groups-rings-fields

[languid trellis]

hmm

[coral steeple]

Anyone able to discern what a is in part b?

[rocky cloak]

An integer

[coral steeple]

Ah wait we know that since r1 is in I1, there is a so that alpha is in I. I think that's how a is defined

[rocky cloak]

That is indeed how they defined I1

[coral steeple]

Although there may be different a so that alpha is in I1, so it's not clear how we know what a or alpha is...

[rocky cloak]

It's just any such alpha

[coral steeple]

Ah, ok. I am not used to such terse definitions

[rocky cloak]

Yeah, it's a bit terse

[sonic coral]

Can i get a hint on how to show that [Q(zeta_12) cap Q(3^(1/12)) : Q] = 2

i’ve realized that Q(zeta_12) contains sqrt(3) which i think is helpful but im not sure how to construct the minimal polynomial

[dim widget]

Is this a Galois theory class or a number theory class

But anyway I am would use Galois theory

Although there is also a nice proof with ramifications theory

But the point is that Q(zeta_12, 3^1/12) is galois

[sonic coral]

it’s a galois theory question

i am really after showing the degree of the splitting field of x^12 - 3 over Q is 6*phi(12)

i know that the degree of x^n - a over Q is either n*phi(n) or (n/2)*phi(n), so i need to show that the degree of the intersection over Q is 2

[dim widget]

Okay well what are all the roots of that polynomial?

As complex numbers say

[sonic coral]

12th root of 3 times zeta^i _12 for i ranging between 0 and 11

[dim widget]

Okay

I see your issue

Well what is the galois group of Q(zeta_12)?

[sonic coral]

i think it’s Z_4

[dim widget]

No

[sonic coral]

klein 4 then

[dim widget]

Yep

And what element is complex conjugation?

[sonic coral]

i mean that will have order 2 so couldn’t it be any of the three

[dim widget]

Well yes

But why don’t you fix an identification of the galois group with Z/12^* explicitly

This will make everything easier to describe

Although it doesn’t really matter I suppose

The point is that Q(3^1/12) is contained within the fixed field of complex conjugation

Do you see how to conclude from here?

[sonic coral]

i don’t think so

[dim widget]

okay well the intersection of these two fields is totally real right?

since one of them is a field consisting only of real numbers

so $Q(3^{1/12}) \cap Q(\zeta_{12})$ is contained in the fixed field $F \subset Q(\zeta_{12})$ of complex conjugation

[cloud walrus]

Math_Discord_Final_Girl

[dim widget]

And thus F has galois group K_4/c \cong Z/2Z

So the intersection is either Q or a real quadratic field

and you have already seen that in fact the intersection contains Q(3^{1/2})...

[sonic coral]

this follows because Gal(Q(zeta12)/Q) is K4, which only has elements of order 1 or 2?

[dim widget]

yep, and complex conjugation is not trivial since Zeta_12 is not a real number

[sonic coral]

so is that what says the intersection is a quadratic field, or is it that along with the fact that sqrt(3) is in the intersection

i see that the degree should be atleast a quadratic extension since sqrt(3) is in the intersection

i think i get it, it follows from the fact that the intersection is atleast a degree two extension, but i have already shown that the intersection is at most 2 in a previous problem

[dim widget]

yep you got it

most useful technique in math is that if $x \leq y$ and $y \leq x$ then $y = x$

[cloud walrus]

Math_Discord_Final_Girl

[south patrol]

Source

[coral spindle]

This is the foundation of all analysis p sure

just shuffle the inequalities until it works

[topaz solar]

That and summing countably many things

[languid trellis]

#1217259452621783154

[white oxide]

kind of rusty on algebra - here $\mathbb{H}$ denotes the upper half plane. what do they mean to identify the two matrices? consider $SL_2(\mathbb{R}) / \pm I$ where $I$ is the identity matrix?

[cloud walrus]

okeyokay

[delicate orchid]

consider the algebra you get by quotienting out by the relator M ~ -M

[coral spindle]

"Identify" in math means "to make identical." So indeed they are trying to describe SL_2(R)/{\pm I}

[delicate orchid]

this terminology is more common in topology where u can quotient by whatever you want

[coral steeple]

Is it true that a+bi is irreducible in Z[i] if and only if a^2+b^2 is prime or =p^2, p prime and congruent to 3 mod 4?

I am seeing somewhat confusing results when looking online

Or is this correct?

Hang on isn't a Gaussian prime different from a prime element of Z[i]?

[delicate orchid]

no they're the same

and those two characterisations you posted look the same

[south patrol]

yeah

[dim widget]

yes but it is not trivial to show

[dull ginkgo]

Does this use quadratic forms

[dim widget]

i mean a^2 + b^2 is a quadratic form

but otherwise no

[delicate orchid]

3blue1brown video

[coral steeple]

Bombed my final

[coral steeple]

Thanks

[barren sierra]

it happens

you'll bounce back

[coral steeple]

Yep. Such is life

[hollow mica]

does every function f: Z -> R have an extension to a polynomial?

[cobalt heath]

I don’t think so, like, perhaps you can think of concrete examples

[lusty marlin]

Any function that takes some specific value infinitely many times but is not constant cannot be extended to a polynomial

[hollow mica]

oh I see, if c is taken on infinitely many times then f - c has infinitely many zeros

[lusty marlin]

Yes

[tough raven]

Nope.

[tough raven]

Exactly. For another example, f: n |---> 2^n cannot be given by a polynomial because it ‘grows too fast’.

[tough raven]

However, IIRC (although my memory is a little rusty) that any function f: Z -> R can be extended to a function given by a power series (that converges everywhere); this follows from the Mittag-Leffler's theorem and Weierstrass infinite products in complex analysis.

(It is essential for this that Z is closed and discrete: there is no point "approached" by integers.
This is analogous to the fact that functions on finite sets can be represented by polynomials.)

[torn shadow]

I am so lost, can anyone help?

[gray frost]

Thats more suited for linear algebra, but id suggest looking the vector spaces axioms up and then start proving my dude

[crystal vale]

If K is a subgroup of G(finite) whose order is divisible by p, then K is contained in a Sylow p- subgroup of G, right?

[rocky cloak]

I guess you can also just construct the power series directly:

f(0) + x(f(1) - f(0)) + (x/2)^2 (x-1)(f(2) - f(1) + f(0)) + (x/3)^3 (x-1)(x-2) ....

[tough raven]

How does one know that the coefficients converge?

Also, negative integers.

[rocky cloak]

The coefficients converge, because each coefficient only depends on finitely many terms. And yeah, I only did positive integers here, but same construction works for any countable set.

[tough raven]

👀

Doesn't the constant coefficient depend on every term?

Or x-coefficient.

[rocky cloak]

The constant coefficient only depends on f(0), since everything else is a multiple of x.

But I see now that you might get some alternating series, when you evaluate this at x other than 0. And then I guess you might get some well-definedness properties

Like does
1 - x + x^2 - x^3 ... vanish at x=1? Hard to say

So yeah, I retract my statement, you might need fancypants analysis stuff

[tough raven]

But the x-coefficient is then n! or something like it.

Thrs does actually sometimes work in p-adic settings, and you can use it to show that every function from N to Z_p which is p-adically (uniformly?) continuous (a purely number-theoretic condition) extends to a power series with Z_p-coefficients.

[rocky cloak]

Well, in what I wrote above the x coefficient is just f(1)-f(0), but as n gets bigger the x^n coefficient becomes more complicated yes

[tough raven]

Hmm, but still doesn't converge with no conditions.

[rocky cloak]

What do you mean?

For any family of polynomials fn (or even power series)
Sum x^n fn
Is always a power series

[tough raven]

When I started writing this message, I was going to say you can always interpolate if you use Z_p instead of R, but you can't.

(But maybe if you use Q_p?)

[tough raven]

Conditions on the original function.

[sonic coral]

Let $F = \mathbb{Q}(x,\sqrt{1-x^{2}})$. Show that $F$ is a purely transcendental extension of $\mathbb{Q}$ by showing that
[
F = \mathbb{Q}\left(\frac{\sqrt{1-x^{2}}}{x+1}\right).
]

[cloud walrus]

Jaxon

[sonic coral]

Is the idea to do subset arguments, like clearly if I have x and $\sqrt{1-x^{2}$, then I have $\frac{\sqrt{1-x^{2}}}{x+1}$

[cloud walrus]

Jaxon
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[sonic coral]

for the other direction. I want to right x and $\sqrt{1-x^{2}$ as some combination of $\frac{\sqrt{1-x^{2}}}{x+1}$?

[cloud walrus]

Jaxon
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[rocky cloak]

That's right

[sonic coral]

I think that i could write $x = \frac{1-t^2}{1+t^2}$ and get $t$ back

[cloud walrus]

Jaxon

[sonic coral]

and then from that, I can get $\sqrt{1-x^2}$

[cloud walrus]

Jaxon

[dull ginkgo]

I am confused with what I underlined (like shit)

So is (a,b) ~ (c,d) if bd_1 is not equal to db_1 vacuously?

