PSE Lectures - Ep 10 - Field classification

Transcript

is to classify um finite Fields all finite fields and and this should be in a Sim ilar fashion to what we did with the finite AB bilan groups now the the first step in this classification you you could say is a um to classify all the the additive structure of of finite

Fields H and in particular we will see that um finite field if of characteristic P ER must have a p to the N elements for some

n now what I I I planned originally I mean and this this is kind of the the standard textbooks ER prove this H this fact in Star H by by showing that um a finite field is a vector field over a over FP so we saw that f is a vector space um over

FP and one can show I mean this is what is typically done in the first year in a bachelor of mathematics ER that you show that every Vector space which is finitely generated in in the sense we talked about last time is isomorphic to the field to the power of of n for some n ER and this would imply a I mean if we could show this it would imply a star right ER so my original plan was to

follow the the standard textbooks in linear algebra that prove this fact I mean this is actually this this fact you can regard as the classification of finitely generated Vector spaces but of course you need some uh some Theory to to prove this fact in particular the the notion of Dimension and so on and I expected this to um to take a couple of sessions um now it's not it's not bad to know about Vector spaces but in our case we have a a like a shortcut ER so if we want to prove star

um we can we can use the the theorem of a classification of finite a billion groups and I will soon show you the argument but it's it's not a couple of lectures is like 10 minutes so it is much better I think to put aside the the stuff on Vector spaces H for now because we don't we don't have a a good motivation for it and and do the shortcut ER so I apologize for the inconvenience but it's kind of a occupational hazard I mean we

are taking a non-standard path and sometimes I will find shortcuts but sometimes I will not ER I cannot promise a this would not happen again but in any case in either way I think the road is is also contributing to to Future material okay so any any questions about this issue I will I will now show the the argument

okay so my proposition um let F be a field of a characteristic P then the the additive group I mean f plus zero is isomorphic to

um the direct sum of zp with itself and times put something in and uh the proof as you can expect um by um classification of finite ailan groups well I should say let F be a

field finite field um so by the classification of a finite a bilan groups the the additive group of f is isomorphic to um integers modul P1 to the alpha 1

plus integers modul PK to the alpha k for not necessarily distinct primes P1 up to PK and with um alpha alpha 1 up to Alpha K some some integers greater or equal to one

now I claim that in fact in this case we must have that H all the all the Primes P1 and up to PK must be equal to p and all the exponents alpha 1 up to Alpha K must be equal to one why is that um if for some I Pi I is not equal to P then by L gr Sorry by

Kushi there is an element of order Pi which is different than P but the additive order of every element in a field of characteristic p is p this we showed um but the additive order of a each X in

F is is p because I mean of course you have P * x equal to Zer this we showed last time and you cannot have a um I mean if n * X is also equal to zero then you would have 1 + 1 n * times x equal to zero and I remind you in a field you

don't have zero divisors so if a multiplication of two things is is zero one of them must be zero and so if x is is a nonzero element and n * X is equal to zero it means that n must be greater or equal to P because the the the notion of characteristic is defined to be the minimal number such that if you add ones exactly this this number of times you get zero so by the definition of characteristic you cannot have NX equal to Zer for n smaller than p and in fact

the only way you can have it is if N is a is [Music] a I mean if P divides n okay so this is all to say um the additive order of every element in F is is p and in particular you cannot have any Prime different than P appearing in this a direct Su factorization here um in blue okay so so all the so

P1 must be equal to PK which must be equal to P now also if Alpha One is a is bigger than one or let's see if for some I Alpha I is bigger than than one strictly bigger can anyone complete the argument what will how how can we get a contradiction in this case

nobody um what we have if Alpha I is bigger than one we we have a um an isomorphic copy of Z pi to the alpha I sitting inside inside the this direct sum right and I remind you a billion finite a

billion groups admit the full Converse of lrange theorem or if you wish the the generalized kosi theor that is every number that divides the order of a finite AB billian group ER will imply that you have an element of disorder okay so H and by uh the converse the granch

theing for finite AB bilan groups there is an element X in F of order in P well p i to the alpha I which is just a a I mean p to the alpha I and if Alpha I is bigger than one then we have an

element which is not of order P it's of order some higher power of p and again this cannot happen I mean the additive order of every of every element in a field of characteristic P must be p in contradiction and to the fact that um order of X must be P thus also alpha

