PSE Lectures - Ep 12 - Field classification

Transcript

okay um so last time we talked about um a way to construct fields from a um modul arithmetic of pooms let me remind you um you take a field and um in a prime

polom I remind you that a prime polom um it means that um f cannot be written is let's say f equals a * B maybe do it with the variables FX equals a x times a BX with

um degree of of a and degree of B A strictly bigger than zero I mean so they are not a constant pols ER and we looked at we denoted that is f sub f of x I guess in our notation last time maybe it's it's better to keep it ER we took Fe for the the prime

polinomial and and then we defined F sub Fe to be the set of all you can call it remainder pols H so let's say RX equals um sum of a i x to the i i equals 0 up to K where the AI are elements in F and K is strictly smaller than n so this is like remainder pols

MERS of um mod Fe arithmetic and we showed so this in fact can be defined for any fee without requiring that it's it's a prime polom but if it's a prime polom um f sub Fe is field with

um I mean mody addition and mody multiplication now how many elements does a this field have can anybody say if suppose f is a finite field so

if the number of el elements in F is some it doesn't matter I mean let's say it's it's a some T how many elements does this F sub Fe ER field is no one this a it's supposed to be a very easy question

so what are all the possibilities to to get a pols of this of this form are I mean you you need to choose um basically you can write um you can always write R is a some i = 0 up to n minus1 AI x to the I where the AI is in F but possibly some of them are

zero so if it is a polinomial of degree um two and N is equal to four then you can extend it to a er I mean to a form of this way where you take A3 to B zero is it P to the r power n yes so so the number of elements in in F sub Fe is T to the N okay is this clear you

were quite quite slow in answering [Music] this so you know what maybe maybe because I feel that you are not a exactly um on the same page let me do an example we did it last time but H it may

be good to to do it again so I take F to be the field of two elements and and Fe is the polinomial x² + x + 1 I claim that Fe is prime Fe is prime since well Prime of course you need to specify over which uh field

over F2 um and the reason is that it has h no roots I mean this you can check by hand H you substitute zero into Fe you get one you substitute one you also get one so it has no roots um and every non-trivial decomposition of a polinomial of degree two must be two polinomial of of degree one which would imply that it it would have a root

so in this case a polinomial of degree two is prime if and only if it has no roots this it's not true in higher degrees but and so so now I I can look at f subv F sub Fe actually as a set it does not depend really on the the specific ER the specific definition of Fe only only on the degree of Fe because I take all the remainder

pols RX equals we can drop the sum so here um I have a0 plus A1 X um where a z and A1 are some coefficient in the the ground field right this is all the the possible remainders of doing a a long division of a polom by

Fe so but let's be more specific what are the the possible pols I mean I have a a zero the zero polinomial then I have the polom one and then I have the polinomial X and the polinomial x + one and that's it right so you see as a set you could choose any polinomial of degree two and the set F sub Fe would be the same but of course if you choose a

different polom fee then the arithmetic on N Sub F sub Fe would be different because the arithmetic is a is doing long division by I mean by okay so um so let's construct the multiplication and addition table addition table is a um you don't need to do any modu because um this set of remainders when you add two remainders H the degree of the result is still a strictly smaller than than the

degree of Fe so whatever you whenever you do mod fee for something with a degree strictly less than the degree of fee you get you get the the thing itself it's like again the analogy with integers I mean if you you take some K smaller than n then K is equal to K mod n you don't need to do any any modul in this case so when I do addition of two pooms it just I do addition as I'm used to in polinomial

so let's say 0 1 x x + one 0 1 x x + 1 and you would get H for example if you do x + x + 1 then this is 2 x + 1 but 2 is equal to Z so this is one okay and H you can fill in the the addition table the multiplication

table is a bit more interesting ER I guess I don't want to write all the multiplication table but ER note that if you take a x square then it is equal to minus X+ one modu of because I remind you that V is x² + x + 1 so this this would mean

