groups-rings-fields

[rustic crown]

i.e. given a pair i, j there exists a sigma in Gal(L/K) such that sigma(a_i) = a_j

[formal ermine]

what does "act on L" mean here?

like a group action? but how is the action defined?

[rustic crown]

yep, a group action.
g in Gal(L/K) and l in L then g * l = g(l)

[formal ermine]

ah

how is this true?

what if sigma maps alpha_i to alpha_k and alpha_j to alpha_l where k > l?

[rustic crown]

it doesn't matter, sigma will just permute the (sigma(a_i) - sigma(a_j))^2

so in the end you still have the n choose 2 pairs

[formal ermine]

but i < j

[south patrol]

yeah you can equivalently write this as half of the the sum over all pairs (i,j)

[formal ermine]

oh

ah

[south patrol]

provided field not char 2 etc lol

[formal ermine]

yeah ok

[south patrol]

But yeah liek the i < j just stops you from overcounting

[formal ermine]

if it will permutate i < j to k > l it will also permutate some k > l to some i < j

[south patrol]

As det says, you'll just permute tings

Well, another way to do it is to say that the sum is over all (unordered) pairs (i,j) with i not j

[rustic crown]

you could forget about that, and that's precisely why we had a square in (sigma(a_i) - sigma(a_j))^2

[south patrol]

like it's clear that it permutes them ig

[formal ermine]

when's the discriminant useful?

[rustic crown]

pretty useful when you're dealing with small extensions

[lament dawn]

Discriminant test can tell you the galois group sometimes

[formal ermine]

ah like

uh

wait

if the discriminant is a square in K

[lament dawn]

S_3 vs A_3

[formal ermine]

then the galois group is A3

yeah

otherwise S3

assuming f is irreducible and separable

oh wow the proof for the inverse galois problem over an arbitrary field for finite groups is so smooth

[rustic crown]

i don't think that should be called inverse galois problem tho

[formal ermine]

yeah lmao

I just like using fancy words

[rustic crown]

given G, it's easy to construct L/K such that G = Gal(L/K)... but the problem is we can't ask it for a fixed K

[smoky ivy]

I’m struggling to show this: \
Let $R \subseteq S$ be an integral extension. Show that $Q \cap R = J \cap R$ where $J = \bigcap_{\substack{m \in \operatorname{Spec}_{\max}(S) \ Q \subseteq m}} m$. I tried a bunch of things with lying over and going up, but didn’t really get any useful results (one inclusion is obvious and follows immediately)

[cloud walrus]

lewis

lewis

[rustic crown]

i don't think it's true in general. consider the case where Q n R isn't a radical ideal in R.

R n m would be maximal ideals containing Q n R, so they would also contain rad(Q n R)

is Q a prime?

[smoky ivy]

oh yeah, sorry

it is prime

prime in S

[rustic crown]

what if R = S = k[[x]], and Q = 0?

J would be the jacobson radical in S, which is just (x) as S is a local ring

[smoky ivy]

okay wait, let me send the exact task

maybe i’m misunderstanding things

[rustic crown]

any maximal ideal in R is given by contracting some maximal ideal of S (by going up ig), so J n R would be intersection of all maximals containing Q n R

[smoky ivy]

yeah, that’s what i got as well

[rustic crown]

but R is a jacobson ring!

which you weren't using before

[smoky ivy]

Yeah, thats probably where the misunderstanding is now 😅

[rustic crown]

so do you see how to continue now?

[smoky ivy]

not too sure, i don’t know how to use the fact that R is Jacobson

I don’t know how to show that J intersect R is a subset of Q intersect R

[rustic crown]

J n R would be intersection of all maximals containing Q n R
Q n R is prime... so what does R being jacobson tell you >.<

[smoky ivy]

it is an intersection of maximal ideals

[rustic crown]

okie lemme elaborate this a little
say m is a maximal in R containing Q n R, then by going up you get a prime p containing Q such that p contracts to m. now pick any maximal M containing p, this M will also contract to m, as m was maximal

so it's the intersection of all maximal ideals (containing Q n R)

[smoky ivy]

yeah, i had this result, but i didn’t know how to continue with that fact

oh wait

if Q intersect R is the intersection of maximal ideals

those maximal ideals also contain Q intersect R, no?

oh yikes, then it quite follows immediately

😅

[rustic crown]

yea lmao

[smoky ivy]

thank you

[prime gale]

What is the dimension (cardinality of a basis) of {continuous functions from R to R} as a vector space over R? (I know an answer, but I wonder if there is a better way.) You can assume Choice.

[long geyser]

well, it has a countable spanning set right?

you'd have to prove it can't be finite dimensional, rest is forced

[prime gale]

Answer I get is bigger. Though I don't like my solution--too complicated.

[formal ermine]

take (f_m)_m as a sequence of functions with f_m being nonzero in [2^(-(m + 1)), 2^m] and zero otherwise

all of these are linearly independent

[next obsidian]

I don’t think it should be countable

That seems way too smal

[long geyser]

hmm

oh true, you can't assume finite support

so you can't just take bunch of kronecker deltas

[prime gale]

Assume compact support if you want. I don't think it makes any difference.

[long geyser]

wait, sequences in R has an uncountable basis right whoops

so it should be same as that

[prime gale]

I agree. It is the same as dimension of sequences in R. But what is that?

[long geyser]

this should be in roman I think, sec

[next obsidian]

R^N should be R no? I forget how exponentiation works but like

[prime gale]

Only proof I can think of uses weird multilinear algebra. Is there super easy solution?

[long geyser]

yes

as cardinality

[next obsidian]

Yeah so

[prime gale]

I agree. |R^N|=|R|.

[next obsidian]

If you can come up with |R|-independent functions

Then that has to be the size of a basis

[long geyser]

definitely there should be a way easier one, you don't need multilinear stuff for this

[next obsidian]

Because if it was bigger your space is too big

[prime gale]

I agree. Maximum size of basis is |R|.

[next obsidian]

And it has to be at least |R|-big

[prime gale]

Is there elementary reason why it must be |R| big?

[next obsidian]

So now use the explicit functions Illumi defined

Illumi gave you a family of independent functions indexed by |R|

By just fudging the support

[prime gale]

I see. That is easy way to see. Thanks!

[next obsidian]

Or well that might’ve been countable but like

You should be able to just partition R up into |R|-disjoint subsets

Then just make a function whose support is contained in just one of those subsets

Then you have your independent functions

[prime gale]

Well, let m be real. That is, f_m is a function with support [0,m] with m real.

[next obsidian]

Oh sure you have to make sure

That each function is non-zero in [m-1,m] tho

If you do that then they’d be independent

Or no wait

I still disagree

[prime gale]

Doesn't work?

[next obsidian]

m is real, not an integer lol

Uhhh

Idk maybe it does lol

[prime gale]

Much simpler than the crazy thing I came up with before. Thanks@

[rotund aurora]

For each real number r, define the sequence (1,r,r^2,r^3,...)

[warm wyvern]

How do I prove maximal ideals are prime without appealing to quotients?

[next obsidian]

Why

[warm wyvern]

It's an exercise in my textbook

[next obsidian]

Rip

Okay well it’s not too hard I think

If ab is in m but a nor b is

You should be able to look at

(m,a) or (m,b), and hopefully prove one of these is a proper ideal

Cuz they properly contain m

That’s my idea

[warm wyvern]

Yeah, i thought about that

It didn't take me too far

Maybe i just didn't think about it long enough, hmm

[next obsidian]

Okay well

If they’re both (1)

Then(1) = (m,a)(m,b) = m^2 + am + bm + (ab) < (m,ab)

So (m,ab) = (1) but ab in m so (m,ab) = m

This should work, the ideal arithmetic probably makes you uncomfortable, but just write out generic elements and you’ll see what I said is true

[warm wyvern]

That...

