PSE Lectures - Ep 16 - Field classification - Algebraic Closure

Transcript

um okay so I would like to start today again with the giving you space for questions and and discussion er do you have any any topic you want to revise so um is it okay we we proceed with the new things I mean h what I have for today is a ER the the section on algebraic

closure and possibly er a few other things so yeah I think we can give it a go yeah okay so let's let's nevertheless do a recap um I mean we have

we have a um for any um Prime power I will write it q q is is p to the N for some H for some n uh there exist um a field fq with um number of elements of

fq is is q p to DN of course this this fq must have a characterist P [Music] um so any field of characteristic P um contains a let let's call it a copy of um of FP right I mean you take a

[Music] one 1 + one and so on and you get something isomorphic to uh to FP um yes sorry so I remember that the characteristic of a field is as you said the amount of times you can s you can plus one your identity until you get back to zero yeah but H how like how is this relating q and B here like so Q is a a power of

P right ER remember that um I mean the a field a field can only have a a characteristic P for some prime P sure yeah so an extension and and of course um the I mean fq with addition and zero is is an a billion group this means that the

ER the order of any element X in fq must divide by by the grung it must divide um P to the I mean it must divide Q which is p to the N right for any EX yep yeah yeah many X in in fq um so er so it means that h x plus X at some point it becom zero right it

it cannot be otherwise um and thean I mean so the the the order of the order of X it must be um P to the k for some for for some K smaller than n right yeah and so since we know so it cannot be any so the the the of course this kind of expression is simply can simply be written as

1+ + one time x yeah okay right and if this is equal to zero right that's that's what it means that the order of of I mean let's this is p to the k then it means that 1 + 1 since X is not zero ER it means that 1 + 1 + 1 P to the K times must be equal to zero

because in a field you you cannot have you don't have zero divisors I mean if two element two elements multiply into zero one of them must be zero yeah okay so okay so now this this means that the characteristic of of fq must be p in fact because it cannot be for for any other Prime time you will not have 1 + 1 + 1 equal Z agree and and that is does that have

something to do with the fact that Q I don't know if that's correct to see with it's an extension field based on P yeah that's that's another way to see it it is I I described kind of a Elementary View but H fq is also a splitting field alternatively fq is a splitting field of the

polinomial X to the P to the N minus X over FP so splitting field is is an extension of uh so it it must it must have characteristic pin thank you thank you very much and and and moreover what we saw it's essentially because of the the

uniqueness ER of splitting Fields up to isomorphism um so up to isomorphism or fields and field um f with a Q to the I mean p to the N elements is isomorphism of fields um between F and and

fq okay so in this kind of um situation I mean you could Define fq in various ways I mean even when I say fq is is a splitting field of a of this polinomial again splitting field is not is not unique on the nose it's it's kind of it's Unique up to isomorphism so to to be kind of administratively

clear we choose one construction ER for fq one one possible way is a we defined explicitly how to do a a a splitting how to construct a um a splitting field from from a given polinomial and for us this this would be the the definition of of fq but of course you could do it differently and you will get something isomorphic so just to kind of disambiguate what we mean I mean fq is a specific

construction and possibly of course a a a many situations can H give rise to to other fields of p2n elements but they will be isomorphic um if you think of it it's a bit like a um I mean we defined the FP as a as integers module p i mean zero up to P minus one with addition and

multiplication modu P but H we didn't have to do it I mean we could also do we could also take the integers modulo the subgroup generated by all multiples of P um and this is this is isomorphic to to zp so I could Define I could Define FP to be Zed modul pz um and it is isomorphic but again to be kind of a clear on what we mean in

kind in the the point set level we fix one One Construction as as the the one we call FP okay so what we want to do now after classifying all finite Fields is to H classify all finite field extensions so um this is just a terminological issue um we say

that um for uh subfield inclusion um let's say F in K we say that K is an extension of f okay so it's kind of two two points of

view either f is a is a subfield of K or k is an extension of F and um what we want after classifying all finite Fields um is to classify all um finite field extensions that means um given a finite

field let's say fq um what are the in the possible finite I mean well sub Fields F inside fq okay so and this will help us in h in what we construct and

um of course one I mean one quick answer is that um FP must be a subfield of fq so we at least we have one subfield of fq um to get to get a bit better view let's um let's have the following LMA I claim is there a