that and now what is the negative of I mean minus x - x + one of course in FS2 is simply x + one okay so I get that x * X is equal to x + 1 in the the mody arithmetic and uh similarly you can do a x Cube and you will get H you will get one mod of

here okay so this this is how you ER you work with these uh these fields any questions about this I I feel that uh you're not completely fluent but I need to hear what is not uh what may not be clear okay

um so it may be good um to introduce a a new notion I'm I'm trying to be very careful in a doing the these things ER but I think here is is a it's it's a good I mean it's convenient to have it as a terminology so a set are um or let's

say a ring is a set R um with a a specified element 0o and one and a a two binary operations H here plus from the cartisian

product to R and multiplication that satisfy all the axioms of the field of a field all aums of field except well possibly um existence of a multiplicative

inverse and well the the easiest example is H the integers right the integers you have addition you have multiplication but H you don't have a multiplicative inverse you have an additive inverse but you don't have a multiplicative inverse in fact if you take the integers modul n with a um addition modu n and multiplication modu

n then this is also a ring right you have a a an additive inverse for every H for every element but you don't necessarily have a multiplicative inverse and yes can you provide some intuition uh why it is useful for us to have something that is almost like a field but without multiplicative inverse okay so that that would be the next example um if f is a field h f x I mean pols over over f is a

ring um right because you can H you can add two pols you have an additive inverse I mean if you you take a polinomial f then you just do minus over all the coefficients that's the additive inverse but you don't have a multiplicative inverse and if you I mean if you want the the the kind of the conceptual reason that it's a it's it's good to have this terminology in our toolbox what we did if you

remember when we proved that um zp or or it's some denoted as FP is a field is the way we worked about it is um we first looked at integers modu n and we saw that you have addition and multiplication that are associative and all the the usual axioms are satisfied except possibly the existence of multiplicative inverse and and and then in order to get to to zp we

said okay what do we need for a multiplicative inverse and and the condition is that n n is a prime so it's convenient to um to have an intermediate notion right because if you look at this definition of rings then of course um every field every field is is a ring so the the set of all Fields um is contained in the set of all

rings and every ring is in particular an a billion group and um so rings are kind of an intermediate notion between fields and and a billion groups um that allow us on one hand to talk about a um a structure with two binary operations addition and multiplication but slightly weaker than a field so that they include for example these these three uh the three

examples that um that are not not not a field but nevertheless have multiplication now maybe the more interesting example is you take a field f and [Music] um Fe a polom that is not prime okay so then f

sub3 is not a field is a ring well let say not a field but it is a ring with of course with Mod Fe arithmetic anybody remembers what why in this case when when Fe is not a prime is not a prime polinomial why is this not a fi because we can't construct this

multiplicative table um where where we can uh find a multiplicative inverse for each element but why why don't we have a a multiplicative inverse for every element I mean there is a very simple argument that again you should think of this as in the analogy to the to the integers how do we prove that um Z I mean integers module n is not a

field we know that in a field you cannot have zero divisors so if you if in a field you have a a two elements X and Y such that x times Y is zero then since you you I mean if you're in a field ER then you have a multiplicative inverse to X and Y if they are not zero so the only way that x * y will be zero is a a if x is equal to Z or Y is equal to Z this it follows from the field

axioms and in in integers modu n i mean if n is not prime then it is K * L for some K and L and then k k * l in in ZN would be zero So Lisa right you understand this why uh why at least integers modu n is not a field what's the argument here uh I I understand that uh if it is

not prime you can factorize it and uh represent as a product mhm um of of the product of um a of a prime polinomial with something else yes it doesn't matter even that it would be a product of prime polom I mean the fact if it's not prime then it is a product of two um of two polinomial with a a a um I mean with non-trivial degree right um but

then A and B belong of course A A and B have degree that is strictly strictly a lower than the degree of Fe right because ER that that's that's what it means to be a non Prime polinomial so if this is so then both of them belong to F sub Fe I mean both of them can be viewed as remainder by by long division with Fe and a * B is equal to zero modu of

