 Alright, so you see uhhh what we did last class was uhhh uhhh to define the notion of a morphism uhhh uhhh and the idea of a morphism is that uhhh it is a continuous map which should pull back regular functions to regular functions okay that is the uhhh that is the definition of a morphism and then uhhh I stated uhhh uhhh I stated a few results let me recall them, so uhhh let uhhh x uhhh be any variety any variety in our case uhhh it means either it is uhhh an affine variety or quasi-affine variety an affine variety as you know is an irreducible closed subset of affine space and we are working over an algebraically closed field okay and a quasi-affine variety is a non-empty open subset of an affine variety and uhhh variety for us at the moment means either an affine variety or a quasi-affine variety which means either an irreducible closed subset of affine space or an non-empty open subset of an irreducible closed subset of affine space and uhhh we saw we saw that uhhh uhhh if x is affine okay then uhhh the uhhh then a x and o x are isomorphic okay. So a x is the uhhh the coordinate ring of x uhhh the ring of uhhh uhhh the ring of polynomial functions on x which is gotten by taking the ring of all polynomial functions of the affine space in which x is sitting inside modulo the ideal uhhh of x which is a prime ideal and o x is supposed to be the ring of all uhhh regular functions on x okay and uhhh we we saw that if x is an affine variety then there is an isomorphism like this alright. And uhhh mind you uhhh this isomorphism is not just an isomorphism of rings it is an isomorphism of k algebra okay it is an isomorphism of rings which is also an isomorphism of k vector spaces okay so k algebra isomorphism and uhhh of course uhhh uhhh o x was defined to be the ring of regular functions on x and uhhh regular functions were defined to be functions which were locally given by quotients of polynomials okay. Now we then uhhh I told you that uhhh the the uhhh we defined uhhh then the notion of a morphism of one variety into another variety and what we did was uhhh uhhh we showed the set of morphisms of varieties from x to a1 uhhh or let me put y to a1 is isomorphic uhhh uhhh or let me let me again put x uhhh or let me put y okay y to a1 is isomorphic it can be identified with oi uhhh for any variety for any variety y namely uhhh every regular function which is a map from y to k okay is is the same as a morphism from y to a1 where k is thought of as a1 uhhh which means that a1 is just k with these are zerski topology and the zerski topology there of course is you know it is a complement finite topology is a topology for which the open sets are complements of finite sets the closed sets are only finite sets okay uhhh. So and I told you that this uhhh uhhh uhhh this can be generalized that the statement can be generalized uhhh so you know this statement can be written as morphisms of varieties from uhhh uhhh x so let me let me go back to uhhh uhhh so let me again write y uhhh to a1 is isomorphic to this can also be written as morphism of uhhh k algebras from uhhh o of a1 k to o of 1 because this o of a1 k is actually isomorphic to kx o of a1 is just the uhhh uhhh o of a1 is the same as a of a1 okay and uhhh uhhh and a of a1 is k of x it is a polynomial ring in one variable okay. So if you want uhhh if you want I can write uhhh I am I am I am using the fact that you know uhhh o and a are the same for a fine variety so if you want I will write this is this can be identified with a of a1 and uhhh uhhh and that is uhhh that is isomorphic to kx okay. Then the all possible k algebra homomorphisms from the polynomial ring kx to any any k algebra is in bijection with the elements of that k algebra okay because this is the universal property of the polynomial ring in one variable okay namely any k algebra homomorphism from kx to o of y is completely controlled by what by the image of x okay it is controlled by the element of o of y that x goes to okay and therefore you will have as many k algebra homomorphisms as there are elements in o of y. So this set is the same as o of y okay so I am just translating the statement like this using the universal property of the polynomial ring in one variable and then I told you that uhhh more generally you can also write it as uhhh what I have written for a1 you can write it for an and not only for an you can write it also for an any affine variety okay. So uhhh this statement is morphism of varieties from y to an can be identified naturally with morphisms of k algebras uhhh from o of a from uhhh a of an to o of y uhhh and uhhh what I can do for a1 uhhh uhhh and I have done for an I can do it also for an irreducible close subset of an namely an affine variety. So I can also write it as morphism of varieties y, x is isomorphic to is natural is in a natural bijective correspondence of morphisms of k algebras from a of x to o of y okay where of course x is uhhh where x is of course an affine variety okay. So this is the most general statement this is the most general statement okay. So in this general statement if you put x equal to an you get this statement if you put x equal to a1 you get this statement which actually so if you put x equal to a1 it boils down