PSE Lectures - Ep 17 - Field classification - Equivalence Relations, Partitions & Cosets

Transcript

okay um yeah uh I think today what I mean you of course welcome to to ask er er for new topics but I wanted to to discuss something that I left out in the um in the first section of the of the notes ER which is called equivalence relations and equivalence relations as

you will see it gives you uh a kind of a a unifying framework for taking questions so this H this will uh this will put the constructions of of quotients that we did in a in a new a new light and hopefully a some something helpful for your understanding um it is not strictly necessary H because you could somehow you could Define quotient ER without equivalence relations this is what we

did so far but ER it it can be useful to to have other points of view um but before I get into it um are there any questions on the a previous material okay so one thing I I want to remark on um on last time er a

construction recall that I defined the algebraic closure for example of FP to be H you take you take the tower FP goes to a FP square or sorry um yeah FP 2 factorial which is FP Square um and here FP and factorial and you define the algebraic closure as the union of all

these and you have natural natural operations of addition and multiplications coming from the fact that all of these guys are field inclusions so um so you could you could do the ER the addition and multiplication at any finite stage you like and indeed if you take a finite number of elements they have to all belong to some finite stage um and I also defined more generally

fq well you take the same Tower but now you start with with fq so you can do fq to the two factorial where Q is now a power of p and so on you have at the end end stage you have F q n factorial and H each of these H inclusions is again a a subfield inclusion so I could I could similarly Define fq bar to be the

union of um of all the fqs um as I defined it here the two sets here are are simply equal uh FP bar is simply equal to fq bar this is because um every every element I mean if you want to to show equality of sets you need to show a double

inclusion ER if I take an element X in this Tower then it belongs to some some finite stage but once it it belongs to some finite stage say FP to the N factorial this FP to the N factorial will be um um a a subfield of some fq to the the K factorial um and and vice versa if you take if you take an element y here then it will belong to some finite

stage but then once I go high enough in in the tower in the the upper Tower um I I will know that why is included in the ER in in this in this stage in the upper so the two sets are in fact equal and moreover this factorial con factorial formula is not really necessary all all we want is that um each of the the stages is a subfield

inclusion so I could ER I could do ER let's say FP bar I take FP then FP squar or um or maybe I choose K yeah so let's say I take a FP Square um FP to the 4 and so on

right here I apologize here the nend maybe it's better to to take a a and then uh then there is no collision with the the other end but I mean I could Define also FP bar as taking FP FP Square FP to the four and in general FP to the the n 2 to the n and this is all also subfield inclusion because we had a LMA that ER I

mean if let's say f p to the S is a subfield inclusion of f p to the T if and only if s divides T so you you can choose any sequence of exponents that satisfy that each consecutive step ER the smaller number divides the bigger number and this will be another legitimate choice of the algebraic closure and it will be it will be simply equal to the the construction we had and

all all that matters is that we have a copy in the algebraic closure we have a copy of every finite field I mean up to isomorphism and one once we have that it means that every polinomial must split because every polinomial um has to have coefficients in in some finite stage and and splitting field of each finite stage is is a finite field and splitting field of a a polinomial over a finite field is again a a finite field

so so everything is is included in the algebraic closure um is this point clear any questions or remarks on on that okay so um ER concerning equivalence relations I

mean as a motivation you can you can recall that um we Define the natural numbers as as sets um I mean we defined a zero to be the empty set H one to be the set containing a unique element being the empty set and a a two is the set containing the empty set is an element and another element being the set

containing ER the empty set as a unique element and so on so this is kind of a foundational ER construction we usually in in this course we don't care too much about foundations but ER you could ask yourself how how do people in the foundations of mathematics or or logic how do people Define um the integers I mean assuming that I have these natural numbers I have I have something corresponding to

addition because I can simply take a I can take Union of sets and it if you do it properly it it gives you addition um and let's say even you you had integers how would you define the rational numbers I mean for for rational numbers you really have an issue because half is um is also two two quarters or three three over six and so on so every element has infinitely many

possible representations which should all be equal so you know let's say you define the the integers and now you want to define the rationals you need to to have a way to somehow declare elements as being a as being equivalent and and that's that's the role of this this equivalence relations so a relation on a set X is simply a

