and we just restrict them to y to become functions on y. And if we have somehow seen that before we could describe morphisms in terms of the pullbacks of the coordinate functions. So we take the pullbacks. So that fi, I mean this is supposed to be the pullback. So if I take phi of yi this will be then some element in O x of x. And we put the map given by these coordinates, by these functions. So the point p is set to this m2. Now this is certainly a map which is defined on x. If I take a point p in x I send it to somewhere. But a power r it only goes to, we have to see to A m. Because we wouldn't want to say it's a map from x to y. But if we just write down something like that we don't know whether all the points of x lie in y. We have to check that. So we want, so we have that. So then this is a morphism. So we have constructed a morphism of which we still might want, however we want to have a morphism to y and not just to A m. So to see it is a morphism to y we have to see that all the points of x are mapped to y. We had seen the last time that if we have a morphism to a larger variety which actually maps to a sub-variety then viewed as a map to the sub-variety it's also morphism. So therefore it's enough to check that. Well, and so how do we do that? Well, we take a polynomial in the ideal of y and we see that it's inverse, it's pullback vanishes on x which will be what we need. So let h be an element in the ideal of y. So if I take h composed with phi this is h of f1 to fm. We had seen this before. The composition is done by putting this into it. The coordinates are replaced by these functions fi. This is h. So according to this, so fi is phi of the small yi. But as phi is a k-algebra morphism we can pull the phi out. So this is phi of h of y1 to ym. But yi, the yi are the coordinate functions. So this is, if I maybe write large yi the coordinates on a m this is yi restricted to y. So it's the coordinates of a m restricted to y. Or if you want it's the class of the coordinates on a m modulo the ideal of y. But so what we have inside here is the class of the yi modulo the ideal of y but h lies in the ideal of y. So that means if we apply h to this we actually get 0. So what we find is that h composed with phi is 0. So that means that if we take the image of x this is in the 0 set of h. And so h was a general element of i of y. That is the phi of x is contained in the 0 set of the ideal of y. So phi of x is contained in y. So we have indeed found in this way a morphism from phi from x to y. And then you know one wants to check that this is inverse. So show, so this I just go on exercise show these two maps. So phi maps to phi star. And phi maps to this thing which I here just called phi are inverse to each other. So for instance if we take start with phi star if we do this construction we should get back to phi. Maybe I should I mean it's not anyway. And on the other hand if we we should also see that I mean there is an inverse in both directions so that we get in the one. And you know this is both pretty straightforward to check. You know you have it now everything is written down and you just have to check that it's the same thing. Okay so I want to use this as an example to see that you can have a bijective polynomial map which is not an isomorphism. No one could have thought that the map is bijective and the morphism than it is an isomorphism but that's not the case. So example bijective polynomial map need not be an isomorphism. So we have this caspital cubic so C is equal to the zero set of y squared minus x to the three in A2. And so if you just look at the real points this we had this picture before it somehow looks like this. Well it's symmetric but you know anyway and here we have the point zero. And so the claim is that this we have a bijective morphism from A1 to this which is not an isomorphism. So you can somehow imagine if this is A1 I just look at the real points and you can somehow project here. And this will be a bijective morphism or rather maybe it's I don't know which way it now goes. Yeah you can kind of parametrize it like this but which would correspond to both. So this will be the projective morphism then the question is whether you can also go back. So and the claim is that you can't. So we have phi equal to actually I didn't give the map correctly. Phi equal to t squared t to the three from we want to somehow parametrize it. This goes from A1 to C isomorphism. So you can you know it's obviously morphism because it's a polynomial map. It's given by polynomials and it you can easily see that it does map to C. Because if you have if this first coordinate is if this is equal to t squared this equal to t to the three then we'll get zero. And it's actually easy to see that this bijective. In fact I can one can give the inverse map and the inverse. How is it so now we take two points here. So a point with two coordinates I say the inverse I call G from C to A1. So I take G of AB is defined to be P divided by A. Obviously if A is different from