to this statement that regular functions are nothing but morphisms into a1 okay regular functions are no different from morphisms into a1 right. So what I am going to do is uhhh this is a very very important statement okay because as I told you at the end of the previous lecture that this is the statement that gives an equivalence of categories between uhhh on the one side the category of affine of affine varieties with morphisms being morphisms of affine varieties on and on the other side the category being the category of uhhh finitely generated k algebras that are integral domains with morphisms being k algebra homomorphism. So what I am going to do is I am really going to uhhh focus on proving this fact okay. So that is what I am going to do now it is a very important fact. So uhhh uhhh proof of uhhh uhhh proof of the statement morphisms let me rewrite it again uhhh varieties from y to x isomorphic uhhh here of course this means uhhh there is a bijection of sets okay is this what you have here is a set what you have there is a set. But the point is that this side represents the geometric side the algebraic geometric side this side represents the uhhh commutative algebra side okay and as usual uhhh whole point about uhhh algebraic geometry is that you go from the uhhh you know from the geometric side to the commutative algebra side uhhh this translation is very important to algebraic geometry okay. So what I am trying to do is I am trying to prove this uhhh this is the algebraic geometric side morphism of varieties from y to x and here is a commutative algebra side which is morphisms as k-algebra of uhhh A y A x to O y. So this is this is what I am going to prove okay. So uhhh I am going to prove that there is a bijection like this there is a natural bijection like this. Now how do I do that? So so define alpha from morphism so again let me write this down I define a map like this first and a map like this which is uhhh if you give me a morphism uhhh phi of varieties from y to x this is an element on this side I have to send it to something here and you know uhhh uhhh what I am going to send it to here is uhhh a very natural thing it is it is a thing that characterizes what a morphism is a morphism is a continuous map which is supposed to be uhhh as uhhh as I considered as a pull back map something that pulls back regular functions to regular functions. So what I am going to do is I am going to just take the uhhh pull back pull back via phi of regular functions and what does that mean it means so the notation for that is phi upper star which is a map from regular functions on y to regular functions on x okay. So this is my map so uhhh what is so what is it that is what is it that is happening if you give me uhhh if you give me a regular function uhhh f uhhh on y I get uhhh the pull back of that regular function to x is a regular function on x uhhh this is the defining property of a morphism apart from its continuity and what does this mean this is just composition. So you know you have you have y uhhh to x you have this map uhhh you have this morphism phi and uhhh f is a regular function on y uhhh I have uhhh yeah I have messed up x I have I have mixed up x and y this should be x and this should be y. So you have a regular function on x f is a regular function on the target so it is uhhh if you want f is a morphism into a1 uhhh or it is a regular function into k okay and then uhhh what is the pull back it is this map which is a composition this is first apply phi then apply f okay this is uhhh this is a this is otherwise called as phi upper star of f this is the pull back and the fact is that if you have a regular function on the target uhhh by composing with the morphism you should get a regular function on the source the target is x so give me a regular function on x I am by composing with phi the pull back will be a regular function on y this is the definition of a morphism okay this is part of the definition of a morphism and mind you that this o x can be this o x can be uhhh uhhh can is it can be canonically naturally identified with a x in fact I should even put equality okay o x is the same as a x alright uhhh so here also when I say that if x is affine then a x and o x are isomorphic uhhh in fact you can say that the in fact they are equal okay if you think of them as functions into k uhhh with into a1 with which is k with this k topology they are equal actually this is actually an equality okay and so it is very clear that I have a very natural map given any morphism look at the pull back map which takes regular functions to regular functions and it is very easy to see that uhhh this pull back map is of course a k algebra homomorphism because you know phi upper star of f f1 plus f2 will be phi upper star of f1 plus phi upper star of f2 and uhhh phi upper star of f1 into f2 will be phi upper star of f1 into phi upper star of f2 you can easily verify that this what you get here is not just a map from a x to o y but that is a k algebra homomorphism because functions by the definition of functions uhhh everything is done point wise some of two functions uhhh is defined point wise uhhh product of two functions is defined point wise multiplying a function by a scalar is a