 So, we have already seen this that if x is a variety over an algebraically closed field k and you take a point small x in capital X then we have defined the local ring of the variety capital X at the point small x ok and so this is the local ring of capital X at small x. Then we have also defined k x which is the so called function field of x function field of x ok and see we have we have seen that k x consists of equivalence classes of rational functions namely functions are defined regular functions which are defined not on all of x but on an open subset of x ok and of course the equivalence is that 2 functions are equivalent if they coincide on the intersection of their domains of definition. On the other hand the local ring is supposed to be gotten by doing the same thing but concentrating only attention to regular functions in a neighbourhood of the given points small x ok. So, and we have seen that you know O x which is the set of all global regular functions on x this is a sub ring of of course of the local ring of every local ring and that is contained in the quotient field and in fact of course k is of course sitting inside regular functions as constant functions ok and in fact this k x is this k x is actually the quotient field of O x x the quotient field of this is exactly this. In fact if you go to any open set U then which is non-empty then k x is the same as k u and O u x is the same as O x x ok. So, I let me also write that this is this is also equal to O u x and this is also equal to k u where u is an open subset of x and a small x is in capital U and so local ring does not change if you go to smaller open set and the function field also does not change if you go to a smaller open set because so called smaller open set is actually not so small ok any open any non-empty open set of a variety is reducible and in fact it is dense. So, what you are throwing out the boundary is just a very small set ok. So, and we have seen that the local ring already there is lot of information stored in these things in fact so I should also mention that this O x so I should also say that O x is this is equal to A x if x is affine. So, that is also you also have that O x is the same as A x if x is affine and then and in that case this is also equal to the quotient field of A x of course when I say when I write A x I this is defined only when if x is affine that is x is isomorphic to a an irreducible closed subset of some affine space and of course there are so many other things of course if x is projective we have seen that O x is this is equal to this ok if x is projective then O x is just k alright then because there are no non-constant global regular functions on a projective variety right. Then we have also seen that this is local ring stores lot of information that so in fact the dimension of x the topological dimension of x is the same as the Krill dimension of the local ring at each of its points and this is also equal to the transcendence degree over a small k of the extension capital K x capital K x is a function field of x is a it is an extension of k it contains small k after all this k is a field these things this is a this is a finite I mean this is a k algebra and this is a local ring this is a local k algebra local ring which is a k algebra whereas this is again a field this is a large and everything is contained in this big field ok and small k to capital K x is a field extension ok because it is a bigger field containing a smaller field and then the transcendence degree of this over this namely the by that one means the cardinality of a transcendence basis of this over this ok which measures the number of maximum number of algebraically independent you know transcendental elements over small k ok that is a measure of the dimension topological dimension ok and the dimension is also shows up as a Krill dimension of the ring ok and you also know that this is also equal to the Krill dimension of a x if of course this is if x is a fine ok. So already there is so much information but then my the object of this lecture is to tell you that this local ring is really very powerful thing it contains out it stores lot of information. So let me tell you the first thing so it is about so it is about the yeah so alright. So let me talk about first about local rings and morphisms ok. So you know suppose x and y are varieties and f is a morphism of varieties you know that that means that x and y could be both x and y could each of x and y could be a fine or quasi a fine or projective or quasi projective and the fact that f is a morphism means that f is continuous and f pulls back regular functions to regular functions. So so the point is that you know if I take a point small x here and it goes to the point small y which is the image of x etc f and you take an you take an open set u an open set here which contains the point x and you take well rather I will do it the other way around let me take an open set v which is a neighbourhood of the point y and let me set u equal to f inverse v then of course f being continuous the inverse image of p is an open set I call that as u and then of course f