PSE Lectures - Ep 13 - Field classification

Transcript

yes ah okay um let's put it like this um we denote the the quotient map from FX to FX mod of Fe um by pi so Pi of some polinomial f of x is simply f of x

modul and I I can I can consider a specific uh element um which is a pi of the polom X and and now um if I consider F capital X in K capital X um I can substitute Alpha and I get zero because I get a f

of pi x assuming that the degree of Fe is is more than one then pi x is simply the remainder polinomial X so this is f of x module Fe which is simply zero so so Alpha is a root of of Fe um it means that I can write Fe of capital x as xus Alpha times some some other polinomial P ofix now what is this

H ER this field FX mod suppose so it is it is a field containing a a root to the polinomial fee but you may ask ER could I could I er er constructed some smaller field that still contain a a root for Fe and so suppose

[Music] um e is field um such that two things happen one is that e contains the field F and two that Alpha is an element in so what I mean I would like to know what is the minimal field containing f and and

Alpha um let's call it you know what let's call it f of alpha so so so what has to happen in F of alpha

um clearly um if if I take any polom and I substitute Alpha in this polinomial I have to have a I have to get an element in F of alpha because F of alpha is a is a field it has to contain Alpha therefore it has to contain Alpha

square and Alpha Cube and so on but it also has to contain a a a constant say Lambda from F times Alpha and it has to contain Lambda time Alpha plus Delta time Alpha squ and so on so it has to contain all pols in Alpha what I can also expect if I have a field containing F and containing

Alpha is that well Alpha has to have an inverse in this field so for any um let's take a a nonzero polom G [Music] um one over G of alpha also has

to has to belong to F of alpha right because G of alpha has to be an inverse sorry G of alpha has to be um an element in in F of Alpha and therefore it has to have an inverse this means so if F of Alpha and one over G of alpha has to belong to F of alpha it means that for any f and g pols in F the quotient F of

alpha over G of alpha assuming that g is not a zero polinomial let's say G of alpha is non zero or or rather G of X than zero H this quotient F of alpha modul G of alpha has to be in in F of Al um okay so what is the the minimal the minimal construction I can do for

um having a field containing F and containing Alpha um I I can take f f of alpha to be simply the set of all quotients F of alpha modul G of alpha such that f and g are pols over the original field f and g is non zero right of course you can multip you can

multiply ER two elements in the field you can H you can also add to elements in this in this set and the inverse um of f Alpha modu G Alpha is simply assuming that this is a non zero element it means that F of alpha is non zero then I can do simply G Alpha the mod fpha is there any question uh is f Alpha um does f Alpha

equal uh F small Alpha over G Alpha just because we chose it in this way right uh can you repeat F Alpha you mean the cap the big F yeah I mean that big uh F Alpha equals small F Alpha over G Alpha just because it is our choice so big F Alpha this one is the set of all quotients of the form F Alpha modu G Alpha right it's the set and the reason is that I

mean I want a minimal I want a minimal field with these two properties and I I kind of I reason about it um in two steps I mean first I assume that it exist and if I assume that it exist I can I can conclude that for any two pols the quotients F Alpha modul G Alpha has to be in this field oh makes sense makes sense okay and then I say okay but this set of all quotients is already

a field because I I have an inverse to to an element I mean it's very easy to construct to to see why the other axioms of a field are satisfied so you get that F like this this guy F capital F Alpha is already a field so it has to be the minimal field containing F and containing Alpha right is this is this clear yes thank you okay now um now take this uh take generic pols f

and g and do long division by by this Prime polinomial Fe so it means that I can write f of x as um FX q1x plus R1 and g x is fix Q2 plus R2 2 x um for for some polinomial q1 Q2 R1 R2 where

um where the degree of of R1 and the degree of of R2 are strictly smaller than the degree of Fe now I substitute in in this decomposition I substitute Alpha F of alpha is zero so I get that F of alpha is R1 of Alpha and G of

alpha is our two of alpha this means that F of alpha modu G of alpha is simply R1 Alpha modul R2 alpha or in other in other words this is R1 Alpha times R2 Alpha minus one but remember Alpha is just a is just

a convenient notation for us for um for pi of X right Alpha is is simply an element in this guy which is which is X I mean if you think of a of this H this field K as the set of remainder pols then Alpha is simply the remainder polom X ER so R1 of alpha is is the same as what we denoted in K as R of X I mean R1 of

