 Okay so let us continue with our discussion which is whose aim is to define what a regular function at a point is okay. So let me again go back to this definition, the definition says that you know you have function f with values in k the base field and it is defined in an open set containing the point small x of course that open set could be that open set is being considered as an open set in the affine variety or quasi affine variety in which x exists okay and we say the function f is regular at that point if it is if it can be written as a quotient of polynomials okay that is there are two polynomials whose quotient will define a function a quotient of polynomials will define a function into the field the base field scalar value function wherever the denominator polynomial does not vanish that is actually so this a function of the form g mod h where g and h are polynomials will always define a function on dh on this basic open set dh which you know in its own right is actually an affine variety and the fact is that you will you will just have to find a point a h such that dh intersection x capital X contains small x and which is possible because it is basically because any open set is a finite union of such dh's okay for various h's right which is a fact that we have already seen. In fact any open set here is the complement of a closed set and that closed set is given by the 0 set of an ideal and that ideal is finitely generated and therefore the that closed set is given by the intersection of the 0 sets of the generators of the ideal and therefore its complement is given by the corresponding union of basic open sets namely the loci the open sets where the corresponding generators of the ideal does not vanish individual it is union of that okay that is how every open set is a finite union of basic open sets and therefore this is my definition and this definition seems to be correct for if you look at a basic open set a good function on this is of that form okay but the point is that whenever you make a point wise definition what you are actually doing is you are gluing okay and the problem with gluing is that it can produce new objects if you glue objects of a certain type it can produce a completely new object okay. So if you glue functions locally you might get something which is very different from what an ordinary function is for example if you glue topological spaces in a funny way you might get a topological space that looks very different from the ones that you started with okay we are going to see later that if you glue affine spaces finitely many affine spaces you can get a projective space okay. So for example if I take two copies of the you know if I take for example the usual topology and take two copies of the unit disk on the real plane and you know if I make them look like two open two hemispheres of a sphere and then glue them like this I will get a sphere okay and the point is that the new topological space I have got which is a sphere is very very special because it is compact which is not a property that is shared by either of the two open disks I started out to glue it. So the problem is that gluing of objects in mathematics can produce new objects which have properties are completely different okay the same thing happens also when you try to glue functions and what I am saying is that whenever you make a definition point wise you are actually doing some gluing and then for all you know you might ask the following thing can happen if I look at all the functions on affine space itself which are regular at every point they could be very well different from polynomials what is the guarantee that they could not they are no different from polynomials but the answer is that you are not going to get anything new okay and that is what that is a surprise and it is a pleasant surprise and that is what we are going to prove okay. So let me try to explain that so just to stress this gluing business let me write out so let me make a few you know let me make few notational conventions so what I am going to do is X in An, K and affine variety or a quasi-affine variety so let me do the following thing let me let me do it one by one let X in An be an affine variety I define O of X to be the regular functions set of regular functions on X okay and so you see this is a new notation I am using this is a calligraphic O okay for regular functions because as yet I do not know anything about a global regular function except that is locally a quotient of polynomials okay and certainly I do not expect it to be anything I cannot expect it to be anything unless I prove something about it okay. So and similarly you know if U inside X inside An so this is irreducible closed and then U is an open inside this irreducible closed and U is a quasi-affine variety so a quasi-affine variety is supposed to be just an open subset of an affine variety then again O of U is defined to be regular functions on U okay and this is these are global definitions okay and how do these global definitions come they come from local definitions point wise definitions okay and so let me so let me write down what is what is the definition of so for example if you right suppose I write Phi belongs to O U means what means okay Phi is a map from U to K is a map that is regular at each small x in U okay and of course I am writing this for U but you know I can as well write it for the same thing will also work for X where X is actually an affine variety okay but I