Makes sense

This shouldn't have taken me so long

[agile burrow]

bump

[solar glacier]

is the trick for this question to map (h,k) to (h,f(k)) which gives the desired iso

going from H \semidirect K of \phi \circ f to the one with \phi

and can someone give me some questions regarding semidirect prods to work on

and also, a semidirect prod is abelian iff both components are abelian and the map is trivial, right?

in which case they equal their usual direct prod

like classical exam like questions for semidirect prods

[chilly ocean]

These are Keith’s notes?

Yes I remember it being something like this, though maybe it was (h,k) to (h,f^{-1}(k))

The details work themselves out nicely in the computation

[solar glacier]

yep

[solar glacier]

i just worked it out, its just f(k) it checks out!

[chilly ocean]

The semi direct product is the direct product if it’s the trivial map

[solar glacier]

his notes are pretty well written lol

[chilly ocean]

There’s a similar problem instead with composing in the other order which works out with f inverse, I’ll look for it if it’s not In there

[primal tusk]

im trying to prove/disprove that c*gcd(a,b) = gcd(c*a, c*b) for integers a,b,c

but im stuck

i get that gcd(a,b) = av + bu for some integers u and v

and so c*gcd(a,b) = cav + cbu

but then i dont know where to go from there

[rustic crown]

to show that the two gcd's are equal, it suffices to show that each of them divides the other... do you see why this is the case?

[primal tusk]

@rustic crown yes i do ill look into that

[chilly ocean]

What's the sign mean

[solar glacier]

semidirect product

mult works as follows

[cloud walrus]

MyMathYourMath

MyMathYourMath

[solar glacier]

hope that clears things up

[solar glacier]

homomorphism

[solar glacier]

binary operation defn in semidirect product

question

if my proof of in UFD irreducible implies prime correct

[barren sierra]

@solar glacier did you mean $pu = p_1$?

[cloud walrus]

Spamakin🎷

[barren sierra]

Yea you did

[solar glacier]

oh yes

sorry typo

[smoky ivy]

how does one determine the hilbert function of such ideals

[smoky ivy]

this is how we defined it btw

[glossy crag]

Say you have a K-algebra A (K a field), some elements a_1,...,a_n and an element of b in K[a_1,...,a_n] that's symmetric in the a_i. Do we know definitively that b=f(a_1,...,a_n) for a symmetric f in K[x_1,...,x_n] or could it happen that b is symmetric in the a_i by virtue of some relations between them, not because it can be written as a symmetric polynomial in them?

[cloud walrus]

Ocean Man

[chilly ocean]

need help with part b)

[tender wharf]

which part exactly

for i) you do the subgroup test

[chilly ocean]

ii) I need help with, should've specified, sorry.

[tender wharf]

do you have anything so far?

what id try to do is find an element with infinite order

[chilly ocean]

nope

the index is defined as the amount of left/right cosets

[tender wharf]

ye ik

[chilly ocean]

umm

[coral spindle]

Try and find a defining property of each coset

[chilly ocean]

can I use lagrange's theorem

[coral spindle]

It will not be helpful here.

[chilly ocean]

$|G| = [G : H] |H| \implies [G : H] = \frac{|G|}{|H|}$

[cloud walrus]

isomorphism

[chilly ocean]

I see

I thought maybe we can do some trick with cardinality or smth

[chilly ocean]

how would this help?

[formal ermine]

all elements of a finite group have finite order

[chilly ocean]

that's trivial

[tender wharf]

err lagrange's theorem is on finite groups

yes which means if we find an element of infinite order your group is infinite

[coral spindle]

The same statement holds in cardinal arithmetic.

It is both for finite and infinite groups.

[tender wharf]

ok interesting

[coral spindle]

It just doesn't help here at all

[tender wharf]

TIL

[chilly ocean]

i'm still not sure how to proceed

[coral spindle]

I already gave you a hint

[chilly ocean]

this?

[coral spindle]

Yes

[chilly ocean]

ok

[solar glacier]

Silly question but to show a Boolean ring is commutative we first show a=-a for all a in the ring then expand out (a+b)^2 to conclude ab = ba?

In showing a=-a we expand (a+a)^2?

[chilly ocean]

if the index [Z^3 : H] is infinite, that implies that all cosets are disjoint, correct? @coral spindle

because otherwise they would repeat

so all I have to do is prove that all cosets of [Z^3 : H] are disjoint

[delicate orchid]

cosets are always disjoint

[chilly ocean]

they are either disjoint or the same

no?

[delicate orchid]

Ok fine *pairwise disjoint

[chilly ocean]

i don't know how to solve this question bruh

[delicate orchid]

But the cosets are always disjoint so I could say “Z is empty => cosets are disjoint” and it would be correct, your reasoning doesn’t make sense

Try finding representatives for the cosets

[chilly ocean]

we have $(u, v, w) \in \mbb{Z}^3 + H$

[cloud walrus]

isomorphism

[chilly ocean]

so...

uh

[delicate orchid]

More like (u,v,w)+H

[chilly ocean]

well yes

[delicate orchid]

Z^3+H = Z^3

[chilly ocean]

uh

(u, v, w) + H yes

okay

[delicate orchid]

So what does the coset actually look like

[chilly ocean]

(u + x, v + y, w + z)

and x + y + z = 0

[delicate orchid]

Yeah

[chilly ocean]

with u,v,w in Z

[delicate orchid]

So pick some element of Z^3 that’s not in that coset to find a new one

And see if you can figure out why you can always find a new one - I.e. there aren’t a finite number of them

[chilly ocean]

pick an element of Z^3 that is not in the coset and find a new coset

okay

wait if $G/H$ is an infinite set, then the index [G : H] is infinite

[cloud walrus]

isomorphism

[delicate orchid]

Yes that’s what we’re showing

[coral spindle]

All cosets are disjoint no matter what

[chilly ocean]

im stuck

can i get another hint

[coral spindle]

You've fixated totally on something different

You've not acted on my hint.

Let me be more specific

[chilly ocean]

I don't understand your hint, I'm sorry

[coral spindle]

Pick a coset, any coset. Look at what it is. Try and describe it in a similar way to how H was defined.

[chilly ocean]

(1, 1, 1) + H

H was defined such that its coordinates summed to 0

[solar glacier]

Can someone verify this plz

[formal ermine]

let $\mathcal{G} = \on{Gal}\bQ(x^p - 2)$ where $p$ is prime. every $\sigma$ is uniquely determined by its action on $\zeta$ and $\sqrt[p]{2}$. because blabla we have that $\sigma(\zeta) = \zeta^a$ for a $1 \leq a \leq p - 1$ and $\sigma(\sqrt[p]{2}) = \zeta^b \sqrt[p]{2}$ for a $0 \leq b \leq p - 1$. let [ \varphi : \mathcal{G} \to G \coloneqq \Set{\mtrx{a & b \ 0 & 1} | a \in \bZ_p^\times, b \in \bZ_p} \subseteq \on{GL}{\bZ_p}(2) ] [ \varphi(\sigma) = \mtrx{a & b \ 0 & 1} ] I know that if $f$ is separable then $f$ irreducible $\iff$ $\on{Gal}(\bL/\bK)$ acts transitively on its roots where $\bL$ is the splitting field of $f$ over $\bK$. can I therefore conclude that $\varphi$ is surjective?