X ER and of course K already contains inverses to H to every every po every remainder polinomial R of X so this is all to say um FX modu Fe is simply equal to F of alha and the

proof is f of alpha is the minimal field H containing F and Alpha H so this guy k k it I mean k must contain F of alpha

right on the other hand [Music] um K is is simply the set of remainder pols R of alpha so K is contained in in F of alpha you can take a what we denoted here R ofx and uh think of it as a r of alpha module one

right so there is a dou inclusion we get um k equals F of alpha is this point clear this um this gives us a a a very nice characterization of what it means to divide by a prime polinomial okay

now we had the definition of a splitting field um or you know what perhaps yeah okay I I'll I'll give the definition then um some uh some other example so I start with the field F and a a a polinomial f of

x which does not need to be a prime polinomial a splitting field um for f over a big F is a field e containing F such that F

splits into linear factors so you could write it for example as f is and just write both both ways F splits to linear factors over over e i e in f of x is a some constant Lambda if if f is not

monic I have to multiply everything by some constant um times xus say alpha 1 up to xus alpha n where a Lambda and Alpha 1 up to alpha n must be well they are elements in e and two is minimal uh with uh this

property ie for any um let's say you want to put someone in the middle between F and e um any field um is such that F splits um over K we must

have k equals to e okay so you you cannot do better I mean so the splitting field probably the the the canonical example that everybody knows is that um

C is a splitting field of um the polom x² + 1 when I think of this polom as a polom over R um it so how do I see that c is is indeed a splitting field of this polinomial um first I want to see that this polinomial splits but that's easy I mean

x² + 1 equal x - i x + I in in C of x so so the polinomial indeed splits and B if um K contains a contains R and um x² + 1 splits over k then

a k must contain I right I mean this polinomial e s plus one if it's split if it splits then K contains an element such that this element H square is equal to minus one or this element square plus one is equal to zero and this yeah I think there's a question in the chat from Lisa um yeah okay let me says will minimal e be unique for f I'm not sure I understand the question um you meaning the definition of a

splitting field I suppose uh yeah I I mean in uh in the second uh in the second Point uh when you explained that uh K should be equal to e we must must uh must be equal to e um my question is like will this e be uh the unique minimal uh for f well youi okay we will show that this is that splitting fields are unique up to isomorphism but for now we don't uh

we don't assume it okay okay so this but but the intuition is of course correct I mean this kind of a condition number two is is very strong so you expect that anything that satisfy condition one and condition two will be essentially unique okay so just to return to this example of the complex numbers um if I have how do I show minim I mean

splitting is is very easy how do I show minimality well if I have a field that over which x square split then this field has to contain I and any field containing are and I must contain C

right because H it C is simply the set of all let's say a plus b i where A and B are real numbers now the second example is ah and uh maybe just to to connect it to the the other definition C you can view it as R is the minimal field

containing R and I right now I can do the same thing that I defined C I can do with the rational numbers so um take let's call it k k is the set of all a plus b i such that A and B are now rational

numbers and then I claim that K is a splitting field of um the the same polinomial x² + one but now viewed over the rational numbers okay so you know you could you could start with the field f equals q and say

well what is the splitting field of this polinomial x² + 1 over Q so of course e x s + one splits over the complex numbers but the complex numbers is not the splitting field of this polinomial x² + 1 over the rational and the reason that the complex numbers in this case is not the minimal field with this property the minimal field with this property is is is

K okay so I will not repeat the argument but but it's a maybe you can take it as an exercise that K is a splitting field indeed of of this polinomial and also K is the minimal field containing q and I okay so we have we have like a a basically three ways to to view this right

you have q of I you have um the uh splitting field of um x² + 1 in QX and you have have what we denoted as um QX modul this polinomial x² + 1 so this is all H this is all

equivalent equivalent definitions is this point clear this I think this is really the heart of the matter and after that the the propositions are uh very easy to see carus what do you think yeah for for me it is yeah but anyone that I mean I I always assume

