PSE Lectures - Ep 8 - Introduction to Fields

Transcript

this call is now being recorded okay so I know that I that many of you could not attend the the previous session um so before we we move on to other uh topics ER have you watched the the the previous sessions do you have questions or remarks about it I watched a bit but I still need to go like deeply through so I would count as as I watch it basically okay I mean another option I

could do like a a sketch revision of a of what happened in the would be helpful but I don't know if people here was uh if people in this was already here last week then U it's no issues like you can just continue yeah I think only only David was David do you have any some opinions um no it's fine for me okay

um okay so so I I will H I will spend a little bit of time on on H sketching what happened last time although we will not rely on it but I think it's an important er er part ER to your general knowledge and uh and we will'll definitely use it later on when we H ER start talking about elliptic curves so so this is basically um the so-called classification of of finite AB bilan

groups um and what it says is that um if G is a finite ban group then you can um of course you can divide the the order of G or you can decompose the order of g into a product of a of prime Powers so let's say a P1 to the N1 up to PK to the

NK and and what the classification tells you is that h g is isomorphic to um Z P1 to the S oh let's say yeah P1 to the s one some P1 to the S

one H L1 plus Z and PK to the s K one up to Z PK to the s l k where the sum of the S eyes ER let's say

i j j equals to from one to l i so for any for any I uh this sum it must be equal to n i so you see this is a kind of one component and then you have you have K different components H but the the bottom line is that g is

isomorphic to a direct sum of CC groups right I mean each of these these groups H is integers modu modu some I mean particular Prime power so we saw that this is a cly group um and and this means I I I can decompose g into a sum of of CLI groups and one coroller from this is that if D divides the order of G then

there is a subgroup age of G such that the order of age is precisely G is precisely D okay and this this is what is is sometimes referred to as the the converse of lrange theorem right because LR theorem tells you that um if you have a subgroup of G then the order of the subgroup must divide the

order of G but this Converse tells you that in finite a billion groups er every every divisor of the order of the group admits actually a subgroup of of precisely this order so this this is a very strong result that comes from the the classification and we will not use this classification today so you safe but I I highly recommend that you go over this uh the episode

um what what it it helps you understand is is a um kind of the the general program of of what we are about to do also now with with Fields ER namely we want to understand all mathematical objects of a of a given type I mean in this case finite a billion groups so we we develop the necessary Theory to to classify to classify them and and in case of a billian groups the the result

of the I mean of the classification is this H this kind of structure and um and it it is very helpful because then later when when we stumble upon a a an elliptic curve over a finite field we know that it's a finite a bilan group so we know it must be of this form so this is a this is roughly speaking what what happened last time but of course ER this is without any proof so

and today I want to talk about Fields now ER I guess you could you could say that a field is a well let let me put the the kind of long definition so field

is set f with a two binary operations which I denote by um plus and a multiplication H

and uh two specified Elements which we denote zero one if you want you can put z f and one F these are two elements in the field and we require them to be so this is the data and now the conditions are a kind of the natural

thing you would expect from a from um multiplication and addition of of let's say real numbers H so one is that both operations need to be associative so X let's say for any x y and Zed h x + y + z is equal to x

+ y + z this is associativity of addition and also h x * y * Z is equal to x * y * Z and the second axom is commutativity so uh for x x and y f x + y is = to y +

x and x * y isal to y * X um then we have let's say neutral elements axom so uh for every X in F 0 * sorry 0 plus X is X and 1 * X is also

and then we have a a inverse inverse axum so for any X in f h there exist an element which we denote as minus X also in F such that x + - x is equal to Z and for any nonzero element so I write it like this F minus this the set with one element containing the zero

element so for any nonzero element there exist an element which we call xus one such that e x * X - 1 is equal to 1 and the last the last axom kinds of tells you the compatibility required from the the two operations I mean so far all the axioms can be stated separately for for addition and for

multiplication and the fifth one says that for any x y and Zed in F if you do x + y and you multiply it by Z you get x * z+ y * Z this is called distributivity sometimes okay ER

and I so this this is kind of a long but I could I could do a compact definition with what we already know about groups and so equivalently a field is a f with a a binary operations plus

times and and specified elements zero different than one in F such that only two axom one is that if I take the field I mean the the the underlying set f with addition and the zero element or if I take the the underlying set of the FED then I I remove

zero and I take multiplication and the the unit element both of these are obedient groups and the second axom is the distributivity I mean the the compatibility between addition multiplication so for any X Y and Z in f x + z is XY sorry x z that's