1 up two Alpha K must be equal to one and we're done right I mean so so f plus Z is isomorphic to zp time zp for some in okay this this is the end of the proof any questions about this uh

argument okay so so the coroller of course is a in in our case in this case the number of elements in F must be P to the N so if you if you just think about what we proved here we essentially classified all the possible addition tables of finite fields of characteristic p i mean the

addition table is is simply given by the the direct sum of a of zp with itself end times okay now of course um what is missing is the multiplication table and ER for this we do need a little bit of theory but really a very very simple one H so let me just check if I didn't miss anything

okay so so next next is a understanding multiplication tables and for this we need to talk about the pooms so I guess it's better to to do it in a

general context of fields because much of what I'm saying is does not depend on the specific characteristic of the field um so so let F be a field a polinomial over f is a formal expression H FX

equals a n x to the N plus a nus one x to the N minus1 plus A1 X plus a z where the AI are and let's see for each I AI is an element in F now what do I mean by a formal

expression okay before that um just terminology these guys ER are called monomials so you could say that um a polom is a sum of monomials and yes for for each monomial the coefficient a what on what field like how is it expressed is it an element of a field that is smaller than

the what we are trying to represent or no no no a polom over f is a formal expression such that the coefficients AI are I don't so okay F enough F enough okay um and the set of all polinomial oh okay um f X is the the set of all polinomial and here I

include um the zero polinomial so the zero polinomial is a um when all the coefficients are zero and um the degree of f is is n if

um a n is non zero and is the maximal and n n is the Maxima index and such that such that n is is indeed different than zero H so there is a a

um a convention to say that the degree of the zero polom is minus infinity this is by convention a this convention is useful in order to unify statements about degrees of polinomial you will see later um but it's not it's not any crucial part it's just to wrap statements about this these issues more

conveniently okay so when I say that a polinomial is a formal sum it implies something about a equality of pols this is the actually the content of this h of of saying that it's it's a formal Expressions that is 2 Pol nomials let's say F and I I will abbreviate ER by by writing this summation that I hope you you know from before so this is some AI i x to the

I where I runs from0 to n and let's say G is the sum of b i x to the I where I runs from zero to M um and we say that f and g are equal if well if it's if and only if of course I mean it's a definition um if n is equal to

M and for each i a i is equal to B okay so they really need to be the same formal expression yes I have one question related to not the not the to polinomial equality but to the previous definition is the degree bounded somehow by any characteristic of the field like the the characteristic itself of the field or I don't know the the order or something like that no no no I you could because you could it's

a formal expression so you could have um just even take the the polom X to the N okay you can have x to the N for any n and it doesn't depend on the characteristic of the field and so on you will always have the polinomial X to the N okay okay okay however and and this is what I want to to say here um every polinomial H let's

say FX in pols over F defines a function from F to Itself by taking a into F of a right I mean I can simply substitute h instead of X I can

substitute a I mean note that the operations that are needed to Define polinomial are simply addition and multiplication right so I can evaluate any polom as as a by by an element in F and this defines a function right we can denote it a I don't know yeah yes uh yeah I just have one small question um does this apply only two pols of one variable like X or the scope could also be like polinomial of more

variables X Y or XYZ so for now only polinomial in one variable we will get to more variables later but there is a there is a split in the theory I mean some things you can do for one variable you cannot do for more than one variable H so so for now we we're focusing on the one variable case and ER and and a I mean the the function defined by or induced by the

polinomial it could be equal for two different pols um so yes um I have one question like uh when G G and uh G is and F is on over the DF so do that mean do a2d um if if uh we um that that to polinomial going to equal if a to the I equal to B to the I but what if like b2d plus P I'm not sure I understand I mean I