factors like this this is just f restricted to u this diagram commutes ok and now you have a map from o you have a map from o v to o u which is actually o of f inverse of v and the map is just the pull back of regular functions. So the notation is f hash and what does it do if you give me a regular function g on o v what I do is that I mean what this does is that it pulls back g to f hash g and what is f hash g f hash g is just first apply first apply f restricted to u and then apply so f restricted to u followed by g so this is what so this is just pull back of functions ok so basically what is happening is that giving a g on o v means that you are giving you are giving a function from v to k which is being thought of as a1 and you know if there is ketopology on k you have a1 and this g is a morphism so a regular function is just a morphism into a1 so g is a morphism into a1 and then you compose it with this and then you get a morphism like this and this is precisely f hash of g this is the pull back of g by f ok and the condition for a morphism is that not only f should be continuous but whenever you start with a regular function on a target on an open subset of the target when you pull it back you should get a regular function on the pull back of that open set on the source ok that is part of the definition of a morphism and one of the important things is that what this leads to is that from f this induces a map f hash y which will go from the local ring of y at small y to the local ring of capital X at small x ok you have a map at the level of local rings right and this map is of course again it is induced by the same thing namely what you do is what is what is an element here an element of the local ring is just an equivalence class of a function which is regular on an open neighbourhood of the point y. So you have something like v, g this is what an element here looks like it is an equivalence class where g is a regular function v, g belongs to ov and v is a neighbourhood of y, v contains the point small y and what do you do it is very simple you simply send it to the class that corresponds to the pull back of that function which is also regular function. So I take f hash g and I take it on u which is f inverse of v and I take this pair and then I take the class. So this is how this map is defined so all I am trying to say is that if you have a morphism varieties that morphism induces for every point small x and you take its image in capital Y you get in the reverse direction you get a morphism of local rings okay and you get one morphism like this gives you a whole bunch of morphisms of local rings going in the other direction okay one for each point here and its image there right. So you have this situation and now you can see that it is rather easy to see that you know if f is an isomorphism f is an isomorphism then you also have a map going in the other direction okay and that map will induce a map in this direction now and you will see that if f is an isomorphism then all these maps on local rings are all isomorphisms okay. So in fact I must tell you that all these ring maps that we are talking about here are not just ring maps they are in fact K algebra homomorphisms okay they are their identity on K and they are K linear okay so it is clearly if f is an isomorphism then f hash y is an isomorphism for all small y in capital Y so you have this this is very clear okay because you just have to f induces f hash y in this direction and then f inverse will induce a map in this direction which will be f inverse hash x and you will have f inverse hash x will be the inverse of f hash y okay so let me write that since f inverse hash x will be just f hash y inverse you will get this okay so because you know from this diagram if f is an isomorphism g pulls back to f hash g under f then f hash g pulls back to g under f inverse okay it is quite obvious so but the beautiful thing about local rings is that you can decide whether something is a morphism using the fact that it is a homomorphism it is a morphism which is a homomorphism okay that is a topological isomorphism plus it induces at the local rings isomorphisms so that is the big fact so here is the theorem the theorem is a morphism f from x to y of varieties is an isomorphism if and only if f is a homeomorphism and f hash y is an isomorphism for all y and y so here is a very powerful theorem that tells you about the power of local rings so what it tells you is that if you want to check a morphism is an isomorphism then you first check it is a topological isomorphism okay then check at the local rings the isomorphisms induced at the level of local rings they are all isomorphisms then the morphism is an isomorphism so for the proof of this you see one way one way we already discussed that if f is an isomorphism then of course f is a homeomorphism and f y hash is an isomorphism for every y that is something that we already seen okay you have to prove the other way around okay so one way is already done so assume that f is a homeomorphism f is a morphism which is a homeomorphism and f hash y is an isomorphism for all y and y okay so what do