new function which is also defined point wise. So you can easily check that this is a this is certainly a k algebra homomorphism okay what we need to prove is that this map uhhh so uhhh I have written alpha but uhhh well so this is alpha of phi but the better notation is phi upper star okay uhhh but since I have written alpha here I have also write it here alright and what will I have what will I have to prove I will have to prove that this map is injective I have to prove this map is adjective that is all I have to do alright. So let us uhhh let us try to prove the injectivity let us try to prove this injectivity. So uhhh I guess the injectivity is very easy so uhhh so what is the situation uhhh suppose there are uhhh suppose I have two morphisms which give rise to the same k algebra homomorphism uhhh upon as pull backs of regular functions then I have to say that these two are the same okay. So suppose uhhh alpha of phi 1 is equal to alpha of phi 2 which is the same as saying that you know phi 1 upper star the same as phi 2 upper star okay I will have to show phi 1 is the same as phi 2 right we need to show uhhh uhhh we need to show that phi 1 is the same as phi 2. So you have to show that these two morphisms are one and the same and what do you mean by equality of morphisms by equality of morphisms means equality of the underlying maps as sets okay equality is just a set theoretic maps. So it is a set theoretic checking that you have to uhhh do okay. So uhhh so how do I check two maps are equal I check two maps are equal by evaluating them at every point and showing that they are equal. So what I do is uhhh well so here is my situation. So I have so I have why uhhh so there are these two maps phi 1 and phi 2 which are morphisms into x and mind you x is an affine variety x is an affine variety which means that uhhh uhhh x is sitting inside x is an irreducible go subset of a suitable a n. So let us consider that it is it is sitting inside it is an irreducible closed inside some a n okay and uhhh let us let us uhhh uhhh let us put let us put coordinates on this a n okay uhhh let let me say with coordinates uhhh uhhh uhhh t 1 etc t n let these be the coordinates on a n alright. So the when I say a n with coordinates t 1 etc t n what it means is that I am just saying that if you take the if you take the uhhh ring of uhhh polynomial functions on a n I am identifying it with k polynomial ring in t 1 etc t n okay this is what it means. The choice of variables in the ring of polynomials in a n those variables uhhh the choice of variables will also give you the uhhh labels for the coordinate functions okay and uhhh uhhh what happens is that you know this is a this is an irreducible closed subset therefore uhhh uhhh this if I go mod the ideal of x which is going to be a prime ideal I am going to get a of x okay. So it is going to be uhhh uhhh so this is the this is the polynomial functions on a n restricted to x okay uhhh to get the ring of functions on x I will have to go modulo the uhhh ideal of x which means I will have to identify two functions on x uhhh if and only if they differ by uhhh a function that is 0 on x and that identification corresponds to going modulo I of x okay uhhh and uhhh so you know under this uhhh uhhh under this map uhhh you know all the every xi any any ti will go to some ti bar okay if you want uhhh thought of as uhhh an element here okay and mind you ti bar is uhhh geometrically ti bar is just the coordinate function ti restricted to x. So this ti bar is just ti restricted to x okay actually what is this quotient map given any polynomial uhhh given any polynomial consider it as a function on the affine space and restricted as a function on x that is this quotient this restrict this quotient map is actually restriction geometrically it is restriction of polynomials to a closed subset this closed subset x okay. And uhhh uhhh how do I check these two maps are equal I check uhhh these two maps are equal by uhhh looking at the image of a point. So let me take a point small y okay the point small y will go to uhhh under phi 1 it will go to phi 1 of y uhhh uhhh and this is a point it is going to be a point of x and under phi 2 it is going to go to phi 2 of y which is a point of x again okay and both phi 1 of y and phi 2 of y are going to be points in affine space. So they are going to have coordinates so uhhh let uhhh phi 1 of y be well uhhh lambda 1 etc lambda n because it is a after all phi 1 of y lies in x and after all it is a point in n space it is given by n coordinates okay and phi 2 of y is say given by mu 1 etc mu n these are the coordinates of this point okay. And what I will have to prove is that if phi 1 star phi 1 upper star equal to phi 2 upper star then I will have to show that these lambdas are the same as the mu's which will say that the point phi 1 of y and phi 2 of y are one in the same point and if I and since y was arbitrary it would tell you that phi 1 and phi 2 are one in the same map okay. So uhhh so the uhhh the the the point that one has to understand is that uhhh what is the meaning of uhhh phi 1 upper star equal to phi 2 upper star. So, phi 1 upper star is the pull back map from the ring of functions on the target the regular functions