groups-rings-fields

[hidden haven]

Also one of the coefficients is p so might be better to use something else instead of F_p lol

[hot lake]

I think we are

[hidden haven]

Yes xD

[hot lake]

I was talking about the characteristic of the finite field

the prime p, not the coefficient of x in the polynomial

so you have a finite field Fp

an irreducible polynomial f whose coefficients I won't write

of degree 3

so you can define Fp³ = Fp[x]/(f)

it will be a field

it will have a root of f

x

[hidden haven]

But it may not be a polynomial over Fp

[hot lake]

and it will have in fact 3 roots of f

x, x^p, and x^p²

[hidden haven]

It's a polynomial over any finite field

[hot lake]

f is a polynomial with coefficient in Fp

[hidden haven]

That's not given I think?

[hot lake]

oh you say it might be Fp² or somehting ?

you're eright

[hidden haven]

Ye

[hot lake]

doesn't really change much I think we should pick a better name for the size of the finite field

so you have a finite field Fr, where r is a prime power
an irreducible polynomial f = x³ + px + q of Fr[x]
of degree 3
so you can define Fr³ = Fr[x]/(f)
it will be a field
it will have three roots of f, x, x^r, and x^r²

and x is not in Fr

also you know from the start that f is irreducible over Fr so it can't have a root in Fr

[latent anvil]

I don't see this

oops

Should've checked if the channel was in use, sorry

[next obsidian]

[hidden haven]

Why are the 3 roots x, x^r, x^r²?

Is that always true in finite fields?

[hot lake]

x is an obvious root

[hidden haven]

x^r = x right?

[hot lake]

and x -> x^r is an automorphism of Fr³ that fixes Fr

absolutely not

[hidden haven]

Oh

Fr³

Right

[hot lake]

well since we are doing finite fields, yeah the splitting field is a degree 3 extension

if we were in other fields the splitting field could have degree 6

that's why finite fields are your best friends

so anyway, the roots a,b,c are in Fr³ (not in Fr)

and so a square root of the discriminant, (a-b)(a-c)(b-c) is in Fr³

the problem is to show that it's in Fr

[hidden haven]

||So is the solution that the Galois group acts on the roots as A_3 so the determinant will be a square?||

I don't remember if that was the theorem actually lol

[hot lake]

I think you can be much more straightforward

if you know uuh basic stuff about finite fields ?

[hidden haven]

hmm let me think

oh

Got it lol

||adjoining sqrt(disc) will be a deg 2 extension contained in this deg 3 extension so not possible? But that's not using finite field properties ||

[chilly ocean]

i think there is a command like ,texspoiler blah

[hidden haven]

Squirtle why are you spoiler walling it lol I was spoiler walling to not give you the solution

[hot lake]

yeah you can do something like that

[cloud walrus]

squirtlespoof

[hot lake]

no you don't know Delta is a square in K

ah wait

is K the bigger field

[hidden haven]

Ye

[hot lake]

k nevermind

[hidden haven]

Same thing as what I was doing

[hot lake]

yeah I think that should be good

you can also apply the frobenius automorphism to (a-b)(a-c)(b-c) to get (a^r-b^r)(a^r-c^r)(b^r-c^r) = (b-c)(b-a)(c-a) = - - (a-b)(a-c)(b-c) = (a-b)(a-c)(b-c) and so it is fixed so must be in Fr, but I guess that's just what moldilocks was saying with his theorem

you can also say that you know the structure of the multiplicative group Fr³* and use that to show that squaring sends stuff not in Fr into stuff not in Fr

Fr³* is a cyclic group of size r³-1

so if you identify it with Z/(r³-1)Z

then Fr* inside it will be the multiples of (r³-1)/(r-1) = r²+r+1

and squaring is just multiplication by 2

and 2 is coprime with r²+r+1 because r is odd

[hidden haven]

Right, I didn't remember the proof

That's neat

[sinful mirage]

are U(n),SU(n),O(n),SO(n) (or any of them) torsion? I am asking this because I am confused about what the character of the dual reo would be in this case

[hot lake]

by torsion do you mean that all of their elements have finite order ?

[cloud walrus]

ProphetX

[sinful mirage]

yes

[hot lake]

then no unless they are a finite group

like SO(1)

[sinful mirage]

there are infinite groups,which are torsion

infinite product of Z_n for example

[hot lake]

yeah but they aren't a manifold of dimension > 0

take a rotation by an irrational angle in SO(2)

[sinful mirage]

so there is no manifold with dim>0(lie group for ex),which is torsion?

[hot lake]

take any nontrivial element of the lie algebra

exp of multiples of it will be either R or R/Z

and R and R/Z both have many many nontorsion elements

[cloud walrus]

ProphetX

ProphetX

[hot lake]

well you can still define the trace of a matrix

even if that matrix has infinite order

I don't see what being torsion or not has to do with anything there :s

[cloud walrus]

ProphetX

[sinful mirage]

cause G was finite

[hot lake]

how do you define the dual representation ?

[cloud walrus]

ProphetX

[sinful mirage]

where the start on the RHS is the dual of a map

my issue is this proof would not hold if the group is not finite nor torsion,and idk how I could say anything about the character of the dual rep for infinite groups

[hot lake]

if the eigenvalues are on the unit circle it will still work

[sinful mirage]

yes,but is this the case for U(n)/SU(n)?

it should not be the case for SO(n)/O(n)

[hot lake]

well it works for their standard representation I think ?

[sinful mirage]

for that yes. but then in general, one can not relate the characters of the dual of any rep of O/SO/U/SU (n) to the character of the rep? like there is no relation (maybe a generalized one of the above?) at all?

[hot lake]

I think it should still work for any of their representation

[sinful mirage]

how could one see this?because the fact that the eigvals are roots of unity is crucial

[hot lake]

because of their compactness

[sinful mirage]

so for any compact lie group the eigenvalues of any representation are roots of unity?

[hot lake]

no but I think they must be on the unit circle

[cloud walrus]

ProphetX

ProphetX

[hot lake]

unit circle means |z| = 1, so |z|² = z * conjugate(z) = 1

and so 1/z = conjugate(z)

[sinful mirage]

aren't all roots of uinty on the unit circle?

[hot lake]

yes

[sinful mirage]

so what's the difference?

[hot lake]

the unit circle has many points that are not roots of unity

[sinful mirage]

ah right

[sinful mirage]

then to conclude, how could I see this?

is this a more difficult/deeper result?

[hot lake]

well suppose you have a g that has some eigenvalue not on the circle

then either g or 1/g has an eigenvalue larger than 1 in absolute value

so their powers g^n or g^-n will have their matrix entries that grow as O(lambda^n) for lambda > 1

but if the lie group is compact

then there is some g' in G that is in the closure of the {g^n} or the {g^-n}

that g' 's matrix would need to have infinitely big entries

[sinful mirage]

how does this follow from the def of compact(every cover has a finite subcover)?

[hot lake]

suppose that for every h in G there is an open neighbourhood of h with only finitely many of the g^n

the reunion of those cover G

so there is a finite number of them that cover G

and so {g^n} is finite

which is uuuuh

impossible

so there must be someone with infinitely many members of {g^n} next to it

[sinful mirage]

it is impossible because this:
so their powers g^n or g^-n will have their matrix entries that grow as O(lambda^n) for lambda > 1 ?

