 Okay, so what we will do now is continue with the importance, continue discussing the importance of local rings. So again you know just to recall we started the variety capital X and the point small x, then you have the local ring of capital X at small x which contains the ring of regular functions, global regular functions namely regular functions that are defined on all of capital X and which in turn is contained in the function field of X, okay which turns out to be the quotient field of the local ring as well. And now the local ring so the point is that local rings contain a lot of information that is the idea. So what we saw in the last lecture is how local rings capture isomorphisms. Of course an isomorphism induces isomorphism at the level of local rings and conversely if a morphism induces isomorphism at the level of local rings and it is a homeomorphism then it is an isomorphism, okay that is what we saw. Now what I am going to show is I am going to discuss local rings and regular functions I am going to say that if you take an element here namely an element which is a rational function that means if the function is regular on an open set and suppose this element is here for every local ring then it is actually defined on all of X, so it is a regular function on all of X. So it is very clear that if you take a regular function on all of X then regular function is of course a rational function because after all for by definition a rational function is just a regular function defined on an open set and since if I take a global regular function is defined on all of X which is also an open set. So any regular function is always a rational function and any regular function is going to live in every local ring namely that regular function is going to give you its germ in each local ring. So something here is going to be in each local ring and of course it is going to be here as well but something here when is it here the condition is that this if that lives in every local ring then it is here, okay. So let me state that that is another important thing so to check that a function is defined everywhere if you can somehow check that it belongs to every local ring then it is defined everywhere, okay and the reason why I will do that is because all the local rings sit inside the quotient field, all the local rings sit inside the quotient field. So let me write that here local rings and regular functions so here is so here is a theorem so theorem is let X be a variety an element of KX that is in the sub ring OXX for all small X in capital X is actually in OX. So this is the theorem it says that you take an element here namely a rational function if it is in every local ring then it has to be a global regular function, okay. So for proof so what I want to tell you is that see if you think of it philosophically it is quite you can even think that this is obvious or even total logical, okay. Philosophically what you are saying is that you have a rational function namely function is defined on an open set but you are saying that it is equal to the germ of a regular function at every point which means you are saying that it lives in a neighborhood of every point and that is the same as saying that it is domain of definition that is a maximal open set where it is defined is all of X which means it is actually the open set where it is defined is not just a proper open subset of X but it is actually all of X which means it is a global regular function, okay. So that is the philosophy but then it involves a little bit of commutative algebra and that is what I want to tell you see the important thing is that you know trying to say that rational function is defined as a regular function in a neighborhood of a point is the statement that you know that this actually belongs to the sub ring, okay. The fact that an element here belongs to a sub ring like this is another way is actually the commutative algebraic way of saying that the rational function is actually defined in a neighborhood of the point small x, okay that is or it can be extended to a neighborhood of the point small x, okay. So you know trying to say that the rational function is defined at every point is if you want to say it commutative algebraically how do you say it you say it like this you say that the element of the rational function defines an element of kx and you are just requiring that it is in each of these sub rings I mean if it is in one such sub ring then that is equivalent to saying that it is defined in a neighborhood of the point small x, okay. So the problem with a rational function is that it could be meromorphic, okay namely it could have it could basically it could be a quotient of regular functions and it may not be defined where the denominator function is vanishing, okay and therefore it need not extend to all of x but the fact that it extends to a point is reflected by saying that it belongs to the sub ring, okay and that is the content of the theorem. So let me do the following thing so what I would first like to say is so we will use this fact namely any variety is covered by affine varieties, okay any variety is covered by open subsets and in fact finitely many of them each of which is isomorphic to an affine variety. So what I will do is I will go from x to you know I will first take the case when x is an affine variety, okay I will first solve the theorem for the case when x is an affine variety and then I will say that therefore it also solves the case when x is a general variety because a general variety is a finite union of affine varieties, okay. So first assume x is an affine variety say x equal to max spec A of x, A of x is equal to A say. So you know this is how