subset are of the cartisian product of X with itself and a when you have a pair X comma y that belongs to the relation we denote h x relation y ER or or even a better notation h x

till the r y now um one one famous example of a relation is simply a function if you take a function F from X to X then you can define a I mean this function gives rise to a

relation um where I declare X in relation to Y if and only if um f of x is equal to Y okay um but we are not interested ER in just any relations I mean you can see from this uh this

notation that we chose that at the very least we we are interested in something symmetric namely that X is in relation to Y if and only if Y is in relation to X that's that's one of the the the intuitions we have about what we want to H I mean about the notion we want that will have help us identify things that are the same and there are two two other natural natural axioms we want one axom is that we want that ex

will be in relation with ex for any ex um and two that we can H we can move in kind of a a transitive way but let me H write it formally so a relation r on X is called an equivalence relation

ER if it satisfied the the following conditions the first condition is called reflexivity and this means that for any element X in the the big set x h x is in relation to itself and the second condition is called symmetry which means that for every x and y and

x x is in relation to Y if and only if Y is in relation to X and the third is called a a transitive and this means that for any x y and Zed if I know that X is in relation to y and Y is in relation to zed then it must hold that X is in

relation to Z so this is kind of a a triangle condition right if you have a if you have three elements and you have a some relation between X and Y and some relation between Y and Zed then you should have a relation between x and z yeah uh how can be the case that X is

not reflexive um take take for example this function take take a function which is not not the identity right so so I mean ER let's say example um if F from X to X is not the identity not the identity on X that

is there exist some small X such that um f of x is is something else than x then the relation X in relation to Y if and only if f of x is equal to Y is not reflexive right because there will be

some element for which f of x is not equal to x a a a this would mean that X will this specific element X or if you want we can call it X zero so x0 will not satisfy that X x0 is in relation to itself so so o overall this H this quantifier H for all X I mean it will not hold is it is it clear it yeah but then it means that like uh it will hold only for identity

functions yeah but I didn't say that I didn't say that relations only arise via functions oh okay okay and indeed there are many in the interesting relations will not arise via functions but I wanted you to have a um a slight more General notion of a relation and keeping in mind that we we focus on a particular type of relations and and for these particular ER relations actually functions are not the

the main example okay yeah so okay but but if if functions do not give us usually equivalence relation what what does give us equivalence relations and this is the following example um I take x equals the integers and n some natural number

and R is um equivalence mod n that is um two integers X and Y are equivalent in this relation X equivalent R to Y if and only if e x is equal to Y mod n now I need to check that this is

indeed an equivalence relation according to the definition so remember reflexivity Symmetry and transitivity so of course for reflexivity um that's very easy because every integer X is a equivalent to itself mod n H and symmetry is also very easy for any two two

integers X is equivalent to Y module n if and only if Y is equivalent to X Mod n and for a a transitivity this you can check yourself this is slightly less trivial but it's very very easy um for any x y and Zed if x is equivalent to Y module n

and y is equivalent to Zed module n then X is equivalent to Z mod n okay and this this should be the the toy example that you have in your head for a equivalence relations now given given a an equivalence

relation I can I can make a a kind of natural construction ER so let let's call it um I will simply denote ER instead of this uh this R notation I will denote it by Tilda so let Tia be an equivalence

relation on a on a set X um if I take an element in the set I Den I can denote or I can consider what is called the equivalence class of x which is all the elements y in the the ambient set such that X is equivalent to

Y this is a this is called it the equivalence class and um let's note something about ER about equivalence classes so so the picture you should have in your head is um okay maybe in in the to example that we have

ER so let's say x is equal to zed and a Tilda is a mod the mod n relation ER then what is the equivalence class of er of one the equivalence class of one is ER well one of course is equivalent to one because this is a a reflexive relation but also n + one is a equivalent to

one and 2 n + one and so on and you also have the negative guys I mean you have um 1 minus n 1 minus 2 N and so on so this is the equivalence class of one and similarly you you can do it for a for any other number and H these equivalence classes I mean if a and b are elements in the equivalence class

of X then I have that a is equal to X and B is also equal not equal equivalent to X Now by a by symmetry I I get that X is equivalent to B and by transitivity h i get that a is equivalent to