guess I could say two polinomial over the same field yes otherwise of course it doesn't make any sense and these you treat it so this is a usually you say a sub I because it's just an index it's not a power of of a so you have a sub I and B sub I and these formal Expressions I mean these two pols are equal if the formal expressions are exactly the same so it means there first they are of the same length so this is Nal M and second each of the

coefficients Ian for each I between zero and and N the coefficient AI is equal to the coefficient B but if I take the the fs and then I plus the A2 di with the p so do I have this uh like Echo polinomial okay this now you ask I'm not sure but of course this equality is interpreted in F in right that's the only the only place

you can sensibly talk about equality of of these coefficients because of course these coefficients leave in F so it could be that say f is a a field of characteristic p and I could write I mean so then this equality is equality in in characteristic P right I mean if you take F to be FP then um I mean then p is equal to zero for example so it could be that AI is equal to p and bi is equal to zero that's just another they are still equal

it's just another way to to express the same element in FB I got it got it okay so and this maybe this this example will clarify a say f is equal to the field of two elements if F2 um and F is the polom x² and G is the

polom X so f and g are not equal but I claim that the function they Define over F2 is the is equal F hat is equal to G as a function from F2 to FS2 because what you do I mean you take zero you take it either to 0er square or to zero and one you take either to one square or to one it's the same function

right so zero basically the identity function but we consider them as two different objects okay so it's important to to make this dis uh distinction now we we have natural operations on on polinomial so if F yes yeah to to define the equality between F at and G you say the equal

because you only look at the evaluation table you say when the input is zero it gives zero input is one it gives one but you don't look at the how you would Define uh the monomials of f and g hat no because I don't I mean this is equality of of functions as as functions and you Define equality by just looking at the evaluation table in this case yes because a function this you could go back to the really the first session we did a function is simply a a subset of the cartisian product of the domain and the target so

it's it's the set of all pairs um okay I mean I I don't want to repeat it's okay maybe I because I join later on so maybe I miss this one okay so so I mean to this is maybe a general remark to anyone who joined the later I I am relying on the things we did previously but it's really okay to H to ask about it at any time so when yeah so I mean just to answer shortly on your question a function between two finite sets is just a table I mean you tell

me you can think of it like a function you can think of it as as a as a a function in a typed language so you have the you have to declare what is your input space and what is your output space and once you declare it you you have to specify for every element in the input space a unique element in the out output space so if if the the the the input and the output are finite sets it's just specifying a table and here so fot and G hat are simply tables and they are equal as

tables thanks okay so but I don't want to I mean this fat and notation is not is not a standard one just to um just to stress this this issue um in terms of operations if you have a two polinomial FX being the sum AI x to the I and a

GX being the sum h b i x to the I then the a and without loss of generality and say m is smaller or equal to to n h then FX plus GX is you simply add the coefficients when you can so in summation annotation it will be you sum um a i + b

i x to the I from zero to n and then possibly if n is indeed strictly bigger than M then you have also the sum of a i x to the i i equals uh m + 1 up to n okay but but intuitively you simply sum the coefficients H what this means is that the

degree of f+ G is simply the maximum of the degree of n F and the degree of G and uh you can also multiply polinomial so so this is a addition and if I want I have these these two polinom f and g I want to multiply

them FX time GX then you multiply the the formal Expressions just as you you know from from school but the the the formula in summation is is a is a bit involved I will write it down so you want to you want to multiply the the sum of a i x to the i i equal 0 up to n with the sum b i x to the I or let's say maybe we take a j

here h j equal 0 up to m and this will be H the sum of some coefficients CK x to the k k equals z up to n + m and the CK are given by the for formula some over all pairs I and j i runs from zero to

n j runs from zero to n and I + J must be equal to K okay and then AJ B IUS J so this this is what you get I mean I can I can explain a little bit how you get this formula um I mean just write H for example we have a n x to the

n a plus maybe we do the n minus one nus One X to the N minus one up to a a z and I want to multiply this by a BM x to the M plus b0 so I look at the monom I I go monomial by monomial the monomial in of degree n plus m is a n * BM time x to n plus M right so

that's the first expression now I want the monomials of degree n + m minus one how can I get there well I have um I have a n minus one time BM and and this will be a monomial of of degree n plus M minus one but it's not the only one

