PSE Lectures - Ep 11 - Field classification

Transcript

okay so H last time we we talked about long divisions of pols so if H yes I think it was some wrong wri in the chat only not rise hand okay let's see ah okay um

so if we have two pols let's say F NG G in coefficients in some field F so this this is like the set of all pooms we proved that there exist unique pols um q and R in FX such

that f is equal to GQ + r and two the degree of R is strictly smaller than the degree of G right and we call this a polinomial quotient and and this guy is the remainder and um I didn't do any examples so so let me let me show you by example how

how it's done let's see what I have here so suppose I want to to have um f is 6 x Cub - 2 x² + x + 3 and G is um x² - x +

one so I will use the notation of long division of of integers um so I put uh F here and here I put x² - x + 1 and so now I'm looking for a um a monomial to multiply a g with such that ER after I um

substract the monomial times the times g i i cancel out the top monomial of f so this is in fact a 6X if you multiply 6 x by uh x² - x + 1 you get um 6 x cub - x² sorry - 6 x squar and and plus 66 and after I subtract I cancel out the

top monomial so I get 4 x² uh - 5 x + three and now I I take this remainder this you can call maybe R1 and I I do the same thing so now what what monomial do I need to multiply G such that when I subtract the result from R1 I I get rid of the top monomial of R1 well this is a four so I add four here and if I multiply 4 by GX I get 4

x² - 4x + 1 and I subtract I get um - x uh minus X plus two ah I multiplied everything by four sorry so the constant is to be four and then I get minus one here okay so uh

so so the result is that um and you can check it 6 x Cub - 2 x² + x + 3 is equal to X x² - x + one multiplied by the result the quotient 6 x + 4 h plus the remainder so I guess you could say minus x +

1 okay is a is this example clear yep yep all right ER so I remind you the slogan maybe we put it here um pols um are analog analogous to integers

okay so and and the one component of this analogy is that you can do long division but in fact a lot of the facts about ER about integers follow from long division and um well a prime prime integers right um and we have a good analog of of prime integers in the realm of polom and this is the following definition a

polinomial f is Prime and sometimes Prime is a is replaced by irreducible um if it cannot be decomposed is f equals a *

B with um degree a and degree b um strictly bigger than zero so bigger equal to one okay so so a polinomial is is prime if I cannot express it as a as a product of two non constant polinomial I mean of course if I allow constant pols then I can always express a a polinomial as in in I mean in many ways typically as a

two product of as a product of two polinomial one one would be a constant and the other would be just modified version of the the polinomial F so this is this is ruled out in this definition so if you take a linear polom any polom of degree um f of

degree one is prime right I mean by by degree reasons H if you present it as a a product of A and B then then the degree of one of them must be I mean if they're are not not constant you you have a contradiction to the degree ER so so all pols of the form um Alpha X Plus beta are

prime now we can take F to be the field of three elements F3 um and I claim that this polinomial FX equal x² + 1 is prime um if it wasn't then it would be a product of two

um of two linear polinomial and and since f is is a has a one as it's a coefficient of the leading monomial I would have xus a xus B for some A and B and F but you can check that h f does not have um does not have any any root

I mean 1 squar + one is a is two which is non zero and two square is four plus one is five which is again non zero now if you take the same polom over um over a larger field

um if you take F to be F5 um and and if the same polinomial as before then f is not prime for example I can find I can find a ER root I mean in fact a you can H so 3 S + 1 is going going to

be 10 which is 0 modu of 5 and H 2 s + 1 is five which is 0 Mod Five so so F or x square + 1 is x - 2 x - 3 is not Prime okay so of course the the the

expected property about about prime pols is that you can decompose every polinomial into a product of prime polinomial this is true but we also want uniqueness and for uniqueness we have a problem with the possible coefficient of the leading monomial so what we do is is [Music] um we restrict attention to monic po pols

a e where the coefficient of the leading monomial is is H one and and then we have this proposition that