I have to prove I have just have to prove that f is an isomorphism which means I have to show that f inverse is a morphism that is all okay how do I show f is an isomorphism just by showing that f has an inverse but f already has a set theoretic inverse because it is a homeomorphism f inverse is a continuous map f has a the continuous map f inverse okay which is also a homeomorphism I just have to show that this f inverse is a morphism once I am done once I once I prove that I am done okay so we only have to show f inverse is a morphism but you see f inverse is a homeomorphism so f inverse is already continuous okay so the only thing you have to check is that it pulls back regular functions to regular functions okay how do you check something is a morphism you have to check two conditions one condition is that it is continuous the second condition is that you have to prove show that it pulls back regular functions to regular functions so but f inverse is already f is already a homeomorphism so f inverse is also a homeomorphism so it is continuous the only thing I have to therefore check is that f inverse pulls back regular functions functions. So since f inverse is already continuous you only have to show f inverse is f inverse pulls back regular functions to regular functions I just have to show this ok. So uhhh so what do I do to do that so you know I have this uhhh uhhh uhhh so you know I have the situation is like this uhhh so here is x here is f inverse and here is y I just f inverse already of course is a homeomorphism I have to show it pulls back regular functions to regular functions. So what I will do I will take I will take a I will take an open subset u u here uhhh open subset uhhh of course whenever I talk about open subset I am uhhh I am not mentioning it but I am only interested in non-empty open subsets ok. So then I take uhhh I take f inverse inverse of u which is actually f of u which is v and this is an open subset here because after all f is an open map f is a homeomorphism so it is an open map so f of u is an open subset so. So f inverse inverse of u is just f u which is just v and uhhh well f takes u to v so I have I have two arrows one arrow this direction uhhh which is an isomorphism uhhh topological isomorphism which is f restricted to u and I have map like this which is f inverse restricted to v uhhh ok I have this what do I have to show I have to show that start with the regular function here I have to show that uhhh f in uhhh uhhh f inverse pulls back a regular function to a regular function. So I will have to look at uhhh o u which is a regular these are regular functions on u and from here uhhh I will get o v these are regular functions on v and this is f inverse hash ok it is I have to check that this f inverse hash which is pulling pulling back regular functions uhhh by f inverse uhhh I actually show that this has to have to show that this lands into this ok. So uhhh so uhhh so let me do something uhhh if I write like this then I am already assuming that it is landing inside uhhh o v so I have to do something I will have to write uhhh uhhh o so I will I will just write uhhh uhhh maps uhhh uhhh from uhhh v to k ok and uhhh mind you I start with uhhh uhhh uhhh a h which is regular function on u ok. So I have h like this uhhh it is a morphism into a1 ok and what I will have to do is the pull back of h under f inverse hash is just composition of h with f inverse ok. So I uhhh so what I will get is I will I will just have to apply uhhh I will have to apply f inverse to v I have to take f inverse uhhh restricted to v I have to first apply that and that will take me from here to here and then I have to apply h this is what it goes to ok. And uhhh now so this is so this is so the composition is this so it is a it is a map like this this is f inverse hash uhhh of h ok it is just this f inverse followed by this h. I pull back the regular function h to a function on v and I will have to show that this function on v this function from v with values in k is not just a function I have to show it is regular. So the first the first thing that I want you to notice is that it is not just a map from v to k it is actually it actually goes into the subset of uhhh maps continuous for the Zariski topology from v to a1 suddenly it is a continuous map because it is a composition of f inverse which is a continuous map and uhhh uhhh the map h uhhh and the map h is continuous because mind you uhhh any regular function is continuous after a regular function of the morphism into a1 and a morphism is always continuous ok. And so this is a composition of continuous function so it is continuous right but what I have to show so so this is not just here but it actually belongs here but what I have to show is that it actually belongs even in ov this is what I have to show I have to show that this is actually here I have to show it is regular alright and that is a smaller subset not every continuous map is a regular function ok. So uhhh but then the point is how do I check something is regular how do I check a map is regular see the notion of regular is something that is very local ok so to check