PSE Lectures - Ep 9 - Intro to Field classification

Transcript

welcome everybody I hope you got a chance to see the the videos of previous sessions and generally I I want to encourage again people to ask questions it's helpful not just for you but also for me and er er I think what will generally happen is er I will post a uh a link to the updated notes Here in the chat ER so you could follow pretty much a an updated version

every time all right so without further Ado um let's let's start by ER recalling what what we had last time is that ER yes yeah so so to ask a question um just raise a hand and ER I can I can hear the the Ping and I will uh I will stop and

then give time for it so before I start maybe any any questions about previous I have one related to the to the previous uh to the previous exercise that you left uh which was related to using uh forat theem yeah so I spoke with Rafael who who was very kind and kind of showed me the proof um on where the two was appearing from the wild but it's true

that you started talking about binomials to to actually justify or prove where the two from the equation was appearing from so I'm I'm a bit curious if you could uh give kind of the solution for the exercise that you that you sent okay let me yeah let me try to to repeat it ER not with the binomials I will get to the binomials soon okay sure so suppose you have um actually I'll choose K here

um is a field of a characteristic P this means I remind you that a 1K plus itself P times is equal to zero or to Zer K and what we showed is that H within K

you have um let's call it f i take a zero then 1K then I will denote as 2K simply 1k+ plus itself and this goes on until ER what you can denote as P minus1 sub k h which is right it's just uh 1 K plus

itself P minus one times and I I claim that of course this is a a subset of K it contains the the zero element and the the unit element H it contains a a an additive inverse for for every element because H um the additive inverse for minus a sub

K is simply P minus a sub k okay of course this if you add these two together you get a p subk and p subk this is p subk is is zero because we are in characteristic P so you have an additive inverse ER and now I mean so this if if a is is a is a number between

zero and P minus one then you have an additive inverse to a a sub K but also you know that in FP in FP you you have a multiplicative inverse for a I mean there exist u in FP is such that

um a * U is equal to 1 mod P which is equivalent to say that there exist some V again between between zero and P minus one such that a u plus h PV is equal to one but but then you can you can add add subk to this

equation right I mean what is a in in in FP a in FP is simply 1 + one a times and the same for for u p and v so what I can deduce is that a sub K times U subk plus P subk time V subk is equal to 1 subk but P subk this is what what is written here P sub K is zero so what I get

is that I have a multiplicative inverse to a subk right there exist there exist U such that a * U is equal to one this of course here I need that a would be different than zero so is this clear yep okay so so what I get is that this this

F is a subfield of K it's it's a subset containing zero and one if you wish I mean it's of course it's closed under addition it's closed under multiplication and contains the the inverses of a I mean the additive inverse and the multiplicative inverse so so f is a subfield okay now as you can see from this reasoning about the multiplicative

inverse the elements I mean the the the operations in in F behave exactly like like the ones in FP I mean the only thing is that I have to formally to for add this this subscript K because it's not it's not like real natural numbers it just it's kind of a adding the the unit element of K to itself this many times so so of course I can construct an isomorphism um from F to

FP if you wish you can construct in this direction ER you send you send 1K to one and AK again you send to a and this respects a this is a homomorphism because it respects addition and multiplication and it's clearly a one to one map because I mean I have a I have exactly the same number of elements in each side and I'm I'm I'm doing a one to one course respond so so

in fact this is an isomorphism isomorphism of fields and and the corer of of all this is that a a a field of characteristic P contains an isomorphic copy of of FP is this clear okay so but but observe

that K if K is is a field of characteristic p um it is in particular a billion group with respect to addition I mean K plus Zer is an a billian group and in a billion groups we a we also

have um for for any integer actually we defined um and the and any let me call it V in K we def find a * V to be V + V A times where if a is negative you take a a you take minus V plus itself a

times okay so so now potentially we have a conflicting conflicting notation it's not exactly conflicting if we we H if we pay attention we could write a k v right because a a k is an element in K so I could multiply a k by V via the multiplication of of elements in K and so what do you make of it I have two two things that look the same but are they the same so I could

write this expression is 1 K time V plus itself a times and by the H distributive axom of fields I am allowed [Music] to ER to write it as a 1 K plus

itself a times times V and and this of course by definition this is our a subk times B so you see the two the two notations coincide now a a a what I can do in this in this situation is I have this H

um I mean for any A and B in FP I can do and and for any V in K I can do a plus b in FP and then multiply it by V and this will be equal to a * V plus b * V I should say I guess from this equality what you can deduce is