because another thing I can do here I have um BM minus one time x to the N minus one and I can multiply it with a n x to the N so it will be a n b n minus one x to the n + nus one okay and then you continue but you can see already from the the Second Step that H that this is equal to a n BM x to the n +

m plus here I have a a sum a n minus one times a BM Plus a n b m minus one and I'm out of space

okay and and there are more Expressions but this this guy is a a c of n + m minus one so C of n plus n minus one is is indeed this H sum of all the all the multiplications of two coefficients whose index sum sum up to to the index of I mean to n plus n minus one okay and there in this case there are there are two options I mean either

you take a coefficient of a the M coefficient of of G multiplied by a n minus one and or you do a n times BN minus one but in general of course there will be more coefficients here I mean for a general K you will have a you will have more than two coefficients in the in the summation of CK is this clear uh it could be a bit uh involved

it is I think and the example helped a lot definitely yeah so so let's do an example but of course you you I think you know it from school I mean ER let's say I take a I meant it helped so the the the stuff that you already wrote like if you don't want do the example is is is completely fine like whatever you yeah yeah no but let's do an example I mean I I take just F2 it's my toy example H FX is H x² + x + 1 or maybe minus one and

GX is equal to x + one then I want to do FX * GX so the first thing I do is like in high school I open everything this is like X X2 + x - 1 * x + 1 this is going to be equal to X Cub plus x² um well actually plus 2 x² um - x -

one but sorry plus x - x - one so this is equal to X cub and + 2 x² minus one but sorry m one question if you are in F2 how can you have a two as a CO efficient so this is answered by the elastic quality oh sorry sorry okay sorry no no it's it's it's okay I'm just like I'm trying to kind of H break it

down to the very the the most Elementary steps I mean what I'm saying you do everything like you are used to in high school I mean you open up the the for the the multiplication of the formal expressions and and then you you take the coefficients you I mean you reduce modu mod P if you're in FP right so I mean in general these two represents 1 + one H whatever it is uh 1 + one in in our field we we replace

it okay so this is a this is addition and multiplication of pols yes um I have a question uh I was just thinking of these extension Fields right and this is something different is that correct ER different than what yeah I mean we don't have to do some some sort of reduction step I think I'm mixing something up here at the moment I mean you are certainly mixing ER stuff because we didn't talk about extension Fields ah okay ER we will get there I promise but for now just I just want to have like

the kind of a a short introduction to what is polinomial ah okay I see see okay thanks okay so a maybe the last last notion we want about pols is a so let F be a polinomial over a field uh capital f and and T an element in

F then T is called a zero or a root of f if um f of of t T well let me let me just write it f f of T is equal to zero now

if M K into f is a sub fielded inclusion so if K is is a a larger field than F then I could take a then for for a given polom in FX I could consider it as a polom in

KX right because all the coefficients of f are elements in in in the field K but note that these are two different objects I mean we denote it by kind of abuse of not in the same letter but this is a different animal I mean this is a different animal than this okay in particular I mean the degree is going to be H equal but the number of zeros need not be the same so

EG uh you take FX to be uh x² + 1 and and the field F you take to be the real numbers then the the then then F has no zeros but if you look at f over the complex numbers the element

I in C is a zero okay so really pols if even though we we could consider them over a different larger Fields H we H we regard it as two different objects I mean okay so so it's kind of you have to decare what is the the field for which you consider the polinomial over all right

so yeah um so can we say that the biggest field has equal or more zeros than the smaller one or does those not apply always yes so so the set of a I mean in terms of a um I mean in terms of size the set of polinomial else so you could say that F if f is contained in k then you could you could say that FX is contained in