[topaz solar]

No you need ad1 = cb1

The bd1 = db1 line is saying these are things which realize the right common multiple property for this pair

[dull ginkgo]

So it's (a,b) ~ (c,d) iff bd_1 = db_1

(a,b) ~ (c,d) implies ad_1 = cb_1

[topaz solar]

No that’s the condition to define ~

It sure implies it because that’s literally how you define ~

[dull ginkgo]

I'm saying is [(a,b) ~ (c,d) implies ad_1 = cb_1] a byproduct of common multiple property or necessary to define (a,b) ~ (c,d)

[delicate orchid]

it's the definition

like I don't know how to make it clearer without just writing it in set builder notation

[dull ginkgo]

So (a,b) ~ (c,d) <=> bd_1 = db_1 AND ad_1 = cb_1

[delicate orchid]

$(a, b) \sim (c, d) \iff [((b_1, d_1) \neq (0,0)) \wedge (bd_1 = db_1) \Rightarrow ad_1 = cb_1]$

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

wait what the fuck is this notation

my brain kind of just smoothed out the fact that a and c fucking vanish

oh no

there they are

what I wrote is correct

[spice whale]

funny uh
field of fractions but weird

[dull ginkgo]

so what the fuck am I showing after that then

independent of b_1 and d_1 is literally WHAT HE IS SAYING

[delicate orchid]

how on earth is that independent of b_1 and d_1 apriori

they're right there, three times each

[hot quartz]

Hey, I have this problem where G is an abelian group and g, h are elements of G such that ord(g) and ord(h) have no common divisor and I'm supposed to show ord(gh) = k = ord(g)*ord(h). The thing I'm struggling with is why (gh)^n is never 1 for n < k.

[rocky cloak]

So (gh)^n = g^n h^n, so if this is the identity
g^n = h^-n
Now think about the order of g^n

[hot quartz]

I'm not sure how the order of g^n would help, but it has to be the gcd(n, ord(g)), no?

[dire siren]

gcd*
and that has to also be equal to the order of h^-n

[rocky cloak]

It would be ord(g) / gcd(n, ord(g)), but yeah follow Filips hint

[dull ginkgo]

I am still confused as shit

Let me phrase what I am confused about better

I am wondering which does he mean:

1. for a FIXED (x,y) ≠ (0,0), (a,b) ~ (c,d) if and only if bx =dy implies ax = cy.

2. (a,b) ~ (c,d) if and only if **for each ** (x,y) ≠ (0,0) we have ditto

3. (a,b) ~ (c,d) if and only if ** there exists an ** (x,y) ≠ (0,0) such that ditto

[rocky cloak]

(2)

[dull ginkgo]

… then how the fuck is it dependent on the specific d_1 b_1?

[rocky cloak]

Well, if bx does not equal dy, then the implication trivially holds

[dull ginkgo]

What “choice” are we making if it’s a forall statement

[dull ginkgo]

Trivially or vacuously?

[rocky cloak]

Well, same

[delicate orchid]

I feel like you're overthinking this a tad

[dull ginkgo]

I am confused by where there’s a dependence in how it’s stated

I don’t see one

[rocky cloak]

If it's supposed to hold for all x,y, then it's independent of which x and y you look at right

[dull ginkgo]

Yeah, which is how he stated it, so why is he asking me to show the way he stated it is the way he stated it????

[delicate orchid]

I agree it's a bit stupid

[rocky cloak]

Their asking you to show it's independent though.

So why should there be a dependence

Can't be both dependent and independent

[delicate orchid]

jagr that's not what miz is saying

[dull ginkgo]

The way he stated it is without a dependence in the first place

[delicate orchid]

^

[dull ginkgo]

that’s what’s frying my brain

[delicate orchid]

the exposition is just garbage, the author should've labelled the relation according to a fixed pair (b_1, d_1) first and then shown that all of the relations are isomorphic (or asked the reader to show)

[dull ginkgo]

So the way he stated it was by fixing a (b_1, d_1)

[delicate orchid]

no, he did what you said

[dull ginkgo]

So basically (1) => (2) here

[rocky cloak]

I think actually just removing the phrase "show it's independent" is the cleanest

Cause it's sort of irrelevant

[dull ginkgo]

Fuck it moving on to the next part of the problem

[delicate orchid]

correct and based

I would've done that hours ago

[rocky cloak]

Well, I think you should still do the problem

Just worry about it being an equivalence, and not what "independent" means

[delicate orchid]

how do you do reflexive without a unit

nonunital chads win once again

[rocky cloak]

Wait, are these nonunital rings?

No, pfew, we're safe

[dull ginkgo]

Next part of the problem, I.e that it actually is an equivalence relation

Well, I’m doing a slightly more general problem

[hot quartz]

I think what I'm doing isn't exactly the intended solution. Is it just something very simple that I overlooked like "ord(g) and ord(h) are relatively prime so ord(g^n) can't be ord(h^n)"? That doesn't sound right at all though

[dull ginkgo]

*domain

Fuck

[topaz solar]

The definition as stated used (1) and he’s saying it didn’t matter what elements that satisfy the property of b_1, d_1 (I.e (2)) is the exercise

[dire siren]

no, the argument leads to ord(g) | n and ord(h) | n
but the use of the order formula is not really necessary:

by raising g^n=h^-n to the power ord(h), you get g^(n * ord(h)) = e, which implies ord(g) | n * ord(h), so ord(g) | n (because ord(g) and ord(h) are coprime), and in a similar manner you get ord(h) | n

[delicate orchid]

Dunno what the problem is, but I just thought I’d mention that statement in quotations is true if both g^n and h^n aren’t the identity - it follows by Lagrange if you define order of an element to be the size of the cyclic subgroup generated by that element

[hot quartz]

yeah this makes a lot more sense. since both ord(g) and ord(h) are coprime the least common multiple has to be ord(g)*ord(h), which is what I needed

[muted comet]

If the order of my group is Prime, it implies the group is cyclic. By contraposition, this means that if my group is non-cyclic, it implies its order is non-prime?

[delicate orchid]

yup

[muted comet]

Thanks!!

[dull ginkgo]

I see, thanks fam

My confusion lies with the fact that the way it’s stated, I cannot tell if $d_1b \neq b_1d$ implies $(a,b) \sim (c,d)$ vacuously

[cloud walrus]

Foghorn (*BWAAAA-UNNNGH*)

[topaz solar]

I think that’s the wrong order, but uh

[dull ginkgo]

Probably is, point still stands though

Because otherwise it’s just two equalities

[topaz solar]

You pick b_1, d_1 such that db1=bd1

[dull ginkgo]

This whole shit is confusing the ever living fuck out of me

So how does he define the fucking relation

[topaz solar]

Like just imagine you have a function on pairs, so you give me (b, d) and I give you (b1, d1) which satisfy this

This works via choice and your ring property guaranteeing such pairs exist

Show that any other choice function gives the same definition I guess?

[dull ginkgo]

?????

Then how the actual fuck is the relation defined this fries my fucking brain still

Aaaaaaaa

[delicate orchid]

I'm starting to side with mizalign on this one. What is going on

well written book!

[dull ginkgo]

This is case 3 as I mentioned

[topaz solar]

No?

[dull ginkgo]

Fuck it just going to prove the “forall” one is an equivalence relation

Makes more sense from the get go anyway

[topaz solar]

Real

[chilly ocean]

something something roots

[topaz solar]

When is R[x]/(f) not a field

[teal wind]

Could someone help me with this problem?

anyone?

[last spoke]

H is a subgroup so think about what elements ah look like and whether there are any ones that are easy to figure out

Think about what aH = bH means

[teal wind]

Thank you. This is what I have thusfar. What are your thoughts?

[zinc plaza]

How does G being commutative imply that all of these elements are distinct. Couldn't it be that one of the g^{-1} is one of the g?

Maybe this is more of a confusion about the semantics of the notation

The way I'm interpreting this is that there are these elements g_i and these elements g^{-1}_i, and g^{-1}_i is the inverse of g_i and vice-versa. In total there are 2m elements apart from f and e. But couldn't it be the case that the inverse of, say, g_3 is the same as g_7 and in this case there are actually 2m - 2 of these elements?

[sonic coral]

if g is not the identity and g = g^-1 then g has order 2, but f is the only element of order 2

[zinc plaza]

I think you misinterpreted my question.

I'm saying that some element $g_j$ might be the inverse of some element $g_i$, in which case $g_i^{-1} = g_j$. And in this case we would have a set ${e, f, g_1, ..., g_i, ..., g_j, ... g_m, g^{-1}_1, ... g^{-1}_m}$ that does not include $g_i^{-1}$ and $g_j^{-1}$ because they are just $g_j$ and $g_i$ respectively

[cloud walrus]

Sayu

[zinc plaza]

Actually, I just realized that I misinterpreted the notation

Or maybe I didn't...

[dull ginkgo]

@delicate orchid

IT WAS (3) FUCK

Every resource says it’s this definition

Problem is it’s not immediate that it’s symmetric

[wise hawk]

sorry if this is a stupid question but is Z_n the same as Z/nZ? i am seeing some sources saying that they are just alternative notations of the same thing, and some sources saying that they are not equal, but isomorphic and it's confusing the hell out of me

[topaz solar]

It’s all the same

I think?

