PSE Lectures - Ep 14 - Field classification

Transcript

so um welcome back and uh ER I I would like to to start today's session ER with a an example on splitting Fields so um suppose

you you would like to get um a polom I mean a splitting field of the the polinomial X to 4 minus 2 over Q okay so the the potential roots of this polinomial I mean when you factor it over the complex numbers I mean over the complex numbers this we didn't h we didn't discuss but over the complex numbers every polinomial splits completely to to

linear factors and and this polinomial is no different ER specifically you you could write um x minus or let me say the roots of f are one a pair of roots is a a fourth root of two but it could be a a positive or

negative and another pair of roots is a a fourth root of of two * I okay and since we know that a polinomial cannot have a um oh wait um actually it this would be here it would be yeah it should yeah it should still be

it should still be fourth root of two and and this this is the complete set of roots but none of these roots belongs to Q so over Q um f is prime um in qek okay so what do we do we we have a a recipe to H to con construct the splitting field of f we we look first at Q of x modul

f and this we know by by General kind of abstract reasoning that this is a field um with um a root let's call it Alpha of um of f i mean but if considered as a polom in K1 of X

now again this Alpha is kind of an abstract a root that that was derived from a a meta trick but and we could choose either of these uh these roots to H to to be Alpha I mean um from the point of view of the this first extension K1 it doesn't matter because Alpha is kind of a I mean K1 would be q a jointed with Alpha namely the minimal field

containing q and containing Alpha and Alpha is is just some kind of creature that if you raise it to the four you get two right so so say Alpha is equal to square root sorry fourth fourth root of two then I look at and now I look at the h q of fourth root of of two this is this is like you doing the complex

numbers right I mean you you look at all the the expressions of the form a + b 4th root of two where A and B are are elements in q and this is at least isomorphic to K1 so o over K1 I can say um that FX is um x minus Square fourth four root of

two X Plus plus 4 root of two but there is another Another Part which is not yet [Music] um decomposed or or splits and this is x² + square root of two okay over I mean over K1 so what do we do I I take the the

part of F that is not linear and it is assuming it is a prime I mean of course we can see that this is prime because a a the roots of this needs to be something with with I and I does not belong to this H this first extension K1 so what I need to do is to further divide I I Define fub1 to be x² plus Square < TK of two and now I look at K1

X modul x² + root of two okay and here I am guaranteed to have a um I mean F1 um has root say alpha 1 over um this new field I mean k K2 and it means that

K2 is the minimal extension of K1 containing Alpha One now again there is a flexibility I mean alpha 1 would be either um either plus or minus fourth root of 2 * I but it doesn't matter because um the minimal field containing K1 and fourth root of 2 * I is the same as the minimal field containing um K1 and and four root of two * * I * minus

one right so you see once once you have either of them you get automatically the the other and so um F1 splits over K2 namely I could write fub1 equals um x minus 4 root of 2

I X+ 4ot of two I and and since F1 splits then I get automatically that that the original polinomial F splits and we get that K2 is the splitting field at least as splitting fi of f over

Q okay and and K2 you could write it as a basically all you need is to adjoin the fourth root of two and the element I once you join these two you have a a you have everything you need I mean and that's the minimal the minimal thing you can do ER minimality also follows um from minimality in each of the stages I mean here um or

maybe maybe here since K1 is this Q adjoin fourth root of two K1 is the minimal field containing q and fourth root of two and similarly K2 is the minimal field containing K1 and four root of two * I so all in all ER K2 is the minimal field in which F splits okay is this a is this clear I have a question uh when we um

introduced QX as over f as K1 by QX did we mean uh polom whose coefficients belong to rational numbers or we just call it through X yeah yeah yeah yeah that's the notation I mean if I write FX it means pols whose coefficients are in F in a field F yeah but uh at some moment we also had I but we didn't introduce uh c x yes but

um well this this is all kind of under the the the umbrella of the the fact that a c is contained in sorry A Q is contained in C okay right ER so I mean this this this example is simplified intentionally ER so I I know I know in advance what are the roots I'm looking for and that's why it's easier to explain but the the general construction does not care

