PSE Lectures - Ep 15 - Recap and Q&A of Fields Lectures

Transcript

ER okay so I would like uh to start today with a kind of an open open section answering ER questions you you may have on previous material um the I I think the way to to uh to go about er er this ER I mean in general the mathematics you should try to have kind of an airtight situation with the arguments namely that that you you

understand every er every step ER of reasoning ER I mean you want it to fly so it needs to be either airtight or close I mean I mean you can have a little bit of leaked leaked air but too much you cannot fly ER so I encourage you to go over the notes it's a they are supposed to be a carefully written so you you should be able to follow every step of course I could make some mistakes or there could be some gaps but if you see any ER you you should ask me and

then I'm happy to answer um beside you can you can also ask me to revise some some section so and that's also an option but er um are do you have any any questions about previous material I I do have one um which is related to the to the definition of fields and field homomorphisms I remember that for field homomorphisms we discussed the the sum and multiplication tables of them so

kind equivalences but can you give again the formal definition of what a field morphism is in an example like just really briefly like no yeah that's that's it's actually very easy to uh to revise I mean if we have a um let's say a um f and K two fields

what does it mean to have a field it means that I have a multiplication in F I would call it dot subf and and addition that I would call plus subf and similarly you have a multiplication in K and addition in okay and you also have the neutral element with respect to addition this is let's say 0 f and0 k and the neutral element with respect to multiplication so this is

one um and okay there are different axioms H connect I mean the the multiplication and addition need to be associative and there is also an axom connecting a a multiplication with addition this is called distributivity it means that if you do a Time B plus C it's the same as AB time a sorry plus AC um and now what is a field homomorphism which you can you can I

mean I I typically call these things um a field map I mean this this kind of a I can envelope V oping terminology that you can call I mean a group homomorphism is a group map field homomorphism is a field map ring homomorphism is a ring map and so on we will have a more structures in in the future and for each of these structures we want to have a a notion of a a a map that is compatible with the structure I

mean what is all you you can think of all the Notions we talk about as in the in the the scheme of a a certain certain data satisfying certain conditions I mean what is f f is a set together with the data ER of binary operation and I mean two binary operations and two ER special elements one and zero H satisfying certain axioms so it's like

a data plus conditions and what you want in in in the notion of a a map between these these structures I mean between two of instances of these structures is a a some that is compatible with with the data I mean so um so a field homomorphism is a function let's say F from Big F to Big

K H such that I need to again to be compatible with the structure it means that first I map Z to zero H then I need to map one to one and then I need to be compatible with the two binary operations so for any a a B in F if I do F of a +

b it's the same as F of a plus F of B and if I do F of a Time B it's the same as doing F of a times F of B and notice I I am being a little bit H ER sloppy notation but this this you should get used to I mean the plus here is the the Plus in F and the plus here is the Plus in K right of course it has to be because

um the input of of of the function f is a is the field I mean elements in the field F and the output is a elements in K and the same goes for um for the the multiplication yes so after seeing this then of course when there's a field fism it means that F and K although might be the same they they are they aren necessarily the same field so they can be different fields so is can you put any simple example of different fields that are homomorphic to each

other homomorphic you should distinguish homomorphism and isomorphism okay that's that's a very important distinction and again this it it h pops out um in basically all the structures we will consider I mean you have a group map for example um I mean you take a um integers to integers modu n you do the mod n this is a group map right because H if you do a mode

n and you you sum it up with a with B mode n it's the same as as doing a A plus b in in Zed and then do mode n sure so this is a homomorphism a group homomorphism or a group map I mean it's compatible with the only the only operation h i mean the only operation that I have in a group which is just a in this case the addition or addition mod n in the these groups are not the

same you have a homomorphism but they are not the same in order for them to be the same in the sense that we consider you need to have an isomorphism namely a a a a group map or a group homomorphism that is both injective and subjective okay okay and this this you cannot possibly have a a injective and subjective map between between these two H these two groups because one is finite and and one is infinite sure sure sure sure sure sure

so so this is really so you can think of it this way in groups and again this this will h kind of be be the same in fields you take two groups g& AG you have their a a let's say there are a billion groups um so you have the addition table H I didn't say I mean you don't have to assume that they are finite but let's just for the example let's assume they are finite so you have some um some elements I don't know G1

G2 G3 and here you have the the addition I mean whatever it is G1 plus G2 and so on and if you give me a a group morphism let's say F then what I get I mean of course I can simply write F of G1 F of G2 F of G3 right and also

here okay now this is not a a full list of the elements in h because I did not assume that that the homeomorphism f is subjective right so possibly there could be other elements here like I don't know H1 H2 and so on but for the the elements that are obtained as an image of the the function f what I say because f is a homomorphism is I mean here there should be um f of G1 plus F of G1 but F of G1 plus F of