[delicate orchid]

Good lord

[hot lake]

phi seems to be quite clearly not surjective to me

[formal ermine]

it must be surjective because my homework wants me to show it's bijective (or rather an isomorphism)

[hot lake]

ah wait

I thought phi's codomain was all of GL2(Zp)

[formal ermine]

oh

yeah I mean that subset before it of GLwhatever

[delicate orchid]

Aff_1(F_p)

[formal ermine]

is it called that?

[delicate orchid]

Yeah that’s that group

[formal ermine]

ah

interesting

[hot lake]

I'm not sure how the fact that G is transitive on its p roots would imply that the map is surjective

[formal ermine]

ah wait

lol I thought too complicated

the galois group is of order p(p - 1)

so there are p(p - 1) sigmas

therefore every a and b is 'reached'

so the map is surjective

[hot lake]

ah yeah if you know that phi is injective too

[formal ermine]

yeah I either gotta show that it's surjective or injective because the sets have the cardinality

oke thanks

[formal ermine]

every group of order 245 is abelian because it has only one 7-sylow subgroup and 5-sylow subgroup with $H \cap K$ and $G = HK$ so $G = H \times K$?

[formal ermine]

(cuz H and K are both abelian G must also be abelian)

[void cosmos]

yes

that is if you meant H cap K is trivial

[formal ermine]

yeah

[smoky ivy]

How can I show that if R is an integral domain and R^* the set of all elements in Frac(R) that are almost integral over R, then R^* is normal?

I got the hint that for $s_0, \ldots, s_{m-1} \in R^*$, I should show that there is a $c \in R \setminus {0}$ such that $cs \in R$ for all $s \in R[s_0, \ldots, s_{m-1}]$, but not sure on how to show that or use it

[cloud walrus]

lewis

[smoky ivy]

(does it suffice to show the above for only the generators s_0, ..., s_{m-1}? if so, then doesn't that follow immediately due to those elements being almost integral over R

[void cosmos]

y

[rustic crown]

show for every monomial i think

what does the notation [h]_H and [h]_G mean?

[agile burrow]

I've usually seen it refer to conjugacy classes in H and G respectively

so for (ii), suppose h_1 and h_2 are conjugate in G but not conjugate in H. Then there exists g in G \ H such that gh_1g^-1 = h_2, meaning h_2 is in H and gHg^-1, so h_2 = e

[delicate orchid]

fusion discussion :wholesome100:

[agile burrow]

I'm not really sure why they use equality of the size of the centralizers rather than just equality of the centralizers in (iii) but it's essentially the same argument there

well actually maybe not

Like we have C_H(h) is a subset of C_G(h), right?

[delicate orchid]

yes

[agile burrow]

oh, well if g is in C_G(h) but not in C_H(h), then certainly g is in G \ H

ok yeah, it's the same argument

[delicate orchid]

ngl I don't really know what we're talking about because there are FAR too many words for me to read but I will agree none-the-less

[agile burrow]

group theory is cool

[delicate orchid]

yeah... maybe if ur a nerd

[smoky ivy]

then again, how do i use that to show normality of it

[rustic crown]

say you have an element t in Frac(R*) which is integral over R*, then it's also integral over some R[s0,...,s_{m-1}], from this can you show t is almost integral over R?

[smoky ivy]

do you have a hint?

[rustic crown]

t is integral over R[s0, ..., s_{m-1}] hence also almost integral. i.e. you can find an r* in R[s0, ..., s_{m-1}] such that r* t^n in R[s0, ..., s_{m-1}], can you change r* a little so that it shows t is almost integral over R?

[smoky ivy]

you could evaluate it in 0

[rustic crown]

what does evaluation mean here >.<

[smoky ivy]

oh wait, that makes no sense

so essentially, we need to show that r* is in R, no?

[rustic crown]

it may not be

but we can easily bring it to R

[smoky ivy]

but isn’t that the definition of being almost integral

oh

[rustic crown]

(by that lemma)

[smoky ivy]

not sure which lemma you are referring to

[rustic crown]

this one :3

[smoky ivy]

ohh

so cr* \in R for some c

[rustic crown]

yee

:3

[smoky ivy]

or well more like cr*t^n \in R for some c is better

with the lemma

thank you tho :)

[rustic crown]

yee that is important, because you want the same c to make cr* in R and c(r*t^n) in R

[smoky ivy]

yup, realized :D

[celest cairn]

Is the Galois Group of $x^{3}+x^{2}+x+1$ equal to $S_{2}$?

[cloud walrus]

Sapphire

[formal ermine]

no

why do you think so?

[celest cairn]

The splitting field is $\mathbb{Q}(i)$ over $\mathbb{Q}$. So I thought that the Galois group would be $S_2$ since the degree of the minimal polynomial over Q is 2. But I guess that’s not the case. So it’s $S_3$?

[cloud walrus]

Sapphire

[formal ermine]

Ah yeah you're right I misread that as a ^4

[south patrol]

x^3 + x^2 + x + 1 factors properly anyway yeah

[celest cairn]

Ok so it is S_2?

[formal ermine]

yes

iso to S2

or Z2

there's only one group of order 2 up to iso

because 2 is prime

[celest cairn]

Ah Ok, thanks. 🙂

[tribal niche]

are submodules of submodules submodules?

[formal ermine]

submodules of the entire module?

[tribal niche]

yea

[formal ermine]

yeah

[tribal niche]

submodules of submodules of a module are submodules of the module?

oh ok thx

[rustic crown]

[tribal niche]

it makes sense but for some reason i can't work it out properly

maybe i'm trolling whatever

[barren sierra]

???

what part are you stuck on proving

[formal ermine]

what about "a subgroup of a subgroup is a subgroup of the entire group" is that clearer

[tribal niche]

idk i'm doing it in my head

[barren sierra]

[tribal niche]

it's part of a different problem

so that shows submodules of noetherian modules have to be noetherian

[tribal niche]

yeah

its probably bad that i relate modules more to vector spaces than groups lmao but it is what it is

[formal ermine]

do relate them to vector spaces

it's just a vector space without division by scalars

[tribal niche]

yeah

[formal ermine]

you can define a submodule via subgroups of the modulet group (the group where the elements come from)

[tribal niche]

vs are modules over fields ofc but i mean when i think about problems relating to modules i think hmm how would this work in linear algebra

[formal ermine]

a submodule is just a subgroup with the scalar multiplication being closed too

[tribal niche]

yeah makes sense

thx

If M is a module, and M' is a submodule, show that M is Noetherian iff M' and M/M' are Noetherian.

^ that's what i'm actually trying to do

something something basis theorem

[tribal niche]

is the homomorphic image of a finitely generated module also finitely generated? this seems very true, just take the image of each generator under homomorphism, right?

[rustic crown]

yee

[tribal niche]

thanks

i don't really get this. what is the difference between these two things. specifically, why were both stated and why is the proof of the second one longer than the first? like if i prove the first doesn't it imply the second or do i have a significant misunderstanding regarding algebras over rings

[south patrol]

Being f.g. as an algebra is different to being f.g. as a modulel

[tribal niche]

i was under the impression that an R-algebra is simply an R-module with a product

[south patrol]

Yes, but saying "f.g." is different

[tribal niche]

oh waiy

i'm tripping

[south patrol]

e.g. Z[x] is a f.g. Z-algebra

But not a f.g. Z-module

But also like, you can be an R-module without being an R-algebra, so 1.4 doesn't follow from 1.3

[tribal niche]

yeah

i'm tripping

[south patrol]

Though 1.4 should be easier to prove than 1.3 tbf

[tribal niche]

R-algebra is a special case of R-module

[south patrol]

Or rather, than Hilbert basis theorem

[tribal niche]

1.4 is technically easier to prove i think cuz 1.3 is a corollary to hilbert basis theorem

[south patrol]

Yup yeah lol

That's what i was saying aha

But yeah

[tribal niche]

i think i just understood 1.3 easier than i did 1.4 cuz i kind of "get" polynomials but don't "get" modules as easily

[south patrol]

Well you just need to know that quotient of a Noetherian module is Noetherian

right

[south patrol]

Yeah fair enough

[tribal niche]

thx

[tawdry crystal]

What does it mean for S to be closed under the inverse operation?

