 Alright so recall from the previous lecture that we were trying to define what a morphism of varieties is okay so we have pair of varieties, two varieties they could be either affine or quasi affine varieties and then we want to say when a map between them is a morphism of varieties okay and we use the analogy that I gave you in the last lecture namely the analogy from the topological spaces okay so if X and Y are topological spaces and F is a continuous map of topological spaces of course you know that continuous map pulls back continuous maps to continuous maps it is just a V statement of the fact that composition of continuous maps is continuous and but the more beautiful thing is that if you have a set theoretic map that pulls back continuous maps to continuous maps then it follows that that set theoretic map is itself continuous the proof is deceptively simple and you could say more or less topological but this is the but it is the importance is with the philosophy that you define a morphism as a map which pulls back good functions to good functions. So let me let me make this definition of what a morphism between varieties should be so here we go let X and Y be varieties by a morphism from X to Y, Phi from X to Y we mean a continuous map from X to Y such that for every open subset V in Y the pull back map the pull back of functions of regular functions I should say the pull back of the pull back of maps V of Phi takes regular functions on V namely O V to regular functions on U namely O U so the diagram is like this here is X, here is Y, here is Phi and here is the open subset and open subset V of course whenever I say open subset it is of course non-empty okay and in all these situations we are not going to look at the case when the open subsets are empty because we do not want it define regular functions on an empty open set okay. So of course I should tell you that where U is actually F inverse V okay so here is V and here is U which is F inverse V and this is also open that is because Phi is a continuous map of course X and Y are varieties so they have the Zariski topology therefore you know what open sets mean and you take an open set here the inverse image of an open set is an open set because and I should use not F I have messed up the notation should be Phi inverse here should be Phi inverse there as well okay and I have this map from Phi inverse V to V. Now this map from Phi inverse V to V will give you a map from O V to O U and what is this map from O V to O U what is an element of O V it is a regular function on V okay and element of O V is a regular function on V so what is a regular function on V it is a map from V to K a regular function on V is a map from V to K which as we had defined in the earlier lectures is a function which is locally given by quotients of polynomials where the polynomials are considered in the right number of variables in which V or Y is the right number of variables which is equal to the dimension of the affine space in which V or Y is sitting inside okay. So well here is V and here is Phi this is U so U takes Phi Phi takes U to V and then the pull back will be this map you pull back essentially means composition so this is this is first apply Phi then apply F and this is called as Phi upper star of F so this is F going to Phi upper star of F the pull back of F which is by definition first applying Phi and then following it up by F okay so this is the requirement the requirement is you take an open subset of the target per IT take a regular function on that if you compose it with Phi you should get a regular function on the source okay this is the condition this is just the condition that the map Phi is continuous and it pulls back regular functions to regular functions okay this is the definition of what a morphism is what you must understand is the following that if you go down from the category of varieties to the category of topological spaces that is you forget the variety structure and just look at all these things as and the underlying topological spaces with the of course with the Zariski topology then what will happen is if you give me a continuous map if you give me a morphism Phi of course V is continuous and if you give me a V and if you give me a regular function on V this F is of course going to be continuous because I have already proved to you that regular functions are continuous okay a regular function from a variety is always a continuous map continuous with the target K being thought of as A1 with the Zariski topology this is something that we proved last time okay so since Phi is continuous and F is continuous is very clear that F circle Phi which is the pullback of F by Phi this is also Phi of star F it is clear that this is continuous okay but what our requirement is it is not just a continuous map from U to K for the Zariski topology it should actually be a regular function from U to K for the Zariski topology that is the requirement okay that is what has to be singled out okay so the moral of the story is a continuous map such that if you take a regular function on the target on an open subset of the target and take the composition okay what you should get on the source a subset of the on the right subset of the source is not just a continuous function it has to be more it has to be itself a regular function namely it has to be locally given by quotients of polynomials that is the requirement okay so it is a philosophy that it is a map of varieties is a map which is continuous and which pulls back good functions to good functions and the good functions here are regular functions okay. Now so here is the so here is the so the point I want to make is the following so here is a so here is a nice little theorem the theorem is if you look at if you take a if you take a variety X and look at all the possible morphisms into A1 you will get exactly all the regular functions on X okay so morphisms of varieties from X to A1 is actually isomorphic to O okay so the regular