am taking a quasi-affine variety right. So what is a regular function on a quasi-affine variety it is a function which is values in the field which is just a map but with the property that it is regular at every point that means what so for every so for every X in U there exist polynomials GX HX in K of X1 etc Xn where of course your U is being considered in U is being considered inside X which is being considered in some in An okay your U is an open subset of X capital X it is a quasi-affine variety and it is being considered in some affine space it is an irreducible closed subset of some affine space and there are polynomials such that D of HX contains X and Phi is the same as GX by HX on DHX this is the you see this is the definition on DHX or rather I should say DHX may not be all of on an open neighbourhood of small X contained in D of HX this is what it means for give me a point X I can the function Phi in an open neighbourhood of that small point small X is the same function that I would get if I evaluate a quotient of polynomials which that with the denominator polynomial lot vanishing and the polynomials being taken in the property number of variables in which dimensional affine space you are considering U and X okay so you see see this definition is you see this is a problem with this definition the problem with this definition is it is so arbitrary in a certain sense see I could have you know the same space the same affine variety X could sit in so many affine spaces see if I if X for example is a two plane then X could be A2 or X could be two plane in A3 may be the X Y plane or it could be the two plane in some AN it could be anywhere and depending on where it is I have to take polynomials in as many variables and then I have to take a quotient so you see it is a there is so much arbitrary in this definition there is so much arbitrary in this definition and that sometimes is a little scary okay but the point is that you have this for every point X so now you see note that the DHX where X is in X where X is in U is an open covering of capital X right because I said for every point small X I am getting this HX says that the affine open set DHX contains X okay and this affine open set basic affine open set DHX is considered in this affine space that affine space of that dimension in which is that number of variables in which number of variables you have taken the polynomial HX okay and of course since I say this such a HX X is for every X in U it means that all these DHXs they cover U okay but then what do we know about the Zariski temple we know that it is quasi-compact therefore what it will tell you is that out of this out of this collection of DHX I can just be content with having only finitely many okay so by the quasi-compactness by the quasi-compactness of U okay U is after all a subset of X which is internal subset of AN and then you know when you take subsets and take the induced topology the noetherianness goes down it is hereditary so U becomes a noetherian topological space and you know a noetherian topological space is quasi-compact as we saw in the last lecture. So U is quasi-compact and that means every open cover has a finite sub cover so what it means is that there exist X1 etc Xm in U such that U is DHX1 intersection sorry union U is contained in the union of all these DHXI HXM and well of course Phi restricted to Phi is GXI by HXI in a neighbourhood of XI contained in DHD of HXI for every I this is what it means. So you see what it means so what it means is when you say that you are having a regular function what it means is that you are actually taking quotients of polynomials and finitely many such quotients such that these quotients they agree on the intersections you see if you take the intersection of DHXI with DHXJ then in that intersection Phi will locally be both GXI by HXI and it will also be equal to GXJ by HXJ so what you are saying is you are saying that a general regular function is simply gotten by taking finitely many quotients of polynomials with the property that these quotients of polynomials agree the functions that they define upon evaluation agree on these intersections okay and this is exactly what gluing is gluing is you take functions locally and then so many functions with certain property glue to give a bigger function if they all give on the intersections the functions should agree. So what you are saying is just take finitely many quotients of polynomials such that the locus of the non vanishing of the denominator polynomials covers your space that is U in this case and such that these quotients wherever the loci where the denominator polynomials do not vanish intersect the quotients evaluate to the same function okay such this is what a regular function on U is okay it is got by locally so what you are so a global regular function a regular function on an open set is actually gotten by gluing finitely many quotients of polynomials so it is a gluing process and now so this is how it is for U it is the same definition instead of U I can put capital X also the same definition works for any subset in fact if you want of affine space but particular in fact instead of U I could have put any subset of affine space okay and I could have said regular functions on that set but we are interested only in regular functions on either open sets or on closed sets either they should be regular functions on irreducible open sets they are open subsets that is either they should be we are