B so you see that any two elements in the equivalence class of of a given element must be equivalent themselves and and this this is indeed the case I mean n + one is is equal modul n to 2 N + one for example [Music] um so that's that's one property of this equivalence classes H another property

let's say I have two equivalence class classes ER that um are I mean represented by by some some pair of elements X and Y that are not equal well I mean of course you you see that um in this example with integers modu n the equivalence class of one is going to be equal to the equivalence class of

n plus one right and so on I mean I have I have infinitely many ways to to represent the the equivalence the same equivalence class I mean the same set of integers as equivalence class of of different elements um but let's suppose I have two equivalence classes that are not equal ER now if I had an element a in the intersection I mean equivalence classes

are simply subsets of the the original set right X is a is always a subset I mean the equivalence class of X is always a subset of the ambient set X and if I had two equivalence classes that are not equal and you give me some element in the intersection of H the two equivalence classes

um then by definition X is equivalent sorry a is equivalent to X and A is equivalent to y but then a a is equivalent to X and Y is equivalent sorry X is equivalent to a and a is equivalent to Y which implies by transitivity

that X is equivalent to Y but if x is equivalent to Y then it must be that the equivalence class of X is the same as the equivalence class of Y because every element you take some element I don't know T if T is an element in the equivalence class of X that is to say t is equivalent to

X then um I mean since I also have X is equivalent to Y ER it follows that a t is an ele is is equivalent also to Y that is to say the t is an element in the equivalence class of white and and conversely I mean the the

role of X and Y here is symmetric so I would get that a a two equivalence classes um I mean one thing to note here is two equivalence classes are equal if and only if their elements I mean the representing elements X and Y are equivalent in this relation so so X is equal to Y if and only

if the the element X is equivalent to Y and and moreover if x I mean the the equivalence class of X is different is not the same as the equivalence class of Y as as subsets I mean as subsets of X then I I simply cannot have any common element in these two equivalence classes because a common element will give me by this

triangle um in principle I mean if if I had a common element a that is a is equivalent to X and A is equivalent to Y then I I would immediately get that X is equivalent to Y and that would mean that the two the two equivalence classes are simply equal so I'm saying if they're not equal they must be disjoint okay and

um no I I can look at the collection which I call Q it's the the it's a set of sets an element in the set Q is it's is an equivalence class of X that is a subset subset of the ambient set X and I I run over all all the elements x and x this

collection is um I mean the way I write it you need to remember that when I Define a set I am allowed to to to have duplicate elements in the the syntax but eventually the the the set the set itself h um you omit repetitions I mean like the set 01 is can could also equivalently be written as 011 okay

so so here in the set Q I have potentially I have a lot of repetitions because ER Whenever two elements are equivalent ER via their relation ER then their equivalence classes are are equal um but the collection here is a is a partition of the set X um and when I say partition I mean

that the union of all the all the subsets when I take x in x this this gives me the amb Universe X and um for any two elements X and Y um either X is equal to

y or X is this joint to Y so this this is the notion of a partition what you you should have in mind is is is kind of a picture you have X and you partition it into um let's say three parts this could be the equivalence class of X1 and this could be the equivalence class of X2 and the one of x3 to say that it it is a partition

means that the union of of all the all the sets um give you the entire the entire Universe X and two elements I mean two two sets in this uh in this collection are either equal or ER or disjoint yes what is the difference between the hat and the collection um I'm just like choosing another word to ER to avoid duplication but collection is is is

set I mean yeah there are like several synonyms you can call it a collection or a I mean this is collection means a collection of sets sometimes people call it a family upsets and so on okay but it is I mean it is a set of sets all right is this this idea clear I I I chose Q for for a reason I mean this Q is going to be the

quotient but let's do an example remember we have a running running example that X is the integers and um the the the equivalence relation is mod n then what I can do I mean of course if I just go by definition I could write this is all the the x is in Zed I mean all the equivalence classes of X such that X is in