[hot lake]

yeah

also if {g^n} was finite then it means g would be a torsion element

and so its eigenvalues are roots of unity

[sinful mirage]

oh right,that's way easier to get a contradiction

[sinful mirage]

and this contradicts that the assumed lie group was compact

therefore the statement |lambda|>1 can't be true

right?

[hot lake]

um I don't think so

we used compactness to show that there must be some h in G with infinitely many g^n in any open neighbourhood of h

[sinful mirage]

and how does this lead to contradiction?

[hot lake]

but since representations are continuous the matrix entries of such an h would need to be infinitely big

because the matrix entries of g^n get larger and larger

[sinful mirage]

why is that an issue?

[hot lake]

it's an issue because for any real number, there is another real number larger than it

so a real number can't be infinitely big

wait there is much simpler

[sinful mirage]

but wait,here we assumed the eigenvalues are complex

[hot lake]

if G is compact

then its image in GL(V) is compact

so the entries there are bounded

and so the eigenvalues are also bounded

[sinful mirage]

why does this follow?

I know the theorem of Heine Borel which is similar to this,but that's on R

[hot lake]

well GL(V) is an open subset of C^(2n)

and compact subspaces of C^(2n) are just the closed and bounded subsets

[sinful mirage]

by using heine borel for R^4n?

[hot lake]

I don't have a clue who that result is named after

[hot lake]

yes

[sinful mirage]

we need to use this replacing n with 4n,because C is iso to R^2

thanks

[cloud walrus]

ProphetX

[hot lake]

well for (R,+) it doesn't hold

[chilly ocean]

It can seem trivial but why is -1 = 1 in Zp. where p is prime?

[hot lake]

-1 = 1 only in Z2 ...

[chilly ocean]

I saw something like -1 = 1 in Zp where p is any positive prime when proving Wilson's Theorem

(p-1)! = -1 mod p

Ah right

you're correct

thanks 🙂

[carmine fossil]

I think your proof involved pairing up of a number with it's multiplicative inverse

With the only numbers without a pair being -1 and 1

[chilly ocean]

Right that's it

[chilly ocean]

But they are inverses of themselves in Zp?

[carmine fossil]

Yes

[chilly ocean]

Where does -1 come from in the product of (p-1)!

Is it from the p-1?

[carmine fossil]

Yes

[chilly ocean]

Awesome, thanks for confirming

[sinful mirage]

@hot lake the proof we did show that they cannot be outside of the unit circle. what shows that they are on the unit circle instead of being somewhere inside the unit circle?

[hot lake]

then you look at g^-1

if g has an eigenvalue inside the circle then g^-1 will have an eigenvalue outside

[sinful mirage]

oh,so we showed that g can't have eigenvalues outside,so g^-1 can't have inside

therefore g must have them on

thanks

[hot lake]

g can't have eigenvalues outside and neither can g^-1 so g can't have eigenvalues inside

[sinful mirage]

wait, from the fact that g can't have eigenvalues outside,neither can g^-1 have outside,why does it follow that g can't have inside?

i thought we can prove similarly that g^-1 can't have eigenvalues inside,but this would be a consequence

[hot lake]

the eigenvalues of g-1 are the inverse of the eigenvalues of g

[sinful mirage]

right,but why does this lead at all to talking about the inside of the circle?

we would need to assume <1 instead of >1

[hot lake]

the inverse of something inside the circle is outside the circle

[sinful mirage]

ah right

[hot lake]

and vice versa

[sinful mirage]

but i'm missing something. showing that g is outside would imply g^-1 is inside,but we showed that g^-1 is outside

[sinful mirage]

how can both statements be true?

[hot lake]

suppose g has an eigenvalue inside the circle

then g^-1 has an eigenvalue outside the circle

then {g^-n} has elements with arbitrarily large eigenvalues, which contradicts compactness

[sinful mirage]

ahh so we reduce it to the case we proved before

[hot lake]

yes

[chilly ocean]

Prove that if A is a set with at least three elements, then the symmetric group (S_A, o) is nonabelian.

Hint please

[sinful mirage]

you can prove that $S_3$ is not abelian by direct computation ( take elements and just multiply them,i.e. $\sigma_1 \sigma_2 \neq \sigma_2 \sigma_1$

[cloud walrus]

ProphetX

ProphetX

[chilly ocean]

I don't know what Cayley's theorem is

That wasn't covered

[cloud walrus]

ProphetX

ProphetX

[chilly ocean]

But you have random specific alpha, beta?

Not arbitrary

[cloud walrus]

ProphetX

[sinful mirage]

having a counterexample already show that it is non-abelian

[sinful mirage]

and this works for arbitrary n>=3

so for any n>=3,you have at least one counterexample, which is enough to show that it is not abelian

[chilly ocean]

I thought we have to show it for all n >= 3?

[sinful mirage]

right

[chilly ocean]

Not just n = 3

[sinful mirage]

the above example works for any n>=3

[sinful mirage]

here we did not assume n=3

in particular,setting n=3,this will give a counterexample for n=3

setting n=4,gives a counterexample for S_4, and so on

[chilly ocean]

I see thanks

[chilly ocean]

I have a question regarding the proof of the primes in the ring Z[i]. Suppose a+ib is some prime in Z[i]. Looking at its norm, we get N(a+ib) = a^2 + b^2 which is an integer. So we can write out its (unique) prime factorisation a^2 + b^2 = p1p2...pr. We have a^2 + b^2 = (a+bi)(a-bi) so a+bi divides p1p2...pr so a+bi divides some prime pj. How can we infer from here that a^2+b^2 divides pj^2?

[hidden haven]

a+bi divides pj -> a-bi divides pj (conjugation is an automorphism)

[chilly ocean]

Haven't looked at automorphisms yet so I'll take that for granted !

[hidden haven]

well you know that a conjugate times b conjugate is a times b whole conjugate?

[chilly ocean]

Yep

[hidden haven]

so a+bi divides pj

ie (a+bi)c = pj for some c

take conjugates on both sides

and distribute the conjugate on the left over the product

[chilly ocean]

Ahh gotcha

We're still dividing in Z[i] right?

[hidden haven]

yes

c is some Gaussian int

[chilly ocean]

Now a^2 + b^2 | pj then since pj is prime we have a^2 + b^2 = p or a^2 + b^2 = p^2. In the former case, that's easy to see that the primes a+ib are either associates of 1+i or a+ib such that a^2+b^2 = p where p congruent to 1 mod 4

In the former case a^2 + b^2 = p^2 so (a-ib)(a+ib)= pxp

I'm stuck here

It looks like p = a-ib and p = a+ib so p = a? But the solution says smthng like p= a,b or associates

[hidden haven]

how do you get p = a-ib = a+ib from that?

[chilly ocean]

Actually my guess is: if p is an associate of a+ib and p is an associate of a-ib then we still get pxp = (a-ib)(a+ib) right?