YY okay and these are really equivalent definitions because if you look at the axioms one two three and four all say that H that you have an ailan group structure with respect to addition and multiplication uh in the case of multiplication you need to remove the zero element okay ER and last remark you can observe or you can think about it as an

exercise um it automatically follows so if f is a well first I claim that the the additive and multiplicative inverses are unique can anybody say

why I mean I claim that you cannot have a a I mean you take an element X in F you know that you have an element minus X such that x + Min - x is equal to zero but you cannot have another element different than minus X that satisfies this property can anybody say why I would just maybe this is not the reason but I would just write the same

thing with the other candidate yes and you can subtract X to both sides and you see they're equal but what do you mean to substract we're trying to uh ER to prove kind of that you could substract I mean to subtract is to add the the inverse okay that's the but remember that we Pro this fact groups right we we showed this you you you can go back to the the chapter of groups we showed that the the the

inverse in a group is unique you cannot have a you cannot have another another one and this is a this is a group so so the the additive inverse must be unique and and similarly this thing is a group so the multi implicative inverse also has to be unique by the same reasoning that we that we had in in groups so this unique uniqueness of of

additive inverse for example tells you that that Y is = to- 6 so this is one observation the second observation is well what happens um when when you multiply by an element by zero so I claim that for any X in F 0 * X is zero and this if you look at the axom let's say you look at the

compact definition it does not appear here right but it follows automatically from a from the axioms may I ask a question yeah sure um for the first item additive and multi multiplicative in versus are unique do you mean that um they are [Music] different I mean for each build element the um the inverse of it sorry the additive inverse of it and the

multiplicative inverse of it there are number different fi element I I mean what I mean okay so like additive additive inverse we denote as minus X right but but like hypothetically it could be that if you take an element X well the axum guarantees that you have an additive inverse but maybe there is another element

Y in F such that that that it kind of functions is another additive inverse right namely X Plus Y is also equal to zero and I claim that in this case y has to be equal to to minus X this is the what I mean by uniqueness and the the the analogous statement would hold for a multiplicative inverse I mean if you right you know by the axioms that x * X - 1 is equal to 1 but if you would

have any element y such that x * Y is also equal to 1 then it would imply that Y is equal to x - this is a uniqueness of the of the multiplicative inverse I see thank you thanks for clar okay and and and of course this is a this is something we we proved for a for groups so you can go back it's in the notes for example and see this disclaim for groups

and it follows I mean of course because of the compact definition we have a a two group structures ER so we can we can automatically deduce this uniqueness okay and as for the the the the second bullet point you see I mean I express zero as 0 plus Z then I use distributivity to deduce that 0 * X must be 0 * x + 0 * X and then I if you wish I add to this equation minus 0 *

X cuz I mean a prior I don't know what is 0 * X so I I add to this equation the the additive inverse of 0 * X and I get that 0 * X must be zero any questions about the definition and the the bullet points can you me that does my sound work I'm sorry can you hear me now hello yes okay sorry yeah I don't know what's going on today with the headset and everything but I have a

question it to me it feels a bit weird that in order to prove uh like thetive multiplicative uh Properties or in that the inverses are unique we are just doing operations by zero like it kind of feels like cheating uh I don't know if cheating but I'm not sure like it doesn't logically prove prove to me that they are unique or at least this is this is not a logical proof this logical proof I mean The Logical proof is something we did in groups but but let me repeat it you know what um let's put it

here let G I write it in the general notation so G multiplication and E be a ER and and X an element in G the axom of the groups guarantees that there exist x minus one in G such that x * x -1 is e if in addition

