groups-rings-fields

[crystal turtle]

Second to what

oh wait duh

[alpine island]

If first isomorphism theorem is discussion's greatest moment, second isomorphism theorem must be discussion's second greatest moment.

[charred flume]

how to prove this?

the highlighted one

[lapis latch]

pretty easy with the fundamental theorem of finite abelian groups

[coral shale]

unneeded.

lagrange n sht would do

[chilly ocean]

Could somebody please hint to me what $1_{R / ~}$ is (is it a function? function composition should yield a function) , and further what a normalization map is? I tried googling the latter and I couldn't really find anything discernible to me

[delicate orchid]

yeah just induct or somesuch

[cloud walrus]

tmpr

[delicate orchid]

it's the identity map on R/~

[chilly ocean]

Thank you, that makes sense. I wasn't sure if this is maybe the identity element in that quotient ring, but it wouldn't make any sense

[barren sierra]

Yea you're composing two functions, output is a function

[charred flume]

what Group to use?

to apply burnside theorem

[delicate orchid]

different up to what? permutation of the colours?

otherwise this seems purely combinatorial

[charred flume]

i think this is combination

wait order matters

[rocky cloak]

presumably Z/12, the group of rotations of the bands

[charred flume]

so permutation

[delicate orchid]

yeah it's either Z/12 or S_12

I think the problem is I don't know what a necktie is is it a circle?

or D_12 if they're in a dodecagon for whatever reason

[coral spindle]

I don't see how the lemma that is not Burnside's™️ is applicable here. There doesn't seem to be any symmetry involved.

If it's just a tie, then it's a straight line down.

[delicate orchid]

yeah that's what I thought

so it's just ||4^12||

[rocky cloak]

Right, I misread it as being a wristband somehow

[coral spindle]

Seems like a question to trick you into thinking that Burnside is applicable when it is not.

[delicate orchid]

a wrist band would make sense

[coral spindle]

Well I mean it is applicable. To G = 1

[delicate orchid]

we could flip the tie around, Z/2Z

[charred flume]

this is from Judson Abstract algebra Exercise 14.4.9

[coral spindle]

But a tie isn't the same back to front; there's a frontside and a backside and they are distinguishable via stitching

[delicate orchid]

no you turn it switchways not flip ways

[coral spindle]

Maybe that was what was intended, idk, but it's horizontal stripes still isn't it

[delicate orchid]

I'm picturing a rectangle with partitions here not an actual tie

but you are right irregardless of the way of the flip

[abstract rock]

judson sometimes just has completely irrelevant exercises

that was one of them

[delicate orchid]

burnside's lemma for the trivial action!

[abstract rock]

there's a few exercises that walk you most of the way there and drop you off a few steps from the solution

[cursive spindle]

I'm gonna raid this channel with analysis

[abstract rock]

algebraic analysis?

[lapis latch]

mfw there is analysis in my algebra

[white oxide]

when lang says "representing $B$ as a factor ring of a polynomial ring" does he mean $B \simeq A[x_1, \dots, x_{m - 1}]/\mathfrak{a}$ for some ideal and then just use the projection map as ur epimorphism

[cloud walrus]

okeyokay

[white oxide]

given induction

[mighty spade]

You just need to embed B into A[[X]]. You then get the factor ring and the epimorphism for free.

[white oxide]

i'm too lazy to think about what the embedding would be given by

i also don't really understand what he means by algebraically independent. does he mean that for each i, \varphi_i given by f \mapsto f(x_i) is injective?

like what's the relation between the evaluation map and x_1, \dots, x_n

ok i understand the nontrivial part

i'll try to prove equivalence of the conditions later but i can't if i don't know what the former condition even means

[mighty spade]

Here’s one:

https://mathoverflow.net/questions/117554/polynomial-embedding

[white oxide]

wait what is R[[x]] again

power series with coefficients in R and variable x

or something

[mighty spade]

yeah

[white oxide]

oh okay i feel like there's a canonical embedding here for some reason

i'll take a think

(well when is there never a canonical embedding lol)

[mighty spade]

You can probably use the universal property of monoid rings to do that

lang mentions it in the group/monoid ring section of the ring chapter

[white oxide]

ooh ok yeah i feel like i'm going on a scavenger hunt in this book

[mighty spade]

although you’d need to generalize to formal power series

[white oxide]

this book is holy

ah ok

just wanted to make sure i'm not going insane; so in accordance with this definition, if we take n = 3 and f(t) = a1t1t2^2 + a2t3 + a3(t1 + t3) and sigma = (132) then sigmaf = a1t3t1^2 + a2t2 + a0(t3 + t2) right

all you're doing is just permuting the indices?

[mighty spade]

Yes

[white oxide]

btw you're probably the best person to ask

how long did it take lang to write this book?

it's fucking massive

my prof said something about ppl kept on asking him to add stuff

so he had to go back and add stuff and that's why it's massive

[mighty spade]

“Learn from the masters, not their students.”

[white oxide]

what does that imply

[next obsidian]

Dude what the fuck this is absolutely not what’s going on

[next obsidian]

Yes

[white oxide]

lmao

[mighty spade]

Does it not work?

[white oxide]

chmonkey coming in hot

[next obsidian]

x1,…,xn are not algebraically independent they’re elements of the algebra generating it

And even if you did embed into A[[x]] how do you descend Noetherianity to a subring?

[crystal turtle]

[next obsidian]

Okay I apologize I was too harsh

I thought you were actually trolling

But I see how you may have mistaken this for a polynomial ring, upon reading the proof tho and seeing it says finitely generated algebra I think it’s clear those aren’t variables

But the problem still exists of how embedding in a Noetherian ring gives Noetherianity

Because that actually isn’t true

[cobalt heath]

x_i's are not variables?

[next obsidian]

No they are elements of B which generate it as an algebra

Meaning the map A[X1,…,Xn] -> B sending Xi to xi is surjective where Xi are indeterminates

[cobalt heath]

Huh

[mighty spade]

fuck, you’re right.

I thought for some dumb reason that an embedding A->B implies an epi B->A

[next obsidian]

Rip

Bro been working in Vect too long

[cobalt heath]

What does Vect category look like?

I wonder if it's like ordinals

[next obsidian]

I think objects are in bijection with cardinals

But there’s a lot of maps

[cobalt heath]

Ah, cardinals. ..also I forgot maps

[next obsidian]

Maybe they are like ordinals…

I guess it depends on if you think all bases are ordered

[cobalt heath]

Hmmmm

[mighty spade]

In hindsight, if it was a polynomial ring you should just apply hilbert over and over

god, what was I thinking

[crystal turtle]

cardinals are in bijection with ordinals

[next obsidian]

What

[crystal turtle]

what do you mean what

[cobalt heath]

Can this be proved with ZF?

[next obsidian]

🤫

[crystal turtle]

Uhh I don't think so? Not 100% sure but I think you should need that every set can be well ordered for this

Or maybe I made it up to begin with We'll never know

[next obsidian]

I believe so

As in I agree with you

[alpine island]

the first thing that comes to mind would be infinitesimal calculus. That's kinda algebraic I guess.

[lusty marlin]

How is it algebraic?

[rocky cloak]

Probably depends how you define Cardinals without choice

[cobalt heath]

Hmm

[spice whale]

no in ZFC it's easy
for κ = aleph_η a cardinal ≥ aleph_ω², κ |→ η
for κ = aleph_η < aleph_ω² infinite, κ |→ η+1
for κ = n finite, κ |→ ω•n

that's a bijection Card→Ord

i don't think it'd be possible in ZF since you potentially have non-well-orderable cardinals

[abstract rock]

algebraic analysis is a thing, its more so about considering analytical things from complex functions

[rotund aurora]

arent you just using here that cardinals are labeled by the ordinals?

[spice whale]

i think you can use that every cardinal is either a successor cardinal or the union of a set of smaller cardinals

[tardy hedge]

Why doesnt Sym(Z5) have 5! Elements

I thought that just meant the set of permutations on Z5

Z5 have 5 elements so isnt it 5! Choices

[sly crescent]

It does

[tardy hedge]

Ok, then i think my prof misunderstood my question

[spice whale]

they probably thought of Aut(C_5)

the group of automorphisms of C_5

[tardy hedge]

We were just doing cayleys theorem stuff. This was showing how Z5 isomorphic to Sym(Z5) i guess

[spice whale]

which has 5 elements i think

[tardy hedge]

Showing the pictures of the elements of Z5 permuting around

[sly crescent]

4, actually

[spice whale]

yes

[tardy hedge]

And i asked something like these permutations are a subset of Sym(Z5)?

Cuz these are 5 perms

But SymZ5 has 5!

[spice whale]

Sym(5) has 5!

Sym(Z5) is confusing

[tardy hedge]

Didnt u guys literally just say Sym(Z5) does have 5!

[sly crescent]

Wait

[crystal turtle]

While you can talk about the symmetric group of any set technically, stuff like Sym(Z5) is avoided since it's confusing yeah

[spice whale]

it's unclear whether you're talking about just permutations, or group automorphisms

[sly crescent]

@tardy hedge by Z5, you mean a group of order 5, right?