about what you do I mean what you know in advance you simply mod out by a prime polinomial you're guaranteed to have at least one root after you mod out by this Prime polinomial and if it splits completely you stop if he doesn't you keep doing it so it doesn't matter eventually I mean you don't need to know the roots in advance to know that this construction eventually terminates into a splitting field of the the polinomial okay so this is this is just to to to be a bit more concrete I I I

wrote the the the roots in advance okay so this is a this is how you go about splitting fields and and what we proved last time is that a splitting fields are unique up to isomorphism splitting field for if

a is unique up to isomorphism a i e if e Prime H is also a splitting field if

then we get an isomorphism between e and e Prime yes there was a question uh yes I think this is what you hinted before but for K2 you you define it very generically with just a potential root Alpha and then you directly said that Alpha is I or something like I I just miss the step what you define it generically from K2 to the this this step here so you define K2 K1 aded with alpha one but I no no no no no that's the definition yes K2 is you take K1 you take whatever is left after

you you split off linear factors from F yeah in K1 you take whatever is left we call it F1 and you and you mod out by it sure okay that's the definition and what we showed last time is that this kind of construction H is always isomorphic to um to Simply adjoin I mean take K1 and adjoin some kind of abstract root to the the polinomial you mod you mod out with okay so because you're saying this

when alpha 1 can only be uh I uh can only be Alpha 2 I square root of two * I I mean Alpha 2 Sor alpha 1 it needs to be a new root yeah yeah right I mean of course a a fourth root of two is also is also a root of this polinomial sure it's already containing K1 so it's already it's already in K1 in order to since I know that K2 has a new rule yeah then then I know that alpha 1

is is either four root of two * I or or minus of it makes sense I mean this thank you okay so and the splitting field is is unique up to isomorphism a ER Now using this uh this effect we are going to classify all finite Fields very easily but I need a I need one more ingredient let me call it an observation so uh let let F be a

field and um let's say I have a polom of the form some a i x to the i i equal 0 to n root Alpha of f is a is called repeated root if um not only that you can write FX equal xus

Alpha times GX but you can write it at least with a power of two I mean potentially more but but at least with a with degree two okay for some h g ofx and what I claim is that there is a simple a criteria to to see when a polom has a a a repeated root

um and consider the formal derivative DF which is simply the polom sum I time AI * minus one I equal Z up to up to

n okay this is like a it's like doing derivative that you know from a from school H to this this formal H expression I mean you know how to do derivative to each of the monomials and you you sum it all up and I claim that ER F has a repeated root or let's say if if

f is a repeated root then um f and DF have a common factor factor of a degree bigger equal to one right that that should be obvious

because if have I mean if F has a repeated root then I can write FX equals X - Alpha squar * GX and this would mean that DF of X is H you know 2 x 2 * X xus Alpha times some G Prime of X right H so so all in all I get H that xus Alpha is a common factor of of F and

DF okay is this is this point clear y okay so now we H we get to the the main goal and this is the uh I guess you can call it structure theorum of finite fields and the claim is

um you you fix a prime then for any and in N there is a field um fq with um Q

equals P to the N elements uh moreover fq is unique up to an isomorphism okay so this this is the the result we were aiming for and the proof now should

be ER it should be very easy to uh to follow H since we we talked about all the ingredients so the statement of course it has two parts existence and uniqueness so let let's first do existence I consider the polinomial x to the Q minus X in fpx

and I let e be um a splitting field of f now clearly the characteristic of e it must be P because e has to contain FP and and once once a field contains

contains FP it has to have a characteristic P this we discussed um and I I claim so we may View f is a polinomial over e

now since the derivative of f is um Q * x q -1 - one but Q is a power um yeah I I didn't say it but okay yes I guess I did I did say it here so Q is a power of p and therefore a q is equal to zero in any in any field of character istic P so this is minus

one and since I know the derivative of f is is is simply a constant then it cannot have a common factor with with f so I know that F has no repeated roots okay now I consider now the set of all roots of

f let's call it K this is all the um all the A's in a in E such that a to the Q minus a is equal to Zer and I claim that K is a subfield of E ER now how do you see it so first of