V right right because a * b equals to Fe so what I get is that I have two nonzero elements in F sub Fe whose product is is the zero element and this this phenomena cannot happen in a field so I can kind of it's kind of by abstract reasoning I know that this this cannot be a fi is it clear yeah yeah okay but it is a ring I mean I do have a a um I do have addition I do have

multiplication I don't want to throw it all away and just consider it as as an a billion group because I care about the multiplication so this is maybe the the justification for why it is useful to to talk about Rings although we will not get into I mean deep into the the theor Rings so in this case f sub Fe is a ring

now there is a natural notion of what it means to be a ring map um for two rings and say r and s a ring homomorphism is a

function f from R to S such that well it has to preserve the specified elements and it has to preserve addition and multiplication so F of a + b is equal to F of a plus F of B and F of a * B is equal to F of

a Time F of B and this is for for every A and B in R and if you wish you can throw also the the axom that but it really follows from from these axioms that it commutes with a additive inverse and multiplicative inverse this is for every a in R and of course multiplicative inverse

only exist for nonzero elements so this is a ring homomorphism it's like basically it's it's the same axioms of what it means to be a fi homomorphism um and F is called an isomorphism if it is bjective

and just like in fields um f from R to S is a ring isomorphism if and only if there exist a a ring homomorphism F minus one from s to R such that H let's say

fus1 composed f is the identity mark on R and F composed F minus one is the identity map on S and this you can prove this is kind of a recurring theme um this is a property that will always characterize isomorphism of every basically of every notion that we will encounter I mean groups Fields Rings Vector spaces and so

on so it's really a it should not come as a surprise we saw this before you can you can prove it as an exercise um now what happens in rings is just like in groups you can talk about quotients and um the notation um we will denote if sub

Fe is um FX modul you can find it in the literature sometimes as FX modul The the thing generated by by F okay and this is analogus a to the notation of integers modul the subgroup of integers multiples of of n right this is a this is this is the

notation for a quotient group so I do not of course as I said I don't want to get into the theor Rings too much um but but I do but I do want to use this notation I think it's useful for you as well um and and a little bit I mean eventually what we will need in what's coming up is just this notion of isomorphism of rings and this this will be enough

So Lisa do you feel that it's motivated enough yes thank you okay so don't be afraid it's not it's not complicated ER again it's like it's like Fields um it just that we are we are dealing with a kind of creatures like this polom pols that we want to capture something that has I mean we want to to be able to talk about a a a structure that has the the

addition and multiplication ER but possibly without inverses in order to H to kind of reason about this pols and pols modul mod another polom okay so now there is a a funny a funny metat trick so

consider a field F and a prime polinomial a v now if the degree of Fe is equal to one then FX what we called f sub Fe and now I remind you we have a new

notation this is by definition all the remainder pols of of doing a a mod fee ER but since Fe has degree one the remainder polinomial have to have degree smaller than one so they have to be constants so it's really all the a z such that a z is in F which is of course just another another way of saying F so this is not interesting I mean we don't care about a a mod mod F arithmetic if Fe is is of degree one and of course if Fe is of degree zero that's

that's even even less interesting so suppose that the degree of Fe is strictly bigger than one okay that we know that the number of elements of FX modu of Fe is simply the number of elements in F to the power of the degree of the degree

of right this is a this is what we said here where was it right if a the number of elements in F is is T then H the set of all possible remainders is T to the N when where n is is the degree of okay so H so we know the number of elements but we can

look um we can consider the ring homomorphism I go from pols over f two pols over F modu and the map takes a polinomial f of x and sends it to f of

x modul of I mean just take the remainder of long division of of f by and we we talked about it last time that this is indeed a a a ring homomorphism because if you you take multiplication I mean if you do f * g f * G would be sent to f * G modu modu but it's it is equal in fact to F modu of