on the target to the regular functions on the source. So, it is from A x to A y uhhh 2 of y of course A x is the same as O x because x is affine okay. So, the pull the this is the pull back okay. And what does the pull back map do uhhh given a function uhhh given a regular function on x I compose it with this to get a regular function on the source uhhh that is how the pull back map works. Take a regular function on the target compose it with the map you get a regular function on the source. So, uhhh so in particular you know the the ti's okay restricted to x they are also regular functions on the target. Mind you A x is A x is uhhh the same as O x because x is affine. So, this ti bar they are regular functions on x the coordinate functions they are just the coordinate functions on the affine space restricted to the irreducible cross subsets okay. So, if I take each of this ti bars which are which is the same as ti restricted to x they will go to this it will go to phi 1 upper star of ti okay. And uhhh what is given to me what is given to me is phi 1 upper star is the same as phi 2 upper star. So, what it what it means is phi 1 upper star equal to phi 2 upper star means that this is the same as phi 2 upper star of ti bar for every r this is what it means this is what is given to me okay. Now just think for a moment what is the meaning of phi 1 upper star of ti bar phi 1 upper star of ti bar is the regular function on y which is gotten by pulling back the regular function ti bar on x. And what does what does pull back do pull back is a composition with phi. So, what this means is it means that uhhh uhhh phi upper star of f as you see phi upper star of f is f circle phi okay is just composition. So, if you translate that you will get t uhhh ti bar composition phi 1 is equal to ti bar composition phi 2 this is true for every i this is what it means now now this is an equality of regular functions on y. So, apply it to the point small y. So, what you will get is ti bar of phi 1 of small y is equal to ti bar of phi 2 of small y. So, this will tell you that ti bar of phi 1 y is equal to ti bar of phi 2 y okay. But phi 1 y is uhhh uhhh uhhh phi 1 y has these lambdas as the coordinates, phi 2 y has mu's as the coordinates. And if you take a point like this the action of ti bar on that point is to give you the ith coordinate okay it is just ti the function ti on affine space picks out the ith coordinate of the point when evaluated at a point. So, if you do that what I am going to get is that I am going to get lambda i is equal to mu i and this is going to be true for all i. And this implies that uhhh phi 1 of y is equal to phi 2 for all y in y and that tells you that phi 1 equal to phi 2 okay. So, uhhh so if alpha of phi 1 equal to alpha of phi 2 then you get phi 1 equal to phi 2 and that is injectivity okay okay. The slightly more uhhh involved uhhh part of the proof is a surjectivity namely to show that if you give me a k algebra homomorphism from A x to O y it arises from a morphism okay. And of course when once you prove it arises from a morphism we are already saying that that morphism is unique because we already proved uniqueness that is the injectivity okay. So, uhhh surjectivity is what we have to see next so how does one do the surjectivity. So, uhhh so what I am going to do is I am going to start with so here is my situation uhhh I have these morphisms of varieties from y to x I have this map alpha and on this side I have morphisms of k algebras from A x which is the same as O x to O y okay. And I am trying to prove surjectivity of this map I have already proved injectivity so what I do is I start with I start with uhhh uhhh uhhh with the k algebra homomorphism zeta okay. So, zeta I start with an element here zeta is a k algebra homomorphism from A x to O y I start with the zeta and I am trying to show that there exists a phi such that alpha of phi namely phi upper star is the same as this zeta this is what I have to prove okay. So, so but but what is A x A x is if you look at it carefully A x is just k t1 tn mod I x okay and so this A x is just k t1 etc tn mod I x okay this is what it is. In fact uhhh so it is k t1 bar etc tn bar it is it is generated uhhh by the images of the ti's okay it is it consists of just polynomials in the ti bars that is all okay. And uhhh so in so so in particular you know you have these elements ti bars which as you know are just the polynomial functions ti restricted to x okay and what you do is when you take their images under zeta they will go to some uhhh uhhh you know uhhh let me look at it for a moment uhhh they will go to some f i's and these f i's will lie in O y okay. See after all uhhh the ti bars are regular functions on x okay and zeta is a ring homomorphism it is a k algebra homomorphism. So, it is going to take an element here to an element here. So, if I take ti bar it is going to give me an element there and what is an element there it is a regular function on y. So, f i is the image under zeta of the ti bar for every i okay. So, the modular story is very simple the modular story is the moment you give me a k algebra homomorphism from k A x to O y the n coordinate functions on A x which are actually the n coordinate functions in the