 So, we were talking about products of quasi-projective varieties and we had this Sager embedding which was from Pn times Pm to Pn, some n tuple A0 to An B0 to Pm was sent to all the products. And now we had seen that this map is injective, the image is closed and we then wanted to say we identify the Pn times Pm with its image and so identify in P large n and also if x and y are quasi-projective and say x lies in Pn and y lies in Pm we identify x times y with image in the corresponding Pn. And we had seen that this makes some kind of sense, we had proven that if we define the product in this way it fulfills a universal property so if z is any variety then the morphisms from z to x times y precisely those maps such that the two, if I compose the map with the projection to x and with the projection to y both are morphisms. So now we will want to soon talk about these two properties that one defines using products which were separateness and completeness and they both have something to do with topology on the product so I have to tell you what the topology on the product is. I mean there's a risky topology on the product. So this is, I don't know, lemma so I want to tell you what are the closed subsets in Pn times Pm. It's actually quite simple so the closed subsets in Pn times Pm are the zero sets of sets of polynomials Cf which depend on the variables on Pn and on the variables on Pm and they are supposed to be homogeneous in these coordinates and homogeneous in these coordinates but not necessarily of the same degree so P homogeneous, pi homogeneous in the xi and the xj. So every monomial of f will be some, will be a product of a monomial in the x's and a monomial in the y's and the degree in the x's will always be the same and the degree in the y's will always be the same but these two numbers don't have to be equal. So now I want to prove that. So we have to remember that we had defined this Pn times Pm was identified in the yj. So we have to remember how everything was defined so we had identified Pn times Pm with its image in P large n and so that a subset here is closed means by definition that its image in Pn is closed or that it's the inverse you know as it's an isomorphism of this image that it's the inverse image of a closed subset in P large n. So let W in Pn times Pm be closed then W can be written as sigma to the minus 1 of a sigma was the sigma embedding for a in Pn closed. So a is a closed subset in Pn so we can you know it's a zero set of some polynomials. So in the variables on the P large n. So we can write, so a can be written as z of say f1 depending on the variable cij that we have on the large Pn until say some r depending on the zj's fi polynomials homogenous polynomials in the variables zij on Pn. So now how does one take the inverse image we compose we have to compose with the map you know we take the map sigma the inverse image on sigma so the inverse image of the zero set is the inverse image of is the zero set of the polynomials which we get by composing the fi with sigma and what does that mean we replace zi by xi times yj. So it follows that W is equal to z of f1 of xi times yj until fr of xi times yj. Now if I you know fi was maybe this is something called k as I have i and j so fk is a homogenous polynomials in this variable cij I replace zij by xi times yj so afterwards I get the polynomial which is by homogenous in xi and yj even of the same degree because it has the same degree each monomial has the same degree in xi as before it had in zij because I just in the xi's as it had before in the z variables and in the y variables also the same degree as it had before in the z variables so the fk of xi times yj are by homogenous in the xi and the yj even of the same degree but we don't care so conversely there would be several ways how to do it I'm not sure if this is you know assume that z is indeed the zero set of polynomials like that so f1 of xi so of all the xi's and all the xj's until maybe I call it g these are now different from before gl of all the xi's and all the yj's so these are all the variables and so assume w is of this form for gi gk by homogenous so in principle one could more or less make these steps backwards but I do it in a different way so I can also use these open cover by these sets ui uj I had before then if I take if I take say cannot call them i and j so fl and yeah okay so here we I mean it doesn't matter very much you have here x i and yj are i and j are somehow running through all of them but now I fix i and j and so then I say if I take a fi times fj of w intersect it with ui times uj remember that fi times fj is an isomorphism from from your i times uj to a n plus m well how do I get this here I put and ui is where the ith coordinate is nonzero and fi is obtained by dividing everybody by the ith coordinate and you know taking the corresponding coordinate so this is so if I take the image and this is just the zero set I claim of you know g1 of x0 and then I leave the ith coordinate I put equal to one and then the other variables run and y0 and the jth coordinate I put equal to one and the other coordinates are just you know the the map is just the zero set of you know so this is the locals so you know let's try to retrace this and the same for the other ones no okay so let's go to the if I ui on ui times uj is where here the ith and jth coordinate are not zero the fi i sends a vector with coordinates a0 to a n to the one where I you know where from a0 from a n where the ith coordinate is nonzero sends it to the locals where I did sense it to the quotient a0 divided by i1 a0 divided by ai until a n divided by ai and so you can see that if I do this if before it was a zero set of this polynomial then afterwards becomes this polynomial where I put this variable equal to one it's just what is and so we get this thing but this obviously is a closed subset in a n plus m so as fi times phi j is a isomorphism which from ui times uj to n plus m it follows that a w intersected this is closed so it follows