Zed so you know I have like a the equivalence class of a um a zero or minus one and so on and but also the equivalence class of one h two and so on ER but this H this writing has a lot of repetitions H because remember the equivalence class of for example zero is going to be equal to the equivalence

class of of n so I could write it if I want to remove all repetitions I could write it as take the equivalence class of zero the equivalence class of one up to the equivalence class of n minus one and that's it because the equivalence class of of n is going to be equal to the equivalence class of zero now this is not the only option of

writing it I could also write it as a I mean for let's say the equivalence class of n equivalence class of n + one up to the equivalence class of 2 N minus one that's another option and there are many many other options these These are kind of a organized ways to to write it but you could also find something more random

but this this representation is is somewhat appealing I mean because it is similar to um to constructions that we we already had so you see if I take this rep representation of Q then I can think of a of Q as corresponding to integers mod n integers mod n was defined to be the set of all numbers Zer up to n minus one H but I might as well

H take the equ equivalence class of these numbers mod n so H another notation for this this Q given a set x with a an equivalence relation

Tilda we the note X modul Tilda is simply the set q that is the the collection of all equivalence classes of elements in X and col it the question set

okay so you you can see that um we have a notion of quotient already on the level of sets and what you can expect is that I mean all the all the Notions we worked with so far I mean groups Fields rings and so on on ER were basically set with an additional structure so I mean since I can Define quotients on the level of sets I can um I mean I can think of this I mean I

can try to upgrade a a quotient set into a a quotient group or a quotient field or a quotient ring ER so so the in other words the question is always the once you have an equivalence relation the quotient is always defined on the level of sets and if you work with a particular structure let's say a group then what you need to verify is that a a the quotient on the level of sets have natural operations that gives it a the structure

of a group or a field or a ring so what we are really interested in is is cases in which we have uh an algebraic structure which is a set with some operations satisfying axioms and we have an equivalence relation that um once we take the quoti set we get again the same the same type of structure that we considered in the first place

um Let me Give an example in in sets just H just another simple example I mean you have a let's say the the set with a seven elements one up to seven and let me instead of writing it all um let me draw it so let's say five is equivalent to seven but also to

six then four is equivalent to one and three is equivalent to two this is clearly an equivalence relation and so the quo set you can represent it in many ways but one way would be to uh to take the represent resentatives of one two and five the equivalence classes of 1 two and five but of course you could also take the equivalence class of four

three and six okay um before we take a break are there any questions on uh ER on this on what we discussed so um and let's take a 10 minutes break

um can we start again yep okay so we said that um an equivalence relation on a set X gives rise to a partition but if um P let's maybe let's let's do fantasy P if p is a partition

of of a set X IE um p is a some collection of of of sets ER let's say UI where I is an index in some index set ER and recall a partition it means that um for any two two sets I mean two sets are represented by two indices I and

J um if if I is not the same index as J then H UI intersection UJ is is the empty set I mean UI is disjoined from U UJ um plus the union of all the uis where I run over all the index ER indices in the index set

is the the entire Universe X so that's that's a partition in general and from a partition I can I can construct a an equivalence relation on x equivalence relation on X by setting um for for two

elements X and Y in the universe x x is equivalent to Y if and on if there exist some index I such that X and Y belong to a set in the partition UI to the same set and and again the picture you should have in mind is H is what we had before I mean if I now give you some partition of the set

X then I can define an equivalence relation by saying that all elements in a given in a given set UI will be equivalent and in another set UJ you have the same I mean the two elements in in the set UJ are equivalent why is this uh an equivalence relation I mean I can be formal but I think you can get the idea without like a full formal treatment I mean first we need to check

reflexivity that is every element is equivalent to itself but that's that's very I mean clearly if x x must belong to some UI because the U Union of all the UI is the entire universe so if I take X it must be somewhere in in this division ER and then by by definition X would be equivalent to X and um of course symmetry is is obvious I mean this

this definition is is is symmetric by Design and and transitivity is also is is also obvious I mean if X and Y belong to to some set UI and Y and Zed belong to the same set UI then clearly x and z are in UI so x x will be equivalent to Z so what we have here is a a correspondence in mathematics we say one to one correspondence

between um equivalence relations on x I can look at the set of all equivalence relations I mean remember equivalence relation is simply a subset of the cartisian product of it x with itself so I can I can consider the the the collection of all equivalence relations on X and on the other hand I can consider

the collection of all partitions of X and it's a a I mean it's a two-way it's a two-way street if you give me an equivalence relation I IO you you automatically get a partition and and vice versa okay of course if if two equivalence relations are not the same then they will give you partitions which are ER which are not the

same and so on so you actually have this one to one correspondence is is just another name of a bje between two sets okay so this this is one one level obstruction of of the discussion we had okay so how how does this ER involve ER the questions that we we discussed in a in let's say in

groups if G is an obedient group and um age is a subgroup then we defined a a for an element X in G ER the age

