 Let me continue with this discussion that I ended with last time what I wanted to say is that when you take the dimension of x to be r okay so you are taking x to be an irreducible closed sub variety I mean irreducible closed subset of An and affine variety in An. So it is a common to use the word sub variety when you look at a variety inside another variety so since x is an irreducible closed subset it is a variety and An you already know is a variety so we say that x is a sub variety closed sub variety of An okay and the point is that if you start with if you take any chain strictly increasing or decreasing chain of irreducible closed subsets of x and if you take if you start with indexing it with 0 and go on up to m and if you take the maximum supremum of all those m's that is going to give you the dimension of x okay and what you must understand is if that dimension of x is r and here by dimension of x I mean the topological dimension of x this is the topological dimension of x okay that is how the topological dimension is defined if that is r then you know I will that will correspond to a strictly increasing chain that is it will start with 0 and it will go on up to r. So you know if dimension of x is r we have a maximal chain z0 properly containing z1 and so on zm zr okay and what you should understand is that z0 has to be a point okay because if z0 is not a point and it is a maximal chain I can get a contradiction by making this chain bigger by putting a singleton point in z0 okay and that will contradict the maximality of the chain so this is a maximal chain z0 has to be a point and zr has to be x okay zr has to be x. So again for the same reason if zr is not x then I can add x here and I will get a bigger chain okay so in so corresponding to this if you look at it in the if you look at the corresponding diagram in the polynomial ring namely the ring of functions and affine space what is happening is the following you have in kx1 etc xn you have ifx so this from x to this point of x that translates to an increasing sequence I of zr-1 properly containing I of zr-2 and so on and it goes on up to I of z0 this how it goes on and mind you since z0 is a point I of z0 is a maximal ideal because points correspond to maximal ideals okay and the smaller the irreducible closed subset the larger the ideal okay and therefore this point should correspond to a maximal ideal so this is a maximal ideal in fact this maximal ideal you know it is generated by x1-lambda1, x2-lambda2 and so on where lambda I are the coordinates of this point you know that already okay and then what is the what is the height of Ix if height the height of Ix is going to be you start with Ix and then you get a strictly decreasing chain of prime ideals and the height is supposed to be the one that gives you a chain of maximal chain okay and so in fact what will happen is that this will be you know p height of Ix properly containing p height of Ix-1 properly containing and so on and the smallest one will be 0 okay because 0 is a prime ideal so the smallest one has to be 0 and the fact is that since this is a maximal chain okay and this is also a maximal chain you put these two together that will be the maximal chain that you can get for the polynomial ring and the whole thing will add up to the cruel dimension of the polynomial ring so this whole thing will be this will be a maximal chain in k x1 etc xn so has length n which is a cruel dimension the cruel dimension you know of a ring is a supremum of the heights of its prime ideals okay and of course the height of the prime ideal because the height is being measured from the 0 prime ideal the ring here is an integral domain so 0 is a prime ideal so you are you are measuring you are starting from the prime ideal you are going down all the way to 0 so if the ideal becomes bigger the height becomes bigger so you can imagine the height is maximum for the maximal ideals okay so this is the maximal possible height and that is the cruel dimension of this ring and that is equal to n okay and therefore but you see if this is r this part is r then this has to be n-r so this is why we say this is what essentially tells you is that this part is corresponds to the fact that dimension topological dimension of x is r this is the part that tells you that the height of ix is n-r and this whole thing is n and this n-r with r adding up to n is what this formula says where r is the ring of polynomials in n variables and what is r mod p p is of course the ideal of x here and what is r mod p r mod p is the ideal of I mean it is a ring of functions on x okay so what you must understand is that this height ix this part this is also equal to the sorry so what I want to say so I wanted to say that this is also equal to the dimension the cruel dimension of the functions on x okay so this is exactly what is happening in this case okay. So now what I want to say is I want to also look at some special cases of sub varieties okay and but let me make one more statement here let me make one more statement here you see you know ax is what this ax is actually the polynomial ring modulo the ideal of x okay so if you give me if for example if you take this maximal chain and you go mod if you go if you go mod ix if you take the image of this whole chain in the quotient okay this is a maximal chain in the polynomial ring and this is a quotient of the polynomial ring okay if you take the image of this chain there you will get this will become 0 ix will become 0 in when you quotient out by ix and this will still be a maximal ideal in the quotient because you know given a ring and its quotient the maximal ideals in the quotient correspond to the maximal ideals in the original ring which contain the kernel. So this maximal ideal iz0 in the quotient