F be a monic polom then there is some K and monic pols monic Prime pooms A1 up to

a such that f is equal to A1 up to a k and A1 up to a are unique up to um reordering um is this statement clear

especially the the last part of uniqueness okay so how do we prove it yes uh just to make it extract clear the F you multiplies all the Primes for is A1 to a it's not like composition it's just multiplication no no no multiplication yeah yeah multiplication um I'm trying to to ease the notation with the not writing the the variable ER I think it it stresses the analogy to

integers better but okay um so how do we prove this well existence existence is basically a the same the same argument as you do in integers I mean h um if f is not um I mean if f is not prime then then it is a product of two primes I mean oh sorry it is a product of two pols which are of smaller

degree of F and if they are not they are not prime then each of them is a product of of two lower degrees and since you reduce the degree every time this process has to stop so and when it stops you you are left with a a decomposition of F into a a product of prime pols so existence is clear

for uniqueness yes yeah sorry again so so when you say if F if not prime it must be the the product of two pooms but in this case there could be a reminder there could be error X which is not new no no no if you look at the definition of prime prime means that it cannot be decomposed as a product as a non-trivial product let's say okay so not prime means it can be decomposed as an nontrivial

product okay right if you wish if it helps I can I can write it down in detail for the Existence I mean no it makes sense though okay it's really just like in just like in integers if you you have an integer if it's if it's Prime you're done I mean you want to prove that an integer is a product of primes if it's Prime you're done if it's not prime it's a product of two of two integers both of them are smaller than this original integer if so one of them is prime you look at the other and you you keep

repeating this So eventually you have to stop and this this gives you a factorization makes sense thanks yeah now for uniqueness suppose I can write F in two different ways A1 up to a and um B1 up to BL remember I mean POS if I if I do not know uniqueness then possibly F could be a a

product of prime pols of different length right I mean different number of of prime pols so this is the K and K and L here um where um AI bi I are monic uh primes now I

claim A1 it cannot appear here canot appear um on the on the left hand side um okay

so um I suppose sorry let me amend the argument um I suppose that I have a polinomial with two different factorizations that f is of minimal degree H such that you have two different factorizations

okay so if if it is of minimal degree then a A1 cannot appear on the the left hand side because then I could pencel it out and I would get a um a polom with the I mean with lower degree than F that still has two different representation of I mean product of prime pols so A1 cannot appear on the left hand side since otherwise we could cancel it and a a

contradict the minimality of f similarly B1 cannot appear on the right hand side okay so without loss of generality let's suppose that a the

degree of B1 is a smaller or equal than the degree of A1 um well okay then I can do long division of a or A1 by B1 so there exist a a q andr such that A1 is equal to B1 * Q + r with the degree of

R strictly smaller than the degree of B1 okay now um since the degree of a of R well R has a a a prime factorization so um R has a prime factorization let's say R is equal to

some constant Alpha times R1 up to RM and here I have to add a a some constant Alpha because I cannot assume that that R is monic and in fact Alpha is non zero right otherwise this whole becomes triv um and moreover a

B1 does not divide any of the r since B1 has higher degree because of degree reasons well okay now I I substitute this this identity and this identity

and so I get let's say a q B1 plus Alpha R1 up to RM this is basically A1 right so now I multiply by the the rest of the A's so it was A2 up to a and this has to be equal to B1 up to

BL now ER let's let's set F Prime to be this product R1 up to RM H times A2 up to AK and I can write F Prime is if you check it it's going to

be the inverse of alpha times H B1 up to well B1 times H B2 up to B minus Q A2 up to AK okay this I obtain um I mean from from start right so I

I I just renamed a something as F right I have here a this guy in in pink is f and then I I just uh I move move things to the other side take a a common common factor I get this double star um but of course our one up to RM

A2 up to AK is another so so I get so we get two factorizations of f Prime ER right I mean of course you can uh you can further factorize this