functions are actually morphisms into A1 there is no difference okay order elements of O X they are they are they are maps into A1 which is just K with the Zariski topology alright and which are locally given as quotients of polynomials that is what a regular function on X means okay and what is a morphism of varieties from X to A1 it is also a map from X to A1 which is a map from X to K but then the condition is that it has to be continuous and it should pull back regular functions to regular functions okay and the beautiful thing is that there is no difference between a regular function and a morphism into A1 so regular functions are exactly the same as morphisms into A1 okay so let me write that that is regular functions are so in fact I should not even put isomorphic I should put equal to or morphisms into A1 so of course is A1 K where K is a fixed algebraic closed field over which we are working okay so it is a very nice theorem it says it tells you that your morphisms into A1 are the same as regular functions into A1 okay so so let try to prove this so you know so what you do is so what is a map of course this map is it is identity map okay it is the identity map but only thing is that when you take a regular function on X the target is taken as K and the target K you do not care about the topology on the target K but if you put a topology on the target K and call it A1 then you know a regular function is continuous so it is certainly a continuous map from X to A1 but the fact is it is more it is actually a morphism from X into A1 okay that is what this theorem says so let us prove this let Phi from X to A1 be a morphism okay what do I have to show I have to show that Phi is a regular function alright now so this means Phi up a star from O of V to O of Phi inverse of V for any open set Phi inside A1 okay so what is a morphism a morphism is a map which is continuous and it pulls back regular functions so if you give me an open set in the target which is an open subset of A1 then and you give me a regular function on that open set on A1 which is an element of O of V then applying Phi up a star the pull back should give me a regular function on Phi inverse V okay now what I am going to do is I am this way is actually very easy you put V equal to A1 itself put V equal to A1 okay then you get Phi up a star will go from O of A1 to O of X okay because Phi inverse of A1 is X the inverse image of the target space under any map is the whole source space okay I will get this okay and then what you and then we will use the fact that you know in O on O of A1 see I have the identity map see the identity map on A1 is a regular function on A1 see the identity map on A1 is a map that sends every point of A1 to its every point of A1 namely every point of K to itself and what is it you know O of A1 if you want is A of A1 the ring of functions on A1 which is KX okay and what is the identity map correspond to it corresponds to the polynomial X identity map corresponds to the polynomial X the polynomial X evaluated at lambda gives you lambda okay so O of A1 is the same as A of A1 okay that is something that we have already seen O and A coincide for a fine varieties alright so O of A1 is same as A of A1 and identity map in O of A1 is actually that corresponds to the function defined by the polynomial X okay so this is so identity map is certainly a regular function identity map on A1 is certainly a regular function and what does it go to it will go to phi upper star of the identity map on A1 but phi upper star of the identity map on A1 is just phi, phi upper star of the identity map on A1 is by identity map on A1 composition with phi which is just phi and but then you are saying that this is here so you are just saying that phi is in OX so this will tell you belongs to OX so it is very clear that a morphism from X to A1 is certainly an element of OX okay so phi as an element of morphism phi as a morphism of varieties from X to A1 is certainly also in OX right so this is very trivial alright I have to now do things the other way around I will have to say that I have to start with a regular function and I will have to start with a regular function on X and say that that is a morphism okay so let me do the other way conversely start with an F in OX start with an F in OX right I want to show that F is actually a morphism into A1 so you know my F in OX means that F is a regular function defined on X its target is K and we have already seen that if this target K is given the Zariski topology and is thought of as A1 then this F is of course continuous okay we have already seen already seen that a regular function that regular functions are continuous so that means that F from X to A1 is continuous so what are the two defining conditions for a morphism it should be it should be a continuous map and then it should pull back it should pull back regular functions to regular functions so we will have to check we will have to check to check for every open set V in A1 and every psi in OV F up a star psi which is just psi F followed by psi is in O of F inverse V okay this is what you will have to check you have to check that it pulls back regular functions to regular functions alright so now you see what you must understand is that V is an open subset of A1 of course I am certainly not looking at empty sets alright so you know a non-empty open subset of A1 is actually a basic open set okay because you see what you must understand is that the open sets in A1 are complements of finitely many points okay and therefore the complements of finitely many points and there is always a polynomial with roots exactly at those points and it is the 0 set of that polynomial which is the complement of this open set and therefore this open set is defined as the basic affine open set given by the non-vanishing of that polynomial okay that polynomial which vanishes at those points in the complement of the open set you must understand that the important fact we are using is