will also correspond to a maximal ideal and you will get a from that ideal you will go to 0 okay but what is that that is precisely the height of a maximal ideal in the quotient ring but that has to be the cruel dimension of the quotient ring and that is exactly r because this length is r that is what is being reflected here this is equality that is what I want you to understand okay. If you read this whole chain mod ix in the quotient ring this will become 0 and you will get from 0 to this maximal ideal which is gotten by dividing each of these by ix that will give you a maximal chain in the quotient ring so its length has to be the cruel dimension of the quotient ring and therefore this is also equal to r because this length is r okay. So I want you to understand this okay now so let me come back to what I was looking at there are special sub varieties that we are interested in in some sense in algebraic geometry there are many theorems the many questions that are proved by just looking at the case when you are looking at the locus of a single equation the locus defined by single equation namely you look at the zeros of a single equation okay so and this is called the hyper surface case so the so many theorems in algebraic geometry can be proved by first looking at what happens to hyper surfaces. So the point is that you know from the point of view of commutative algebra why this is nice is because this corresponds to studying one equation at a time because hyper surface is supposed to be the locus given by a single equation right. So that is the importance of studying hyper surfaces right so but I will give a different definition of hyper surface so what I will say is a hyper surface is let us give a definition which comes from dimension okay so I will define a hyper surface to be an x like this whose dimension is 1 less than the dimension of the big space okay that means it has co-dimension 1 in the big space so co-dimension of a subspace in a bigger topological space is just the difference of the dimension of the bigger topological space minus and the dimension of the smaller topological space so hyper surface is so x is called a hyper surface if x has co-dimension 1 so x is called a hyper surface if dimension of x is n-1 okay of course whenever I write dimension of x I mean topological dimension okay so I am not going to keep writing the subscript top you must always remember that whenever I say dimension of a topological space it is always topological dimension that is something that you should not forget okay so and why the word hyper surface is because you know well if you are if you are in if you are in one variable then there is not much because in the one variable case you are looking at a1 and the only closed subsets are finite subsets of finite subsets of points you know that very well so hyper surface is just a if you want just a single point okay that is all you will get so it is not really anything very interesting. If you go to more than one variable if you go to two variables then you get a curve okay if you go to two variables then essentially you are in a two dimensional space and you are looking at the zeros of a one equation the dimension should come you are essentially trying to look at zeros of one equation that is what you expect to happen the dimension comes down by one so in a two dimensional space a hyper surface is a one dimensional closed irreducible closed subset and of course a one dimensional object is always called a curve so when n is equal to 2 you get a curve in two space okay when n equal to 3 you actually get a surface in three space okay and if n is greater than that you do not you no longer call it a surface you call it a hyper surface so the word hyper is reserved for n greater than 3 okay of course if n is 3 you call it is just a surface okay if n is 2 then it is actually a curve in two space okay. Now you see see there are so again this is another very important thing in algebraic geometry you can make a definition from the on the geometric side you can also make a definition on the commutative algebraic side and then the question is the natural question is are these two definitions equivalent. So if I want to define a hyper surface as co-dimension one irreducible closed subset then that is the definition on the geometric side on the other hand if you want to intuitively use the fact that something that has one dimension less than the bigger space has to be given by single equation then that will tell you that the you know the commutative algebraic definition will be you are looking at the zero locus of a single polynomial okay. So I can give a commutative algebraic definition okay that this x is a hyper surface in the commutative algebraic sense if the ideal of x is generated by a single polynomial okay so you see now I have two definitions so this is definition one then let me write let me also put it like this x is called hyper surface in the commutative algebraic sense if ideal of x is equal to is generated by single polynomial f for f in the polynomial okay. So these are two ways of defining what hyper surface should be this says hyper surface is given only by one equation the fact that you are using one equation is says that you are using only one polynomial so it is commutative algebraic and the topological or geometric idea of a hyper surface is that you are cutting it is a dimension based definition you are cutting by one okay and you can ask whether these two are the same the answer is yes the answer is yes and that the proof of that involves significant amount of commutative algebra and I will tell you what the results are that leads to the proof of that so you see suppose x is a geometric hyper surface suppose x is a geometric hyper surface so my geometric hyper