ER this component um but in in whatever you have um where um B1 appears in one factorization but not in the other and the degree of f of f Prime is strictly smaller than the

degree of f because uh remember f Prime is this is this product and I I removed I removed some I mean it is of a lower degree than F so this is a contradiction to the minimality to a minimality degree

F okay um any questions about the the proof now is the time is it are you comfortable with the proof or I can I can explain again in a in a sketch what is going on yeah for me maybe it would be useful okay

so um what is I mean I want to prove that um if I have a monic polom I have a a unique factorization up to reordering of of this polinomial as a product of primes ER now suppose so I I assume by contradiction or towards contradiction that this is not the case if it is not the case it means there exist at least one polom let's call it f with two different factorizations but of course if there exist such I I can I can look for the

the polom or or let's say a polinomial with a a a two different factorizations that is of minimal degree right I mean the yeah yeah yeah yeah yeah okay so so I assume that f is such but with a minimal degree of course it does not mean that f is the only one it because possibly there could be two different polinomial fub1 and F2 that have a a different factorizations and both of them are of the of equal degree

right but all I need is one so I assume f is such now because of the minimality of f A1 that appears here cannot appear anywhere here because if it if it would then I could cancel it out on both both sides and I would get a a a two different factorizations um of a polinomial of a lower degree than F right so so this would I see yeah yeah yeah yeah yeah I see I see I see so and for the same

reason B B1 cannot cannot appear B1 cannot appear on this side okay because again I I could cancel it out and contradict the the minimality of f h so so so now I do so I assume that H just for convenience that the degree of B1 is is a smaller or equal than the degree of A1 and then I do long division I do A1 modul B1 I I get some quotient and remainder and the remainder is guaranteed to be of

a lower degree than B1 ER now I take the remainder and I decompose it into a a product of primes here because I cannot assume the remainder to be monic then I write it as a um as some constant times a product of monic Prime pols let question okay monic monic prime factorization but you have this this

uh this coefficient Alpha that you have to add um and and I note that because of degree reasons I mean because of because of this B1 cannot divide any of the ri I mean B1 is of degree bigger than R so clearly it cannot divide any of the r makes sense Mak sense Mak sense okay so so now I just sorry can you go again why you can write R such as this so um I I can I mean we said that the existence you can always so if R is not

monic then let's say Alpha minus one * R is monic for some Alpha I mean namely the the alpha appearing in the the coefficient of the the top monomial of R right and if R is a just think of like if R is a a x to the N plus something yeah then of course you can write a a minus one of R this will be a monic polom yeah and and we said that monic

pooms have factorization into a product of a monic Prime polinomial this is the exist existence part right right okay so I can write R in this way um and now I just I just plug it in into the the equations right because I have a A1 up to a k equals B1 up to BL but instead of A1 I I plug in this right and now I I Define f Prime to be

a basically the the product of these purples right R1 up to RN times A1 up to a k and I claim that this this equation star in fact gives me two different factorizations of f Prime well how do I see it I can write F Prime as a this constant Alpha minus one time B1 times whatever is left and I can also write right F Prime as it was originally defined as a product of R1 up to RM A2 up to

AK now this blue part you can is a product of prime polinomial prime in fact Prime and monic polinomial a h um that that is equal to F Prime what you have here let's say the the the purple part you could also I mean you could also Express the purple part as a product of prime polinomial I mean you

take whatever is in the brackets and you factorize it into a product of prime pooms what you will get is some constant Alpha minus one times B1 times some um product of possibly non monic pols H but what you get here is that on the purple side you have B1 and on the blue side you cannot have B1 right I mean because B1 cannot be cannot be one of the a A2 up to

a ER because of the the original argument and B1 also cannot be a a one of the R1 up to RM because it is of higher degree so what you get is that you get two factorizations of a of f Prime in which in one factorization you have B1 and in the other factorization you don't have B1 so this is two different so you you can h i mean you can extract from from the equation double star two different factorizations of f Prime into a product