[hidden haven]

yes

[chilly ocean]

That's not easy to see so I guessed from the solution

[hidden haven]

notice that (a+ib)(a-ib) is a prime factorization of a^2 + b^2

and so is p^2

[chilly ocean]

Aha and Z[i] is a UFD so we get unique a factorisation up to associates?

[hidden haven]

yep

[chilly ocean]

Cool thanks

[hidden haven]

[chilly ocean]

Also, when coming across the notion of maximal ideals, I've seen a statement saying that Q[x] / <x^2+2> (where Q[x] is a PID and <x^2+2> is a maximal ideal since x^2+2 is irreducible) is isomorphic to Q[sqrt(2)]. The proof of this was rather fast in my notes. The example given was by taking the coset x^2 + <x^2+2> and noticing that it is the same as the coset 2 + <x^2 + 2>. Two elements are in the same coset if and only if they differ by a multiple of x^2+2. Noticing that, and since every coset of p(x) + <x^2+2> can be reduced into a linear polynomial ax+b + <x^2+2> we can construct an isomorphism from Q[x] / <x^2+2> to Q[sqrt(2)] given by f(p(x) + <x^2+2>) = a+bsqr(2).

Can anyone confirm? Thanks 👍

[hidden haven]

Ye that works but nicer way to do it is to take map Q[x] to Q[√2] by sending x to √2 and fixing all the rationals then invoking first isomorphism theorem

[sturdy mirage]

i think the coset x² + <x² + 2> is the same as the coset -2 + <x² + 2> no?

not +2, but -2

[hidden haven]

Ye

[chilly ocean]

Hmm do you subtract x^2 - (x^2 +2)

?

[chilly ocean]

👌🏻👌🏻

[sturdy mirage]

x² = 1*(x² + 2) - 2
so the first term here is in <x² + 2>
then you subtract 2

-2 is the remainder after doing polynomial division

you divide x² by (x²+2)

[chilly ocean]

Got it so x^2 and -2 both differ by (x^2 + 2)

That’s how I see it in terms of cosets being equal

[sturdy mirage]

i can't quite wrap my head around that, but maybe!

yeah that should work

[simple charm]

Sorry if this is in the wrong channel but could someone explain the difference between a unitary magma and a monoid to me

[hidden haven]

Unitary magma need not be associative

It's just a binary operation that has a 2 sided identity

[simple charm]

So you could say that all unitary magma are monoids but not vice versa?

[hidden haven]

All monoids are unitary magma

[simple charm]

My bad, thank you

[hidden haven]

[chilly ocean]

p(x) + <x^2+2> = q(x) + <x^2+2> iff p(x) - q(x) is in <x^2+2> ie the diff is a multiple of x^2+2

that's just how equality of cosets holds ie aH = bH iff b^-1a in H

[craggy berry]

Anyone have familiarity with SLPS?

[chilly ocean]

What's that?

[craggy berry]

Straight line programs

[chilly ocean]

I guess quite a similar argument works for showing that the fields R[x]/<x^2+1> and C are isomorphic right?

[oblique river]

yes

[delicate orchid]

you'd send x to i right?

just making sure I can remember this proof

[oblique river]

you send x to whatever root of x^2 + 1 you want

you dont have to use that poly though

C is also isomorphic to R[x]/<x^2 + 4>

[delicate orchid]

wait lemme think through why

oh yeah duh C's a field so (2i) is just C

makes sense

[oblique river]

more generally, C = R[x]/p(x) for any quadratic polynomial p(x) with no real roots

[delicate orchid]

would it work with a quartic?

[oblique river]

no

all quartics are reducible over R

[delicate orchid]

well if the quartic factorises into a quadratic with no real roots it would be equivalent

[oblique river]

no

R[x]/(x^4 + 1) is not a field

because x^4 + 1 is reducible

[delicate orchid]

oh yeah I didn't mean in general

I'm kinda just reasoning through it outloud

oh yeah cause you'd also quotient out (x^2-1) which would break it

[oblique river]

what do you mean

x^2 - 1 doesn't have anything to do with x^4 + 1

[delicate orchid]

x^4+1 = (x^2+1)(x^2-1)

[oblique river]

no

that's wrong

[delicate orchid]

oh yeah that's x^4-1 lol

ok then I don't see how x^4+1 is reducible over R at all, it has 4 complex roots?

[oblique river]

not having roots doesn't mean it's irreducible

it factors into two quadratics

every polynomial of degree 3 or larger is reducible over R

x^4 + 1 = (x^2 - sqrt(2)x + 1)(x^2 + sqrt(2)x + 1)

[delicate orchid]

ah thank you

I could not see that at all

[unkempt comet]

I wonder how long mutes last.
there's been an oversight in writing abilities on the Advanced mathematics category
Edit: Honestly I'd much rather have perms for #bots than this if it's temporary, if it's not temporary then the discord mod meme is real

[delicate orchid]

#measure-theory

[unkempt comet]

Nice one
Interestingly, if you block someone's permission to write on one channel, it blocks their ability to edit previous messages
Though simply removing that person's abilities to see the text channel has the same effect

[delicate orchid]

to atone for my embarrassment earlier I worked through and manually calculated what R[X]/<x^4+1> is and realised that the image of the polynomials x^2+sqrt(2)+1 and x^2-sqrt(2)+1 were zero divisors so it couldn't be C and also that the kernel of the map f(p(x)) -> p(sqrt(i)) for any root of i is an ideal generated by one of the two factors of x^4+1 so you don't get R[x]/(x^4+1) and can't use the first iso theorem so I now understand it very well lol

[oblique river]

yes that';s right

if p(x) is reducible over a field F then F[x]/p(x) won't be a field at all

cuz the images of the factors of p(x) will be zero divisors

in general, F[x]/p(x) will be a product of fields

one corresponding to each irreducible factor of p(x)

(assuming that p(x) has no repeated factors)

for example, R[x]/<x^4 + 1> is isomorphic to the direct product C x C

R[x]/<x^3 - 1> is isomorphic to R x C

[delicate orchid]

ahhh that makes SO much sense

[oblique river]

because x^3 - 1 = (x-1)(x^2 + x + 1)

or something like that

the R corresponds to x-1 and the C corresponds to x^2 + x + 1

if you have a repeated factor it's a little more complicated

[delicate orchid]

rip

[oblique river]

for example, R[x]/<(x-1)^2> is not R x R

[delicate orchid]

awww I got really excited there

[oblique river]

it just... is what it is

[delicate orchid]

anyway that was a fun exercise

[oblique river]

👍

[delicate orchid]

I did know that R/I isn't a field if I isn't irreducible but I never really internalised why

but now I have!

[oblique river]

do you know the chinese remainder theorem in number theory?

[delicate orchid]

yeah

I've also been taught the ring theory version but kinda forgotten it

it's something to do with the quotient of the union of disjoint ideals being something

ah I think I remember?

[oblique river]

yeah

the most general statement isnt super important here

[delicate orchid]

R/(i1i2...in) iso to R/i1 x R/i2 ... R/in iirc

[oblique river]

i just wanted to yoint out that that's what was going on above

with the R x C

and C x C

[delicate orchid]

cause if the ideal is generated by an irreducible element it's maximal so has to be disjoint to every other ideal generated by an irreducible element so the chinese remainder theorem applies?