question yeah uh when you tell uh classify all uh all samplings do you mean by classify somehow describe all possible in our case uh field extensions ER yeah I mean ER classify is kind of a soft notion I mean it's a soft statement it's not like a a a rigid mathematical claim ER like like we have in the statements of Lama's propositions and and Theorem but what what it usually means is a to have a a condense or a a

short description or neat description of of all possible objects of a given type okay thank you okay so and this this this is a a recurring theme of we had because H we we talked about classification of a a billion groups finite a bilon groups I mean a a finite a bilon group is originally is defined abstractly right as like something with the binary operation satisfying some Maxum and and the classification of finite Fields oh sorry of finite a bilon

groups was in fact every finite ailan group is a direct sum of of cyclic groups um so this this is considered the classification because it's a neat it's a neat description of all objects of a given type and and the same the same goes for finding Fields okay so the following LMA is going to be useful it's a kind of a number theoretic statement if p is a prime

I claim that P to the m minus one divides P to the N minus one if and only if M divides n um the way to go about it is there is a famous identity um if n x is kind of a polinomial identity I can write x to the N minus

one is um x - one times x to the N minus one plus and plus this term and uh I mean the way to see it you get I mean things cancel out I mean you you do x times this you get x to the N ER then you multiply a a then you get a bunch

of of other stuff like x to the N minus one up to x times times 1 which is X but then you do 1 time x to the I mean -1 * x to the N minus one so you get minus x to the N minus one and you continue doing you you basically cancel out um all the Expressions um minus X but then the the

last expression does not cancel out the last expression is minus one * * 1 which is minus one okay so this this is indeed e x to the N minus one um okay let let me just ER I can I can write it like this x to the N minus one is x - one times the rest of it and this is

a this is the key the key fact for the the proof of dilemma um now how do we proceed so we have an if and only if here ER let's do One Direction um if M divides n so let's say um n is equal to M * k then I write P to the N minus one as h p to the m to the K minus

one and now taking in this identity x equals P to the m what I get is like P to the m minus one times um P to the m or let let me write it shortly as sum P to the m to the J where J is is from 0 to K minus one yes if you if you did the

substitution of x equals P to the m but you kept writing everything with P to the m is that like is that intended or yeah no I didn't do the substitution I I just I mean I'm just saying H I'm using this box star okay fir enough fir enough okay I see okay I see thank you um and from this it follows of course that P Tod M minus one divides P to the N minus one conversely

um if P to the m-1 divides P to the nus1 um and I write um n equals and let's say mq plus r i mean I do long division of n by m then of course this R is a

is smaller than M and bigger equal to zero and what I can write I mean p to the N minus one divide divided by P to the N minus one this is we assume to be an integer but it is also equal to P to the mq+ r minus one or let me write it slightly differently um here I have P to the m minus one and P to the m q + r minus P

to the m q a plus P to the mq minus one and now I just I just split it so this is a p to the mq plus r minus P to the m q ided p to the M minus1 plus P to the m q minus one divided by P to the

m minus one um yeah so let me call this uh this thing a t um of course t t must be equal to I mean t must be an integer sorry actually um we write it as a p to the mq um P to R -1

/ um P to the m -1 plus T um now T is in Zed t t is an integer because um we we already proved that um I mean since m m divides M * Q we know that um P to the m q minus one is divisible by P to the m minus one

this is the first part of this LMA um if T is in Zed it means that m i mean p to the m q time P Tod rus1 / P to the m minus one is also in Zed right because the initial expression was an integer um and this means that P to the m minus1

divides P to the r minus one but I mean r is smaller than M right R was the remainder of doing a long division of n by m so so we must have our equals z and this means of course that M divides n any question about the the Lemma I have a question so in this last

step how do you get that R is zero like okay it must be uh smaller than M but why zero and I mean I that this means that m is is it's one but I'm not sure I mean if m is bigger than than than R then P to the m minus one is bigger than P to the r minus one yes so so it's it's clear that R must be smaller than m but what I don't see is why this leads to R being being like

zero because I mean how can a bigger number divides a smaller number the only way I mean the only way that a bigger number can divides a smaller number if the smaller number is zero I mean then okay everything can divide zero okay okay I I I'm sorry thank you thank you very much yes uh can you explain a bit uh again these manipulations with uh powers that help us to understand that something is an integer ER you mean this this