all ER zero is clearly contained in in K I mean zero is is indeed the root of the the polinomial X to the Q minus X now suppose I do I take two roots and I uh um I add them together I want to know that if I add two Roots together I still get a root I mean I want to know that K is closed under addition but I also want to know that K is closed under additive inverse so there is a quick way to H to do both of these axioms in in one

go um by doing the following if a b are elements in k then I can use this say remember this freshman expon exponentiation I mean if you do a minus B to the Q it's the same as a to the Q minus B to the Q I mean we had it for p in any characteristic p H field a minus B to the p is a to the P minus B to the P remember

this and by by a trivial induction you get the same result for any power of p i mean you you I mean take take Q equals P squ for example then you simply applied apply this principle twice and you get the result is this clear so of course if I have this equality then since a a and b are roots of of the polinomial F then this is simply a minus

B so a minus B it is indeed an element in K I mean a minus B to the Q is a minus B it means that a minus B is a root of the the polinomial F so here I in one go I checked two of the axioms of a field namely or of a subfield namely that it is closed under addition and under a additive inverse and similarly I can check in one go that it is closed under multiplication and under a

multiplicative inverse and that I do with a and the following I mean I take some A and B in K let's say a B is is non zero and I do a * B minus1 I want to show that it's it's again an element in K so I raise it to the power of q but if I raise it to the power of Q it's the same as as

doing a to the Q times B to the Q minus one and since A and B are in K namely roots of of the polinomial F this is this is simply a * B minus one so I get that a * B minus one is again an element in K okay so so K is a field a a and it's a field contained in in

E yes sorry maybe it's very simple but why again do you can you simplify a to the k equal a a to the Q equal a a to the Q sorry a to the Q equal a because a is in K the K is defined to be the set of all roots of the polinomial ER f over over e right e e i take e as the splitting field sure of F and I know that F does not have repeated roots in fact and so now I look at the set of all

Roots we know that it it I mean the set has to be of size Q right M and now if a is an element in k then by definition a to the Q is equal to a because a to the Q minus minus a is equal to zero oh sorry yes okay okay FR no that's that's a completely fine I mean asking silly questions is the the fastest way to to learn perfect yeah yeah makes sense I I had a second one so

I can just jum in why because you said you mentioned it before but I'm I'm not sure I quickly see why there is charactertics p in E why e has charactertics P because uh well because it e is an extension is an extension of FP yeah once it's an extension of FP it has to it has to have curent characteristic P because I take 1 + 1 + 1 P times since it's zero in in FP it has to be zero in E I mean yeah okay makes sense okay thank you okay so I know that K is a subfield

of of e ER but F splits over e sorry over over K because K contains all the roots of F and since e is the the splitting field if you recall it's the minimal the me in which F splits it means that K has to be equal to

e and thus I get that I have a a field of a um a a q or or P to the N elements okay so this this is existence um so I take e field with

um qals P to the n elements then of course um e must be of characteristic P right um so I have an isomorphic copy of of FP inside e

okay now um I take a look at er what I call E star this is simply you take e you remove the zero element and you consider it as an a billion group with respect to multiplication and the the unit element one so this is an AB bilion group

um of size Q minus one so for every A in E A to the Q minus1 must be equal to 1 which is completely equivalent to say that a to the Q minus a is equal to Z this for every element in thus e is

contained in a splitting field of um this polinomial F x to the Q minus X over over the ground the ground field F since we know that e has Q to the or I mean p to the N elements

ER already then it must be a splitting field itself this this was what we showed in the existence and uniqueness now follows from uniqueness of of splitting feets

okay this H this is the end of the um of the classification of finite Fields any question about the the proof is it is it clear okay one question so can you defin f of x as uh X to Q minus X but could like this could be any set of polins because you can sum any

constant here uh of degree zero and this is a complete a complete maybe not a completely different polinomial that characterizes F so is is there any issues on that the fact that the polinomial is not unique or it just simply doesn't matter um I mean of course I you're saying that I could write I could take FX to be some Lambda times time x to the Q minus X for yeah for instance yeah yeah why is it not an issue first the the potential I mean the the roots of f let's call it f Prime

okay oh I see that's yeah okay the roots of FR Prime are the same as the roots of of f yeah now all I'm saying is that um I mean there is a splitting field for f prime or for f whatever you choose yeah and this this splitting field has to be unique up to isomorphism um and what this this proof shows is that in fact every field of Q to the I mean p to the N elements it has to be a it has to be this dis spitting field because I take I mean in any field of Q