CET of X is a is the following set it's denoted is is x times H I guess um I don't know what you prefer but in general I I take for for groups I take a multiplicative notation and in multiplicative notation the cetes are a are denoted as x times H and this is a the set of all I mean or all

products x times AG where age is in ag yes for elliptic curves will you also be using multiplicative notation no for elliptic curves I'm going to use this H um oam um this is really um it's a matter of a it's a matter of choice of taste if you wish but generally the literature of abstract group theory is is typically written in in multiplicative notation okay ER and then

for a billion groups ER you this depends on your choice I mean many many authors immediately ER take additive notation for a billion groups um but you know you could you could also do multiplicative notation I I think it's good to to get used to to both and be able to H to do a switch in your mind this it also kind of forces you to ER to look at this ER in in an abstract sense I mean it's a binary operation we can call it multip multiplication or addition it

doesn't it doesn't matter eventually Fair thank you madam um and so yeah so in additive notation you will denote the cetes as X Plus H right because the whole the whole point of this notation is to indicate that I take all the the binary operations of of x with an element in h now we defined G modu age to be a simply the

set of all coets and we said recall that um I mean this this kind of writing it potentially has a lot of repetitions um because X time H is equal as a cette to y * H if and only if x y inverse belongs to age okay we had a LMA yes um ju just to check this uh notation

uh G SL H has nothing to do with G without elements of h no yeah this this H this you should read as a g modu age G without elements in age is a is written like this it's like a it's like minus sometimes really people just just denote G minus H okay okay so we had a LMA this is a key LMA in all this this cette

business that two cetes are are equal if and only if when you take the the the multiplication or the binary operation of one element with the inverse of the other you get a you get an element in the subgroup AG and what I claim now ah and maybe maybe a just another another point to recall

um g mode age has a an a billion group structure given by ER the unit element is e * H but of course e Times age is simply

age and you take look at this definition you take all the all the multiplications of e with a with an element in age you simply get all the elements in h and if I now want to multiply a two elements then I simply multiply their representative and then I take the corresponding cette corresponding coet of course is the set of all expressions of the form XY AG where

AG is in age okay which of course by associativity it's the same as saying it's all the expressions of the form x y h such that H is in ag and so on and okay we H we took it took us some effort to H to see that um that indeed this gives the structure of an AB bilan group I mean the inverse so this is a multiplication the

inverse of an element X age is you take the inverse of the representative xus one H now I claim the following the collection of all all coets is a

partition of of G so what do I need to show first I need to show that the union of all the sets in the collection is is giving me the the entire ambient Ambient set in this case it's the set G and

indeed if um let's say G is an element in the set G in the universe then um G must belong to g h because a a G is equal to G * e and of course since since H is a is a subgroup it must contain the neutral Element

e so so G * e belongs to to H the set GH and I chose the generic element G it means or I could maybe maybe it would be better to call it ex just to to be consistent so every ex you take is an element in some piece of dep partion and and so so the the the union of all

all the cetes is G and I now want to show that um two sets in the ER in the collection are either equal or disjoint so suppose X and Y are elements in G and I take two cetes x h and y AG that are not equal um I want to show of

course that the intersection of of the two cetes is is the empty set that is I want to show that if two sets are not equal then they must be disjoint H so suppose well towards contradiction that I have an element let's call it a in the intersection

what does it mean it means that there exist two elements let's call it a Ag and H Prime in Age such that on one end a is x * H and on the other hand a is x * H Prime if I have this then a

um sorry on one hand a is equal to xh and on the other hand it's it's equal to y h Prime okay that that's what it means to to belong to the cette YH and what I can do is if I have this equality I can express a ex as um y h

Prime H inverse and I can also Express y as um x h h Prime inverse but the way it's written I mean it is y times an element in h and here it's x times an element in h so this so I get that X must belong to Y age and Y must belong to X