furthermore um there exist Y in that x * Y is equal to E then well what do I do I multiply all this equation by H xus one actually I multiply it well since I m I multiply all both sides on the left by x minus one

then I get x -1 * x * y if you wish formally I this is the case e xus one sorry x -1 * e but of course by associativity I could write x -1 Y is equal well and then the the the right hand side is something times e is always this something so this is x minus one and by okay the axum of the group Y

is equal to X thank you very much for this reminder that was useful okay so so the the inverses are are unique in a field just just as they are in okay famous examples one the the the origin of of all these generalizations is the real

numbers ER but you could even go a bit a ER a bit backwards and consider just the rational numbers so Q I remind you is the set of all quotients A over B such that A and B are integers and B is not zero so if you wish all the the generalized fractions this is called the rational

numbers and er you know from elementary school that you can H add two fractions and and get another fraction and you can multiply two fractions get another fraction H and these these operations are associative and so on you have an inverse a I mean the additive inverse is just putting minus and the multiplicative inverse of of a generalized fraction is you you invert the the order of A and B so a over the inverse of A over B is is B over

a h so so this is this is this is another field of course we saw that you take what we denoted as as integers modul P but now I will also denote it as FP this is all the numbers zero up to P minus one of course p is Prime with the operations um addition modul

p and multiplication mod P and to to check that FP is a field I remind you um to see that I mean the associativity and the I mean the associativity of of of modu addition modul P multiplication is a is straightforward also compatibility of of a I mean what we called here distributivity Axiom is a is is easy to

to verify the only the only axom that is not easy to verify is that you have a multiplicative inverse for any nonzero element but this this is what this is the content of for example FMA little theorum which we also showed so so multiplicative inverses also exist ER okay so of course you could go also to the complex um in one of the

presentations of complex numbers we talked about this um um Polar Polar representation so you take r k Theta where R is some non negative number Theta is a some angle between 0 and 2 pi and uh you could you could multiply and you could add and you can also find the the I mean of course the the

additive inverse is simply a taking minus here and uh you can find the multi duplicative inverse so yes so let's say if if Zed is R ke Theta then and let's say that that it's it's not zero so so R is uh is strictly positive and then the inverse of of Zed R inverse

K minus Theta okay and because we we showed that if you if you want to multiply two two complex numbers ER you multiply the you take the for example the polar representation and you multiply the radiuses and and you add the two angles and what happens when you do it for these two H these two elements you get one times kiss of zero which is okay and maybe an interesting example

we will of course we will soon restrict attention almost only to finite Fields but it's worthwhile to see I mean what can happen here in between in between q&r so first we need to at Le at least identify uh one nonrational number so proposition I claim that square root of two is not a rational number and this is a classical proof

from a Greek mathematicians ER is suppose towards contradiction that it is a rational number and write square root of two is a quotient of two integers but

of course these quo they they they they're not unique right because for example one half is also equal to two quarters and so on ER but I could demand that the gcd of A and B is one I mean if it's not one I simply cancel all the common all the common divisors of A and B until I get uh this is called a reduced fraction and of course A and

B are nonzero integers okay but then um I simply raise this equation to I mean to the power of two and I get that 2 b² is equal to a squ now if a is even

then since I assume that the gcd of A and B is one then B must be odd but but then I have a a a problem in the parity of this equation because a squ is also even and er okay so a square is also

so and in fact not only that a square is even if if a is even let's say a is equal to 2K then a s is equal to 4 k s um and this implies that

um b² is equal to 2 K squ and this cannot happen because H because b b is odd so b square must must be odd it cannot be it not to be two times something okay um b square is odd and that's a contradiction and if a is

odd alternatively odd and then also a square is odd but then but then a squ needs to be two times a 2 * b s and this cannot happen we cannot have

contradict and the the parity a okay so in both cases we get we get a contradiction and this means that we cannot have a square root of two H being a being a rational number okay and and this brings me to to a a new

