PSE Lectures - Ep 18 - A Recap & doubt solving session over all Field Classification lectures

Transcript

yeah makes sense makes sense yeah I was just like wondering um yeah about this case but yeah yeah it's a let me just write it for a second I mean let's say you have a field F and you look at at a polom FX over f h and suppose that this polinomial already splits this means that I can write FX is Alpha times h x - alpha

1 times x - alpha n where the alpha and Alpha 1 up to alpha n are in the original field right that that's what it means for a polinomial to split if it if it is a monic polinomial then you can you can replace Alpha by by one because ammonic polinomial it means that when you multiply all the

X's I mean you have you have x to the N with coefficient one right okay so so you see if H if the polinomial splits already over F then I can write it in this form and where the the all the coeffic I mean all the the scolars alpha and Alpha 1 up to alpha n are are in the original field and then of course you can open the expression but the expression will of

course be a polinomial over the original field um simply because if all the alphas element in the field then all the multiplications and additions I mean all the operations you you can cook from the alphas are also elements in the field then if if K is a field containing F then I can simply write it just as before okay okay yeah yeah makes sense makes sense thank you and the the pro so you

you should really think of it as the the the reason that pols don't don't always split over a field is that you don't always have a root I mean like x² + 1 over over the real numbers does not have a root right and uh so so to go to I mean to split this polinomial I need to go to a larger field and in in if I think of it as a polinomial over the complex

numbers then it will it will actually split as x minus I X+ I okay so the the the problem of of splitting is that you you don't have enough roots in in the in the original field okay yeah this makes sense so so basically like the use case of splitting field is that you have a polinomial over a field and you don't have the roots in that field so you need to extend it to to have the The Roots yes okay and the way just to I mean to give you the in I

mean the splitting fields are probably the the most abstract part that we did so far but but the intuition should be fairly simple I mean it all follows from this case that you start with let's say you start with FP and you consider a prime polinomial over FP fix then we we showed that if we do FP X modul this Prime polom FP Remember

by the analogy FP X is is like the integers I mean pols over a field behave like integers in the sense that can decompose every polinomial into a product of primes and I have a gcd I have long division and so on so but in any case when I divide the by the prime polinomial I get a this this is the set of all remainders right just like when you do a

zed modulo modul n it's like the set of all remainders modul n right and so in this this polinomial over field I I divide by a prime polom as a set it's it's the set of all remainders RX equals sum a x to the i i equal 0 to K where K is a is strictly smaller than n and the AI are elements in the

field and we saw that this this we can call k k with a a addition mod and multiplication mod is is is a field and then the the the interesting phenomena is that when I consider pols over K so of course k

contains a contains a FP it contains FP because I can always choose the a a remainder polinomial to be simply a constant so if you give me some AI in in FP I can think of the AI as as a as a polom with no I mean as a constant polom right and H and so I have FP contained in K this means that that I can think of a of Fe this original polom I can think of it as

a as also a polinomial overk but since I mean just as a ation issue we already have an indeterminant with the lowercase x so if I want to disambiguate the situation I need to I mean when I talk about polinomial in K it's better to to have a another notation for the the IND determinant for the variable so I consider a um pols over K with a capital x as as the

the variable and then I can I can simply think of Fe as a polinomial in this in this setup right because Fe I mean whatever it is Fe is some sum AI x to the I then this this Fe will be the same will be Su AI but but now the the variable I changed to capital x and what we saw is that in the setup P suddenly have has a root in in K so so if it has a

root it SP I mean this means that it splits as a xus Alpha times some some other polom let's call it P ER and the whole business of splitting Fields is relying on this this metat trick I mean of course Fe originally was a prime polinomial over FP so it cannot I mean Prime polinomial do not have any root in the in the original field because if they had a root I mean I could write Fe f

x as e x minus this root times some some other polinomial PX but this cannot happen because Fe is prime right it cannot be expressed as a a non-trivial product of of a of other polinomial H so so I went to this H this construction and I got that suddenly I have a root but once I have once I can find one root I can basically repeat the the same trick ER the subtlety is that c May no

longer be prime polinomial over this bigger field K so I will split it into whatever a prime factors Prime decomposition it has over K and I will take each of the the factors of the Prime decomposition and I will I will do the same so I will get a um from FP I will get k and then from K I will get let's say K1 and I will get a finite sequence of extensions of the

field each of them guarantees that at least I I added one root to the the original polinomial and since a polinomial can only have at most n Roots I am guaranteed that this process will H will stop it after finite number of steps and this this is the splitting field ER is this clear that's kind of a high level explanation yeah yeah for me yeah this helped a lot and also clarified some gaps that I didn't know I have but yeah

yeah like this meta trick especially is kind of the hard thing to think of um but but yeah I think it's clear now for me yeah let me try to even further CL clarify by introducing a a new notion that ER you probably saw but but we didn't talk about before this is a definition ER so let F be a field um a

polinomial in two variables over f is a an expression of the form now I have a a a two two variables let's say X and Y and I can do powers of