[topaz solar]

This is scuffed af looking again ngl, look at how asymmetric af that definition is. This seems rather different to the original b_1, d_1 definition

I mean it’s not actually asymmetric since if I swap the two around ~ why not just swap bar s bar r

[next obsidian]

It’s sorta like, not but also is

[topaz solar]

But it reads weirdly sorted

[dull ginkgo]

It’s the standard Ore construction

[next obsidian]

They’re isomorphic absolutely, but Z_n is usually just “cyclic group of order n”

[topaz solar]

Makes sense

[dull ginkgo]

But they go “lol it’s an equivalence relation” when it’s not at all obvious that it is

[next obsidian]

Which is the set {1,g,…,g^n-1} with multiplication given by g^a•g^b = g^{a+b mod n}

It’s kinda bleh

[dull ginkgo]

Giving me a damn headache, I’ve been on this problem for 2.5 hours

I can’t even show symmetry of the relation

[next obsidian]

Hey, at least you can blame someone else

Now wait until you get stuck for 2.5 hours on something because you forgot some assumption or misread something

😎

At like 3am

The day of a presentation

On that topic

[topaz solar]

So, take \bar s, \bar r witnessing those \exists, then \bar r s1 is in S

[dull ginkgo]

Oh my fucking god yeah

[topaz solar]

Use the common left multiple in S between s1, s2 shenanigans

As for the right hand part about r1, r2, I dunno offhand immediately but you should try using that Ore condition and trying out whatever you get?

[crystal vale]

If the group is generated by {x,y | x has order n and y has order 2 and x^(i)y = yx^(-i) } , how can I show that there are exactly 2n elements, how can I prove that uniqueness of elements?

It is a dihedral group, I want a algebraic explanation

[dull marsh]

I think you can use induction on the length of words generated by x and y to show that they are elements of the form x^ay^b; From there it's obvious

[dull ginkgo]

I still can’t get symmetry

For the life of me I can’t cancel out enough stuff, the common multiple shit just keeps appending extra terms to the end I can’t get rid of

assuming s_1 r* = s_2 s* => r_1 r* = r_2 s*

I can’t show s_2 r’ = s_1 s’ implies r_2 r’ = r_1 s’

If I use common multiple stuff, I have extra terms I keep adding to the right but cannot cancel out

[topaz solar]

Might be possible to punt both r1, r2 into S

With a multiple

[dull ginkgo]

That’s what I’m trying to do

[topaz solar]

And from there things work nicer maybe

[dull ginkgo]

But getting back to the implication is what I get stuck at

Because then I have extra terms to deal with

[topaz solar]

Ah wait, see the Ore condition, that lets you pull things nicer than just in S

Because you get s in S, r in R

[dull ginkgo]

what

[topaz solar]

Which I mean is strictly weaker but it’s already asymmetric

Def. 2.1

Can get common multiples not just of the s terms

[cobalt heath]

Wiki states that ore condition lets you move around elts of S

[dull ginkgo]

Well when I do that I just get extra factors I can’t get rid of to get the implication I want on the right side

[cobalt heath]

What kind of factors?

[topaz solar]

Well we don’t need the same witnesses to be witnesses, we can have em be whatever

[dull ginkgo]

?????

[topaz solar]

Also yeah idk what factors exactly arise

[dull ginkgo]

I mean the terms that give the common multiples

[topaz solar]

Like, how does it look if you try to work it out

[dull ginkgo]

They linger on the right end and I cannot cancel them out

Like for example, start with s_2r’ = s_1 s’

[cobalt heath]

You just need to find one r' and s' satisfying the property.

[dull ginkgo]

Quantify r’s* = s’r*

But multiplying both sides of the above equality by either s* or r* gets you fucking nowhere

[cobalt heath]

Hnn

[dull ginkgo]

s_2r’ = s_1s’ => s_2r’s* = s_2s’r* = s_1s’s*

Fuck this

It’s so confusing to type out

[cobalt heath]

Wait, where is this implication from?

I don't see it from the definition

[dull ginkgo]

There we go

s_2r’ = s_1s’ => s_2r’s* = s_2s’r* = s_1s’s* but now we’re stuck

s_2r’r* = s_1s’r* = s_1r’s*… stuck again

Can’t do shit with that

[cobalt heath]

I can only see "exists s, r: .. and .."

No implication

[dull ginkgo]

Read it more carefully

I have been on this for 3.5 hours

And it is past midnight

how am I too fucking stupid to solve this

“Trivially it’s symmetric” is what another source says

Cool.

Very helpful. Not a single goddamn resource on why

[cobalt heath]

It is still "exists s, r: ss2 = rs1 and sr2 = rr1"

[dull ginkgo]

I am trying to find another one that satisfies the criterion

But swapping 2 and 1 assuming the first

But I can’t

[cobalt heath]

Another condition?

I mean

[dull ginkgo]

Because I am accumulating too many terms on the right I can’t do anything with

[cobalt heath]

This does not make sense to me, it seems you are confusing specific case exercise and the general definition

[dull ginkgo]

rS cap sR being nonempty just doesn’t seem helpful because when you add more terms to the right you literally can’t cancel them out when using the relation in the one direction

[dull ginkgo]

I don’t even know anymore

[cobalt heath]

Sorry for annoying you, maybe you are exhausted

[dull ginkgo]

I’ve been on this problem for like several days too

I just can’t do it, it’s extremely embarrassing

It’s a simple textbook problem

[cobalt heath]

Well, average noncommutativity exercise

[dull ginkgo]

It holds for monoids.

I can’t even find a goddamn resource on it

They all just imply it’s trivial which is borderline insulting

[cobalt heath]

My approach for showing commutativity of Theorem 2.5:
||Suppose "s0 s2 = r0 s1 and s0 r2 = r0 r1" for some (s0, r0).||
||One has "s3 r0 = r3 s0" for some (s3, r3). Then,||
||s3 s0 s2 = s3 r0 s1 = r3 s0 s1,
s3 s0 r2 = s3 r0 r1 = r3 s0 r1.||
||Take s' = s3 s0, r' = r3 s0 and you are good.||

But I dunno if you are solving the exercise or making sense of the theorem

[dull ginkgo]

This doesn’t show anything

[cobalt heath]

Wdym?

[dull ginkgo]

I don’t think it shows commutativity of it

Just another solution?

[cobalt heath]

Oh wait

Hnmm

[dull ginkgo]

See my struggle here

To get an element of s in front of the s_1 is where I get completely lost

[cobalt heath]

Let me think for a while

[dull ginkgo]

Because we’re still going to have an r at the end of our product so it can’t be in s unless we show the product ending in r is in S

[crystal vale]

How can I show that if x^(a)y^(b) = x^(c)y^(d) then a-c=0 mod n and b-d = 0 mod 2

[dull marsh]

x^(a - c) = y^(b - d)

The element on the left belongs to <x>

The element on the right belongs to <y>

What's the intersection of <x> and <y>?

[crystal vale]

Identity e

[dull marsh]

Right, so x^(a - c) = e and y^(b - d) = e

[crystal vale]

Oh got it, y not belong to <x> , that's an assumption I forgot

Thank you ❤️

[dull ginkgo]

@cobalt heath I think I have another route

[cobalt heath]

Oh

[dull ginkgo]

Instead showing equivalence to a more “obvious” equivalence relation that is more obviously symmetric

Also something of note

assume st is in S

[topaz solar]

Does this not work?

[dull ginkgo]

then there exists a t* s* such that
(st)t* = (s)s*
so tt* = s* in S

[cobalt heath]

It did not swap 💀

[dull ginkgo]

WAIT THIS IS THE IMPORTANT BIT

[topaz solar]

Ah rip

[dull ginkgo]

multiply both sides by t*, we have that the tt* part is in S

Satisfying our criterion

[topaz solar]

Ye

[dull ginkgo]

Fucking hell FINALLY

Of which I stumbled across by accident instead trying to prove transitivity

tldr if there exists an s such that st is in S, then there is a t* where tt* is in S

[cobalt heath]

Hmm, how does this help?

[dull ginkgo]

multiply both equalities to the left by t*

[cobalt heath]

Like, t* s s2 = t* r s1 ?

[dull ginkgo]

Assume $s_1r’ = s_2s’$, then $s_1r’ \in S$. However we know that $(s_1r’)r^* = s_1s^* \Rightarrow r’r^* = s^* \in S$ \\
Thus $s_1r’r^* = s_2s’r^$, and $r’r^ \in S, s’r^* \in R$

[cloud walrus]

Foghorn (*BWAAAA-UNNNGH*)

[cobalt heath]

But why is s1r' in S?

[dull ginkgo]

S is a multiplicative set,

s_2s’ is in S

[cobalt heath]

Ah wait, do we assume multiplicative set here?

[dull ginkgo]

Of which it’s equal to

[dull ginkgo]

He’s

*yes

[cobalt heath]

Ah, it is given by iteration, duh

[dull ginkgo]

In other words the relation is symmetric

However we have the beast known as transitivity now

[cobalt heath]

Why are s1 and s2 on the left?

[dull ginkgo]

Oh, I was using the right version

Same but everything is swapped lol

[cobalt heath]

Ahh

[dull ginkgo]

Actually transitivity might be easier

[cobalt heath]

Clever, making r'r* into S is wild lol

@dull ginkgo thank you for trying this stuff, cautionary tale for me to never go into noncomm territory

[muted comet]

can anyone give me a proof in affirmative or negative, that for a group. there does not exist an uncountably infinite number of subgroups?

a link would suffice.