zero and which means if the point is not zero zero and I put it zero for AB. To see at least from this definition it's not obvious that this would could be described as a polynomial map. It doesn't look very much like it. But you know in theory it could still be there would be another way to write it. I mean anyway you can see that this is the inverse if you just put you know look what happens. If you have you know you get you know if A is t to the third and is t squared and B is t to the third then B divided by A will be t. Not very surprisingly so you get it back and also works in the other direction. And the point you know the only point where the first coordinate is zero is the point zero zero. So you can also complete it like this. So but I claim this is not a morphism. So so phi is not an isomorphism. In other words G is not a morphism. Well for instance we can see it like that. I forgot something but now I first finish it and then we'll see this we could also have seen earlier at any rate. So because if we look at phi star from A of C to K x y so which is K x y model y squared minus x to the three. It's not completely obvious but I claim that the ideal of C is the ideal generated by this thing. So the pullback goes to KT and so so what is the pullback? Well the element x is mapped to the composition with this map. So to the first coordinate here so t squared so it maps x to t squared and y to t to the third. And you see phi star is not an isomorphism clearly because it is not directive. So one polynomial which is obviously not in the image is t. If it's not constant then at least we have to have a t squared. We had seen earlier we were not actually using now this result we just proved. We had seen the previous time that if we have a morphism and have an isomorphism then the map on global regular functions is an isomorphism. The pullback on regular functions is an isomorphism and here it isn't. So phi is not an isomorphism. And if one believes in this picture one also maybe wouldn't want this to be counted as an isomorphism because they don't look so very similar. We do have some point here which is very special, this cusp. And we also can see that it's precisely that point where it goes wrong because obviously on this on the open set where we leave out the zero this is a morphism inverse map because the quotient of two regular functions and denominator is nonzero. But at this point it fails to be morphism and it cannot be made into one. So now we wanted to somehow now we we had defined to be an affine variety to be a variety which is isomorphic to a closed sub variety of air. Now one could think that this is still the same as the closed sub varieties of air. But in fact it's not the case for instance one thing that can happen is that you have a closed that you take an open subset of a closed sub variety of air and it's still a fine. That's isomorphic to a closed sub variety in some other affine space. And that's actually quite again it's not so it doesn't look so exciting but it's very useful later because it somehow implies the fact that if you have any variety and any point on any variety it has a neighborhood which is a fine. And so if you want to prove anything about varieties which you can prove by proving it locally you can only you only have to prove it for a fine variety. And that is actually something which the more advanced one is in the subject the more often one does it that way. Okay we are not going to use it so much but it is an important fact although here it looks a little bit you know not so exciting anyway. So let X so I first find that X in N a closed sub writing and we take F some polynomial in this and then so which does not you know so so maybe does not lie in the idea of X doesn't vanish on the whole of X. Then the principle open defined by F is XF which is just defined to be X without the zero set of F. So that we can certainly define but the claim now is XF is always is an affine variety. So if we take any close sub variety of a N and we take the complement of a hyper surface it will always be a fine. Okay so in fact you can find it as a close sub variety in a N plus one. So let's see so we put that to be so say the zero set of the idea generated by the ideal of X and one more polynomial. Which might remind you of the proof of the Nulstern sets. X N plus one minus one so we have here one more variable X N plus one and this is a zero set in the N plus one. So we take the elements in the deal of X we add one more equation in an extra variable which says that and you know as you can see here this condition. If this thing vanishes it means that F cannot be zero at that point. So somehow we will see that this thing will be isomorphic to XF. So so that is a close sub variety of the N plus one isomorphic to XF. Okay so so we can define a map from XF to N plus one which has this as an image. So let Fee I just write it down X1 to XN. So the XI are just the coordinates on N. So this is kind of the identity on the first N coordinates and then