X and multiplied by powers of Y and all that multiplied by a coefficient so the coefficient I will I will have a double index for the coefficient which correspond to the the the exponent of the X and the exponent of the Y ER and here um I mean i j may run from a um I mean some I mean from from degree zero you could up to degree

n for some n um where the a i j are simply elements in the field so for example I mean over R you could do um this we can denote as fxy and overall I could say fxy is H 3 x² y + 2 y² plus

X that's that's a PO in two variables right so this this three is what I denote as the coefficient of a 2 one and this uh two is the coefficient that I denote as a02 because the power of X here is zero and here the the coefficient one is what I in my kind of bookkeeping I I call it a A10 okay and if you think of

it um when we talked about um this this k k itself as a set is a set of pols and you could say that when I take K with with the new indeterminant x capital x then what I actually do is I consider pols in two variables this is kind of technically true I mean on the level of sets I am considering a a pols in two

variables um but the the issue is is that first right so so I mean what is KX with with capital indeterminant it's all the the expressions of the I mean maybe I'll just write one typical element here a typical element in in K with capital x is a um let's say r0 X Plus R1 x times capital X

Plus RN x x to the N where the the r the r i r0o up to RN are simply pols in I mean simply elements in K right this this blue box is a typical element in in K with capital x so you can think of it as Pol I mean if you just H change a notation call um capital

X Y then it will be a polinomial in two variables but the ER the difference is with the way we do operations I mean on pols in two variables on so this the set of all ER of all pols in two variables is denoted F XY um and on fxy we have addition and multiplication as usual right I mean how

do you you let's say I had a here another polinomial G XY ER which H which is maybe of the form a 5x or let's do y 2 + 5x then the way I do I mean the way I I add them is is just like you you would imagine I mean I take all the the monomials now like the kind of basic building blocks of a of pooms in two

variables are are these guys right so I take all the monomials ER in one and I compare them to the monomials in the other and in the case that I have monomials um of the same type I simply add the coefficients if I want to add them so if you if you will add these two polinomial you will get F + G um is 3x² y

+ 3 Y 2 + 6 x and and similarly you can do multiplication just the formula is a bit a bit more complicated but notice that in K in K capital x the operations are not a um are not the the same as the operations of a a um addition and multiplication of polom in two variables because when I add the coefficients I mean when I take

two expressions of this form and I add them together or I I multiply them together I need to add or multiply the coefficients that is these these RS I need to add them according to the addition multiplication in K and addition multiplication in K is is a a modul of Fe so this is another example of a of two sets that are I mean that seem to be a a one is contained in each other but the algebraic structure is radically

different different so we don't think of them of them as as the a Sub sub algebraic structure like Sub sub field or sub Rings right so that's that's really the the whole Crooks of of this h of this construction is that it's not really polinomial in two variables because addition and multiplication are defined differently um any any more questions about this okay

um last time we defined find the notion of of equivalence relation so for for a set X an equivalence relation uh is a a subset are um in the cartisian product of X

with itself so you you can think of it as as like a r is some a collection of pairs let's say A1 B1 up to a n BN it doesn't have to be finite but just for illustrations and um with when with the note um

a equivalent to be if if the pair a is in R this is just a just a a more convenient way to to denote it h um and this this H this subset of pairs is an equivalence relation if it satisfies one um

reflexivity that is for any X um the or for any a maybe the pair a comma a belongs to X sorry belongs to R which is equivalent to say I mean it's it's the same as saying a is always equivalent to a and the second condition is called the symmetry

and it it says that for any a in X for any sorry a n b in x a is equivalent to B if and only if B is equivalent to a and three is the the transitivity which means that for any three elements a b and c if a is equivalent to B and B is equivalent to