V time G modu of right and the same for addition f+ G module of e is equal to F mod and and plus G so so this is a ring homomorphism I call it Pi this some sometimes called the quotient the quotient map now I look at

um so Fe or maybe put it like this um f is contained in FX modu X is the set of a constant a remain pols right I mean here remember we have this

H all remainders of the form sum AI x to the I with some limitation on the degree of course you can just take a a the remainders that are just the constant a Zer and this is this can be identified with f okay so if I call this guy K let me call it k

a we can consider we can view p is a polom over K right I mean k is a field so we can always talk about pols over any field here it's a bit it's a bit perverse because we talk about pols with coefficients in pols but this kind of meta trick is a is useful ER

so I mean it means that um I can view the set of all pols over K now I put it with a um with a big variable X to to differentiate from the small variable that comes here and I can view Fe as as a polom over this

um over this this new field K right I mean whatever Fe was I just now declare decare the variable to be capital x instead of lowercase x and that's my polinomial is is this clear yeah I have a question so because I don't see why like so you have uh you define K as f of x over V of X like the the set of polinomial uh well yeah the field over the Fe of X but and then you say that

um you can view uh Fe as the polinomial over K but if you evaluate V over K which is f of X over V like in my mind I was following but but that is step it was like okay Fe is getting cancell out because it's it's okay so that's that's a I mean how does typical element in k k of X looks like it's um you can write it as um let me call

it r0 plus R1 x + R2 x² plus um let's say RN x to the N such that r0 up to RN are coefficients right they they need to be elements in K so but K is is the set of all remainder pols I mean so so R r0 up to RN they

need to be pols you could write yeah the they would be pols over F in in the the the small X intermediate or in determinant let me write it slightly better RN x to the N um such that r0 up to

RN are pols over f with degree actually n is maybe not good here because we already took n so let me let me [Music] do let's say m such that the degree of each of the r i is strictly smaller than n because they need to be in K and K is the set of all remainder pols R

ofx um equals to Su a i x to the I I equals 0 up to K and K is strictly smaller than n right that's the that's the set k k as a set is is the collection of all possible remainders by a a mod Fe so pols over K are these formal sums now I I just took the variable and made it a bit different if you wish it's kind of you can write it also it would be a bit a bit combersome but you could also write

r0x plus R1 x times the capital X Plus R2 x * x² and so on okay is this clear now how how to to view nomials over thank you very much because this this uh made me see uh the thing I mean I was not seeing the right difference between the two like the lower and capital x yeah thank you yeah yeah so and and now of course you have fee fee with lowercase x is some kind of a let

me just write it with coefficients a zer A1 X up to a n x to the N right we said that Fe is a polinomial of degree n now a z A1 up to a n they can they can be viewed as as remainder pols so Fe with with capital x is the same thing as 0 A1 but now I replace X by capital x and and I view all these guys a a z A1

up to a n in a in K I mean I view them as constant remainder pols okay and now there is a funny trick yeah and in a fe with small X all a were just filled elements exactly so Fe with small because Fe is so A1 a a z up to a n are elements in the field but we said that the field can

can be included into the set of all remainder pols right yeah yeah yeah okay now watch this I take remember we have this map PI from f of x to f of x modul p of X this this is all small X right this we call K and I take Pi of the

polom small X this is this is a assuming that Fe is of degree bigger than one this is X right because if he is of degree one sorry of degree bigger than one and X of course is a polinomial of degree one then doing the module of Fe you don't do anything just like if K is smaller than n you don't do I mean modu k modu n is simply K now so I I view this pi

x as an element in K and as an element in K I can plug it I can substitute it into a a fe with capital x so Fe with capital x substitute Pi of x it means that I do F of Pi of X right this is and

this of course since Fe capital x is an is a polom in K of x capital x when you plug in an element in K into Fe when when you substitute an element in K into Fe you get a a again an element in K right just like you take a polinomial over F you substitute an element in it and you get an element in the field so now a a fe with pi of X is simply Fe of x lowercase x now this is

an element in K supposed to be an element in K but in k f of x is zero so this is this is equal equal to zero in K so what we got is that pix is a root of f capital x i e in in