[hot lake]

if 0 is in S then every pair is related to each other by taking s'=r'=0 and the theorem is trivial. So then we assume 0 isn't in S.

if (s1,r1) ~ (s2,r2) we have r,s' such that s' s2 = r' s1 (and s' r2 = r' r1).
s' s2 is in S, so there is another pair r'',s'' such that r'' s' s2 = s'' s1.
Then s'' s1 = r'' s' s2 = r'' r' s1. Since R is a domain and s1 is nonzero, you get s'' = r'' r'.
now picking s''' = s'' = r'' r' and r''' = r'' s', we have s''' s1 = r'' r' s1 = r'' s' s2 = r''' s2 and similarly s''' r1 = r''' r2, which shows (s2,r2) ~ (s1,r1)

[rocky cloak]

If X is a set, then in Z^X you can think of the subgroup generated by (1, 0, 0, ...), (0, 1, 0, ...), (0, 0, 1, ...) respectively. There are |X| such subgroups, so Z^X has at least |X| subgroups (in fact it has many more).

And of course |X| can be as big as you like.

[long obsidian]

Is there a specific type of morphism for group rings that respects the structure? Or is it just like, well you can have a morphism of groups between them and you could also have a morphism of rings between them

I guess you could say the ring of scalars is compatible? Otherwise I can't think of the structure of a group ring that would be natural to preserve

[rocky cloak]

Well, any morphism of groups induce a morphism of rings, so one is just a special case of the other there.

A group ring is a hopf algebra, so I guess you could look at homomorphisms of Hopf algebras, but I think that will just end up being the same as group homomorphisms.

[muted comet]

so we can have uncountably many?

[rocky cloak]

Yes, just pick X to be uncountable for example

[long obsidian]

Oh so then are the laurant polynomials a hopf algebra then? What is the structure maps that make this ring a hopf algebra?

[rocky cloak]

Yeah, the Laurent polynomials are just the group algebra of Z.

A hopf algebra is an algebra together with a comultiplication and antipode that satisfy certain relations ( https://en.m.wikipedia.org/wiki/Hopf_algebra )

For a group algebra kG, the comultiplication is given by
g |-> g(x)g
and the antipode is given by
g |-> g^-1

[hot quartz]

I need help in solving this problem:
If G is metabelian and ϕ : G → K is a group homomorphism, then ϕ(G) is metabelian.
I am supposed to use the homomorphism/isomorphism theorems, but we've never worked with metabelian groups or group homomorphisms before so I don't know what to do here or how to even approach this.
For reference the only definition of a metabelian group we were given is as following:
A group G is metabelian if there exists a normal subgroup A < G such that both A and G/A are abelian.

[delicate orchid]

Phi induces morphisms on A and G/A via restriction and composition with the canonical surjection respectively, now you just need to show that the image of something abelian (or normal) is abelian (or normal). This seems easier than the isomorphism theorems

Or just use the presentation of a group extension and one line it :trollface:

[rocky cloak]

I mean there aren't so many things you could do.

Just start with A and look at at phi(A), then see that it works.

[ebon pine]

Hello, I need some help with this question -
Let R be a commutative ring with identity. Let M be a maximal ideal of R and R* be the group of units. Then prove that M is a unique maximal unique ideal if and only if R = M union R*

I looked around for help and it seems that using the fact that, "Every ideal is contained in a maximal ideal" almost trivialises it. I am wondering if there's a proof that doesn't use that as the book hasn't introduced it yet.

[spice whale]

you can use Zorn's lemma to prove that every ideal is contained in a maximal ideal

then go from there

[ebon pine]

True but like is there no way to not use Zorn's Lemma

[spice whale]

hm

that's actually an interesting question

[hot quartz]

hmm... I'm still unsure how to show that phi(G)/phi(A) is abelian, does that derive from G/A being abelian or is it something else?

[ebon pine]

Well where I looked, all proofs use that fact

[hot quartz]

what does it even mean for G/A to be abelian? Is it g1g2A = g2g1A for all g1, g2 in G? If yes then the rest should be easy

[delicate orchid]

same thing it means for any group to be abelian

[rocky cloak]

Every ideal being contained in a maximal one is equivalent to the axiom of choice. And it wouldn't surprise me if the statement is no longer true when you remove choice. So I wouldn't think so.

[delicate orchid]

all rings are artinian and noetherian buster

[rocky cloak]

Indeed, if we assume the negation of choice, then there is a ring R without any maximal ideals. Then for a field k, kxR has a unique maximal ideal 0xR. But (1, 0) is a non-unit not contained in it.

[ebon pine]

Interesting... Are there any examples of such a ring

[rocky cloak]

Well, the axiom of choice is equivalent to "no such rings exist"

[ebon pine]

I mean if we do not assume choice then is there a ring which doesn't have a maximal ideal

Like a concrete ring

[south patrol]

You won't be able to write it down concretely without disproving choice

Lol

[ebon pine]

[rocky cloak]

I guess you want a ring, where is independent of ZF whether it has a maximal ideal.

I'm sure things like that exist, but I wasn't able to find anything by googling.

[ebon pine]

I found out that there exists rngs which do not have maximal ideals. Not sure about rings.
Also you mean unable to find?

[rocky cloak]

Yes, that's what I meant

[chilly radish]

@rocky cloak

https://math.stackexchange.com/questions/317028/a-confusion-about-axiom-of-choice-and-existence-of-maximal-ideals

relevant asaf karagila post

[rocky cloak]

I guess a maximal ideal of R^X modulo finitely supported functions is exactly given by a non-principal ultrafilter on X. And the existence of non-principal ultrafilters is a weak form of choice iirc. So that would be a pretty concrete ring

[chilly radish]

That would work I think

You can obviously never do this constructively tho

[rocky cloak]

Im not sure what it would mean to do it constructively

Like the bijection between maximal ideals and ultrafilters feels pretty constructive to me

So if we assume N doesn't have any non-principal ultrafilters, then we have constructed such a ring

[ebon pine]

Thanks, this cleared up the confusion

[topaz solar]

Good explanation

[topaz solar]

Well, you can’t prove it doesn’t have a maximal if you don’t prove X has no such ultrafilters etc, but yeah this seems fairly constructive up-to-acceptance of “X has no non principal ultrafilters”

But ya know, it’s strictly weaker than choice etc etc

But with AD, you have no such ultrafilters on N

So like, look internal to HOD which models AD iirc, maybe?

[chilly radish]

Karagila is goated

[celest furnace]

I’ve heard people hate these type of questions

If AC is false then there is a finite set with an infinite subset or something like that

Math gets suuuuper weird

[rocky cloak]

You have to be a little careful with statements like this. Just because a statement is independent of ZF doesn't mean that it follows from the negation of AC.

Also, you need to use a slightly more unusual definition of 'finite' for this to work. Namely, Dedekind-finite.

[celest furnace]

👍 So if you don’t assume AC. This isn’t the same as assuming not AC?

Wack

[dull ginkgo]

Axiomatic consistency

[celest furnace]

Also I had searched for that result for a long time, thanks for telling me about Dedekind finite lol now I can tell people this with confidence

[rocky cloak]

That's right. But also, even if you assume the negation of AC you cannot prove the statement you gave. That would only be the case if the statement was equivalent to AC

(which it isn't)

[celest furnace]

Ah

[dull ginkgo]

We don’t have enough information/axioms to answer it one way or another basically

[celest furnace]

Yeah this stuff suuuuucks

[dull ginkgo]

this rarely happens in algebra with ZFC apparently

[rocky cloak]

Well, if you assume AC then you can determine it (in the negative)

[celest furnace]

Assuming AC can show that finite sets do not have infinite subsets ?

That’s good.

[scenic saffron]

hi im not sure on how to check these

[topaz solar]

Or countable choice, even

[random pasture]

is showing that [x] doesn't always have an inverse sufficient for a proof?

[topaz solar]

No

Unless you can show x never has an inverse

[topaz solar]

You aren’t trying to show it’s not always a field, you are trying to show it is never a field

[random pasture]

i see

[topaz solar]

When is R[x]/(f) a field?

[random pasture]

when f is reducible

[topaz solar]

that is like the opposite of true

[random pasture]

mb i meant irreducible lmao

[dull ginkgo]

cubic… intermediate value theorem…

[languid trellis]

I'm pretty sure the more general criterion is R/I is a field iff I is maximal, which in this case corresponds to the ideal being generated by an irreducible polynomial

[topaz solar]

Yes

[south patrol]

yes

[delicate orchid]

For A, consider the map sending (12) to the reflection in D_3 and (123) to the rotation (this is natural if you label the corners of a triangle and think about D_3 acting on it)
B) is just computation of how {1,3,5,7} multiply mod 8
C) let (1,1) in G, show that this has order nm and thus generates all of G (hint: bezout’s lemma)
D) H is cyclic and G isn’t

[scenic saffron]

how does one send (1 2) to a reflection in D_3

not sure how to think of that one

and overall how do u actually check if two groups are isomorphic to one another

also i havent learnt bezouts lemma yet

[delicate orchid]

Construct an isomorphism as I just did for D_3 and S_3

[random pasture]

for b i keep getting that there isn't an inverse for some reason

[scenic saffron]

what should i ponder to try and construct an isomorphism

for this question, i guess its easy bc of what you said for D

that I cant

But in general

[delicate orchid]

(x+4)(ax+b) = ax^2+(4a+b)x+4b = -3ax-a+4ax+bx+4b = (a+b)x+4b-a = 1, now solve for a and b

[delicate orchid]

Idk just try the obvious things? D_3 is defined by permuting the three vertices of a triangle, and S_3 is defined as permutations on a 3 objects, so send elements of D_3 to their corresponding permutations

That’s the only tricky one

[south patrol]

The inverses are 1/45 and 1/(x+4)

[sonic coral]

don’t you mean -45 and -(x+4)?