affine space in which x is sitting restricted to that uhhh x. They automatically give you a bunch of n functions on O y okay what does what does it mean to give you n regular functions on O y it means that you are giving n maps uhhh into k n scalar maps into k on y but that means what you are trying to do is if you think about it for every point of y if you evaluate this n maps you are getting an n tuple of points okay. So, you can see that this should give you a map from y to a point to an and that will be the map that will be that will be the morphism that you are looking for okay that is how you get this ejectivity. So, we will have to use these f i's to define a map from y to an okay by simply evaluating a point of y at uhhh these f i's and putting them as coordinates okay. So, you will get a map from y to an and then we will easily see that the map will actually factor through x and we will prove that the map is a morphism and we will prove that the alpha of that map is zeta okay therefore we will get this ejectivity of zeta okay. So, so let us do that. So, what I am going to do is let me rub this side off so I mean what you should remember is uhhh that the the moment you are given uhhh uhhh n regular functions on a variety you must always remember this whenever you are given n regular functions of on a variety you are actually uhhh given a map of that variety into an this is the idea okay. So, what so let me write this down define uhhh phi or let me use psi from y to an by y going to f 1 y f n y okay this is a this is a this is a well defined map because the f i's are all regular functions on y. If you evaluate a point of y at any of the f i's we are going to get a scalar you are going to get an element of k and this is an n triple of elements of k. So, it is a point in an which is just k n with a rescue topology and uhhh here is my map alright. Now, the first claim is is that uhhh psi uhhh maps into the first claim is that psi maps into x this map is not just going into an but it is going to go into the irreducible closed subset x of an it is going to go into it is going to go into uhhh this irreducible closed subset that is because it came from zeta as you will see. So, how do I show psi maps into x? So, it is very simple uhhh so how do you so what does it mean to uhhh how how does one verify this you have to show the give a point of given a point of small y alright given a point small y of capital Y the corresponding point psi of small y is a point in of an which actually lies in x this is what you have to show but how do you show a point of an find space lies in irreducible closed subset you will show that by showing that that point is in the common zero locus of the ideal of x how do you show a point of an lies in uhhh closed subset you will just have to show that it is a common zero of the ideal of that closed subset. So, all I have to do is I would take a point small y in capital Y I will take its image under psi which is this okay and then I have to verify that this point the point with this coordinate these coordinates satisfies every uhhh equation every polynomial in the ideal of x okay. So, that is what I have to do so so let uhhh g belong to uhhh ideal of x okay let g be in the ideal of x mind you g is a polynomial in n variables t1 tn g is a polynomial in so many variables okay then uhhh g restricted to x is actually g bar is is 0 in ax which is just k t1 etc tn by ix obviously an element in the ideal if you take its image in the quotient it is 0 but then you see uhhh from ax I have this map zeta I have started with the map zeta which is a k algebra homomorphism from ax to o y okay zeta is what I started with and zeta is a k algebra homomorphism in particular it is a ring homomorphism and ring homomorphism will take 0 to 0 okay. So, moral of the story is that if I take 0 which is g bar under if I apply zeta I will get zeta of g bar is 0 in o y this is what it means okay. So, so if I write that down what I will get is I will get zeta of g bar is 0 which means that uhhh but so this is if you write it carefully it is g and what is g bar it is g of uhhh uhhh so you know uhhh now I am going to use the fact that zeta is a k algebra homomorphism okay zeta of g bar is the same as g of t1 bar etc uhhh uhhh tn bar uhhh so I will have to say something so so let me write write that again zeta of g bar of uhhh well if you want okay g t1 bar etc tn bar is 0 okay mind you g is a polynomial in the t's g bar is a polynomial in the t bars okay g bar is a polynomial in the t bars and zeta is supposed to take this 0 but what how is zeta what does zeta do to the t bars the zeta maps the t bars to the f f's so so this implies that g bar of f1 etc fn is 0 this is what it means you see zeta takes t1 bar to f1 okay and it takes t i bar to f i then it will take a polynomial in the t i bars to the corresponding polynomials in polynomial in the f i's okay maybe maybe I should just remove this maybe I should just remove this bar because the bar is already inside I should just remove g bar is g of this just the image of g okay I remove this bar that superfluous okay so what this translates to is that g of f1 etc f1 is 0 on and what is this see this is what is g of f1 etc f1 it is it is