that something is regular I just have to check it at every point right so you know uhhh so I will have to just check that this function is regular at each point of v that is all I have to check right and so for that we do the following thing you know uhhh uhhh so now comes the this business about uhhh uhhh so now comes this now I will have to use this hypothesis that f inverse uhhh hash uhhh I mean this f y hash are all isomorphisms for every small y in capital Y I will have to use this. So what I will do is I will start with the I will start with the y in v so uhhh I have uhhh so I have uhhh uhhh you start with the small y in v and you take uhhh small x in u with f of x uhhh is equal to y which is same as saying uhhh uhhh x is f inverse y ok. So uhhh small x small y goes to small x and this y is f x f of small x ok and uhhh and of course I have chosen y in v. So you know I have I have this o v is regular functions of v then I have I have this o y y which is same as o v y because as I told you uhhh that the local ring does not change if you go to a smaller open set alright and you know this is the regular functions are sitting inside the local ring of course ok and uhhh on the other hand uhhh uhhh I have also uhhh o o x x o capital x small x which is a local ring of capital x small x is the same as o u x because after all u is a smaller open set which contains a point small x and the local ring does not change if you go to a smaller open set and this is also contained inside this ok and the point is that this f inverse hash induces f inverse hash x uhhh uhhh yeah. So f inverse uhhh uhhh so I will have to I still have to say that f inverse is uhhh uhhh uhhh yeah. So I need to I need to do something here so I will not use this I will rather use the map that is in the other direction which is given to me what is given me given to me is that f y hash is an isomorphism. So you know so there is a map like this which is f y hash ok and this is an isomorphism this is given it is given that this is an isomorphism alright it is given that this is an isomorphism and mind you uhhh this f y hash is induced by f which is actually going in this direction ok. So uhhh it is given to me that this uhhh uhhh you know this f y hash is an isomorphism for every small y ok I have to use that. So well uhhh so what do I do I do the following thing so you know see this this h is a regular function on u ok. Since h is a regular function on u it gives me uhhh germ at the point small x namely what is the germ it the germ is just you take the equivalence class u, h uhhh you take the pair u, h and take its equivalence class that is an element here ok. This is how you have a map from the regular functions on an open set to the local ring at a point on that set namely you take this pair consisting of the regular function and that open set and take the equivalence class and such equivalence classes uhhh up to equality on intersections uhhh is what constitutes local ring at the point ok. So I I have this but then you know f hash y is an isomorphism so this this comes from something here and that is here that is in the local ring. Therefore what it means is that uhhh so let me write this down uhhh since uhhh f hash y is an isomorphism there exists uhhh a germ here which goes to that and how do how is a germ here represented a germ here is represented by a regular function on an open subset of y ok. So uhhh so what I have is well I have uhhh that exists uhhh w, uhhh let me use some other symbol I am not I am not use g let me use g that exists w, g which is a germ an element of the local ring of capital Y at small y uhhh uhhh uhhh which is of course the same as o if you want o v y uhhh so you know that since w can I can even think of w as it inside v uhhh such that in fact this is unique because it is an isomorphism such that uhhh f hash y of uhhh this w, g is uhhh uhhh is this class u, h ok. So this this happens this because this has a pre-image here this is an isomorphism ok. So what I get is I have I have a w, g here which goes to this under f hash y because under f and that w, g is of course a unique germ because it is it is uhhh because a f hash y is an isomorphism KLG by isomorphism. Now now what you will have to notice is that uhhh uhhh see if if you if you now look at uhhh this g g is defined on w and this w is uhhh uhhh an open see after all w is an open subset of y and uhhh in fact uhhh uhhh it it it contains the point small y so it all you will also intersect v you could even think of w as an open subset of v but in any case if you want to think of it as an open subset of y you can always intersect it with v. So what happens is that you have you have o of uhhh uhhh w uhhh intersection uhhh v you have this ok and you have you have uhhh you have this fellow here namely you have uhhh uhhh you have this g g is a regular function there ok and what is happening is that uhhh well you know if I take if I take uhhh if I take the image of this under f inverse ok then I will get uhhh I will get f inverse of w intersection v ok uhhh uhhh and I will get o of this and this will be f hash I have this f hash ok this f hash uhhh you