example you consider this is the noted Q with a joining square root of two and what you do here you you simply consider all all the elements of the form x + y Time Square < TK of 2 where X and Y are rationals and addition and multiplication is is very intuitive I mean h a it's if you think of it it's

analogous to defining the complex as you take all the the elements of the form X Plus+ y i where I is like the square root of minus one okay and uh you can so of course you can you can H you can add two of these and you can multiply two of these when you

multiply of course when you multiply y * sare < TK of 2 with Y Prime time sare otk of two you get 2 * Y which is a a legitimate element back in the field and maybe an exercise a um what is the the multiplicative inverse of this element like a generic okay ER so I I leave it for you H but

but of course since we proved that H square root of two is not a number then we get something in between right we get Q ad join square root of two is somewhere in between the rational numbers and the ring numbers and of course I can join many many more Roots like I can jooin square root of of three and of five and and so on any questions about this uh this

example not yeah it's ah it's still it's still recording Okay so um observe the following I claim that a

F has no this is called zero divisors and what I mean is that if um for x and y in F you have x * y equal to zero then

either X is equal to Z or Y is equal to Z or both but but or in in in mathematical statements we mean inclusive or okay and why is that well I mean if XY is equal to zero then I can multiply XY by x -1 and this must be equal still to zero because I can multiply Zero by xus

one H so if if x is non then there exist a multiplicative inverse x xus one and I can multiply all this equation and get that Y is equal to and of course if Y is is is not zero then I can do the same let's put it in the same format uh x

y - one is is also equal to zero and this would mean that X is equal to Z okay this is this is a very very trivial observation but but from it we can already deduce something useful H so I take an integer well a natural number n and I consider integers modu n but now well of course I have a zero

element a unit element and a two operations right addition modu n and multiplication modu n so I have all the data that is needed for a field but the question is are the are the conditions of I mean are the axioms of the satisfied and I claim that this is fi if and only if and it to P is a prime prime okay so One Direction we saw I

mean if if n equal to P we that integers modul p is a and I remind you in terminology we call it the the prime field okay so so the the the real new content of this proposition is that this is the only case for which integers modul n is is a

field so so the the converse direction if n is equal to K * L with K and L bigger than one so so n is not a prime in in in integers mod n k * l is equal to well to to n which is ER and K and L are strictly bigger than

one so they cannot be zero H so they constitute a what we just called zero divisors right and and a field cannot have zero divisors so so ZN is not a field um a main a

main notion in the ER in the theory of fields is is what is called the characteristic of a field so so let F be a field the minimal m n such that 1 +

1 n * is equal to zero is called and the characteristic of F and we denote a a Char of f equals to

n if no such n exists of course it could happen H we say we write the characteristic of the field is okay and um of course the the characteristic

of the real numbers is a a is also the characteristic of the the rational numbers is also the characteristic of the complex numbers and this is and the characteristic of the Prime field FP is is p but potentially we we may have other examples

proposition we prove next it says that this is the only this is the only two possible examples H for any if either charf is equal to zero or Char F equal to p with a p Prime well how do you prove

this again this follows automatically from the axioms of the so we Den note the characteristic of the field by if n is not well if n is different than and it is not prime so it is it is a product of K * L with K and L bigger than one then I can do the following I I do 1 +

1 + times but so so this on one hand this must be equal to zero because n we we chose n to be the characteristic of the field and N I remind you the characteristic is the minimal integer that this equation holds but if n is equal to to K * L I can write it as 1 +

1 K times 1 + one L times right I mean uh this is end okay and and what I get is that K * L equal to

zero which means since in a field you don't have zero divisors H that either K is equal to zero or L is equal to zero but n is the minimal number such that 1 + 1 + 1 n times is equal to zero and K and L are strictly strictly smaller than than right H so so this this is contradiction to the

minimality um is this clear are there any questions clear okay so so you see we are like moving moving in small steps but um what we are after I mean ER let's

see okay let me give an example ER this is in the notes this is 5.5 so I consider F this is this is something s Den not is as GF four and as a set it's just the elements 0 1 2 and