[cobalt heath]

Well, multiplicative inverse

[fair lantern]

Hi! could someone help me with this?:
Let $H,K\leq G$ and let $H'\unlhd H$ and $K'\unlhd K$, then $H'(H\cap K')\unlhd H'(H\cap K)$

[cloud walrus]

Gkn

[crystal vale]

Use if H and N are normal then HN is normal , so H' and H intersection K' are normal in H intersection K

So both are normal in H'(H intersection K)

[crystal vale]

I want an example G is a non-empty set closed under binary operations and associative property hold. But only one cancellation.

So if let G = {a,b} and a•a = a, a•b = a, b•b=b, b•a=b then it is closed, associative and only right cancellation hold but it is not group.

Is it correct?

[dull marsh]

Yes

[crystal vale]

Okay, thank you

[hidden wind]

people doing abstract algebra in the 1850s be like

[clear shard]

Hello guys,Im trying to understand koszul duality, I've showed dualoty for symmetric and exterior algebra, and now I wanna find dual algebra of Weyl algebra,and get stuck,ars there any reference for this,or hints?

[sonic coral]

do f(x) and f(-x) have the same galois group? It’s just a reflection over the y-axis so my intuition tells me the automorphisms mappings will just pick up a negative but i’m not sure

[south patrol]

they have the same splitting field

assuming you mean galois group of the splitting field

[scenic saffron]

why if A | n and B | n, does that mean AB / gcd | n?

[lusty marlin]

AB/gcd(A,B) = lcm(A,B)

[dull ginkgo]

Once again struggling to interpret a jacobson problem

can "then" be replaced with "and" here

oh shit

misread it

nvm this problem isn't horrible

[next obsidian]

Lol

[dull ginkgo]

like finding the arc of the fucking covenant for jacobson

$b_1 b_2 = (b_1 a_2) (a_1 a_2)^{-1} (a_1 b_2) = (b_3 a_4) (a_3 b_4)^{-1} (a_3 b_4) = b_3 a_4$

[cloud walrus]

Foghorn (*BWAAAA-UNNNGH*)

[dull ginkgo]

i think it was a mistake and $b_4$ was supposed to be $a_4$

yeah it was looking later at the problem at the equivalence relation

[cloud walrus]

Foghorn (*BWAAAA-UNNNGH*)

[dull ginkgo]

Okay Malcev's construction is kinda sick

Freeness vs what happens when it's already in a group because embedding it in a group basically forces the extra relation which destroys the injectiveness

that's awesome tbh

okay thanks jacobson for the actually fun problem

[next obsidian]

You’re welcome

[dull ginkgo]

the problem before it was absolutely fucking abysmal

wait I realized the first part is fucked

$a_1 a_2 = a_3 b_4 = a_1 b_2$ in a group implies $a_2 = b_2$

[cloud walrus]

Foghorn (*BWAAAA-UNNNGH*)

[dull ginkgo]

i don't think that was intended

I am going to take a wild guess that the intended relations are $a_1 a_2 = a_3 a_4, a_1 b_2 = a_3 b_4, b_1 a_2 = b_3 a_4$ so that $b_1 b_2 = (b_1 a_2) (a_1 a_2)^{-1} (a_1 b_2) = b_3 b_4$

[cloud walrus]

Foghorn (*BWAAAA-UNNNGH*)

[dull ginkgo]

@next obsidian actually how would I "formally" prove the free monoid doesn't satisfy the last relation

[next obsidian]

What last relation

[dull ginkgo]

$b_1 b_2 = b_3 b_4$

[cloud walrus]

Foghorn (*BWAAAA-UNNNGH*)

[next obsidian]

What are these lol

Like it’s true if b3 = b1 and b2 = b4

[dull ginkgo]

Let $M = \langle a_1 ... a_4, b_1 ... b_4 : a_1 b_2 = a_3 b_4, b_1 a_2 = b_3 a_4, a_1 a_2 = a_3 b_4 \rangle$ as a monoid.
Let $\phi$ be a map from $M$ to group $G$. Then $\phi(b_1 b_2) = \phi(b_1) \phi(b_2) = \phi(b_1) \phi(a_2) (\phi(a_1) \phi(a_2))^{-1} \phi(a_1) \phi(b_2) = \phi(b_1 a_2) \phi(a_1 a_2)^{-1} \phi(a_1b_2) = \phi(b_3 b_4)$, so it CANNOT be injective

[cloud walrus]

Foghorn (*BWAAAA-UNNNGH*)

[dull ginkgo]

b ecause the free monoid doesn't have the relation b_1b_2 = b_3 b_4

but idk how to formally prove that

[next obsidian]

I mean that’s true because they’re formal now

But I don’t know if this construction is implicitly modding out by that relation

Cuz otherwise it doesn’t really make sense

I feel like it depends on what’s written on p 68

[dull ginkgo]

what do you mean

[next obsidian]

Like if it’s just the free monoid

[dull ginkgo]

M is the free monoid with the relations given

[next obsidian]

You turn these things into formal objects which only satisfy the most trivial relations

[dull ginkgo]

that's basically it

[next obsidian]

And then b1b2 ≠ b3b4

But if it went through the effort of saying it started with the given relation

[dull ginkgo]

but i need to prove that b1b2 = b3b4 is NOT implied

[next obsidian]

I expect that you’re also just forcing those relations

[dull ginkgo]

it's saying that there is no injective map into a group

[next obsidian]

Idk man

I’m sick

[dull ginkgo]

because phi(b1b2) = phi(b3b4) for any phi to a group

[next obsidian]

So I don’t have the mental energy to explain what I mean

[dull ginkgo]

fair

[next obsidian]

No that’s not true is my point if this is JUST the free monoid

[dull ginkgo]

get some rest and hopefully you feel better

[next obsidian]

But I suspect this is the free monoid mod out by the relations that ai bj have coming from G

And then you mod out by even more

[next obsidian]

This is cuz if this wasn’t true, there’s literally no fucking point in specifying that the elements satisfies relations

[dull ginkgo]

Coming from G?

[next obsidian]

Like for example modding out by [b1][b2] = [b3][b4]

Where [-] is indicating it’s the formal object inside the free monoid

Like my point is okay

Let’s say g in G and g^2 ≠ g as an object of G

Then when you take the free monoid generated by g and g^2

[dull ginkgo]

The fact is that those three relations without the fourth implies that there is no injective monoid morphism into a group

[next obsidian]

You have an element [g^2] and an element [g]

But [g]•[g] ≠ [g^2]

[next obsidian]

Yeah I don’t believe this

Because if you don’t mod out by the relations that the bi had coming from G

I don’t think those are ever of relevance ever again

Idk maybe I’m smoking pack

Okay maybe it’s true

But then I hate this problem because

The only way it would be true what you’re saying is if the b1b2 = b3b4 relation comes about from the other 3 relations

Which I guess is what your computation says

But umm

Yeah it shouldn’t have specified at the start you pick elements satisfying that 4th relation too cuz it isn’t a real assumption

anyway

[dull ginkgo]

It’s true

[next obsidian]

Ummmm

[dull ginkgo]

Look at phi

[next obsidian]

I mean to me it’s kinda clear idk

But I guess if you really wanted to prove it

You come up with a monoid M

And a set map {a1,…,a4,b1,…,b4} -> M

This defines the map from the free monoid to M

And you want to cook it up such that im(b1b2) ≠ im(b3b4)

[dull ginkgo]

Yeah

[next obsidian]

So I would just grab a group

[dull ginkgo]

I know some relations together can imply other relations

It’s showing that the three by default don’t imply the fourth

[next obsidian]

And pick 4 elements which don’t satisfy b1b2 = b3b4

[dull ginkgo]

But I guess you’d need to construct a monoid where the fourth doesn’t hold

[next obsidian]

Like pick Z/2Z and Z/2Z

Map b1,b2 to (1,0)

Map b3 to (0,0)

And b4 to (1,1) or whatever

The ai go to literally anything

Okay tadah

[dull ginkgo]

But the other 3 do

[errant wedge]

What's the difference between a free abelian group and a finitely generated abelian group?

[topaz solar]

Every finite abelian group is finitely generated for one

[errant wedge]

Just that the basis for a free ab grp need not be finite?

[topaz solar]

And free abelian need not be finitely generated

Z x C_3 is finitely generated, infinite, and not free abelian

[errant wedge]

Whats an example of a non-free finite abelian group? Sorry

[topaz solar]

?

Uhhh, any of them? (Except 1)

C_3 (I.e. Z/3Z)

Q is neither finitely generated nor free also

[errant wedge]

Why could you not choose {1} as your basis?

[amber wraith]

the representations wouldn't be unique

basically finitely generated abelian groups always have a free part and a torsion part, and the free abelian groups have no torsion part (https://crypto.stanford.edu/pbc/notes/group/fgabelian.html)

[topaz solar]

Free abelian has to map to everything doesn’t it?