to the N elements H I have a to the Q minus a equals z for every a m yeah okay so it means that the polom the polom splits completely so it's it's it's splitting field and hence I I have no I mean up to isomorphism I have no choice thank you okay a let's do something uh short before the

break so we we know that a um we can construct a a finite field using this a splitting method but we also know that we can construct a a finite field by modding out a a pols by some prime polom um and the question is I mean if

um Fe is a prime polom of degree n then M FX modul Fe is a field

and the the number of elements in this field is simply the number of elements in F to the power of n right because K I remind you our definition is it's the set of all remainder pols by taking long division with three so this is all the set of all let's say r x equals um a a0 + A1 X plus a k x to the K where the AI are in

F and K is is some integer strictly strictly smaller than n and you have exactly F to the n h ER possibilities for for RX so potentially I could say um if if I take F to be FP then um fpx modul Fe

is a field of p2d n elements right and We Know by the general the structure theorem we know that this this field will be isomorphic to the splitting field I mean there is only one one field up to isomorphism of size P to the n this field is is a often denoted as as fq you can also denote

it as a Gala field of P to the n now the question is can I find um a a polom which is prime over FP of Any Given degree if I can do that then I have a very uh well I have an alternative construction right but we don't know how to to find er er Prime po pols of every given

degree ER I mean there are algorithms and so on but we we haven't discussed it it's it's quite involved to look at it but what we could do is to H come up with this proposition let P the prime then for any integer

n there exists some n bigger than n with or and a prime polom Fe a of degree n okay so the proposition tells you we

might not be we might not know at least at the moment how to construct a prime polinomial of every given degree but I can go as large as I mean I can go infinitely in infinitely High degrees and the proof is actually very simple um assume by contradiction

that uh the claim is false ER if the claim is false it means that there are only finitely many prime pols over FP and let's enumerate them um I call it F1

up to FK um but then you take the product and you add one to it and you get a new Prime polinomial let's call it let's call it Fe a SE

a a fi does not divide Fe for every I okay and that's a contradiction and if you think of it what I did here is simply taking the proof that there are infinitely many prime prime numbers and doing the analogous proof for a prime pooms okay so at least this proposition

guarantees that you can find a prime polom over FP of arbitrary degree let me just mention um why does why if I assume that the claim is false I know that there are only fin many prime polinomial because FP is finite and if the I mean so the set of all polinomial of a given degree ER

D over FP is is a finite set so if I know that the degree cannot be I mean the degree of prime pols cannot exceed some some number n then I know that there could be only I mean only fin many pols up to degree n in particular there will be only Finly many Prime polom okay is this uh is this clear are there any questions all right so let's take a

break and uh come back ER to another another claim of H finite fields and then some kind of a revision that I postponed until now all right so we we have a classification of um finite Fields but there is

still ER the question well um we know that um the additive group of of a finite field is isomorphic to um the the direct sum of

a of copies of FP um and so if Q is is p to the N then we know that the number of copies of of FP here is exactly Q sorry exactly n right we we did this with a the the classification of finite AB bilan groups but if you recall the classification of Fin anabil and groups

had two versions and and now I claim I mean using using the second version of a classification of finite ailan groups we can also pin down the the multiplicative group I mean the multiplicative structure of er of the field fq so this is a this is the next theorem let PB Prime and take Q some

fixed power of P then the multiplicative group which we denote as fq Star right this is a you take fq you remove the zero element and you look at this set with the multiplication given in in the field fq and the unit element

one and I claim that this group is cyclic that is there exist some some element ATA such that the order of ETA is precisely the the order of the group which is a of course Q minus one okay this ETA is called a

primitive okay so how do we H go about proving this claim I let N to be the maximum and the maximum of all orders of elements X in the field fq okay or or rather in the multiplicative group fq star Now by L Gran the order of of every

element needs to divide the the the size of the group which is in this case Q minus one or P to the N minus one so in particular n must be smaller equal to P minus well P minus one p to the N minus one and now applying the second