Ag and from here I think you you can get a feeling that um X age would would have to be equal to to Y age I mean if if x age contains contains Y and Y age contains X then it's natural to expect that the two two sets are equal and indeed I take take any general element um let's call it Alpha in a

xh so let's say Alpha is equal to um xh1 in xh then since X is already in y age it means that um Alpha is equal to if you want the notation I mean it would be y h Prime hus one times H1 but this of course this is y times

some element in a so so it belongs to Y and similarly if um beta is some y * H2 an element in YH then um I can write beta as since I can write y as x times an element in in age then I would get x

h h Prime inverse times H2 and this is clearly an element in in xh so so I I get d inclusion I mean X age is included in y age and Y AG included in X Edge which which means that the the two sets are equal okay and and this completes the

proof I mean I mean we wanted to show that we that the collection of all sets all cetes is a partition we showed that on one hand the union of all cetes is a the original group G and any two cetes are either equal or disjoint so here we got a we got a contradiction mean we assumed that see we assumed here in blue that X age was not equal as a cette

to to Y age I mean we assumed by contradiction sorry the the the purple ER apologies we assumed by contradiction that there is some Er some common element and H what we got is that ER the two I mean the two sets are are equal so this this contradicts the ER the blue the blue assumption

here okay yes I'm not sure I understood the first part um like how a unit element helps us to prove that the union of all H cetes is G well I mean what do I need to to prove in order to see that the union of all of of a collection of sets is is this ambient Ambient set G I take a generic element in the ambient set let's call it

ex and I need to to show that it belongs to to one of the pieces of the Union if I show this then of course if it it is an element in one of the pieces then it belongs to to the union I mean you can think of it this way like since X age is always a subset of G the union of all X ages it must also be a subset of G right there is no other way so in

order to show that the union of X ages is equal to g i I only need to show the the opposite inclusion so here I have a question mark um how do I show the opposite inclusion well I take an element X in G and I need to show that it belongs to the union I mean but belonging to the union is is belonging to one of the pieces of the Union makes sense and so now I need to show that X belongs to some some piece of the

Union I take the piece of the Union that I take is is exactly X age and if if I take this this cette x age then clearly X is in this H in this cette because I can write X is x * e and e is in H I see I see I see okay okay cool yeah so all in all we get that the we have a partition of of the set g into into cetes and

we go back to the toy example I take g equals to the integers um I take age to be um all the multiples of integers times some some given natural number and if I take the the quotient I mean this H this is a bunch by definition it's a bunch of

cetes um here now I have a um in this concrete example I have a additive notation so the cetes I I will have are um 0 + NZ 1 + NZ up to n minus one plus NZ that's that's one option to I mean a to write the collection of all cetes without

repetition but the claim I had is that the collection of all CET I mean by the claim g mode age in the collection of all cetes is a partition of the ambient the ambient Group G so so it's a partition of the integers and a partition

um partition of the the integers gives rise to an equivalence relation so what what is the equivalence relation that I get from this specific partition anybody that's it's it

shouldn't be a hard question it's n times h i mean I need to I'm asking what is the equivalence relation you get out of this this partition I mean here you have zero 0 plus NZ 1 + NZ up to n minus one plus NZ is a partition of um of the set of integers right I mean

every integer is either a a a multiple of n that is it belongs to 0 plus NZ or it has a a a remainder one when you do long division with n or it has a um remainder you know two up to n minus one when you do long division with n that and that's that's the only I mean it's a an exhaustive description of all the integers must satisfy I mean must belong to one of

these cetes so and and of course any two cetes here are a this joint in this particular description so it's a partition and we said that a partition and you know you could take nals 3 then what I get is is a partition of the integers um in one set I have the a let's call

it a zero + 3 Z and here I have 1 + 3z and here I have two plus 3 Z this is the name of the cetes but what's inside here is here I have I will have the number zero the number three and so on here I I will have one of course also minus two here minus 3 and and so on so this is