that if no one jumps in uh everything is clear but however you prefer okay also try to say something yes no yeah thanks uh I'm a bit confused like which is the relation between splitting field and a field extension so yeah sorry ER yeah you want to complete the question no no no it's it's it's that so sorry so a field extension we did not actually mention that too much but field extension is simply a a two Fields I mean it's like what we call the sub

field inclusion so if you have a so the terminology field extension is the same as a a subfield [Music] inclusion let's say F into K right it means that K is a field and F is a subfield contain I mean f is a

subfield so er so this field extension is a very general [Music] um is a very general term right a a splitting field is a particular kind of field extension so for example in this in this H in this case I mean of course C is a field extension of Q because it's a field containing q ER such that the the operations are the same but C is not a splitting field of of x s + one

because well it's not minimal for example yes carless this made me this made me question uh in the properties of of field extensions and so field inclusion basically uh if I record correctly the multiplication and addition table for the sub field which is included and the field on which it on which it is included into are shared up to the up to the modulus of the of the of the field that is included is that correct or not not up to modul they they are the the the multiplication tables in

and addition tables in F are s simply restriction of the multiplication tables and addition tables in K to the elements in F right I mean exactly okay okay yeah yeah yeah yeah yeah exactly so they are simply a restriction sure thank you thank you Madame for the reminder okay so so this is this is what we're the kind of landscape we are a um dealing with I mean we have a a a a particular kind of field

extension that we call the splitting field and the proposition we showed last time um for any F in F and a splitting field there exist splitting field

of a f over a big F and uh let me repeat the proof because I think ER it was not completely clear last time [Music] um let's say without loss of generality f is monic I mean monic or not monic it it doesn't change the ER the possible roots

of F and splitting field only depends on the the possible roots of of f so if it exist for for monic it exists for H for every polinomial uh right yes are there any properties in um for these kinds of splitting Fields such that I don't know we can model them on on on some way with polinomial or something similar such that we can disclose which ones are and which ones are not splitting like um how can I say it there was a time where we saw for

example primitive polinomial how were they able to generate spaces and and I'm wondering if there's any smart way to actually check uh splitting Fields uh rather than just decomposing the entire thing into the roots and and checking ER well we H we will get to it ER I guess okay okay answer okay then it's fine but you can see from this example for here that one one way to get the splitting field of

a um x s + one was simply to H to mod out by x sare + one and the reason is that um I mean x s + one when it's splits it it splits into xus I X plus I um and um so and once you have a field containing containing I you already immediately split the polinomial x s plus plus one because if you have I you you have to have minus I I see okay so this in a way

you take a prime polom um all the all the roots that can I mean if you add one root and I mean and all the other roots are kind of a follow from the axioms of the field that they must H they must be included in the minimal field containing your root then you got immediately a splitting field that's kind of okay okay I kind of get the intuition but but don't want to take over more time so if we we get i'll patiently wait that's okay it is natural not to understand a a fully a notion I mean of

course we didn't prove anything about it yet and and even after you prove you only know understand it formally so you have to be patient I mean sure it comes comes over time makes sense thank you Madan um okay so I write FX as um fub1 X up to FK of X where um fub1 up to FK are

Prime but Prime you always need to to have a reference fi so Prime [Music] over over F over F FX okay this is the kind of the Prime decomposition ER analogous to to the prime decomposition of a of an integer [Music] um now I consider FX

modul F1 this I call K1 um then um F1 has a root let's say alpha 1 in K1 so um over K1 I can write now I'm not going to carry out this h capital x um

but if you want you you you could make it I'm just going to write F1 of x equals um xus Alpha um f prime one of X in in K K1 of x and this means by by the discussion we had is that K1 is in fact the minimal field containing F and alpha alpha one

sorry right um now F1 X well if F1 Prime X splits over K1 a um we we take well maybe if it doesn't

split um over K1 we can take K1 modu the prime factor um say fub1 Prime of X is equal to