I think?

[oblique river]

yep

[delicate orchid]

epic

[novel parrot]

$p^2 = \sqrt{ \frac{m}{4d}}$

[cloud walrus]

ActiveChapter

[novel parrot]

If m and d are integers, d being squarefree. Why does that mean that the square root of m/d must be a perfect square?

[obsidian sleet]

#elementary-number-theory ?

[chilly ocean]

Quick question, in a field with 5 elements 0,1,a,b,c it doesn't matter what 1+1 is as long as it's not 0,1 right?

It will just have different operations performed later down the line

[magic owl]

If you're just giving names to elements then no, but you have to make sure to choose them consistently

like if 1+1+1=a and 1+1+1+1=b then you can't have a+1=c

[chilly ocean]

right

what if 1+1=b and 1+b=a and 1+c=0

oh wait

1+a=c

nvmmm

Id have to think of what the others are

Yeah it doesnt matter

They will just be the same field

Just a and b are switched

wait are fields unique??

🤨

{0,1,a,b,c} = {0,1,b,a,c}?

[magic owl]

Finite fields are determined up to iso by their cardinality

[chilly ocean]

For any two fields, if they're of the same cardinality then they are isomorphic

[chilly ocean]

Also the cardinality any finite field is a prime power

[magic owl]

This is not true without the word finite

There are infinite fields with the same cardinality that are non-isomorphic

[chilly ocean]

"cardinality" implicitly assumes finiteness

but you are right

[magic owl]

no it doesn't

what

[chilly ocean]

If a set is infinite, what is its cardinality?

[magic owl]

Depends on its cardinality thats the whole point

[old lava]

lmao

[magic owl]

Cardinality was designed as a way of discussing sizes of infinite sets

in some meaningful way

finite sets weren't the reason for developing a theory of cardinals

[old lava]

there would be no purpose whatsoever in even using the word cardinality if you could not apply it to infinite sets

[magic owl]

you a word

[old lava]

whoops

[chilly ocean]

What is the cardinality of R then?

[magic owl]

Uh there are like symbols people use to discuss these things

but R is kind of a proptypical cardinality

It's the cardinality of the powerset of N

Basically you have N and anything in bijection with it

these are called countable

and then you can take powersets

and powersets always increase cardinality

[chilly ocean]

In almost all proofs I've seen where cases are distinguished between fine and infinite sets, it goes as follows: either the set is infinite, or it is finite and we assume its cardinality is some positive integer

[magic owl]

yes

that is a good technique

but the infinite stuff also has a cardinality

[old lava]

yes, because the cardinality can be infinite

[magic owl]

its normally just less useful

[chilly ocean]

I see

haven't come across that

[old lava]

and also kinda weird to work with

[chilly ocean]

isn't like an exponential growth?

[magic owl]

Uh thats a decent way of intuiting i guess

but like

these things stop making a ton of sense

[old lava]

it's a way of intuiting it, but infinities don't really behave like numbers exactly

[chilly ocean]

2^n every time you take a set of size n

[magic owl]

Okay so

given a finite set A

the powerset has carinality $2^{|A|}$

[cloud walrus]

MaxJ

[magic owl]

so if A is infinite

people often write $2^{|A|}$ for the cardinality of its powerset

but its not like

[cloud walrus]

MaxJ

[old lava]

but yes, the cardinality of the real numbers is often referred to as $2^{\aleph_0}$, where $|\mathbb{N}| = \aleph_0$

[magic owl]

the same things exactly

[cloud walrus]

F[x]-module

[magic owl]

there is also a theory of cardinal arithmetic

but its complicated and doesn't work exactly the same

for example

I can keep throwing copies of N into a basket with R

and I'll never increase the cardinality

[delicate orchid]

imagine assuming the continuum hypothesis

[oblique river]

Ultraproduct moment

[molten silo]

hi guys

[delicate orchid]

hello mr dirac

[latent anvil]

wait when did someone assume the continuum hypothesis

no I'm serious lol

I don't see it in the discussion above

[molten silo]

lol

[hidden haven]

It's often assumed for results in cardinal arithmetic

And that was being discussed

[latent anvil]

sure it just seemed like it was responding to something directly in the discussion

[magic owl]

no one assumed it

dw

youre not crazy

that comment was just random

[next obsidian]

Max is back

Max is back

Max is back

Max is back

[final pasture]

not sure who's max but ig

[magic owl]

me

[final pasture]

[cloud walrus]

squirtlespoof

[hidden haven]

oh I've solved this exact problem in an assignment

Brb while I do the zero effort look up

[final pasture]

[hidden haven]

You have to show that a the quotient (let's call it N/H) is isomorphic to that other galois group (call it G)

Use first isomorphism theorem, define a map from N to G, surjective, which has kernel H

Yep

So given an automorphism of L/F which normalises all automorphisms of L/K, can you get an element of G(K/F) from this in a natural way?

(in particular you have an automorphism of L/F and need an automorphism of K/F)

isn't that just restriction?

Wait what do you mean by all the roots in K?

what does roots of K mean?

problem is K isnt a Galois extension of F

yeah you have the right idea somewhat

But notice what youre doing is just restriction of the automorphism to K

(youre looking on the action on generators, then extending it back to all of K)

you see this?

right so thats why youd guess restriction will be the operation because thats pretty much all we can do

only problem is, the restriction of an automorphism of L/F may not be an automorphism of K/F

it fixes F

but it may move K around

and this is where you use the fact that the automorphism normalises Aut(L/K)

so the claim is: restriction of automorphisms of L/F that normalise Aut(L/K) are in Aut(K/F), and this gives a group homomorphism with kernel Aut(L/K)

Try proving these 2 things, the kernel thing should be much easier

yep

Not element wise

but elements of K go to elements of K

you dont get it immediately

but its not too hard

by dont get it immediately I mean its not by definition or anything

${\sigma\in G(L/F) \mid \sigma^{-1}G(L/K)\sigma = G(L/K)}$

[cloud walrus]

Moldilocks

squirtlespoof

[hidden haven]

yup

[cloud walrus]

squirtlespoof

[hidden haven]

Correct

Succinctly, automorphisms that restrict to identity on K are exactly the ones that fix K elementwise

right

for all elements x in the normaliser of a group H you have xHx' = H

x' being the inverse

Take fixed fields on both sides

and you maintain equality

Try continuing from here?

fixed field of G in L = L^G = {all elements of L that map to themselves under all elements of G}

yes

yes

you wont be conjugating the field with sigma

you want to show that sigma fixes L^H elementwise if sigma normalises H

(and K is a field, so L^K isnt defined)

thats "sigma normalises G(L/K)"

but then why must it fix L^G(L/K)?

[hidden haven]

the point of this will be the H = G(L/K) gives L^H = K and youll get exactly what is needed

[hidden haven]

That restriction is a group homomorphism?

Yeah the composition of the restrictions is the same as composition of the original maps, restricted

[ivory dust]

hey guys anyone know how to approach this?