of prime pols but F Prime is of degree strictly smaller than F so this this is a contradiction to the minimality is this clear now it is for me at Le yeah okay so once we have this uh this kind of anchor [Music] um definition let f and

g be two pols a greatest common divisor and gcd F G is a polom h that divides

both f n g and is of maximal degree um okay now where well using this this previous proposition so if FNG G

harmonic gcd of FNG G is unique because you can decompose f and g into a product of monic Prime pols and see what is the what are the prime polinomial in each of the the the decompositions that are common I mean it's like you do again it's like think of f and g as as two integers you write it as a as some product of primes P1 up to PK and I

mean q1 up to ql and you look for I mean the gcd is just you take the common primes that appear in the factorization so you you basically do the same thing for f and g and since for monic pols the factorization is unique then the gcd is unique for nonm monic polinomial if F or G are not

monic um a gcd of f and g is unique up to a multiplication by um invertible element Alpha in F so a nonzero element okay because of course if if H

um if F let's say f f is not monic and G is monic just take some constant out of f to make it monic so it will be a product of a constant and ANM monic polinomial and you you will do the the the gcd to G and to this modified f h and this will be this will be your gcd it will be up to a constant is this clear like this subtle point of a uniqueness here

anyone yep we put thumbs up maybe you don't see but yeah okay okay um now we're going to ER to actually employ the the analogy between pols and integers in order to get um an analog of the the Biza so let me remind you I mean we actually didn't do a a proof of the BMA but the the statement we did I mean

um the statement is a a for positive integers let's say a a and b there exist some integers u and v such that Au U plus BV is equal to the gcd of

unv now there are many ways to see what why this is true of course you could describe an algorithm and show that it it stops and so on this I leave to you H to read it other places actually it's I think it there's even something in the notes um but the proof that this is this is true can be done without any algorithm so [Music] um consider the set K to be a the set of all expressions of the form Au Plus b v

where u and v are integers okay this of course it's a nonempty set because can pick u and v to be one for example and let K lower case k be the smallest positive element okay now since lower case k is in this

capital K set there exist u and v in Zed such that K is equal to Au U plus BB um we can also do long division of a of a by K and we can write you can

write a is um K * Q + r with a a r strictly bigger sorry strictly bigger than strictly smaller than K and bigger equal to zero ER and if I can write this I can plug it in so I get a r is equal to a minus qk which is

um a minus Q Au U + b v which is a a a 1 minus u u plus B minus q v so so I get that R is also an element in k but R is strictly smaller than

than than K and K is the minimal minimal positive element that I mean that that is in this set capital K so I get that R is equal to zero that's that's the only non-contradictory way yeah why cannot R be negative because R is a remainder so it is a right yes yes sure right sorry yeah thus a is equal to Q * K so Q divides a sorry K divides

a similarly and K divides also B so K is smaller equal than the gcd of of A and B right I mean so so K is a common a common divisor of A and B so it has to be smaller or equal than the greatest

common divisor now now since the gcd of A and B must divide a and a must divide B um and K is equal to some Au U plus BV I get that a gcd of A and

B must divide K right because it divides a and it divides B so the gcd of A and B is smaller or equal to K and I had the other inequality here I E K is equal to the gcd A and B okay so so I get that the gcd of A and

B is some Au plus BV as as I wanted any question about the the proof this is the the the BMA and I I remind you that the coefficients u and v are typically not unique but we only care about existence in this uh in this case so um try to to go go over this in the

break let's take a 10 minutes break and er what I'm going to do after the break is to adjust the argument of this LMA or the proof of this LMA ER to get the same effect for pols okay so see you see you in 10 minutes see you okay um can we start

again all right so the the polinomial version of this uh ER last LMA is the following I mean uh um let F be a field and A and B two pols then there

exist pols u and v such that a Au U plus BV is equal to the gcd of A and B if you want we can we can make a and b monic but it's not it's not necessary so H consider K to

be the set of all expressions Au U plus BV such that a u and v are are pols in the set I have elements of a smallest degree I mean possibly I have more than one element I mean possibly we can have me several elements of a of smaller degree but I have at least one so let let K in