[Music] um I don't know F F1 Prime X up to uh v s Prime x with um P1 Prime up to FS prime prime Poli over K1 take

um K1 um let's say K2 being K1 modul the first Prime polinomial F1 Prime then K2 will will simply be K1 a joint with some um Alpha 1 Prime where alpha 1 is a

root of um V1 Prime okay so and you continue in this way repeat this process since um f can have um only finitely many many

roots [Music] um um we terminate at a some let's say k k n in which or let say

sorry K in which F splits completely we get a sequence of field field inclusions we started with f and then we added one

root um once we added this root we continued until H some KL let's call it k ER and each k i + 1 is the minimal

field field containing K i and a a some some new root so yeah Alpha I + 1 of of f in fact so since in every step of the way we had a a a the minimal field containing containing the the previous

field and another root it means that K is K the last one is the minimal field containing a containing F and all the roots all the roots of f of of this original polinomial small F and K is the minimal field containing um f

and all roots of a of f i e k is splitting field f okay so so this is the proof H is the argument clear I I uh I try to to make the exposition a

bit yes yeah hi uh so can you go back to the definition of K1 the first one when you explain this um yeah here uh yeah so maybe it's a it's the I I missed something on the definition or whatever but why do you say F1 as roots alha one in K1 so this is the the preliminary discussion we had in this H the beginning um whenever you mod out by a prime polom the resulting field has a a root this is the the the heart of the all of

this H discussion is I take a prime polom right Fe yeah a prime polinomial and I look at FX modul that is FX modul of Fe is the set of all remainder pols are are X right with the mod now I look at the map from all pols to a pols modul this this map is the the quotient map or the

the mod fee map so I take a polinomial f to a f modu okay in particular I can tell take the polom X this this is a polinomial in f of x right I can take Pi Pi of X assuming that V is is of degree bigger than one Pi of X is simply X is the the Remain the polinomial X right yes h um but now I can consider

Fe as a polom over K right because K contains K contains a copy of of f namely the constant remainder polinomial the ones only with a zero right right so if all the coefficients of are are in F then they are in particular they are in K and I can consider a fe is a polinomial over K pols Over K are are kind of um a weird creature because they would be like a

r let's say RK x to the K but the RK is itself a polom right but what happens when I substitute in this F capital x I substitute Pi of X which I I I for convenience I called it Alpha okay I substitute Pi of X I get a fe lowercase x but I mean of course I I take a

polom over K I substitute an element in k then I have to get some element some new element in K right this new element in K is fix modul which is zero so so the the conclusion is that F of alpha or F of Pi of X is zero in in K of X okay it's it's kind of a metat trick but

ER what I get is suddenly that I have a root of a of v in in this uh in this fi that's the the the heart of the matter really okay makes sense okay and now I I simply I take a polom in this proposition I take a polom i decompose it into a bunch of prime pols yeah the rest was clear it was just this this prob that was like a straightforward thing from okay but the was pretty clear for me thank you okay H other question before we take a

break all right so um let's meet here in 10 minutes sure so we now come to the the second part of a um the the splitting Fields ER properties namely ER we want to show uniqueness but um it will be easier to this this is a phenomena that happens a lot in mathematics that ER you you want to prove something but

it turns out that if you formulate the statement more generally I mean if you in fact if you try to prove something something stronger then the proof becomes easier because you you kind of abstract a inconvenient details um so this is the the following theorem uh suppose Raw is a field

isomorphism and let f of x be say a n x to the N plus A1 X plus a z and F Prime I denote simply you apply row on all the coefficients of f um that is row of a n x to the n

row of A1 X row of a z with of course a a f a polinomial over the first field and F prime a polom over the second field let e be a splitting field um of

f over a big F and E prime a splitting field of f Prime over big F Prime then there exist uh field

field isomorphism let call it Sigma from E to e Prime um and not only that it is an isomorphism IT restricts to when I when I restrict a sigma to to f i get the original isomorphism row right here inside of e we must have a f and inside of

a inside of e Prime we must have a f Prime and we started with a map row so the the CL of the the theorem is that you get a map a field isomorphism Sigma that extends the original isomorphism RW okay it's a that's the the theorem we want to prove so do way to to go about it is H

is by induction um by induction on on the degree of f which is n yeah um the note this is this is an abuse of notation but it's it's very natural to do raw from f x to F Prime X

um if you take a polom ER whatever some a i x to the I then raw simply applies the original Row the original row between F and F prime it applies it on the coefficients this is sum I row of a i x to the I in particular when I apply row on