[south patrol]

S is closed under the inverse operation iff for every s in S, s^-1 is in S as well

[tawdry crystal]

Thanks

[south patrol]

np

More generally, to say something is "closed under [...]" means that you won't leave the set if you do/apply the thing in question

[smoky ivy]

I'm struggling very hard on how to compute these..., I tried a bit, but I can't seem to find an easy closed form, some of it required some tedious combinatorics

[rustic crown]

could you remind me the definition?

[smoky ivy]

this is something i did i guess, but obviously not a proof or anything, just testing

[cloud walrus]

lewis

[smoky ivy]

maybe considering the degree of elements in I somehow? not sure

[tribal niche]

can anyone help on the only if part? i proved the if part already

[delicate orchid]

which way does the arrow point for only if

[tribal niche]

if M' and M/M' are noetherian show that M is noetherian

[delicate orchid]

ah, the hard direction

[tribal niche]

yeah

💀

[smoky ivy]

consider a chain in M and use the canonical homomorphism M -> M/M'

[delicate orchid]

could always prove the contrapositive

[tribal niche]

this is the other direction

[smoky ivy]

and you can also use M -> M' by intersecting with M'

[delicate orchid]

show that M not notherian => neither are M/M' and M'

[smoky ivy]

nop, you can do that for that direction as well

[rustic crown]

i think k[x,y]_{<=d} --> I_{d+3} given by multiplication by (y^2-x(x^2+1)) is an iso of k-vector spaces

[tribal niche]

how? because the homomorphism is not injective, right?

[smoky ivy]

so essentially, I_{<=d} isomorphic to k[x,y]_{<=d-3}?

[rustic crown]

yea

[smoky ivy]

i don't quite get the intuition behind it

[rustic crown]

so dim(A_{<=d}) = dim(k[x,y]_{<=d}) - dim(k[x,y]_{<=d-3})

[smoky ivy]

so consider the epimorphism M -> M/M', get a chain in M, applying the epimorphism to the chain yields a chain in M/M' and intersecting the chain with M' yields an ascending chain in M', use noetherian property now, the claim is then M_i = M_n for all i >=n with n, coming from being noetherian

[smoky ivy]

yeah, that's obvious then

[smoky ivy]

may i ask how you got there

[rustic crown]

was thinking degree d elements in I, initially thought that the non-homogeneity of the generator of the ideal might be a problem, but it was okay

[tribal niche]

ah this makes sense. the issue was i was trying to do this with finite generation of submodules and wasn't getting how the homomorphism idea works

[smoky ivy]

ah, i see hmm

[rustic crown]

now i'm wondering what would we do if there are multiple generators like in the third part

[tribal niche]

i feel like there's a way to do it given the simple equivalence of definitions, but that is pretty clean ngl

[smoky ivy]

yup, that's my thought as well

i think there's no way of not counting a bit

but considering the generators like in the last part, you can probably stop counting pretty soon at some point

[smoky ivy]

ah, i see

yeah, the proof is quite elementary if you look at it this way

[rustic crown]

maybe some type of inclusion-exclusion can be done

something like
dim(I_d) = 3dim(R_{d-4}) - 2dim(R_{d-6}) - dim(R_{d-8}) + dim(R_{d-8})

does this sound reasonable?

[smoky ivy]

i dont quite see how that holds haha

[rustic crown]

it's like how to get a degree d guy in the ideal

you get it from a degree d-4 guy multiplied with each of x^4, x^2y^2, y^4

but then there is a double counting

stuff which is multiple of x^4y^2, x^2y^4 should be subtracted

and so for pair (x^4,y^4)

then stuff which are multiples of all 3, i.e. multiples of x^4y^4 again should be added back in

so finally something like this
dim(I_d) = 3dim(R_{d-4}) - 2dim(R_{d-6})

dim(I_{<=d}) is just the summation of the previous equation

ig i could make the argument because the ideal was homogeneous

[smoky ivy]

okay, wait, let me reread it and try to digest it

[rustic crown]

feels correct lol

idk

[smoky ivy]

could you expand on that

[rustic crown]

stuff in the ideal is x^4(**) + x^2y^2(**) + y^4(**)

[smoky ivy]

yeah

[rustic crown]

to get a degree d things, the ** should be degree d-4 polynomials

ofc you can mix match different degrees, but that's just going to make the combi harder

i hope i'm not making things up

.<

[smoky ivy]

Oh okay, I think that makes sense to me a bit somehow

[rustic crown]

(don't believe me completely)

[smoky ivy]

it's harder to understand when there are more generators

[rustic crown]

yea, maybe there is a nice simple procedure to handle these for graded algebras

[smoky ivy]

not sure if i was able to understand it fully, it's hard to write it down formally, imo

[rustic crown]

yea combi is weird

maybe the correct way to write it is via some isomorphisms again

like start with the surjection
R_{d-4} ⊕R_{d-4} ⊕R_{d-4} --> I_d

and try to do the inclusion exclusion thingy with vector spaces

this sort of looks like the koszul sequence

[smoky ivy]

(never heard of it)

[rustic crown]

maybe that exact sequence will help you write it nicely

[smoky ivy]

even then, inclusio exclusion is hard to write it nicely in this context

[rustic crown]

but i think what i intuitively wrote above should give it

maybe wait for someone else to answer >.<

[smoky ivy]

i alr tried earlier haha

[rustic crown]

[celest cairn]

How do I find the degree of $\mathbb{Q}(\zeta_{4}^{3})$ over $\mathbb{Q}$?
I know that $\mathbb{Q}(\zeta_{4}): \mathbb{Q} = \phi(4) = 2$.

[cloud walrus]

Sapphire

[south patrol]

What is an easier way to write \zeta_4

[celest cairn]

i?

So degree of $\mathbb{Q}(\zeta_{4}^{3}) = 2$?

[cloud walrus]

Sapphire

[delicate bloom]

can you confirm this formula for me by evaluating the terms on the right, here $i^2=-1$. $$[\bQ(\zeta_4):\bQ] = [\bQ(\zeta_4):\bQ(i)][\bQ(i):\bQ]$$

[cloud walrus]

Merosity

[solar glacier]

I just want y’all to know I think I passed my algebra prelim

Thanks to the help I received here

Hey Z[x] isn’t a PID right cause (2,x) isn’t principal

And gcd(-2+6i,-5+5i) is 3+i right lol

There’s gotta be a calculator out there for GCD of Gaussian integers

I wanna make sure I got those right

[chilly ocean]

,w gcd(-2+6i,-5+5i)

[cloud walrus]

Results provided by WolframAlpha

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

Requested by TTerra#5291

[solar glacier]

And a group of order 351 is not simple@right? And a group of order 144 abelian has 10 iso. Lasses right

[woeful sage]

it's the same as F[x] right? the ring of all polynimials over F?

[chilly ocean]

only if F is infinite (responding to neamesis)

[woeful sage]

i see

[solar glacier]

Yea but Z isn’t a field

I’m assuming here F is a field

[woeful sage]

also is every ring of polynomial over an infinite field a vector space over that field?

[solar glacier]

Oh ur asking a question 😂

I thought u were fixating on what I said

[chilly ocean]

infinite fields have nothing to do with the set of polynomials forming a vector space

defining polynomials over a field F as functions taking a certain form or as formal expressions both give you a vector space over F

they agree when F is infinite

[solar glacier]

We’re mine correct ?