you recover an affine variety from its co-ordinate ring. If x is an affine variety you have Ax it is affine co-ordinate ring and if you take the maximal spectrum then you get back the affine variety, okay so in fact so you know if you want Ax is sitting inside some affine space as an invisible closed subset and Ax is the co-ordinate ring of x, okay it is the it is the regular function of affine space namely polynomial ring in as many variables as a dimension of affine space modulo the ideal of the closed subset, invisible closed subset which is a prime ideal, okay. So A is an integral domain which is a finitely generated k algebra, right and now what happens is that you have so you have situation like this you have Ax which is equal to O x that is A for us, okay because if x is affine Ax is same as O x and that is A and that is sitting inside well O x x local ring, okay and mind you this local ring well I will rather write I will write it here, okay and this is sitting inside Ax this is the function field, right and how does all this translate well O x is Ax which is A and then the local ring is Am, okay where M is the maximal ideal of A that corresponds to small x. So after all small x is an element in capital X which is which can be thought of as a maximal ideal in Ax, okay which is A. So this is just Am, okay this is the expression for the local ring at a point on affine variety where the point corresponds to this maximal ideal M, okay and what about the quotient field this will be you will get Q of A you have the quotient field of A and that will be the same as the quotient field of X the function field of X, okay. So this and mind you Q of A is the same as Q of Am if you have an integral domain then the integral domain every localization of the integral domain is also an integral domain and the integral domain sits inside as a sub ring in its localizations and all the localizations have along and the integral domain itself they all have one and the same quotient field, okay. So this is A is an integral domain Am is localization of the maximal ideal M which means that you invert everything outside the maximal ideal and both A and Am have the same quotient field, okay Q of A, okay and that identifies with the function field of X, okay and now what is what is happening is that I have a so what is given to me is a rational function which so this is a little bit of simple commutative algebra but the point I want to make is that whatever commutative algebra little simple commutative algebra that we are doing is beautifully enough to give as this result, alright. So you start with so you start with the rational function here I mean you start with the rational function here by definition it is here so it is of the form f by g, okay so I start with an f by g here, okay I start with an f by g here because it is see because this rational function is an element here and this is identified with this I am easily able to write as f by g because it is the quotient field of A where f and g belong to A and what is given to me is that this f by g belongs to Am for every maximal ideal M that is what is given to me, okay so f by g is equal to well an element here, okay and an element here well let me write this as f sub M by g sub M in Am for every M maximal ideal, so max spec is the set of all maximal ideals away and for every maximal ideal I have this local ring and this f by g belongs to the local ring means the f by g is equal to an element in the local ring alright and an element in the local ring is written like this where the denominator is not in the maximal ideal, okay this is given to me what I will have to you know what I will have to do I will have to show that this f by g is actually in A, okay I have to show that this rational function I started with is actually here, okay so I have to show that f by g belongs to A which means I will have to show that you know I have to actually show that g I have to show that g divides f, okay I have to show I have to if I show that g divides f, okay then I am just that is the same as saying f by g is an element of A and that will show that f by g belongs to A, so I will have to show that g divides f alright, so we need to show f by g belongs to A that is that well g divides f in A or in other words you just have to show that f belongs to the ideal generated by g this is what we will have to show f is a multiple of g, so f by g is an element of A, okay that is what you will have to show. Now you see now let us analyze this of course you know if g is a unit there is nothing to prove g is a unit there is nothing to prove because then f by g is f by a unit, okay and f by a unit is also an element of A, okay so assume that g is not a unit, so assume g is not a unit since g is not a unit it means that you know every non-unit is contained in a maximal ideal, okay so there exist maximal ideals m which contain g, okay there exist maximal ideals m which contain g and so you know for such a maximal ideals m you look at this equation, now what you will get is you will get f gm is equal to g fm for such m implies what does this imply, so you see g is in the maximal ideal m, okay and therefore this right hand side is in the maximal ideal m therefore the left hand side is in the maximal ideal m but maximal ideal is prime and therefore one of them has to be in m but gm is not in m therefore f has to be in m, okay so this will tell you that well f will have to be in m for every m maximal which contains g you have this, okay and anyway but what is very very important is this condition even if I take an m which may or may not contain g, okay perhaps this is just an observation that comes from this equation but what is more important