see f is in this direction so f hash will be in this direction. So f will f will go from f inverse w intersection v to w intersection v ok so f hash will be pulled back of regular functions in the opposite direction namely a regular function on w intersection v goes to a regular function on f inverse of w intersection right and uhhh so you know if you take if you take if you take this g here what will happen is that I will get f hash g that will be here so this f hash g will be a regular function on f inverse w intersection v and uhhh uhhh and of course uhhh f inverse of w intersection v will intersect u anyway because uhhh uhhh x is x is in w intersection v and x is also in u ok and the point is that uhhh see this it is this f hash which actually induces f hash y because you should remember that the morphism the the morph the the k algebra homomorphism the local rings are induced from the given morphism by pull back of regular functions. So the point is that you have a competitive diagram like this so you have if you go to the local ring of x at small x and you go to local ring of y at small y then this is sitting inside this this is sitting inside this and you have this you have this f hash y and this f hash y is precisely going to take see the germ of f hash g uhhh uhhh see to whatever it is going to go to ok that is the same as taking the germ here and applying f hash y ok. But if you know if I if I take the germ here g is going to uhhh w uhhh it is just going to this germ ok see it is just going to the germ of w, g ok and that so so maybe I will write little below so that I can have more space well this guy is really going to go here to to uhhh germ of w, g and this fellow by our uhhh by our definition of f hash y is going to go to uhhh well uhhh u, h germ of u, h and that is precisely what this f hash is f hash of g has to go to ok f hash of g has to go to u, h alright. So uhhh so the moral of the story is that uhhh you see f hash of g uhhh f hash of g and see but f hash of g uhhh will go to what? f hash of g will go to w intersection v intersection u uhhh, so let so let me write that down uhhh correctly. So, what you will get here is uhhh so let me have some more space to write. So, what is this going to go to is just going to go to uhhh if you want w uhhh f inverse w intersection f inverse of w intersection v, f hash of g uhhh and that is the same as uhhh this fellow here uhhh u, h as germs because this is what goes to that okay and you have something like this this diagram commutes okay. Now so so the moral of the story is that this f hash g and h coincide when are two germs the same when when are two germs the same when are the germs of two pairs the same by the definition of the local ring two pairs define the same germ if the functions coincide on the on an intersection. So, so what this tells you is that f hash of g coincides with with h in a neighbourhood of uhhh y of of x okay f hash of g coincides with h in a neighbourhood of x that is that is because these two germs are the same alright. But then uhhh but what is f hash of g? F hash of g is just uhhh you know uhhh you apply uhhh you first apply uhhh f and then you apply g. So, f hash of g is just uhhh g circle f uhhh g circle of course f restricted to uhhh f inverse uhhh v intersection w this is what f hash of g is f hash of g is just uhhh the pullback of the function g. So, g is a regular function on w intersection v and uhhh f takes f inverse w intersection v to w intersection v and you first apply f then you apply g the composition is precisely f hash of g. So, what you are saying is that uhhh g circle f restricted to f inverse w intersection v coincides with h but that is the same as saying that you know uhhh because f is a homeomorphism it is the same as saying that uhhh g coincides with uhhh h circle f inverse v in a neighbourhood of y. So, this so so so this implies that g coincides with uhhh h h circle f inverse restricted to v in a neighbourhood of y because because you know f uhhh f is a homeomorphism on whichever neighbourhood f hash g coincides with h if you apply uhhh uhhh if you apply uhhh f inverse on the right you will get that g will coincide with h restricted h circle f inverse v. So, you see what is f hash g f hash g is g circle f restricted to f inverse w into f f restricted to f inverse of w intersection v okay that coincides with h okay then if you apply if you apply f inverse on the right you will get g coincides with h circle f inverse restricted to v in a neighbourhood of y because you just apply f inverse okay. So, so what have we proved what we have proved is that this h circle f inverse v in a neighbourhood of the point small y is equal to a g but what is g g is a regular function. So, what you have proved is that this function this function that is defined on whole of on the whole of v is at every point equal to a regular function that means it is locally regular and a function that is locally regular is regular because the definition of regularity is local okay therefore we are done. So, this implies that uhhh uhhh h circle f inverse v is actually in ov okay. So, uhhh so that ends the proof that