C then a is equivalent to C and these axioms if if you think of them um it's it's the natural at least the the kind of minimal properties that you would expect from a from a a framework that declares things to be equivalent right I mean an element always have to be equivalent to itself if I mean and if a a is equivalent to B also this notation

is suggesting symmetry so I would expect that b is equivalent to a and then this transitivity I mean it's it's very natural properties to H to ask for and we saw several examples but what I want you to get is the the the intuition that equivalence relation of course um so if let's say Tilda is an

equivalence relation on X we denote um let's say I take an element we Den note a with brackets is the set of all um the set of all B in X such that a is equivalent to B and this is called the equivalence

class of of the element a and and the quotient of x by the equivalence relation is simply the set of all equivalence classes so the intuition of of this um this definition is that when I when I take the quotient this is quotient set when I take the quotient I I Collapse

all the elements that are equivalent to each other I Collapse them to to one element to one point ER and you can think of it as as kind of Glu gluing or collapsing it's it's a um similar points of views ER and we had this example I take X to be the the integers and H the equivalence relation is is mod n let's say n is equal to

three then the equivalence classes I have are um I mean I I can represent them in in many ways but one canonical way to do it is to take the equivalence class of zero the one of one and the one of two and Z modu this equivalence relation is simply a set with a three elements the blue one the purple one

and uh and the yellow one okay so I I took an infinite set the integers and I after I I Define the quotient I I collapsed it into into a set of three elements but what is happening here in this for example in this uh integers modu n is that I have I had in I had operations on on the original set I had

addition and and multiplication and in fact I have induced operations on the set of equivalence classes because so Z modul this equivalence relation modu n has a it is called induced operations um of plus and multiplication right because I can take a let's say

zero okay zero is is Trivial but actually I could do it for any I mean I take a the equivalence class of a if I want to add to it the equivalence class of B I simply take the equivalence class of a + b and that's the result and if I want to multiply I simply multiply the the two Representatives

now it's not always true that you will get induced operations ER on the quotient set I mean when you take when you take the quotient you you glue things together so you might destroy um the the properties that you had ER on the original set I mean when I Define this operation on equivalence classes what I really need to check is that this is a well defined function in check

that a well defined operation what I need to see is that if h a is equivalent to a prime and and B is equivalent to B Prime which of course means that the equivalence class of a is is equal to the equivalence class of B and the equivalence class of sorry the

equivalence class of a prime and the equivalence class of B is equal to the equivalence class of B Prime if I have this situation to have a well defined operation of of plus and and multiplication I mean potentially I I I would have Collision because I have a plus b and this is a this is what I defined as as addition of two equivalence classes but I could also write the equivalence classes is a

prime plus b Prime and it's it's supposed to be the same I mean it is the same expression right um but by my definition I Define addition as addition of Representatives so I do I do AE plus b Prime and potentially these two things I mean what I need is the the the purple purple equality between the two expressions for any A and B I mean for any a b a prime and B Prime that

satisfy this condition I need I need the purple equality this is what it means that my my operation is well defined and but from basic properties of of modulus we know that it's a it's it's indeed the case right because I mean in equivalence mod n it is er so I mean it is satisfied because a is

equivalent modu n to B if and only if I can write a as let's say R sorry q1 um n+ R1 and oh hi um um so if a is equivalent to a prime then I can write a as let's say q n+ r and a prime is Q Prime n+ R

Prime and to say they they are equivalent it means that R is equal to R Prime and if that's the case then of course when I do a um I mean when I take a and the of course the same goes for B so B is equivalent to B Prime um let's go okay I I I don't want to to load up to not too too many notations

but this would be um I mean because because of the nature of this equivalence relation modul n h we get that the the the operations I just defined ER are indeed well defined on the on the question set so what is happening here is that yeah uh when when you tell uh well defined you mean that we can take any elements uh from the set or from from from group or from the field and uh

something like for example uh addition will work absolutely for all the elements right I mean when I say well defined in the the the formal sense what I need when I provideed I mean when I claim that I have an operation a binary operation on a set it needs to be a function right I mean let's let's put it like this more formally a well defined as above

means that we have a function from a integers I mean the quo set of integers modul of this equivalence relation times itself to itself that's what it means to have a binary operation Plus it needs to be a function in a function for any particular input for any fixed input it has to have one exactly one output it cannot have two

outputs so if I give you the input A A and H B this operation we defined it is a plus b but now I just I give you the same input but I I um I take another presentation of this input namely I I give you the input of [Music] um a prime and B Prime where I mean as before a is

equivalent to a prime and B is equivalent to B Prime so this as elements in the quotient set or as pairs of elements in the quotient set the these two are equal right but the way I Define the Operation I I do a prime plus b Prime and then I take the equivalence class because so in other words the way I Define the Operation plus depended depends on or at least I

mean the way I defined it it depends on the the ER particular presentation of the elements and if I actually claim that I have a function I need I need that for two two different presentations of the same input like uh like so I have I get the output I get is is a is the same output okay so well defined really means that I have an actual function makes sense

okay H is is it clear this is a this is an important point and we will H we will use it later on of course um so okay so what I'm saying here is the the particular example of modu n and you can think later of modu modu it has a similar description is is an operation I mean it's an equivalence