KX is this your question um yes uh but also like the zeros are going to be equal or greater as well for K than for f yeah yes okay yes but you you need to have a subfield inclusion for this okay H it is not true like what you need to be careful I mean you could view for example FP as a subset of of the real numbers right because FP is just all the elements is zero up to P minus one but this is not a subfield it is a subset

inclusion but it's not a a subfield inclusion so then the what what we said does not does not apply okay so you really need to to have the same I mean the the operations of addition and multiplication in F to to extend to the the addition multiplication in K okay makes sense thank you all right ER if there are no other questions I suggest we take a a break sure so you the minutes yeah oh

okay um just to sum up what we did H we defined so we defined two operations on polinomial one is a plus and the other is multiplication

and of course these these operations are associative associative and moreover we have a neutral element H with respect to addition and a neutral element with respect to multiplication so zero is neutral with respect to plus and one is

neutral with respect to multiplication we even have a inverse with respect to addition so minus FX is the additive inverse H to to f ofx but we don't generally have a multiplicative inverse for a polom so

um I claim if well FX in F polom over f has multiplicative inverse if and only if the degree of f is precisely one sorry precisely zero

IE FX is some some con Conant and this constant is is non zero now why is this why is this true here you can H you can use a degree argument I mean if GX in well in F satisfy

FX GX is equal to the polom one then this equality of pooms in particular implies equality of the degrees so the degree of f * G must be equal to the degree of of one the degree of the the polinomial one is is indeed zero but the degree of the multiplication of two pols is H the sum of the

degrees and so we cannot have the sum of the degrees equal sorry degree is yeah so so the only way we could we could have it is if the degree of f ER and the degree well the degree of f equal to zero and the degree of G is equal to zero so they really need to be that's the only way you can have a a

multiplicative inverse is if you have a nonzero constant being one polinomial and and then the inverse would be the the polinomial the constant polinomial with the inverse of the the inverse of C okay so so this is all to say FX is a um almost a field right we have all the axioms of the field are

satisfied except H possibly for H this um the the multiplicative inverse of an element ER yeah uh why is there no such thing as uh colloquially let's say 1 over F ofx so that uh if we just multiply F ofx and one over F ofx we get one yeah so this is a this is one possible extension of the notion of polinomial is what is called the Lauren Lauren sus and

in Lauren sums indeed you you are allowed to have one / X it's like it's like polinomial but the monomials could be x to the N where N is a is possibly negative and then you will have a ER you will have a multiplicative inverse but it is a different notion and we are focusing at the moment on polinomial which have you know interest in their own right so yeah so that one possible extension is is to add this negative powers of X and then you

will have inverses um but we will take a different uh approach okay so a a long division the slogan here and I guess throughout this this chapter of of pols um pols in one

variable are analogous to integers okay and the um this manifestation of the Logan one of it is that you can do long division of polinomial so how do you do it um so so let F be a field and uh

suppose I have two two pols FX is equal to a n x to the N plus A1 X plus a z and GX is um BM x to the M plus B1 x + b0 ER with without loss of generality I suppose that m is smaller or equal to M

now of course um so the the degree of f is n the degree of G is M and and this really means that a a n and and B and BM are non zero now I set n0 to BN and I take c0 to be a n over

BM and now what I can do I can take FX minus c0 x to the n0 minus n times GX okay and what this does for you is is removing the the top monomial of f because the top monomial of f will be cancelled with the top monomial of the of the second term of this term okay so this this will be

um let's call it our one of X and it will be a n minus oneus b m -1 c0 multiplied by x to the N minus one h plus some some other monomials so the degree of r one is strictly smaller than H than the degree of then the degree of f which is n z okay it

doesn't have to be n minus one because these coefficients might might might might be zero I mean ER it could be that both a n minus one and b m minus one are zero or it could be such that the the difference here is zero so we are not guaranteed that this is a a polinomial of degree n minus one but it is a polinomial of degree strictly smaller than n okay uh so I call the degree of of R1 I call

it N1 and now I do the same similarly you can write a I I take now R1 instead of f and I substract so okay um I should say um

if N1 is a smaller it could be smaller than m m i remind you is the degree of g h then we don't we are done well otherwise I can I can do the same trick with f replaced by R1 so I do R1 minus some constant

C1 times um x to the N1 minus M time GX H and and C1 is chosen such that the top monomial of of R1 will be cancelled this this I call R2 um and as before I have that the degree of R2 is strictly smaller than the degree of R1 which is a N1 so this I call