three and I need to tell you what is the addition and multiplication I I tell you it with tables so 0 1 2 3 0 1 2 3 let's say addition ER you know what it will be simpler it will be simpler to to Simply say

that as an additive group I want F to be integers module two direct some integers module two and as a group I want a a f minus 0 to be isomorphic to integers modul 3 ER let me write the the multiplication just for

completeness so of course 0 multiplied by anything is zero and then we have H one two 3 uh two 0 2 and 32 1 okay and it turns out that this satisfies the the axioms of the field I mean what you need to

check sorry gf4 has distributivity this remind you it's it means that x + y * Z is x z+ y z okay so and once you prove this it means that this is a

however so okay and and of course the What is the characteristic of GF anybody say uh two two okay so because of course the the characteristic is completely determined by the the additive structure I mean the additive group structure of er of the field and and indeed if you you add two

elements in this group by the way this is called the decline if you add two elements in the client group you ER I mean you get a I mean if you add sorry the unit element in the in the client group I mean you add one comma one to itself you you get zero so what we have here is a finite field of a prime characteristic

right but it is not of prime order and this should be a kind of a motivation for you or for us to investigate what are all the possible options and how do I how do I figure out what are the the the addition and multiplication table of a of all possible finite Fields this this will be yes what do you mean by the possible options uh because with with a Gala field uh of four elements like this is the only multiplication table that you can get so

what are the but but how do you know it I mean how do you prove it okay fair enough yeah no CL okay so we we this this this is what it means that we we do a rigorous account we we prove what we claim and this kind of this approach forces us to really understand what is going on I mean and later on you will be convinced that this is the only option yeah okay like does it have anything to do with the fact that um this is the

only field or group I don't know how to mention it right now but that it's isomorphic to R R2 D plus R2 for the sum and to R3 for the multiplication I don't follow what do you call R is the real numbers no yeah the real numbers yeah sorry but so what is the question sorry if you want to prove that this is the unique um the unique multiplication table that you have uh for for a G field of four elements does this imply using isomorphisms in between groups and subgroups or something like that in order to prove it for

instance well it's a it's more involved okay I mean the the issue is like in a in four elements you could say I don't need all the theory ER I just I just uh run all the options and I can I can see immediately well not immediately but after a little bit of effort I can see option okay like it's not I I can see this this is the only up to isomorphism gf4

ER will be the only the only field with four elements ER and you could say that you don't you don't need Theory to prove it because you can I mean you can just try all the possible option for multiplication and addition see what satisfies the axioms and uh and conclude that there is only one option up to right up to a change of symbols here there is only one option for a field with four elements okay but what will you do when I give you a big Prime

like a a prime with the 100 bits yeah yeah you can go that way yes sorry um sorry I didn't follow why um the cor characteristic of gf4 is two okay ER let's see so so I claim ER I mean this as an ailan group it's it's isomorphic to integers modul 2 times itself let's write uh let's write it in a

explicitly 0 1 2 3 0 1 2 3 and of course zero time zero plus anything is this is the same so 0 1 2 3 0 1 2 3 then 1 + 1 is zero and H 1 + 2 is three and 1 + 3 is

two and then here I have um two three 3 0 1 and 3 2 1 0 okay so if you wish this this addition table would would correspond to the addition in um a Zod 2 direct some Z 2 where one one would correspond to 1 comma 1 h

two well either to a 0 comma 1 or to H 1 comma 0 I need to check okay and in three would correspond to to the opposite so either to 1 comma 0 or 0 comma 1 okay and and indeed if you you do one comma 1 + one comma 1 you get you get 0 comma 0 in in the CLI but alternatively you

could just say the the addition table is is simply defined as this and and from the addition table you see immediately that 1 + 1 is zero so the characteristic to is this clear um yes thank you so so it's not I mean I I advise you this this is a good exercise to kind of convince yourself that this satisfies all the the axom of of the field so I gave you now explicit addition and multiplication table it's a bit this but not not too much it's certainly feasible you can

check all the axioms of of a thank okay now well I say that this is the only field up to isomorphism so but I didn't Define really the notion of of isomorphism between Fields so let's let's do it now let F and K be Fields