relation that is particularly well behaved in the sense that the the algebraic operations I had on the original set are induced on the I mean yeah induced on the quo set which a prior for General equivalence relations does not need to happen I mean there is no reason I I can cook very easily example in which I simply identify a bunch of elements in Z let's say you could Define a let's say a non

example um Define on uh z h the relation let's identify zero and one zero is is identified with one and for any a of course I need to say that a is equivalent to a okay and now this this very simple

this is an equivalence relation it's very easy to to check but the quotient set Z module of this equivalence relation does not have any I mean the the the operations of addition and multiplication are are completely destroyed right so in general equivalence relation when you take the quo set you potentially you you destroy a a all the additional structure all the additional algebraic structure and you need to be very careful in choosing the equivalence relation so that a you you still have a

induced algebraic operation on the question set but this is the the particularly nice property of of modu N and and in fact also of modu okay H how about we we take a break now or if if there questions we can take I can take it now okay so uh let's let's meet here in 10

minutes um can we start again yep okay so just to um to wrap up what uh what we talked about er er in the the first half um let's let's call it proposition um the integers modul the equivalence

relation of modul n is um a ring with respect to H the induced binary operations of a addition mod n and um and multiplication module and

moreover it is isomorphic to the Ring of integers modu n we we talked about integers modul I mean z modu n remember this in in our original definition simply the the the numbers zero up to n minus one with we talked about it as a billion group but in fact it is a ring because you can also do multiplication um and multiplication

will be associative and it will have a a neutral element being one and so on it will be compatible also with addition I mean it will satisfy the distributive law um and this guy here is just another way to to write it I mean because I can I can represent all I mean all the equivalence classes as as so right I mean ER the equivalence class of zero the equivalence class of one up to the

equivalence class of n minus one this is an exhaustive description of all possible equivalence classes when when I divide by the the modu N relation and on this equivalence classes on this collection of equivalence classes I have a a addition and and multiplication and this addition multiplication it defines a ring which is isomorphic to this ER to this ring I mean if you want to define the the is

orphism you for example in this direction you simply take the equivalence class of a a and you map it to the the element a or I mean of course here a you could you could map it to a modle n okay and this this is a a injective and surjective map and it respects I mean it's compatible with addition multiplication of course

intuitively you can just think that I I I simply chose another notation for these elements I mean now they are equivalence classes but the way the operations are defined is is exactly the same I mean again from a computer point of view it's the same the same structure is this proposition clear because I want to now generalize it to pols okay so in

pols I basically have the same story so remember we defined um f is field and we take um a polinomial let's let's CH uh change the notation because I don't assume now that the the polinomial is is prime now I can Define I

mean I I can Define polom modu of this polinomial this uh this F um just as I did before I mean I take all the remainder pols let's say the degree of f is n then I take all pols whose degree is a strictly h smaller than n and you see I have operation of a

modul of Fe sorry modul f um addition and multiplication the fact that that my my polinomial f is not prime does not prevent me to define a a modu f arithmetic because module F arithmetic only relies on long division of pols so if I want a a I want to multiply R1 with

R2 then um I write um I take the ordinary multiplication of the polinomial R1 * R2 and by long division I know that I can I can write it as a FX times some polinomial QX plus some remainder RX and and then I

Define um R1 R2 modu F I simply Define it to be R this R okay so I I I do the same as I did with with integers I do long division and I take the remainder and this operation of of multiplication and and addition modu F this

defines a ring on um a ring structure on the set that we called FX modu modu f right on the set of all remainder pols now alternatively um f induces an equivalence

relation on the set of all pols and namely let's say alpha x is equivalent to Beta X if and only

if Alpha is equal to Beta modu F that is to say um if I write Alpha I do long division of alpha by F then I get um Alpha equals q1 f plus R1 and beta is let's say q2f plus R2 H then I mean Alpha is equivalent to

Beta if and only if R is simply equal R1 is simply equal to R2 okay so now I I once I Define an equivalence relation on all pols then I can I can take H the quotient set I can do FX modu this equivalence relation that I denoted till F and the f is the set of equivalence

classes of pols with respect to this modul F relation and the same story for integers again remember the the analogy for pols over a field is that I mean you can you can do very similar things you can do with integers ER basically all the things that are that rely on Long Division it can be transported to to polinomial H so here as a set remember it's it's this H the set of all Representatives I mean

sorry all equivalence classes represented by by some element in the original in the original set right so here Alpha note Alpha does not have any degree a limitation it could be of a degree bigger than than the polinomial F or smaller but once I do long div I mean once I do long division of of Alpha by F the remainder that I get by definition