G1 by the homomorphism property it's F of G1 plus G1 right so I I could actually write here F of G1 plus G1 if I if I know that the the function f is a a group homomorphism I see I see I see I see okay so so a group homomorphism takes the addition table of G and copies it to some Sub sub addition table of of H now if if F was a a an isomorphism it means that it's injective

and subjective then I would really have I mean this list F of G1 F of G2 F of G3 would be an exhaustive list of the the elements in age right because because the map f is is subjective and and there will be no repetition because the map f is is injective so you could really if if f is an isomorphism you could really think of the the addition table in age as simply a copy of the addition table of G with just different names of the

elements I see see okay so this is why we say that isomorphic structures I mean isomorphic groups are the same right I mean they're not the same as sets they could be different sets but if you for instance if you input it into a computer the computer doesn't care you you know you call it X or you call it y doesn't matter all all all that matters is the the the addition table I mean right now the same goes the same goes for Fields I mean if now I have two Fields again they do not need to be

finite but in our case we we mostly deal with finite Fields so you have the addition table of F and you have the multiplication table of F and you have the addition table of G oh sorry of K and you have the multiplication table of K and and a homomorphism is something that does something similar to to what we had with groups but for both the the addition table of F and K and the multiplication tables of F and

K right yeah okay and if you have an isomorphism between F and K if this is an isomorph ISM then I could I could really identify the addition table of f with the addition table of K and the multiplication table of f with the multiplication table of of K up to a change of names and the change of names is given by the the map F right by by this function [Music] if okay so so again we will

say that two isomorphic fields are this I mean are the same from the point of view of a computer so this clearly explained my and cleared my D then that this went to a place where I think this is related to the other question that I have so I I'll try to introduce it which is uh from the standpoint of splitting Fields um I am so I I went through the Dos again and I see the properties uh s that we need such that we can have splitting Fields but is there an intuition behind having an

anamorphism or isomorphic uh fields to the properties required uh to have a splitting fields in the sense of do we require to explain um splitting Fields any intuition of theomorphism SL isomorphisms uh to to to understanding I'm trying to just grasp a view of a splitting fields that is just not an enumeration of properties but rather some kind of geometrical um point of view put on some sense I don't know to say yeah so let me start with the end I mean um there's no geometry in splitting Fields I mean at least not not for

now splitting field is an algebraic notion with no apparent geometric intuition because and especially in our case we have finite Fields so it's very hard to I mean we will do some geometry of finite Fields later on but this it needs to be understood in a different sense and splitting field is is an algebraic notion ER now isomorphism and homomorphism I mean this is fundamental notion in any in any structure we will consider so if you talk about Fields you will always

consider I mean there would almost always be some kind of isomorphism of fields involved in the either in the statement of the theorem or in the proof it's it's almost impossible to avoid it and the same goes for any of the other structures we consider I mean can you imagine that we would do the section of groups without the notion of group isomorphism it would be insane right and so so of course splitting Fields they also involve the notion of of field isomorphism and the the Deep reason is

that actually splitting field is is a turns up to be a a a kind of field that is unique up to isomorphism of fields so what happens I mean you start with the field ER f it doesn't have to be a finite field right it's like a most of the the stuff we do on splitting fields are are true for any fields and you take a polom over this

field which is prime you remember the the definition of a prime polinomial yeah yeah primitive Pol it means it it cannot be decomposed into a product of into a nontrivial product of two pols I mean of course could be it could be a a product of a constant and another polinomial or something like that but it's it cannot be a a product of two polinomial of degree bigger than zero okay so and when we had this H um I mean okay

so in particular um Fe has no roots in F right because if it I mean if it had a root then you you could write f x equals x minus the root times some f Prime X so then F would not be Prime polinomial yeah okay so so a prime polinomial you can be sure it has no roots but now I would like to to construct a a bigger

Universe in which Fe splits completely to linear factors right yes so if I could I I mean I I call this universe a splitting field I mean I would like to to obtain a field that contains a um so I want some field k um that contains if

and of course if it contains F then I can consider Fe um I can consider Fe over the bigger field I will still write I will still write Fe right it's just a it's basically the same polinomial but now over K and um and over K we have a complete split of of the polinomial fee so I write it like this Alpha times um product let's say x minus

Alpha I I equals a one to n n is the the degree of Fe where Alpha and Alpha 1 to alpha n are now elements in the field K yeah you write Alpha in front just because in case it's not a monomial in case like the leading factor is not one yeah exactly okay thanks it's called the monic sorry yeah monic yeah yeah okay so I I I would like to I mean if I could I would like to obtain a perfect

Universe ER at least a perfect Universe with respect to H to Fe I mean it has all roots and so one thing I mean eventually I would like to say ER this perfect Universe I mean I would like to be minimal as well because I don't want to add too many stuff that I don't need in this sense I mean [Music]

um this this is the third property um if K Prime um is let's say okay let's call it l if L is a field that a a on one hand sits between

F and K and on the other hand um Fe splits over over L then L must be equal to K and you cannot do you you cannot do anything er anything I mean smaller than the than K yes there was a question it was a

question to uh yes I added a question on the chat so yeah I I was wondering so if this extended field does not contain all the possible roots from the polinomial doesn't mean that it's not considered a splitting field or it could still be a splitting field yeah yeah a splitting field must contain all the roots because I mean you see this H this formula in in purple it means I mean it means that alpha 1 up to alpha n

are all the possible roots of Fe okay okay right because first of all I mean first of all you there are there are certainly roots of fee because if you substitute instead of X you substitute Alpha I then you get zero and second they must be all the roots of Fe because um this is I mean you yeah how how do how do I say it um if you had any if you have any other element a let's say beta and beta is

different if beta is different than than any of these guys in fact if it's only different than alpha 1 up to alpha n you plug in beta into into Fe you get something that is non zero right because I mean you get beta minus alpha 1 beta minus Alpha 2 and so on each of these Expressions is non zero in a field you multiply nonzero elements you get a nonzero element so you cannot have a any other root for Fe

yeah are there any pols that for some reason you can you can't find the splitting field for no no I we proved that a splitting field always exist okay exist and and unique up to isomorphism um so okay so let's do an example we we I just I think I I had a like a good example in the notes okay

example this you can ER you can find it what is the number yeah it's like five 57 I think thank you so I take I I start with a the ground field I start with Q okay the rational numbers and a a fe f of x is X to the 4 let me call it f of x just just simpler f of x is X x to 4 minus