F I get F Prime what we denoted is f Prime now let's make two observations one is that a a row is a ring isomorphism I mean you clearly you get a um and raw respects a addition and

multiplication because you simply apply the original original row on the coefficients and H the original row is bjective so so this extended row is must also be bjective mean is this clear why why this extended row is a ring isomorphism okay so and then the second second observation is a bit more

involved I say that if if PX is a prime factor of f of x then when I apply a row on P I get yeah a prime factor of a a row applied on on F

which as as said this is fime X now why is this second observation true um raw is a ring homomorphism so it commutes with a multiplication of pols so um if I write FX I mean PX is a prime factor of f so I can write um FX equals PX * QX for some some polinomial QX and then I apply

Raw on f of x it's the same as applying Raw on PX time QX and since row is a is a ring homomorphism I get that this is row PX times row QX um and this is of course f fime of x so what I get is that a row PX is indeed a

factor of um fime of X now why is this a prime factor um if it's not a prime factor so let's say if a a row of a p equals to A X time b x or maybe we call it a prime X time B Prime X um with a degree of a prime and degree of of B

prime a bigger than bigger than one or sorry bigger ER or equal to one and of course a prime and B Prime are pols over F Prime then remember raow is is a ring isomorphism so it means that I have a map in the other direction this map in the other direction I I know how it looks I mean assuming that the

original row was like from from F to F Prime I had the original row being an isomorphism then it means that I have a map in the other direction again a field a field isomorphism which we typically denote as row minus one such that the composition in each uh in each order is the identity map and I if I have row minus one on the level of fields I have row minus one on

the level of pols right like we would have row minus one in in this direction between polinomial where you simply you simply apply row minus one on all the coefficients of a of a given polinomial in in frime of X and and we are guaranteed that row minus one is also a ring homomorphy so now if

H if raw of PX was not a prime polom it would be a product of two I mean a non-trivial product of of pols a prime and B Prime in in F Prime of X and then I apply row minus one or on row of e of PX I get well on one hand since row is the inverse I get P of X right and on the other hand since row of

PX is a product of two things I get that this is row -1 of a prime X time r - one of B Prime X and row minus one is simply acting on the coefficient so it preserves the degree whatever the degree of a prime and B Prime was the degree of of row minus one of of them is the same and a degree of r - one a

prime and the degree of Rus one B Prime is a a is bigger or equal to one um which cannot happen uh since PX is Prime in a okay is this uh is this point

clear we we claim this is the claim that if you give me a prime factor of a of F and you apply row you apply the the map between a a polinomial rings that is induced by the original isomorphism raw then if you apply it on PX you get immediately a prime factor of a um of raow of of of the original polinomial F of

okay so this this is the the basic observation H now and suppose first that F splits to linear factors um sorry F splits to linear factors over a big

F then we we this is the ER the best case we don't have to do anything um then I take e to BF this is a split it's already a splitting field um or if and e Prime equals F Prime is a splitting

field of f Prime and then I need to Define that's kind of the the ER the claim of the proposition I need to define a map Sigma that restricts to the original map row but here I in fact I have equality uh so set

Sigma equals and and of course Sigma is an isomorphism of of splitting Fields okay this is if if the original polinomial F splits over over the the original field the capital F and otherwise let a px be a prime factor of um f of

x with um of course with degree bigger than one I mean if it's a it's a prime factor of degree one then it means that I mean f splits at least into PX times another thing so if we assume that F does not split to linear factors there is at least a a one Pol Prime polom of degree bigger than one

um I let fub1 be pols over F modu PX I think our notation is this modul P um since row of p is prime

in F Prime x h we also have a field which I will denote as a F1 Prime this is you take F Prime and you mod out by row of P which we know is a prime polinomial

note that we have fi isomorphism I'll denote it as as Sigma or maybe Row one it goes from F1 to F1 Prime and it is induced by by row I mean

h ER you take you take any remainder polinomial in in F and you apply row on the coefficients and you get a remainder polom modul R of P okay ER if you want we can h I mean we can check the details of H of why this is a field isomorphism I mean we need to show

that that it respects addition and multiplication ER for R and S to remainders remainder pols in in F1 again F1 is f of x modu p then I take row of r + S it's simply a a row of R plus row of s this is by by definition of row I mean

and and the same would go for a I mean this is kind of a topological map I mean I started with a an isomorphism of fields and now I extended to pols and I extended to polom modu another polinomial and so on it it it always stays in isomorphism I mean if you you want an inverse to this map you get you can construct it by a by taking the inverse of of of raw and apply it on remainder pols in fime of weeks and so