is always of degree smaller than than F so I could just as well what are the all the possible results of a division by I mean long division by F it's the same as a I mean I take all pols are whose degree is H is smaller than the degree of f which we called n okay so it's it's just another way to

to um describe the same set because whenever I have a polom of degree bigger than n or bigger or equal than N I can do long division and the remainder polinomial that I get I mean the equivalence class of the remainder is the same as the equivalence class of the the original polinomial and of course on on FX modul F we have a ring structure defined by um

taking let's say a representative Al I mean an equivalence class represented by Alpha if I want to to add to it the equivalence class represented by Beta I simply do alpha plus Beta And I take the equivalence class and if I want to do alpha equivalence class times beta equivalence class I simply do alpha times beta and then take equivalence class and just as we had in the integers we need to show I mean we need to show

that this this in indeed induces well defined binary operation that does not depend on on Representatives but the same the basically the same proof I mean if you think of it when we prove that the the modul N operation is er is well defined on equivalence classes of integers modul n what we used is long division and so you can transport the the proof to this ER to this case of of of pols and you get that the operation

is well defined and the op and and the summary is that um FX modul um the equivalence relation induced by a polinomial is a ring and it is isomorphic as a ring to what we denoted

as FX modul modu F right remember this this guy as a set it was defined as the collection of all remainder pols R um R belongs to FX and the degree of R is smaller than the degree of f and here potentially I have a different a different set but as I said above here I can represent the quotient

set as the collection of all equivalence classes of all pols are with um with degree less than the degree of f so again from a and the operation of course the the addition multiplication ER is essentially the same I just uh from a computer point of view I get the same multiplication and addition table okay is this uh is this point clear

great um I kind of uh Wonder ER if we should go to this example but let's try so uh this this this is another another way of using equivalence relations and this is called the projective line so um and

consider the uh the punctured plane I take the real numbers square so so the the plane and I remove the origin this I call X and Define a relation on

X as um I take two points let's say XY and I I declare that this is equivalent to X Prime y Prime if and only if there exist some a nonzero scalar Lambda such that XY is equal to Lambda X Prime Lambda y Prime I claim that this

relation Tia is in fact an equivalence relation ER and the way to see it I mean first reflexivity of course you take Lambda equals to 1 then X comma Y is going to be equivalent to X comma y a a symmetry I need to show that

um of course if if x comma Y is equivalent to X primea y Prime there exists some Lambda non zero such that XY is equal to Lambda X Prime Lambda y Prime H but then since Alpha is is a nonzero scolar I can invert it and I I can get that h x Prime y Prime is equal to 1 / Lambda

x 1 / Lambda y so X Prime y Prime is equivalent to XY yes under what assumption can we assume that Lambda is in invertible like we can only invert Lambda if it's part of a of a field right it is the real numbers are field okay sorry okay sure yeah yeah yeah my bad my bad yeah that's true that's true my okay so and I mean I I this by Design I only declare a two points in the so you you

can picture it geometrically I will soon elaborate but you take a point here XY and it is equivalent to all the points that sit on the line that passes between this point and the origin yeah that makes sense that makes sense okay but by Design I I I only allow two points to be equivalent if there is non zero scolar so that one is a multi multiple of the other and once it's a nonzero

scalar I can take the the multiplicative inverse okay yes so this is this is a a symmetry and a transitivity let's say I have a X Y is equal to a Lambda X Prime Lambda y Prime H and furthermore I have that X Prime y Prime is equivalent

to uh xou Prime y Prime this means that there is some Delta such that these two points are are the same then if I simply substitute instead of X Prime y Prime this Delta xou Prime and Delta y Prime I get that X Y is equal to Lambda Delta Lambda Delta xou Prime Lambda

Delta y Prime so if I have that X Y is equivalent to X Prime y Prime and I also have that x x Prime y Prime is equivalent to xou Prime y Prime that's a that means that there exists a Lambda with the first equality and there exist a Delta with the second equality then I get the third equality

which which means that XY is equivalent to xou Prime y Prime so this is transitivity right if h i mean if I have this and this then I get H I get this okay is this is this clear so a definition the projective

line is the quotient set you take R2 minus the origin and you mod out by this equivalence relation and what I what I claim is okay we saw that um for some equivalence relations you um if you had like an

algebraic structure you have an induced algebraic structure on the on the quotient set What's Happening Here is that on the um on the original set I have not an algebraic structure but a geometric structure and in this equivalence relation induces a a geometric structure on the quo set so how can we represent it

um so to represent the quotient geometrically remember I mean we can think of I have I have this h i mean I start with the punctured a 2d plane

and now a an equivalence class of a of a point is the set of all points that belong to the line passing passing between this point and the origin so this the the whole collection of of purple points is is not one equivalence class and I have many of these right I have another line I take just a point here and maybe also a green

one so each of these colors is is one equivalence class H how do I represent it I take I mean I can take a a for each a for each line I can take only the points um that whose distance from the origin is is exactly one then I get a I get the circle which would have been a a good