H minus 2 one 3 Etc and then 2 5us one and so on and this this is a complete it's a it's a partition of the the set of integers right and we said that a partition gives a gives us an equivalence relation it's it's an equivalence relation on the

integers and my question is what is this equivalence relation I mean we we did see it before something like K plus n z equivalence relation in order to Define an equivalence relation you need to tell me I I give you two elements in the the ambient set a and b and you need to tell me when exactly a is equivalent to B you need to fill in this question

mark exactly the equivalence relation we get is the mod three equivalence relation a is equivalent to B if and only if a is equal to to be mod 3 is this uh is this clear so you see what is happening here that is that we had a construction of this quotient

group the question group was was a a bunch of a bunch of cetes and I could Define the Operation on these cetes ER but the collection of all these cetes is a partition and as a result it can be equivalently described as simply an equivalence relation on on the set ER I mean on the underlying set of the group G an equivalence relation I can take simply the the quent the quent set I

mean the collection of all equivalence classes and the collection of all equivalence classes is is simply I mean it's just an alternative description of the question and more generally this is the typical way to to construct a quotient of of objects we Define some equivalence relation such that I mean when I look at the equivalence classes I get I I get again the same I mean the same type of structure that

that I started with um as another example I take um a field F and I look at at polom over this field now suppose we take a we take some polom Fe in a I mean o over the field

F then we defined a FX modul Fe is the collection of all remainder Pol polinomial H let's call it RX equals sum a i x to the I or I equals to one up to a let's say the degree of V is is n then the remainder polinomial go up to up to level n minus one so let me just write it slightly

better it's the collection of all RX RX is a a z plus A1 X plus a n minus1 x to the N minus one where all the coefficients the AI are elements in the field f so this is as a set as a set FX modu of Fe is simply the collection of all remainder pols and what I claim so we we saw if

um if Fe is is prime polom then a a mod Fe arithmetic um defines uh a field structure on um on FX modul but it I

could also take um the ER the equivalence I could Define an equivalence relation via the following let's take take two two general pols let's say f and g pols over the field a the field F and of arbitrary degree not necessarily of a degree less less than

n i could declare f is equivalent to G if and on if f is equal to G modul and it's an exercise you should you could you should be able to show that this is an equivalence relation and now I look at the

quotient yes yeah sorry m one question when we do maybe not this partition we do F ofx of the field um and we divided by this fee uh what all of the cets that we are getting are like how can we reason about whether they are splitting fields of the original F uh or not ER I have a hard time making sense of what you asked because so what what I'm yeah if you if you have a partition over the field um f defined over a prime

polinomial whatever uh the resulting coets of this would they be like sub fields of f or like how can we understand it when we translate that to Fields ER you're asking how can we understand the example of the groups in the s in the setup of fields yeah exactly yeah yeah yeah and whether is it related to to splitting fields or not ER so the answer is that it's just the the construction as we did it you you could not um I mean there is no notion no sensible

notion of cetes in the the setup of fields I mean cetes they come up in groups in fields you have the notion of of deals but I I did not H I did not discuss discuss it instead if you're looking for a unifying framework I mean a unifying point of view that capture both the this quotient object and the the quotient of groups then what what you should consider that's that's my uh my point in this session

what you should consider is is equivalence relations so so if you want like a to to to view both of these constructions I mean the quotient group and this quotient field in the same in the same way that you should say that you have an equivalence relation that defines the quotient and on on this the quotient set by the equivalence relation you have natural operations in one case of of an abent

group and in the other case of the of a field okay did may I ask also now that you mentioned thing uh why over fields we cannot do this reasoning and in just over groups is because we have a extra restrictions over what is defined as a field such as having inverses or stuff like that or it doesn't matter and it's something else yeah I mean what what we are working over is not actually a field I mean this this creature is a ring oh okay fair enough so it doesn't have a it doesn't have inverses for all elements in fact it doesn't have

inverses for any any polinomial of degree bigger than zero does not have inverse a multiplicative inverse ER so we yeah we couldn't play the exactly the same game as before okay okay but what we do instead is we we look at a a mod fee arithmetic of course you need to assume that Fe is prime otherwise you won't have multiplicative inverses but yeah so this is the story of and

it's it's related to splitting fields in the sense that this is the the construction that allowed us to find one root of Fe I mean in this field Fe if if I if I look at it as as a polom with a new variable capital x then it is xus Alpha times some c x in other words it has a root in this H in this new field K and the whole the whole the whole

story of splitting fields and the and so on relied it was essentially just a repetition of this PR principle I mean once you are able to find one root of of a polinomial then you can repeat the construction find all the roots so but that's that's another another topic thank you okay so what I claim here is that you could realize I this this construction of a a polom modul prime polom as a a a def defined by an