K I can write f of x or in in in K capital x I can write f of x to be xus pi x times some other polinomial C of X right that that's what it means to that a polom has a root so you see what happened here is that Fe was a prime polinomial it means in particular it has no roots over over the field f and now I took this

mody arithmetic field K and suddenly over this field K Fe has a root and this field K is is an extension of the field F I mean it contains a copy of f inside it so what I did in fact is like a um is something very similar to when you pass from R into C I mean okay but the ER right I mean here in R you can look at at the

polinomial x² + 1 and once you this this is a prime polinomial over R but once you you add a root I mean in C it decomposes as xus I X+ I so you can ask how how do how did people think about I mean about c one way to view it is this H this kind of meta trick so I know it's a delicate matter

so yeah I'm I'm glad you I'm glad you like it um yeah any questions AR now what do you think you you follow yes thank thanks for the clarification and and like this last thing well it was reminding me a bit of something that we saw but I don't remember exactly the name like this thing of like that you can um decompose things uh like I mean now this this like you you have this f of x equals to x minus P of X multiplied by this other symbol which I don't know this other

symbol you can keep doing this like recursively yeah I we this not in not with polinomial but with with uh I mean not with with with this setting but with did something similar to prove I don't remember what like that that we could decompose things uh into Prime so I guess that we are hting toward the similar Direction but in this new setting but okay but it's important that you get this H this meta trick I mean that in this in this new field k I suddenly have a route for Fe Fe Fe did not have any roote in the original field

and now I I enlarge the field in a way that forces a a a forces Fe to have a root it's kind of yeah that's very cool okay ER and so may maybe now is a good time for a break we'll come back after the break and and we'll see some application of this all right so um the field F to be the real numbers and you take the polinomial fee to be x² +

1 then Fe is prime because it doesn't have any root in in r H and now I consider pols over the real numbers modul x² + 1 and this as a set it is simply the set of all possible remainder pols I mean a0 plus A1 X where here I allow a a z and A1 they possibly they could be it could be zero

so in general this is all the Expressions a0 + A1 X or a Zer and A1 are elements in R and now I consider the I mean the element X in this in this set and if I do so if I do x² well x * x h then this is equal

to minus one modu right because x² + 1 is equal to0 modu of so what I get is that it's the set of all Expressions a0 + A1 X such that x square is equal to minus one and of course this is the same as I mean it's just a a different notation for what we know to be the the field of complex numbers namely the set of all expressions of the form a0 plus A1

I a Zer and A1 are re numbers such that of course the I sare is equal to to minus one okay so this what we we did before the break in in this example would would give you a a the at least one way to obtain naturally the complex numbers from ER from The Ring numbers any questions

why do sorry why do we prefer talking about complex numbers and instead of his Fields instead of finite fields of this one sorry can you repeat the question uh why do we talk about the complex numbers instead of talking about RX ided by x s + one in that case uh it's isomorph I mean it's it's a a a yeah no I can I can see it it just yeah well I mean ER you know the way people learn about the complex numbers is a usually it's in earlier stages of the the mathematical

porpose so it's easier to just H Define complex numbers like this with this imaginary number and Define the addition and multiplication and start doing stuff with this but but in fact from a a more advanced mature H perspective what we're doing here is we're moding out by a prime polinomial and this means that we're adding up a root of this of this polinomial yeah yeah okay thanks

okay so this this brings us to the following a following definition um let F be a field and FX a polom over

f a splitting field um of f over of course capital F is a field K or actually is a field maybe we call it e