2 now X to 4 minus 2 is prime over is H Prime over over Q you can check it I mean there are different ways to check it but but for example you cannot have a a I mean it's very easy to see that you cannot have any root of of f over the rationals because a root of f would be a um fourth root of of two something like that right and fourth root of two is not a rational number

just we we we showed that a square root of two is not a rational number and you can in in like in similar ways you can show that also fourth root of two is is not rational okay and of course also you could do I mean minus 4 root of two this would would also be and in fact um if I think of of Q as contained in in the complex numbers then I could figure out all the all the possible roots of of F this would be plus plus minus 4th root of Two And A plus

minus itimes 4ot of two okay and none of them belong to Q so over over Q I I I cannot find any root so now um I I consider K1 let's say this is a a q X I mean pols over Q modul the the modul the the polinomial f

I mean if you recall we had this construction that you can consider all remainder pols this I mean this set QX module F we we also denoted it as a QX subf this is kind of an analogous notation to a ZN right and this is analogous notation to Z modu NZ okay but but ER in terms of a elements this is all

pols are X which are a [Music] um in our case they must be of degree three or less so it's like a um a AK x to the k h plus A1 X plus a Z where the coefficients a a z up to a k are in the original field I mean in Q and K is strictly smaller than the

degree of the the polom we we take modulo with so in this case k is a smaller smaller than four okay this is a set right and on this set I have a binary operations let me let me write it slightly better it's all the RX AK

x to the K A1 x a z such that um AI is in Q and K is strictly smaller than the degree of f which is four okay and on this on this set K1 we have a um a modu f arithmetic I mean I could do if I have R1 and R2 in

K1 then I could do R1 + R2 modu F and I could also do R1 times R2 modu F in fact if you think of it um when I do r1+ R2 I don't need to do module F because R1 and R2 are of degree smaller than the degree of F and when you you add two pols you you don't change the I mean the the the the

resulting degree is the maximum of the degrees so it would still be a degree that that is smaller than the degree of F and therefore the the by the long division of pols you don't have to do any any any operation to get the modulus right it's like a like in integers if I do um three modu of five it's it's still three but for multiplication you do need to do a module F and these operations they this what we we showed that because f is

Prime and this is kind of analogous to I mean if you take integers modu modu a prime you suddenly get a field I mean you get a multiplicative inverse for every every element and so K1 is a field now K1 of course contains um contains Q because you you can consider the constant pols I mean the the ones with only a a free coefficient a

z ER okay so so now what we are guaranteed is that um this is kind of an abstract construction right this QX modu F but we are guaranteed that that there will be a um K1 I mean I can consider of course um f of x let's call it like this F of of big

variable X in K1 with with big X right I I I I introduced this capital x because I want to distinguish it from this er er this x that uh that we have I mean so in K K1 X an element in K1 X let's say um I don't know G it means that you write it as a

[Music] um RK small X time x to the k h plus h RK minus 1 x times x to the K minus one and so on okay and the way to add two of these guys I mean if you have a g Prime um say R RK Prime

yeah sorry this step is is a bitcar to me because now you're defining a poal with two variables small X and big X so for me this is B polinomial and like you're only putting one variable in J the big X and so I'm like kind of confused how do you evaluate the small RX coefficients now I mean for me the r the r coefficients are simply elements in a field yes okay I mean so this is what I wanted to say here if you now want to to multiply these two together so in particular you will have you should have RK x times RK

Prime x times x capital x h to the two H 2K yes okay you're just coefficients and then you don't evaluate them just but in fact what I need to do here is the multiplication here it needs to be multiplication mode F because I the way I multiply coefficients of pols is I multiply them in the field they they came from right so this is this needs to

be multiplication mode F so you multiply RK and RK Prime and then you do long division by by the polinomial f is this clear of course there are many other terms here okay so this is if you want this is a a a a it's like a a polom whose coefficients are pols mod f so yeah oh modu F just means uh divided by F like long long division by F

polom yes I mean module F if if I if I want to do R1 * R2 module f i I write it I mean I am guaranteed that there is some a polinomial q and polom R such that I have this equality right with with r i mean the degree of R is strictly smaller than the degree of f it's like in integers right yeah yeah makes sense okay and and then I I I throw away this this part I don't care I

just say R1 R2 is equal to R R mod f okay so and and this is what I do in K1 K1 remember it was defined to be all pols of degree smaller than the degree of f with addition and multiplication being a module F and we showed that this is a field and since is a field now I can consider pols over this field how these these creatures behave I mean you if you

have two of them say G and G Prime you you do multiplication of polinomial but when you need to multiply the coefficients you which I mean which are these RK RK Prime and so on you do the modu F okay is this this clear this this is really like a touching a a like a central point in in this splitting Fields so now okay I can consider f with capital X in K1 of X how can I do it because F had