equivalence relation on the set of all pols and when you look at the the equivalence classes I mean the you take the the ambient set take the ambient set FX and you mod mod by the equivalence relation this means recall by definition it means that I take all equivalence classes of elements in the in the ambient set but this this writing has has

repetition so if I want to um to express it without repetition ER I I need to take um an exhaustive set of of Representatives and and this would be in fact one way to to do it would be to take a the equivalence classes of all pooms are such that the degree of R is a less or equal than n minus

one or if you wish the degree of R is is smaller than n that is the degree of P and the reason is that the the reason we have dise equality is that recall F I mean the equivalence class of f is equival is is equal to the equivalence class of G if and only if f is equivalent to G in our case equivalence is equivalence modu of Fe so f is equivalent module of Fe to G if

and only if they are equal modu of Fe but if they are equal modu of Fe I can substract any I mean any any multiplication of a of Fe from H from both of them right I could I could write f as let's say q1 f plus R1 and I could write G as Q2 Fe plus R2 that's that's just by by long division and

er f is equal to G modu Fe if and on if R1 is simply equal to R2 as as pols with with degree strictly smaller than the degre of Fe okay so so in fact the collection of all of all equ equivalence classes under the the modul of relation can equivalently be described

as the I mean the set of all equivalence classes of kind of remainder pols and any remainder polinomial is is going to appear in this uh in this set and um if if the degree I mean if you take two remainder pols two pols of degree less than n and you know that they are not equal I mean if R1 is not equal to R2 then

obviously the equivalence class of R1 is is going to be different than the equivalence class of R2 CU they're if they're not equal and they are already in degree smaller than n then they cannot be equal modul so in particular the equivalence class is going to be different so in other words if I take this a quotient set I mean I take pooms with inter in determinant X and I mod out by this equivalence relation of

modul then a the collection of all coets um r with degree R smaller than than n it has no rep repetitions this this guy and no repetitions so I you know it's it's really what what I want you to get er to

get here is it's really analogous to I mean we find the integers modul n as simply the set of all numbers zero up to n minus one with a a module n addition let's say but we saw that you could Define Z divide by the modu N equivalence relation and this is a the set of all

um of all equivalence classes zero up to n minus one at least this is one way to write it and on this collection of equivalence classes you you can Define um you can define an operation I mean um you take two equivalence classes A and B and if you want to add them you simply do a + b you I mean you could take simply the the the equivalence class of A and B which is it's since we Define the

equivalence relation to be module n it's going to be equal to a + b mod n and then equivalence classes and the so this is the story on on integers but the same story can be can be told for for pols I mean when we took pols modul a prime polom we had instead of a a a the elements Z up to n minus one we had all the remainder pols and but alternatively I could

Define an equivalence relation on the set of all Pol nomials and um when I take the quotient set by this equivalence relation one way to represent it is is all the remain I mean all the equivalence classes of remainder pols and just like I have a modu of arithmetic here I could find modu of arithmetic

here right I mean you give me R1 and R2 let's say you want to add them CL r two and take the equivalence class the taking the equivalence class here at at the end is simply doing module Fe and the same would be for for multiplication I mean you take an equivalence class of R1 an equivalence class of R2 you want to multiply them well you multiply the

representatives and you take equivalence class again taking equivalence class is is doing moding so in other words this this is an alternative description of a of the construction of the field okay we we're over time but ER but I want to leave some some room for questions because this point is a I think is important for for later ER H how do you feel about it is it h is

it clear is it okay I mean because it as always it sounds clear but when you revisit it alone after the session something kind of get a bit messy so maybe it ising the questions the beginning session yeah this is a I mean this kind of approach is the the reason I postponed the issue of equivalence relations

ER to to now because it's it's a bit more abstract yeah and and typically typically the abstract the abstract approaches are harder for a for newcomers it depends on your taste I mean some people actually get along with abstract stuff more easily um and yeah so but but the the the advantage of of this obstruction is that you can you can view both the the

construction of a quotient groups and the construction of this H [Music] um pooms divided by a prime polinomial as as being instances of the same of one same kind of procedure ER so that that's that's what I I was trying to say here list is now much clearer than um when we uh took uh the partition for the first time it

it was like I think fifth or sixth lecture or something like that yeah yeah yeah so ER but but you feel that this is ER this is clear now yeah yeah