interested in functions on either irreducible closed subsets that is functions on affine sub varieties or we are introduced we are interested in functions on open subsets of such affine sub varieties okay so U is an open subset of X and of course whenever I am considering a subset I am certainly not looking at the empty set so U is a non-empty open subset and mind you a non-empty open subset is both irreducible and dense okay because capital X is both irreducible and dense I mean because capital X is irreducible okay. So you see what we have defined as regular functions is something that is very strange it is these are gotten by gluing finitely many quotients of polynomials so now I can ask what will I get if I put O of if I apply this O to An itself what will I get if I apply O to DH what will I get if I apply O to X what will I get if I apply O to DH intersect X what will I get the answer is very beautiful the answer is you will get exactly what you go what you will get if you apply A okay that is the beautiful thing okay and the fact is that is the sophisticated way of saying that is that that describes that is why all the four are actually affine varieties okay. So in more general algebraic geometry you can define A the A for objects which are affine objects okay and then you can define the O for any general object and then the theorem is that a general object is an affine object if and only if the O is the same as the A okay and that is exactly what is happening here alright. So I mean that is saying it in very loose terms okay to understand the exact import of that statement you should study a scheme theory which is which should be a second course in algebraic geometry okay but nevertheless does not do any harm in my stating it here. So if at all you go ahead to study scheme theory you can come back and try to remember this statement okay so let me so let me make that statement let me make let me make this remarkable statement so here is theorem O of An is equal to A of isomorphic I should say okay let me put isomorphic. So here is the theorem the theorem is that what you define as regular functions are going to give back exactly these functions which are given for which are given when you apply A to these objects either affine space or affine varieties or basic open subsets of affine space or basic open subsets of affine varieties okay this is the statement right. So the technique of proof is literally the same it is basically it is a technique in commutative algebra. So what I will do is I will let me first prove the first one and in fact you can see that you can deduce the remaining if you are a little careful and just apply the same philosophy as the proof of the first okay. So what I will do is let me prove let us prove O of An is isomorphic to A of An let us prove this alright let us prove this so how does one do this. So what I will do is I will do the following thing. So I will define a map define a map A of An to O of An by very very simple map take the A of An is just the polynomial ring in n variables okay and simply send it send a polynomial G to the function G from An to K so it is a very very simple map. So what you do is take up take an element here what is it it is a polynomial in n variables over K and a polynomial in n variables over K is a function it is a function from An to K by evaluation you can evaluate the polynomial at every point and that is in that is in certainly a regular function because a regular function is something that is locally given by quotients of polynomials and this is globally given by single polynomial and the single polynomial G can be written as G by 1 if you want you can think of this G as G by 1 and the locus where 1 does not vanish is everything okay and therefore it is also a regular function so all I am just trying to say is that every polynomial is certainly a regular function there is no doubt about it a regular function is something that locally looks like a quotient of polynomials but something that is globally a polynomial is also a regular function because it is a polynomial divided by 1 if you want and 1 is a constant polynomial which always makes sense okay. Now the fact is that you see just like all the polynomial form a ring you can add two polynomials you can multiply them and then they form a vector space over K and then so the ring of polynomials is a K algebra okay it is a finitely generated K algebra. In fact it is a free polynomial algebra in so many variables the same way the ring of regular functions is also the set of regular functions is also a ring that is the first thing you have to realize because you see you take sum of two regular functions okay the sum will also be regular because if you take a regular function is locally given by a quotient of two polynomials and if you take two such regular functions and add in a suitable neighbourhood the corresponding quotients of polynomials the sum of quotients of two polynomials is again a quotient of two polynomials okay in the correct neighbourhood where the denominator does not vanish and the product of two quotients of polynomials is again a quotient of polynomials right. So the moral of the story is that when I define this O of something whatever it is the regular functions on that thing that is a ring in fact that is a K algebra it is it has addition it has multiplication it is a vector space over K you can easily see that if you take a regular function multiplied by a scalar the result is again a regular