function F from F to K is a field for morphism well first I want that um f of Z is equal to zero and F of one is equal to

two for any X and F I want that f of x + y is f of x + f of Y and I also want that f of x times Y is f of x * F of Y and so this is this is a homomorphism and and a a f is called an

isomorphism if it is in addition inject ER bject um in fact

yes so does this mean does this mean that if we go that way and we try to find an isomorphism where either F of K is GF of4 the other field has to be F of GF of four also like there's no other way in which from another field can have for this function of right well I mean to be precise isomorphism does not does not mean equality of sets so if I defined GF of4 in this manner it means that I I

committed to a specific set of symbols 0o one two and three and and you if you want to Define an isomorphic field to gf4 you do not need to commit to this set of symbols the only thing you need to commit is to the the addition and multiplication tables so you could call it a I don't know a a b c d where a b c and d are some different objects than 0o one two and three but you define the multiplication table just like this just like I did here I see I

see and this will be isomorphic right it doesn't have to be equal and this this is why we we have to have this flexibility of of isomorphism because things in reality are not equal I mean you denote it in different letters you get something that is is not equal as a set but it has from it has the same structure of addition and multiplication tables thank you okay so this this is on the on the kind of naive level but in in slightly more sophisticated

examples you you can you can look at certain certain sets of for example of pols and and and find out that this this specific set of pols has a structure of a field which will then be a a isomorphic to this Gala field okay so it can I mean it can come in in many different flavors but the the the bottom line is that you cannot I mean this this will be the classification of finite Fields up to isomorphism there is only one okay

awesome okay now we we talked about it in in the case of groups so if if you think about it a field homomorphism it needs to be a group homomorphism with respect to the additive group structure and a group homomorphism with respect to the multiplicative group structure this is equivalent definition and what we had in group oops is that a if f is homomorphism I will say just a a a a

field map just like we said in groups we said a a group map then for any EX F of minus X is automatically minus f of x and for any non zero x f of the inverse of X is the inverse of f of x

okay this essentially it follows from the fact that we have it for groups okay so example h of course we can we can look at the inclusion as a function so I take rationals and I include them as the real numbers and this is a field a field

homomorphism and I can also take the real numbers and include them into the complex numbers and these two are build Maps now a non example I claim that there is no field map from the prime field FP into the

if there was then I could write zero as I mean of course zero is f of zero let's say this this field map is f h but zero in FP I can write as since FP is is characteristic P I can write as 1 + one p times but f is a field homomorphism so

it must commute with a with addition so this would mean that that F of one itself P times is equal to zero but again f is a field homomorphism so it must send the neutral elements to the neutral elements in particular F of one is one so I get in in q 0 is equal to 1 +

1 B * and this this is a contradiction okay and I would say more generally this this I give you as an exercise if you have a um two two Fields with a two different Prime

characteristics let's say the characteristic of f is p the characteristic of q and P is not equal to Q um there is no fi map F from from the field F to the field K okay and this is done similarly to

this uh to the second example so you see a a Fields kind of have a a a more rigid structure and this ER this is quite natural to expect because we put more conditions so we put more conditions there would be less objects it would be harder to satisfy the conditions we put so there would be less objects that actually satisfy it any questions about the about this part

yeah um when you mention there's um the field structure um is it better to have um I would say that so is if using the field in cryptography um is it better that this the the structure is simpler or um for example compared to group so if we want to use a um we want to consider using group or bill um would that be better to use

group well so it really depends on the what you are after I mean what happens in a typically in finite groups We Know by the classification theorem that you essentially get a I mean you get a direct sum of of integers modu modu n for various n right and um as an a billion group you have um I mean the the the the cryptographic

difficulty would depend on factoring a a factoring an integer to let's say a product of primes right now what turns out to be it may not seem like this from a first glance but there is a a quite consistent advance in the the efficiency of algorithms to decompose a an integer into a product of primes ER so your security ER is is Con constantly diminishing I mean as the the

