groups-rings-fields

[formal laurel]

What I should have say, I guess is that you can descompose the field extension by this two fields. But they are not linearly disjoint

[south patrol]

I feel honoured

[rocky cloak]

Well, was I right?

[rapid cave]

@formal laurel I think I get it

The main idea is that the witt vectors gives a equivalence of categories between rings R which are p-adically complete, p-torsion free rings on which R/(p) is perfect, and perfect rings in positive characteristics
just the way you phrase this here, are finite fields not p-torision free

[formal laurel]

No

they are the opposite

[rapid cave]

right

oh I read what you wrote wrong

like R is the witt vectors and the perfect rings are the finite fields

[formal laurel]

indeed

This correspondance is very cool since using some more computations you can show the perfectoid correspondance.

And this is also related to the Fargues-Fontaine curve

But, yeah, the main example to keep in mind is unrammified extensions of Qp

[rapid cave]

right, I have no clue about the other two examples

[rapid cave]

you are welcome to add anything if you want to

anyway thank you guys @formal laurel @rocky cloak @south patrol

[rocky cloak]

KevLee

[supple ice]

you absolutely deserve it

[tardy hedge]

we love KevLee

and jagr and potato and radu and everyone

Lol

[knotty badger]

for part $b$...

$o(a^k) = d$ if and only if $\frac{n}{\text{gcd}(n, k)} = d$

Write $n = \alpha \text{gcd}(n, k)$ and $k = \beta \text{gcd}(n, k)$ with $\text{gcd}(\alpha, \beta) = 1$. and $n = \gamma d$

then we have $\alpha = d$. And $k = \beta \text{gcd}(n, k) = \beta \frac n \alpha = \beta \frac nd$, so $r = \beta$. and $\text{gcd}(r, d) = \text{gcd}(\beta, \alpha) = 1$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

for part c:

$o(ab) | \text{lcm}(o(a), o(b) )$ if and only if $(ab)^{\text{lcm}(o(a), o(b) )} = e$. but $a^{\text{lcm}(o(a), o(b))} = e$ since $o(a) | \text{lcm}(o(a), o(b))$, and $b^{\text{lcm}(o(a), o(b))} = e$ since $o(b) | \text{lcm}(o(a), o(b))$. since $a$ and $b$ commute, this implies $(ab)^{\text{lcm}(o(a), o(b))} = e$

then, $\frac{\text{lcm}(o(a), o(b))}{\text{gcd}(o(a), o(b))} | o(ab) \iff \text{lcm}(o(a), o(b)) | o(ab) \times \text{gcd}(o(a), o(b)) \iff o(a) | o(ab) \times \text{gcd}(o(a), o(b)) \text{ and } o(b) | o(ab) \times \text{gcd}(o(a), o(b))$

$\iff \frac{o(a)}{\text{gcd}(o(a), o(b))} | o(ab) \text{ and } \frac{o(b)}{\text{gcd}(o(a), o(b))} | o(ab)$

$\iff o(a) | o(ab) o(b) \text{ and } o(b) | o(ab) o(a)$

$\iff a^{o(ab) \times o(b)} = e = b^{o(ab) \times o(a)}$

[elfin wraith]

[knotty badger]

now, $(ab)^{o(ab)} = e$ means $a^{o(ab)} = b^{- o(ab)}$, and you can substitute this to deduce the last line

[cloud walrus]

Pseudo (Cat theory #1 Fan)

Pseudo (Cat theory #1 Fan)

[knotty badger]

what was your strategy for solving it wai?

[south patrol]

Qw lol thank u

[tidal torrent]

Is the external direct product of the abelian group of rationals under addition and the abelian group of rationals under multiplication isomorphic to the field of rationals

I don’t think they are isomorphic idk

[south patrol]

Consider the number of elements of order 2

[south patrol]

And ig when you say field you mean abelian group

[tidal torrent]

What structure is this called when an abelian group is both under addition and under multiplication over a field F that satisfies all the properties of a vector space

Is that even a thing

[tidal torrent]

Oh that’s nice

[tough raven]

WDYM? An abelian group is only equipped with one multiplication.