we have one over F. So this is a map from XF to N plus one is a morphism. So it's a morphism because it's given by an N tuple of regular functions. So the first coordinates N plus one tuple the first N are just the coordinates and they are certainly regular. The last one is one over F and this will be regular whenever F doesn't have a zero. But XF precisely contains of the points of X where F is not zero. So one over F is regular on XF regular functions. Okay so and I claim it also maps to Z and Fee of XF is equal to Z. So I think that's you know that's easy to see that you know if you just look these are the points X1 to XN where the first N you know if I look at Z. This is Z is the set of all points A1 to AN plus one in AN where the first where A1 to AN are point in XF and the last AN plus one is equal to one over F of the other ones. And this is precisely the same as saying that it is the image of this map. Just by definition it is this and it's also clear and it's clear Fee is objective. I mean the inverse map is just the projection to the first N coordinates. So now, so first of all you know in order to be in principle we have to first show that Z is a closed sub variety. So it's not a closed subset but it's also irreducible and this follows from this cause so as XF is irreducible. Also Z is irreducible namely if I just write if Z is equal to the union of two closed subsets then XF is the union of the inverse image of the Zi's and according you know as the Zi are not equal to Z and the map is subjective also inverse images are not equal to XF also closed because Fee is continuous. And so it would follow so it would follow that XF would be reducible. So now all in another sense we have this. So if we have this then we have this is this but we know this is not the case because XF is irreducible so also Z is irreducible. Incidentally this argument obviously works in complete generality. I've just proven that if you have an amorphism from some variety to some other variety then the image is always irreducible and that's what this argument proves. So thus we find that Z is a closed sub-variety of the N plus one and now we are allowed to talk about morphisms starting at Z. And you know we can see that phi to the minus one which already had told you what it was and it's given by just the projection to the first N coordinates is also morphism. So we find that indeed XF is isomorphic to Z. Okay so this was about this thing now. So I had given some kind of description of the morphisms to sub varieties of the fine space in terms of kind of the coordinates of the morphism. Now we want to do something similar for morphisms between quasi-projective varieties. So again we want to say that if we have a morphism then its components are either given by polynomials or by elements in the or by regular functions. So let's see how this works. Maybe I first briefly talk about a special case. So morphisms of quasi-projective varieties. So first I talk about polynomial maps which I can also define for quasi-projective algebraic sets. Definition at Y in Tn, say X in Tn, Y in Tn, Tm quasi-projective algebraic sets. So a map phi from X to Y is called polynomial map if it's given by polynomials. What does it mean? If there exist polynomials f1, f0, so fm in kx0 to xn. So as many coordinates as the source, as many of them as we have in target which are homogeneous of the same degree. And they have no common zero on X. So if you take the zero set of all of them then this doesn't intersect X. Such that for all points p and x we have phi of p is equal to f0 of p until fm of p. So we write phi is equal to f0 of fm. So again the components of phi are polynomials. Then we call it the polynomial map. So again what I mean here is f0 of p is not very defined. You know this is just a polynomial. If we multiply p by some non-zero constant we get something else. But if we replace here p by any representative a0 to an and take the same representative for all of them. Then if we take another same representative for all of them we get the same result. Because the whole thing is only this class is only up to multiplying by the same constant. So this is very fine. So anyway you can easily write down any example that you want. But so it turns out we'll see in a moment that so these are morphisms which actually occur. So it often happens that morphisms between quasi-projective or projective algebraic sets are just given by m-tuples of polynomials. But there are more morphisms because to have a morphism we actually only need that it's locally a polynomial map. So that for each point in x there's a neighborhood so that on that neighborhood it's given by an m-tuple of polynomials. So it's a more general concept and this is really not good enough. But it is something you know the easier morphisms are like this. So now before doing this we want to first study and anyway are going to use. We have seen that an is projective to an open subset of pn. Just this set which I called u0 or something. So we have this