N2 um and um of course I can continue this process until um I mean until the the N the n i becomes strictly smaller than M and we continue until ni I is smaller than M this this process must be ER must be finite right because we we continuously

H decrease the degree and at the end we get well I have FX which you can you can call R zero of x and FX is by definition um c0 x n0 minus M time GX plus R1 of

X then I have R1 of x equals to C1 x to the N1 minus n GX plus some R2 and this goes on until uh let's say the last step is a RS minus one X is equal to some CS minus1 x to the NS - oneus M

GX plus some RS and since this is the last the last step I have that the degree of RS is strictly smaller than M okay now I I sum up all of these equations and what I get I mean if you

look at it this R1 is cancelling with this R1 and this R2 will be cancelled with the R2 that that will be here so after all the cancellation what I mean of course r0o is not cancelled and rs is also not cancel so we get a FX which is r0 equal to GX times um

c0 x to the n0 minus M plus plus many things plus h CS minus one x to the NS -1 minus M plus rsix with a the degree of RS strictly smaller than the degree of G so we we call this this guy a q of

X and this guy we call R ofx so what I can I can I can write is FX is equal to um GX QX plus RX and this guy is called the quotient where this guy is called the remainder any questions about this uh

this procedure this is called the the ukian idian division sometimes ER I I imagine this at least in notations and indices is a bit involved so uh um you're most welcome to to ask about

it anyone is this completely clear I mean I would be happy of course but uh yeah so does it mean that uh at every step we first select the C by looking at the uh highest degree coefficients in the like the next the next iteration of the polinomial yes so look at what we did in the first step ER I mean I have these two H these two pooms H and I assume that N is a

bigger or equal than M so I want to get rid of this uh this monomial in order to get rid of it I multiply I take c0 to be this a n over BN and I multiply GX by by c0 time x to the n0 minus n so what you get get here just to write the the top monomial of this uh this expression will be exactly I mean a n over BN

times x to the N minus M times so so this is this is c0 time x to the n0 minus M and then I multiply it by the top monomial of G top monom of G is BM time x to the m and you have other things right but you see this this the top monomial of the of the purple polom is precisely a n x to the N

right Alex yes thank you okay so what I did I simply every time this is a this is R zero right so I choose the coefficient C c0 such that the top monomial of r0 will be cancelled and I can always do it I mean right this this only it only this definition of c0 for example only uses the fact that BM is is a nonzero element so I always I can always find

some some coefficient in G which is the the coefficient of the the highest nonzero monomial and then I simply do the same procedure I mean I could write the formula for C1 ER but it would be it would be a bit combersome the principle is the same I I take I want to know I mean I I take the the the highest monomial in R1 and I I want to cancel it so to cancel it I need to multiply G by by some some monomial

and and there is such a monomial such that the difference will cancel the top monomial of of R1 and I I do it repeatedly until I I get something of degree less than M okay now this procedure it it kind of spits out ER the long division of of h f by G this is sometimes denoted as a f%

g so f% G would be this remainder um but of course you may ask I mean is this the only way I mean at least are these the only pols that satisfy this relation so Theory um for FG a polinomial over a field H with um degree f

bigger equal than degree G there exist unique pols a a Q and R such that FX equals H GX QX plus RX okay so we proved a existence and now I want to prove a

uniqueness so suppose also a q Prime and R Prime satisfy uh FX equals GX Q Prime X plus r Prime X then so we have

this equation star and we have the equation double star if I combine them together um I get that GX QX minus r Prime X is equal to GX Q Prime x minus r of x uh or maybe let me put it better as GX

um QX minus Q Prime X um equals RX minus r Prime X but by definition by assumption or by Design the degree of R and the degree of R Prime need to be strictly smaller than the degree of