[Music] part I think they are they are very similar that's why like any uh any any any part for example like uh but the the second one you say that like now T is in that uhhuh yeah I I claim that okay so H I claim that t is an integer because M divides M * Q I mean we had the the the first part of the claim was that if M divides n if n m divides n then we have this purple division

relation yeah right okay so I mean I use it when I claim [Music] that t is in Zed I mean M divides M * Q so P to the m minus one must divide P to the mq minus one it's like you put you you can call this guy n or is it is it like a a rule and like I I understand that like M divides mq but I don't understand why uh P to the power mq minus one will divide uh

will be divisible yeah yes I mean this is the first part of the claim if you have two integers M and N or yes yes yes yes okay okay okay okay okay I understand okay yes so so yeah so T is in Zed and since T is in Z then this guy must be in Zed right because the whole I mean we started by the assumption that a p to the N minus one divided by P to M minus one this guy is in Zed is is an integer so if T is an

integer is this clear you have any other doubts no okay okay okay it's okay now okay one one thing I didn't say is that I mean why why do I I mean why does it have to be if this is an integer why does it have to be that P to the m minus1 divides P to the r minus one I mean couldn't it it maybe it cancels out with P to the m minus k Sorry with P to the mq but but the gcd of of a A pmq and P to the m minus one is one I

mean they they're not divisible by I mean they don't even have a a common factor right clearly I mean if you do ER you do long division you get a you get something non zero I mean you get a a so um so okay this is a kind kind of a technical issue but it's going to be useful for us ER later or soon um and this is the following

Theory let a q be a a Prime power um and then for every M that divides n there is a unique subfield

F of fq with a exactly a p to the m elements um and moreover every subfield of fq is

obtained is of this form is is of this form okay so this this you can regard this this statement is like the classification of um finite extensions is is the statement clear it's

a um it's a bit long so yes so to to summarize what you're trying to say here is for every root of M so for every divisor of M that we can find that doesn't give the residue also uh we can derive from there a a subfield for which the original field can be seen as an extension fi off exactly so what the statement it tells us is I mean you can view it this way I start with FP then I could extend it to FP squ for

example and I could also extend it immediately to f p Cube but I don't have an inclusion here because two does not divide three y however I could go for f p to the six and FP to the 6 does include FP to the 3 because 3 divide six and it also includes I mean f f p to the two because two divid

six so yes yeah sorry but now that you're saying that it also made me wonder if we are extending Fields what's happening with the Primitive pooms over which the field lays on like I mean the prime pooms the notion of a a a prime polom is intimately related to the field in which it is defined over so I could have a polinomial here F FX which is prime but but here it may not be

prime right exactly exactly okay and once it is not prime I mean if it completely splits I mean if if F splits here then it also splits in any in any extension because but but but it's a bit unfair because you can always split right like you can always find a splitting field you just you might need to go to the complex numbers or whatever but you can always split it somehow yeah yeah I can eventually I can I can always split but I mean I don't know

that necessarily it splits in FP Square what could what could happen and what typically happen is that um you extend you you you after one extension let's say ER this one after one extension you find a one root or two Roots but it doesn't completely split so you need to go further to another extension okay and then you get more roots and you continue until you get all the roots okay that's basically how we constructed the splitting field right we

did it in stages okay but what I want you to to to remember from this this picture the splitting issue is not is not the the main the main thing but rather that we don't have a subfield inclusion or a field extension in this case in this purple case yes sorry mad don't want to stay here for but I have another question yeah U so so far we are proving this inclusion of non inclusion

basically by uh division without reminder so exact division uh but depending on the Prime on the Primitive polinomial that fp2 and fp3 are defined over these fields it did could could be like could have this inclusion no Okay so is it true true it's true so this is a this is important when I write here in included I mean not included here H it doesn't mean they they are not they could be there could be an inclusion of of

sets for example FP in in our definition is just the number numbers Z up to P minus one so you could say it is included as a set it is included in the real numbers right the real numbers contain Z up to P minus one but this is not a subfield inclusion okay I see I see I see I see see and that's the important it's really it's really crucial ER we we care the most about I mean we essentially only care about sub inclusions because the

fact that the two sets are included um is not very helpful if the the operations are different yeah okay so so depending on our definition on our kind of Point set construction it could be that FP square is included in FP but they will not be according to this theorem they will not be subfields I mean or field extension okay thank you thank you very much okay so um let's prove the theor is