projection. Now the claim is that this projection is actually an isomorphism. So that you know we really are allowed to identify an with this open subset. And so this will be quite useful and this also will allow us to deduce this fact that we just said. That if we have a morphism between a quasi projective varieties then it's locally given by m-tuples of polynomials. We deduce it from the fact that we have a similar statement for sub varieties of a fine space. Once we have the statement that open subset of that this u0 is isomorphic to air. So let's see. So we had defined an open cover of an by open subsets ui from i equal to 0 to n. Now ui is the set of all a0 to an in pn subset ai is non-zero. If you want you can then normalize ai to be 1. And we had shown we had seen that we have a bijection. This was phi i from an you know it's the other direction from ui to an which sends a point a0 to an. So the corresponding quotient where every element is divided by ai and ai divided by ai is left out. So this is a0 divided by ai and it goes on. When we arrive at ai divided by ai we throw it away and then we get to an divided by ai. And the inverse was ui from an to ui which is obtained by taking an n tuple which I write as a0 to an. However the ith one is not there so it's just n instead of n plus one. And we map it to the same thing where in the ith position now we put one. But now obviously in projective space. So this we had seen these are these two maps are obviously inverse to each other. And so ui is bijective is in bijection to an an. Now we want to show that phi i is an isomorphism. So that according to our definition of morphisms really can say that you know there's no real difference between ui and the n. We have an isomorphism between them. So we use so to be an isomorphism it's also a homomorphism so we have to be able to talk about continuity. And relate the close subsets and one and the other and so one tool for this we use for this is a dehomogenization so definition. So the dehomogenization of a polynomial k0 in x0 to xn actually homogenous polynomial but doesn't play a role in the definition. Is just obtained by putting x0 equal to one. And so if f was a homogenous polynomial in x0 to xn then this fA will no longer necessarily be homogenous but it will have one verbal less. And so we do this here because if we take a close subset in say u0 then its inverse image in an if it's a zero set of f its inverse image will be the zero set of fA. And this will show us that this is continuous so let's see. So now we have this segment theorem. Well it's not I mean it's not really strictly necessary for defining this as homogenous but we'll only consider it when f is homogenous. We can also say I mean if you know we can always put x0 equal to one whatever we have but we are only considering homogenous polynomials here so we can assume that f is homogenous. Okay theorem so maybe I will just do the first part. So if I take fi from ui to an the map that we just had is an isomorphism. So let's see anyway we will so obviously whichever I we choose the statement is the same. It's just a question of which index we have here so we can assume that I is equal to zero. So we just look at fi0 from u0 to this is the same thing. And I just write fi and u so u0. So now we have so what is fi? First one to see that fi is a morphism but that's obvious because what is fi? Fi is just equal to x1 divided by x0 until xn divided by x0. Because after all it's a map which sends a0 to n to a1 divided by a0 until n divided by a0 so that's just this. And this xi divided by x0 is a polynomial is a quotient of two polynomials of the same degree obviously. And the denominator does not vanish on u0. Because u0 is the point where then at first the zero's coordinate is not zero. So this means that the xi divided by x0 is an element of this regular function on ui or u0 for all i. So that's it follows that fi is a morphism because we had said that if you have a map to any sub-variety of some affine space which is given by n tuple of regular functions then it's a morphism. So that's the easy direction. Now in order to find that fi is an isomorphism we have to see that the inverse map is also a morphism. And now unfortunately we have to use a definition because that we don't have a criterion yet how to say by looking at the components whether something whether a map to a projective space is a morphism. So let's just do it. So remember that u of a1 to an is equal to 1 a1 to an. So to be a morphism it has to be first continuous. So that means we have to take a close subset of u0 and show its inverse image under u is closed. So let w which I can therefore write as z of f1 to fm intersected u be closed in u. So where the fi are some homogeneous polynomials, homogeneous. Okay, now what is the inverse image under u of this? So u to the minus 1 of w is what? Well this is a set of all a1 to an in an such that if I apply u to it I lend in w. But u of this is just 1 a1 to an is in w or equivalent. So okay and to be