another regular function on y see no notice that the f i's are all regular functions on y okay and what is g of f1 etc f1 it is some polynomial in expression in the f i's with coefficients in k okay and you know if you write a polynomial in a bunch of regular functions the resulting thing is also a regular function because a product of regular functions is regular a sum of regular functions is again regular and a regular function multiplied by a scalar is also regular function because scalar functions are regular functions which are constant okay constant functions are always regular okay. So this is certainly a regular function and it is 0 in o y so that means if I evaluate it at every point of capital Y I am going to get 0 so this will tell you that g of f1 of y etc fn of y is 0 for all y small y in capital Y this is what it means a regular function is 0 means if you evaluate it at every point it has to be 0 but what does this mean this is this is the same as saying that the point f1 y with coordinates f1 y through fn y this point is actually in the 0 set of g okay. So what I have proved is but what is this point this is just psi of y I have proved that for every small y in capital Y the corresponding point psi y is a 0 of every function g which is an ideal of x and that tells you that psi lands in x okay. So therefore this map that you got from y to a n actually factors through the closed subset x so you actually have a map into x okay. Now the next part of the proof is to show that this map is a morphism and I will have to show that this map is a morphism and I have to show that the pullback map that it induces is actually your zeta and then I would be done alright so let us see how to do that that is also pretty easy to do okay. So here is an exclaim the exclaim is psi is a continuous map the exclaim psi is a continuous map okay so situation is like this I have y I have psi this goes into x which is sitting inside a n okay and I want to show that this is continuous okay and mind you if I show that it is continuous I am actually more or less done showing that psi is a morphism because the movement psi becomes continuous okay then the only thing I have to check to check that psi is a morphism is that it pulls back regular functions so regular functions but the way we have defined it it will follow very trivially that zeta is actually psi star that is alpha of psi that will be very simple to see okay. So actually this is the crucial thing that needs to be verified that it is continuous okay. So the well the to see that that is also pretty easy let me think for a moment yeah so let me so it probably involves showing that the pullback of the pullback under psi is actually zeta I need that. So let me do that so let me write that here psi induces zeta and psi is a continuous map okay how do I explain that psi induces zeta it is very very simple. So you see alpha of psi which is psi star is going to be a map from ax to oi okay. In fact I should not say oi you know strictly I should say I can define for any map from if psi is any set theoretic map from y to x then given any set theoretic function on x I can pull it back to give a set theoretic function on y. So you know what I will do is I will just write maps from y to k okay and what is this is the usual pullback map pullback map exists for any set theoretic map even at the set theoretic level okay. So you see if you give me if you give me any polynomial h if you give me any element h which is a polynomial restricted to x actually then h goes to psi upper star of h and what is psi upper star of h is just first apply psi and then apply h this is the map okay and what I want you to understand is that what is this map see if you take a point of y h circle psi of y is h of psi y okay this is by definition psi y has been given a definition this is just h of f1 y and so on fn y because that is what psi y is okay but this is if you look at it h of f1 y fn y if you look at this calculation it is actually zeta of h operating on y because you see zeta of g is g of f1 etc fn so similarly h of f1 etc fn is zeta of h by the same argument. So moral of the story is so this will tell you that psi star it will tell you that and mind you so this will tell you that psi star is actually zeta and it will also tell you that psi star goes into goes just not into maps from y to k it goes into regular functions on y o of y is a subset of this is a set of all possible maps from y to k okay just all possible functions they need not even be continuous okay whereas o y is very special these are all morphisms of y into k they are the regular functions on y they are very special okay and the fact that psi star is zeta which you have verified point wise will tell you that and since zeta takes values in o y it tells you that psi star also lands inside not just in this but in this subset okay. So this proves that psi induces zeta okay so the only thing to show that the zeta is started with comes from a psi here is to show that actually psi is here namely I should show that psi is actually a morphism okay and what is the defining property of a morphism the defining property of a morphism there are two conditions the first condition is it has to be continuous the