I want:
(2+x)(ax^2+bx+c) = 1 in Z3

given x^3+x^2 = 1

basically trying to find inverse of the element (2+x) in quotient ring Z3[x] / (x^3+x^2-1)

[woven delta]

Well let's solve for a, b, and c

[ivory dust]

i tried solving and ran into so many contradictions 😭

cuz its over Z3

i got like c = 0 then c= 2

if i tried in diff metho

method*

[hidden haven]

What were your equations?

[ivory dust]

this

[woven delta]

Well first reduce the cube

[hidden haven]

And you set this equal to?

[ivory dust]

= x^3+x^2

[hidden haven]

It could be a multiple of that

So if that gives contradictions

Try equating with 2(x³+x²-1)+1

[ivory dust]

oh ok ill try rn

i also tried

x^3 = 1-x^2 = 1+2x^2

theres only 27 elements (- 2 elements that are trivial) so i tried brute force and still couldnt find lol

well not comprehensive brute force but tried a bunch

[hidden haven]

Brute force

[woven delta]

Okay using x^3=1-x^2, we get ax^3+bx^2+2ax^2+cx+2bx+2c = bx^2 + ax^2 +cx + 2bx+2c+a=1

So this gives us (a+b)x^2 + (c+2b)x + 2c+a-1 = 0

[ivory dust]

bx^2 + ax^2 +cx + 2bx+2a+2c = 1 ?

[woven delta]

Oh sorry

[ivory dust]

bx^2+(2b+c)x+2(a+c) = 1

is what that reduces to

i thnk?

Also side note, there should be an inverse no? its Zp[x] / irreducible ideal of

Zp[x]

[woven delta]

Okay I corrected my above work

Sorry am writing on my phone on a train lol

[ivory dust]

no problems

[stone fulcrum]

Indeed, this should be a field because maximal ideal

[woven delta]

So that gives us 3 equations, a+b =0, c+2b = 0, 2c+a-1=0

Wait sorry I made another mistake lol

I wrote a 1 where I should have written an a

Anyway yeah, 3 equations 3 unknowns, you know the drill from here

[ivory dust]

wait

oh

ok i got a=2, b=1, c=1

[woven delta]

Nice

[ivory dust]

but when i multiply (2+x)(2x^2+x+1) = 2x^3+5x^2+3x+2 = 2x^3+2x^2+2

is 2x^3+2x^2+2 = 1?

[woven delta]

Reduce mod x^3+x^2-1

[ivory dust]

omfg

it is

😩 thank you

thats prob why i kept getting tripped up

vevery time id expand it to check answer id think its wrong

[hidden haven]

[tall jay]

I was wondering if someone is available to help me with this proof.

[sharp sonnet]

write the prime factorization of a and b in terms of d, then consider the prime factorization of ab
for further help #elementary-number-theory

[cloud walrus]

bim

[gritty sparrow]

Can you think of an open cover of G that might have a finite sub cover? (Remember: G0 is open)

[vestal snow]

Let U, V be open affine subschemes of a scheme X. Given a point p in U \cap V, show that there is an affine open subscheme W containing p which is contained in U \cap V

Here's the problem

Never mind I figured it out

[chilly ocean]

Is there a typo here?

|x-s| <= 1/2 and |y-t| <= 1/2?

[sturdy marsh]

you can prove something a lot stronger

Nike's trick flashbacks

[rustic crown]

so x and y are actually what you're thinking as |x-s| and |y-t|

[gritty sparrow]

No, s and t are integers x and y are the distance from the integers

[chilly ocean]

Ah right, the notation slightly confused me

thanks 💯

[vestal snow]

Lol I think I remember this discussion from last year

I'm trying to test my intuition for locally ringed spaces. The stalk at a point p measures the properties of elements when restricted to an "infinitesimal" distance around p, and taking the field of the stalk measures the value of the element at just p.

Does this sound correct?

[lime skiff]

Hi, I'm trying to understand the structure of roots of unity in finite fields

to begin with, I'm tasked with showing that gcd(q,n) = 1 iff there exists a positive integer m such that n|q^m-1

(we're looking at finite fields F_q)

one direction is okay

suppose such an m exists, then kn = q^m-1 => q^m-1 * q + (-k) *n = 1

[carmine fossil]

You know Lagrange theorem?

[lime skiff]

so that gcd(q,n) = 1

err which one

[carmine fossil]

Order of subgroup divides order of group

[lime skiff]

ah yes

[carmine fossil]

Consider the group of integers coprime to n

By that,I mean look at (Z/n)^*

[lime skiff]

ah okay

well I know this has order phi(n)

but other than this

[carmine fossil]

q^(phi (n))=1 mod n

Implying n divides q^m-1

[lime skiff]

yeah that's fermat or something

OH

w e w

ahhh I see

you took m = phi(n)

that makes a lot of sense

another question is the following

suppose gcd(q,n) = 1, then all the roots of x^n-1 in F_q[x] are distinct

do I also need to use something like what we did above?

I know that x^n-1=(x-1)(x^n-1 + ... + x + 1)

so I thought this would have to do with cyclotomic polynomials

but I'm not fully sure what to do

mmmh I must've seen a result like this when I skimmed through galois theory

wait but mm

well n divides q^m-1 right

[carmine fossil]

Yea

[lime skiff]

I guess I just need to do some algebraic manipulation in the exponent

because I know x^q - 1 = 0 for all x in F_q

err

might be missing a minus 1 sorry

I meant to say x^{q-1} - 1 = 0 is true for every x in F_q

so in particular the roots are distinct

and now x^n should divide this or something

at least that's my intuition

maybe it's simpler than this

okay, I don't really know what to do I think

this question came before the other one so I guess I don't need to use n|q^m - 1

since n is coprime to q = p^r I have ap^r + bn = 1 so I thought I could use that in some way to modify x^n-1 but I don't see how immediately

x^q-1 / x^n gives me a fraction that I don't know how to simplify either

maybe that was the right idea? not sure

[chilly ocean]

I feel like I'm missing something when solving the following Diophantine equation: x^2 + 4 = y^3. I'm told to work in Z[i] where I can factorise x^2+4 into (x+2i)(x-2i) so that the equation becomes in factorized form (x+2i)(x-2i) = y^3. Now using the fact that Z[i] is a UFD, we can get the factorisation of y into irreducibles a_1a_2...a_t so that (x+2i)(x-2i) = (a_1)^3(a_2)^3...(a_t)^3. Notes say that there are then two cases to consider: either x+2i and x-2i are divisible by some irreducible a_i or x+2i and x-2i are associates of cubes. Can someone explain to me as to why those 2 cases are considered?

[rustic crown]

if (x +2i) and (x - 2i) are coprime and their product is a cube... then they themselves are (associates of) cubes

[chilly ocean]

it has nothing to do with the uniqueness of factorisation in this case?