on now ER since a px is a root in F1 PX is some x minus a times P1 of X for a sum a in in F1 ER

similarly well now I apply I apply row on this whole polinomial so I get um raw of P of x it has to be equal to a row of xus minus a times row of P1 of X and row of some polinomial is simply acting on the coefficients so this is x

minus of a times R of P1 of x so a a is a root of a a w of PX um now f of

x was a um I mean was a px times times something and PX decomposes into xus a times times another thing so all in all I I get that f of x is x - x a Time some fub1 of X and fr Prime of X is x minus row of a times FS1 Prime of X well F1

Prime is simply applying Raw on F1 with of course the degree of F1 and the degree of F1 Prime are strictly smaller than the degree of F I mean let's say degree of F1 is strictly smaller than the degree of F degree

of of F1 Prime is strictly smaller than the degree of f Prime in fact these degrees are equal because F and F Prime are um I mean f Prime is obtained from F by just a doing something to the coefficients so degrees are in fact equal but since the the degrees of fub1 and F1 Prime are strictly smaller than the degrees of F and F Prime we can use the induction hypothesis so by induction by induction

hypothesis there is oh maybe do it like this um let E1 and E1 Prime B the h b splitting fields

of F and F Prime respectively then by the induction hypothesis well of course over o over F1 and F1 Prime so by induction I have a a a an isomorphism of of

fields between E1 Prime sorry between between E1 and E1 Prime Sigma that extends the original isomorphism r okay so and now I claim that in

fact E1 is a splitting field not only for F1 one but already for for f over over the original field big F why is that I mean it it contains the ER it I mean clearly f h splits into splits over over E1

because E1 is a splitting field for for F1 um o over um and E1 is a splitting field for this polinomial F1 over over the field big F1 but over the field big F1 F already splits into x - A Time h times this lowercase F1 so so all in all I get that over E1 F splits into linear factors that's

that should be clear so that's clearly um f splits to linear factors over uh E1 and why is E1 the minimal field in which F splits into linear factors I I mean I get I have F and then I have F1 which is FX modul um

modu PX um and this is the minimal we we this this is the discussion we had in the the the previous hour this is the minimal field containing the root a and then what I did I I extended it into a splitting field E1 um which is by definition the minimal field containing all the rest of the roots I mean all the roots of F1 so all in all so

E1 is a the minimal field or a minim the minimal field in which F splits okay so this is for E1 and the same goes for E prime one similarly E1 Prime is a splitting field

field um of of f Prime over big F Prime okay and er we completed the proof right er er so so we get some Sigma from E1 to E1 Prime they both contain F and they are splitting Fields with respect to F and F

Prime and the sigma extends the original isomorphism okay so this this was probably the the most involved proof ER or one of the the most involved proofs we we saw um any questions let me draw a

coroller a a splitting field uh for f over f is unique unique up to isomorphism how do you you show that well you simply take in the previous proposition in the previous theorem you take row to be the identity map between F and itself

of course the identity map is is an isomorphism and um row of f is simply F so so we get that any two splitting Fields e and e Prime um of f any two splitting Fields must be isomorphic via VIA an

isomorphism that restricts to the the identity map okay so so we get that a splitting field exists and and unique up to isomorphism which is basically the best one one could hope for of course uniqueness kind of on the nose is a is too strong of a property to expect okay uh I'd be happy to hear your

feedback ER what may not be clear ER Carlos what do you think uh from my cite all good uh at least for now maybe when I do a review later I I'll find some things but but for now everything okay it's yeah I mean the the argument is very simple it just H it was kind of a the formulation of the the theorem was convoluted a bit but yeah agree I I think it requires

some uh offline time to to to Revis it prior to yeah s said the same yeah okay so you it's good that you review you review the proof again in the notes there are slightly more details ER and you're always welcome to ask questions in in the next sessions so sure but but this this is I think the the the idea the idea of the proof should be quite simple and so what we get is

um is that splitting fields exist and unique and unique and we are now very very close to to classify all finite Fields using this this notion of splitting Fields but this will have to wait for the next time okay so have a good rest of your day thanks you thank you and take care