Circle if I I had better drawing skills and in on this Circle H I actually Identify two antipodal points and

so so if if I identify two antipodal points you can think of the the projective line as simply one half of the circle in which I identify um H the two ends so um I take one half of the circle because I want to have a um a set of Representatives which with no repetition so I take one half of the

circle let's say the the upper half and now I I identify these two points right because these two points are antipodal I mean they sit on the same line passing through the origin so they they are in the same equivalent class but if I identify two points what I get is again a circle maybe I can write it like this these these two points let me call it Infinity minus infinity but they are I glue them

together to um to Infinity so the reason it is called the projective line is that what I actually did here is you can think of it as as I took the the line of real numbers and I introduced another new Point H which I call Infinity and this point is a I mean it's kind of a it it's in the end in the the the left hand and in the the

right hand ER I mean if you go a ER up in the positive direction of the real numbers ER and you continue as much as possible you eventually you get to Infinity but also if you go in the negative Direction you go ER you go to minus infinity and this is a this is declared to be the same point okay so this is this is a way to represent the quotient and it it is a I mean of

course if I wanted to be rigorous about I mean formal about what it means to have a geometry it would H it would take us more more effort but just on the intuitive level you can you can see that um the equivalence relation defined a gluing recipe and and once I I do the gluing recipe I get another geometric model yes sorry I don't really see the link between the more algebraic notation you had before with uh A and B are equivalent to a prime B Prime if only

there is a Delta multiplier between them and this and I'm missing the link between the geometrical representation and what you explained before okay so um let's take a I mean think of of this as XY when is it when we declared it to be equivalent to Lambda X Lambda y for any Lambda that is non zero right yeah ER but what are these points Lambda X Lambda y you take the the line

with a slope Lambda that passes through the origin I mean the this blue line Y every I mean Lambda X Lambda Y is some point sitting on this line Y okay and all the blue all the blue points are in one equivalence class right correct yeah and the same goes for the the green and the purple so now just like we did before with equivalence classes I I mean of course on the formal level the quo set is simply the set of

all equivalence classes but to H to get a better ER presentation of the question set I would like to to have unique representation or as unique as possible representation for the equivalence classes just like I did with I mean integers modu the equivalence relation of modu n i I wrote it as the equivalence classes of zero up to n minus one yeah and I I want to do the same here so for a line I I want to represent a

line with just one point so I could I could take the the point on the line whose distance from the origin is one and let's say the the the point that sits on the on the upper the upper half of the of the plane okay so what I get is the is this half circle but I need to be slightly more careful because this point is identified with this point right this they see it on the same

line I mean this this is a one comma 0er this is minus one comma 0 and they are indeed I mean take Lambda equals minus one it shows that they are equivalent by by the definition of equivalence I I I set right y okay so so these two points also need to be so I almost got a a unique representation of equivalence classes except these two guys the these two guys I now need to to

identify so so I get you know maybe I should have taken the the lower half and then when I identify them I get I get again the circle okay is this um I think I think I understand the the problem of having the two points but but it it is the same for the purple points and the green points you have you have two points at at distance one from the center yeah I and I'm just saying look I mean you can do it in many ways but a so to I mean if I want to to have a

uh if I want to have something like I had in in integers module n in which I I have one equivalence class I mean like a a a list of equivalence classes that does not does not have repetitions then I take I take the upper okay okay yeah I take the upper half of the circle h i could I mean I could simply not take this point and and and take this point yeah and then I have then I have

this unique set I mean exhaustive set of Representatives but if you think of it geometrically since this point is identified with this point even if I didn't take the if even if I didn't take this point the the geometry is such that when I go in this direction and I get to this point and I kind of continue going in this direction I start getting into this right yeah okay so this this is why I I I guess I guess to be consistent you

could uh you could draw it like this and here here here is the the new point at Infinity yes Carlos yeah I was just trying to abstract this to any kind of Dimension so basically we could also say that R2 is an equivalent class of R3 so a plane a plane inside of R3 is an is an equivalent class and a plane passing through the origin yes why does it have to pass through the origin that's what I don't get because

what you will have in so er so I will I mean the projective plane I now take our three and I remove the origin and I declare XY Zed to be equivalent to Lambda X Lambda