version of a the classification of finite bilon groups we get that the multiplicative group of fq it must be isomorphic to um integers modu M1 direct sum integers modu M2 direct

sum integers modul Mr ER where each of these Mis divide the the next one and now okay what you need to remember is that ER here fq is is given in multiplicative form this is the left hand side is

multiplicative and the the right hand side is given in additive form okay so um but what what we can conclude from the the left hand side this guy what is the the the order of I mean what is the maximal order of an element in the the left hand side sorry in the in the right hand side um the maximum order

of an element in a the the right hand inside is is simply Mr Right you cannot you cannot do better than than Mr so so it means that that that n is equal to Mr Right isomorphic groups have precisely the same H ER the same distribution of of order of

elements for any X in in fq or in fq Star a the order of x must divide um Mr right because again look at the right hand side ER since each of the Mi divides Mr the order of every element in this in

this sum must divide must divide Mr so so this is true also for um for the left hand side ER and what it means is that h x to the N minus one is equal to zero um write it it in different forms the

polinomial X to the N minus one when I think of it as a polom in fq right this is supposed to be capital x um admits at [Music] least a q which I mean sorry Q minus one which is a p to the N minus one roots and if if it admits a p to the N

minus one Roots um it means that that N is a um I mean you see we know that n n was a less or equal to P to the N minus one but this this part gives us that n is greater or equal to P to the N minus one because the polinomial of degree n

can have at most n Roots so so I get this this fun inequality which means that n is equal to P to the n minus one but n was defined to be the the the maximal order of some element in in fq or fq Star so we we conclude that there

exista in fqx such that ETA to the Q minus one equals to one which H which means that fq star is cylic okay so this is a this is the basically pinning down the the

multiplicative structure of of a finite field any questions about this uh this result I get I get a feeling that something is maybe not completely clear or or is it a yes okay yeah I'll go ahead for the

last for the last point of the proof something I'm not sure can you go down to the uh sorry say again sorry when you go down to the last point of the proof I I most of the proof I think but maybe the last point is not clear like how do you uh sorry no maybe when you say order of x equal n on on top of the screen but you have now uh then you say that the polinomial X to the N minus one ad at least Q minus one Roots uh this is for me I'm not sure how do you see this because n is just defined as a maximal order but we don't

I'm not sure why you say the polinomial x^ n minus one admits Q at least Q minus one Roots yeah so I I take any element in fq maybe it's better to call it in different names okay I call it a okay ER I look at this H this side of the the isomorphism a a has an analog here right let's call it maybe a bar okay right a a is typically is is originally chosen here but using the

isomorphism I can find a a a bar here with the kind of the same properties now every element here on the on the right hand side a has an order that divides Mr yep okay so so so I know that the order of a also on the left hand side must divide Mr but Mr is n okay okay if if the order of a divides n it means that a to the N is equal to one yes or or a to the N minus one is equal

to zero yeah so the polom X to the N minus one has I mean every element in F QX if fq star is a root of this polinomial and and I have a q a q minus one elements here okay yeah makes sense okay so so the polinomial X to the N minus one admits at least well at least I mean admit exactly well okay so admits q q minus one

Roots so or or P to the N minus one Roots so P to the N minus one it has to be smaller or equal to n because you cannot have more Roots than the than the degree of the polinomial yeah the rest where the rest was clear it was just this point I think it's it's clear enough okay thanks okay it's it's a very simple H it's a very simple reasoning I think the the the crooks is that I use this the second version of the classification of finite ability groups so this kind of it shows you why why it's good to have this

flexibility of the two versions all right so um I think in the remaining time I would like to do a revision of what happened in this H this whole track of classification of finite Fields next time we will start with a what is called the algebraic closure um and this will be important later on but but for now let's kind of crystallize what we did

so the the the main the main issue with this classification of finite fields and this is something that of course will be useful later on is this analogy with between integer and a pols over a field in one variable right so here you have the you can say the integer n it may correspond to a

um to a degree n polinomial and then there is this notion of a prime and and and we we have this this analog in pols of of prime polom ER then we had a long division so if I want

to um let's say I I want to write I want to divide K by by n then I get some h some quotient and some remainder with the remainder strictly smaller than n and the same is true for pols so if I get a um let's say a I do G is equal to f * Q Plus R where R is of degree