surface means it is a hyper surface in a geometric sense according to this definition okay. So again let me reiterate this is what I mean by geometric hyper surface something that has dimension one less than the dimension of the ambient space right I would like to show that x is commutative algebraically also a hyper surface namely that x is defined only by one equation why is that true because of the following thing see I of x is prime of course because x is irreducible so I of x has to be prime we are only worried about the varieties okay. Then you also know that height we have just now seen the height of I of x plus the dimension of x is n right the height of I of x plus the dimension of x is n which is the dimension of the Krull dimension of the whole polynomial ring which is same as dimension of the affine space okay. So let us write that height I of x plus dimension of x is equal to n we know this alright of course we know it means it is because of this theorem this formula I have written down is a theorem if r is a finitely generated k algebra namely a quotient of the polynomial ring in n variables over a field and suppose it is quotient by an ideal is a prime ideal so that the quotient is actually an integral domain then the Krull dimension of the quotient plus the height of the prime ideal by which you gone modulo to get the quotient should add up to the Krull dimension of the ring okay that is a theorem okay so I am that is what is being used here. Now what is given is since this geometric hyper surface dimension of x is n minus 1 so dimension of x is topological dimension of x is n minus 1 will tell you that height of I x is 1 okay. Now this is a theorem in commutative algebra okay so theorem a noetherian integral domain is a unique factorization domain it is written UFD for short or sometimes it is also written as factorial ring in some books okay if and only if every prime ideal of height 1 is principal okay this is a theorem from commutative algebra okay. So what it says is you start with the ring commutative ring with 1 which is noetherian okay which means every ideal is finitely generated or the ideal satisfy ACC as a mean chain condition and assume that it is also an integral domain that means it has no zero devices it is the same as saying the zero ideal is prime then to conclude that it is a UFD unique factorization domain okay namely that there is a notion of irreducible elements prime elements and any element can be uniquely factored into a finite product of irreducible elements to certain finite powers and this factorization is unique up to permutation of the factors and up to units okay. So an example of unique factorization domain is of course the polynomial because you know polynomial cannot always be factored okay this so the condition that a noetherian integral domain is a UFD is equivalent to every prime ideal of height 1 being principal that means you take a prime ideal if it is height 1 then it has to be generated only by a single element okay this is a theorem now if you use this theorem use the fact that this Ix is prime and it is a prime ideal in this polynomial ring the polynomial ring is a UFD and this is a prime ideal it has height 1 so it is principal that means the ideal is generated by a single element okay and that element has to be irreducible mind you because of that if that element breaks up as f1 f2 then Ix will then x will break up as 0 of f1 into 0 of f1 union 0 of f2 it will so it will not be irreducible. So the fact that x is irreducible which is equivalent to the fact that Ix is prime tells you that this f which generates Ix has to be an irreducible element okay so put all this together so this will tell you that if x is equal to f where f is an irreducible polynomial and of course non constant it has to be non constant because you know if it is constant you know if it is a non zero constant then it is a unit the ideal generated by that will be the whole ring okay and the whole ring is not a prime ideal and if it is the zero constant polynomial then I if you take the zero prime ideal its height is zero because there is nothing smaller than that. So f has to be a necessarily an irreducible polynomial it has to be a non constant polynomial so what this tells you this tells you that x is committed algebraically a hypersurface okay so what this tells you is that if you require a close sub variety to be a hypersurface the geometrically namely it has to have one dimension less than the ambient space the ambient affine space then it is then it means that it has to be also commutative algebraically a hypersurface namely it has to be defined only by one equation and that one equation has to be an irreducible non constant polynomial okay. Now we can go the other way also the other way is probably a little easier conversely let x be a hypersurface in the commutative algebraic sense start with this so ideal of x is f it is defined by hypersurface in the commutative algebraic sense means it is defined by a single equation. So ideal of x is f I will again reiterate that f has to be an irreducible polynomial if f is equal to f1, f2 then x will become x which is 0 of f will become 0 of f1, f2 which is actually going to be 0 of f1 union 0 of f2 and x irreducible will tell you that this cannot happen one of them has to be the whole space I mean it has to be x itself. So x irreducible will imply that f is irreducible x irreducible as a topological subset will imply f is irreducible as a polynomial okay. Now what do I want to prove I want to prove that x is geometrically hypersurface that means I have to show that it has height I mean it has dimension n-1 what is dimension of x again you use this formula dimension of x is equal to n-height of ix okay. So now we need another deep theorem