[rustic crown]

didn't get you... we need unique factorization to conclude that. their product is a perfect cube so each irreducible appears a multiple of 3 times. and if x+2i and x-2i don't share irreducible factors, then each irreducible would occur a multiple of 3 times in each of x + 2i and x - 2i

if this was the case then we could say x+2i = u(a + bi)^3 where u = 1, -1, i, -i

(but since each of the units in the case is a perfect cute itself... i^3 = -i and (-1)^3 = -1, so wlog you may assume x + 2i = (a+bi)^3)

[lime skiff]

okay I figured it out

[chilly ocean]

Hmm I sort of get it. So equality of products holds (x+2i)(x-2i) = (a_1)^3(a_2)^3..(a_t)^3 is (x+2i) is the cube of some of elements of the fact of y and (x-2i) is the cube of the other products

[rustic crown]

only when they are coprime... you need to look at the case when they do share some divisor

notice if there was an irreducible factor dividing both x + 2i and x -2i then it would divide their difference 4i and so would be associates with 1 + i

[chilly ocean]

thats how I proceeded actually

if some irreducible a_i divides both x+2i and x-2i then it would divide 4i = (1+i)^2(1-i^2)

[rustic crown]

yea 4i ~ (1+i)^4

[chilly ocean]

so a_i | (1+i)

right right i get it

(1+i) = k*a_i so k must be a unit as 1+i is irreducible so they are associates

[rustic crown]

yep

[chilly ocean]

How can you see that so quickly

[rustic crown]

a_i is an irreducible factor of 4i ~ (1 + i)^4 which has only one irreducible factor (upto ~) which is 1 + i

[chilly ocean]

When doing the exercise I've written something else and I wonder if its correct

With a_i | 4i = -i * (2i)^2 so as -i is a unit, a_i and -i are coprime so a_i | 2i

[rustic crown]

a_i and -i are coprime
this is a weird thing to say... you're saying like in Z, 100 and 1 are coprime

my eyes don't even look at the units lol

[rustic crown]

but yea its correct

[chilly ocean]

Hmmm

Got it

the formulation is a bit dodgy

indeed

ah also if the irreducible a_i divides -i x (2i)^2 then since a_i doesn't divide -i (if it did then i = a_i*k would mean that both a_i and k and are units contradiction), it is the case that a_i divides 2i and 2i is irreducible so a_i and 2i are associates so 2i | x + 2i so 2i | x which means -4 | x^2 and x is even which is a contradiction (assuming x and y are odd)

[rustic crown]

2i isn't irred

[chilly ocean]

ah true true you're right

[rustic crown]

remember we said if x +2i and x - 2i were not coprime then an irreducible common factor would divide 4i ~ (1+i)^4, so that irred factor would be 1 + i, now it has to divide x + 2i. and it already divides 2i which means 1 + i divides x. this only happens when x is even. (there are plenty of ways to see this, one way is to notice that N(1+i) divides N(x)... so 2 divides x^2 which means x is even)

[chilly ocean]

actually 4i = -(1+i)^4 I just computed it explicitly 🙂

[chilly ocean]

yeah so x+2i is a multiple of 1+i of even norm so x+2i has even norm too ie x must be even

[rustic crown]

yep

This emote is sooo cutee

[chilly ocean]

[rustic crown]

(assuming x and y are odd)
you can't just assume this right?

or did you already handle the case when x was even?

[chilly ocean]

the question asks for when x and y are both odd

[lavish spoke]

is xy^4 + x^3 y^3 - 12 irreducible in Q[x, y] = [Q[x]][y]? The solutions say p = x works with Eisenstein's with R = Q[x], but surely x doesn't divide -12?

it doesn't even seem like they're talking about the same polynomial

[cloud walrus]

det

[lavish spoke]

why?

the working makes a lot more sense in that case tho

[rustic crown]

say f(x) in R[x] has degree n. Then consider x^n f(1/x).
if f(x) = g(x) * h(x)
then x^(deg f) f(1/x) = x^(deg g) g(1/x) * x^(deg h) h(1/x)

x^n f(1/x) this operation is like reversing the coefficients

[lavish spoke]

ohhh ok

that was never pointed out explicitly in my course but makes sense

[rustic crown]

[delicate bloom]

maybe that's what they meant by 'inverse eisenstein's criteria'

[chilly ocean]

A nonempty subset H of a group G is a subgroup of G if and only if (1) ab ∈ H for all a,b ∈H and (2) a^−1 ∈H for all a ∈ H.

If we want to prove => direction, we can assume that (H,*) is a group right? Since (G,*) is a group

And we get the => direction for free, since every element of a group has a unique inverse and a group must be closed under multiplication

[rustic crown]

you also need to show that e_H = e_G to be able to say that unique inverse imply a^-1 in H.

[chilly ocean]

Oh okay

Anything else?

[rustic crown]

rest looks good

[chilly ocean]

For showing that e_H = e_G, we know that e * f = f in G, and ff = f in H where e = e_G and f = e_H

Can we say that therefore e * f = f * f?

[tough raven]

Yes, subgroups must be groups.

[delicate bloom]

since you know a and a^-1 are in there, and so is their product a*a^-1, you know it has e

[chilly ocean]

It does have e, but I think det is right that we need to show that it is the same e

Since it might not be true for all elements

[tough raven]

For the forward direction, don't you assume the multiplication, identity and inverse are the same for the subgroup?

[chilly ocean]

But don't we need to prove that they are?

Also what do you mean the multiplication?

[tough raven]

Well, IG it depends on how exactly you defined subgroup.

Never mind me.

[rustic crown]

yea e_H satisfies x^2 = x in G, but x^2 = x must mean x = e_G

[rustic crown]

yep both are equal to f.

cancelling f gives e = f

the very simple thing i should point out is that f * f = f in H

but we also get that f * f = f in G

[chilly ocean]

Why do we also get f * f = f in G?

[rustic crown]

okay what is a subgroup of (G, *)?

its a subset H, such that the restriction of * : G x G --> G to H, ie *|_H : H x H --> G has image in H and (H, *|_H) forms a group.

f was an element of H, so it's also an element of G. and f * f in H is defined as f * f in G.

i feel like i'm making it more complicated ._.

[chilly ocean]

I guess so

No it makes sense

I just had to make sure that f * f is really defined the same as in G

@rustic crown Why isn't *|_H : H x H -> H though?

[rustic crown]

i just restricted the function (*) G x G --> G to the subset H x H. we require this restriction to map into H which same as saying we require H to be closed under *.

like you said closure is literally free, both sides assume that H is closed under *.

[chilly ocean]

Right

But don't we define *_H as H x H -> H

[rustic crown]

yea... but going from * you only get a map H x H --> G. without any further assumptions this is all you get...

[chilly ocean]

Because all subgroups are groups, and therefore we could treat them as just that, without "extra" information

Why not a map from H x H to H though lol

[rustic crown]

its just saying for an arbitrary subset H, product of two elements of H might also lie in G and not necessarily in H.

[tough raven]

You define the function _H as acting the same way as it does for * : G x G -> G
i.e.
h1 (_H) h2 = h1 * h2 necessarily for all h1, h2 in H
So you can't "choose" the values of *_H to make them in H

You need H to be special enough that it will always be in H

Which is closure under multiplication

[tough raven]

which is the same requirement on both sides

[chilly ocean]

I see

So a subgroup is a restriction of the map that we already have

It's not a new "thing"

[rustic crown]

the operation is a restriction.