that A1 has the only closed sets in A1 are finitely many points and this is basically commutative algebraic reflection of the it is a reflection geometric reflection of the commutative algebraic fact that every ideal in the polynomial ring in one variable over a field is generated by a single element okay it is a PID alright so the fact is that so let V be D of G okay where G is an element of if you want A of A1 which is identified with Kx G is the polynomial in one variable right and V is D of G it is a basic affine open defined by G right then what is O of V then O of V will be O of DG and what is O of DG our definition as we have checked it is just Kx localized at G okay the regular functions on DG are the same as the A of DG and that is defined to be the localization by G okay so O of V is just this okay and so what is an element of O of V of the form it is of the form some H by G power M where M is a positive non-negative integer okay an element of O V looks like H by G power M alright this is how an element in the localization looks like and you are thinking of this H by G power M as a function on as a regular function on V and every regular function on V looks like that right and now if you so let me draw this diagram so I have X I have A1 and here is F and here is D of H okay which is my V and then I have F inverse V this is open of course F is a regular function it is continuous so the inverse image of an open set is open so F inverse V is open and then I have I have on D of H I have a regular function okay and that regular function is given by H by G power M this is an element of O of D of H and then I compose it with F to get the pullback this is a regular function into K okay sorry this is D of G should be D of G I am sorry should be D of G I think I have messed it did I mess it up somewhere there no I did not okay I mess it up here alright should be D of G right so yeah so I have H by G to the M which is a regular function on DG and then I compose this to get the pullback function which is F upper star of H by G power M which is by definition equal to first apply H and then apply first apply F and then apply H by G power M and so it is going to be well so if you calculate it is actually F's H circle F by H circle G to the M this is what it is okay so because this is what it will be when you evaluate it if you take an X here then it will go to F X and when you evaluate H by G power on M on F X you will get H of F X by G to the M F X so the effectively this is H circle F by H H circle G to the whole to the M alright and here of course and now the point is that what do I have to check I have to check that this is I have to check that this is a regular function on F inverse V I have to check that this is a regular function F inverse V okay so it is something wrong sorry this is G sorry you are right it is G circle okay it is G circle oops yes G circle F to the M right yes yeah please check that it is F followed by H divided by F followed by G whole to the M yeah thank you. So I will have to check that this is a regular function I mean if you think about it for a moment you will not hesitate to realize that this is a regular function okay but then if you want the way to look at it is the following the way to look at it is that you see the locus see F is a so let us look at what F is so F is a regular function on X okay. So you know I want to say the following thing I just want to say that if you if I tell you that H circle F is a regular function on F inverse V and G circle F to the M is also regular function on F inverse V that does not vanish on F inverse V then the quotient will also be a regular function on F inverse V I am just using the following thing I am using the following thing namely that suppose you have a regular function that does not vanish then its reciprocal will also be a regular function because the only condition for a regular function is that it is locally given by quotient of polynomials and if that quotient of polynomials does not vanish it means that the numerator polynomial does not vanish if they are in the if there are no common factors and if the numerator polynomial does not vanish then the reciprocal of the quotient is also a valid quotient of polynomials and that tells you that the reciprocal of a regular function that does not vanish is also a regular function alright. So therefore if I claim therefore it is very clear that it is enough to show that H circle F is a regular function it is enough to show that H circle F is a regular function on F inverse V G circle F to G circle F is also regular function of on F inverse V and G circle F does not vanish on F inverse V okay it is very clear that G circle F cannot vanish on F inverse V because if G circle F vanished at a point on F inverse V it will mean that at the image of that point under F G will vanish but then the image of that point is supposed to lie in DG where G cannot vanish so it is very clear that G circle F cannot vanish on F inverse V so the only thing therefore I have to prove is that H circle F or G circle F are actually regular functions okay but so let me write that down it is enough to show that H circle F and G circle F are regular functions on F inverse V okay this is all I will have to show okay now this is again something that is very very easy to see because you see see F, the F I started with was a regular function on X so F is given by a quotient of polynomials in the right number of variables the number of variables being the dimension of the affine space in which X sits alright so F is locally a quotient of polynomials okay so H circle F will also be a quotient of polynomials the same number of variables because H is just a polynomial in one variable when I take a polynomial in one variable and substitute for that variable another polynomial in some m variables the resulting thing will again be a polynomial in m variables it is obvious therefore the moral of the story is that it is very clear that since F is locally