H containing containing F such that um when view FX in now in E of x it decomposes into linear factors and so linear

factors I really mean that um then F of EX over e would be some constant let's say Alpha times um x - alpha 1 time x - alpha alpha n where n is the degree of of right so a so Al Alpha and Alpha 1 up to alpha

n are elements now in E I mean there are roots of this polom that possibly did not exist in in F but they do exist in e h so a splitting field is is a field extending F such that the polinomial decomposes to linear factors and E is minimal with respect to this property

and this minimal means that if K contains F and the polom f of x decomposes into linear factors over k then e is minimal so so K must contain e okay it's a bit a bit long definition

but uh um I hope it's clear H any question question s and we have two example of course and C is a splitting field x² + 1

over R yes uh in the bracket uh K is the same that was e above right um there are many brackets so the last one the last one the last one where where it is like if uh uh f is in K and f f ofx de composes into linear vectors over K this is what in the abstract above K is what in the abstract above we called e ah yeah yeah e no I mean yeah okay in a way but this

is just explaining what does it mean that that e is minimal with with respect to this property I mean if if you have any other field containing F I mean if you have any other field K containing F such that the polom FX decomposes into linear factors over this k then k must contain E I mean you cannot you cannot do a a I mean e e is minimal right minimal with respect to inclusion

okay okay so here um so maybe it's worth seeing this example I mean I claim that c is a splitting field of x² + one this is this f of x um o over R now you need to be a bit careful in in a in reasoning about it because well the

EAS effect of course a oversee f of x you can write it as X minus I X+ I so over over C indeed F splits into linear factors but that's that's still not not showing that it is a splitting field because we have to show minimality for minimality

if a k contains R and F splits over K of course since f is is a monic poomi i e f f of x would be equal to let's say xus Alpha and of course the the other thing is going to be X Plus

Alpha then H Alpha square is going to be one sorry Alpha square is going to to be minus one because Alpha is a root I mean alpha s + 1 = 0 is equivalent to say that Alpha square is equal to minus1 so this Alpha is a I mean as far as the the addition multiplication goes this Alpha is the same as as I

and um and since K is a field ER it must contain all expressions the form um let's say a

z plus A1 Alpha for a z A1 in R right because a field is closed under addition and under a multiplication so in fact this would mean that K contains an isomorphic copy of C I mean essentially C so so C is really minimal with respect to this property

is this clear what do you think Lisa kind of kind of er well it's better if it it would be completely clear I mean look this Alpha I just call it Alpha for a distinction but it really has to be it it really has to be I I mean it is an element satisfying Alpha square + 1 equals z i mean or or Alpha square

equals minus one so Alpha is is just another name for I right and uh and once I have this a f containing I and containing R so so maybe it's better to call this Alpha I and then I mean of course if if I have a field containing I and containing R it has to contain all the expressions of the form a0 plus A1

I yes it's it's clear so it must contain C whatever it is whatever K is it contains I and it contains R then it must contain C so this tells you that c is the minimal field with respect to inclusion that a a for which this polinomial FX splits into linear factors okay and this really show this this shows that that c is the splitting field of the polom x² +

1 over R okay we need remember we need two things splitting field is is a is a field for which the polom splits to linear factors and it is the minimal the minimal field that that this happens ER other questions what um I'm trying to think what what might not be

clear here okay let's try a proposition now of course this this definition is nice we saw even an example but we are not guaranteed that a split field always exist right ER and this is the the content of the next proposition um let F be a

field and FX some polinomial and then there [Music] exist a splitting field over of f over capital F

okay so um let us compose F into prime factors over over the original field the big F so I write f of x equals fub1 of X multiplied by F let's say k of X now if all of these factors are

linear if um fub1 up to f FK are all linear then uh then the original field is a splitting field of a of f right I mean it is a field in which F splits into linear factors and it is the minimal field containing F because I mean of

course you cannot you cannot do anything smaller and otherwise ER without loss of generality is suppose the degree of of F1 is bigger than one um consider

K1 to be Pol over f modul fub1 um and in this K1 we know that f1x has a root um in K1 let's call it let's call it Alpha for

or maybe yeah say Alpha right this is a this is what this is the meta trick we we talked about before the break this means that I can write F1 as xus Alpha times Su H G1 Su G1 X in

k1x now I can consider um we can consider FX um over is as a polinomial in I mean there a polom Over