K be an element of um smallest degree so K is is equal to some Au U * b v now I do long division of a by k h a is equal to K * Q

plus r with of course degree of R is strictly smaller than the degree of K and of course bigger or equal to zero so I can write r as a minus qk a which in turn is a minus um Au U plus BV

q and again I can write it as a 1 minus quu plus b minus QB and so I get I get again that R is in the set K capital K but R is of a of degree strictly smaller than K and K is supposed to be the the element of smallest positive degree should say

so it follows that the degree of R is zero um well in fact ER I I want to show that R is is really the zero polinomial so let me take um K to be um the the element of smallest positive degree of smallest degree not equal to to the zero

polom so if I choose K like this then I get as an immediate query that that that R must be the zero polinomial because it is a it is a polinomial that can be expressed as a time something plus b time something and its degree is indeed strictly smaller than k okay so so a is equal to K * Q

um and as before um so what I get is that a divides K and similarly B divides K so the gcd of A and B has to divide K ER well actually um oh what I want

is so so K is a common divisor of A and B so the degree of K has to be smaller equal than the degree of the gcd of A and B right because it divides I mean a a and b oh I'm sorry here it should be

reversed so I get that K divides A and K divides B so so K is a common divisor of A and B hence its degree must be smaller or equal than the degree of the gcd but a gcd of A and B divides of course divides a and divides B so gcd of A and B has to divide

Au U plus BV which is K and in particular the degree of gcd a B has to be equal to the degree of K well I mean has to be smaller or equal but but this of course means that the two degrees are equal and so uh yes so um the they are I

mean in fact um so if we assume AB harmonic then then by uniqueness of the gcd we get that K is equal to well on one hand AU plus BV and on the other hand to the gcd of A and

B and otherwise otherwise um well um this will be true up to a constant so so you know what let's just assume let's just assume that A and B are monic I apologize for all the all the

patches but is this is this proof clear it's it's really a it's really like a a a a demonstration of of the the power of this analogy for me most of it is clear only the last step where you enforce that A and B harmonic and first you can say the last the last step here okay so let's so this is why I modified the statement let's suppose that A and B are monic okay if A and B harmonic then in the last part of the argument

um I get a I mean k is a polom that divides a and divides B and is of the same degree as the gcd so by the I mean the gcd of A and B is unique if A and B harmonic ER and it is the minimal it is the the the common divisor of A and B with minimal degree so if K is equal to the degree of I mean if the degree of K is equal to the degree of the gcd they must be equal right by okay yes uniqueness harmonic to say that the gcd

is unique you need this yes yes f so maybe here since we assume but what I want you to take from this is is really a the power of the analogy I mean the analogy does not work completely right I mean you could say polinomial are different uh object than than integers ER and factorization into prime factors is not unique unless

unless you talk about monic polinomial so the analogy is not perfect but nevertheless it's it's very helpful and you can really like plug it into an argument and ER and see that you get the same I mean the same argument is applicable for the the polinomial case yeah it's uh so interesting that the argument is very very similar just yeah yeah and ER I don't want to spoil the surprise but the analogy will help us further into constructing a finite Fields

so er do I have an example here I guess not so the next thing is a modulo arithmetic modul arithmetic of the pols so let Fe we a polinomial in coefficients in

f b of degree n now if you do you take a polinomial over F and you do modulo I mean you do long division with Fe the remainders the possible remainders you get are are a a basically all the pols with degree less than degree of of so I can look at the set I call it f sub Fe is the set so we

Define I I'll need more space so F sub Fe is the set of all r x equal to um let's put it like this a n minus one x to the N minus one plus h A1 X plus a z where the the AIS are a elements in