[tardy hedge]

I meant integers modulo 5

[sly crescent]

Okay

That’s a group of order 5

[tardy hedge]

Ok

So what am i missing

Idk like i just thought writing Sym(Z5) is the same as S5

[delicate orchid]

you're not fixing the identity judging by ur little petagram

but again Sym(Z_5) is very non-standard, you have to clarify what that means

[sly crescent]

There are 5! permutations of Z5’s elements, but most of them aren’t automorphisms

[delicate orchid]

^

[spice whale]

Sym(Z5) could also potentially mean the group of automorphisms of Z5
which are isomorphisms Z5 → Z5

i.e. bijective functions f such that f(x+y) = f(x) + f(y)

[rocky cloak]

Z/5 is not isomorphic to Sym(Z/5), but it is isomorphic to a subgroup. Cayleys theorem says that any group G is isomorphic to a subgroup of Sym(G).

[delicate orchid]

oh so that's what's going on

[tardy hedge]

How do i do 13

I can see how they have the same multiplication table

But idk how to create the “function”

[charred flume]

maybe use the determinants for each matrix and map it to an element of S_3

not sure

[delicate orchid]

map (1,0)(1,1) to (123) and (1,0)(0,-1) to (12)

an isomorphism is just a relabelling of whatever you've got on the top of your multiplication table so just write down that relabelling

[tardy hedge]

Yeah i was thinking of literally just manually writing out the relabeling

But to show the operation is preserved do i really need to go thru each pairing to show it works?

[delicate orchid]

if you've written out the table then you've already done that

[tardy hedge]

I see

[charred flume]

yeah right you can just do this, you just explicitly states the map for each element

[delicate orchid]

or do it on the generators

and then show that the relations in G are satisfied by whatever you've sent them to

[tardy hedge]

Im definitely getting lost in the sauce now

Going too fast , i want more time with each topic before moving on

Like when this course is over i just wanna reread the whole book over again lol

[topaz solar]

I mean, what exactly is meant by cardinal is a key question too imo

[rocky cloak]

Unhelpful advice: PGL2 acts (faithfully) on the projective line and G is the isotropy group of infinity. And PGL2 acts 3-transitivitely, so G acts 2-transitively.

Helpful advice: if you know they have the same multiplication table you can just litterly write what every element maps to. There's only 6.

[tardy hedge]

Haha at what point would one be able to understand and appreciate the “unhelpful” advice

[delicate orchid]

give it a year

who calls it the isotropy group tho.... strange!

[rocky cloak]

Good question. When I've TAed into to abstract algebra I've given as exercise to compute the isotropy groups of GL2 on the protective line as exercises. So around the end of a first abstract algebra course I guess

Not so obvious that it's 3-transitive I guess, but not very hard to prove

[rocky cloak]

Sounds a little more fancy then stabilizer.

[delicate orchid]

true!

[rocky cloak]

If we didn't use these fancy words it would all be obvious and I'd be out of a job

[delicate orchid]

all of alg-geo

[cloud solar]

Hello

I have a problem

Let A be a finite non trivial ring. Show A is not a field if and only if the equation x^n + y^n = z^n has solutions in A* for all natural numbers n>=1

The second case "right to left" is easy. Supposing A is a field then for all x in A* , x^(q-1)=1 , where q is the order of A

And for n=q-1 we get 1=2 so 0=1 for all x,y,z in A*

So we have a contradiction

So A is not a field

For the "left to right" case, i said: since A is a finite ring then we have only invertible elements and 0-divisors. So since A is not a field we have an element a in A* s.t. there is b in A* with ab=0.

We suppose there is n>=1 s.t. The equation does not have any solutions in A*.

Let S={x^n with x in A*} (for that fixed n)

I want to get a contradiction somehow

We know 2a^n≠a^n right?

So a^n≠0

Hm

We also know b^n≠0

Any idea? I was thinking that set S is helpful somehow

[rocky cloak]

If you can force ba to also equal 0, then you can just consider (a+b)^n

And ||if ba is nonzero, then (ba)^2 = 0||

[cloud solar]

And how this helps

[rocky cloak]

Like what do you get if you expand (a+b)^n ?

[cloud solar]

a^n + b^n

If ab=ba=0

[rocky cloak]

Exactly

So that would give you a solution

[cloud solar]

Oh

So the problem is to show ba=0

[rocky cloak]

Yes, or deal with the case ba \neq 0 seperately

[cloud solar]

If ba is not 0

(ba)^2=(ab)^2=0

[rocky cloak]

Exactly, and you already dealt with the case where you have nilpotent elements

[cloud solar]

Can you explain

[rocky cloak]

If a^n = 0, what can you say?

[cloud solar]

a nilpotent

And that is a 0-divisor

[rocky cloak]

Yes, but what can you say about the equation

[rocky cloak]

You said it here

[cloud solar]

We get a contradiction

[rocky cloak]

Sure, or rather a^n + a^n = a^n would be a solution

[cloud solar]

Oh yeah. So if we have a nilpotent element the equation has a solution

So ba nilpotent

So (ba)^n+(ba)^n=(ba)^n if ba ≠0

If that n is >= 2

And if n=1 it is trivial

So if ba=0 then a^n+b^n=(a+b)^n
If ba≠0 we know (ba)^2=0 and (ba)^n+(ba)^n=(ba)^n.

But for the ba=0 this works only for a≠-b. If a=-b we have -a^2=0 so a^2=0 and again a nilpotent element and we are done

I asked this problem because this is the only solution i found on aops

And at the moment I dont know what is the jacobson radical or about artin-wedderburn theorem

[rocky cloak]

Yeah, a bit more heavy machinery then necessary

Maybe you should post your solution to aops (idk how that forum works)

[stark helm]

Condition:if I define ei with 1<=i<=n be the standard basis for R^n, and give a permutation sigma in Sn, define A_sigma: R^n to R^n as linear transformation such that A_sigma(ei)=e_sigma(i). Confusion: 1.Does standard basis for R^n means a n vector. 2. What does A represent, is it a matrix or a vector or something else? And I didn't see any relatoin between symmetric group and A here 3. What will e_sigma(i) look like here? I still want to know how will it related to symmetric group?

[rocky cloak]

1. The standard basisvector e_i is the vector with 1 in position i and 0 everywhere else. For example for R^2, the standard basis consists of (1,0) and (0,1).

2. A is a matrix yes, or a linear transformation anyway. It maps the basis vector e_i to e_sigma(i)

3. e_sigma(i) will be another basisvector. For example if n=2 and sigma is the permutation that swaps 1 and 2, then e_sigma(1) = e_2

So A_sigma is just the linear transformation that permutes the basis vectors according to sigma.

[stark helm]

I think I have understood that, since i is in the interval [1,n], if I say sigma(i)=a where a is in the interval [1,n], and another alpha in Sn st alpha(i)=b, where b is in interval [1,n]. And it produces e_sigma(i)=e_a = a standard basis while e_alpha(i)=e_b=another standard basis? Is that true?

Actually I have anothe question, given the previous condition, what does sgn(sigma) represent, I don't know what does sgn mean here?

[night onyx]

Well whatever it is, it's some number associated to a matrix

what are some possible answers?

[stark helm]

I think it must be an integer? but I still don't know what does sgn be like, Yes it must be related to a matrix as I mentioned A above

[night onyx]

yeah the sign is either 1 or -1, depending on how many swaps the permuation does

what's a number you associate to a matrix which swaps signs when you swap columns?

[stark helm]

Got it, really appreciate that

[stark helm]

Question: If there is a cyclic group G with exactly 3 subgroups, then i consider the subgroup of the group G must be a cyclic group and say identity can be one of them. The other subgroup is just {<g>} and determine g has the smallest order 4. In that case, I can construct {e}, {e, g, g^2, g^3, g^4} and {e, g^2, g^4}, I exclude all possible number that is not one of <g> because it doesn't satisfy closure. And smallest order larger than 4 can produce more than 3 subgroups, Do you guys think this is the correct statement?

[stark helm]

What does the number associated to a matrix mean? But I actually feel wondered if sigma means k from sigma(i)=k?

[chilly radish]

He's asking if you know of a function that assigns a number to a matrix such that it changes sign when you swap a pair of columns

[stark helm]

I know that, it should be related to determinant?

[stark helm]

Then I think it should be related to determinant of A_sigma st A_sigma(ei)=e_sigma(i), so I am wondering if sgn(sigma) means the result k produced by sigma(i)?

[night onyx]

No, sgn(sigma) can only be 1 or -1, it isn't the value produced by some sigma(i). sigma is a permutation of the numbers {1,2,3,....,n} so $$\sigma (i) \in {1,...,n}$$

[cloud walrus]

kryojyn

kryojyn

[night onyx]

so in particular if sgn(sigma) = -1, that clearly isn't one of the numbers 1,..., n

[tardy hedge]

Whats the difference between Sn and Sym(S)?

If S has n elements, is there any difference ??

[lusty marlin]

Sₙ is isomorphic to Sym(S) if S has n elements.

So this boils down to asking what the 'difference' is between two isomorphic groups.

[abstract rock]

Is S a set or a group as well

[lusty marlin]

Does that matter?

[hexed bloom]

Hi, I just want to ensure something: is it correct that if we have two rings $A\subset B$ and an element $\alpha\in B$ integral over $A$, then there exists $n\in\mathbb{N}$ with $A[\alpha]=A+A\alpha+...+A\alpha^n$ ?

[cloud walrus]

Gibzen

[night onyx]

not much, Sym(S) is the group of all bijections of the set S, S_n is in particular Sym({1,...,n})

If S has n elements, say S = {s_1, ..., s_n} then Sym(S) is isomorphic to S_n

[abstract rock]

the set of all bijections between group and itself is not the same as the set of all automorphisms of a group

[night onyx]

so in that case I guess it doesn't matter

regardless of any additional structure on S, Sym(S) is just bijections

[lusty marlin]

I know that, the notation for the automorphism group is usually Aut(G), not Sym(G).