coefficient in q in in the the the ground field and of course the ground field F I mean q is simply a subset of this these guys right I mean so in particular if you know F had a f was a I mean now I write f as X to 4us 2 2 I regard it as a constant polom let's call it r0 r0x I call it

two and and now f is a a a polinomial over the field K1 with a new variable X I mean capital x but what I am guaranteed in K1 is that um f has a root why yeah so I mean this this means that independently of what primitive polinomial over what field you have you can always find this F such that even if it's the most abrac thing in the world you are guant you're guaranteed to find

all of the roots needed to for it to become aing F yeah this what this would be the the eventual conclusion of our discussion but let's let's kind of Build It Up MH sure sure sure yeah yeah um so I I I so I first claim that in this setup um f has a root in K1 now how do you see it I mean remember what is k1 K1 is the remainder

pols um with module F arithmetic in particular in K1 I have the remainder polom I mean X I mean small X let's call it maybe R1 we can call it R1 R1 X being the polinomial X and this is a a legitimate element in K1 because the degree of R1 is is one which is a

strictly smaller than the degree of f now I plug in a this this r1x I plug it in into F capital x so the notation typically for substituting an element in a polinomial is is like this right and when you have a a polom and I mean a polom over a field and you substitute an element in this field into this polinomial you get an

element in the field so this I know with without doing any computation I know that the result must be an element in K1 right this is kind of the the basic mechanics of a substitution um okay but now I I really I really start to to compute I mean it's not much so it means that I get f with low a lowercase x right this is what I I mean this is f of x

substitute X so all I do is I replace the big the big variable x with the with the lowercase x yes I'm sorry I'm it may sound obvious but I'm bance why are you not replacing big X by R1 of X not just X small X well I mean R1 of X is is defined I I just sorry sorry sorry yeah okay it was obious thank you okay so so I replace the the I mean it's not like just a technical issue it's like a

it's kind of a metat trick since I can substitute a um any polom with with a a variable small X I mean any polinomial of degree smaller than the degree of f i I substitute this degree One polinomial X when I substitute X into F capital x I get of course F lower case X which needs to be an element in the field K1 but in the field K1 I have modu F arithmetic so f with lower variable I

mean lower case X is zero in K1 because in K1 I do modul F so this is all to say that F of capital x has a root in K1 one root I mean I don't know anything I just I just have one root is is this clear Carlos it's it's clear like I I agree with you 100% it just feels uh like cheating kind of I mean not cheating but I don't know how to say it like fair enough I agree with what you said but it's like you you are making up this construction such that we can go this way

but it's indeed an abstract construction but but nevertheless it's a you can you can follow every step and you see that it's a it's true yeah yeah yeah yeah fair enough fair enough I'm like Nicholas like I understand now the uh but yeah sometimes these arguments are like I don't know how to say they don't feel real though I don't know no it's it's true that it's it's kind of a metat trick I mean um but that's the only way we can reason about a um because in Q itself okay I mean the thing in we are

like studying a concrete example so this H this could be made a bit more Concrete in our concrete example we know what are the the future roots of this h of this polinomial because we we know how how the the behavior of these things happen in in in the complex numbers but in general you give me a field you give me a polom I don't know I mean if it doesn't have any root the only thing I could and I don't I don't

know how to include this field into some bigger field that I already know I mean then the the only thing I could do is is ADD kind of a new symbol that would would be a root of this polom a being a root of this polinomial it means that this symbol satisfy a certain algebraic relation and that's that's the only thing I can do abstractly so so and this this Construction in fact does exactly that it it adds up a new

symbol that a that a is I mean is a root of the the polinomial F that means it satisfy the algebraic relation F of of this element F of alpha let's call it Alpha this is typically Den noted just just for a like this ambiguation we denoted in kind of a different letter so so I what I did

here okay let me I I could I could even explain it slightly better with this formal construction of adding Alpha I mean this is this is related to what we what we also discussed um that I could take a q let's say and consider the Min I mean I could take you I I take F in our case f f is is X to 4 minus 2

and say Alpha is a root root of f um in some some bigger field I mean if I knew that there is a bigger field let's call it K if I knew that there is a bigger field containing q and containing a root of a root of f then I could consider the minimal

field containing you and containing Alpha how do I do it um I think that would have to wait after the break but but the the it turns out that to do Q of alpha I mean to to H consider the minimal field containing q and containing Alpha is the same as doing H K1 okay it's the same as doing this is going to be isomorphic to Simply do

Q X module f okay okay so what I did here is I I simply added a a some some abstract root of of the polinomial in a minimal way yeah but let's take a break right this is good to uh sure so 1113 we're here yeah okay cheers okay um can we

continue sure uh as for the topic when to finish there's also one question from on the chat we can yeah go ahead with yeah um uh you told that uh we can assume that for the computer two isomorphic objects are the same can you explain it like why it is the same I I understood from your explanation that they are um kind of the same size the same structure but I don't don't understand why they are this safe okay so let Let's uh let's consider

this destroying a of say a billion groups with a an isomorphism F right let's suppose G and H are a billion groups and and F is an isomorphism um I I write the the addition table of of G and then I apply F on the on the addition table of G it means I I apply F on the on all the elements in G and I apply F on um all the all the cells right all the

blue cells but a I know two things I mean first of all I know that f is an is a bjective map that means that when I apply F on the on the elements of G I get all the elements in h right yes it's it's a bjective I I not only that I get all the elements in ag I I get it with no repetition I mean I had three elements in G I apply a bjective map I get three

distinct element in age which are exhaustive I mean they they comprise a full list of the element in AG in what order I don't know but it's a full list now I apply F on a on the cells I mean on G1 + G1 for example this by definition is f of G1 + G1 but f is a homomorphism so this is in fact equal to F of G1 plus F of G1 whatever it is in age I I have no clue but it's it must be f f of G1 plus