that if you you take p and you multiply it by a by V well I mean you get p k multiplied by V and PK is is zero so you get 0 time V which is zero and and this is why this this equation star holds if I interpret a a let me write it explicitly if I do A+ but Plus in FP with B then it's the same as doing a * V plus

b * V now in addition I have um a * V + W equals AV plus a and this is for any a in FP and VW in K okay and what I claim is

that this kind of structure let me put two stars here so this this the essentially these two conditions star and and double star are a useful to to abstract in order to understand the structure of fields of characteristic p and this this should motivate the following definition yeah so let let F be a

field uh a vector space over if is a set v um with a a specified element Zer V and

um well binary operations I have actually two so one binary operation is plus this goes from the cartisian product of V with itself to V so it means you can take two vectors and add them together like okay v v+ w and the other operation is called multiplication by a

scalar this takes an element in the field and uh an element in V in V and gives you back an element in V so you can you can you can imagine that I mean here I take V comma W and the notation would be V + W and here I take a comma comma V and I send it to a * V okay so this is the data of of a vector space But as usually in these

kind of algebraic definitions we we want some conditions that H this data satisfies and the conditions are a pretty pretty easy or pretty intuitive in light of the discussion so one I want that V with this addition and the zero element Zer V is an a bilan group and then I have

um one condition is that I want associativity of this H this is called a multiplication by scalar uh and this is called addition of vectors um so I I want the multiplication by a scolar will be

associative this means that for any a b in F and for any V in V A Time BV is equal to a * B * V you see here the multiplication is the multiplication of f okay so so this is associativity of

multiplication by a scalar and the third axom is compatibility between addition of vectors and multiplication by a scalar and namely we want a distributivity so for any a in F or let's say um any a b and F and VW in v i I can do two things right I can do first a plus

b time V and this should be equal to a plus b V and I can also do a * V + W and this should be equal to AV plus a w okay so this this is a vector space and the main example for us a

a for a field K with with characteristic p k is a vector space over FP um any questions about this this is a of course this is a a new notion

but I hope it's it's a it's more intuitive now that we had definitions of a similar algebraic structures ER now let me let me get back to your question Carlos you asked about this binomial binomial argument sure so for a um for a

field K with a a characteristic p ER for any X and Y in K I claim this is the the sometimes referred to as the The Freshman Lemma um this ER this is what discussed in in

FP which means that x + y to the p is actually x to the P time y to the P like the very naive equation actually actually holds and the reason I mean we actually if you if you remember when we proved h um we took x + y to the p and we decomposed it via the binomial formula so this is a sum of

um p over i x to the i y to the P minus I I equals 0 up to p and we noticed that um if I is strictly bigger than zero and strictly smaller than P you take p over I is like a p factorial divided by I factorial H time P minus I

factorial um but since since p is prime actually um P divides ER this expression so so you could write this a p p over I is p times something let's say t but then as as we saw in the the the discussion beginning of the session ER when you when you multiply by P you get zero in in characteristic P so all

these expressions in the middle vanish and and you get er that x + y to the p is equal to x to the P plus y to the P okay so this is this is the the sketch of a the argument you you could H take it as an exercise to go over the proof of this uh this LMA this LMA was stated in FP previously we proved it in FP but in fact it's true for any field of characteristic

P okay so this and if you want moreover x + y to the P to the N is equal to x to the P to the N plus y to the P to the N well how do you do it you do induction I mean it's really trivial induction for n equals 1 we know it and for n equals 2 well I I'll leave it to you okay okay

so um okay so so this this is an aside but what we really care about because we want to understand a general fields of characteristic p and this structure of a vector space over FP is the the first stepping stone into classifying fields of characteristic and what we really need to to do about this structure of vector space is is is again to

classify ER Vector spaces more specifically what is called the finitely generated Vector spaces but we'll get to that so the plan for this session and probably the next one ER is to to classify finitely generated Vector space which I will explain in in a minute and er use it in order to ER to get a first first ER stepping stone into classifying fields of characteristic P any questions

okay so a a famous vectory spaces well of course um every field f is a vector space over itself right I mean of course you you you can take a in