[solar glacier]

This and

[solar glacier]

These

[woeful sage]

sorry mymathisyourmath i totally interrupted you

[chilly ocean]

i encourage you to 1: give a "polynomial function" on a finite field which is not the zero polynomial, and 2: show that, over an infinite field, a "polynomial function" which is identically zero is the zero polynomial

[chilly ocean]

yes

(2, x) is not principal

[woeful sage]

alright

[solar glacier]

Cool how about the other two lol

It can be shown group of@order 351 has a unique 3-subgroup

[chilly ocean]

yes i googled them and they look right mymathyourmath

[solar glacier]

And 144 is 2^43^2

Oh cool thanks

So number of partitions of 4 is 5 and for 2 it’s 2 so it’s 2 times 5 which is 10 iso classes

Cool cool

I have a good feeling I passed it

[solar glacier]

Thanks 🙏

So what IS a Lie group is it a group with manifold structure

I’m taking a wild guess from what I’ve heard in the past

Or no

Lemme google this

Yeah a group that’s also a manifold

Interesting

What subfield of math studies Lie groups ?

Sounds like something I’m interested in for research

Would it be Alg geo

[solar glacier]

Differentiable manifold, that is

I need to soon decide on an area of research and we have a few guys doing Alg geo

[solar glacier]

Since I like algebra and I like topology/manifold theory

[south patrol]

Ye

You can show that R[x] is a PID iff R is a field

By basically the same argument

Like to show (2,x) isn't principal we are basically using the fact 2 ain't invertible

[tawdry plover]

If R is any ring in general (maybe without unit and commutativity) then for maximal ideal in R is R/M a field

If not what would be a counter example to this

[next obsidian]

Take a division ring which isn’t a field

[chilly ocean]

part ii) of b)

need help

haven't been able to figure out an approach

[rustic crown]

can you think of an infinite family of distinct cosets of H?

[lucid shadow]

H is generated by (1,-1,0) and (1,0,-1)
So isomorphic to $\mathbb{Z}^2$

[cloud walrus]

CollinGao-ALT

[coral spindle]

Missing a lot of steps there

[chilly ocean]

linear independence of the columns

[rotund aurora]

yeah

what I was gonna say

I remember trying to count entry by entry once

[next obsidian]

[chilly ocean]

the first column can be taken to be any non-zero element of Z_2^3. the next one has to be outside the span of the first (how many choices are there?). the last one has to be outside the span of the first two (again, how many?)

[rotund aurora]

high schoolers know about linear independence

also, in math, age is almost irrelevant tbh

https://en.wikipedia.org/wiki/Linear_independence

[chilly ocean]

someone else can explain what i wrote. my homework needs me

you were right the first time

[rustic crown]

did you delete your first try?

how many non-zero things are there in (Z_2)^3?

.<

the 0 vector in coordinates is (0,0,0)

[chilly ocean]

no i miscounted too

ignore this message

i put 2 twice

[rustic crown]

and there are a total of 8 things in (Z_2)^3 right?

i was asking non-zero vectors and not vectors with each coordinate non-zero

so there are 7 such

nah, you've only selected the first column

still need to choose the 2nd and 3rd

to keep the det non-zero the second column can't be a multiple of the first one

so how many choices do you think are there for the second column?

which ones?

say the first column we chose was (1, 0, 1)

(also sorry for writing that as a row :p)

you can't do (0,0,0)

that would make det = 0

what else

yep

so do you think there are still 3? 🙈

oh no, i'm just asking for the second column

after fixing the first column, how many ways can you fill in the second column?

why does (1, 0, 0) not work?

(if the first was (1,0,1))

yep

what are the multiples of the vector (1,0,1)? these are the ones what we dont' want

the only multiples of (1,0,1) are 0*(1,0,1) = (0,0,0) and 1*(1,0,1) = (1,0,1)

so we don't want these two

and rest 8-2 = 6 will all work

not just multiples

here we don't want things were are a*(first clolumn) + b*(second column)

wait i misunderstood this

could you clarify?

yep

the multiples of first or second are already counted here when b = 0 or a = 0

anyway, so how many things do you not want for the third column?

and what's that?

yep, all these combinations are bad

so how many are there?

and what makes you say that >.<

like what was your reasoning for saying 1? >.<

just plug all possible values of (a, b)

yep

exactly!!

what

there were
(8-1) choices for first
(8-2) for second
(8-4) for third

so 7 * 6 * 4 in total

[frank cosmos]

One of my exercises is to find a group smaller than the unit complex circle containing all cyclic groups. Does the group |z|=0.5 work?

This is by definition smaller since pi<2pi and for C_n we have the subgroup of orientation preserving rotations of 2pi/n

[rustic crown]

it's not a group

multiplying two things will land in you inside |z| = 1/4

also in terms of cardinality of sets, it's not any smaller

[delicate orchid]

I believe the notion of "smaller" means "a subgroup of the circle group" rather than like

less magnitude?

[rustic crown]

ig i was distracted by you know which group

[frank cosmos]

i see, thanks

the answer key says the group of all roots of unity and letting C_n be 2^(2ipik/n) for all integers k (i checked this when i thought i was correct) but I'm not sure I understand why this works

[delicate orchid]

2^(2ipk/n)? you mean e^(2ipk/n)?

[rustic crown]

[frank cosmos]

2^(2ipik/n), although the notes could have a typo

[delicate orchid]

that is absolutely a typo

[frank cosmos]

also should it mean Z^+ instead of Z?

[delicate orchid]

doesn't matter, same group

[frank cosmos]

right

so then how is this smaller than e^(itheta)?

[rustic crown]

for starters it's countable

[frank cosmos]

oh right theta could be any real number between 0 and 2 pi while there are (1+2+3+4+...) roots of unity

[rustic crown]

but like wew said, it might just mean you want a proper subgroup which it definitely is

[frank cosmos]

since its a sum of integers

thanks, i think i understand it now

[rustic crown]

it might be useful to see another form of this group

notice that the circle group is isomorphic to R/Z with the map R/Z --> S^1 given by t --> e^(2ipi t)

so under this isomorphism, what does this subgroup look like?

[frank cosmos]

[k/n]? for integers k between 0 and n

[rustic crown]

you're letting it range over all n as well

so these are classes of all rationals!

or simply Q/Z

[frank cosmos]

ahh

[rustic crown]

[toxic zephyr]

so i'm working with the dihedral group, semidihedral group, and generalized quaternion group. i see that they're referred to as the "maximal class 2-groups", but i have no idea what that means. and in this paper i'm reading, they only talk about D_(4n), so is D_(2n) (where n is odd) not in the same category?

[tribal niche]

if S is a ring and R is a subring of S, if I is an ideal of R, then do we have that IS is an ideal of S?

[coral spindle]

Not typically, no.

I can't think of any nontrivial examples of when this happens at all

[tribal niche]

using the same setup i am struggling to understand why this is true

[coral spindle]

I think there is probably something more to that setup that you're missing

[rustic crown]

are you assuming they're non-commutative?

[coral spindle]

No

[tribal niche]

all rings are commutative

[rustic crown]

IS looks like an ideal to me in the commutative setting

[coral spindle]

Take S = Z, I = R = Q

Clearly IS = R.