another deep in the sense rather fundamental theorem it is called the Krull's howped ideal sorts okay it is called the Krull's principle ideal theorem okay. So let me state that theorem Krull's principle ideal theorem it is also written as howped ideal sorts and what does it say it says let if an element f in a commutative ring with 1 is neither a zero divisor nor a unit then every minimal prime ideal containing f has height 1 I just say that the commutative ring is assumed noetherian is assumed let me put that for safety sake okay. So this is another important thing algebraic geometry the fact is that the most sophisticated part of algebraic geometry is supposed to work over any ring not even over noetherian rings but at least when you are doing decent amount of geometry you really want to work only with noetherian rings because you get noetherian decomposition for example okay. So for example you know the noetherian decomposition in affine space which told you that any algebraic set can be decomposed uniquely into a finite union of affine closed sub varieties which are unique if the decomposing is not redundant that is no such closed sub varieties containing any other in the decomposition this comes out of the noetherian property of the topological space okay and therefore it is usual that you always work with at least in the first course in algebraic geometry you always work only with noetherian rings so it is harmless to assume things in noetherian of course it is a matter of technical expertise to see which of the theorems will still go through if you remove the noetherian hypothesis okay. But the point I want is to say is that look at what this says see the beauty with combative algebra is that if you translate it to algebraic geometry it actually has a meaning okay and it is as follows see at least in this case see you see if so what it says is you take a combative ring with one assume it is noetherian if you want take an element f okay assume the element is not a zero divisor okay and assume it is not a unit right. Then you look at the ideal generated by that single element f okay and you can talk about the minimal you can talk about prime ideals which contain that element f there are such in fact any non-trivial ideal in a combative ring non-zero combative ring is always contained in a maximal ideal any proper ideal is always contained in a maximal ideal this is if you want consequence of zon's lemma okay. But therefore you know if you take the ideal generated by f it is a proper ideal it is a proper ideal because f is not a unit alright and the ideal generated by f contains certainly is contained in a maximal ideal more generally you can look at prime ideals also which contain the ideal f ideal generated by f and what the theorem says is that if you look at any minimal prime smallest among these set of primes which contain f the height of such a prime is 1 okay. So it is what it says is that you take a minimal prime containing f then between then from that prime that prime has height 1 okay between that and the smallest possible prime okay f is caught f is caught there in between these 2 okay of course the smallest possible prime will be the 0 ideal if the ring is an integral domain okay in which case you are saying that the ideal generated by f is caught between the 0 ideal and the smallest prime which contains f each of the smallest primes which contain f. So there are 2 facts I want to tell you let us first apply it to our situation in our situation our commutative ring is of course the polynomial ring in n variables okay and here is and the element f is a single non-constant irreducible polynomial okay if you will take the ideal generated by f that is already a prime ideal mind you the ideal generated by a single element is always a prime ideal in fact the truth is that if you take any unique factorization domain and you take irreducible element there is an element which cannot be factored into smaller elements okay into non trivial factors which we tend to call as smaller elements okay smaller factors such an irreducible element if you take the ideal generated by that that will be prime okay. So since the polynomial ring is unique factorization domain and since you have started with an irreducible polynomial which is an irreducible element in a unique factorization domain the ideal generated by that element will be a will be a prime ideal so this is certainly a prime ideal. So if you look at this any minimal prime ideal which contains this it has to be this itself if when you look at the minimal prime ideal generated by an element f it will be different from f it will be different from the ideal generated by f only if the ideal generated by f is not a prime ideal. If the ideal generated by f is a prime ideal then the minimal prime ideal which contains ideal generated by f is ideal generated by f itself. So if you apply Krull's principle ideal theorem what will tell you is that the height of f is 1, the height of ideal generated by f is 1 but the height of ideal generated by f is the same as height of ix so it will tell you, so this will tell you that height that dimension of x is actually n minus 1 which will tell you that x is geometrically a hypersurface it will tell you x is a geometric hypersurface okay. So what you have got is that if you start with a hypersurface in a commutative algebraic sense you get a hypersurface in a geometric sense and you already seen the other way if you start with a geometric hypersurface then it is a hypersurface in a commutative algebraic sense. So both these definitions will say this definition is completely geometric it is gotten by saying that it is one dimension less it is co-dimension one. This definition is commutative algebraic you are saying you are looking at only