suppose M divides n we showed that the multiplicative group if Q star it means you take fq you remove move the zero element and you you look at it as as a as an a billion group with respect to multiplication ER and we showed that this group is I mean

actually there is nothing to show this this is a group is of order Q minus one which is p to the N minus one and by LMA um P to the m minus one must divide P to the N minus one

um now if you recall this was a coroller of the classification of finite a billion groups we had what what we called Converse lrange theorem it means that if you have an a billion group and you have some number divided order of this group then you must have a subgroup of of this order so by the converse lrange e the there exist a

um a subgroup which I will call Kar and this is a subgroup of fq star of order P to the N minus one and now Define k to be this K

star union the zero element of fq um so of course K by definition I mean k Star by definition has a multiplication but in fact it also has addition it has addition because we can induce the addition from it is a subset of fq H so we can use the addition of fq to to add two elements in in K

ER we will observe that um we can restrict addition and multiplication in fq ER to 2 K the question is so so K has a it has a zero element right this by definition it

has a um an element one because it is a a multiplicative subgroup of fq star um and we can add two elements and we can multiply two elements if I If I multiply two elements in K I mean in case star for example then I I know I get a new element in Kar this is because Kar was was chosen to be a a subgroup um but I still need to to justify why if I add two elements in k

um using the addition coming from fq why do I still get an element in K so if a and b are two elements in K well um ah let's let's view it this way um

so this we use lrange Converse lrange but we can use lrange theorem every element in K star and satisfies uh x to the um P to the what is it m minus one equals

one right because the the order of the group K star is p to the m minus one so of course the order of every element must H must divide P to the N minus one if I raise H the element to the the P to the m the order of the group I must get one but of course this is equivalent to say that um for any X in k x to the P to the m minus X is equal to Z this was for any X in in K star but if I

do it for zero that's I mean this trivial um what this means is okay and now I take two elements in h in k then and if I do a + b and I raise it to the P to the m then by the The Freshman expon exponentiation this is a to p m plus b to the P to the

m but by by this identity St are since A and B are in K I get that this is A + B um so A+ B is a root of x to the P to the m minus X and this means that a plus B is again in K because K has exactly P to the m elements and therefore all the roots of

of this polom I mean all the elements of of K must be precisely the roots of this polinomial this this polinomial has a has P to the m Roots so if I if I know that a plus b is a root of this polom I get that a plus b is an element in k h this means that K is is closed under addition of course it's closed under multiplication and you can show maybe it's good it's it's a good exercise um K is is also closed under

um additive and multiplicative inverses yes a question yes um I I'm sorry I messed like the I think I'm not sure about the last step uh so I understand why a plus b to the power of PM equals a plus b but then the conclusion of a plus b is a root of x to the P to the to M minus X I'm not sure I see it and why is it then a field an element in K like the last two the last

two steps I'm not sure I really understand them could you go through it yeah yeah so so the second to last step is fairly easy I mean if something is a root I mean if some something is a root of x to the P to the m i mean X is a root of or let's call it maybe t t is a root of e x to the P to the m minus X if and only if t to the P to the m is equal to T So if I prove that A + B to the P to the m is equal to a plus b it means that it's a root makes sense okay now I

claim the number of elements in K is exactly P to the m right I mean we started with a subgroup a multiplicative subgroup of size P to the m minus one and we added zero so we have P to the all the elements of K satisfy are roots of this polinomial right because all the elements in case satisfy by the gr all the elements in K satisfy x to the P to the m minus one = 1 okay I guess I'm not sure I followed

the step from this one to the X to the P minus xal Z the star well just multiply by X okay right it do work okay so I I know for any element in K star satisfy satisfy e to the m minus one equals 1 yep makes sense then and now it means that every element in K satisfy x to the P to the m minus x equal Z yes so all the elements in K are roots of the polinomial X to the P to the PN minus X but a a this polinomial can have at most P to the m Roots so it

means that all of the roots of these pols are the polinomial are in in one to one correspondence with the elements in K makes sense now okay and now I take some some pair of elements A A and B and K and I add them together I find that a plus b is again a root it means that a plus b is is in K perfect thank you makes sense okay so as an exercise you can prove that a if I take a minus a for example if a is in k then minus a is in