[rustic crown]

R is a subring

[coral spindle]

Oh lmao

[rustic crown]

weird notation lol

[coral spindle]

I read it the wrong way around

[barren sierra]

Yea the backwards lettering also threw me off

[rustic crown]

here i was questioning my existence

[tribal niche]

oh yeah weird notation my bad

[coral spindle]

OK yes, lmao it would be an ideal

[tribal niche]

i was mirroring the notation in the text

S is the larger ring and R is the smaller ring

[barren sierra]

[tribal niche]

my bad

[rustic crown]

i just read it as R --> S, so didn't mix it up

[tribal niche]

wait can someone explain why it is an ideal

[coral spindle]

I think you should try and prove it

it's not a hard exercise

[barren sierra]

IS is an ideal of R, not S right?

[rustic crown]

of S

[barren sierra]

wut

[rustic crown]

it maynot even be a subset of R

[tribal niche]

I is an ideal of R and IS is an ideal of S

yeah it could contain elements not in R

[barren sierra]

FUCK I flipped it in my head again

[tribal niche]

sorry

[barren sierra]

[rustic crown]

[rustic crown]

just be careful with the notation, IS means finite sums of elements of the form i*s for i in I and s in S

it's NOT just the set {i*s : i in I, s in S}

[tribal niche]

wait wat

wdym by finite sums

[rustic crown]

like a general element would look something like i_1*s_1 + ... +i_n*s_n

[tribal niche]

so if you have two rings, R and S, RS would denote finite sums of the form r_1s_1+r_2s_2+...+r_ns_n?

i'm sorry i actually didn't realize this for some reason

[barren sierra]

ye that's the definition

[rustic crown]

assuming it makes sense to multiply stuff from R with stuff from S

(eg. can be done if both are stuff inside a same ring)

[tribal niche]

what type of product is this?

[lethal dune]

induced

(not elaborating)

[rustic crown]

you usually do this only for IM where I is an ideal of a ring R and M is an R-module

in this case the module M was the bigger ring S itself

[lethal dune]

hey det

[tribal niche]

oh

[rustic crown]

you can also do this when M is another ideal of R (as ideals of a ring R are exactly R-submodules of the R-module R)

hewwo ryu

[tribal niche]

oh so what's really happening is that we are taking the product of modules, and the specific modules that we are considering are the ideals of R and S taken as a module over R?

[rustic crown]

product of an ideal and a module

cause usually it doesn't make sense to multiply elements of two R-modules

[tribal niche]

https://math.stackexchange.com/questions/2537341/how-do-we-define-the-product-of-modules

i'm looking at this

[rustic crown]

yea there that fancy frak m is an ideal

[tribal niche]

ok i think i understand what you are saying

this product is only defined for ideals and modules

[rustic crown]

if you know about tensor products, then it's the image of the map obtained by tensoring I --> R by M

[tribal niche]

i see

thanks

[tribal niche]

sorry i have one last question, to clear up all my understanding. this doesn't work because it wouldn't be closed as an additive group, right?

[rustic crown]

yep, that's exactly why we force it to be closed under addition by looking at finite sums of stuff from there

[tribal niche]

yup that's what i understood too thank you very much!

[rustic crown]

[agile burrow]

I just looked it up because I've never heard of this before, but a maximal class p-group is a group of order p^k and nilpotency class k, where p is prime and k > 1. Equivalently, the abelianization of the group has order p^2. The maximal class 2-groups are classified and fall into dihedral groups, semidihedral group, and generalized quaternion groups (as you observed).

The dihedral groups D_{2n} where n is even have abelianzation Z/2 x Z/2, which has order 4 = 2^2 hence they are maximal class 2-groups. On the other hand, if n is odd then the abelianization is Z/2, so these are not maximal class 2-groups.

[toxic zephyr]

oh i see... it looks like that means I'm working with a slightly different/potentially more general class of groups, then.
i really wish i understood groups better because a lot of what you said is stuff i don't really know 😦
looks like i should probably shoot my mentor an email after all, and see what he thinks.
thank you so much, @agile burrow!

[agile burrow]

Don't worry if some of the terminology is unfamiliar! Yeah, it seems like this can be described as a special case of a broader class of groups but getting a good grasp of a specific case (in this case, p = 2) will go a long way. Things like abelianizations aren't too out of reach if you're working with interesting groups like these

[hollow mica]

confusion:

T^k(V) is the tensor product of k copies of V.

T(V), the direct sum of all T^k(V) for k = 0, 1, 2, ..., should be an infinite tuple (v_1, v_2, ...), where v_i is a tensor in T^k(V), and additionally, there exists j such that v_k = 0 for all k > j (this is a part of the definition of the direct product of an infinite number of vector spaces).

Why does Tu write that an element of T(V) is actually a finite sum of the v_i's? Is it because it is common to identify the tuple with the sum?

[chilly ocean]

Is it because it is common to identify the tuple with the sum?
yes

[hollow mica]

Is there any reason for that besides notation?

[chilly ocean]

the multiplication on T(V) will be obtained by just distributing as you would if these were sums of, say, real numbers

like if $$v_1 + v_2, w_1 + w_2 + w_3 \in T(V)$$ then their product is
\begin{align*}
(v_1 + v_2)(w_1 + w_2 + w_3) &= v_1(w_1 + w_2 + w_3) + v_2(w_1 + w_2 + w_3) \
&= v_1w_1 + v_1w_2 + v_1w_3 + v_2w_1 + v_2w_2 + v_2w_3
\end{align*}
where i write the elements next to eachother to stand for the multiplication defined here

[cloud walrus]

what is a Lie group?

[chilly ocean]

i feel like you've already asked this question before and i gave the same explanation

but a loooong time ago

[hollow mica]

yea it was a looong time ago

i only half understood but had to swallow it and move on because class moved too quickly lol

let's see if it sticks this time

[chilly ocean]

this comes from the "and then extend by R-bilinearity" part btw

[solemn garden]

My textbook has an exercise which says something like "Prove that given an abelian group G with two distinct order 2 subgroups, there is a non-cyclic subgroup of order 4 in G". I solved that, and started wondering about whether it would work if G were non-abelian. I think since ab ≠ ba, powers of ab can't be reduced to e, or simplified unless more information is given. Am I right? Also, what are some examples of such non-abelian groups G(that have 2 subgroups of order 2)?

[rustic crown]

distinct order 2 subgroups, just correspond to distinct order 2 elements

and there are plenty of groups with many order two elements, like the dihedrals

the smallest example being S3

[south patrol]

Nice

[nocturne bone]

?Why does #S = 6

[coral shale]

S_X is symmetric group of X? So asking why #X = 3, equivalently?

[swift totem]

In that part of the argument, you're working with N_2=3, ie, X is the set of 3 Sylow-2 groups in G and S_X is the symmetric group of X (a set with 3 elements) of order 3!=6

[glossy crag]

Does anyone know any cool exercises related to PIDs and fraction fields? I'm supposed to work on some with the students, but the ones made available are kind of lame.

[rustic crown]

standard one is to prove gauss' lemma: if D is a UFD then D[x] is a UFD.

[glossy crag]

They had that in the lecture + it would be much too much for them.

I was thinking maybe show them what localisation is, as a generalisation of fraction fields.

[rustic crown]

i don't think localization is a good thing to introduce like that... at least it would confuse them why we call it localization and they won't really be able to appreciate why one wants to do that, like fields are simpler objects so it's nice to consider them but not obvious to me why someone would want to invert stuff in an arbitrary commutative ring with zero divisors

unless ofc you motivate it through AG

which would be too much imo

[glossy crag]

I was thinking of showing only the case for integral domains by defining it as a subring of the fraction field and showing it had a universal property.

Can't think of anything better on that particular subject.