G right because we are dividing by g h so we have this Triple Star equality of pols um we know that the degree of R and the degree of R Prime is a strictly smaller than G so the degree of the the difference R minus r Prime is also strictly smaller than the degree of G so the only the only way this Triple Star equation of of Po omals can

happen and this means that I mean of course the degree of the multiplication is the the sum of the degree since the degree of G times Q minus Q Prime is equal to the degree of G plus the degree of Q minus Q Prime we must

have um Q equals Q Prime right so so q q minus Q Prime must be equal to zero and and from this which follows let um also RX must be equal to R Prime X okay that's and that's the end of the

proof is this clear this H uniqueness argument okay so what we get here the the I mean the the summing up of this uh this discussion is that we can do long division just like we do in h in integers so coroller if

FX The polom and a in F is root of f then I can write FX equals to x - A Time F Prime X for

some f Prime X well of of course fime X must be of degree strictly smaller than the degree of f how do I see it ER I do long division of um FX

by I take GX to be x minus a so there exist some pols q and R such that FX is equal to a xus a q 2 X Plus RX now this is an equality of pols of course implies equality of the the

associated functions so I substitute a I get well F of a is zero because we assume that a is a root and this must be equal to well a minus a g of a plus r of a so R of a is also zero

but the degree of R must be strictly smaller than the degree of what we divide by it must be strictly smaller than the degree of xus a which is one so it means that the degree of R is zero H and R well R is really R is really the constant polinomial zero or in fact in our convention the degree of the

the zero polinomial is is minus infinity so but if I know that the degree of a polinomial is is a is smaller than one then either it's it's a it's always a constant polinomial um if I know that at some point it evaluates into zero it means that it's the zero Pol okay so so if R is zero then it means that FX is xus a times

GX time sorry times QX okay in another corer from this polinomial of degree n over over field over field F can

have at most n roots can anybody justify given an argument for this uh this corer I mean you can previous previous uh corollary and just basically Define the function is a product of you know more than n plus

terms and then come back and say uh that the the product is the degree is much bigger than n exactly but then you can come back reg just talk yes yes so let me just write it down for completeness if FX is a polinomial of degree ER of degree n that that has

with let's say a m bigger than n Roots then FX let's say the roots are A1 up to a m then I could write FX as simply the product of xus a i i = 1 up to M and so degree and and this this is a

contradiction to the degree of f okay so um okay we we are almost out of time but let me kind of sketch what is coming up or know what

um yeah yeah I'm sorry I must have missed something I don't see exactly why how can you define FX like this just from the previous cor can you explain a bit why don't you have G of X so you know so if okay so let me expand a bit um f since A1 is a is a root I can write a FX as x minus A1 times let's call it q1 of X with of course q1 is a is going to be

a I mean the degree of q1 would would be the degree of f minus one right yep yep um then of course A2 is a root of q1 because I mean distinct distinct root right oh right okay yeah makes sense so since they are distinct and A2 is a root of FX it must be a root of of

q1 got it yeah so1 alls yeah exactly so q1 is going to be x - A2 time some Q2 and and I repeat h i get that FX is indeed this uh this product and why there's no final Q m h by degree because once you did it m times well okay you could write the I

mean because the the it depends on how you want to to build the argument but um I mean basically okay there there could be a yeah you could write it like this but it's it would still contradict contradicting the degree the the fact that m is is strictly a okay we we're trying to prove by contradiction so if you can prove that F

has more than n Roots then that's sufficient for has to follow the timate yeah yeah so uh yeah but I mean okay I think the point is clear so uh just a slight matter of formality okay so let me let me sketch what is coming up next I mean once we have we have this a long division of pols I can h i can consider first I can

consider the the gcd of two polinomial if you have f and g I can do the gcd of of f and g this is going to be the the polom of maximal degree that divides both h both f and g it could be that h f and g are are copre so it could be but of course it doesn't have to be I mean take a x - one * x + 2 and x - one time X2 + 2

then typically the gcd of these two would be x minus one ER I will I will go into details on what is going on first we will need to revise the the ukian algorithm for a gcd so what I want to ask you for next time in the notes which you can always find find an updated version in in my website there is a section called A A Elementary number

Theory and if you can read the the part of it with the ukian algorithm for integers that find the gcd of two of two integers then I think we will have an easier time next time generalizing these two polinomial okay other questions or requests for next

time all right so I have a good rest of your day thank you cheers thank you thank you thank you thank you bye bye thanks bye