smaller and then the degree of F and using this long division we we proved that um there is this there is a a gcd I mean here you you have a gcd of K andn gcd of K andn is simply you take a you take the prime factorization let's say P1 to

the A1 up to PR to the AR R and you take let's let's just say that the the prime decomposition is over the same set of primes H sorry then in this if it's in the same set of primes then this is p to the BR and the gcd you you just take the maximum right you take a a a but this is all based on the fact that in integers there is a unique

decomposition unique up to a a shuffling the order there is a unique decomposition of an integer into a product of primes and we proved that in in polinomial over a field you have the same a the same kind of a factorization I mean you have a unique factorization of F into a pols maybe I can write it here in the same in the same

way ER with a pi Prime polinomial and we showed that a a this factorization is unique at least when when f is a is monic Pol is monic polinomial this factorization is unique up to a up to shuffling the order of the of the components I mean of the the the different factors so using using the fact that I

have a unique factorization into into a a product of primes I can Define at least a a a for monic pols I can define a gcd of f and g right and I mean of course if if f and g are not monic polinomial the gcd is is is almost unique it's not unique it's it's Unique up to a multiplication by a constant by an invertible element in the in the field

um so so so this is this is basically a corollaries of the the fact that I can do long division um and and in fact using this long division I mean we saw that you can also lift the the b b um tells you that if you have um any two

integers A and B there exist some u and v such that Au U plus BV is equal to the gcd of of A and B um these These unv are called the bazook

coefficients and they're they're generally not unique but it's enough for us that they exist um in order to to do what we want namely in the case of integers we wanted to prove that integers modul P are are a field and how do you do it well you all you need to all you need to find is

a multiplicative inverse so um for a in in integers modul P um of course the gcd of a n p is one so there exist some u and v such that a u plus PV is equal to one but but this means

that auu is equal to one modul p so so U this first bazo coefficient is the multi is one option for the multiplicative inverse of a but the same almost I mean verbatim the same argument can be applied to polinomial mod Prime polom namely you take a a you take fe uh to be a prime polinomial and you

look at first we we denoted this uh well anyway you look at a a um polom modu and you have addition and multiplication modu P so so it is a it is a ring and all you need to to show in order to show that it's a field is that there is a multiplicative inverse but um let's say Fe just for Simplicity Fe is is a monic polinomial and I take another monic

polom a then ER I know because p is prime I know that the gcd of A and P is one and the Biza that has a version in in terms of pols guarantees that I have pols u and v such

that um Au plus PV is equal to one and this means that a sorry p p was a p here is a is denoted as Fe so I get that that Au is equal to to one modu and this means that U is a multiplicative inverse in in FX modu of of course if if Fe is not

monic then it is a constant times aonic times a ammonic polinomial so you can simply apply the same argument for the Demonic polom that you extract from from Fe and the multiplicative inverse will then simply be the the constant times the the multiplicative inverse with respect to demonic polinomial so all in all FX modul Fe is a field okay so this this shows you how

far the the analogy between integers and and pols can take you it's a I think it's a it's quite a quite surprising that you could use this this analogy for for that long um we have more steps uh Main regarding the splitting Fields but let's say on this part is there any point you you want a more clarification I think from from the

reactions of everyone it was it was pretty clear okay um so so once we have this then we showed that in in FX yes now I don't know was just a message I didn't want to interrupt sorry I was thanking you for linking all the links yeah sure but somebody else raised the

hand or something I heard a a ping I think the Ping was from the message okay so um so the next step was to show that in FX modul Fe ER which we can we can call k um a fe has a root this is the the heart of the matter really so I'm happy to repeat this kind of

argument I mean can we sorry madam is it will it be possible uh because I think as it happened on the uh um on the last minutes on the last session uh when we went also through a pretty long proof or extensive proof uh is it possible to review this during the week and at the start of the next session uh if anyone has uh the review of the entire thing you mean the the uniqueness of a splitting Fields up to isomorphism yep yes sure ER let me try