[rustic crown]

right, but they won't really see a use of that i think, so probably not a good idea

how about giving an exercise to prove some universal identities thingy?

for example if you wanna show det is multiplicative, it suffices to show it in the ring Z[x_ij, y_ij] and for matrices X = (x_ij), Y = (y_ij), and further reduce it to show for the fraction field of that... so wlog it suffices to prove such identities over fields where we can do a lot more linear algebra

other examples are like vandermonde det, or some stuff with resultants >.<

but meh, these are kinda computational, so not that fun

[coral shale]

det is multiplicative

[glossy crag]

Thanks for trying, but this has little to do with the topics they're studying atm, so I can't in good faith present that.

[chilly ocean]

[glossy crag]

The subrings of Q (or more generally Q(R) of some domain) are precisely the localisations of Z (or R), right (the set of denominators of reduced fractions in the subring is multiplicative)?

[glossy crag]

Found a good one btw, show that if R is a PID and R<=S<=Q(R), then so is S.

[rustic crown]

forget what i siad, i think it's heavily flawed lmao

[formal ermine]

what does <= mean here

[chilly ocean]

sometimes people use <= in algebra to mean subset with extra structure (sub-group, sub-ring, sub-space, etc.)

[rustic crown]

i think it would be true for bezout domains i.e. integral domain where every finitely generated ideal is principal. Given any subring of the fraction field, consider the multiplicative subset S = {s in R | 1/s lies in the subring of Frac(R) containing R}. the claim is this subring is S^-1R. to see that, if a/b was in the subring, then say (d) = (a, b) so ax+by = d, which means a/b * x + y = d/b = 1/b' is in the subring. so b' in S which means a/b in S^-1R. other inclusion is easy.

[glossy crag]

How come the usual proofs of this I see are so complicated. Once you have the prerequisites (primitivity, equivalence of irreducibility over R and Q(R)), it's just a matter of showing the irreducibles of R[x] are prime (you have factorisation into irreducibles by induction on degree). I feel like the proofs I usually see are more convoluted and less conceptually clear than that.

[rustic crown]

yea, i think people don't usually prove that UFD iff (a) irreducibles are primes (b) ACCP holds.

[glossy crag]

I feel like this topic is often poorly presented.

[rustic crown]

i like to phrase the argument with this lemma, f | g in D[x] if and only if cont(f) | cont(g) in D and f | g in Frac(D)[x]

[glossy crag]

I usually think of it in terms "primitive divides in K[x]=>divides in R[x]", which is the dame thing basically. Jacobson's presentation of this is the best imo.

[coral shale]

(pls use =<, and not overload <= 🙏)

[rustic crown]

overloading good

[glossy crag]

Actually that's what I mean to write, but I'm sure whoever's reading got it.

[rustic crown]

=< looks like a sad face, while <= is happy

[coral shale]

R<=S<=Q(R) is sus

[glossy crag]

<= looks like a pointy-hooded Klansman to me, or Pyramid-Head.

[abstract spear]

What is a Lie group

[coral shale]

just missing a P

[rustic crown]

[chilly ocean]

this isn't the place or time

[tawny pine]

<= is also consistent with some programming languages :>

[coral shale]

[abstract spear]

Lol

[chilly ocean]

please stop pinging me

[rustic crown]

=< this is my snippet for \impliedby lol

and <= for \le ofc

[glossy crag]

It's like one of those "ASK ME ABOUT X" bumper stickers.

[tawny pine]

very slick notation

[abstract spear]

Bro just trolled me

[formal ermine]

what is a Lie group?

[abstract spear]

Now I feel like a complete idiot

[abstract spear]

Wtf

Where you key come from

[formal ermine]

?

[abstract spear]

Not discrete

Your name

What is a continuous function

[formal ermine]

a continuous function is a function that is continuous

[glossy crag]

"And don't call me Shirley"

[abstract spear]

True dat

Lol

[coral shale]

bruh #chill

this aint the place

[abstract spear]

Uh so uh algebra algebra

[chilly ocean]

it's a simple fucking request

[glossy crag]

You could always turn off notifications, you know

[abstract spear]

Faxs

[coral shale]

Thats not the issue or point

[abstract spear]

What he just said yeah

[coral shale]

ANYWAYS -> #chill

[abstract spear]

Lol

[formal ermine]

let L/K be a galois extension of degree 245

how do I show that there exist intermediate fields with degree 49 and 5 over K

I know that its galois group has 1 sylow 7 subgroup

of order 7^2

and 1 sylow 5 subgroup of order 5

but that doesn't guarantee me that their fixed fields have the same order, no?

[rustic crown]

[L:L^H] = |H| and so [L^H:K] = [G:H]

[coral shale]

iirc the correspondence does mention index and degree

[formal ermine]

Ah

it's index

so [G : 7 sylow subgroup] = 7^2 5 / 7^2 = 5 = [L^7 sylow subgroup : K]

same with the other one

[rustic crown]

nvm, i thought you wanted intermediate field of degree 7 >.<

[formal ermine]

degree 49 and 5

yeah this was what I was looking for

thanks det

[primal tusk]

Im learning about the division algorithm rn and im really struggling because I feel like I dont understand the motivation behind learning it

is there anything I should be thinking about when learning this? Also what is the big idea that I need to understand

[coral shale]

what division alg

[formal ermine]

euclidean algorithm?

[primal tusk]

for all integers a, b b>0, there exists some integers r,q such that a = bq+ r and 0<=r<b

[coral shale]

this is #elementary-number-theory i think?

[primal tusk]

im learning it in my modern algebra class

[long geyser]

I mean it pops up in any algebra book also?

[primal tusk]

ive heard that it extends to polynomials and other algebraic objects

[coral shale]

im sure the generalised one does

anyways

[coral spindle]

Have you learned about modular arithmetic?

[long geyser]

no the baby one does also lol

[primal tusk]

yea i have

[coral shale]

for a motivation, u can compute gcd?

[coral spindle]

Great. The division algorithm gives you a way of choosing representatives in modular arithmetic. It extends to things like polynomials, because we can also define things 'mod' a polynomial, in some sense.

[long geyser]

could it be that you don't understand the big idea because it is so fundamental?

[coral spindle]

You will see this later, so be a bit patient.

[primal tusk]

yea i feel like its just so simple, yet the proof is kind of complicated for me

[coral shale]

not many remember the proof no

[long geyser]

which part felt complicated to you

[coral spindle]

So maybe it isn't as simple as you thought

[coral shale]

most remember the alg as its simple

[primal tusk]

i mean the entire proof is just long and there is no way i would be able to come up with the proof on my own

[long geyser]

really?

[coral spindle]

You would, given more mathematical maturity.

[primal tusk]

i mean to me it is but ive only taken proof based linear algebra

[long geyser]

we take an integer n and want to divide it by k

so we look at n-k, n-2k, ...

there is a least positive one among those

all feels pretty straightforward

the least nonnegative one must be strictly less than k

because otherwise you can subtract off another k

I think that is as intuitive as it gets

[primal tusk]

true maybe i just need to try doing the proof myself a few times and it will click

i guess my big question is, if i find this confusing am i going to suck at algebra

because i feel like this should be super easy but it just isnt

[coral spindle]

Why would you be saying this?

Just be patient

Since when has math been easy lmao

[long geyser]

it tells you you suck at algebra right now, doesn't really say anything about you 1 day later from now

yeah math has never been easy

[primal tusk]

ur right

[coral spindle]

You are literally here to learn

[long geyser]

I sucked at algebra a year ago, and I still suck

but I suck much less relatively

and I'm happy w/ that

it's a process

[formal ermine]

why are the sylow subgroups here disjoint?

like $U_7 \cap U_5 = \Set{e}$

[coral shale]

suppose they werent

uh nvm idk what U5 U7 is

but think generators maybe

[formal ermine]

the 5 sylow subgroup and 7 sylow subgroup

[coral shale]

yh idk sylow

is it just a group of order 5

and a group of order 7

[formal ermine]

group of order 5 and group of order 7^2

[coral shale]

ie. C5 and C7

[formal ermine]

maximal p subgroup

[coral shale]

ok either way

lagrange

[formal ermine]

wdym

[coral shale]

think about orders of elements in the 2 groups

[formal ermine]

ah U5 is cyclic so every element has order 5 but the only divisors of 49 are 1, 7, 49?