F of G1 so what I get when I apply G sorry I apply f on the addition table of G I get the full addition table of H I see I see I see so the order might differ but uh the elements will be the same exactly so I I get okay yeah clear clear I get some representation of the the addition table of AG again I don't I don't know in what order I mean this it could be that g G3 would would have been the the unit element in in G so let's say zero

g and then I would know that since f is a homomorphism I would know that F of G3 this guy is is in fact must be the zero element in in age right but I prior I don't care all I all I all I know is that I I get a a a representation of the multi table in in age and so from a computer point of view it doesn't matter I mean whether you know whether you write the the the zero element in the third place

or in the fourth place doesn't doesn't make a difference Okay C C thank you so I mean again from a computer point of view all you care is you have a bunch of symbols and you have an addition table or a a I mean you have a binary operation between these symbols and you have a a table of how to comp I mean how to convert binary operation of two symbols into a into a new symbol on the list and the same the same goes for an isomorphism of fields only that in an isomorphism of fields you will identify

both the addition table and the multiplication table so again from a computer point of view two isomorphic Fields they are the same I mean you just you call the symbols in a different in a different name okay is this clear yes this is why the notion of isomorphism is so important to us because we cannot guarantee I mean we we make construction especially with this splitting field is a very good example we make a

construction that introduces a new set of symbols um this in this case this remainder pols and so on and possibly there could be another construction that does the I mean that has the same properties and what we want to show is that the these two construction will result in isomorphic fields so this is kind of the I mean one one I mean one use of of this notion of

isomorphism a but before the break we spoke I mean I I told you about a a uh an alternative way to interpret um KX modu F I mean we saw that KX modu f contains a contains a root of f which just for clarity I I denoted as Alpha but Alpha is really is really this this remainder polom X okay thought of is in in

this in this in this field K1 now just for maybe like slightly simpler example I mean suppose you you um you had a a GX being e x² + 1 as a polom over Q as a polinomial over Q it's it's Prime because you don't have any any any

rational number satisfying x² + 1 equal Z you don't have any root of these pols and it's a polinomial of degree two so it must be prime uh and now suppose I I didn't know I mean of course a post priori since we already know about the complex numbers we know what should be the roots I mean the root should be plus and minus I but suppose I didn't know anything about the complex numbers right at some point in the history it must have been the

case h so what I do I I introduced a new symbol Alpha and I I would like to consider the minimal field ER containing q and containing Alpha such that Alpha satisfy a relation Alpha square + 1 equal Z so what do I do I consider and this let me just find it in

the notes you can then follow better yeah okay so this is a slightly before um proposition 5 52 I consider Alpha sorry Q Alpha is the set of quotients

um let's call it um okay let's just for compatibility with the notes let's denote this polinomial as Fe okay and and now I consider Q Alpha so I I want a a um and I want to ad jooin a new element Alpha that will be a root of of Fe a root of simply means it satisfied this algebraic relation Alp square + 1 equal Z and the way to do it I consider

all the quotients F Alpha modul G Alpha such that f and g are pols in the field ER Q I mean over the field the ground field Q and G is not the zero polinomial now it what I I mean I claim that this is a field so of

course you can you can add two ER two two things together and if you you have a f of alpha modu G of Alpha and you you add you want to add to it f Prime of alpha G Prime of alpha then you do what you you know from elementary school I mean you do f f Alpha G Prime Alpha plus F Prime

Alpha G of alpha divided by um G Alpha G Prime Alpha and this is again a legitimate element in in Q of alpha because it's a quotient of two um I mean of two pols when you substitute Alpha and the um I mean the inverse of of f of alpha mod G of alpha the multiplicative inverse you simply invert the order I mean G of alpha modul F of alpha of course when

you multiply these two together you get one yeah so you claim that this is a field so q q of alpha but so is it guaranteed because of how it's is it constructed or what's the way to prove that this is inde a field okay so so to show that this is a field I need to h define addition and multiplication operation exactly I I need to have a a zero and one I mean neutral elements with respect to addition multiplication and I need to have

a well associativity and and distributivity and and also inverses yeah okay so addition and multiplication I Define here I mean this this is addition H multiplication you can expect the same I mean if if you want to multiply F Alpha modul G Alpha with fime Alpha mod G Prime Alpha then you multiply F of alpha times F Prime of alpha if you want you can regard it as you you take the polom

f you multiply it as a polom with f Prime whatever it is and you substitute Alpha and the same you do in the in the bottom right I mean f and g remember f and g are pols okay so addition and multiplication I Define of course it's going to be associative because we know that addition and multiplication of polinomial are associative and here I I just do it in the I mean in the

nominator and the denominator yeah is there a question yeah my question is about the definition of q of alpha as well um you say it's the uh set Q adjoin with the alpha elements but when you write concrete definition like this I don't see there's a new element well I mean you take the the polom you take the polom you take the polinomial FX equals X and then in particular you would have F of