research advances and I mean by now we already have sub exponential efficiency ER to decompose integers into product of primes and this would mean that for the same level of security you need to choose higher I mean longer Primes ER in fields there is also an advance in so the security in fields we we still didn't get to a point where we can talk it sensibly but the security in fields

will be a slightly different and and the security in elliptic curves will be a a a third thing okay so so a a the let's say the answer to your question I cannot give in details but generally speaking it's it's a matter of security I see so that that's one reason and of course from from the point of view of course ER we want to use elliptic curves and and pairings and elliptic curves cannot be I mean pairings of elliptic

curves and elliptic Curves in general cannot be defined over groups you need the additional structure of the I mean you need the two binary operations in order to to talk about elliptic curves because remember that an elliptic curve is like a polinomial right it's like a y² = x Cube plus a X plus b in order to make sense of the the solution for this polom you need to have both addition and multip and multiplication you need to live in in some word where you have

multiplication and so this is kind of the the the natural place to to look is field ER and you you could not make sense of this polinomial equation over a group because you what is I mean what is y * y right I mean okay if you if you take the the operation of the group to be a multiplication you can make sense of y * y but then what will be a I mean a X plus b and you you wouldn't have addition So eventually you need you need two binary oper ations to make sense of of in general of a polinomial

equations okay I see thank you uh all right maybe we can squeez in something er something else okay I claim

that if you consider this this example of the the galwa field gf4 of course you can view um the field with two elements as as a subset of the field with the of the galwa field with the elements and it's not only it's not only technically a subset because this is just a matter of a how I I call my symbols it's also compatible with the addition and

multiplication right I mean look the the the multiplication of the elements 01 in in the Galla field multiplication table is precisely the multiplication you would have in the field with two elements and also the addition in the the mul in the the Galla field with four elements restricts to the addition of a o of the the field with two elements and so what it means is this is very similar to what we had in in groups

I I would call it a subfield let me let me phrase the precise definition um let F be a field a subset K of f is a subfield well it has to contain the zero

element and the and the unit field with respect to the binary operations plus an multiplication of of f okay so so in particular if K is a

subfield of f then then K is a field in its own right and the the the main example of course is that the field with two elements is a now adding adding to this H

rigidity statement about about Fields ER I claim the following let P be Prime and um and f a field with a characteristic P

chains subfield that is FP okay so how do I uh how do I see it and consider the K well I I take a a

1 1 + 1 and and or maybe it's better I take one 1 + 1 up to 1 + 1 P times but 1 + one p times because the characteristic of of f is p this this is equal to so all in all I have a um I have a set with P and of course I um I can I can multip

since since this is a ER since f is a field I can multiply and add to H two elements in K If X and Y are in k h say x is equal to H 1 + times and Y is equal to 1 + times then of

course I can ER I can multiply these two and I still get by by this will be also I mean it's going to be a a sum of ones so so it's also an element in the in the set K and uh in fact this will be of

course K * L * and of course if I add then ER I also get a also get a sum of ones so so this is also in in the in the set inverse so you see where where this is going of course the inverse of of X in additive I mean if x is is a um 1 Time 1

* sorry 1 + 1 + 1 L times then minus X will be 1 + 1 P minus L and you can verify yourself that that x minus one is going to be x to the P minus 2

okay because this is this is essentially from from my theorem and x to the P minus 2 of course is also a sum of ones because you take 1 + one one p minus 2 is some some sum of ones so it's in okay so this is all to say that a k is a subfield of F and I can Define quite naturally a map

from K to FP by sending one let me be just a disambiguated one of f which is an element in K I send it to one and of course 1 f one F I send to 1 + one and so on and this this is a is a field field isomorphism okay um so so the the the crooks of the

the proposition is that you if you have a field with prime characteristic it may not be the prime field that we we know and love but it contain an isomorphic isomorphic copy of of the Prime field um excuse me yeah when you wrote minus X didn't you make a mistake isn't it P minus K the number of ones ah yes yes yes uh yes thanks I apologize thank you K yeah okay but this is this is the same