So they weren't talking about automorphisms, if we assume this notation

[abstract rock]

yeah but its important to clarify in case the assumption isnt justified

[rocky cloak]

yes thats right.

[stark helm]

Can anyone see if that's true?

[rocky cloak]

The statement is a little hard to parse, but it is true that Z/4 has exactly 3 subgroups. Z/4 is not the only such cyclic group though, if thats what youre saying

[terse crystal]

I see no question. You typed “question: …” but there is no question after it

Cyclic group of order n=Πp^r, number of its subgroups, equals number of factors of n, (=Π(r+1))

[steel pulsar]

Exercice : Let G be a group and let H be a subgroup of G such that G/H has Exactly two elements. Show that H is a normal subgroup in G
Correcion :

What I don't understand :
To be honest, all the reasoning, I need someone to explain it to me step by step, please, because I'm lost!!

[delicate orchid]

Let gH, H be the two cosets. So Hg is either gH or H, assume Hg = H, find the contradiction etc. etc.

[steel pulsar]

?

[rocky cloak]

Cosets partition the group. This means that any element of G is either contained in H or gH.

Similarly every element of G is either contained in H or Hg.

Do you see why this means gH = Hg?

Do you see why the fact that gH = Hg means that H is normal?

[tardy hedge]

How can we tell whether the group units of Zn is cyclic or not

Group of units Z5 is cyclic, but group or units Z8 is not

How can we check without just computing powers of elements and seeing if it works?

[steel pulsar]

Because by definition G / H = {H, gH} because the group has 2 elements so that's why we say that each element of G is contained in g or gH, no ?

And for Do you see why this means gH = Hg?Do you see why the fact that gH = Hg means that H is normal? I don't know...

[stark helm]

I actually see the statement that G is a cyclic group of exactly 3 subgroup, then order of G is square of prime number, so I use phi(n)=4, but it turns out that no n can be the square of prime, can you see some problems here?

By the way, I actually can not find other kind of cyclic group that has exactly 3 subgroups, could you give me an example?

[rocky cloak]

Z/9

[stark helm]

I actually want to know if my idea is correct, and could you explain how to know about number of its subgroups in details? Πp^r means r from 0 to n?

[rocky cloak]

Square prime order are the only ones yeah

[stark helm]

got it, it seems work

[steel pulsar]

seems so hard, 2 days trying to figure it out... 🥺

[rocky cloak]

So if we multiply both sides of the equation by g^-1 we get
gHg^-1 = H

Which I'm guessing is your definition of normal

[steel pulsar]

but here H and h is not the same

[delicate orchid]

The subgroup lattice of C_n with n = p_1^k_1…p_m^k_m is the tensor product of chains of length (k_i+1). Not 100% relevant but very cool

So with n = p^2 we have a chain of length 3, so 3 subgroups

And since 3 is prime these are the only cyclic groups that can have 3 subgroups

[steel pulsar]

I don't understand H and h is not the same in my correction

They say we have G/H = {H, gH} with g ∉H. Let's show that gH = Hg. Let h ∈H, we have ghg^(-1) ∈H . Indeed :

ghg^(-1) ∉H => ∃h' ∈H : ghg^(-1) = gh' => g = (h')^(-1) h ∈H (contradiction)

We have now : gh = ghg^(-1)g from where gH ⊆ Hg

As Hg is disjoint from H (the complement of gH) we have gH = Hg

This is what the correction says (I can't understand why I'm here)

[rocky cloak]

So your definition is that ghg^-1 is contained in H for all h?

If so, just not that ghg^-1 is in gHg^-1 which equals H

[steel pulsar]

Sorry I don't understand ... 🥺

[tardy hedge]

Ok so i recall that if n is prime then its cyclic because all groups of prime order are cyclic

Nevermind^

Its not n thats prime

[delicate orchid]

Do you know about the fact that U(n) is isomorphic to Aut(Z/nZ) yet

If not it’s very easy to see, U(n) acts by multiplication on Z/nZ which you can prove is a group automorphism

[rocky cloak]

Okay, let's try to decode your picture instead. So assume for the sake of contradiction that ghg^-1 is not in H. Since every element is either in H or gH we must have ghg^-1 in gH

So ghg^-1 = gh' for some h'. Cancelling g we see that hg^-1 = h' and thus g^-1 = h^-1 h'. But h^-1 h' is in H, while g^-1 is not. Contradiction!

[tardy hedge]

Not yet, we will prob see that later tho

[steel pulsar]

OK so many questions I have : 1. You said that each element is either in H or in gH why ghg^(-1) is not in one of the two? In reality he is in one of the two but as you suppose you say to yourself that it is not, right?

2. Why do we have to have ghg^(-1) in gH and not in H?

[tardy hedge]

Ok so in this question i see that U(7) and U(14) should be isomorphic and i wrote the function to just send the generator in 7 to the generator in 14

[steel pulsar]

3. Why is h^(-1)h' in H and g^(-1) is not in H?

[tardy hedge]

My question is that is checking that operation is preserved trivial because both groups have the same operation in the first place?

[steel pulsar]

4. Having this contradictory conclusion, what does this say about the final answer? (Show that gH = Hg)

[delicate orchid]

If g^-1 is in H then because H is a subgroup it’s closed under inverses so g^-1^-1 = g would be in H

h^(-1)h’ is in H because H is closed

[rocky cloak]

Every element is in some coset, H having index two means that there are only two cosets. One is H and if g is not in H then the other will be gH (as that is the coset containing g)

2. If ghg^-1 was in H for all g, then H would be normal, so then we would be done.

3. h^-1 h' is the product of two elements in H, thus is in H. g is assumed not to be in H, otherwise ghg^-1 would be in H, and we would be done.

[rocky cloak]

I can't read french, are you saying your definition of normal is that gH = Hg?

[wooden fulcrum]

Hello I have a quick question (don't want to interrupt the co nvo), is there any semigroup homomorphism that doesn't map everything to a non-unit idempotent element that isn't also a monoid homomorphism?

[delicate orchid]

I think such a map is essentially the zero map into a monoid

[rocky cloak]

You mean you have two monoids and you want a homomorphism of semigroups between them? Sure only the identity has to map to an idempotent, other stuff doesn't have to

[delicate orchid]

You’d have f(xy) = f(x) for all x, y in your semigroup, not sure how you would consider this a monoid map though

[rocky cloak]

For example r |-> (r, 0) from R to RxR

[delicate orchid]

Oh I misread

Ignore me

[rocky cloak]

(R being real numbers under multiplication)

[wooden fulcrum]

I mean that the counterexample I saw (with 2 monoids G and H) regarding why a homomorphism of semigroups wouldn't be a homomorphism of monoids was that it'd map everyting to some idempotent in H, then it'd be a homomorphism allright but wouldn't be preserving identity

[rocky cloak]

That's one counterexample sure

[wooden fulcrum]

but if it doesn't map everything to an idempotent I thought
f(a . e_G) = f(a) = f(a) * f(e_G)
would have to hold for all a and f(e_G) hence wouldn't that make f(e_G) identity under the new group?

ah I see it now I think

the issue is that it's not surjective so there can be some non-identity that does it for all the images of the a s

right

[rocky cloak]

That's right, if it was surjective then it would map 1 to 1

[wooden fulcrum]

epic

thank you guys

[reef trench]

Hi, a quick question: this is basically Schroder Bernstein theorem but for algebraic structures. If there exists an injective homomorphism from A to B, and an injective homomorphism from B to A, then does that mean A and B are isomorphic?

[steel pulsar]

ghg^(-1) ∉H => ∃h' ∈H : ghg^(-1) = gh' => g = (h')^(-1) h ∈H (contradiction)
We have now : gh = ghg^(-1)g from where gH ⊆ Hg
As Hg is disjoint from H (the complement of gH) we have gH = Hg

this part

We told ourselves that g^(-1) = h^(-1)h'

[rocky cloak]

Free groups should give you a counter example

[steel pulsar]

so we have contradiction

but after the contradiction they conclude as what

gH ⊆ Hg and gH = Hg

[reef trench]

Could you elaborate please?

So we have two free groups A and B

[steel pulsar]

In fact, all this progress that they have made in the correction is to show that gH = Hg, do'nt know why

[rocky cloak]

Every free group on countably many generators embeds into the one on two generators

(and the free groups on differently many generators are not isomorphic)

[rocky cloak]

Since ghg^-1 is in H, ghg^-1g (=gh) is in Hg

[rocky cloak]

What is your definition of normal then?

[steel pulsar]

if ghg^{-1} ∈ H

[daring nova]

basically to show that normal subgroups are the only groups for which the quotient set is a group. But with H having index 2

[rocky cloak]

Alright, but then you can stop the proof immediately when you reach the contradiction

[steel pulsar]

But why we said ghg^(-1) is not in H but in gH, to say at the end that ghg^(-1) is in H

ah

it's supposition

It's not true

[rocky cloak]

If you assume something, and reach a contradiction. Then it's false

[steel pulsar]

Ok so we assume that ghg^(-1) is not in H to prove that it is in H

?

[steel pulsar]

and after we say this, the proof is end ?

[rocky cloak]

There are two possibilities. Either ghg^-1 is in H or it is not in H. If when we assume that it's not in H and it turns out that's impossible, then the only possibility left is that it is in H.