F okay um this is a the set of all possible remainders um of long division and let's say F long division

fee for um for general polinomial f right I mean the clearly the remainder has to be of degree smaller than the degree of Fe um and any any polinomial of degree n minus one or lower can be obtained as as long division I mean for example take this polinomial R and do long division with v then by definition ER the remainder is is R itself because you start from something with a degree strictly smaller than the

degree of Fe the the long division terminates at the beginning so so F FX is the the the set of all possible remainders now on a fix we have I call it mod fee arithmetic namely you take um let's say you take R

ands in a f f you can do r + S mod right I mean in fact this this would be by definition it will simply be R plus S as as pols and you can also do R Time s you do R * s and then a mod mod Fe is simply defined to be R Time

s i mean the remainder of R * s in long division by and these operations are a are going to be associative so um let's H let's see why I mean well actually uh not sure it's

worth um let me put it in Brackets you you can check it yourself but it's really associativity of addition and and multiplication of polinomial give you um give you associativity of the of these modular operations I mean for example if you do um so let's say um so

if you have a r equal to F modul Fe and you have S equ Al to G modu of V then you can write r as sorry f is um quotient time V plus r and you can write G is let's say t * V plus s

but then f+ G is going to be equal to m q + t p plus r + S so you see that that f+ g module Fe is equal to r + S right so so in fact you I could have defined this uh this set f f i mean f FX as the set of all pols it's like I do

and again this analogy I mean how do you define integers modul n you can look one way is to look at all the integers and well you define kind of equivalence classes you say if I take any two integers I the addition and multiplication I I just do a regular addition multiplication and I I do long division by n and and take the remainder and another way to define the integers module one is to say okay I look at all possible remainders 0 up to

n minus one and I have addition and multiplication modu n on this on the set and again this is given by the long division so what I get in in this polinomial case is really completely analogous can do addition multiplication mod and and so okay you you can you can see similarly that if you take F time G in this case

right f * G is going to be equal to R * s modu H and so what we have here um well we we have almost a field right we have addition we have

multiplication maybe before this theorem of course so F FX is almost a field and and and I mean if you want the the additive inverse of a the additive inverse of R ER is is simply minus r

um zero is is neutral to addition and one I mean the polom the constant polinomial one is neutral uh to multiplication um but we don't possibly we don't have an inverse for a for an element in in this uh this funny set f f sub3

if Fe is prime F PX in fact I guess the we can say it's an only if and only if if fix is prime is is sorry is a field if and only if three is

prime um before I go to the proof any questions about this modul Fe arithmetic all right so but the proof is really really easy we actually we did we did all the preparations ER so it's going to be quick um if Fe is not Prime and write um feals

to f * G with um degree F and degree g bigger or equal to one and and of course strictly smaller than the degree of then since f and g are of a smaller degree than Fe I can really look at them or view them as elements in this F sub Fe and I get that if I do modul

arithmetic f * G is equal to zero mod so I get two nonzero elements whose multiplication is zero and this cannot happen in the field you remember we had a we had a lar like a observation really that in a field if you multiply two elements that are non zero you cannot get zero so f x has zero divisors and cannot

be okay now conversely suppose Fe is prime ER and let R be some arbitrary element in this uh set of remainders um

then the gcd of a r n Fe is one right right because f is prime f is Prime and R is is of degree smaller than Fe so the gcd has to be one hence there exist um polinomial a and b in a weeks such

that a r sorry pols u and v if I'm consistent in the such that r u plus FV is equal to one i e um are youu is equal to 1 modul so U is a multiplicative inverse

inverse h to R okay hence F VX is a field here I should say R not equal to zero of course a ER say say monic I could do it a first for monic polinomial are but if a monic polinomial has a multiplicative

inverse then all the non-m monic pooms have a multiplicative inverse because I can always express a polinomial as a constant times a monic polinomial cor if f is equal to FP and um Fe a prime polinomial then

then F sub Fe is a field with a p to the N element um where the degree of I mean where n is equal to the degree of okayy this this cor really does not require a proof I mean you just count what are the possibilities of all