is this condition that for every maximal ideal I am getting an element outside that maximal ideal and the beautiful thing about that is that you know whenever you give me one element out of each maximal ideal then this collection of elements actually generates the unit ideal, okay that is a crucial topological fact it is the same as saying that all the basic open sets defined by all these gms they cover the whole space, okay see now so let us not worry about this observation for the moment, okay let us look at this see you take gm does not for every for every m in max spec a which is actually x there exists gm which is not in m that is what you have here, alright and now the claim is the set of all you take all these gms where m belongs to m is a maximal ideal of a the ideal generated by this is actually a it is a unit ideal, okay and the answer is I mean why this is true is very very simple because you see if you take the ideal generated by all these gms the ideal generated by gms if it is not a if it is not the unit ideal then it is a proper ideal, alright some an ideal which is not the unit ideal is a proper ideal and you know any proper ideal is always contained in a maximal ideal therefore this ideal will then be contained in an m knot, okay but then gm knot will be in this ideal which is contained in m knot but by definition gm knot is not supposed to contain to m so not supposed to contain be contained in m knot that is a contradiction, okay so let me repeat it if this is not a then this is contained in an m knot where m knot is a maximal ideal, okay and then the problem is that gm knot is in this collection and that will be contained in this which is contained in m knot but gm knot is not supposed to contain be contained in m knot you get a contradiction therefore this ideal generated by all these gms is a, okay this is this is this is just equivalent to so let me also say that this is equivalent to x is given by the union of all the d gms where m belong to max by k because what you should understand is you see you see if gm is not if gm is not in m, okay you look at d gm what is d gm d gm is the basic affine open set where the function gm does not vanish, okay so it is same as saying that the point corresponding to m is in the basic open set defined by gm, okay so but I have covered all the points I have taken all the maximal ideals so I have covered all the points and for each point I have this d gm so you know if you draw a picture this condition is actually like this I mean it is a bad picture which is very only set theoretically descriptive but it is not correct picture but it is correct enough to think a little so you have to give you some basic idea of how things are so you have x given any point m, okay you are thinking of a maximal ideal in A as a point of x, okay and for this every point you are able to find this d gm which is an open subset which contains that point, alright that is what it means to say that gm is not in m that is what it geometrically means and since you have covered every such point the union of all these d gms will be x, okay but then you know Zariski topology is quasi-compact therefore only finitely many gms have to cover x but the fact that finitely many gms cover x is the same as saying that those finitely many gms will generate the unit ideally, okay so which is if and only if x is equal to union of i is equal to 1 to l d gmi so here I will have those g it is ideal generated by d gm1 etc up to d I mean up to g gms of l. So if something is a unit ideal then 1 can be generated by finitely many gms call those gms as gm1 through gml then the ideal generated by gm1 through gml contains 1 so it is the full ideal a but then writing a is equal to the ideal generated by gm1 through gml is topological equivalent to writing max spec of a which is x is union of all the affine opens defined by these gmis that is what is the topology that is involved, okay so you can see side by side algebra and topology going hand in hand for the Zariski topology now if you look at it carefully what you get is well therefore what you get is therefore there exist fmi so let me use something hmi in a such that sigma hmi gmi is equal to 1 i equal to 1 to l this is what it means to say that the gmis generate 1, okay that means ring linear combination of the gmis is equal to 1 for a suitable ring linear combination, okay. Now you know but then you see now I can use this I can now use now I can take f by g is equal to f by g times 1 where you know I am doing all this computation here in the quotient field, okay whatever equations I am writing now are valid in the quotient field so f by g is f by g times 1 and the 1 the 1 in the a is the same as the 1 in the am is the same as the 1 in the quotient field, okay so this makes sense also in the quotient field so it is f by g sigma i is equal to 1 to l hmi gmi because after all this is 1 now I push it inside I will get sigma i is equal to 1 to l f by g times hmi g sub mi but you see I have f gmi is g fmi, alright so this is going to be sigma so you know if I can write a sigma i equal to 1 to l f gmi hmi by g but that is equal to sigma i equal to 1 to l f gmi is actually g fmi by g and well g cancels out and what you get is I get sigma i equal to 1 to l fmi hmi which is actually in a this is an element of a, okay so what I have done is I started with f by g in kx and I proved it is actually in a x so it is actually a regular function, okay so this settles the case when x is affine, okay now when x is not affine you know that x can be covered by finitely many open sets each of which is affine and what this argument will tell you is that on each of those affines the rational function will be a