K I mean a is in K I claim that minus a is in K and also if a is not zero um then a minus one is in K I mean this part actually you you don't need to show because um K star was a a a multiplicative subgroup so it contains all the multiplicative inverses ER the way to show that you have a an additive inverse I mean that additive inverses are a also

in K is I mean the additive inverse would also be a root of this polinomial okay so H so it must be in and this is all to say so to sum up we have a a a subset of f which is a field and so this is is is a sub field of size of size P to the m okay so this but of course this this

relies on the fact that M divides n um so so now we turn to uniqueness so if f is a subfield I mean the claim of the the theorem I remind you is that there is only one I mean for every M dividing n there is exactly one subfield of fq of size P to the m and

and no more I mean so so so to show this this this second part of the frame I mean the uniqueness let's suppose I have a a subfield then um clearly by gr the order of f i mean f is in particular a an a

billion group under and it if it's a subfield of fq it it in in particular it will be a a subgroup of the additive group of of fq so it means that the order of f must divide the order of fq which is p to the N but of course so so the order of f is some some power of P P let's say p to the M for some n for some

M um yeah ER now the multiplicative group of f is a a cyclic of order P to the N minus one um actually the cyclic is not

necessary so its element are precisely uh the roots of x to the PM P to the m minus X or actually uh since we're talking about F star I guess we could say it like this so H so the

elements of f are precisely the roots of the polinomial X to the PM minus X and well I mean over over fq and this this Define so this defines uniquely ER the set the set f um that's if

is unique you Sor um is this clear this this uh completes it was a bit combersome way but is this this issue okay

hello yeah yeah I think I put up and no one said anything so it's clear okay so let me just quickly revise and we can then take a break um I mean basically what you you see is that H the only way the only way to get a subfield of fq is to have um something of size P to the m where M divides n this is because of the LMA I mean it's because of the LMA tells you that ER that P to the m minus one divides P to

the N minus one if and only if M divides n so by kind of lrange argument that's the only way you get sub field and and then you you see that any such subfield must be the roots of the polinomial X to the P to the m minus X and this this essentially pins down the only option you you get I mean it means that there is a unique subfield of this form okay and then so this this this picture here is uh is the way to view

it let's uh let's take a 10 minutes break and uh we come back to to define the algebraic closure and the the defining property um can we start again y okay so what we have from the the classification of a of finite field extensions is um we can construct a tower starting from

FP going well if I if I do FP Square I could do FP FP square but then I do not have I do not have a field inclusion into FP Cube so instead what I'm going to do is um I mean I I do FP square and it is included in FP 3 factorial

so I regard it like this of course I mean n minus one factorial always divides n factorial so I have this Tower and and this this can go to Infinity right and in fact I don't have to start with FP this is FP I could also start with fq and I know that fq is

contained as a subfield in fq squ of course q q here is as usual a power of a power of P and F Q sare is going to be included in f h Q 3 factorial and so on up to Infinity so what I'm going to do I

mean the algebraic closure of F Q is H simply you take the union over all these I mean of all this Tower this is fq to the N factorial where n is an integer so is a natural number so maybe

n goes from one to infinity and the the the important property of the algebraic closure so of course the algebraic closure is no longer finite field but nevertheless it's very useful for us in studying a a finite Fields um well first in proposition I claim that fq bar is a

field the way to see it is a yeah soorry I'm a bit confused by the definition of the alic closure because if you start by uh example if you put n equal one the first so you have fq to the factorial of one and then you adjoin fq to the factorial of two but we know that fq to the factori of one is already included in fq to the factori of two so

like you don't add new stuff because only the last I mean it's not the last because it's Infinity but only the last step will each time you increase then you you already contain the previous field that you defined before yeah that's exactly I mean that's exactly the reason I need to take it I need to take the infinite Union because in any finite piece okay let me just maybe say where we heading what I'm going to prove is first that this algebraic closure is a field and second that every polinomial splits over the algebraic

closure so and and of course you cannot expect that if I give you a polinomial of degree D and if D is bigger than than n or if D is bigger than n factorial or or if if D is bigger than a q to the N factorial then F cannot possibly split over a F Q to the N factorial right because there is only Q to the N

factorial elements and the degree of f is bigger than that so it cannot have a all the roots in in fq to the N factorial so what happens is that if I want kind of a unifying framework in which every polom split then I need to go up to in the ladder up to Infinity I mean I kind of push my my problem to infinity and then it gets solved so that's the reason I need to I need to take the infinite union now of course you could say okay I mean if you're only interested in F to