w intersected ui times uj is closed in ui times uj and you know the ui times uj form an open cover of pn times pm so if that if the intersection of w with every open subset of a cover is closed then w is closed okay and so that's this statement so now we come to some of these these two properties so the first statement the first property that the variety can have and which actually according to our definition all varieties have is that they are separated and the statement there is that the diagonal is closed so I have to say what the diagonal is and then so definition so x should be a variety so the diagonal so delta x so is so is delta x which is just obviously the set you know all the points in the agony so the set of all points pp p and x which is some subset of the product of x with itself and you also have the diagonal morphism so I don't one doesn't need that very much but anyway the agonal morphism which is um say delta x from x so x times x p is mapped to pp so this is a morphism so I can write it as according to our definition as the identity on x and the identity on x no so it stands for p2p and so this is a morphism with image delta clearly the image is there and now so the first claim is that the diagonal is always closed in the in the product so lemma delta x is closed in x times x and then one also gets that delta x from x to delta x is an isomorphism but the last statement is kind of automatic once you know that this is closed and therefore it makes sense to talk about this so yeah so it's it seems maybe kind of obvious but it's actually needs a tiny trick um so which one to the main thing we want to show is that delta x in x times x is closed so every variety according to our definition is isomorphic to a locally closed sub variety of some projective space no so therefore we can we can replace x by an isomorphic variety and so we can assume that x is a locally closed sub variety so well and now we have to you know if you want to say that the diagonal is closed we have to somehow write down the equations no but we can make it a little bit easier by noticing that it's enough for us to see that um p in in projective space itself the diagonal is closed so we have that delta x is equal to the diagonal of p n intersected with x times x and so if we know that the diagonal in p n times in p n is closed then this will be closed in x times x so if delta p n is closed in p n times p n then it follows because x times x carries the induced apology that delta x is closed in x times x so we can assume x is p n so that and now we just have to see we just write down the equation for the diagonal so i cling that you just can write down the formula the diagonal in p n is the zero set of i take xi times yj minus xj times yi so where you know say x1 x0 to xn are coordinates on the first p1 first pn and y0 to yn are the coordinates on the second factor so where i and j run through all possibilities okay maybe i different from anyway it doesn't matter so i claim it's this so so in this case it's like it is indeed you can see that this is indeed a determinant of some of some minus so let's see so why is this if i lay take a point a0 to an and b0 to bn in pn times pn then this will lie in the zero set in the zero set here if you know all these terms are zero when i put ai and bj and what does it mean so if and only if if i take the matrix a0 to an p0 so so the entries here are zero a1 b1 and so on and so on a n bn if i take this matrix if this matrix has ranked one because you know the minus of the matrix are precisely ai bj minus bj bi aj so and the rank is one if and all two by two minus are zero because the the rank is at least one because these are non-zero vectors we know that so so this lies in the zero set if and only if the rank is one but that the rank is one means that these two vectors are put or proportional so this is if and only a0 to an is equal to b0 pn okay so therefore we have found indeed the equations for the diagonal in pn and so we know that the diagonal is closed and now the last statement is obvious so now we can talk about the morphism from x to delta x so as x is a variety i mean subjective morphism delta x will be a variety as you know because the map is subjective and it is clear that then delta x from x to large delta x is an isomorphism because obviously the projection to one of the factors is the inverse okay so this is this property so you know so you see it's not particularly difficult but we see that we have somehow explicitly used that somehow this is called a projective so that we can go into a projective space and in fact it is true that this property that the diagonal is closed does not hold in arbitrary generality so if you have a more if you look at schemes or something which would be a generalization of what we have here or abstract varieties which is something which locally looks like and fine like a in a fine variety then it will not be true in general that the diagonal is closed you know it's a property you can ask for you call the variety separated which we'll now say if this is true and we have just shown with this definition that quasi-projective varieties are always separated so I will just write this as a mark so then so that the fact that so the fact that delta x subset x times x is closed it's an important property and actually as I tried to mention before it somehow replaces for us the host of property okay so we know that algebraic varieties will never be housed off I mean except when they are points or something like that but this is something which implies some of the consequences of the host of property and so therefore we can sometimes argue as if we had the host of I mean we can get results which similar to what one gets for host of spaces so therefore it's a very useful property um and as I said you know a variety but is called separated if delta x subset x times x is closed and now according to if we take our definition of variety then all varieties are separated so that's not so exciting but as I said