remainder pols I mean over FP you basically have to choose H the coefficients a z up to a n minus one ER and they could be anything so uh and so you get you get P to the N options for all the remainder pols ER any questions about this uh this theorem it's basically existence well it's more or less existence of fields of of p2d n

elements I get a feeling that um it was not completely clear but I need to hear what is a what is missing Carlos what do you think for

me ER okay so so it's clear but again the take what I want you to take from this is really if you think of it this proof is is almost the exactly the same as the proof that that modu I mean integers modu Prime are a are field I mean you you use the bazo the bazo coefficients to to find the multiplicative inverse the multiplicative inverse is the only kind the only non-al Axiom that you need to check in

order to verify that integers modul modul Prime are filled or in this case polom modul a prime polinomial so let's do an example maybe this would uh help a bit so um let us construct a

field with a four elements so I take a polinomial of degree 2 over FS2 so let F be FS2 and a f is the polinomial x² plus x + one and then a fe

has I mean has no roots over F2 hence pH Prime I mean how do you see it has no roots in this case you you simply you can do a Brute Force I mean you substitute Z instead of X you get something non zero you get one and you substitute one you get 1 + 1 + 1 which is still not

zero so I I took a prime polinomial over FS2 now the set of all possible remainders FX is basically it's all pols of degree one or less so it is a um I mean x square sorry h x x + one H one and zero right this is

the and the possible possible remainders let me reorder it say 0 1 x and x + one now I can do I can tell you the addition and multiplication tables so I I do 0 1 x x + 1 and 0 1 x x + 1 and uh if you do the calcul I mean of

course ER I get here zero one sorry 01 X and x + 1 but when I do well 1 + 0 is one is 1 1 + 1 is z 1 + x is x + 1 so you see the addition the addition is really trivial I can complete it for convenience but this is this is X and I have X x +

1 0 1 and x + 1 x one and zero so this is the addition the multiplication is slightly less trivial but it's very very simple ER here well at the zero I get

zero and now I will get a 1 x x + 1 one um 1 x x + 1 and the only the only non-trivial thing is to notice that x² is equal to x + one modu modu so here it's x + one and uh here I I'm going to get

one and here x + One X is 1 and x + 1 x + one is going to be X so right I mean you can see just just to justify some of the I mean let's say to justify this uh purple cell I mean x² is going to be equal to minus x + 1 modul modul F but I mean f is x² + x +

1 right so this is this is the same the same thing as saying like e x² + x + 1 is equal to 0 module I simply moved something to the other side but minus x minus x - one over FS2 is simply x + 1 okay ER so this is the this is the the um addition and

multiplication table and we get a field of four elements ER you can check that it's it's actually a a isomorphic to the the field of four elements that we kind of defined in the I mean a couple of sessions ago with the the elements zero up to I mean 0 one to three so it's another presentation but it's a it's a Concrete presentation right however the the problem well at least one problem is that we don't know that Prime pols exist in every degree over any

field at least from what we saw it doesn't follow so the existence of a field of P2 the N elements for any n is not yet guaranteed however um there are pols are prime polom of arbitrarily

large large degree ER let's say o over um over FP otherwise if a the prime pols can only exist up to a certain degree ER there would be would be finitely

many many prime pols over over FP say a we can call them A1 up to a k but then you take A1 and you multiply it up to a and you add one is also Prime right um

because because for each I AI does not divide let's call it a does not divide a right it has a a remainder one if you do long division of a by this Ai and we said that a A1 up to a k are all the prime polinomial so yeah yes okay so it it what you get from this argument which is again it's it's a a and so this

this is a contradiction it means that I I have to have infinitely many Prime pols and and so of course um I I would get a a um a Fields um Fields f f sub Fe ER or fields of P to the N elements with n as large as I wish but but possibly I don't I I don't know that for every integer n i can get a field of P

Tod n elements okay I guess this is time to wrap up so um any questions about what we we did or a request for next time thank you all right H so see you next week bye thanks by right thank you thank you bye bye you see you