a quotient of polynomials H circle F, F is locally a quotient of polynomials on X okay then H circle F is also locally a quotient of polynomials on X okay so let me write that down this is obvious for locally F is equal to P by Q where P, Q are polynomials in the right number of variables which is a of a n with X sitting inside a n okay and you know and if H is equal to H of X is equal to sigma ai X to the i i equal to 1 to say some T then H circle F is locally H of P by Q which is sigma i equal to 1 to T ai T power i by Q power i okay which is a polynomial in K of X1 etc Xn divided by some power of Q so you are done. So the moral of the story is that if you really think about it it is very clear that a regular function is also a morphism so there is no difference between regular functions on a variety and morphisms into A1 there is absolutely no difference okay so the beautiful thing is that to define a morphism we started with the regular functions and then found that regular functions are themselves special cases of morphisms okay in that sense it should not be very surprising but you must understand that you see right from the beginning this is again you know amplifying more and more the philosophy of Felix Klein that you know the geometry of the space is completely controlled by the functions you choose on it so you know we started with affine space we started with n dimensional space Kn we chose the functions to be polynomials using the polynomials we define the Zariski topology on Kn okay and then from the Zariski topology we started translating back into commutative algebra and we came down to defining we first came down to guessing what should be regular functions okay for example on basic open sets and then generally what should be regular function what the regular function should be on a general open set okay and then after we defined regular functions once we defined regular functions we found that if you take affine varieties the regular functions are just polynomials okay and we also found that using these regular functions we could define morphisms and then it is not surprising that finally regular functions themselves turn out to be morphisms okay alright so that is a very nice thing it is a very nice result but now what I am going to do is that I am going to say that this is the statement here is only it is still a special case of a very very general statement okay so let me let me make that statement so again let us look at that statement morphisms of varieties so this is morphisms of the category of varieties from X to A1 is identified with OX okay and you know and I just want to say that this is isomorphic to homomorphisms of K-algebra from A of A1 which is which can be identified with KX to OX okay so you see now what I am going what I am trying to do is I am now trying to translate things see on this side the morphisms I have is in the category of varieties now I want to go this is geometric part to the geometric side I want to go to the commutative algebra side and I want to translate morphisms of varieties into morphisms of rings in this case morphisms of K-algebras and this is the this is the way I do it what I do it what I do is I think of OX as the as the I interpreted as being bijected to the set of all K-algebra homomorphisms from the A of A1 which is KX to OX see you know there is this universal property of the polynomial ring in one variable which says that a K-algebra homomorphism from a polynomial ring in one variable is completely dictated by the image of the variable X okay so any K-algebra homomorphism from KX to OX from KX to any ring is simply controlled by what you are sending X to so more generally I am so let me recall universal property of KX of RX okay where of course R is a commutative ring with one what is the universal property the universal property is that if B is if A is any R-algebra then A home R-algebras from RX to A can be identified with A by simply sending a map F from RX to A to FX to the element FX of A because it is a substitution it all R sending a map from RX to something means that you are substituting X for something in every polynomial in X with quotients in R so it completely is controlled by what you are substituting X with if you substitute X with some lambda then you have to substitute every polynomial in X with X in place with lambda in place of X it is controlled by that this is the universal property and in fact this universal property is also valid in several variables more generally homomorphism as R-algebras from RX1 through Xn to A is isomorphic to A to the n where you simply send a map F from RX1 through Xn to A to the tuple FX1 dot dot dot FX to the n tuple that is a map an R-algebra homomorphism from a polynomial ring in n variables is dictated by the images of those variables this is these statements are restatement of the universal property of the polynomial ring over a given ring okay. So and I am just applying instead of R I have put the field K and instead of A I have put OX so the homomorphism from K-algebras from KX to OX is just OX and that is what this is okay and now let me tell you how this statement generalizes the statement generalizes like this okay it is very beautiful so it generalizes like this. Homomorphisms from X to A1 can be identified with homomorphisms of K-algebras from A of A1 to X okay and morphisms from X to An can be identified with homomorphism of K-algebras from A of An okay and morphisms from X to Y okay where Y is a fine okay so instead of taking the target to be A1 I can take the target to be An instead of taking the target to be An I can take the target to be an affine variety. So the morphisms from any variety into an affine variety can be identified with homomorphisms of K-algebras from the affine coordinate ring of that affine variety sorry this should be a Y and mind you this AY is the same as OY there is no difference because Y is affine and the in particular you see in particular if X is affine the last statement