if one doesn't more generality than sep mark one cannot even read it at least I can't separate it so according to by dilemma with our definition all varieties are separated but there are more general definitions of varieties and then they aren't okay and so some simple consequences so um for instance so corollary if we have two morphisms from uh so let's say phi both phi and c from x to y the morphisms of varieties then the locals where these two morphisms are equal is closed in x so then the set of all p in x such that phi of p is equal to psi of p is closed in x so in particular if you have two morphisms which coincide on an open subset they are equal in a non-empty open subset for a non-empty open subset of x then it follows that phi is equal to psi because an open subset a non-empty open subset is always dense as x is irreducible so proof is quite simple so what is this set so this set that we have here is well the set where these two map to the same point so it's the inverse image of the diagonal by these two maps so this is phi comma is it like this okay so phi and c are morphisms so this map phi c no phi c is a map from x to y times y you know is a morphism delta y is closed in y times y so that's it follows that this is closed in x as the inverse image of a closed set by a morphism we can also see that the graph of a morphism is always closed it's like a some kind of closed graph theorem like you have in function analysis so the graph obviously is what it always is so let p from x to y the morphism of varieties so the graph of phi well it's just as you rule it's gamma phi which is the set of all pairs of a point at its image where p is a point in x so this is some subset of x times y and the claim is that corollary that first that the graph is closed in x times y and well it's also isomorphic to y so so I have that say p1 from so if I take p1 restricted to the graph by p1 and maybe p2 restricted to the graph from gamma phi to y is an isomorphism is it really obviously it's the other projection which is an isomorphism so p1 like this no it's kind of so if you have a a map you can write the graph and obviously if you project back to where you start it will be the trajectory projection and I claim it's an isomorphism but the main point is that this is closed and this is it follows again directly because it's again the inverse image of some closed set so I mean just gamma phi what is it it's the set of all points p phi of p with like this so I can take again a map like that I can take say I take for one thing the map phi and I take the identity on y so it's times identity on y with a minus one of the diagonal so this map phi times the identity on y is according to our definition a map from x times y to y times y which on the first coordinate is phi and on the second is y and then you know this says precisely that you know this is a set of all points p in x and you know q and y such that q is equal to phi phi of p and that's what it says so and this is so it's closed because this thing is a morphism and delta y is closed in y times y and that this is an isomorphism is kind of immediate we can take for instance we can see that the inverse of p1 restricted to gamma phi can be taken as so I say it's the identity on x comma phi from x to gamma phi this is anyway how you normally parametrize the graph no you on the first coordinate you you send a point to itself on the second you put it to the image and this should be an this isn't an isomorphism and the inverse is just a projection down to where you started with and you can see these these these are now both morphisms and they're obviously inverse to each other okay so this was this separatedness now we come to the other property that one can you know of morphisms and varieties that one can express best in terms of products which is completeness and whereas the separateness is replacing the house of property for us the completeness replaces compactness and you know as I had told you that we somehow want to view projective varieties as compactifications of you know a fine variety so we should have that the projective varieties are complete and this is what we want to prove and in fact according to our definition a variety is complete if and only if it is projective so if a variety is quasi projective and not projective or just a sub variety of refined space which is not a point then it will never be complete okay so let's try so completeness and this is a somewhat more subtle property so we should have that so we want that so as I said completeness replaces for us compactness and it's again a property one can describe in terms of of the of products so I will now introduce at the moment so first I start with something else definition so that you first you know so if you have a map say f x y of topological spaces and so such a map is called closed if well the image of any closed subset is closed closed in y for any or closed subsets z subset x okay so for instance yeah so somehow you know if it's continuous then the inverse image the inverse image of any closed subset is closed but now we have the image is closed okay and now completeness is a somewhat well strange statement so the statement is a variety x is called complete if whenever I do the when if the second projection x times y to y is a closed map for all varieties x for all varieties y okay so whenever I have a closed subset of x times y then the image in y will be closed and somehow you know that is not automatic it somehow is so let's see first some example but it's not true so for instance we have that a one is not complete and you can actually easy see that we'll see later that as I said if I find varieties if they are not points will never be complete but we just want to see it for a one a one is not complete so there are for instance we can take as the other one also a one so we take if you take z the zero set of x y minus one in a two which is the same as a one times a one so the graph of the real point somehow looks like this and we take the