the Q3 factorial you shouldn't care about this and about I mean about the previous the previous pieces but what I really care about is not any of the finite pieces but rather the the union of all of them is this a answering your question uh yeah kind of at least I see the why we need to go to Infinity to have this any for any poal makes sense yeah so uh okay but we I I will prove it it's not a I mean it's not yet rigorously

shown ER okay so let's let's first prove that the algebraic closure is a field [Music] um so first FP is contained in in the algebraic closure it is contained in F Q to the N factorial for any n and this means that it is contained in

the in the infinite Union this is a subfield of course and and it means that it is contained in the in the union which which we Define to be the algebraic closure for now we only we only have I mean we only know that it's a set inclusion we cannot possibly say that it's a subfield inclusion because we didn't prove that fq bar is a field that's that's really the the claim um but in particular so uh

so we set the zero element in fq bar and the the unit element in fq bar to be a those of of FP 0 FP one FP now how do I um how do I Define addition and multiplication in fq bar to

Define addition and multiplication let X and Y be two elements in fq bar if an ele I mean if I have an element in the union it must belong to some finite piece right I mean then there

exist n and K let's say such that X belongs to F Q to the N factorial and Y belongs to F to the Q K factorial now I don't know maybe n is bigger than K maybe K is bigger than N I don't care because these guys must be subfields of f to the Q to the N factorial K

factorial so it means means that X and Y are elements in this bigger field and in this bigger field I can multiply and add X and Y I mean possibly I don't know how to add X and Y if I think of them as elements in these H in these fields because it could be that n does not divide K and also K does not divide n so If X and Y do not live in the same place I extend it enough so that they will live in the same field and then I can add in and and

multiply them ER and so so addition multiplication is defined in this way ER clearly if h X is a a an element in fq let's say x belong to the the piece fq to the N factorial then x - one and also minus X would also have to belong

to H to the same field f Q to the N factorial hence they belong to the the algebraic closure okay so this this basically now associativity all the axioms say associativity ER commutativity and uh

distributivity of the plus and multiplication in fq bar follow uh from and the corresponding axioms in in each of the finite pieces

okay so so what I what I conclude is that fq bar again the union of F Q to the N factorial is a field ER is this uh this proposition clear yes so it makes complete sense to me I'm just wondering uh doesn't having a characteristic of a field being

infinite uh cause any kind of problems or trouble uh when when formalizing fq uh fq of I'll say something factorial again Carlos can here yes so what I meant is if you have um you're basically saying that fq of in factorial has an infinite characteristic a thing no no no no that's that how it doesn't have an infinite order it has an infinite order but not infinite characteristic recall that a characteristic of f is the minimal number oh you're right

okay I see you're right you're right such that um 1 + 1 + 1 equal Z if this minimal number exist of course if it doesn't then the characteristic is zero so any field containing since fq bar sorry fq bar contains f FP it means that fq bar is of characteristic P this you have no choice because the the yeah but it is of infinite order it's kind of a it's a new

phenomena for us a field of finite characteristic but of infinite order okay but a same since it follows the aums it doesn't prevent us from saying that it can actually apply all of the corresponding operations all the has an infinite hence we don't know the mod that we are applying yeah I mean we never we in all our development of the basic theory of fields we never restricted the definitions to include only finite Fields I mean and indeed we talked about the real numbers the the rationals and so on so a field does not need to be

finite it just it's new phenomena I mean until now the only positive characteristic fields that we saw were finite but in fact this this this new construction gives us an infinite field of positive characteristic understood thank you maybe I can I can add that if a field is of characteristic zero then it is automatically infinite because it has to contain one 1 + one 1+ 1+ one and so on and and all of them cannot be equal to

zero so they are all different because I mean if yeah if if 1 + one would equal to one then it means that one is equal to zero and so on yes so uh just I want to make make sure I understood because um what you just like the the fact that I think it's it's the thing you already saw before but the fact that the characteristic of f so sorry the algeb the character the characteristic of the algebraic closure is the same as the characteristic of FP is because the one you define is the one from FP but you could really Define a

one from from another extension field you could Define a new one right ER I could I mean it's to be more precise it's because I defined the one in in fq bar to be this this here in the blue box but I also Define the Operation to be a the operation in induced from any of the finite pieces any of the finite pieces contains FP p as a subfield so it means that if you you do addition of the of the element one in any of the finite pieces it's