C_3 -> C_5 is obviously never surjective

So this is a problem

[errant wedge]

Ahh ok

[amber wraith]

free abelian groups are basically Z^n, but you can also add on a bunch of cyclic components and they'll still be finitely generated

[topaz solar]

Don’t conflate finite generation with a basis, since we can also represent every integer with multiples of {2, 3} also

Even though Z is in fact free

[errant wedge]

thank you

[topaz solar]

I have trolled my algebra prof with lines like “there are only countably many free abelian groups”

“Up to forcing”

[dull ginkgo]

wtf is forcing

[topaz solar]

Goofy

Ofc Kryojyn disappears for this

[amber wraith]

forcing is very strange

take a model of set theory, get a new model of set theory, CH holds in new model

or ~CH

[topaz solar]

It’s a way to make new models

Generally speaking

And specifically the usual ways make a new model which preserves “what is an ordinal” and such

[amber wraith]

roughly though, you start with a model of set theory, within that model you can construct sort of "virtual models of set theory", and you also have a way of collapsing a "virtual model" into an actual model, and depending how you do it, you can "force" statements to hold in that new model

this is how is was shown in the 60s that CH is independent of ZFC

for which Cohen won the fields medal

lol had your algebra prof heard of forcing?

[topaz solar]

He had read Hodges’s building models by games a long time back

Nilpotent groups were the reason though

I don’t quite get how to make the generic ultrafilters for the Boolean valued approach still though

But trying to learn forcing is why I’m doing the CST thing

[amber wraith]

yeah I prefer the approach which doesn't use boolean valued models, I admit that the boolean valued contruction looks nicer, but the whole jump from the poset to the CBA generated by it is I dunno, abstract isn't the right word

something lol

it's too "something"

[topaz solar]

I mean, the regular open sets then it’s dense iff order isomorphic or wtv yada yada

Just take the limit of your finite embeddings

I don’t know the way to sidestep V^(B)

[amber wraith]

like how to do it without V^B?

[topaz solar]

Ye

[amber wraith]

you define M^P directly, instead of partial functions V^B -> B they're relations on M^P x P

I dunno, the V^B approach is nice honestly

lol I waffle on which I like better

but the two approaches aren't like, hugely different in spirit

and in the end M[G] is the same thing

[topaz solar]

Yeah

[amber wraith]

one thing though

I don't get the whole "name-value" terminology

[topaz solar]

Uhh names not being unique moment?

[amber wraith]

I'm surprised that it doesn't use some sort of "collapsing" terminology, it feels like a sort of Moskowski collapse sort of thing

[topaz solar]

No idea, hopefully I’ll better know after Halbeisen shenanigans (this is an ad :3)

[amber wraith]

I literally just mean why they use the terms "name" and "value" rather than like

G-sets

or like P-sets and G-collapse

I dunno lol!

[topaz solar]

Idk on that tbh, probably because labeling or something?

[amber wraith]

oh is this an alt-account for you?

[topaz solar]

what

[amber wraith]

are you also sagesharp or is that someone else

[topaz solar]

Same account

[amber wraith]

oh just different name

[topaz solar]

Also smh didn’t join

[amber wraith]

lol I'm going to

I didn't know by your message if you meant "don't bother" lol!

[topaz solar]

Anyhow, I mean the name-value rather than just G-set or wtv is maybe since they’re not quite sets in that they’re only “up to Boolean valued” shenanigans maybe?

Nah I meant “this might be stuff you already know”

Since like, literally talking about forcing and the like right now even

[amber wraith]

yeah, I'm still new to the subject, I want to mainly reinforce ideas, make sure I have good intuition

I hadn't heard of the name Jech or Kunen before October lol

[topaz solar]

Anyhow, combinatorial POV might be good anyway

[amber wraith]

yeah, chapter 9 in Jech I haven't read yet

[topaz solar]

Still been meaning to read more of jech

[amber wraith]

it's big

I gotta buy halbeisen, should be here in time

[topaz solar]

There’s always the contingency

[dull ginkgo]

[topaz solar]

I’m rather poorly versed in set stuff though oop

[dull ginkgo]

[hollow mica]

I assume the first k should be a K

But does anyone know what (K^x)^2 or D could mean

[sonic coral]

I think D is the discrimenant. It’s trying to say that if the Disc(f) is a square then Gal(L/K) is a transitive subgroup of S3 contained in A3, so is A3. If disc(f) is not a square then Gal(L/K) is a transitive subgroup of S3 not contained in A3, so is S3

[boreal inlet]

wait, is f a cubic polynomial?

[boreal inlet]

and yes, this is exactly it

[crystal vale]

I want to prove that if G is a non-empty group which has no non-trivial subgroup then G is a finite set of order p.

So let a≠e belong to G(infinite )then <a> = G and similarly a^2≠e because if a^2 = e then G has a non-trivial subgroup. So <a^2> = G.

So a in <a^2 > which shows that a has finite order , which contradicts that G has no non-trivial subgroup.

Now let G have order mn(both are greater than 1 ) then let a≠e then a^m ≠ e but it shows that (a^m)^n =e which contradicts that G has no non - trivial subgroup. So G must be the order of some prime p.

Is it correct?

[south patrol]

(note every group is non-empty, so you should say non-trivial)

[south patrol]

But yes, this is great

[crystal vale]

Yes, thank you

[boreal inlet]

This is cool

[minor dragon]

Does someone familiar with the theory of finite monoids here ? I’m reading a book on it and the author wrote « a finite monoid is a collection of finite groups loosely tied by a partially ordered set ». But he didn’t explained exactly what he meant by that.

[delicate orchid]

he could be referring to some kind of idempotent decomposition (with the normal ordering on idempotents), or is trying to (vaguely) describe partial groups?

[smoky barn]

Is it possible to represent a polynomial as a sum of two squares iver a finite field

[minor dragon]

What do you mean by partial groups ?

[delicate orchid]

I was misremembering the definition of a partial group so don't worry

is it true that in a finite monoid we have "Islands" of elements that are all mutually invertible connected by non-invertible elements? This is sort of how partial groups work but without inverses

no that clearly isn't the structure - all units are connected through the identity

[dire siren]

not necessarily
in characteristic 2, the square of any polynomial has the coefficient of any X^(2n+1) equal to 0
in particular X cannot be expressed as a sum of two squares

[minor dragon]

I don’t understand what you mean by « mutually invertible connected by non-invertible elements ». Are you talking about pseudo-inverse, i.e., about inverse monoids ? If it is the case, I was wondering the same, can we describe a finite monoid by a set of submonoids such that they are inverse monoids.

[stiff kraken]

Can someone help me with a problem? Let K be an extension of F and a an element in K but not in F that is algebraic over F, b an element in K that is transcendental over F. I have to show that there does not exist an x in K such that F(a,b) = F(x). I am struggling to see where I have to begin? Can anyone point me in the right direction? I've been trying to do something with tower rule and degrees of extensions but I am not sure that's the right way.

(sorry if it's the wrong channel)

[dire siren]

@stiff kraken if F(a,b)=F(c) for some c in K, then c cannot be algebraic over F (because of b)
prove that c being transcendental over K produces a contradiction with a∈F(c)-F

[stiff kraken]

trying that, I've already concluded that c must be transcendental due to tower rule but now I'm stuck, currently trying to see if maybe I can get a contradiction due to the fact that F(c) would be isomorphic to F(b), but not sure if that's the way

[dire siren]

you can prove that the elements of F(c)-F are transcendental

[stiff kraken]

Is there a characterization of the elements in F(c)-F? Is there a way to write them other than f(c)/g(c) for some polynomials f,g?

[stiff kraken]

I don't understand what I get by knowing that a is in F(c) but not in F, how does that help me compared to just knowing that a is in F(c)? If a is algebraic and it's in F(c), then there exists a polynomial p(x) such that p(a)=0 and a = f(c)/g(c) for polynomials in F[x]. Then p( f(c)/g(c) ) = 0 and I can conclude that if p(x) = x^n + c_(n-1) * x^(n-1) + ... + c_1 * x + c_0 then for h(x) = f(x)^n + c_(n-1) * f(x)^(n-1) * g(x) +... + c_1 * f(x) * g(x)^(n-1) + c_0 * g(x)^n, c is a root, therefore h(c)=0 and that would be a contradiction, as long as I know that h(x) is not the zero polynomial, and I assume that's where the assumption that a is not in F comes handy, I just can't figure out how.

[dire siren]

@stiff kraken we only care about the characterization of the elements of F(c), and we get a contradiction with the fact that a is not in F at the end
so, since a is in F(c), we have a=f(c)/g(c) for some f,g in F[x]
this implies f(c) - a*g(c)=0
so c is a root of the polynomial h(x)=f(x)-a*g(x), which is a polynomial in F(a)[x]

||if h is not the zero polynomial, then|| ||h(c)=0 implies that c is algebraic over F(a),|| ||but a is algebraic over K, so it follows that c is algebraic over K, contradiction||

||therefore h is the zero polynomial||

||in particular h(s)=0 for some s in F with g(s)=/=0 (we can find such an s because g is not the zero polynomial), which implies a=f(s)/g(s)∈F, contradiction||

[stiff kraken]

need a few mins, not reading spoilers yet, will try to figure it out on my own, thanks for the help and pointing me into the right direction

[dire siren]

btw, is F assumed to be infinite? I feel like we need this for the existence of s in the last spoiler

[stiff kraken]

it is not assumed to be infinite no

Why would we need it to be infinite?

[dire siren]

if g(x)=0 for all x in F, then we can conclude that g is the zero polynomial provided that F is infinite (because 0 is the only polynomial with infinitely many roots)

otherwise, if F is a field with q elements, then g(x) = x^q - x is zero for any element in F, but it is not the zero polynomial

[stiff kraken]

Isn't the fact that we know g(x) is not the zero polynomial enough for us to conclude that there is an s in F such that g(s) is not 0?