second projection this will map this thing the image will be everything without the old okay so p2 um from so p2 of c is equal to a one without zero of course this is a second projection this would be this is a set of all possibilities for the second coordinate and you know if you know if y is equal to zero there's no possibility to find an x so that you get one and so this is certainly not closed in a one it stands okay maybe let me see so you know one reason why completeness is very important is that you know it has some property that you also have when you have a compact variety so if you have a complete variety then any morphism starting from a complete variety has its image closed it's like if you have a compact topological space and you map to a house of space then the image of the compact space in the house of space would always be closed under any continuous map and it's the analog of that so this is a yeah maybe I call it proposition so if let x be a complete variety um and let um say phi from x to y be um be a morphism of varieties so then phi of x subset y is closed and the the proof is very simple because um we can use the graph of the map so let um we have gamma phi subset x times y is a closed subset and by definition the image of phi is the second projection of the graph um so thus if x is complete phi of x is closed okay so it's a very simple fact but it somehow shows again the analogy with the house of things so the image and so this actually does give us a new way to construct varieties so we don't have to explicitly write down the equations just if we have any morphism from a variety to another variety we always know that the image is a variety okay but now we want to show that um all projective varieties are complete that's not that's a theorem or projective varieties so first we make some so the main step is actually it will turn out that it's enough to prove it for to prove that pn is complete then it follows very easily for projective varieties and actually the main step is even slightly less than that so the main step is to prove one main step is that if I say for some reason I change p1 into p2 into p1 before because I find it easier so the first projection from pn times pm to pn is a closed map okay I mean in principle it has to prove that now as I have exchanged one and two that the projection the first projection from pn times x to pn is closed for any x but you know it's the start to do it if the okay there's a set this is once one has done this the rest is formal so you know then we have to do it so we take a closed subset here so it's an algebraic set so we can write x as a zero set of say some polynomials f1 of x y I now make my life even easier until f r of x y where here x is an abbreviation for x 0 for the variables x 0 to x n no on pn and y stands for y 0 to y m the coordinates on pm and then I want to make some assumption I claim that we can assume that all fi all these polynomials so it's a zero set where you know these fi are bi homogeneous and x and y because that was the lemma we proved in the beginning so we can so in particular each fi has a certain degree in x and a certain degree in y and I claim we can assume that all the fi have the same degree in y and assume all of i have the same degree which are called d in y why is that so assume I mean most of the time obviously it will not be the case so one of them will have the largest degree one of the degrees one of the degrees one of these polynomials fi will have the largest of these degrees in y you know I take the maximum the other ones have possibly smaller degrees and then I claim I can replace them by other polynomials so that the zero set is the same and they have the same degree so if fj has a lower degree I mean then the maximum of these then we can replace it by other polynomials namely I say so assume the degree is the maximum degree minus e then I replace it by y zero to the e times fj y1 to the e times fj and so on until y n to the e times fj so if this had degree e less than the maximum degree now all of them have this maximum degree d and the zero set is the same as before because what is the zero set of all these polynomials it is the the the union of the zero set of the fj's and the common zero set of all these but the common zero set of all these is empty so it is just the same as the zero set of fj which has the same zero so by possibly making the number of polynomials we are looking at here much larger I can assume that all of them have the same degree in y so okay until now we are only we are still kind of so somehow I will eventually have to prove that the image that the second projection of x the first projection of x in p n is closed so you know that it's closed means that it's a zero set of some polynomial so somehow I have to create these polynomials out of nothing and so you can somehow see that this cannot be so easy because we don't see them so we somehow and that's why it's slightly tricky so we have to go about it slightly slowly so we fix a point p in p n and we want to investigate what is the condition for p to lie in the image so then p will lie in the first projection of x but if and only if there is a point in x whose first coordinate is p so that means that if I write it now like this if I take the zero set of these same polynomials as before f1 until whatever and now I just write f1 of p y as you know where I put instead of x the coordinates of p so there's comma so one does it lie in the image if there is this is the set of all points in x whose first coordinate is p so this this set must be non-empt this is some subset in in p m and we have to see that this is a closed condition so now we use the projective nulsternsatz so we are using a somewhat big theorem so what does the projective nulsternsatz say it says that the zero set is the empty set if the corresponding zero set of some polynomials is the empty set if the ideal generated by these polynomials contains all monomials of a certain of some degree for some degree it