[steel pulsar]

OK, thank you very much for your help, it's all a little clearer thanks to you. It's all a bit complicated, I have the impression of not being able to get through it when we don't apply " method" in mathematics

[steel pulsar]

Do you have any advice for thinking carefully about abstract algebra on questions like that?

The only definition to know is that of the normal subgroup and then?

[summer path]

I mean many of the group theory problems you see in a first course are kind of just follow your nose and use the definitions

But you get more used to it through practice

[austere escarp]

What are groups actually used for?? I’m a couple months into my group theory class but honestly I can’t see how they’re actually applied. We’ve covered dozens of theorems and definitions and properties and structures, but how is it actually used? And I don’t just mean like “in quantum mechanics” or smthn, like literally what are you doing/whats an exact problem it solves and how

[rocky cloak]

One combinatorial application of group theory is Burnside's lemma, which can be used to solve problems of the form:

How many ways can you color the sides of a cube (or any other symmetric shape) using some fixed number of colors.

So you have a quite complicated counting problem, that is reduced quite a lot by looking at symmetry groups and group actions

[alpine island]

What's the 'closest' a ring can be to a field while still having zero divisors?

[rocky cloak]

Commutative artinian I guess

Edit: so things like products of fields or F[x]/x^2 and things like that

[rocky cloak]

There are also a few groups that appear in cryptography. You can look up Diffie-Hellmn key exchange for example

[rocky cloak]

Noether's theorem in physics gives a bijection between preserved quantitees of a system and "infinitesimal" symmetries. So you can use group theory to compute preserved quantitees

[alpine island]

I was thinking about dual numbers too lol

[rocky cloak]

Guess it depends on the setting you're interested in, what "closest" should mean. But at least in representation theory going from finite dimensional to finitely generated over commutative artinian ring is a natural "minimal" step

Of course you can add even more adjectives if you like

[alpine island]

I am taking my first round of algebra courses, and I just think zero divisors are a neat concept. Generally though we talk a lot about integral domains, so I was wondering how much you can do without that restriction

[rocky cloak]

I see, well maybe F^2, F[x]/x^2 and Z/p^2 can be useful examples to keep in mind

[alpine island]

Z/p^n is good for number theory

[topaz solar]

Superstable semisimple rings are pretty close

Direct sum of a finite ring & finitely many matrix rings M_n(F) over algebraically closed fields

If it’s commutative, that’s pretty killer

(They’re furthermore of finite Morley rank, so \omega stable too)

Infinite superstable fields are algebraically closed

“Close,” as jagr said, is a bit less than objective I feel like, since this is clearly different than his idea

[crystal turtle]

Groups are also very useful in cryptography. For instance, basic cryptosystems might use the multiplicative group of a finite field to encrypt information, and they would use some of the basic theory such as Lagrange's theorem to decrypt that information. You can talk also about other cryptosystems such as in elliptic curve cryptography, where to a polynomial y^2 = x^3 + ax + b you assign a way to "add" points on the curve, giving a group. I don't know as many details about this second approach though lol

[void cosmos]

diffie-helman

elgamal

all of this uses finite cyclic groups of order p

basic group theory is so useful for basic crypto

[crystal turtle]

Yup.

[white oxide]

how exactly is this clear/non-tedious to show? (in particular, closure under multiplication)

i tried writing out a proof but it seems very tedious

are there some elementary facts that i'm missing

[coral spindle]

You literally do not need to care about the fact it's polynomial

This is totally the wrong approach

[white oxide]

:(

[coral spindle]

You have that sigma f = f for all sigma. Show that that is preserved even when you multiply and add maps.

This has nothing to do with polynomials.

[white oxide]

ok wait now i'm confused

bc we have to show that if f and g are symmetric

then fg is symmetric

no?

[rocky cloak]

When in doubt go simple, what happens when n=2

Then symmetric means that f(x, y) = f(y, x)

[coral spindle]

Let h = fg. What does h(x,y) look like.

[white oxide]

ohhh right

🤦‍♂️

lmao okay thanks that clears things up

[coral spindle]

Great

[white oxide]

that was embarssing

sanity check: $f$, $g$ symmetric implies $fg(t_{\sigma(1)}, \dots, t_{\sigma(n)}) = f(t_{\sigma(1)}, \dots, t_{\sigma(n)})g(t_{\sigma(1)}, \dots, t_{\sigma(n)}) = f(t_1, \dots, t_n)g(t_1, \dots, t_n) = fg(t_1, \dots, t_n)$ right

[cloud walrus]

okeyokay

[white oxide]

also is there a formula for the elementary symmetric polynomials? lang only gives s_1 = t_1 + \dots + t_n and s_n = t_1 \dots t_n

[coral spindle]

The wikipedia page has a nice list of the first few and a general pattern: https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial

[south patrol]

You can also nicely define them via $\prod_{i=0}^{n} (1 + x_i t) = \sum_{i=0}^{n} e_i(x_1,\dots,x_n)t^n$ which is listed in the properties section

[cloud walrus]

potato

[south patrol]

(i say this because the formula is also interesting for other reasons)

[tardy hedge]

Super dumb question incoming

Why does “it follow” that N is a divisor of ord(G)?

[next obsidian]

Definition of lcm

[steel pulsar]

Hello

Question Construct an isomorphism between the quotient C* / {-1, 1} and C*

Correction

We consider C* → C* given by z → z². It is a homomorphism of surjective groups, with kernel {−1, 1}. By first isomorphism theorem, it realizes an isomorphism C/{±1} ≃ C

My questions :

1. Why in this problem do we seek to find surjectivity?

2. Why φ(z) = 1 <=> z² = 1 ? And why we choose the value of 1 ?

3. I don't understand why say that z² = a then z ≠ 0 helps solve the problem (last step)

These are my 3 questions ...

[coral spindle]

1. because we need it for the first isomorphism theorem to give us what we want. I suggest you look at the statement of the first isomorphism theorem now.

2. phi(z) = z^2 by definition. And we choose the value of 1 because 1 is the identity element of C^*.

3. You're right, it is irrelevant to say that z is nonzero.

[steel pulsar]

2. Why phi(z) = z² by definition ? I don't understand

3. So the conclusion of my teacher is not good ?

[steel pulsar]

Yes but it's the correction

not written in the statement

[coral spindle]

So? It is still z^2 by definition

[steel pulsar]

Statement : Construct an isomorphism between the quotient C* / {-1, 1} and C*

[steel pulsar]

Yes I believe you, I want to understand why

[coral spindle]

It is z^2 by definition because they defined it to be so in the answer. You could have defined it to be something else, but this is irrelevant. There are other solutions, but this is one.

[steel pulsar]

And what criterion must this choice respect ? We can't take everything

[coral spindle]

That it is an isomorphism.

[steel pulsar]

Isn't this related to C*?

[coral spindle]

Well I should say that there's no obvious reason there shouldn't be other solutions. I'm now not going to think about whether or not there are others.

[coral spindle]

Idk what you mean by this. Indeed we are looking at the group of units of C

Oh perhaps I see the confusion wrt point 3.

In fact yes, I do see

Let me correct my evaluation

3. Recalling that C^* = C \ {0}, we need to argue surjectivity by finding a nonzero z such that z^2 = a, so indeed we do need to point out that z =/= 0.

My mistake in not seeing this.

[steel pulsar]

And why after saying that can we conclude directly with the final answer? It bothers me, like the feeling that something is missing

z ≠ 0 then ∃ z ∈ C* , φ(z) = a

Then C* / {-1, 1} ≈ C*

[coral spindle]

Because solutions to nonconstant equations always exist in C.

This is called the fundamental theorem of algebra. A bad name, but memorable.

[steel pulsar]

My problem is to know why we can directly use the 1st isomorphism theorem after having said that

[coral spindle]

We've calculated the kernel and shown the homomorphism is surjective, so we just substitute this information into the statement of the first isomorphism theorem.

I hope it's clear that phi is a homomorphism

[steel pulsar]

it's a bit complicated

Thank you!

[white oxide]

why exactly are we allowed to apply the induction hypothesis here? since it may be the case that f(t_1, \dots, t_n-1, 0) is of degree d still right

also if we're applying induction shouldn't the weight be <= d - 1?

[coral spindle]

remind me what weight means?

[coral spindle]

Thanks

[white oxide]

ye

[white oxide]

since the term with the highest degree need not have t_n in the product right

[coral spindle]

The induction hypothesis is for n, not for d.

[white oxide]

wait wdym

he starts inducting on d in the last paragraph of the photo i sent

[coral spindle]

Oh right.

Yup, still applies.

We're applying the hypothesis for the lesser n, not for the lesser d.

[white oxide]

ahh, i see

so if it's less than n then we can consider any degree d right

"induction with respect to n"

[coral spindle]

Yes. If you need to, write down the induction hypotheses explicitly. There are two at the point you underline.

[white oxide]

lang proofs are mad confusing since he's not explicit and what he's doing but it's kinda fun to figure out

[coral spindle]

I think this is perfectly explicit, double induction is simply alien to many.

[white oxide]

indeed you are a lot better than me at math

[coral spindle]

No, I have simply seen double induction before

OooOOoh understanding is simply recollection in disguise oooOoooOOh

[white oxide]

lmao

why does f_1 necessarily contain t_1...t_n as a factor? i understand that now we're applying the induction hypothesis w.r.t d (i.e. we can say that f_1(t_1, \dots, t_n) = g(s_1, \dots s_n) so that s_n = t_1....t_n appears in at least one term, but how do we know that it appears in every term?