Principle as as what you what you know in the prime field and all you use is the the distri distributivity axom that must hold in the field F any other questions yeah yeah I have another one um well in two uh so the first one is if we go above um when you have basically mentioned uh the the property between G2 and gf2 and

gf4 one being a subfield of the other one why is like for GF three will I assume it will not be a subf gf4 right uh is it because the multiplication table is not sure yeah the multiplication table is a look this this comes back to the exercise ER that I wrote here if you have two two Fields with different characteristic you you cannot have a a field map and you need to remember that if you

have a subfield inclusion then in particular you have a filled map yeah okay right so um so to prove it you use the same thing the the same thing we we did here I mean if you had a fe if you had a field map from a field of characteristic k h sorry a field of characteristic P to a field of characteristic Q then you could write zero in the Target zero in the Target must be F of zero okay okay see which is f of 1 plus

one + one p times so you would get that well and f f of this is is just one time one time one p times in the Target in in the Target Field so you would get that the characteristic of the Target Field cannot be Q it is p yeah I see right that makes sense so yeah so this is this relates to this H rigidity rigidity of the this Prime Fields they you you cannot there are many things you you simply

cannot do the the other thing is uh maybe next week we can start by revisiting how you applied for the mat theorem to get the P minus two on the sum of ones uh on the example at the end well this I can I can tell you already now I mean okay um you know I mean what is X x to the P minus 2 is like a you have a binomial

um X is is we said it's like a a 1 + 1 + 1 K times yeah okay so this is in in if you remember in groups ER we denoted if you you add in a billion group you add an element to itself K times we denote it as K times the element sure okay and so the same the same is true

for Fields I mean I can denote K * one and and so I I can simply denote it as K this is like K * one okay so so now I really have like a complete analogy with the ER with the the integer mod I mean for example right because when I will do so when I will do a k times l in the in the general field in this

kind of abstract field I it will be exactly as a a K * l modu p right because you just you just open up you open you you write K as a sum of ones you write L as a sum of ones sure okay that makes sense and and you use the fact that the characteristic of a of K is is P that is a sum of ones p times is is zero MH this this is essentially the property of of a saying

that addition multiplication correspond modul yeah but but that doesn't at least for me of course I'm probably not foll but why how does this justify the minus two so the P fine okay but the minus two exactly so now okay so so now but if these correspond to multiplication module p then I could say if you want you could Define a um a a map who Define a map between K and FP that sends a a k to

K okay yeah ER or if you want you you can also Define in the inverse direction from FP to K you you send K into K okay okay yeah so or X into but and this map is compatible with addition and multiplication as we defined it in K and in FP right sure yeah so so now I know that in FP if I do

x to the P minus minus 2 I get x minus right and x to the P minus 2 must go to well to x to the P minus 2 I mean there is no I mean this is like a this this is like the identity map it's just that that in k i i i mean I have kind of different notation because it's a an abstract field I mean if I would want to be precise in in what I write I would need to write I mean k is a subfield of f

right so the element one is 1 f 1 f+ 1 f as many times as you do right this is but it's just a matter of notation like K is really it's like an isomorphic copy of FP so x to the P minus 2 will be the same here it would it would okay so I realize it's not a it's not a formal proof but actually it would be good for you to to try it as an exercise yeah I'll definitely try I don't feel to um positive on getting a good outcome but

but I'll definitely try and ask on on the Discord otherwise okay alternatively you could so if you remember in the field to prove that inverses exist you have two Alternatives one is the for little theorem and the other is the the baz identity right because if a if K is is smaller than P then gcd KP is equal to one and this means that there exist this bazook

coefficients such that UK a plus um VP is equal to 1 so it means that u k is equal to one modu p so this is the inverse right so so U is equal to K one I see okay okay and this again this you can whatever I did here with integers I can do here with the the kind of fake integers yeah I see

okay okay okay okay that's insightful thank you thank you madam sure okay so uh we're a bit over time but I hope H I hope you're okay with it we we'll continue next time with the the classification it would take us a while thank you very much and apologies for the recording issues and like the mic and all all that stuff computer is cive today all it's okay