K1 and as such it has at least a one linear Factor right because because I can write a I mean whatever I could write in F I can write in K1 because K1 contains F so I could write in f um the polinomial

f of x being the product of fub1 up to FK and now over um over K1 I can I can write FS1 as x minus Alpha times G1 and and the rest I leave the same so F2 of X up to FK of and by doing this I mean I I can repeat now I I take I mean if GX if

g1x is not linear H we repeat repeat the process to get K2 that contains K1 in which I would have a a another at Le at least one other root let's say

beta of f um since F can have at most and finite number of roots this process

terminates so it terminates in some a a in some field K containing f such that indeed F decomposes into into linear factors so I can write really Alpha time x - alpha 1 up to x - Alpha 2 sorry alpha n where um the

alphas are elements in K and um now for minimality um observe that um I mean

K1 K1 is the the minimal field containing a containing F and containing Alpha because remember what is um what is alpha alpha is this Pi of X right this a ER and let me let me just uh scroll back to this uh this part right I took a I took a polinomial fee

let's say it was Prime and I constructed K and then suddenly a fe when viewed as a polinomial over K had had a root Pi of X this I mean this k is the minimal is the minimal field containing the original field F and the root Pi of X because what is this h i mean what is this k k is all the possible

remainders and once you know that K contains I mean Pi Pi x is simply X so once you know that you have a field containing F and containing X it must contain X squ it must contain X Cube and so on it also it must contain um Alpha time x and Alpha times x square for for Alpha in F because it is a field so in fact it must contain all the remainder pols so so K is the minimal field for which H for which this

happens um and so the same goes here I mean for minimality observe that K1 is the minimal field containing F and and Alpha okay so this H and in every so so

K all in all k would be um would would be by induction the minimal field containing um all containing F and all Roots all the

roots of um of FX okay so so this this means that ER this K would be a a splitting field ER for a for for the polinomial

okay so I I I understand this is a kind of a more abstract than what we uh what we done what we've done so far in the in the previous lectures um how do you feel about it it's clear the the the minimality is is also okay okay so let let me just I I I

repeat the the the argument for minimality I repeat it here in this uh in this example that we did here I mean I took a I took a field F I took a field F and I took a prime polinomial ER and and now I constructed this ER this field K that is the set of all possible remainers

of mod long division and this this K this we showed that this K contains an element namely the the polom H so do I have like I mean k is the set of all remainder pols so all expressions of the form um a0 plus A1 x uh up to a k x to

the k for some coefficients I mean a z up to a K in m in F and in this in this field this new field K we can look at the polom X and X suddenly is a is a root of Fe ER o over over this this field K but if you look at it this construction is the minimal thing that you could have done if you wanted a field that contains F and contains a and

contains X because once it contains f it contains I mean all the constant remain all the the constant remainder pols and once it contain X it must contain all scolars times x and all powers of X and all sums of of these of these Expressions so it must contain K so in fact K is the minimal field for which you I mean for which which contains F and and contains the the root pix and by this argument I mean you you

apply it in this proposition I mean in this proposition every time we we split a factor we know that what we did is the the minimal thing that you could have done so this this shows you that all in all the resulting the resulting field ER K1 was the minimal for which the first route was constructed and and eventually the resulting field K is the minimal is the minimal one that that can contain all the roots of the of the the original polom okay

so now would be a good time for a questions com comments requests for next time and so on all right ER then uh we continue next time with this uh splitting field stuff what will happen I'll just a kind of a overview of what's coming up um we are going to prove that splitting

fields are in fact unique unique up to isomorphism and once we know that splitting fields are unique up to isomorphism we will able to very quickly very easily classify all finite Fields so that that's a that's what what's coming up all right so uh hope you enjoyed and have a good rest of your day thank you thank you thank you thank you thanks

bye