[next obsidian]

It’s symmetric

[coral spindle]

[next obsidian]

It contains each ti as a factor

So the product

[white oxide]

huh ok

[coral spindle]

f = g * t_n is symmetric. Apply a permutation.

[steel pulsar]

just last question: why phi is a homomorphism?

(i know the definition of homomorphism)

[coral spindle]

phi(xy) = phi(x)phi(y), that's all that is required.

[steel pulsar]

Is there no need to apply it specifically to exercise?

[coral spindle]

I don't know what you mean by that.

[steel pulsar]

Well, maybe it seems obvious

[coral spindle]

We do indeed need phi to be a homomorphism, otherwise the first isomorphism theorem does not apply. Is that what you were asking?

[white oxide]

am i dumb? why does the bottom polynomial not contain X_1X_2X_3 as a factor then?

[coral spindle]

Because it's not divisible by any X_i

The theorem does not claim that every symmetric polynomial has a factor as you say.

[white oxide]

hmm ok

[cobalt heath]

Yea basically you need to gather the factors.

[white oxide]

okay i'll come back and have a think it seems trivial enough but for some reason i can't understand it

thanks

[next obsidian]

@white oxide if you believe f_1 is symmetric this is what’s going on

f_1(t_1,…,t_n) = f_1(t_1,…,t_i-1,t_n,t_i+1,…,t_n-1,t_i) because it’s symmetric

Then, because f_1 is divisible by t_n we write f_1 = t_n•g

By the relation above, we have also that f_1 = t_i•g(t_1,…,t_i-1,t_n,t_i+1,…,t_n-1,t_i)

So f_1 is divisible by t_i for all i

This means it’s divisible by the product t_1•…•t_n by general UFD stuff

[steel pulsar]

For this question : Show that the group U of complex numbers of module 1 is
isomorphic to the quotient group R/Z

In my correction he chooses x -> e^(2πix) why ?

[stark helm]

Question: show that the center of the group GL(2,R) is the group Z:{aI:a in R} and i menas identity. So I pick arbitrary A in GL(2,R) as 22 matrix with real entries, and show that (aI)A=A(a*I), but the comment writes it does not show every element of the center must lie in Z, but I don't understand what does the comment mean? Could anyone explain that?

[lusty marlin]

You have shown that every element of Z belongs to the center, by showing that it commutes with an arbitrary matrix in GL₂(ℝ). You have not shown that every element that commutes with every matrix in GL₂(ℝ) is in Z.

[stark helm]

So I need to show that only form of a*I can be center of G while other form can not be the center of G right?

[lusty marlin]

Yes

[stark helm]

Then suppose a form not aI, and let aB where B is another form of 22 matrix, and because BA is not generally equals to AB, so B must not commute with every matrix A, hence getting the fact that only aI work? Is that correct?

[remote nymph]

i am so sorry to interrupt this but im dying on a pset due tonight

[lusty marlin]

No. The statement 'because BA is not generally equal to AB' isn't a valid step in a proof. You need to consider an arbitrary matrix B that is not a scalar multiple of I, and demonstrate an example of a matrix A in GL₂(ℝ) that does not commute with it.

[stark helm]

Then I consider arbitrary matrix A[a b c d] and B=[3 1 2 7], and AB=[3a+2b a+7b 3c+2d c+7d] while BA=[3a+c 3b+d 2a+7c 2b+7d], only 2b=c will satisfy AB=BA, then pick another matrix C=[1.5 2 1.4 1.7] and BC does not match CB, which means arbitrary matrix B can not commute with matrix in GL(2,R)

Is that correct?

[hasty shore]

can anyone help me with this?

Since it is given a algebraic system(a general case).
I cant use associativity and idenetity right?

How should I start this proof

[lusty marlin]

It follows directly from the rules that the given operation satisfies

[hasty shore]

r you saying that it is like whatever the first term answer will always be the second tem?

[lusty marlin]

It would appear so, yes.

[hasty shore]

ahh got it ty

[stark helm]

Is it correct?

[stark helm]

question

[lusty marlin]

You just showed that this specific matrix B does not commute with every matrix of GL₂(ℝ)

You need to show that for each matrix that is not a scalar multiple of I, there exists a matrix in GL₂(ℝ) which does not commute with it.

[terse crystal]

Hint: A in center then A commutes with I+Eij, then A commutes with Eij where Eij is the matrix with (i,j) element 1, others 0

[last spoke]

Okay so I am completely lost on how to determine the discriminant of a quadratic number field (\mathbb{Q}(\sqrt{D})). We've got to prove the whole (d = D \leftrightarrow D \equiv 1 \mod 4) and (d = 4D \leftrightarrow D \equiv 2,3 \mod 4)

[cloud walrus]

StarvinPig

[terse crystal]

It follows from definition. Recall that the ring of integers of Q(sqrt(D)) is Z[sqrt(D)] when D=2,3 mod 4
is Z[(1+sqrt(D))/2] when D=1 mod 4 and there are only two embedding (identity and conjugation)

[last spoke]

I do not recall

[terse crystal]

In algebraic number theory textbooks

Check any of them

[last spoke]

Not in our one

I'm vaguely recalling it from lecture but like, I've been lost for over a month

[last spoke]

Reading the question we might be asked to prove this

[terse crystal]

So we prove it:

An element a+bsqrt(D), satisfies x^2-2ax+a^2-Db^2=0. So this element is integral iff 2a, a^2-Db^2 are integers. Meaning 2a, 2b are integers

So two cases: a, b from Z, or a=(2r+1)/2, b=(2s+1)/2. The latter case a^2-Db^2 from Z gives you D=1 mod 4. So if D=2,3 mod 4, ring of integers is Z[sqrt(D)]

If D=1 mod 4, (1+sqrt(D)/2 is integral (r=s=0) and ring of integers is Z[t] where t=(1+sqrt(D))/2 since sqrt(D)=2t-1, and (2r+1)/2+((2s+1)/2)sqrt(D)=r+s sqrt(D)+t

QED

[last spoke]

I kinda get that

No way in hell I could do it on my own but I get what you're doing

[terse crystal]

A classical result. this proof I have seen in textbooks

If people generally can prove it easily it would become an exercise rather than a theorem kind of things

[last spoke]

I think for us its an exercise

[terse crystal]

Lol

[last spoke]

For context, this is what I gotta do

Technically due a week ago but like, its exam season and this lecturer don't care

[terse crystal]

Interesting it does it backward

It proves the discriminant thing using some other method then use that to obtain integral basis. Interesting, I don’t know how though. I obtained integral basis first

[last spoke]

I wouldn't be surprised if going backwards was the correct method

[terse crystal]

Probably

[last spoke]

I have a 55% take home test on this shit in under 2 weeks and I am not looking forward to that

[terse crystal]

We don’t need to care about lectures either. All things can be studied by ourselves since it’s math

Always tons of books available. We need no lectures

[last spoke]

I can't read for shit, this textbook melts my brain and the lecturer is no better

[terse crystal]

Then read another one.

Professors showing directions, recommending reading materials, people in this server clarifying doubts in our reading and also recommending reading materials. We don’t need lectures, especially for basis stuffs

[last spoke]

So I'm just wondering why just these 2 cases

[terse crystal]

Because 2a, 2b from Z gives you these two cases

[last spoke]

I get how (a \in \mathbb{Z}) implies (b \in \mathbb{Z}) by the same deal as (2a, 2b)

[cloud walrus]

StarvinPig

[terse crystal]

Yeah. a,b from Z gives you the first case

a, b from (1/2)Z - Z is possible only if D=1 mod 4, this gives us the second case

[last spoke]

How do you know no other cases

[terse crystal]

I mean, a,b from Z always gives us Z[sqrt(D)] is contained in the ring of integers. We have some extra integral elements a+bsqrt(D) when a,b are from (1/2)Z-Z. Possible only when D=1 mod 4.

Because a+bsqrt(D) integral iff 2a, 2b from Z…

[last spoke]

I get 2a, 2b

What I'm missing is how (a = \frac{2r + 1}{2} \implies b = \frac{2s + 1}{2})

[cloud walrus]

StarvinPig

[terse crystal]

a^2-Db^2 from Z

[last spoke]

So (\frac{(2r + 1)^2}{4} - Db^2 \in \mathbb{Z})

[cloud walrus]

StarvinPig

[terse crystal]

Yeah

Db^2=n/4. D integer

D integer then b has to be in (1/2)Z-Z

Power of 2 in b^2 is -2, so power of 2 in b is -1

Wait , thinking

Nvm

[last spoke]

(Db^2 = \frac{n}{4} \in \mathbb{Z}) so (Db^2 \equiv 0 \mod 4). Since (D) is squarefree, (D \not \equiv 0 \mod 4) so (b^2 \equiv 0 \mod 4). Therefore (b \equiv 0, 2 \mod 4)

[cloud walrus]

StarvinPig

[last spoke]

maybe?

Nvm it's not in Z

[terse crystal]

It’s just if b is from Z, a^2-Db^2 can’t from Z, contradiction

So b from (1/2)Z not Z, so of the form (2s+1)/2

[last spoke]

yea this way I can prove, because then (\frac{(2r + 1)^2}{4} \in \mathbb{Z}) so (4r^2 + 4r + 1 \equiv 0 \mod 4)

[cloud walrus]

StarvinPig

[terse crystal]

Yeah

[last spoke]

So got the 2 cases

Neat

So second case means (\frac{(2r + 1)^2}{4} - D\frac{(2s + 1)^2}{4} \in \mathbb{Z}) so ((2r + 1)^2 \equiv D(2s + 1)^2 \mod 4)

[cloud walrus]

StarvinPig

[last spoke]

And since ((2x + 1)^2 \equiv 1 \mod 4), (1 \equiv D \cdot 1 \mod 4) so (D \equiv 1 \mod 4)

[cloud walrus]

StarvinPig

[terse crystal]

Yeah

[last spoke]

Okay now I'm just blanking on how to build the ring of integers

[wooden fulcrum]

are there commutative semigroups that aren't cancellative?

if we have y * x = z * x for all x and it's commutative then that means the behavior of y and z is the same for * with all members of X hence they are identical no?