like doing addition of the the element one in FP you see my point yes because my my next question was related to this like why why can't you choose another definition of one in in the algebraic closure you could define the zero FP on FP zero FP square and one FP Square for yeah but because because they all contain FP as as a as a subfield the zero and one in these extensions must be the the original zero and one in FP but you cannot Define zero and one in

from FP Square for you have to choose the same zero and one that are the the one from FP no no I don't have to I mean okay what I'm saying is you could choose a zero fq bar to be a zero fq yeah okay for example but what is zero fq fq contains the subfield FP Y what does it mean that it contains it as a subfield it means that the 01 in FP are the same as the 01 in in fq okay that's the definition of what we

call a subfield then that's yeah subfield it's like a a subgroup a subgroup must contain the unit element of the original the neutral element of the original group perfect thank you okay so in fact here i i b basically had no choice but defining the the the elements I mean the zero and one elements is is is the one coming from from FP yeah I see that now okay thanks thanks a lot okay so this is this is the are there any other questions this is the algebraic closure

ER it's a field and and now I claim that we have kind of [Music] a perfect a perfect setup in the algebra closure in the the sense of the following theorem um let a q be a Prime power ER and I I take some polinomial I don't care what is it

in pols over the algebraic closure right we just prove that the algebraic closure is a field so I can take a um Pol a polinomial over this uh this field then F splits to linear factors and let's say f is equal to Alpha time the product of xus alha

I I = 1 to n of course for for Alpha and Alpha 1 up to alpha n some elements in the algebraic closure okay so how do I prove it what the the the the kind of line of argument that we do in in algebraic closure is we move to a a finite setup

um if we can and what happens in a polinomial I mean f has um finitely many coefficients let's say a a z A1 up to a n each of the if each of the

AI it is in the algebraic closure so each of them separately must belong to some finite piece I mean this the algebraic closure is a union of of of finite pieces so all of them together must belong to some H to some finite piece so there exist let's call it a capital N such that a z A1 up to a n belong to F Q to the N factorial i

e I can write F or FX as a polom over fq to the N factorial okay um so now I I can do yes can you give a bit more intuition about um writing f ofx um within fq of factorial again I'm I'm sure I follow like why is this relevant or what is it

like I mean what I want okay let me just so what I want to do now here is I mean I will construct a splitting field of f uh the splitting field of f a splitting field of f of F and ER is a um f i mean it is f to the q n factorial times um I don't know D

for some D but but this splitting field is of course is just is contained in in just another finite piece let's call it f to the Q in Prime factorial ER and and this guy of course is contained

in the algebraic closure this means so I mean of course if a polom already splits in a sub field then it splits in the the extended field right I mean so F splits in in F Q to the N Prime factorial hence also in fq

bar okay so this let me just repeat kind of the the line of argument um we know that I mean we constructed splitting Fields over finite Fields ER so if I am given a polom over the algebraic closure the way to see that it must must splits in the algebraic closure is first to go back to the finite case the way to go back to the finite case is that a polom is a finite creature it's like it has finitely many coefficients if it I mean because of

this it means that it must belong even though the coefficients are officially in the algebraic closure all of them must belong to some finite stage of the of the Tower and if they are in in a finite stage of the tower they all belong to some finite field and I can I can construct this splitting field over this finite field which will again be a just another another finite field hence contained in some other finite piece of the Tower and once I know that the polinomial is splits over a subfield it must split over over the bigger

field um any questions about this uh this proposition this is really like Central to what we will do later yeah yeah I like I didn't I didn't follow exactly like the Pro itself I would say to be honest um could you if like the intuition from maybe not a different perspective but phrase it on a different way let me try to repeat with like maybe

slight modifications ER look the statement is that every polom over the algebraic closure splits to linear factors I mean splits and splits to linear factors is is the same I mean for us splits means to to decompose into linear factors H now in particular if if you look at the statement of course if if you have a polinomial over FP then in particular

it will be a polom over F FP bar right because FP is a FP is a subfield of FP bar if I have a polinomial over a small field I can think of it as a polinomial over a big field but I can also do but in the case of fq bar I can also go in the other direction how do I go in the other direction F has finitely many coefficients I mean let's say f f is equal to a sum AI X to di I right yeah sure and all and all of the