[coral shale]

yes.

lagrange

orders of elements divide order of group

[formal ermine]

yeah

yeah

ok thanks

another question

is there any easy way to calculate the discriminant of a polynomial

like

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

these are the roots

[coral shale]

i cant even remember the defn of det >.>

its uh

product of difference of roots or summing right

there might be something u can do with vieta

[coral shale]

yh for small degree u can easily compute what it is in terms of coeffs

using vieta

i think

otherwise i dont think theres an easy way
istg i asked the same q 1 month ago

here

[coral shale]

oh yh and u meant discriminant

not determinant

[formal ermine]

typo

mb

what are vieta's formulas

[coral shale]

lazy to explain, lookitup

relates roots and coeffs of poly

[glossy crag]

For low degree polynomials there are explicit formulas, but they become unwieldy already for deg f=4. Degree 3 however is feasible, you can look it up.

[formal ermine]

please provide me a link instead of saying "look it up"

[glossy crag]

If you bothered to google "discriminant" and open the first link that comes up, you'd see it immediately.

[formal ermine]

google results are personalized fwiw

[glossy crag]

I'm pretty sure Wikipedia comes out (almost) at the top regardless, but w/e.

[formal ermine]

fair

thanks tho

[chilly ocean]

why do we call lattices "N_5" or "M_3" ?

are there other naming conventions? I never heard of any, just those two

[toxic zephyr]

in $D_{4n}$ i was able to prove by induction that the parity of the $r$ exponent in an expression like
$$\prod_{i=1}^m(s^br^a)^{k_i}s^{d_i}r^{c_i}$$
is independent of $d_i$ and $b$. but i don't think this would be true in $Q_{4n}$ when $n$ is odd (such as $Q_{12}$, where $s^2=r^3$), since the powers of $s$ can randomly add an $r$ with an odd exponent, which doesn't happen in $D_{4n}$. can anyone give their opinion on whether or not they think this would be true or not?

[cloud walrus]

nilpotent nix

[livid grove]

Is my answer correct?

$A^3=I$ so $x^3-1$ kills A implies minimal polynomial must divide $(x-1)(x^2+x+1)$. As $A\neq I$ this forces minimal polynomial to be $x^2+x+1$. As $A$ is 2 by 2 this is the only invariant factor. $x^2+x+1$ has as eigenvalues $\frac{-1\pm \sqrt{-3}}{2}$ and so jordan normal form is diagonal with these eigenvalues while rational canonical form is $\begin{pmatrix} 1 & -1 &\ 0 & -1 \end{pmatrix}$

[cloud walrus]

Kähler

[livid grove]

Oh I messed up the rational canonical form, that's why I was getting a weird answer (characteristic polynomials being different... which is impossible because it's preserved under similarity)

It is (0 -1 // 1 -1)

[rustic crown]

yee looks right, in D_{4n} moving an s left through some power of r will only negate it and so won't change the parity, and s itself can no way contribute to a power of r. but in Q_{4n} stuff like srs will become s^2r^-1 which is not good because s^2 can contribute something

[toxic zephyr]

damn that's a real shame. looks like i'm going to have to scrap these proofs/lemmas and reassess if the results generalize 😩
thank you sm for looking, i really appreciate it!! ❤️ ❤️ 🙂

[ruby sundial]

how did you get rational canonical form?

[livid grove]

RCF is block diagonal of companion matrices corresponding to invariant factors. Minimal poly is an invariant factor, and is the only one because it's degree 2.

I originally messed up the companion matrix computation having a 1 on the diagonal instead of the lower diagonal

[chilly ocean]

@coral spindle maybe you know? Or could ask some semigroup theory collegues?

[coral spindle]

I don’t know myself, but I’ll ask someone I know in the office today

Is there any context you could add just in case?

I should ping you @chilly ocean

[chilly ocean]

Well it's just out of place that two lattices corresponding to absence of modularity/distributivity have such specific names

No other context than that tbh

[toxic zephyr]

for a group
$$Q_{4n}=\gen{r,s\mid r^{2n}=e,s^2=r^n,s\inv rs=r\inv}$$
can anyone help me find a function $\func{f}{Q_{4n}}{Q_{4n}}$ such that
$$f(x)=\casedef{e,&x=r^a\r^n,&x=sr^a}$$

I'm coming up blank...

[stone fulcrum]

Is that not the complete definition of the function right there?

[cloud walrus]

nilpotent nix

[toxic zephyr]

ah i mean like a word map f(x) with constants sorry 😅
like x^3sr^2... etc.

[next obsidian]

Defining it on generators defines it on the entire group because of multiplicativity

Unless you’ve fixed what a is beforehand or something this defines the function

[toxic zephyr]

yeah but i need a closed form for it in terms of x,r,s 😔

[next obsidian]

You don’t even have a “closed form” for elements of your group??

[coral spindle]

I think you need to be more specific about your requirements

it's not clear what you mean by "closed form"

What you wrote above defines the function entirely

[toxic zephyr]

I'm trying to build a word map in parts. for example, one part is
(sx)^2k
another is
[x,r]^k
etc.
i need a word map which is r^n only when x is in s<r> and the identity on <r>

[next obsidian]

You have to explain way more

Like, how is a being quantified? Is that formula supposed to be true for all a?

Is n fixed?

Is it supposed to be multiplicative?

If the answer is yes to all three, then you’re forced to send r to e and s to r^n

[toxic zephyr]

n is fixed, and it has to be for all a, yes. i don't think it has to be multiplicative (does that mean f(x)f(y)=f(xy)?)

[next obsidian]

Yes

[toxic zephyr]

then, no.

[next obsidian]

Then I don’t see like

What the issue is at all

Just define it as that on those specific elements

And send everything else to just whatever, like send them all to e if you want

[toxic zephyr]

the whole point is to get a closed form. that's the goal 😦

[next obsidian]

What does that mean, and why does that matter?

You can’t even write down elements of your domain in a uniform way

[toxic zephyr]

wdym? s^b r^a works?

[next obsidian]

Then just define it piecewise?

[next obsidian]

Also this expression isn’t unique so

That makes it also difficult to use b and a

Since whatever function you want to cook up in terms of that needs to be invariant under a lot of stuff

[toxic zephyr]

the whole point is to get a closed form non-piecewise defined word map with constants.
ex. x^4 maps
r^a to r^(4a) and sr^a to e
[x,r]^c maps
r^a to e and sr^a to r^2c
which is close, but n isn't necessarily even.
if you don't have any ideas for that, then that's fine. i'll just keep trying 🤷
i've just been stuck on this for a few days now and i thought I'd see if anyone here had any suggestions

[next obsidian]

Why do you need this?

And I feel like you might be able to prove this doesn’t exist somehow, but I’m not sure. Specifically like if you openly allow like, multiplication by r and s, raising x to a power, somehow I feel like you could get some contradiction demonstrating f isn’t well-defined or something but idk. Maybe some theory says this isn’t true because you can define a function piecewise ¯_(ツ)_/¯

[toxic zephyr]

for a research project.
I'm pretty sure it exists. if it doesn't, then thats a lot of progress down the drain

[next obsidian]

I just don’t see why having a closed form is necessary?

[toxic zephyr]

because that's the whole point lol