F and the VF and you can do a time V and it will satisfy all the axioms more generally you can take the cartisian product H this is written as F to the N so you take F cartisian product with itself F time n times and I I remind you the cartisian

product is the set of all tles say X1 up to xn such that for any I x i is in F now on this caran product you have the structure of a a vector Space by defining scolar multiplication by a h let's say let's say V is the X1 up

to xn then I Define a well H let's let's maybe first do addition so if W is a is y1 up to YN you simply do H addition in n F in each coordinate so this is X1 + y1 up to xn + YN and ER for a scalar a you do

a is is simply a X1 up to a x in okay and of course the zero Vector zero FN is the vector with all the coordinates being the zero element in F so okay I mean the cartisian product since f is f with the zero element in addition is an a billian

group we saw that if you take the cartisian product of a a billion groups or even just generally groups you get you get another group with exactly with this operation H so so clearly F FN is a is a billion group with respect to this this addition operation and multiplication by scalar is associative because we do it in each coordinate and in each coordinate you do you do multiplication in F and and

multiplication in F is associative so you get associativity of scalar multiplication and also distributivity since I do it in each coordinate I actually have distributivity because I have distributivity of um in in F so in each coordinate okay now the the kind of the general principle from the

discussion we had in the beginning of the session is that if if you have any field inclusion this is a a subfield inclusion then a k is a vector space over if so for example h r is a vector space

over over q and er C is a vector space over over R but also C is a vector space over Q because key is also a subfield of of c as a non example a a r is not a vector space over FP even though you could of course you

could think of the element in FP as included in R right it's just the the integer is zero up to p p minus one but the the operations in FP are not compatible with the operations in R so when you will try to do let's say let's say you take a um a in FP and h v in r v in

R let's say you take a two elements you take A and B in FP H then you do you may do a plus b in FP and then try to multiply it but but this is not going to be equal to take a multiply it by B plus b multiplied by V because here the multiplication is meant in FP for example if I take a a to be one and B to be P minus one then a plus b is

zero H so multiplied by V is going to be zero always but a * V is V and B * V is is p minus one p minus one time V so so I get a p * V which is not not equal to zero in R yeah um it's like a a na question but we don't um have a requirement for Vector space elements to be vectors they are still like field

elements um for us vectors is just a terminy ology for an element in a vector space okay okay so this is unlike because we we do kind we take kind of an abstract approach ER of course what you can what you usually learn in in like preliminary mathematical courses is that H you take for example the the plane the plane is is simply R2 R2 is is the cartisian product of r with itself right and then you take

a a like an arrow that starts from the origin and you call it a vector right um but this in our language we simply refer to this point H this V is going to be H some point in the I mean the end point of V is going to be X comma Y and for us this is the vector it's simply an element in a in R2 and of course if if the field we are working over has kind of

geometry then we can interpret a a vectors in a in a geomet way but we are really interested in in fields of characteristic p and these guys they don't really have uh any apparent geometry so we we really need to take this abstruct approach and and uh I mean anyway this is the kind of a origin of the terminology is this geometric meaning so it's it's good to to keep in mind is this answer your question your question yeah sure thank

you okay um maybe a less trivial example because we want to do kind of a bit more abstract stuff H take the field with two elements so I take F to be F2 and and consider and the power set let's say um

I take the field of two elements and X an arbitrary set H the power set of X is a the set of all subsets of x this is the set of all s such that s is a subset of X of course including h x itself is a subset of X and also the the empty set

is is a subset of x h now we take V to be simply this power set P of X and now I need to tell you how to H add two vectors and how to multiply by a scalar so actually let's let's first start with the multiplication by a scolar um I mean of course 0 Time s should be zero the the

the zero element of V is the empty set right it's a subset of X so it's it's an element in in V then multiplication by a scalar H I really have no choice because zero times any Vector needs to be zero so 0 * s is going to be z v which is the empty set and 1 * s again I don't have any choice because one times any Vector in a vector space is the the vector itself so it's it's

s this maybe you can you can do as an exercise it follows immediately from the axom of a vector space that if you you take the unit element of the field in this case one you multiply it by a vector by any Vector you get the same Vector okay so this let's put it like this exercise true in any Vector space um okay so multiplication by a scalar is

kind of canonical ER now I need to tell you how to add add two vectors and what is the the additive inverse so um for two vectors S and S Prime in v this really means S and S Prime are subsets of x i define the addition to be the so-called symmetric sum so I take all the elements in s that are not in S

Prime Union all the elements in S Prime that are not in s and a minus s is defined to be the El the the set of all elements in X that are not in s okay and you can you can do is exercise um verify that this is

indeed okay any any questions about this