alpha which is simply Al I mean way maybe a a better way to introduce it would be how would you construct a minimal field containing q and containing Alpha that satisfy this a a purple relation well so you introduce a symbol like Alpa and and so Alpha must be a a a an element in this whatever this field is Alpha must be an element in it but since it is a field I must have alpha squar alpha Cube

and so on they also have to be in this field right because it's a field it's closed under under multiplication of two mean if Alpha is in this field then Alpha time Alpha must be in this field yeah okay but also um if a a a is a rational number then a Time Alpha also must be in Q of alpha yeah and similarly a times a alpha squar also

so all in all for every polom F over the rational numbers if I plug in Alpha F of alpha must be an element in Q of alpha oh I see okay okay now this is supposed to be a field so Alpha minus one there must be some kind of element that is the inverse of alpha so Alpha minus one must belong to Q of

Alpha and similarly Alpha minus 2 and and so on they also must belong to Q of alpha y okay and of course you take IR rational number a then a * Alpha minus one must be also in Q of Alpha and so for any polom G over Q I must

have 1/ G of alpha belonging to Q Val yeah makes a l sense thanks and and of course this means that of course g g needs to be non zero g is not zero um but since F of Alpha and one over G of alpha are in Q of alpha I must must have that F of alpha modu G of alpha is an element in Q of

alpha so so far I I mean assuming that Q of alpha exist I I kind of detected a a necessary necessary elements that must be in Q of Alpha and I claim that this collection of elements I mean the namely all the elements of the form F Alpha modu G Alpha are a suff sufficient list of elements so that Q of alpha will be a field okay so I mean I kind of I I I I did I mean I assume that Q of alpha

exist or I said if Q of alpha exist it must contain all these elements of I mean all elements of this form and now I show that the collection of all elements of this form actually form constitute a field okay so how do I show it um well I need to show you how to define addition and multiplication this addition and multiplication are defined like this h and now I need to show you well associativity of addition and associativity of multiplication they

follow from associativity and I mean of addition and multiplication of polom over Q right and [Music] um the inverse if I want the additive inverse of an element of this form well I I have it I mean this is a this is simply you do let's say you do minus on the on the numerator okay and and of course this by

by the the way we defined addition the addition operation if you do F of alpha module G of Alpha and you you add to it a minus F of alpha modul G of alpha you get the zero element the zero element by the way you can represent there are different ways to represent it but H you can take the the zero polom I mean take the zero polinomial mod any mod any guy that is non zero and if you want an inverse like a

multiplicative inverse that's also I mean it's also very simple you have F of alpha modul G of alpha you take an inverse it's like a g of alpha modu F of f of alpha and what you want to remember so the all I said now is relevant for any abstract symbol Alpha right I could I could actually introduce a new symbol Alpha and consider a a this collection of all pols with the I mean with

coefficients in q but with variable Alpha the variable the new symbol and this would indeed constitute a field but in our case we also want that Alpha would satisfy this relation this Alpha square + one equals zero so that's not a problem um what we do I mean in our case we have f f x is x² + 1 um and so when I have an alpha of

course satisfy this relation Alpha is a root of Fe um so when I do F of alpha mod G of alpha I can it's like of course it's like taking the polinomial FX and substituting Alpha but what I can do is a a first reduce f and g modul Fe so I can always write f as a say q1 Fe +

R1 and G is Q2 p+ R2 and since Alpha is a is is a root of the of the polom fee when I substitute Alpha in F I get a a q1 of alpha F of alpha plus R1 of alpha but F of alpha is zero right so I get so I get that F of alpha is R1 of alpha where R1 is the reduction of f

module okay and the same would be true for a G of G of alpha would be R2 of alha so in fact when I consider F of alpha module G of alpha in this setup it is equal simply equal to R1 Alpha modul R2 of alpha where R1 and R2 are the the mod Fe reduction of of FNG G

okay so in fact um I could rewrite Q of alpha as the set of all R1 X sorry R1 Alpha modul R2 of alpha is such that um R1 and R2 are pols over Q with R2 not the zero polinomial and the degree of R1 and the degree of

R2 are strictly smaller than the degree of Fe which in this case is two okay and and that's it that it's a field we showed that it's a field the the I mean the relation alpha alpha square + 1 equal Z is a a being taken into account um by this by by this consideration I mean this

relation ER would not happen if if Alpha would be some kind of abstract new variable but if if it's a an abstract variable satisfying this relation then then I have I have this equality in in Q of alpha yes so we can Define any Q of alpha on any f of x as long as the degree of f of x is less than uh F or G I mean my my question basically is like we we wrote Alpha square + 1al z but we

could have very very well written like Alpha square + 2al Z AL square + 3 and anything anything right we don't the roots of these things but we just Define this the Q of alpha like this so we know that in this field Q of alpha is going to be yes of course if um if Alpha was already a root in the original field then you don't add anything I mean yes yeah because again Q of alpha you you should think of it as like the minimal field containing

q a and Alpha so if Alpha was and this this by the consideration we we talked about that's that's the way to construct it so of course if Alpha was already in Q then you you don't get anything new but Alpha could be a a a symbol satisfying any polinomial relation being a root of any polom again over over Q now we know so remember this Fe was e X2 + 1 we know because we know the complex