function because locally you are just multiplying the numerator polynomial by that scalar okay in the in the expression locally as a quotient of polynomials for that function okay and it is also clear that that is a K algebra and so on. So these isomorphisms that I have written here they are not just isomorphisms of rings they are isomorphisms of K algebra okay they are isomorphisms they are ring isomorphisms they are isomorphisms of vector spaces also okay that means scalars go to scalars lambda a scalar lambda in K thought of as a constant polynomial lambda goes to the same constant function lambda thought of as a regular function okay. So these are all K algebra homomorphisms okay. So this map this map that I have written here is actually it is very easy to see that it is obvious to see that it is a ring homomorphism because g you know f g1 plus g2 what g goes to and what I mean this association will preserve addition multiplication it is K linear and all that. So this is a ring homomorphism it is a K algebra homomorphism okay and then the fact I want to make is that I will have to prove two things I will have to prove that this is injective I have to prove it is a ejective if I prove that then I get that this is an isomorphism of K algebra which is what I want okay so it is so let me write that it is a K algebra homomorphism that is obvious no doubt about that. Then how do you show it is injective it is injective why is it injective well if you take two polynomials g1 and g2 which has functions are different it means that there is a point in an where the values of g1 and g2 are different that means the difference polynomial g1 minus g2 does not vanish at that point okay and that should tell you that g1 and g2 cannot be the same okay. So it is very clear that it is injective the injectivity is just very severe okay in other words you know I am another way to say it is that I am saying that if you take a polynomial and if I evaluate it as a function and suppose as a function it is a 0 map as a function then the polynomial has to be the 0 polynomial okay. So this is something that is obvious probably it just requires the fact that the field K is infinite okay which is true because the field is an algebraically closed field and algebraically closed field is infinite okay. Of course the problem is over finite fields you can always find polynomials which are non-zero polynomials but which evaluated as functions end up being 0 function okay that can happen for finite fields but small K is not a finite field it is an algebraically closed field and an algebraically closed field is always infinite and for an infinite field if you have polynomial which if it is upon evaluation it is the 0 map then the polynomial has to be the 0 polynomial okay that is the statement that it is injective okay but what is really crucial is the fact that it is so injective so that requires a little bit of proof so let me do that. So that is where a little bit of commutative algebra will come in and you will recognize what is going on if you have done a course in earlier course in commutative algebra and where you have proved that you know the you have proved the quasi compactness of the Zariski topology when you take the prime spectrum offering okay. So let me do this it is surjective so what I will do is I will start with so let phi be a regular function okay take a regular function right so phi is a map from An to K and by definition of what a regular function is there exists finitely many points okay and such that the union is the whole affine space and such that phi is quotient of polynomials in this sense okay. So let me write that down there exist points same such that An is d h1 h sub x1 d sub hxm okay where hx1 through hxm are polynomials in invariables and phi is equal to gxi by hxi on dhxi for every i and of course gx1 through gxm they are also polynomials in invariables okay this is from the definition of what a regular function is okay. Now you see see the first thing is that this has a meaning in terms of commutative algebra the meaning that it has in terms of commutative algebra is that the ideal generated by the hxi is the unit ideal okay so this means commutative algebraically so you know it is a translation of all these geometric facts into commutative algebra okay and you extract some information from commutative algebra so the fact that A is union of all this means that the ideal generated by hx1 etc hxm is one is ideal generated by one namely it is a whole polynomial this is the first thing that needs to be noted and why is this true why this is true is because well if this ideal is not the unit ideal then it is a proper ideal and then we know that every proper ideal is contained in a maximal ideal okay and you know the maximal ideal is a that corresponds to a point okay so what will happen is that if you that point will have some coordinates okay but that point will be here it is a point of A n so it has to be in one of these so that means that at least one of the hxis does not vanish at that point but that will contradict the fact that this is contained in the ideal of that point okay so this is exactly what I am going to prove if not if not hx1 ideal generated by hx1 etc hxm is contained in a maximal ideal which is of the form say x1 x1-lambda1 etc xn-lambdan okay this will tell you that you see it will tell you that so you know if I apply if I apply the Z right you know if I1 is contained in I2 then Z of I1 contains Z of I2 where