contains all monomials of that degree okay so now we have the opposite statement so so this is equivalent but this is equivalent to so for all s bigger than zero if I take the ideal generated by these polynomials these are now polynomials in the y variables this ideal does not contain all monomials of degree s in the variable okay in this condition for given s I may be called star s because I will have to reformulate a few times so now it's clear we have said that the degree in the y variable of all these polynomials is at least y no it is precisely y so in particular we find that it's impossible so this cannot be true so this so the condition star s is obvious trivial so it's trivial 2 for s smaller than d because there's actually this thing does not contain any monomial of degree smaller than d so in particular it certainly does not contain all monomials of degree d so thus therefore it's enough to show for all s bigger than d bigger equal to d if I let if I look at the set of all points p in p n such that p satisfies this condition so that the ideal generated by these things does not contain all monomials of degree s this set is closed in p n so because then so for each s this is a closed set because then so I don't know if I give this some name I call this say a s well maybe not a s whatever x s because then we have that the first projection of x is just intersection of all s bigger equal than d of x s no because the statement is that the image consists of all points p in p n such that this condition star s is fulfilled for p for all s bigger equal to d and so if for any given s this is a closed subset then the image is is the intersection of these closed subs and therefore is closed so so we have to show this statement so we have to fix an s bigger than d and we have to show that this is closed in p n and for this we actually do have to now give some equations for it and we will do this by some linear algebra so we fix s bigger equal to d and now we just you know write down what we have so we denote so I will quite a lot of notation so anyway denote say d by mi of y this is so mi depending on the y so y always stands for y 0 to y m y n so to y m and this so this is from i from one so m plus s choose m these are the monomials in the y variables of degree s anyway so there's certain we have so many we have a number of these monomials can just write them down doesn't matter what the number is there's some number here and we so denote so first this like this then we also denote by say n whatever j of y and now I don't nearly need to again j runs to something the the monomials in y of degree s minus d okay if you can figure out how many these are so these are always i somehow number them in any random way no just have one for every one again in any order so then we find so the elements of degree s in this ideal that we have here f1 of py f r of py are all those which are obtained you know all these have degree d in the y variable so they are just obtained by multiplying these elements by all monomials of degree s minus d they are just they are precisely so they are the linear span so the span in terms of vector spaces of of the set of all products n i of maybe n well now I call them n i I mean anyway sounds over all possibilities n i of y times f j of py so for all i and j so i this was would be m plus s minus d choose if I want m so i from zero this thing j from zero to r so I take these are we have these elements they spend some suck vector space and that's up to vector space are all the elements of degree s which lie in this idea okay and we somehow want to see that this does not contain all monomials of degree s so it's a question about the dimension of the span so I write down so denote again the product all the products I take all the products n i of y times this thing but I actually take it now with the variable x instead of the p so denote so I take all of them so by somehow g k of x y so there's some kind of way how I renumber them know so that I have some case which run from somewhere somewhere in you know from this is from one here so from one to the product of these two numbers and I have all these polynomials in x and y okay I make quite some story you know k equals one say to t where t is the product of these two and this again in any order so now I have somehow made these complicated things so I can reformulate the condition star s now this is equivalent to well okay t would turn out to be by what I just said I mean somehow irrelevant what it is it's just some number but it will be equal to m plus s minus t choose m times r no it's just all possibilities of multiplying them so then this condition is equivalent to the following if I take these polynomials that I have here all the polynomials which are obtained as products so and I now put back the p for the x so g k of p comma x for one smaller equal to k smaller equal to t so this finally many polynomials do not spend the whole space of polynomials of degree s or there was a y here not x because according to what I said the elements of degree s in this ideal are equal to the linear span of these elements here and the statement that this does not contain all monomials of degree s just says that this linear span is not equal to the set of all polynomials of degree s okay but now you know until now it looks like I have really done nothing but now the point is we can express this in terms of looking at the rank of a matrix so let's see maybe I can so we can write if I take this g k of x y so I'm back to the polynomials I don't put the p I can write this as a sum over all I now I wear a i j a i k of x times mi of y mi are the the monomials of degree s so this is would be i from 1 to m plus s to s or m whatever it's the same no because where this a i k from x is homogeneous in the x variables you can easily see that so I mean it just means I have a polynomial in x and y which is homogeneous of degree s in the y variables then I can and it's also homogeneous of some degree in x variables and I can