in w means it says in the zero set of that. So in other words this is equal to a set of a1 to an such that in the n such that if I take fi of 1 a1 to an this is equal to 0 or i equals 1 to m. This is the same as w was a zero set of these polynomials. But this is nothing else than fi a of a1 to n. That's after all how f a was defined with the homogenization. So in other words we find that this is equal to the zero set of f1 a until fm a. So it's closed in there. Okay, so it's just this straightforward thing. And now the other thing we have to prove you know to have that it's a morphism. It must be compatible with regular functions on open subsets. We have to keep on going through the definition. So again so let the v subset u open and h be a regular function on v. We have to show a first star of h is a regular function on u to the minus 1 of v. Just that's what the definition says. So again we do you know as always we can in order to check this. We can make this v smaller. We just have to be able to check it on a neighborhood of any given point in v. And therefore we can assume that h is a quotient of two polynomials homogenous of the same degree. So making v smaller if necessary. We can assume that h of q is equal to f of q divided by g. Let me write p. I don't have a p before. f of p divided by g of p for p in v and f and g are polynomials homogenous of the same degree. Because we know that regular functions can locally be written as quotients of polynomials which are homogenous of the same degree. Well now what is u up a star of h? So this is h composed with u. So this is f composed with u by g composed with u. But we have seen we have been using that here. If we take f and compose it with u. This means we you know send a 1 to a n to 1 a 1 to a n and put it into f. So that means we place f by its the homogenization and g by its the homogenization. So this is just f a divided by g a. So we find that the pullback is again a quotient of two polynomials. And now on a fine space to be a regular function there has to be just a quotient of two polynomials locally. And the denominator must not be 0. So this is and g a has no 0 on u to the minus 1 of v. Because after all the value of g a at the point in u to the minus 1 of v is just the value of g at the image point. And that was non-zero after all. So it follows u star h is indeed an element and phi it follows therefore that phi from a n to u is an isomorphism. Okay this is what we wanted to show anything you can still say. Okay so we proved this. So in particular for instance this says in particular we see that phi from u 0 of mu phi to a n is a homomorphism. It is an isomorphism so it's a homomorphism. So we see that the so thus if we use 0 to identify a n with an open subset with u 0 in p n we find that the usual Zariski topology on a n is just equal to the Zariski topology on u 0. You know the restriction of the induced topology from p n. So that also does somehow seem to justify that we want to sometimes make this identification because it's really an isomorphism. You know it's compatible with the structure we have. And so in particular if x in a n is a closed sub-variety and we view a n equal to u 0 p n we can take the closure of so the closure of x in p n is a projective variety and it's called the projective closure of x. So and in fact there is a so we we can always if you have a in a fine variety we can always consider its projective closure in p n. And I mentioned somehow that going from a fine to projective space is somehow similar to compactification. So somehow one can view this as saying you know at least intuitively that if you have a fine variety we always have a canonical way of compactifying by taking its closure in p n. But I mean I'm not saying that you know if you look at the Zariski topology this is not the compactification. You know the Zariski topology is so strange that this thing is already quite compact but it's never hosed off. But in some suitable sense you can still view it as a compactification. For instance if you are over the complex numbers and use the complex topology then the fine variety will not be compact but this one will be. Okay. What else? Okay. And you know actually here we had this de-homogenization. I wiped it out but there's a way how to you know if you have a close subset in p n you can de-homogenize it. You can it's a zero set of some homogeneous polynomial you put the first coordinate equal to zero to one to get a homogeneous polynomial zero set of that will be the intersection with a n. And you can also go the other round to find the closure that's a bit more complicated. So there's a way of homogenizing a polynomial by adding multiplying by some power of x zero. And then you'll find that the if x is a close sub right you can in terms of homogenization define the you know define x bar is a zero set of homogenizations of the polynomials here. There's an exercise to that effect in the notes. Okay. But maybe that's enough now next time we will continue with these morphisms.