numbers we know that the H the the roots of this polom are I and minus I which one of these is Alpha I don't know and I don't care I mean in fact so Q of alpha would be isomorphic to what is called the gausian rationals ER it's I mean one way to to to view it is all the numbers of form a plus b

i such that A and B are russal but of course you could you could write it as all the numbers of the form a plus I mean a minus b i or A+ B * minus I where A and B are rationals this these two fields are isomorphic right H so which one these is Alpha I don't know because Alpha was simply an abstract symbol satisfying a

relation but the result I get is is the same okay yeah so then also since the since the constructions are like this obstruct you that's why you never have only some of the roots and you always can have all of them basically because you like you know them beforehand so you can always construct some kind of abstract symbol such that you can clean up all of the roots of your of your polinomial right well so so so not quite I mean because in the

general setup I don't H I don't know what are the roots in advance I mean I don't have this a a I mean in our case ER we had these complex numbers in which we can find all the roots yeah of two but you just made up a symbol so you can carry over the computation yeah but but the thing is like let's um we go back to the example of a fix equals X4 plus what it plus two or minus

think let's see minus two minus two H okay so over Q of course um and now I ah and I didn't say but um this Q of alpha is is isomorphic to ER to just take a pols over q and mod out by the prime polom

Fe why because this this specific H blue a blue description of Q of alpha it tells you that it's it's the same as doing a polom modu because what did you do in polom modu of you had a a this guy you had remainder pols right are it's like the set of all remainder pols with addition and multiplication mod but of course if you have our R1 and

R2 since in in this guy in K1 um let me you know what let me write it in a different place just to have more room so we have a on one hand and we have K what we called K1 this was a a q of X modu of v f could be whatever Prime polinomial you want and this was all the remainder pols

are so the degree of R is smaller than the degree of P H and on the other hand we have q of alpha but we we took a description of of Q of alpha as all the quotients R1 Alpha modu R2 of alpha such that the basically the degree of R1 and the degree of R2 are smaller than the degree of Fe now I

claim that these two fields we we showed each separately we showed that each of these sets are is a field but I claim they are isomorphic fields I mean I can construct the isomorphism and show it kind of formally but to get it to get the intuition I mean what you have here if you have R1 and R2 in K1 then of course because sorry this is k1 because K1 is a field

I have R2 of K I mean R2 inverse right this is a a we call it just to be more compatible with the notation H this is r1x and r2x but remember that X is a root of Fe right was it was a root of a fe when I

consider F capital X in K1 capital x but this is just to say that X is a root of of of Fe but X we called it Alpha right yeah and indeed it set in K in K1 X satisfies

um Fe of X let's say or maybe Alp Alpha satisfies Alpha satisfies Alpha to the 4 minus 2 equal to zero because this is simply saying X to 4 minus 2 = to Z and this is exactly f of x lowercase x and in K1 Fe of lowercase x I mean in K1 you do modu of Fe K1 is is with addition modu of and multiplication modu of

Fe is this clear this yeah it is in indeed it I kind of get the intuition that if you find a splitting field over another field you can always find a splitting field over the splitting field and so on and so forth it just never ends well it let I will get to it soon but first I want you to understand this this first step yes least so so what I'm doing here I mean the K1 is like joining a new element Alpha satisfying

a new element Alpha satisfying a relation Alpha to 4us 2al zero and doing it in in a minimal way because it's the same as doing Q of Alpha and we we basically reasoned about the construction of Q of alpha it's the minimal thing that you can do if you want to to ER introduced a new symbol Alpha satisfying the relation F of alpha equal Z so what happened is I mean again we we can we take this example of

a x to the 4 minus 2 now I took QX mod and I know that um Fe has a root this is isomorphic to Q Alpha for a alpha satisfying Alpha to 4 minus 2 equal Z which so I mean this Alpha I mean

we know a post prior that Alpha is either plus minus Square sorry a fourth root of two or a plus minus a i times a 4ot of two this but this I put in Brackets because we don't care we don't know we we for for us Alpha is just a a a symbol I mean a symbol satisfying a relation so and now I I consider so we know that fix in this is

k1 and now F capital x is in K1 capital x and we know that this is it has a root Alpha um so it is xus Alpha times some f Prime and of course the degree of V Prime must be smaller than the degree I mean than the degree of Fe it's it's in fact exactly one degree less than than Fe okay and now I consider K2 K2 okay okay so before I consider it I

mean now if a um f Prime has linear factors or or or maybe I could write it like this um say h f Prime is of of X um is some linear factors x minus Alpha

I I = 1 up to some s times some this is f Prime some fouble prime Prime where a a fouble prime is a well it it doesn't have to be this needs to be double Prime H it doesn't have to be prime okay so let's say let's say it like this ER let fou Prime x

equals um P1 of X up to PS of X where m P1 up to PS R Prime over K1 of X I mean over K1 or Prime in K1 in K1 of X okay some of them may be

linear m aren't you missing a prime in the Fe of B yeah yeah yeah yeah ER so some of them may be linear it means that I drop them I mean without loss of generality suppose um um P Pi I of X ER is not

linear for every a a i bigger than some index in the middle I mean some some J then I take each of these guys I can consider M K1 x modul m p j + one of this I call K2 and by the

construction that ER that we get I mean now I I I I have a root um there exist Alpha 2 a root of um P J + one of X in K2 right this is the construction this this K2 it it needs to be Isom to to K1 ad join some alpha alpha 2 yeah so these AR two questions to me