write it as a sum over you know some these monomials in y times some homogeneous polynomial in x okay so I have this matrix of polynomials but now you see this somehow gives us you know we have a this linear map so then the claim is the dimension of the linear span of say this g k p x which is what we were wanting to know k equals 1 to t is just the rank this matrix I don't know which way we're out of no goes t times m plus s choose m matrix a i k of x no because here we have a you have a basis of this vector space of monomials in y and you want to know you know we write down these things as linear combinations of you know if I put here p this is just a number so you know if I put the p so here p so if I put p this is just a number and so then if I write the you know some vectors in terms of a basis via elements of a matrix then the dimension of the span of the vectors is equal to the rank of the matrix okay and so therefore we see that the condition that the linear span of these things is not equal to everything is equivalent to the fact that this matrix does not have the full rank so so thus this condition star s is equivalent to that the rank of the matrix a i k of p where you know i went from 1 to m plus s choose m and k from 1 to t that this rank is strictly smaller than m plus s choose s no then the dimension of the vector space in which this thing here lives and so this means that if I look at the minus of this matrix of this thing they should be zero so so now I can wipe out maybe the condition so thus the set of all points p in p n p n such that this condition holds for p you know the condition was this is equal to the zero set of all m plus s another word s but if I wrote m although it's the same when we write all the same m times s times m plus s m minus of this matrix so I have this matrix so the minus means I have the determinants of the sub matrices no and so it's a matrix which has different lengths of column and so anyway so it should not have the it does not have the full rank it doesn't have the same rank as the dimension it has in one direction if these minus are all zero well and you know this is a matrix with entries polynomials in the x variables so the minus are the determinants of these sub matrices they are polynomials in the x variables so it is a zero set of some polynomials so thus we have that the set holds for p p is a closed subset is closed in p n and as we had seen we have that the image p one of x is the intersection of all of these for all possible s so p one of x is closed in p n okay and so this proves the first statement that if I take the map p n the first projection p one p n times p m so p n is a closed map okay so now we want to see how one deduces the statement from that to prove that all projective varieties are complete and we haven't even quite shown that p n is complete we have just shown this statement that I just wrote so let's now show first that p n is complete so that we do the general case so first show p n is complete so that the statement is that if I take you know p n times y and project to y no no then this is a closed map so let's see so let for any variety so let y be a variety so we know that every variety is isomorphic to a quasi-projective variety no so it's even if it's either locally closed in in a n or p n for some n and then but if it's a locally closed in a n we can it has isomorphic to to something in p n so we can assume that y is quasi-projective so it lies in some p m and is locally closed there now we take z a closed subset of which we're going to go so a closed subset of p n times y we have to show that the second projection to y of that is closed in y okay so we certainly can take so let z bar be the closure of z in p n times p m that we can certainly do we can close it so then by what we just proved we know that if we take the second projection of z bar bar this is closed in p m because we know that the you know the projection from p n times p m to one of its vectors is a closed map so now I claim that what we want follows from that we have to show that the second projection of z is closed in y so let's see so what is the second projection of z well I claim this is just a section projection of z of this z bar intersected p n times y now because z is closed here so it is it's therefore the intersection of p n times y with its closure in p n times p m it's a section of so of p n times y with its closure here so it's just this z is just equal to this but you know if we project here you know what happens is just by definition this is just the second projection of z intersected with y you know and this says that it's intersection of y with a closed subset so it's closed okay and you know the now we come to the really general now we want to show that any projective variety is complete finally so finally let x subset p n be a closed sub variety and let z contained in x times y be closed you know x times y is closed in p n times y you know so therefore it follows that z is also closed in p n times y because it's closed in something closed and now I can just apply the previous thing if something closed in p n times y then the second projection to y is closed in y therefore by previous step p2 of z is closed in y okay and so with all this we finally have shown that projective varieties are complete so as you see it was a somewhat tricky thing these steps you know it's also a little bit with trick but it's kind of formal you know you have to somehow feel it's a bit like cheating but the most of the work is is done here to actually write down the equations in the case that we are dealing with p n and p m and you know and was and you can see that the proof is sufficiently complicated that we have no idea what the equations are actually you know which is also how it should be because you know we don't know where they come from so therefore the proof has to be difficult okay maybe I stop here and then the next time we will give some some applications of this and then we start with something new okay