[dire siren]

no; see the counterexample above

[stiff kraken]

how do you mean x^q - x is zero for any element in F?

[dire siren]

so F-{0} is a group with q-1 elements, so a^(q-1)=1 for any element in F-{0}

then a^q=a for any element in F (including 0)

[stiff kraken]

yeah but the g(x) = zero polynomial (classes wise) no?

then we are working with classes

you are talking about a quotient ring right?

or field

[dire siren]

no

[stiff kraken]

but what you are saying is true modulo

so we would be working with classes

[dire siren]

we can't really reduce things modulo here due to the elements that are not in F

[stiff kraken]

but a^(q-1) = 1 modulo n for (n,q)=1 and q=prime

isn't that right?

[dire siren]

yes, but the problem arises when you evaluate the polynomial in elements that are not in F

so for example in Z/3Z the polynomial X^3-X when evaluated in any element is 0

but if you extend Z/3Z to some other field K, and decide to evaluated X^3-X to an element from that field, it might not be 0

[stiff kraken]

yea but we used the argument that g(x) is not the zero polynomial in F[x], not in F(a)[x] or in some other extension right?

[dire siren]

that just fails for finite field, unfortunately

say F=Z/3Z and a=1/(c^3-c)

[stiff kraken]

so you are saying there can be a polynomial p(x) with coefficients from a field F that is not the zero polynomial but p(x) = 0 for every x in F?

[dire siren]

yes

[stiff kraken]

that sounds counterintuitive

[dire siren]

yeah, the thing is that equality as functions does not imply equality as polynomials

so in the case of p(x)=x^3-x in Z/3Z we have p=0 as functions, but not as polynomials

[stiff kraken]

maybe

how would you even extend a field like Z/3Z though? Is it even possible to create for example Z/3Z(pi)? I am not even sure how it would work tbh as the addition and multiplication defined in Z/3Z is modulo wise but pi is not even an integer

anyway, thanks for the help

[languid trellis]

It may be easier to consider Z/3Z[x] first

then we many quotient out by an ideal to get an extension

For example, Z/2Z[x] / (x^2 - 2 )

which has the elements 0, 1, x, 1+x, and the relation x^2 = 2 = 0 (mod 2)

So, in some sense this is Z/2Z[ \sqrt(2)]

[dire siren]

it is possible to create Z/3Z(pi)
it's isomorphic to the field of fractions of Z/3Z[x]

[languid trellis]

Is it a general result that if u is transendental over R then R[u] is iso to the field of fractions of R?

[dire siren]

yes

[languid trellis]

hm

[smoky barn]

Hello @dire siren about what i asked before polynomial as sum of squares of two other polynomial in same field
Its possible in real field but how do you do it if its under mod n where n is a prime

https://ask.sagemath.org/question/10062/polynomials-as-a-sum-of-squares/

This post shows how to do it in a real field but
For example how can i represent
(x^16 + 1 ) %p = (P² + Q² )% p
where p is a prime
P and Q are polynomials under same field

How would we find P and Q

[dire siren]

R(u)

and R[x]

[languid trellis]

is it typical to use square brackets for indeterminates and parentheses for a specifed element?

My book uses R[u] and R[x]

[dire siren]

square brackets for rings

round brackets for fields

[languid trellis]

ah gotcha

[south patrol]

usually from context you'll know whether it's an indeterminate or not

though x,y,z,t are typically reserved for that

[languid trellis]

mhm

[dire siren]

@languid trellis I realize my replies were not clear; so what I meant is that the result is "if u is transcendental over R then R(u) is isomorphic to R(x)"
where R(a) means the field of fractions of R[a]

[languid trellis]

I understood, but I appreciate the clarification nonetheless, thank you

#1217259452621783154

[dire siren]

it can't be true over the reals, because the degree of a sum of squares of polynomials is always even
I'm not sure about the problem modulo p

[vivid tiger]

what how

Z[\pi] iso to Z[x] methoughts

[dull ginkgo]

It is

The valuation map of Z[X] into R that sends X to \pi is injective as the kernel is trivial (no poly evaluates to 0). The image, Z[\pi], by first isomorphism theorem is iso to the domain, Z[X]

[vivid tiger]

yeah my point is that this isn't Q[X]

[dull ginkgo]

For integral domain R, R(u) is isomorphic to the field of fractions of R[u] if and only if u is transcendental over R, or the valuation map is mono

[vivid tiger]

I don't understand how this sus lossible

Oh

reading comprehension

Yup checks out

[dull ginkgo]

Honestly the easiest route is the universal property of the field of fractions itself

Yeah

Because the injectivity of that valuation map allows the field of fractions to factor through it

[obsidian crypt]

If E is an extension field of F and $\alpha\in E$ is algebraic over F. For any $\beta\in F(\alpha)$, why is it the case that $F(\beta)$ is a subfield of $F(\alpha)?$

[cloud walrus]

CaratCake

[topaz solar]

R[x] no, fractions of R[x] is same as (fractions of R)(x)

[topaz solar]

Yeah, I think there might’ve been some [] vs () shenanigans

Or like (fractions of R)[x] vs fractions of (R[x]) etc

Mfw associativity failure

[dull ginkgo]

RAAAAA BULLSHIT NOTATION RAAAAAAAAAAA

[topaz solar]

Yeah basically

[languid trellis]

im so confused

idk who to trust

[topaz solar]

C[x] isn’t C

Which is field of fractions of C

And fractions of C[x] (i.e. C(x) here) is not C

Since I don’t think you get a sqrt of x

So not alg closed

[languid trellis]

I mean this example doesn't satisfy hypothesis of having an element transcendental over C, unless technically the indeterminate x is transcendental over C

[topaz solar]

x is indeed transcendental over C

Since I mean, it’s kinda impossibly to satisfy a C polynomial

[languid trellis]

right

hm

[topaz solar]

For an algebraic u, we have like

K[u] = K(u) though

[languid trellis]

I think we also need the minimal polynomial of u to be irreducible over K

else K[u] isn't even a field

[topaz solar]

Well, it’s the minimal polynomial

It better be irreducible

[dull ginkgo]

can i be clear here

[topaz solar]

https://tenor.com/view/ralsei-clear-gif-26477239

[dull ginkgo]

I fucking hate $F(\alpha)$ as a notation if $\alpha$ is an element

[topaz solar]

Yes

Very annoying

[cloud walrus]

Foghorn (*BWAAAA-UNNNGH*)

[languid trellis]

I'm pretty sure ther exist algebraic elements with a reducible minimal polynomial, in which case K[u] is not even an integral domain

[dull ginkgo]

IT'S EVEN WORSE IF IT'S JUST IMPLIED WHAT FIELD IT'S IN

OR HOW

Like imagine going "oh yeah, we have $F(\alpha)$... alpha's a hyperreal"

[cloud walrus]

Foghorn (*BWAAAA-UNNNGH*)

[topaz solar]

Well, u is in a field (since alg closure), and this is a subring, so…

[dull ginkgo]

[topaz solar]

It better be an integral domain

Sub rings of integral domains are integral

[dull ginkgo]

god the amount of notational misery I have about shit like, $\mathbb{Q}(\pi)$

[cloud walrus]

Foghorn (*BWAAAA-UNNNGH*)

[topaz solar]

Unless you’re doing something over some very non-field-like ring, I’m pretty sure minimal polynomials are irreducible also

So uh

I mean, an idempotent does satisfy x(x-1) but that’s not in our field except 0 or 1

[languid trellis]

this is the result I was referencing

I can't think of an example rn

oh

Z/(2) / (x^2 - 2)

which is Z/ (2) [sqrt2]

[dull ginkgo]

I've never actually formally proved this it just kinda makes sense in my head

i wonder how much math I know is actually me just gaslighting myself

[topaz solar]

This just adding a nilpotent x yeah?

[languid trellis]

i mean yeah kinda

x^2 = 2= 0

mod 2

[topaz solar]

I wouldn’t wanna call that algebraic

But I mean pop off ig

It is indeed satisfying a polynomial but

[languid trellis]

sqrt2 is a root of x^2 - 2 = x^2 (mod 2)

a bit wacky

but there you are

[dull ginkgo]

Integral extensions

[topaz solar]

I wouldn’t say it’s algebraic since it’s not in the algebraic closure

[dull ginkgo]

does jacobson cover the proof of the existence of closures lol

its very silly and goofy

[topaz solar]

Appending elements not in any field extension be like

[dull ginkgo]

Oh yeah, we like variables. What if we had as many as the fucking polynomials

[topaz solar]

https://tenor.com/view/crying-emoji-dies-gif-21956120

[languid trellis]

i'd say it's algebraic because its a root of the polynomial

don't call it sqrt2

just call it "x"

such that x^2 = 0

:trollface:

[topaz solar]

But yeah if your u is actually algebraic and can live in a field at all

[dull ginkgo]

oh boy a lot of it is like

[topaz solar]

Then the minimal polynomial is reducible

[dull ginkgo]

what's the difference between isomorphism and actually being an "equal" field

[topaz solar]

Since you wanna include dumb ring extensions here it is quantified more tightly :3

[dull ginkgo]

fuck

[languid trellis]

fuck it unfields your field extension

[dull ginkgo]

Usually you can be like "hehe" $\mathbb{Q}(\sqrt{2})$ just has an element that is the root of $x^2 - 1$ easy. Good luck describing $\mathbb{Q}(\pi)$

[cloud walrus]

Foghorn (*BWAAAA-UNNNGH*)

[dull ginkgo]

i hate notation

[dull ginkgo]

once again, integral extensions

[languid trellis]

Q(pi) = R bc i said so

[dull ginkgo]

You know what else is a scary field

the goddamn p-adics

[topaz solar]

Wrong!