Z is the operation that associates to an ideal it is 0 set. So this will tell you that Z of hx1 etc hxm contains Z of this maximal ideal which is actually the point which is actually simply the point singleton point lambda1 through lambdan okay alright that means what this tells you is that this point is you know it tells you that this point is a common this point is a common 0 of all these hx okay and if this point is a common 0 of all these hx that contradicts this statement because this point is here okay and by definition every point here has to be contained in some locus where a certain h does not vanish but then you are saying that I have but I have been able to find the point which is not in any of these okay it is a contradiction the contradiction to an is equal to dh1 dhx1 union dhxn xn okay right. So the moral of the story is that indeed the idea this condition the fact that the affine space is a finite union of certain basic open affine open sets means that the equations corresponding to those basic affine open sets those polynomials actually generate the unit ideal okay. So what all this will tell you is that this generate the unit ideal so what will tell you is that there exists there exists you know let me give some other names okay so let me use ff's so there are f1 etc fm polynomials in n variables such that sigma fi hxi is equal to 1 I get this okay in other words the one is in ideal generated by the hx so one should be generated by a ring linear combination of the hx and that is what I have written here these ff's are ring elements the coefficients from the ring okay so of course I is equal to 1 to n okay. Now you see now the point is that I will have to show that let us go back to what I started with and what I want to show I started with a regular function on an okay and I am just trying to prove subjectivity of this map so I have started with something here a regular function an and I am trying to find that come show that it comes from here so I have to cook up a polynomial which is equal to that function okay so I will have to use this information to cook up a polynomial g such that if I consider g as a map I get this phi that is what I will have to do okay and that is pretty easy to see so the trick is you multiply both sides by phi okay multiply both sides by phi okay and what I will so let me do that so sigma i equal to 1 to m fi hxi into phi is equal to phi okay I get this just by multiplying both sides by phi and then I use the fact that you know you see phi is gxi by hxi on dhxi but if I cross multiply I will get phi times hxi equal to gxi not only on dxi I will get it everywhere okay so I am using the following fact if two polynomials are equal on an open set if two polynomials treated as functions they coincide on a non-empty open set if the two functions defined by two polynomials if they coincide as a function on a non-empty open set then the polynomials are equal okay so what this equation tells me is that on this non-empty open set dhxi the function phi times hxi and gxi they are the same okay and I want to say from that that they are the same phi times hxi and gxi are the same if you want as regular functions on the whole space so I am using so in fact let me restate it more correctly I am saying that if you have two regular functions which coincide on an open set then they have to coincide on the whole space if two regular functions coincide on a non-empty open set then they coincide on the whole space the reason is the reason is topological the reason is because regular functions are continuous and open subsets are dense it is very simple the reason is topological okay. So all I am saying is that if I have two regular functions okay and if they the fact that they coincide on an open set on an open set means that the difference regular function is 0 on an open set okay but the set of points where a function is 0 is a closed subset because a function is continuous therefore this open subset is contained in a closed set but the open subset is dense which will tell you that the open subset that it will tell you that these two regular functions are the same everywhere okay. So let me write that down so that you know okay so let me rub this side I still want this side of the board so let me write that down so that it because clear to you so what I do is here is a lemma so here is a lemma lemma 1 any regular function function is continuous so here is a lemma any regular function is continuous so you see so what is the proof so the proof is take psi in O of U if you want okay or let me take O of something okay so let me take O of U right then so psi is actually a map from U to K psi looks like g mod h locally okay where you know g and h are polynomials they are polynomials in the right number of variables and you are considering U inside that fine space okay and then you see to show psi is continuous it is enough to show psi inverse of a closed subset is equal to a closed subset alright. So psi inverse of a closed subset of here is a closed subset of K and is a closed subset of U this is what I will have to show alright but then what is a closed you know the Zariski topology on K which is A1 the Zariski topology is just the compliment finite topology namely the only closed sets of finitely many points so to so psi inverse of a closed set is just a sin inverse of finitely many points and to show that that is and psi inverse behaves well with respect to unions psi inverse of a union of sets is the