y Lambda Z right I see the origin crashes crashes this completely okay fair enough is that is that is that why you do it or or or I'm I'm not following I mean so and now I have the 3D [Music] plane and I would like to get I mean a let's say I have a point here XY Z this I mean this we didn't cover in the course but it's

a it's typically things you you see in school that if you take just a a scalar multiplication of of x y and z um okay so it's not it's not Plains H yeah this this is this was my mistake I mean it is still lines so if you take Lambda X Lambda y Lambda Zed this is some some point on the line passing through the origin oh wow I was not expecting that I

thought this what we play no no no it's it's a it's a line so it's can you give some ination on why is that like why it doesn't matter that you that you if you bump up the degree of your kind of Base playground to put it on some sell which in this case was a three get lines so this depends on the way you define the equivalence relation right I mean I see fa enough iine I want to define the equivalence relation in the same algebraic way that I did before but now the way to

represent the projective plane is I take um as Representatives I take the points whose distance from the origin is one that is the sphere yeah exactly yeah that makes sense and this was the second now that you say the sphere because now instead of having just two points colliding you have an infinite amount of them right like the the whole border of the sphere is colliding with itself yeah but okay so now I will take the upper half of the sphere let's say this upper blue half of

the sphere and a a I would I would need to identify the antipodal points H sorry here the antipodal points on the the uh the base of the the sphere okay so it's again you take the upper half of the sphere and you kind of glue you glue let's say the the red

line you glue to the ER in the pink line but you glue it in in an order I mean you glue ER you Glu this point to H to its antipodal point so it's like a you have this this direction here and and the kind of opposite direction here okay okay fair enough and and what you get is again a

sphere if you think of it you glued I mean all the all the points on the the so I took the upper half of the sphere all the points which are not in the base not in the basic the base circle of the the half sphere but they they are they have unique I mean they are not glued to anything yeah correct but the but if I take the base circle of the the half sphere then a a half of the I mean half of this Bas circle is glued to the other half so that would that then be Infinity for you this would be now a line of

infinity a line of infinity yeah fair enough yes okay so and the last question that this makes me wonder is so now the the 00 Z so the the the null point or I don't know to say point I mean this this is how can I say that the zero is now Infinity itself like we don't have zero in this equivalence class like but we didn't anymore if you think of it we didn't have zero in the first place okay that's true yeah yeah that's completely true yeah fair enough fair enough and the reason we didn't have zero is that we we cannot ER I

mean you canot create the equivalence class zero zero would then would then be a would then belong to every equivalent this this cannot I mean cannot be it what what will happen is that you I mean on the formal level H you will not be able to Define an equivalence relation via this formula if you you have the zero yeah you have violation of the the axioms of equivalence relation because you will have that it's linearly independent from the rest I mean it will be it will be equivalent to all other points but if if it's equivalent to all other points then

okay you could technically you could say it's equivalence relation but then it's a trivial one I mean it's all the points equivalent to each other yeah and it's an uninteresting example okay so H we will not I mean ER we will not do a lot of projective geometry but I think the idea of the projective planes ER will be useful to give you intuition on on what's coming up on elliptic curves um I have a question yeah um considering these equivalence classes or equivalence relations it seems that they

are very much related to homomorphisms in some sort of way like they they give rise to some sort of homomorphism or so my question is what what's what's really the difference there or do do I have a misunderstanding here ER yeah I think this it's it's a different type of object I don't I don't see what's uh what's your intuition here H I think didn't we had this example with the modular operation and yes like and it didn't we also had some homomorphisms considering that that uh that you could yeah basically map the

natural numbers onto some uh yeah smaller numbers okay I think I think I understand what you're trying to say I mean we have a homomorphism let's say from Z to Z mode n right yeah this takes it takes a number a and it sends it to a mod n now when I Define the quotient with respect sorry with respect to the equivalence relation of modul N I also have the

quotient map that takes an element and send it to the equivalence class of I mean of itself with respect to the the relation and we said that I have an isomorphism let's say from integers module n to integers divided by the equivalence relation so let's let's paint a picture like this on one hand I could I could go from integers to Z mode

n by sending a to a mode n on the other hand I could I could do the the quotient set and I also have a map this is called the quotient map this goes to a and I also said that there is an isomorphism let's say from here to here it basically takes a to the equivalence class of a right yeah so I think actually I'm

wondering what is the difference between those two is it just some sort of notational stuff or yeah I mean there are you mean these two yes exactly these two that's what I was trying to to say in the the propositions I mean they are isomorphic ailion groups or or rings if you want if if you want to consider ER both the the addition and the multiplication together ER they are isomorphic so what does it mean for two algebraic structures to be isomorphic it means you have multiplication tables addition tables and they are are the same up to a change