says that the morphisms so here of course these several morphisms is varieties on this side it is a geometric side there are morphisms between geometric objects and this side is the algebra side the combative algebra side. So the morphisms of varieties from X to Y can be identified with homomorphisms of K-algebras from AY to AX and mind you I can replace OX by AX because OX and AX are the same because X is an affine variety okay. So this is the these are the two important statements this one and why are these statements important these statements we will prove these statements okay and they are just you know generalized versions of this first statement which is just the simple statement that the regular functions on variety are the same as morphisms into A1 okay it is just a generalization it is a grand generalization of that but how grand it is is that it actually gives you an equivalence between the category of affine varieties and the category of finitely generated K-algebras that is the reason why this is a very important statement. You have category so category so let me do that in the next board so that I have a little bit more space to write down what I want to write down so what I do is now I can really complete the picture that I stay I gave several lectures ago and then I was just you know throwing statements at you so here on this side I take the category of affine varieties okay and here the objects are affine varieties namely irreducible closed subsets of An for some N positive these are the objects in the category and I am going from that to the category of finitely generated K-algebras that are integral domains sometimes instead of writing so much people just say the category of affine coordinate the category of affine rings over K the affine rings means they are actually rings of functions on affine varieties okay so and abstractly they defined as finitely generated K-algebras that are integral domains which means that you are just taking a finite you are just taking a polynomial ring in finitely many variables over K and going more low prime ideal so that you get an integral domain okay and you know I have this I have a I have a functor that takes objects here so what it does is that if you give me an X it takes this X to well A of X okay so it tends it takes X to A of X and it takes An to A of An and it takes this closed immersion to a quotient okay and you know what that map is if you call this closed map if you call this closed inclusion as I X you know what this map is this is just I X star you can check that this closed this map the inclusion map is actually a morphism of varieties and you can check that the pullback I X star that induces that it induces from the O of this to the O of this which is the same as the A of this to the A of this is simply the restriction you can check that okay so this I X gives you this I X star okay and what are the what are the morphisms the morphisms here are well if you have a if you have X and Y to affine varieties I I get you you have a morphism F then that will give rise to a map on this side that will go the other direction so it will go from A Y to A X and you know what that is that is just F star it is a pullback so the fact is that if you give me a morphism of varieties the pullback map is actually a ring homomorphism okay that is something that you can very easily check it is obvious and in fact it is a K algebra homomorphism so the morphisms on this side are K algebra homomorphism they are ring homomorphism which take one to one which respect the vector space structure they are K linear so they are K algebra homomorphism and the fact is that this is this is the functor now A is a functor it is a functor because it goes not only does it associate an object to an object but it associates a morphism to a morphism okay to every morphism varieties you get associated K algebra homomorphism so this is a functor and the fact is that this is a this is an equivalence of categories this is an equivalence of categories because there is an inverse function which is which as I told you is given by max spec the max spec functor starting with a starting with a finitely generated K algebra A I can give you I can look at max spec A that is a variety okay that can be identified with a variety and if you give me a K algebra homomorphism G Phi from A to B then what you will get is you will get a from the max spec to the max spec you will get a homomorphism in the reverse direction namely max spec Phi and this is just pull back of given a prime ideal in B I mean given a prime ideal in B the inverse image of the prime ideal will be a prime ideal in A and the fact is you can check that because these are finitely generated K algebra which are integral domains in K is an algebraic closed field if you take a maximal ideal in B and you pull it back you will get a maximal ideal in A and so it will take max spec to max spec and this is the inverse functor and these two functors are inverse of each other in the sense that you start with an object you go then and come back what you get is an isomorphic object okay and then similarly if you start there you go here and come back you will get something that is you will get an isomorphic K algebra so if you start from here and go and come back you will get an isomorphic variety if you start from there with an algebra go and come back you will get an isomorphic K algebra okay and that is why these two functors are called inverse functors and we say that each of them defines an equivalence between the geometric side which is the category of affine varieties and the commutative algebraic side which is the category of finitely generated K algebra which are integral domains and you know it is easy but and certainly not an exaggeration to say that this is the full blown form of the Hilbert-Nulstall-Ansatz okay with all the other definitions in the right place this is the grandest form of Hilbert-Nulstall-Ansatz okay so with that I will stop this lecture.