zeros of one equation and they are one and the same okay this is again I mean this is what you should always appreciate there is something going on here which has complete translation in this side okay and of course these two theorems here which come into the picture they are very very important and at some point if you have not already seen them in a course in commutative algebra you can you should make it a point to set aside some time for extra reading and if you can when you can do that try to look at a sketch of a proof of theorems like this. But what I want to tell you is that there is geometric significance. So for example suppose f is not so let me explain more generally what the statement is saying suppose f is not an irreducible polynomial okay what does this statement say it has a geometric meaning what is it it is the following let me write that let me try to explain that you see suppose f in the polynomial ring is non-constant and f is equal to let us write f1 f2 etc fm be it is factorization unique factorization okay with each fi irreducible. So this is again the fact that any polynomial can be if it is not irreducible you can break it down into a product of factors each one of which is irreducible and of course I am writing it like this but there could be some factors could repeat okay. So you know let me write it so you know maybe I should put powers to be very accurate so you know if I will have to put something like n so n1 f1 power n1 f2 power n2 fm power n sub n. So if I write it like this then you know I mean that no fi is the same as any other fj okay and these you know these powers are all uniquely determined this is just like the fundamental theorem of arithmetic where you say any integer can be uniquely factored into product of prime powers the primes occurring are unique and the powers of each prime that occur are unique and it is the same thing that is happening in the polynomial ring in several variables okay and that is why it is a unique factorization. Now of course here I can always push in or pull out a constant so this factorization is unique up to a unit which is a non-zero element of k okay. Now if you look at the 0 set of f okay then this you know it is just going to be 0 set of f1 union 0 set of f2 union 0 set of fm this is what it is going to be this is what you are going to get right. Because 0 set of f will be 0 set of f1 power n1 union 0 set of f2 power n2 and you must understand the 0 set of a power of f is the same as this same as 0 set of f itself because the set of points where a power of a polynomial vanishes is the same as set of points where the polynomial vanishes. So when I go to the 0 set all these powers are gone I do not care about the powers. Now if you watch look at each zfi each zfi is irreducible as a subset it is an irreducible closed subset why because each fi is already an irreducible polynomial and you know since each fi is an irreducible polynomial and it is an irreducible element in the polynomial ring which is a uft the ideal generated by an irreducible element is a prime ideal. So the ideal generated by each fi is a prime ideal and the 0 set of that prime ideal is therefore an irreducible subset okay therefore each zfi is irreducible and what and of course no zfi is contained in some other zfj that is because no fi divides any other fj they are all distinct irreducible polynomials okay that is what unifacturization means when you write into product of factors the factors are not repeating certainly okay it is only to take care of repetitions that you put the powers right. Now watch these are all irreducible closed subsets okay if you look at this what is this this is actually the noetherian decomposition of f this the way I have written it this is the noetherian decomposition of f mind you z of f is an algebraic set z of f is not a it is not an irreducible algebraic set because f is not ideal generated by f is not a prime ideal that is because f is not irreducible alright and by and you know the noetherian decomposition is unique noetherian decomposition says that if you have a noetherian topological space you have a closed subset then the closed subset can be written as a finite union of irreducible closed subsets and this decomposition is unique if you assume that none of these subsets is contained in any of the others okay. So this is a noetherian decomposition of course up to a permutation of the of this z f is okay if you watch if you take the if you take the if you go to the go to this case go to what we have proved so far that geometric hypersurface is the same as hypersurface in the commutative algebraic sense what will tell you is that each of these has dimension in minus one each of these is a hypersurface. So what it will tell you is z of f is a union of hypersurface z of f is a union of hypersurface okay it is a union of hypersurface and if you look at let me let me let me look at the following let me do the let me give a tentative definition how did I define the how did I define the coordinate ring I mean the ring of functions on a closed subset I simply defined it as the ring of functions on affine space modulo the ideal of that set okay. Now what I will do is I will just I will just put a of z of f okay so as polynomial ring modulo the ideal generated by f make this definition okay make this definition this is not a very good definition for the reason that since the ideal generated by f is not prime you are going the ideal modulo which you are going is not a prime therefore this crazy thing is not an integral domain this quotient ring is not an integral domain. So for example in this quotient ring f1 f you see f1 bar f2 bar fm bar which are the images of f's the fi's in this quotient you see they are if you raise them to the power these