second thing is it should pull back regular functions to regular functions the second condition is already satisfied the pull back map induced by psi is already zeta and zeta is pulling back regular functions to regular functions it is taking every regular function on x to a regular function on y okay. So it is already pulling back regular functions to regular functions therefore the only thing that one has to check is actually that it is a continuous map okay so that is the reason why I said in the beginning that the essential claim is that it is a continuous map and once you do that you have proved everything you have proved it here alright. So how do I check something in the continuous map I will have to just check that you know inverse image of open sets are open or I have to check inverse image of closed sets are closed but that is a pretty easy thing because you see how do you define the Zariski topology on a on x there what is the Zariski topology on x how do you define the Zariski topology you for example you specify closed subsets given as zeros of polynomial okay. So what is a closed subset of x it is given as a common zeros of bunch of polynomials okay and if you compose those polynomials with psi okay those polynomials are regular functions okay therefore if you compose them with psi the resulting things will be regular functions on y and the inverse image will be precisely the locus of those zeroes of those regular functions on y and what you must understand is the Zariski topology given by the Zariski topology on y coincides with the topology that you get by taking the closed subsets to be common zeros of regular functions you see after all either you can when we started defining the Zariski topology we always took set of common zeros of polynomials but then when we went to a general variety which is which could even be an open subset of a closed irreducible subset then we had to define regular functions and these regular functions were locally given by quotients of polynomials okay and the zero set will be therefore locally the zero set of the numerator polynomial which is anyway closed okay it has got nothing to do with the denominator polynomial okay the zero set of a quotient of polynomials is essentially the zero set of the numerator the denominator polynomial is anyway assumed to be non-zero okay otherwise we would not put it in the denominator. So what you must understand is if you take a variety okay and if you take a bunch of regular functions on that variety and you take the common zero locus of a bunch of regular functions on that variety the result is again a closed subset of that variety just like common just like closed subsets of affine space are given by common zeros of a bunch of polynomials closed subsets of any variety are simply given by the set of common zeros of a bunch of regular functions okay and so if you start with a closed subset in here that will be the common zero locus of a bunch of polynomials okay and its inverse image here will be the common zero locus of the bunch of regular functions that are which are pullbacks of these polynomials and that is again closed. So what this argument tells you is that psi pulls back closed sets to closed sets so it is continuous and we are done okay. So let me write that down psi is continuous as psi inverse of z of some j is going to be psi inverse of is going to be intersection of okay so if you want I can write the full-blown set theoretic thing that you might so z of j is h inverse of zero intersection where h belongs to j right what is a zero set of an ideal zero set is you take this for each element in the ideal each polynomial in the ideal look at it zero which is h inverse of zero and you take the intersection this is the common zero set and you know psi inverse taking inverse is behaves well with respect to intersection so this is going to be intersection h belongs to j psi inverse of h inverse of zero okay and if you see that this is just the zero set in this is just the zero set of first apply this is a zero set of h circle psi intersection of the zero set of h circle psi where h belongs to j okay and this is closed closed in y so this implies psi is continuous and since psi is continuous it is already the map that induces pulls back regular functions to regular functions so it is a morphism and so I have found psi here I have been able to find psi I have been able to find psi which goes to z that gives me the physicality and I am done with the proof okay so proof ends here. So the crucial point that you will have to understand is the following which you can think of as you know as a separate exercise if you want I mean it is actually if you understand it if you have understood it well it is just a tautology but take any variety take a bunch of regular functions okay then the common set of common zeros of those regular functions is a closed subset of the variety okay it is a closed subset because regular functions are locally quotients of polynomials and taking the common zeros of such regular functions actually amounts to taking common zeros of a bunch of polynomials namely the numerator polynomials locally okay and that is the fact that you will have to understand okay so I will stop here.