regular function, okay and what will tell you is that the regular function lives on all open sets of a cover, okay and it coincides and it will of course coincide on the intersections therefore it will give a global regular function, okay so that finishes the proof in the general case, okay so let me write that down this settles the case when x is affine, okay when x is general x is a finite union of open subsets open affine subsets, okay if x is a general variety then we have seen that any general variety is a finite union of open affine subsets and what we by what we just proved the rational function, okay the rational functions will not change whether you consider rational functions of x or rational functions of any open subsets, okay therefore what you will get is that the rational function is started with actually is a regular function on each affine open sets which all of which form a cover for x and of course on the intersections also it will be it will give the same function because everything is finally happening in the quotient field which is the same, okay so basically what will happen is that you will get a regular function on each affine open subset by the previous argument and all these regular functions will coincide on the intersection then therefore they will define a global regular function after all a global regular function is a global function which is locally regular, okay so to define a regular function you have to just define it on an open cover, okay so the case when x is general is also covered, okay and so that is the general case you know if you want maybe I can even write something in symbols so that you are little bit more convinced comfortably so you know x is union i say j equal to 1 to m xi where each xi inside x is open affine which means that each xi is an affine variety each xi is an open subset of x which is isomorphic to an affine variety and then you must understand that you know kx is the same as kxi for every xi, okay and o capital X small x is the same as o capital X small xi, o capital small x capital xi, small x if small x is an xi the you know the local rings and the function fields do not change if you go to open fields, okay so what will happen is that if you start with an element of kx it will belong to and it can be thought of as an element of kxi for each i but because each xi is affine and it is in each local ring by the previous argument it will be in o xi for every i, so you will have an element of kx which is in o xi for every i but then that will mean that it is actually in o x, okay so and a phi in kx is equal to kxi belongs to o xi for every i by affine k is above and thus to o x and that is the end of the proof, okay so after all you know to define a function you have to only define functions on open sets which agree on intersections, okay and similarly to define a regular function you have to give me a regular function on an open cover which agree on the intersections and that is what happens here, okay so this is a very nice fact that to conclude that rational function is actually a global regular function that it is defined everywhere you just have to check that it belongs to every local ring, okay so that is another that demonstrates another important power of the local ring, okay then the next thing I want to tell you about is something that is far more profound it is the following thing see the you can see that somehow the very word local ring will tempt you to think that it somehow encodes all the local information at the point but the problem is local information if you want to think of it is a information on an open set containing the point you must remember the open sets are huge for the Zariski topology because the open sets are after all every open set is irreducible and dense, okay therefore if you take an open neighbourhood of a point it is not a small neighbourhood it is not small in the sense of usual topology it is not like having a smaller and smaller disk or smaller and smaller interval of a point on a real on the real line or having a smaller and smaller disk of epsilon radius of a point on the plane it is not like that every open set is huge therefore the fact that the local ring stores such information should tell you that it is actually having much more than local information so the truth is the answer to that is yes it actually also stores information on a huge subset containing the point so that is the power of the local ring so what I am going to next state is this very beautiful result it says that take two varieties X and Y okay and take a point of one variety take a point of the other variety take a local ring at this point for this variety and take a local ring at that point for the other variety if just these local rings are isomorphic okay then there is a whole open subset containing this point in this variety and an open subset containing that point in the other variety which are isomorphic so if two local rings are isomorphic if the local ring of a variety at one point just the local ring alone is isomorphic as a K algebra to the local ring of another variety at some other point then a huge open neighbourhood of this point on this variety has to be completely isomorphic to a huge open neighbourhood of the other point. So even isomorphism of local rings is actually in this sense pretty global okay it tells you that if two varieties have two points which are whose local rings are isomorphic it means that these two varieties are on an open set on open sets they are actually isomorphic that means the only place where they are not isomorphic is a boundary okay so that is another powerful property of a local ring. So let me state that so local rings so I will call