[rocky cloak]

The statement is not that the equality holds for all x implies y=z

It's that the implications holds for all x, y, z

[coral spindle]

There are indeed commutative semigroups that aren't cancellative. For example the subsemigroup {0, 1} of N, under multiplication.

This is not cancellative since 0*1 = 0*0 yet 1 is not 0.

[open sluice]

you are saying that x -> y * x and x -> z * x are identically equal for all x
this lets you conclude that y * and z * are equal as functions
but I don’t think you can use that to conclude that y and z are equal as group elements

[coral spindle]

Indeed it does not

[rocky cloak]

You can just define the operation as xy = 0 for all x and y for example

[wooden fulcrum]

oh right I understood cancellativeness wrong I think

so in the case where I know it holds for all x though I can cancel, right?

[open sluice]

think about this

[wooden fulcrum]

aren't all elements 0 then though?

[open sluice]

why should they be

[wooden fulcrum]

well they all have the same behaviour under the operation hence there is no difference between them?

[open sluice]

you're just assigning the value 0 to the product x * y for all x and y in your set

it's like defining f(x, y) = 0 for all x, y in C

[wooden fulcrum]

ah I'm looking at this completely wrong

[rocky cloak]

You have just noticed the reason why studying semigroups is stupid, congratulations

(I'm joking, but not really)

[wooden fulcrum]

so 2 elements in a group can be considered distinct even if they behave completely the same, idk why I thought otherwise

xddd

I was thinking of set transformations where you'd consider 2 transformations the same element if they do the same thing

right??

[open sluice]

ye

[rocky cloak]

In a group or a monoid this never happens

[open sluice]

err, that too

[wooden fulcrum]

yeah I know 1 gets rid of it

it's weird

[coral spindle]

Aside: this is why in the version of Cayley for semigroups, we have to look at the action on S^1, not on S.

[rocky cloak]

And the set of transformations of something is a monoid so

[wooden fulcrum]

that's what brought me to this xdd

well kind of I think

I wasn't sure why you couldn't do the injectivity without 1 and I almost asked but then I Realized it would have to be commutative

ahhhhh so ah I get it

if you didn't have 1 in S then you could be mapping multiple semigroup elements to the same transformations riiight

thank you lads and ladettes

if the questions are too simple/spammish feel free to tell me I just sometimes am not entirely sure and I get stuck because I feel like I shouldn't be moving through with the material until I understand the part

[coral spindle]

If people don't wanna answer the questions they just won't, dw about it

this is a fine place to ask simple questions

[tiny jolt]

is x^4+x^3+x^2+x+1 irreducible over F2?

[coral spindle]

,w is x^4+x^3+x^2+x+1 mod 2 irreducible

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by boytjie

[tiny jolt]

cool

[main needle]

what is this group? Where (a,b) is the free abelian group with generators a,b.

[tribal moss]

Subtract 2a+b from a+2b (or vice versa) to learn something interesting ...

[delicate orchid]

You can imagine the quotient group as the group where all of the matrices in H are set to the identity , so each element of the quotient group is determined by the value of the w, hence there are 2022. You can directly show this by showing that you can get from any (z, w)(0, 1) to (z’, w)(0,1) via multiplication by an element in H

[white oxide]

i'm a little bit confused, how can we view D as a function of t_1, t_2? or rather what are t_1 and t_2 here

[coral spindle]

CONNNTEEEEEEEEEEEXXXXTTTTTTTT

It's clear that D has some definition that you've not included here. How else would we be able to prove that D is equal to something?

[white oxide]

hmm ok let me post the CONNNTEEEEEEEEEEEXXXXTTTTTTTT

give me one sec

[coral spindle]

I will burn the world for the sake of context

[white oxide]

here's your CONNNTEEEEEEEEEEEXXXXTTTTTTTT

oh yea

[coral spindle]

Great there you go.

Self-solving problem.

[white oxide]

and t_1, .., t_n are algebrically independent

[white oxide]

wut

[coral spindle]

The t_i don't have to be algebraically independent, by the way. There's no requirement for that

[white oxide]

oh well it was just a condition lang fixed at the beginning of the chapter

or idk

he just said let the tis be algebrically independent

[coral spindle]

Well I think lang is full of shit or you're misunderstanding

Bc there's no reason that should be the case for an arbitrary polynomial

[white oxide]

na he said it at the beginning

anyways

help a poor man? [Please help]

[stone depot]

[Please help]
I want to show, group of order 9 is abelian, my approach is like
if any elem has order 9 then its cyclic and abelian too

ao consider all are having order 3,I read on stack that, if I show there is isomorphism between elements of group and their inverses, then the group is abelian, so clearly here is isomorphism as every element has its square as inverse, and vice versa
Am I correct?

[coral spindle]

so clearly here is isomorphism as every element has its square as inverse
This isn't clear to me, justify it please

[stone depot]

I mean every elem has order 3 so
in every subgroup there is a, a^2, e
so element a inverse is a^2

so clearly there is isomphism as for one element, there is only one inverse its square and vice versa

[coral spindle]

I disagree that it's obvious there is an isomorphism. There is obviously a bijection, but an isomorphism is not merely a bijection.

[stone depot]

yeah but I think it's kernel also consists of single element thats e only

[rocky cloak]

Squaring isn't usually a homomorphism

[coral spindle]

You can take it from here jagr

[stone depot]

yeah, now I am also finding a flaw in homomorphism, can u please provide a better way of proving isomorphism here

what I mean is a function which maps element to its inverse
in this case element a inverse is a^2 and element a^2 inverse is a

[white oxide]

what does lang mean by "hence we can write" is he saying that the fact that cont(bf) = bcont(f) for b in K (where K is the quotient field of A) implies that the displayed text is well defined or smt

like say he didn't write cont(bf) = bcont(f)

what's stopping him from saying we can pull out the content to get a primitive polynomial

[topaz solar]

There’s a non-cyclic group of order 9 I think

[delicate orchid]

Maybe!

Only way I can immediately recall (it’s the weekend. I’m not going to do any thinking) to show groups of order p^2 are abelian is using p-groups have non-trivial centre and G/Z(G) cyclic iff G abelian

[white oxide]

how are gcds in a UFD not unique?

well i guess if you include units in ur factorization or something huh

[coral spindle]

yes

[barren sierra]

yea they're always unique up to unit

the "greatest" depends on the ordering you use

so like for integers you can just order them in the normal way

but if you aren't working in a place where < is well defined

the typical ordering is you say a < b if a | b

and then in this order in say the integers you have that -1 and 1 are both gcd(2, 3)

where greatest is in terms of this divisibility ordering I stated

not <

[white oxide]

i see

yea that makes sense

thanks!

[last spoke]

I finished the question a hot 11 hours later btw

[barren sierra]

it be like that sometimes

[last spoke]

Well I slept in the middle of that

But yea

I have one question left but also like, I have no clue what's going on there even more than the previous one soooo

[rocky cloak]

One approach is to show that any p-group has nontrivial center, so in particular this means every group of order p^2 is abelian

Also, there is a nonabelian group (of order 27) where all elements have order 3. So that should show that your argument doesn't work.

[delicate orchid]

People tend to ask these “groups of order p^2 are abelian for a specific p” questions prior to the introduction of group actions so I’m not entirely sure how you’d prove this

[delicate orchid]

There’s more than one! There’s TWO! Isn’t life just wonderful

[rocky cloak]

I guess since it's only order 9 you can start listing group tables

[delicate orchid]

Oh wait you clarified exponent 3

[delicate orchid]

Life is HORRIBLE

[barren sierra]

what's the question?

[barren sierra]

what directions have you tried?

[last spoke]

I've played with the forward direction a little bit since I feel like I've seen something in class on it but honestly I'm just lost

[rocky cloak]

The case where the ring of integers equals Z[sqrt(d)] should be pretty straight forward I guess

Just check whether Z/p[sqrt(d)] is a field or not

[last spoke]

I mean define straightforward

[rocky cloak]

Like Z[sqrt(d)]/p = Z/p[x]/(x^2 - d), when is a quadratic polynomial irreducible over a field?

[last spoke]

Mind is drawing a blank

[barren sierra]

what would it mean for x^2 - d to be reducible?

in the context of d being squarefree

vs x^2 obviously not being squarefree

[last spoke]

I feel like there's a shortcut here i don't know

[barren sierra]

for any x, is x^2 squarefree?

[rocky cloak]

How would you usually factor a polynomial?

[last spoke]

((x - a)(x - b) = x^2 - (a + b)x + ab = x^2 - d) so (a + b = 0) therefore (d = a^2) contradiction. Therefore irreducible?

[cloud walrus]

StarvinPig

[rocky cloak]

Sort of, the point is that a quadratic polynomial factors iff it has a root

[delicate orchid]

We in Z/pZ[x]?