union of psi inverse of those sets psi inverse behaves well with respect to the operation of taking unions. So to show that psi inverse of a finite set of points is closed it is enough to show that psi inverse of a single point is closed okay so enough to show psi inverse of a point lambda in K is a closed set a closed subset of U this is what I will have to show okay. So but you see but then given X in U there exists GX HX with X belonging to D of HX and psi is equal to GX by HX okay so what this will tell you is that you know it will tell you that so in a neighbourhood of X containing contained in D HX this is what this is the definition local definition alright but then but psi inverse of a lambda intersection this neighbourhood is just GX mod HX inverse lambda in intersection this neighbourhood and this is just equal to Z of so GX mod HX has to be lambda so GX has to be lambda HX so GX minus lambda HX has to be 0 so it transits to Z of GX minus lambda HX intersection this neighbourhood which is of course closed closed in this neighbourhood and you have done because you know to check that a subset of a topological space is closed it is enough to check it on an open cover okay. So what I have done is I have for every neighbourhood every point I have shown that psi inverse lambda intersection that neighbourhood is a closed subset in that neighbourhood okay and if I vary X I get an open cover and now for each of the sets in the open cover I have verified that the inverse image intersected with that set of the open cover is a closed subset okay so this proves the lemma now let me prove another lemma which is the lemma that I want to apply there. So of course you know I am here I am looking at regular function either on an affine variety or a quasi affine variety fine okay so here is lemma 2 in fact if two regular functions agree on a non-empty open subset then they are equal if you have two regular functions they are equal on non-empty open subset then they are equal everywhere proof is two lines proof is trivial two functions agree on a non-empty open subset means the difference function is 0 on a non-empty open subset but any non-empty open subset is dense okay for the Zariski topology. So you have a function that is you have a continuous function that is 0 on a dense open subset so it has to be 0 everywhere so it is obvious so proof is any non-empty open subset is dense and it is just and a function which is 0 on a dense a continuous function which is 0 on a dense open subset has to be 0 everywhere simple topology. So if you apply these two lemma I mean I need this lemma now you look at now you look at this equation that I have written here in this equation you know if I calculate if I look at the product Hxi times phi the product Hxi times phi is Gxi and Dhxi okay but Hxi times phi is also a regular function it is a product of two regular functions and Gxi is also a regular function because you have already seen every problem is regular function so you are saying the regular function phi times Hxi is equal to the regular function Gxi on the non-empty open subset Dhxi therefore by this lemma it is the same everywhere therefore for Hxi times phi I can put Gxi okay. So what I will get is I will get sigma i equal to 1 to m fi Gxi equal to phi and that tells you that phi is actually that polynomial given on that given by the left side that is the polynomial that I wanted okay so now I let me continue so let me rewrite that sigma i equal to 1 to m fi Hxi phi is equal to phi this is equality as regular functions okay but Hxi times phi is the same as Gxi because Hxi times phi is equal to Gxi on Dhxi and therefore on whole affine space because of this lemma therefore I can replace Hxi times phi as Gxi everywhere so what this will tell you is that it will tell you that phi is actually sigma i equal to 1 to m Hfi Gxi and that is the end of the proof I have proved that the regular function is actually the function that you get by evaluating a polynomial what is that polynomial here is a polynomial okay that is the proof. So you see there is a tricky bit of commutative algebra coming inside the proof okay so the moral of the story is there is no difference between the ring of regular functions on affine space and polynomials on affine space okay so every regular function on affine space which is defined locally by gluing polynomials by gluing quotients of polynomials if you take a regular function on the whole affine space which you have gotten locally by gluing quotients of polynomials the resulting regular function is actually a polynomial in other what that means to what that means to say is that it is the function that is gotten by evaluation of single polynomial and that this is the this is that fact written in the form of an equation okay so you do not get anything new okay. Now you can use the same technique of proof to prove the other statements okay all the time you will use this fact that whenever the corresponding space is a union is contained in a union of finitely many basic open affines then the corresponding equations polynomials that occur in those that define those basic opens the ideal generated by that is the unit that is the key and then from that by applying this lemma you can get this you can get the proof for all the other cases okay. So in the case of so in all these cases your regular functions agree with the ring of functions that we defined the coordinate ring of functions that we defined okay so I will stop here.