this local rings and birationality so local rings and birationality okay so let me define this notion of birationality. The two varieties are called birational if there is a non-empty open subset of each which are isomorphic okay so X varieties X and Y are called birational if there exists U open in X of course non-empty and V open in Y of course again non-empty such that we have an isomorphism of varieties U to V. So two varieties are birational if there is an open subset of one which is isomorphic to an open subset of the other see the varieties themselves need not be isomorphic but the point is there is a huge open set of this of each of these which are which look the same okay that is the concept of birationality. Now what you must understand is that this actually this notion of birationality is actually an equivalence because you know if X is if X and Y are birational that is of course symmetric okay X is birational to Y is the same as Y is birational to X and of course an isomorphism is always if two varieties are isomorphic they are of course birational because you can choose the open sets to be the whole varieties themselves. So the notion of birationality actually it is a relation which is an equivalence relation if X is birational to Y then Y is birational to X any variety X is always birational to itself by the identity map which is an isomorphism okay and if X is birational to Y and Y is birational to Z then X is birational to Z okay because any two open sets will always intersect okay so if you have an isomorphism if U inside X is isomorphic to V inside Y and V inside Y is isomorphic to W inside Z then you can get the isomorphism composed in on V intersection V prime which will be non-empty okay therefore this birationality of varieties is an equivalence relation okay and if you go if you take the set of all varieties if you take the category of all varieties and then you go modulo this equivalence you get what is called the set of birational equivalence classes of varieties and actually it is the first step in classification of varieties is to find in each birational equivalence class a variety with say good properties for example a variety which is smooth okay and so you know basically whenever we study objects in mathematics we always ask the question how many non-isomorphic objects are there okay so or in other words you try to describe the set of isomorphism classes of objects okay this is called the classification problem so you can ask the classification question for varieties you can ask you take all varieties and you go modulo isomorphism then you get the set of isomorphism classes of varieties what is that set okay now the fact is that that is not a very easy question okay and in fact the first answer is to instead of going modulo isomorphism which is very strong you try to first go modulo weaker equivalence relation like birationality so you first ask the question what is the set of birational equivalence classes of varieties okay that brings down the problem to a more tacklable level okay so so birationality is very important in the classification of varieties so let me write this down birationality it is easy for you to check that is an equivalence relation and now comes the theorem that I wanted to state theorem is let x belong to x small x be a point in capital X small y be a point in capital Y such that of course x and y are varieties OXX the local ring of capital X at small x is isomorphic to OYY isomorphism as K-algebra so I just have two varieties capital X and capital Y and points small x belonging to capital X small y belonging to capital Y such that the local ring of capital X at small x is isomorphic to the local ring of capital Y at small Y by an isomorphism which is a K-algebra isomorphism okay then the result is that there is an open set U in X in capital X which contains the point small x there is an open set capital V of Y which contains the point small y and there is an isomorphism of varieties from U to V which under pull back of regular functions at the local rings induces V okay so then there exists U in X open V in Y open with X belonging to U Y belonging to V and an isomorphism of varieties so let me use so let me use a different symbol for this so let me call this something let me call this a psi phi from U to V such that phi hash Y is actually psi so it is a very very powerful theorem what it says is that if one local ring of one variety is isomorphic to another local ring of another variety then these two varieties are birational they are the same on huge open sets they are the same okay and that tells you how much information the local ring contains just isomorphism we think of local ring as something that is concentrating attention at a point but this tells you it is much more than that just local ring at one point of one variety being isomorphic to local ring at another point of another variety is enough to say that these two varieties are the same on a huge open set except for a boundary they are isomorphic okay so that is the fact that gives you that the fact that local ring contains a lot of information even more or less global information as in the sense that it contains information on contains information on a big open set okay so one needs to prove this and again the beautiful fact is that this also involves lot of commutative algebra but pretty easy commutative algebra involving localizations and it is very it is proof is not very difficult but the point is that you can just use nice commutative algebra to capture these two open sets as nice basic open sets okay which are isomorphic basic open sets containing those two points which are isomorphic okay so let us do that in the next lecture.