[rocky cloak]

(which you might notice is the exact condition we are asked to consider)

[last spoke]

Okay I think I get the structure of what we're trying to do now

Kinda

[delicate orchid]

||If so it’s a ufd, so a square free integer d can never equal a non-square free integer x^2|| hence ||x^2-d can never be 0||, hence ||the quadratic doesn’t factor||

[last spoke]

Isn't it just a UFD by default

Nope lost again

[summer path]

Wew is saying that if you are indeed in F_p[x], then yes it's a ufd

But also maybe try to articulate what is confusing you?

[last spoke]

I think it was just trying to remember how irreducible shenanigans work

I think I've gotten the Z{sqrt{D}] case now

Ignore the bad writing

For (D \equiv 1 \mod 4) then I think we do a similar thing except swap out the minimal polynomial.

[cloud walrus]

StarvinPig

[last spoke]

And at least (\frac{1 + \sqrt{D}}{2}) is a root of (4x^2 - 4x + 1 - D)

[cloud walrus]

StarvinPig

[last spoke]

Since (D \equiv 1 \mod 4), (1 - D \equiv 0 \mod 4) so (\frac{1 - D}{4} \in \mathbb{Z}). So I can divide the whole thing by 4

[cloud walrus]

StarvinPig

[last spoke]

Okay big dumb moment, is p a prime integer or just prime in (\mathbb{Q}(\sqrt{D}))

[cloud walrus]

StarvinPig

[rapid junco]

for a cyclic group does one have that any hom maps generators to generators

between two cyclic groups

[coral spindle]

No

Z → Z defined by 1 mapsto 0

[rapid junco]

finite*

[coral spindle]

Z/6 → Z/6 defined by 1 mapsto 0

[rapid junco]

Suppose $\varphi_k$ is an automorphism. We will prove that $\varphi_k$ maps generators to generators. Let $g_1 \in G$. Then $g_1 = g^a$ for $a \in \mathbb{Z}$. Then $\varphi_k(g_1) = \varphi_k(g^a) = \varphi(g)^a$ for $a \in \mathbb{Z}$. Since $\varphi_k$ is surjective we have

[cloud walrus]

Brayden

[coral spindle]

Oh, you meant automorphism, not just any hom.

[rapid junco]

trying to finish this proof but im confused as to why I am not done

no I am trying to see why bijectiveness matters

cuz I can't see why

[topaz solar]

Well, obviously you can have f(x)=0 everywhere

Which fails

But also, the image is a subgroup, yes?

[rapid junco]

yes

so IM <= G

WTF

damn

surjective means image is G

fk me

[topaz solar]

[rapid junco]

okay so essentially we have that \phi(g) is a generator for the image

but then surjectiveness guarentees im = G

so we never need injectiveness here

[topaz solar]

If S generates G, then f(S) generates f(G)

But wait! f(G) is the whole thing

Done

[rapid junco]

so we need not need injetivness

we can get a weaker result just for endomorphisms

surjective

[topaz solar]

If it’s finite and surjective Endo

Then it’s an auto

[rapid junco]

non injective surjective hom

this still holds

[topaz solar]

Doesn’t exist if they’re finite

And it’s G->G

[rapid junco]

what if its G->H where H <= G

[topaz solar]

Then that’s not a surjective endomorphism

Unless H is just G

[rapid junco]

right lmfao

okay thanks

group theory is currently weak

we will get better

for infinite groups one has to have injectiveness to show gens get mapped to gens

[topaz solar]

Try showing this

[topaz solar]

No, if S generates G, then f(S) generates f(G), but if f(G) < the codomain, then f(S) does not generate the whole codomain

And if you map to generators of the codomain, then it’s surjective

[rapid junco]

same size and surjective implies one to one by definition of function

[topaz solar]

With or without injective

[last spoke]

Okay for (D \equiv 1 \mod 4) got ((p)) prime if and only if ((2x - 1)^2 \equiv D \mod p) has no integer solutions

[cloud walrus]

StarvinPig

[last spoke]

Just gotta get that 2x - 1 to a y and I'm done

[rapid junco]

one last question

so it follows that an endo that maps gens to gens is surjective, thus an automorphism

[coral shale]

if this is groups, its not clear what the canonical generator or anything like that is

its not clear to me what mapping generators to generators formally means

oh wait nvm its an endomorphism

yh still, im not convinced ur just interested about mapping the same set of gens to itself

[rapid junco]

it should be because you can always find a preimage using hom properties

[rocky cloak]

Hint: ||the quadratic formula||

[last spoke]

Yea I tried that but I feel like I get stuck somewhere gross

[rocky cloak]

Where did you get stuck?

[last spoke]

It's an iff so I separated it out. First part easy enough, second bit gross

I feel like i'm gonna need to use that (D \equiv 1 \mod 4) at some point here

[cloud walrus]

StarvinPig

[topaz solar]

Well, if it’s finite and of the same size

Since bijective hom

[rocky cloak]

Why don't you just say that if y^2 = D, then x = (y+1)/2 is a solution to (2x-1)^2 = d?

[last spoke]

I need to show it's an integer

[rocky cloak]

But you're working mod p...

[last spoke]

But I need integer solutions

[rocky cloak]

Everything is only with integers

[last spoke]

But if y is even

[rocky cloak]

y is an integer modulo p, so the concept of odd an even is a bit out the window

[last spoke]

I think I need the x/y to be integers for it to work

[rocky cloak]

Like for example modulo 5 we have 3=8, so saying a number is odd or even doesnt really make sense

[last spoke]

Given that (\exists x \in \mathbb{Z}) such that (x^2 \equiv D \mod p), show (\exists y \in \mathbb{Z}) such that ((2y - 1)^2 \equiv D \mod p)

[cloud walrus]

StarvinPig

[last spoke]

So (y = \frac{2x + 1}{2}) works if y is an integer

[cloud walrus]

StarvinPig

[topaz solar]

2 is a unit mod p

[rocky cloak]

Like you do division modulo p, so it's always an integer

[topaz solar]

[rocky cloak]

Division by 2 is just multiplication by (p+1)/2

[last spoke]

I am absolutely not familiar with modular arithmetic if that wasn't clear enough btw

[rocky cloak]

Alright, well if x+1 is odd, you can just use x+1+p instead

One of those will have to be even and then you can do division by 2 an get an integer

[last spoke]

Does (x^2 \equiv D \mod p \implies (x + p)^2 \equiv D \mod p)?

[cloud walrus]

StarvinPig

[last spoke]

nvm that's not what we get but it's a similar thing

Wait no I was right

[rocky cloak]

Yeah if x and y are congruent modulo p, you can freely interchange them in any polynomial expression modulo p

[last spoke]

Yea cool just checking

[rocky cloak]

Wait, so your doing algebraic number theory, but don't know any modular arithmetic?

[last spoke]

I know like basic stuff but it's not my wheelhouse where I can do it with confidence

[rocky cloak]

Fair enough

[last spoke]

wouldn't we need (\frac{x + 1}{2} \equiv \frac{x + 1 + p}{2} \mod p) then?

[cloud walrus]

StarvinPig

[rocky cloak]

Guess it depends what you mean by division by 2

If you mean actually integer division then that won't give you integers

[last spoke]

Yea that's what I mean

[rocky cloak]

Then they won't be equal because one will be an integer and the other will not be

[last spoke]

Yea thought so

Though if x is even then use (x + p) instead

Because p is prime so it'll be odd

Therefore (\frac{x + p + 1}{2} \in \mathbb{Z})

[cloud walrus]

StarvinPig

[last spoke]

Add the p to x, then go to y. I was doing the inverse

I DID IT

[torn warren]

I have a question here, if set $E=C, F=Q, \alpha=2^{1/3}, \beta=2^{1/3}e^{2\pi i/3}$, are they conjugate?

[cloud walrus]

WT

[torn warren]

both alpha and beta satisfy the same irreducible polynomial, so by definition, they are conjugate. If this is true, then the conjugate here is different from the conjugate roots in complex analysis?

[rocky cloak]

Yes, they are conjugate.

Complex conjugate / complex conjugation typically refers to elements that are conjugate in C/R, so a special case of being conjugate more broadly

[torn warren]

thank you!

[torn warren]

why are these two quotient rings are isomorphic?

if F=F', these two quotient rings are isomorphic only if alpha and beta are conjugate, right?

p'(x) is irreducible poly in F ' [x]

not the derivative

[rocky cloak]

They can still be isomorphic even if they're not conjugate. But then the isomorphism will be more complicated than just mapping x to x

And phi could be a nontrivial automorphism of F, in which case they won't necessarily be conjugate

[torn warren]

could you please give an example?

[rocky cloak]

So a simple example could be something like sqrt(2) and 2sqrt(2). These are not conjugate, but Q(sqrt(2)) = Q(2sqrt(2)).

An example more in line with whith this theorem could be to let F = {0, 1, s, s+1} be the field with 4 elements and have alpha be a root of x^2 + x + s. There is an automorphism of F mapping s to s+1, so as per the theorem you can extend this an automorphism mapping alpha to a root of x^2 + x + s+1.

[steel pulsar]

**Statement **: Let G be a group and H a subgroup of G. We define the centralizer CH of H in G as the following subset of G

Question : Show that CH is a subgroup of G

Correction:

Let e denote the identity element of G. For all h ∈ H, we have eh = he. Therefore, e ∈ CH. Let g1 and g2 be in CH. For all h ∈ H, we have g1g2h = g1hg2 = hg1g2. Thus, g1g2 ∈ CH, and CH is closed under the group operation.