ends the proof. So, so let me repeat what we did is we started with a regular function h on u and then we pulled it back we wanted to show that this is a regular function on v okay but what we end up proving is that for every point small y of capital v in any in any if you give me any point small y of capital v there is a neighbourhood where this function is equal to a regular function that means this function is locally a regular function a function which is locally a regular function is regular because the definition of regularity is local this function is already continuous okay. So, we have proved that f inverse hash pulls back regular functions regular functions we already know it is continuous therefore f inverse is a morphism and we are done therefore f is an isomorphism okay. So, what you must understand is that uhhh uhhh it is uhhh you must realize that uhhh the uhhh uhhh to check that something is a morphism there are there are two steps one is to check topologically that it is a is continuous second thing is you have to check that it is uhhh it pulls back regular functions regular functions. But the point is that uhhh the checking uhhh that it pulls back regular functions regular functions can always be done even at the level of local rings that is essentially the idea. So, to check that a morphism is an isomorphism uhhh you all you all you have to check is that the morphism is topologically an isomorphism namely a homeomorphism plus you must check that it induces isomorphism at the level of local rings at all at at at every local ring okay. So, uhhh so that is the power of local rings so from now onwards whenever you see a whenever you see a morphism which is topologically a homeomorphism what is an easy way to check that it is uhhh actually an isomorphism you just check that at local rings it induces isomorphism and mind you I told you that there are morphisms which are topologically isomorphisms that is homeomorphism but for which the inverse map is not a morphism there are morphisms like that they are bad morphisms okay. So, uhhh uhhh just because a morphism is a topologically isomorphism need not mean that it will have an inverse morphism it need not be an isomorphism it may be so a morphism can be an isomorphism the topological sense between two varieties but it may not be an isomorphism in the sense of varieties may not be an isomorphism varieties. So, you can have two varieties a topologically the same topologically the same but as varieties they are very different okay okay. So, this is one uhhh important fact about local rings the next thing that I want to talk about is about uhhh uhhh the uhhh the other important fact about local rings is uhhh uhhh you know. So, you know here we have seen that uhhh somehow with the behavior of the morphism is controlled by all the uhhh uhhh ring home morphisms that it induces at the level of local rings right that is what it that is the whole idea. Now, for example the fact that this is an isomorphism is captured from the fact that all the local ring at the at the local ring level it induces isomorphism okay. So, the the behavior of morphism at the local ring level controls everything okay. Now, that is at the level of morphisms what about level of regular functions. So, you know uhhh let me make a very simple statement uhhh which uhhh total uhhh it seems total logical. So, if you if you have a rational function on a variety okay uhhh by definition a rational function on a variety is a an element of the function field it is basically it is a regular function defined on an open set. Of course, there could be a locus where it is not defined okay namely the complement there could be some close set where it is not defined. Now, uhhh when do you conclude that a rational function is a regular function. So, you know uhhh a rational function is an element here if capital X is a variety a rational function is an element here and when do I conclude that it is here okay when when do I conclude that a rational function is regular which means this is a this means uhhh what I am trying to say is I am trying to say that a rational function is actually defined everywhere okay when you say rational function a rational function is only a regular function that is defined only on an open subset it need not be defined on the whole variety okay. But if you ask the question when is a rational function uhhh uhhh actually regular the answer to that is a following to check that something here is here it is enough to check that it is here in every local ring okay. So, uhhh uhhh it is a it is a very uhhh uhhh it is a very nice result to check that a rational. So, you know the idea is like this the idea is that if you have a rational function and if it is in every local ring the fact that it is in every local ring means that you know at every point in a neighborhood of the point it can be extended to a regular function that means you can extend it to a regular function everywhere therefore it should be a global regular function. So, it should be here okay but then how does one prove it uhhh accurately using commutative algebra we will do that in the next lecture.