[chilly ocean]

Okay

[unique juniper]

if R is a domain and a polynomial is reducible in R[x]. it will always be reducible in its ring of fractions right?

idk why need the coeffs of p(x) to have no multiple in common

[carmine fossil]

Yes

[unique juniper]

it says that it doesnt work if gcd is not 1

hmmmm

[carmine fossil]

nvm,it doesn't work

[unique juniper]

i cant think why

[carmine fossil]

2+2x^2=2(1+x^2)

[unique juniper]

it wont

[carmine fossil]

Reducible in Z[x]

Irreducible in Q[x]

[next obsidian]

It’s because you could factor out an irreducible constant in R

But that becomes a unit in the ring of fractions

What Buncho Dragons did is an example, you can factor out 2 in Z[x] and this shows it’s the product of two irreducible things

But in Q[x] the 2 is now a unit

[unique juniper]

AHHH

THANK YOU

[fast pine]

Hello, I have the following question:
For which unique value m contains Z_4 a non-trivial quotient ring Z_m?
Does anybody which m it is and why?

[viscid pewter]

apply lagrange's?

[fast pine]

Mmm I don't really know how applying Lagrange in some way leads to the answer

[viscid pewter]

so what values can m be just by lagrange's

[fast pine]

aha wow i think i get it, thank you! because Z_4 has 4 elements a quotient ring could only have 1,2 or 4 elements because it needs to be a divisor of the amount of elements of Z_4 (which is 4). A quotient ring with 1 or 4 elements would be a trivial one, so the only option is 2 elements, so m=2. Is this true?

[viscid pewter]

yep

[fast pine]

alright thanks!!

[chilly ocean]

What's a good source to learn about rigged Hilbert spaces?

[cloud walrus]

reking

reking

[gritty sparrow]

Even more n divides [K:F]

Do you see how that would help

[sturdy mirage]

i see how that would help, but I can't quite seem why n divides [K:F]

oh wait, i think i see that..

[gritty sparrow]

Nice

[cloud walrus]

reking

[gritty sparrow]

Yep

[sturdy mirage]

Thanks!

[gritty sparrow]

No

Problem

[sturdy mirage]

Oh?

[gritty sparrow]

I meant to say no problem, but it got split into two lines for some reason

[sturdy mirage]

Save him, not hang him

[opal osprey]

How could I show that $(7,3+\sqrt{-5})$ is a prime ideal in $\mathbb{Z}[\sqrt{-5}]$ without much computations?

[cloud walrus]

MisterSystem

[oblique river]

what is "computation" to you? I would use the isomorphism theorems in ring theory

I would write something like

Z[sqrt(-5)] = Z[x]/(x^2 + 5)
and therefore
Z[sqrt(-5)]/(7, 3 + sqrt(-5)) = Z[x]/(x^2 + 5, 7, 3 + x)
but when we're quotienting a ring by multiple elements, we can choose the order we quotient in. So, instead of quotienting by x^2 + 5 first, let's quotient by 3 + x first

Z[x]/(3 + x) is isomorphic to Z, and the isomorphism Z[x]/(3+x) --> Z is given by sending x to -3

so Z[x]/(x^2 + 5, 7, 3 + x) is isomorphic to Z/(3^2 + 5, 7), or just Z/(14, 7), which is just Z/(7)

[maiden ocean]

i need someone to profinitepill me

on why they are cool

or what about the fact that infinite galois groups are profinite is intriguing

[opal osprey]

ooooh

[magic owl]

pro-objects in general are very interesting and often times constructions that make sense for the normal objects

like finite group constructions often give you profinite stuff

like in equivariant homotopy

u should read my reu paper sloth

its about Pro-Spectra

[opal osprey]

And I have proven before that if we have a prime $p \in \mathbb{N}$ such that $p \equiv 3 \mod 4$, then so $\mathbb{Z}/(p) \cong \mathbb{F}_{p^{2}}$, and so since this quotiente we have just found is a field, then $(7,3+\sqrt{-5})$ is maximal and a fortiori prime.

I get what you did

[magic owl]

and pro-hausdorff-groups

[maiden ocean]

Im doing etale things this summer under a prof/with a grad student

but this stuff seems neat

wikipedia says it is connected to cohomological dimension

[cloud walrus]

MisterSystem

[oblique river]

@opal osprey what you wrote isnt entirely correct -- Z/(p) is always isomorphic to F_p. but that's irrelevant here; we just proved that our ring mod our ideal is a field and therefore the ideal is maximal (and hence prime)

[opal osprey]

F_{p^{2}}

that's an exercise from Artin

[oblique river]

moth, from the perspective of galois theory, I wouldn't say it's particularly "interesting" that galois groups are profinite. It's like asking "why is it interesting that homology groups with real coefficients are real vector spaces". they just... are real vector spaces, because that's what they're defined to be

galois groups of infinite extensions are profinite because they are an inverse limit of profinite finite things

it's more like, "homology is nice and we can actually do stuff with it because we get vector spaces which are easy to understand"

[magic owl]

inverse limit of finite things right

[oblique river]

we can actually do things with infinite galois groups because they are profinite and therefore we can actually work with them

yeah sorry

[magic owl]

i forget if profinites are closed under limits

[opal osprey]

Exercise 5.7

It is quite nice

[oblique river]

i'm familiar

[opal osprey]

So yeah Z/(7) is a field

[oblique river]

but that's not what you're doing here lol

here we have Z/7

[opal osprey]

ooh

[oblique river]

not Z[i]/7

as I said before

[opal osprey]

I am sorry

[oblique river]

Z/p is isomorphic to F_p

[opal osprey]

I have read that the wrong way

[oblique river]

that is the definition of F_p lol

[opal osprey]

Wait uh, ok I think we are having a problem with notation here lmao

oH

YEAH

You are right

I was being dumb

[oblique river]

I'm concerned -- how did you solve that exercise?

if you thought that Z[i] was just Z

[opal osprey]

No

For some reason I jsut read what you did and thought

ooh

I did something similar

but nah

I have proven something different lmao

It's jsut what happens when you read something quick

I mean

It's in portuguese

so

meh

I used the fact that Z[i]/(p) is isomorphic to F_{p}/(x^{2}+1)

[oblique river]

that's right

[opal osprey]

And then tried to specify the structure of F_{p}/(x^{2}+1) in the case p = 3 mod 4

[oblique river]

and the way you prove that fact is doing exactly the same thing I suggested for the problem you asked about

[opal osprey]

I thought since we had a weird quotient we had to do more computations and such

I was just being lazy and didn't want to get my hands dirty

but yeah

I get the idea you did

anyway thanks

[oblique river]

np

in my opinion, that kind of computation is "the way" to do these kinds of problems

[cloud walrus]

beeswax

[oblique river]

maybe try #elementary-number-theory

[tall jay]

Gotcha. I'll repost the question

[chilly ocean]

Hi guys, I'm struggling on this question quite a bit if anyone wanna help

So I've found the centraliser

But I'm not sure about how to write down the conjugates of M in H

The product gives

$$\begin{bmatrix}
1 & 1+f \
0 & 2
\end{bmatrix} $$

[cloud walrus]