Now, let g ∈ CH. For all h ∈ H, we have gh = hg. Therefore, hg⁻¹ = g⁻¹h. Thus, CH is closed under taking inverses. Therefore, it is indeed a subgroup of G."

My problem :

One thing I don't understand is about the order. In the solution, it states that g_1g_2 h = g_1 h g_2 = h g_1 g_2. Why doesn't it directly say that g_1g_2h = hg_1g_2 ? Why the need to write g_1hg_2?

Like in statement it's written g h = h g, so why g_1 h g_2 ?

For this 2nd statement : Show that if H is a normal subgroup of G, then CH is a normal subgroup of G.

**Correction : **

here is same, we have g^(-1) h g ∈ H , so x ∈ C_H we have : g^(-1) h g x = x g^(-1) h g which deliver that h g x g^(-1) = g x g^(-1) h
I don't understand why he plays with the order ....

[ivory trail]

it just emphasizes that you are using the centralizer property of g_2 and then g_1

good to emphasize in an early course, but afterwards you can just write it the way you suggest because you are writing for a more advanced audience

if you're a student in an early course, you also want to show the grader that you know what's going on

so you usually want to spell more things out

also here you want that because writing g_1g_2h = hg_1g_2 directly is what you want to show

so you want to show the reader that you are using that g_1 and g_2 are in the centralizer to show that g_1g_2 is in the centralizer

rather than just stating what you want to prove because "it's obvious"

if you're being asked to show the centralizer is a subgroup you need to go into this level of detail

[torn warren]

when you say " automorphism of F mapping s to s+1", do you mean g(y)=y+1, and y takes 0, 1, s, 1+s ?

so g(0)=1, g(1)=0, g(s)=1+s, g(1+s)=s

[steel pulsar]

You don't understand, I don't want to write it directly because it's obvious, I don't understand

[rocky cloak]

No, that wouldn't be an automorphism. Automorphisms have to map 1 to 1 and 0 to 0

I just mean the automorphism that maps s to s+1

[steel pulsar]

My question was not to know if it was useless or not, it's e I don't understand why the reasoning

[ivory trail]

oh i see. so since g_2 is in the centralizer of H and h is in H, by definition of the centralizer, g_2h = hg_2

so we use the associative property and this again with g_1 to get g_1g_2h = hg_1g_2

[steel pulsar]

yes but why g_1 h g_2

why h in the middle

[ivory trail]

to spell it all out,
g_1 (g_2 h)
= g_1 (h g_2) (g_2 is in the centralizer of H, which contains h)
= (g_1 h) g_2 (associative property)
= (h g_1) g_2 (g_1 is in the centralizer of H, which contains h)
and then we can drop the parentheses from the first and last expressions because of the associative property

[torn warren]

do I understand correctly?

[ivory trail]

it shows up as part of the computation i just wrote

[rocky cloak]

Yeah, that all looks right

[torn warren]

does this mean, if choose any two irreducible poly which has the same degree, then the two quotient rings are always isomorphic?

given F to F

[steel pulsar]

for me g_1 (h g_2) and (g_1 h) g_2 is the same it's weird...

[rocky cloak]

For finite fields, there is at most one field of any given order. So there it's true.

In general for example Q(sqrt(2)) and Q(sqrt(3)) are not isomorphic

[ivory trail]

well due to the associative property, they are, so we can drop the parentheses if we feel like it (and we often do)

[steel pulsar]

and for that ?

[torn warren]

your example (0, 1, s, 1+s) is the finite case, so if pick any irr poly with degree n, then the extension field has the order $4^n$, so there is only one extension field with this order (up to isomorphims), hence, when you set the mapping (in my graph is h) $h(\alpha)=\beta$, this is the isomorphism from $F(\alpha)$ to $F(\beta)$, and furthermore, it is to map the two generators $\alpha \to \beta$ of two cyclic groups. Do I understand correctly?

[cloud walrus]

WT

[rocky cloak]

So you can't necessarily realize the isomorphism by mapping alpha to beta no. You may have to map alpha to something else

[ivory trail]

here you left-multiply the first equation by g and right-multiply by g^(-1), which yields the second equation

[rocky cloak]

Or maybe I'm misunderstanding what you're saying. What is h here?

[torn warren]

h is here on my pic...

[torn warren]

I want to udnerstand your example, so I make this pic, here alpha is map to beta

alpha and beta are generators for each extension field

[rocky cloak]

Okay, then I'm confused about what you're asking. Are you asking about an isomorphism between two arbitrary field extensions, or one constructed using this theorem from earlier?

[torn warren]

my op is "there is an isomorphism if and only if alpha and beta are conjugate?" you ans is no and give me this constructed counter example...

[rocky cloak]

Okay, you're just asking about the specific example, then yes. The automorphism will map alpha to beta

[torn warren]

now, I try to understand your ex, so I wrote them.

[steel pulsar]

to prove is normal subgroup is not ghg^(-1) ? He choose g^(-1)hg

[torn warren]

ok, so it has to be mapping the generator to generator...

[ivory trail]

i'm not sure what you mean

[rocky cloak]

I'm not sure what you mean by "has to". Like what is the starting condition you're asking about?

You start with a field F, an extension F(alpha) and an automorphism F -> F. Then you just define beta to be what alpha is mapped to by the extended isomorphism

[ivory trail]

the second equation shows that gxg^(-1) is in C_H, and since x in C_H and g^(-1) in G were arbitrary, this shows C_H is normal in G

[ivory trail]

it's also normal if and only if g^(-1)Hg = H for all g in G

because you can just replace g by its inverse, and you're still checking the equation for every element in G

[torn warren]

1. specify the elements in F,
2. define the automorphism F to F
3. choose alpha and beta to give the ext-field F(alpha) and F(beta)
4. give the isomorphism h:
5. here is my "has to" comes: my question is h(alpha)= y, and y must be the generator of F(beta), surelly maybe more than one generators in F(beta), and set y=beta is one of the choices.

[rocky cloak]

If you choose alpha and beta independently, then there is no guarantee that you can map alpha to beta

[torn warren]

but in your ex, you pick up two irr poly where alpha and beta are roots, you pick up these two irr poly independently?

what is the trick to choose these two poly?

[rocky cloak]

No I did not, they were constructed from the theorem you posted

I picked the one for alpha, then applied the isomorphism to get the other one

It works like this:

1. Pick automorphism F -> F
2. This gives you an automorphism phi: F[x] -> F[x]
3. Pick polynomial p in F[x], and define q to be phi(p).

Then F[x]/p is isomorphic to F[x]/q, and the isomorphism is given by phi

And because of the way phi is constructed it maps x to x. So if we think of F[x]/p as F(alpha) and F[x]/q as F(beta), then the isomorphism maps alpha to beta

[mossy lintel]

Sorry to interrupt, how do i prove F(a, b) = F(b, a), where F(a, b) is first adjoining a then adjoining b to F?

[torn warren]

ah, sorry, I misunderstand, what I want to know is

[rocky cloak]

You just apply phi to the coefficients

[rocky cloak]

Guess it depends on your definition of adjoining. If it just means you take the smallest field containing F and a, then it should be clear that both of these are the smallest field containing F, a and b.

[torn warren]

but why for this Q(sqrt(2)) and Q(sqrt(3)) are not isomorphic? can I still apply this thm to map sqrt(2)) to sqrt(3) ?

[rocky cloak]

This theorem says nothing about this, since sqrt(2) and sqrt(3) are chosen completely independently. They don't come from any underlying automorphism of anything.

As for why their not isomorphic, you can show that one has a square root of 2 and the other doesn't for example

All this theorem is saying is that if you have a field F and a field extension F(alpha), if you take some isomorphic field F', then there is an isomorphic field extension F'(beta).

It doesn't say anything about whether or not unrelated field extensions are isomorphic

[torn warren]

can I say this?

if I force them to be mapped, then we get contradictions

[rocky cloak]

Sure, it's also the case that the identity is the only automorphism Q -> Q, but it's not so important I guess

Perhaps a bit weird to say that sigma(1) will be 3/2. I would maybe instead say that

s(2) = s(1+1) = s(1)+s(1) = 1+1 = 2

[torn warren]

right, this is much better!

[torn warren]

but I am confused by this thm, it says it is isomorphism iff alpha and beta are conjugate, just now you give the Q(sqrt2) and Q(2\sqrt2) that they are not conjugate, surely your ex has different irr poly, x^2-2 and x^2-8, so in your ex, the ismorphism is not defined in this textbook's way. Does that mean the isomorphism is not unique?

[cobalt heath]

I'm reviewing modules over PID for algebra exam. How do I make the proof less annoying to me?

[rocky cloak]

This theorem says that the map sending alpha to beta and fixing F is an isomorphism iff they are conjugate. It doesn't say anything about the existence of other isomorphisms

And indeed there can be several isomorphisms, and there can be isomorphism when alpha and beta are not conjugate

[torn warren]

so your two examples, one is Q(2sqrt2), another one is the finite case (0,1,s,1+s), both of them are non-conjugate, right?

[rocky cloak]

Yes

[torn warren]

thank you so much, really appreciate!

[terse crystal]

Yeah. And remember 1/2 is well defined in Z/pZ. So x^-x=(x-1/2)^2-1/4. In the end it is still about whether there exists y^2=d
A prime number, not prime element or something

[torn warren]

why I feel the field part is more difficult than groups and rings, there are many subtleties and the book didn't clarify....

[mossy lintel]

The definition im using is F(a,b) = (F(a))(b), but i think it can be done by first arguing F(a)(b) is the minimal field contains both a and b.