of notation right just uh that's kind of the the the whole idea of of isomorphism is that from a computer point of view it doesn't matter it just you have you have the same algebraic structure but you you chose different symbols to uh represent it I see okay okay and these this this is this is a very good example of a a a of isomorphism I mean we did two different constructions I mean they're not equal on the nose but ER but they're really the same for

us so I see okay so so this is what and and this two I mean it's not only that these two constructions in themselves I mean these two purple a bilon group say are isomorphic it's also comp I mean compatible with the the quotient maps I mean on one hand I could ER I could take the quent map here on the other hand I could take the quotient map here and under this

isomorphism I get the same I mean if I start from an element a here I could take the modu n and then the equivalence class of of the a of the modu N result and on the other hand I I could just immediately take the equivalence class of a and I get the same result this is a it's called a commutative diagram but I don't want to drag you into this hole too much but all I'm saying is that under these

constructions uh it's not only that they are isomorphic also the the the quo are compatible with the isomorphism ER somebody raised the hand I think it was me in the chat sorry ah no it's okay um is this clear thas yeah I mean I think I I just wondering what what if this is um I mean basically these equivalence classes I mean it's it's some sort of tool you introduce but in a way yeah I'm

wondering uh for what we will be using it later so well I mean okay I think you can you can take this projective line and planes is an example of a use of equivalence classes if you want another example a kind of from a a foundation of mathematics Point of View ER if you remember in one of the first sessions I gave a definition of the natural numbers like I zero is the empty set and one is is the set containing the empty set

as a unique element and a inductively you define ER n to be whatever you had as the Set n minus one Union the set containing whatever you had is the Set n minus one okay so this is a definition of the the natural numbers as as sets right and you can ask how if I if I like how would I Define the the integers for example so this this give you a definition of all the natural numbers

actually natural numbers h plus zero ER now let's just call it n um how would they Define ER the integers so what I say I ER I take I take the set X to be all pairs of natural numbers in the sense I defined above and H and

Define an equivalence relation on X the that is on on pairs of natural numbers H that is I take ab and I declare it to be equivalent to CD if and only if a A+ D is equal to um C plus b you can ask how did I

Define addition on the natural numbers but to every I mean I can always Define a n + one to be ER you take n and you take Union the set containing ER the Set n so you basically by Union of sets you have addition on on natural numbers and now I I declared a on pairs of natural numbers AB to be equivalent to another pair CD if this

relation is satisfied but so and you can think of ab is representing a minus B ER you the the reason you can think can think of it is is a the way we Define the equivalence relation we said that a is is the same as CD if and only if their difference is I mean this is basically if and only if if I had a a a a definition of of a of a

substraction this would be the same as saying that a minus B is equal to C minus d right but if I only have if I only have addition then I use equivalence relation um to kind of speak to be able to speak about a yeah so and and if I take now this module of the equivalence relation I I simply I simply Define in the foundation of mathematics this is

simply defined to be the integers okay because you have so you have like equivalence class is but of course you can represent a minus B in many ways I mean it it is like I mean you can represent a negative number let's say minus three it is a 3 minus 4 but it's also four minus 5 and so on sorry minus one is 3 - 4 but also four minus five and and so on I see okay that's cool so and now now you def that you define the the the

integers you can Define the rationals H you take pairs you define an equivalence relation on the inte on pairs of integers you say a comma B I mean you need to define a a multiplication of of integers but let's say you already already have addition of a of natural numbers H and multiplication is just defined via addition right so I let's say I have multiplication of of integers and now I want to Define

rational numbers I Define an equivalence relation on this a set of pairs of integers to be a a a equivalent to CD if and only if um a * D is equal to B time C which is of course secretly saying that A over B is the same as C over D but right but if I didn't have a defined division

then then this this would be the way to go about H and and and the rationals is simply pairs of integers again here it's like I need that B and D it's actually pairs of integers ER whose second coordinate is non zero to be to be precise and the equivalence relation is simply defined to be I mean the the qu set is simply defined to be the rationals okay H there was a

question thanks ER yeah okay so if you want a reading recommendation you can you can Google it up H but this is like a def I guess piano oxums and um in definition of of arithmetics via VIA set theory this is this is standard way now you can ask how do I find the real

numbers for the real numbers you need a what is called dedin DED I'm not sure how to spell it I think it's it's it's like this dedic Cuts so you you will take a a subsets of a of rational numbers and you will define an equivalence class on these subsets and uh the the question set will be real numbers okay um we are over time so uh

just uh last questions to to see to have a wrapping up the discussion