[languid trellis]

doesn't model theory give us some pretty fucked fields

[dull ginkgo]

i think -1 is a square for p-adics which completely destroys any semblence of an order that works well with its algebraic shit

[topaz solar]

If you consider them as sub fields of where the elements you’re adding lie ig

[topaz solar]

Algebra gives you awful fields, model theory just says bad ones exist

[dull ginkgo]

admittedly ostrowski's theorem is kinda cool and isn't too hard

[topaz solar]

Because Z exists

[languid trellis]

Are there any fucked infinite fields that I might understand

[dull ginkgo]

closure of the hyperreals probably

OH I KNOW

[topaz solar]

F_p((t))

[languid trellis]

because there was a result about finite subgroups of infinite fields being cyclic that i want to be more impresssed by

[dull ginkgo]

Isomorphism between the completion of the closure of the p-adics and C with choice

[languid trellis]

what is this

[dull ginkgo]

is this referring to the rational functions of F_p or

if it's like a field of fractions of like the fucking ring of Laurent series I'm going to implode

[topaz solar]

It is

[dull ginkgo]

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

[topaz solar]

It’s Laurent series, ie fractions of power series

Since we over a field

[dull ginkgo]

oh wait it's a field by default

I HATE ALGEBR

[topaz solar]

Yeah

Anyway F_p((t)) is absolutely atrocious

[dull ginkgo]

is it defined differently than $\sum_{n = -N}^{\infty}{a_n x^n}$ for char not 0

[cloud walrus]

Foghorn (*BWAAAA-UNNNGH*)

[topaz solar]

I think $\int \bQ_p\dd\mu$ is isomorphic to $\int F_p((t))\dd\mu$ as a valued field

[cloud walrus]

Everlasting Sharp

[dull ginkgo]

what measure is mu

[topaz solar]

Ultraproduct

[dull ginkgo]

[languid trellis]

can you think of any finite subgroups of this?

[dull ginkgo]

I was interested for a second

not anymore

[topaz solar]

Anyhow, point being somehow this encodes all the data of the p-adics

[dull ginkgo]

thank god nGroupoid isn't here because he'd raise Adele rings from the depths of hell (langlands)

anyway I have some shit for EE to do so smell you later

[swift prawn]

these are awful lmao

even just F_p(t) is not that well understood

[topaz solar]

I think they’re TP2?

Translation: abhorrent

[swift prawn]

i'm doing a reasearch project this summer on the model theory of analytic functions... i'll get back to you at the end of it about it lmao

but yea as far as I know for non char 0 fields it's pretty rough out there

[topaz solar]

On like, uhh, these analytic function rings like this one, or like weird ones like R with all the analytic functions

If you end up reading stuff like Montenegro explain it to me

[swift prawn]

I think mostly on analytic function rings. I think it will mostly be focused on the type of model theory that you can do when you add the analytic functions (over some field like C I think) as constant symbols. Or at least I think it is along those lines

[topaz solar]

Spooky stuff

[fluid patio]

hi do anyone know
about spherical harmonics and its relation with wigner d matrices

[distant sentinel]

Does anybody know why $\hat{G}$ has to be discrete?

[cloud walrus]

justuswie

[distant sentinel]

$\hat{G}$ is the Pontryagin dual of G here

[cloud walrus]

justuswie

[cobalt heath]

Adele rings are not that bad tho

[rocky cloak]

Since the topology on the dual is the compact open topology, and G itself is compact, it's enough to show that any homomorphism G -> S1 that maps everything close to the identity in S1, must be the identity.

It just comes down to f(g^n) = f(g)^n, and that any nonidentity element in S1, can get far away from the identity by raising it to some power.

[distant sentinel]

alright thank you, that helps!

[stiff kraken]

Hello, I am reading that if c is transcendental over K and K(c) is an extension of K, then "there is a map deg: K(c) --> Z such that if a is not in K then deg(a) >0 ", can someone explain why that's true or point me to this theorem's name if it has one?

and does anyone have to suggest any good textbook on galois theory? Currently studying without one so it's a bit hard

[rocky cloak]

Are there any assumptions on this map? Like you can of course just define deg(a) = 0 for a in K and deg(a) = 1 otherwise...

[stiff kraken]

I am not sure, I found it online on an answer on math stack exchange from 3 years ago and I am trying to make sense of it

[dull ginkgo]

I am confused what you mean by map

[stiff kraken]

I'm not sure either, that's why I thought I'd ask, maybe it's something I haven't learned yet, but it turns out everyone is just as confused

[delicate orchid]

this sounds like viewing K(c) as the quotient of some polynomial ring and then defining deg: K(c) -> Z to be the standard degree function K[x] -> Z but factored through the quotient, so you get a map K[x]/(f) = K(c) -> Z.

maybe this works

I'm also using the weakest variant of "map" to literally just be a set map but this probably has nicer properties as it's coming from a valuation

[stiff kraken]

I wouldn't know, sorry

[long obsidian]

Hey I'm confused about some statement I saw that said the alternating group An is a subgroup of the permutation group Sn but the rep theory of Sn is nicer than rep theory of An. In what sense is it nice?

[delicate orchid]

the representation theory of Sn is very combinatorial and thus has a bunch of easy formulas (hook length formula comes to mind) to work out character values and stuff like that

young diagrams are the best way to view the "niceness" imo

(another nice fact is that, since S_n is ambivalent, all of the characters have real values - which isn't S_n specific but it's still nice)

[long obsidian]

okay I see yeah that makes sense thanks. Does An have representations that come the restrictions of reps from Sn? Or however you could get reps of An from Sn

[delicate orchid]

Restrictions to An is very nice because it's an index 2 subgroup

so not only is it normal, so you can use loads of clifford theory like things only splitting into at most two S_n-conjugate things upon restriction

index 2 is nice for other reasons I can't quite recall properly rn

ah yeah, you're also guarenteed to find every irreducible of A_n by restricting irreducibles of S_n and then taking decompositions

[rocky cloak]

A reasonable guess might be that the answer is just wrong (it has 0 upvotes and doesn't seem to be accepted).

They seem to be confusing K(c) with the polynomial ring. K(c) is isomorphic to the ring of rational functions, which does have a degree map, but it's not true that deg(a) > 0 nor that deg(f(a)) = ndeg(a).

One thing you can do is look at [K(c) : K(a)], it's not too hard to show that this is finite for any a not in K, and that [K(c) : K(f(a))] <= n[K(c) : K(a)].

[wide brook]

Greetings! While navigating through some exercises I stumbled upon the following one. It had three field extensions K over $\mathbb{Q}$.

• $\mathbb{Q}(\sqrt3,i)$, which I proved is a splitting field over $\mathbb{Q}$.
• $\mathbb{Q}(\sqrt[3]{3})$ which I proved is not a splitting field over
  $\mathbb{Q}$
• $\mathbb{Q}(\sqrt[3]{3},i)$ also not a splitting field over $\mathbb{Q}$ (for this I’m not sure but I took the polynomial $(x^2-3)(x^3-3) and \sqrt3$ does not belong in that extension

Now the exercise asks which of these have the property that there exists unique isomorphism of fields K->K. I’m thinking that for the second one, which is the only splitting field there could be multiple isomorphisms since a root can map to its conjugate through an isomorphism. So I’m clinging towards to the result that this one doesn’t have this property since we have at least two isomorphism, the identity and the one that maps an element to its complex conjugate. For the others, since they are not splitting fields and are just fields I don’t see why we could not have multiple isomorphisms. So, does none of the above extensions have the property? Am I missing something here?

[cloud walrus]

Alma221

[hushed flume]

Hey in my notes it asks us as an exercise to prove by induction the formula for the number of k cycles (k-1)!*(n choose k how do we do this?

[delicate orchid]

note that we can form any (k+1)-cycle by taking a k-cycle (x_1, x_2, ..., x_k) and appending an element x_(k+1) to it, where x_(k+1) isn't equal to x_i for any i < k+1

you'll need to count then number of ways you can get a particular (k+1)-cycle like this but that shouldn't be too tricky

[hushed flume]

yes this is the part that I am stuck on

not sure how to do it

I assume that I can take each k-cycle and add the new element to k different spots ( between two elements or in front of the first one)

but then I don't know how many times I get the same cycle

[delicate orchid]

that's equivalent to adding it at the end and then counting the number of ways you can do it

two tuples (x_1, ..., x_k), (y_1, ..., y_k) corrispond to the same cycle iff they are cyclic permutations of one another

like (1234) and (4123) are the same cycle but (1234) and (1324) are not

[hushed flume]

yes I understand this

but let's say that I can choose from two new elements and append it somewhere in these two cycles

[delicate orchid]

well then if you add the element to a different spot you're guaranteed to get a new cycle, assuming k > 2

[rocky cloak]

Try to find all the automorphisms of Q(cuberoot(3))

[delicate orchid]

wait is this true for k = 3? one moment

ah no there's a slight subtly

[hushed flume]

I think its not

[delicate orchid]

appending it to either end will get you the same cycle
(123) -> (1234)
(123) -> (4123) = (1234)

[hushed flume]

I was aware of that

[delicate orchid]

but otherwise they're unique, because the adjacent numbers will be unique