Laïka

[chilly ocean]

But I don't see a generalisation

Also the solution says there are 56 conjugates of M in G but I guess not all of them are in H

Hehe given the size of G that's reasonable

Hmm moldilocks you're not quite happy with that

[hidden haven]

no it just seems like a big number

[chilly ocean]

Yh but I've worked it out it seems legit

[hidden haven]

yeah its probably correct, but big numbers are scary

[chilly ocean]

$$\begin{bmatrix}
a & b \
c & d
\end{bmatrix} \times \begin{bmatrix}
1 & 1 \
0 & 2
\end{bmatrix} = \begin{bmatrix}
a & a+2b \
c & c+2d
\end{bmatrix} $$

[cloud walrus]

Laïka

[warm holly]

did yu compute the inverse multiplication as well? then you can set an equality

[chilly ocean]

$$\begin{bmatrix}
1 & 1 \
0 & 2
\end{bmatrix} \times \begin{bmatrix}
a & b \
c & d
\end{bmatrix} = \begin{bmatrix}
a+c & b+d\
2c & 2d
\end{bmatrix} $$

[cloud walrus]

Laïka

[warm holly]

ah ok

[chilly ocean]

Yeah was about to right that dan

a+c = a implies c =0

then a+2b = b+d implies a+b = d

so we get a matrix like this

$$\begin{bmatrix}
a & b \
0 & a+b
\end{bmatrix}, a \neq 0, b\neq -a$$

[cloud walrus]

Laïka

[warm holly]

so the centralizer is H no?

[chilly ocean]

uh centralizer of M in G

[chilly ocean]

uh in H d need not be a+b I assume

or it doesn't matter?

[warm holly]

I didn't calculate but you wrote that "a+2b = b+d implies a+b = d"

then ye it's an entire H the centralizer

I guess.. If someone could confirm it would be better xd

[chilly ocean]

$$\begin{bmatrix}
1 & 1 \
0 & 2
\end{bmatrix}$$ is in the centraliser but $$\begin{bmatrix}
1 & 1 \
0 & 3
\end{bmatrix}$$ is in H but not in the centralizer?

[cloud walrus]

Laïka

[chilly ocean]

thats whats making me doubt

[warm holly]

Ah ye so it’s not the entire H it might be a sub group of G but if you give an answer explicitely that centralizer= obtained matrix it’s ok that is what the question asks

[chilly ocean]

cool thanks ^^

but for the very last part im unsure how i'd use that product to generalise and get all conjugates of M in H

[chilly ocean]

Nvm, figured it out

[hidden haven]

What was the solution

[chilly ocean]

weirdly they wanted us to get inspired from the product they asked us to compute lmao

the upper left and lower right entries are independent of and b and they're always fixed 1 and 2

I've got another thing I'm stuck on

Let G be a group - an element of G is always-0 if f(g) = 0 for any homomorphism f from G to (Z_2,+). Given an always-0 element g in G and a normal subgroup H prove that g is in H.

That looks trivial if we assume H is the kernel but what if not?

[hidden haven]

You can construct a map which has H as kernel

[chilly ocean]

Hmmm

G -> G/H

f(g)= gH

[molten silo]

How do i show R[x,y] is not a principle ideal domain

[hidden haven]

Yes

[chilly ocean]

show that its not a UFD for example?

[chilly ocean]

Ye but then how do I prove that g* (always 0 ) is in H?

[hidden haven]

gH = 0H by definition of always-0

[hidden haven]

It will be a UFD, since it's a polynomial ring over a UFD

[chilly ocean]

Ah R is the reals right?

[molten silo]

yes

[chilly ocean]

How is that?

[hidden haven]

The ideal (x,y) is not principle, use a degree argument

[hidden haven]

Quotienting is a homomorphism

So the always-0 element must map to 0

[chilly ocean]

the always-0 is defined in the question as this

thats what confuses me

always-0 isn't 0

[hidden haven]

oh homomorphism to Z/2Z

[chilly ocean]

Ahhh

[chilly ocean]

what if we modify this

and get G -> theta(g)*H?

but then H isnt guaranteed to be the kernel alas :/

[hidden haven]

[chilly ocean]

Idk how to make the link between the isomorphism for which H is a kernel and some isomorphism between G and Z/2Z

[tough raven]

What is G though?
GL_2(Z/7Z)?

[chilly ocean]

G is anything

Ah

yeah

you're talking about this

[tough raven]

[tough raven]

What
Only a subgroup of index 2 right?

So the always-0 elements are precisely the intersection of all subgroups of index 2.

[tough raven]

So whatever modification, this will only work if the intersection of all normal subgroups is the intersection of all normal subgroups of order 2

Which seems rather unlikely

[tough raven]

Wait what if H = 1?

[chilly ocean]

you're right

I've been thinking too much about this

[chilly ocean]

yep

[tough raven]

🤔

At least for S_n I think this will be A_n.

n >= 5

[chilly ocean]

Yeah

[tough raven]

Since given two distinct subgroups of index 2

[chilly ocean]

But I want to show that given any subgroup H of index 2 an always-0 element belongs to it

[tough raven]

I think their intersection is of index 4?
And it's normal in both.

[tough raven]

Well, that's the definition + rephrasing.

[chilly ocean]

yeah the proof is still missing

alas

[tough raven]

This is the proof, right?

[chilly ocean]

Well not really

f: G->G/H is a homomorphism with kernel H

but then i'm stuck right

[tough raven]

Are you?

What's the point of assuming that H has index 2?

Use that somehow.

[chilly ocean]

That it is normal

[tough raven]

Oh.

Then it would have been true for any normal subgroup.

But it's only true for subgroups of index 2.

[chilly ocean]

Yeah...

[tough raven]

What is G/H?

Concretely.

[chilly ocean]

all left cosets of H ?

xH for x in G

[tough raven]

And how is index of a subgroup H defined?

[chilly ocean]

The number of cosets

hmmmmm

[tough raven]

[chilly ocean]

|G/H| = |Z2|

But then?

[tough raven]

Well

What do you want to do?

What would suffice to finish the problem?

[hidden haven]

How many groups of order 2 are there

[chilly ocean]

1 only

so G/H and Z2 are isomorphic

[hidden haven]

[chilly ocean]

Right so just to recap

f: G->G/H is a group homomorphism

with kernel H

since G/H has order 2

every element of G is mapped to H or to G \ H (without H)

but G/H and G are isomorphic so the elements mapped to H are exactly those that are mapped to the identity?

Can it be more formal than that?

[hidden haven]

H is the identity in G/H

G/H and G are not isomorphic

You have G → G/H → Z/2Z, where the second map is the isomorphism

Composition of the 2 maps sends only the elements of H to 0

[chilly ocean]

do you have to define G/H -> Z/2Z explicitly?

[tough raven]

Well

You can define
h in H -> 0
h not in H -> 1

And 'directly' define the map from G to Z/2Z

Or you can say G/H has two elements, the identity (all H) and the other (all not in H).
The unique isomorphism of this two-element group to Z/2Z maps identity to identity and non-identity to non-identity.

[tough raven]

which ends up the same as this.