the first one is is there any direct relation between Alpha 2 and Alpha One like like U is Alpha 2 like some sub group or I know whatever you want to call it of Alpha One like a sub group of roots or something like this I mean Alpha 2 and Alpha 1 are both going to be roots of the polom but there are going to be Roots there there are going to be different Roots because um when I considered now actually here it should be F No it

should be F Prime not fouble prime I mean I took I took out the root Alpha from Fe and I'm left with f Prime and now I I decompose F Prime into prime factors some of them may be linear so some of them may be like x minus x minus Alpha I okay ER so it could be that um K1 of X I mean in K1 of X Fe has a more than one root I

mean just as in the example of when you add the the the symbol I then you you get Roots two roots of the polinomial x s + one um but I I don't care I mean I just take whatever I'm left with I mean let's say a a a Pi Pi of X for I bigger than J are the nonlinear factors of of f Prime I take one of them I choose whatever you start with the first one and you take the mo

modul I mean the modul field of K1 of X with respect to this Pol this this is a new field K2 because I moded out by a prime polinomial I get a field again and this field contain a root Alpha 2 of this polinomial PJ + one that makes sense so okay and and of course since since a a um I mean PJ + one is simply a a a a prime factor of of Fe then this root Alpha is

Alpha 2 is also going to be a root of of Fe and and I continue in defin I mean I continue until I run out basically in every step I I reduce the degree and the degree is finite so I know that it's some finite I mean this process terminates now what happens in in this I mean in this example I take a a fe is again is like x to the 4 minus 2 I take Q X

modul um this is some Q Alpha for Alpha being being a root of this polinomial which one I don't know but let's say this is Q to the square root sorry fourth root of two which you can describe as a + b 4 root of two where A and B are rationals right you know how to to add and multiply you simply introduced a new symbol and it satisfy a certain relation of course since you introduced it a a square root of two which which was not

an original element in Q will suddenly I mean must be an element in this extension because square root of two is 4th root of two * 4 root of two so it must be an element in in Q of 4ot of two okay so you are doing you are doing some a a additional stuff when you add a a when you add a root but now okay so now I added I added this root now I look at the polinomial fee I'm not going to write this big variable

I think you got the point but now x to the 4 minus 2 can be can be written as um x minus 4th root of two times x x + 4ot of 2 times if you check if you do the long division you check what is left you get X SAR plus Square < TK of two and x² + square otk of two does not have any

root so um in Q square root of two h x this thisy guy which we can call a f Prime F Prime does not have any root because in order for f Prime to have a root in in this guy I would need some symbol like I I mean I would need some symbol satisfying um the symbol square equals to minus one

y okay so now I continue the process and I do a q square root sorry a fourth root of two pols modul F Prime okay and now I am guaranteed to get a root of f Prime what is this root I mean again a post priori we have the complex number and so on but a priori it would be just a new

symbol satisfying the relation um e square plus square root of two equals z but in practice what I get is is now I have q of square sorry fourth root of two and um let's say I * 4 root of two right because I I this is like the the roots of this guy are plusus I 4ot of

two um okay so how do you write it I mean this this guy is like all the elements of the form a plus a b 4 root of two plus C time I 4 root of two where a a b and c are elements in C and okay and and the the and this is a this is a splitting field this this splits um the polinomial

Fe completely right I once I have I 4 root of two I have also minus I4 root of two and I have exactly four roots of of the polinomial f we can even simplify a bit this is going to be isomorphic to Simply take a Q and A join four root of two and also a join I if you take multiplication of all these guys you get I mean this this by definition is like all the elements of the form a plus um

sorry A Plus B 4un of 2 plus CI okay yeah uh would there be a way to get these fields by adding one extra Element no and well sorry actually the answer is yes but ER this this this is another proposition that er um I don't want to get into yeah there is a

way okay thanks um but for for us is like a a we we do an iterative process we add a a roots to whatever prime factors of the original polinomial that we are left with and eventually we know that eventually the process has to terminate because we we reduce the degree every time and so we are eventually we are left with a splitting field for the polinomial Fe and we had a a proposition that said that a a splitting field is unique up to isomorphism because if you

think of it of course a um I I mean I took a a I took few of I mean I took a fe in Q of x i moded out I added some some abstract element Alpha that is a root of this polinomial is this Alpha Square sorry fourth root of two or is it a i times fourth root of two I don't know so all I can guarantee is that eventually I get some Isom I mean the eventual field a splitting field is is Isom is unique up to

isomorphism but I cannot control in the the process I don't know what I mean what roote did I add at any at any point okay so this is a we are a bit a bit over time but I I still I want want this point to be clear so um any other questions on on this issue my second one was whether this was diverging or converging but since you remove degree it has to converge at some point it's all for me thank you very much by the way M this was really really

useful okay yeah I'm glad this is a and now just to to finish up I mean what do we do we we consider over FP we consider the polom X a a to the H to the Q minus X where Q is the power of P okay this is Fe and so I get a splitting field unique up to isomorphism of this guy and this is called fq so it contains

FP and in this guy a every now so this polom there is a consideration that shows that a all the roots of this polinomial must be different it cannot have repeated roots okay so er er I see that people people are leaving so let let's let's just continue next time with the whatever other questions you have but try to go over the notes and see if you you have any

gaps thank you very much thank you so much thank you thank you see you next week bye