powers and multiply them you will get 0 but individually they are not 0 they are so each fi if the image of each fi here is a 0 device at mind you and but the point is in this ring see in the if you look at so let me write that this is a quotient of k x1 through xn okay. Now take a prime ideal p which contains f okay a prime ideal p contains f if you go down here okay it will give rise to a prime ideal p bar which will contain 0 okay it will be a prime ideal p bar which will contain 0 what you should understand is that for each of these prime ideals I can take the ideal generated by fi's okay you take the since fi divides f the ideal generated by fi will be multiples of fi and f is also multiple of fi so ideal generated by fi will contain the ideal generated by f okay anything which is a multiple of f is also multiple of any fi. So I can take a prime ideal which contains fi okay that will correspond to the prime ideal generated by fi bar in the quotient okay and the fact is that these fi's they will be the smallest prime ideals which contain f that is because of these unique factorization you will have to do you have to convince yourself that the smallest prime ideals which contain f in this ring are precisely the fi's okay and what does Kuhl's principal ideal theorem says it says that in the polynomial ring itself the smallest prime ideals which contain this f have height 1. In other words what it says is if you if you commutative algebraically look at only 0 of a single equation but do not insist that the 0 set is irreducible you will not get a geometric hypersurface but you will get a union of geometric hypersurfaces that is what it says that is the that is the full content of this theorem okay. See you can ask this question right a commutative algebraic surface is if you want to just define it as a surface which is given by a single equation since I already want something that is irreducible that single equation has to be irreducible but if I relax the condition that the that the the locus is not irreducible then it is just an algebraic set. So you are looking at single 0 you are looking at the 0 locus of single polynomial the polynomial is not necessarily irreducible then what it what this says is if it is irreducible then it is a hypersurface if it is not irreducible it is a union of hypersurface that is what it says that is why that should tell you why the statement of this theorem involves the minimal prime ideals which contain f you see they become relevant in this non irreducible case. The minimal prime ideals that contain f they correspond to the ideals that correspond to the irreducible components of the 0 set of f that is the that is the connection the geometric connection to the statement okay and the importance with the importance with this theorem is that you know you can ask more generally this question. So this is also part of algebraic geometry you have something nice happening okay for us we have always started with the polynomial ring n variables but you can work with more general rings if you work with more general rings you can ask the question when will this be true you start with a geometric hypersurface is the same as a hypersurface in the commutative algebraic sense for what kind of spaces will it be true based on the ring of functions on those spaces and that is the answer given by that is the geometric content of this theorem what it says is if your space has a ring of functions which is a unique factorization domain okay if your space is such that it is ring of functions the unique factorization domain there is no difference between a commutative algebraic hypersurface and the geometric hypersurface that is what it says that is the geometric content of this theorem okay. So what you should understand is this is the point about algebraic geometry you have some statements which are completely statements in commutative algebra but if you translate them they translate into something very geometric. So you know how to define what a unique factorization domain is in a commutative algebraic sense it is a you know it is an integral domain in which you have unique factorization every element can be written as a product of powers of irreducible elements in a unique way okay that is what a unique factorization domain is this is a commutative algebraic definition but what does it geometrically mean? So geometrically you can say if you geometrically you always think of rings as rings of functions on some space so algebraic geometrically how to define a unique factorization domain one way is you say I mean look at all you can ring of functions if you want to think of a ring of functions is a UFD then the space must have the property that the geometric hypersurface should be the same as the hypersurface in the commutative algebraic sense it is for those spaces that the rings of functions can be called unique factorization domains okay. So what you must understand is that this unique factorization which is a very you know completely algebraic statement it is a purely commutative algebraic kind of statement that has the geometric significance that define low psi defined by single equations are the same as low psi which have co-dimension one you see that is the that is how you geometrically interpret the completely algebraic definition of what a unique factorization domain is okay. See the whole beauty of algebraic geometry lies in this you take something completely commutative algebraic completely see what it means geometrically and you do the other way also. So this is an example as to how you can make this translation okay so in my next lecture what I will do is I still have to explain how the inverse of this A function is the max spec function so I will have to do that so I will do that in the next lecture.