coefficients a z up to a n they I mean by definition they are elements in the The Big Field fq bar yeah exactly but what is fq bar fq bar is the union of all fq to the N factorial nals 1 up to Infinity if a z is an element in fq bar then it means that there is some h n let's call it n0 such that a z belongs to f Q to the

n0 factorial you see this yeah yeah yeah now I follow okay and the same goes for the the rest of the the coefficients I mean all of them will belong to some finite stage of the Tower yeah H but once all of them belong to some finite stage I can take a I mean oh I see so once once all of them are in a in a particular let's say floor of the tower you can get the maximum floor which will be finite and notore okay okay okay okay I'm so stupid yeah thank

you there will be there will be some n i mean basically you could you could take n to be ER you could take n to be the product of all um all the NIS I equals well I mean you take all the product of all of them ER and you get some some kind of I mean some stage of the tower for which all of the coefficient of the of of the polinomial f are in this stage makes sense okay but if all the coefficients of f are in this stage I

can I can think of f as as a polom over this subfield I mean this this guy is a subfield fq to the N factorial is a subfield of fq bar sure yeah okay so I can think of f as as a polom over the the subfield now this subfield it's a finite field so I I can do do the splitting I have a splitting field yeah correct the splitting field is O of of a polom over a finite field is again is again a finite

field yeah sure yeah right and so it must be contained in some in some stage of the tower oh okay I see I see I see now I see I see that Mak yeah because the tower remember the tower let's say I start with fq then I do F Q to the N factorial sorry F Q uh 2 factorial and so on here I have F Q to the N factorial if you give me a finite field of the form fq to the D of course it is a subfield of fq to

the D factorial so it is always any finite field is contained in some stage of this Tower sure that makes sense including the splitting field of f so the splitting field of f is contained in some stage of the Tower and hence F must must split over the algebraic closure okay I okay the only thing that now I'm missing but I guess we will say it later is how this is useful for us in in later stages like that we're able to prove this exact thing because looks really abstract to hold on some yeah it

it is abstract um but you can think of it as a is kind of a a a a unifying frame work for pols I mean of course I mean for let's say I even I even only care about a pols over FP but they don't have not all all the polinomial over FP have roots in FP oh okay fair enough ER and so of course if I take if I take one particular polinomial if you

you told me in advance what what is the polom you're interested in I I can take a a a splitting field a of of this polinomial it will be some finite field and I can I can look at this pols over the splitting Fields ER and then it will be good for one particular polinomial that you chose in advance but if if I don't tell you anything in advance and I want I want a framework such so that every polinomial you throw at me will have will split then I need to push I kind of I

need to push these constructions of a splitting field to Infinity yeah that makes sense that makes sense okay and and and what it does for you is kind of it unifies the statements you want to make about pols without a a a um without choosing in advance a particular polinomial very thank okay so um other questions about the algebraic closure let me finish the uh the session with

the um what is called the fundamental theorum of algebra and this says that um every polinomial over the complex numbers H splits splits to linear

factors okay so this is a this we don't prove um but it and I also I didn't Define the algebraic closure of a general field but the What's Happening Here is there is a a wider definition of algebraic closure even more abstract than what we did that is applicable to any field and if you take the algebraic closure of the real numbers you get the complex numbers okay and and an even more interesting thing would be to look at

the algebraic closure of the rational numbers the algebraic closure of the rational numbers would contain the rationals but and it will be contained in in the the complex numbers but it's not equal to the complex numbers so there there is a lot I mean understanding the the the algebraic closure of the rational numbers is is like a modern research topic ER so that's that's kind of the slightly wider picture yeah all

right yes Carlos are there any resources like that are visible or accessible to read about the fundamental theorem of algebra about what about what you just explained ah ER yes I don't have something off the top of my head um because we will not see it right yeah we we will not see it because we don't need it we we only need we only need the algebraic closure of finite Fields Mak sense ER we could potentially use I I'm not sure we even use the fundamental theorem of

algebra um but it is useful to know I mean it kind of it tells you that once you you add the the the root of the polinomial X x s + one you kind of automatically add all the roots to all the pols so that's a it's kind and it's a nice property okay thank you thank you Madame all right um next time of course you can you can always come up with questions on previous

material um but I I have something prepared for next time this is a well equivalence relations and May be the projective space awesome okay so H have have a good day cheers thank you bye bye bye thank you