to to do it now even okay the existence you you're talking about existence or uniqueness of existence existence existence so we we we went through the entire thing and we said let's like leave the this week to like study in between and review the entire thing and then go back to it and if anyone has questions on the next session ask as them so I was wondering if we can do the same with uh the the gwa F and and this mapping that you that you just shown um to us what we can link polinomial to to Fields I'm just um I don't understand

what part you want to uh to review so what I want is to have time by myself to go through it again and see if by myself I understand it such that then I can ask questions on the next session or I can just uh like sure this this is always a yeah this is always open I mean you can always review any material we we did and ask a a question in the okay in the next session ER this this you you should feel free to do always okay because I say because it feels that maybe it can

disp the the the the order a little bit because we maybe have transition to a different topic or or something like that and and and it kind of brings back the discussion or whatever so okay so yeah I I see now er a a a question in the in the comments about a existence existence of splitting Fields so let me try to review this part so I take a a k to be the field of FX modu of P right we know that it is a

field and now I look at pols over K so or or maybe before before we do this of course our definition of k was all the remainder pols um with coefficients in F and with degree strictly smaller than n where of course the N is the degree of

of okay and and the addition and multiplication of of elements in case you simply do addition or multiplication of pols and then you reduce modu of so in particular f you is contained in K because you can take the remainder polom we which is just a constant just a Zer and this will amount to every every element in F can be presented as a remainder polinomial of degree

zero now I can look at a a KX but we for this ambiguation we denoted it as K with capital x and the reason is that K with capital x is like a polom a r0 X Plus R1 X and then plus let's say

Rd x to the D where a these guys r0 up to Rd are pols I mean are elements in K but elements in K are are remainder polinomial so this is kind of a a meta now now I look at the at pooms with coefficients being pols but remember that if I if I if I add two two elements in

K if I do for example r0x plus R1 x capital x and I want to to multiply it let's say with um R1 Prime x x then I do multiplic of the pols but so I will

get r0 X R1 Prime X but multiplication I need to do it in K and in K I need to take modu Fe so it will be this and then plus r1x time R1 Prime X modu of times x x squ okay so this is this is how you you

go about adding and multiplying a pols in in K capital X now in K capital x since it it contains F and and Fe originally was a polinomial with coefficients in F I can consider Fe with capital X as a polom in K capital x okay but in K capital X I have um the element which I call

um I can call it RX we call it Pi Pi of X this is simply the I mean the polom X considered in in K capital x I mean okay so it's it's simply an element in K and just like I do in any in any other field if I have a polom over a field and an element in this field then I can

substitute the element I mean I can evaluate the polom at at the element so I can do H sometimes it's denoted you take the polinomial and you evaluated it at the element but what happens now okay I take the polom basically I replace capital x by by small X so I get Fe with small x f is small X as an it must be an element in K right um but f is small X is equal to

zero in K because because in K I do modul so all in all I get that this this element which we later denoted as Alpha a um is is a root of of FX so I can write FX as xus Alpha times some c x over over K okay and this is the the heart of the

matter once I I can find a rule for a prime polom I can repeat this this argument namely I take K1 to be a FX modu um then I know that Fe has a routin in K1 so in fact what we showed is that a K1 will be

simply F joining some abstract element Alpha that that is defined by by the property that it is a root of a of Fe and this F Alpha is the minimal field containing F and some kind of abstract element that functions as a as a root of of Fe ER and now Fe may I mean Fe at least a a I mean I can split at least one linear

factor out of of Fe maybe more but at least one that means that I I am I mean I can look at at at p over over K1 I mean maybe not P because maybe I can split more linear factors but after I split as many linear factors as I can I am I am left with let's say C1 of X okay and I can consider

K1 sorry K2 equals to K1 X modul the remaining part of of Fe so C1 and and again C1 will have a root in in K2 and K2 will be the minimal so K2 will be K1 I mean the minimal field containing K1 at joining some root Alpha One of of C1 and I continue the there there could be only finitely many I mean this

process has to terminate because the degree of fee is is finite and I am left with the with a splitting field okay I mean namely I am left with with a field containing I mean in which I I was able to split all the linear factors that I could from fee so so Fe completely splits is this clear in this existence all right uh so

er again it will be great if you prepare a a questions on whatever we did for the next time it could also be on a previous material on a billion groups if you want ER and we we will start with that next time