[south patrol]

Sounds like an F-algebra

[tidal torrent]

Maybe I meant a field under both operations over a field F that satisfies all properties of a vector space idk

Commutative ring that has the unity and for every nonidentity element has a unit

[karmic moat]

like a field extension?

[south patrol]

Sounds like an F-algebra

Or ye ig if it's a field then a field extension of F

[twilit wraith]

thats an F-algebra but with a little more structure

[tidal torrent]

I’ve never learned that in field theory what is that

Not quite an extension though

[twilit wraith]

so you may know that a vector space is a field acting on an abelian group

[tidal torrent]

I further learned how like a field of rationals is basically two abelian groups of rationals under addition and multiplication something like that

[twilit wraith]

an algebra is a field acting on a ring with 3 more rules about distributivity and compatibility

[tidal torrent]

Oh that’s cool

[twilit wraith]

some algebras youre already familiar with are R3 with the cross product and C with complex multiplication

[tidal torrent]

How come we don’t learn the algebras as a whole separate structure in GRF

Groups rings and fields

[twilit wraith]

though some people say that the vector multiplication has to be associative

in which case cross product doesnt work

[twilit wraith]

they follow more naturally after having learned about modules

[tidal torrent]

Yep under addition

Does this go hand in hand with representation theory

[twilit wraith]

yeah

though modules are more applicable iirc

i havent done much rep theory yet but i will soon

[noble nexus]

the theory is quite similar to the theory of rings really, I do think its probably worth talking about them in a ring theory class

algebras over a ring in particular basically generalize rings

because algebras over Z turn out to be just rings

[tidal torrent]

Fields are like the only commutative algebraic structure that I know of

[tidal torrent]

Hm I guess that makes sense

[tidal torrent]

Will fields be heavily studied in the commutative algebras

[noble nexus]

fields are heavily studied in galois theory

in commutative algebra terms fields are quite trivial

since commutative algebra is usually interested in things like ideals, of which fields have none (other than 0 and themselves)

for fields to be interesting you need to look at extensions of fields (which is basically galois theory)

[thorn jay]

when some questions are easy you just need to ask harder ones

[tough raven]

I would say fields are heavily used in (more advanced than a first course) commutative algebra.

So you have to study them moderately heavily first (e.g. in Galois theory as above).

[next obsidian]

eh

[noble nexus]

yeah that's possible I've only really done a first course

[tidal torrent]

What is there to learn beyond the scope of basic Galois theory

I learned like one thing in Galois theory and that stuff already seems very advanced for me

For finite fields/finite Galois theory

[tough raven]

Beyond in what direction?

There is oodles more Galois theory of a number-theoretic flavour (ramification subgroups, Galois groups for local fields, class field theory, Galois representations, ......).

[tidal torrent]

Galois representations

[tough raven]

There are also various things of a kind of commutative algebra/algebraic geometry flavour (automorphism groups of transcendental extensions e.g. Aut(k(t)/k) = PGL2(k) but for more variables you run into unsolved things like Cremona groups; or inseparable extensions which are kind of wild and I think practically impossible to classify fully; or generalise to various kinds of finite-dimensional algebras over fields e.g. CSAs)

This is just an idiosyncratic list of things I know of.

[tough raven]

Oh and this involves infinite Galois groups

[south patrol]

If you don't agree that fields are trivial then you are a fake algebraist

[tough raven]

The trivial field with one element

[tough raven]

The trivial computation of $\pi_1^{\mathrm{et}}(\operatorname{Spec} \bQ)$

[cloud walrus]

Raghuram

[next obsidian]

On a level that isn’t unknown, but “classical and beyond basic field theory” is everything relating to ètale and smooth stuff

[tidal torrent]

In what sense would fields be trivial

[south patrol]

Minor spelling mistake

[tidal torrent]

Confused on what you mean by that

[south patrol]

I was joking lol

[tidal torrent]

Hm

[south patrol]

Compute galois group of Q

[tough raven]

Copy cat

[south patrol]

Oh didn't see lol

[tough raven]

Copy potato*?

[tidal torrent]

What are Coexter and Cremona groups

[supple ice]

cremona groups are just Bir(P_k^n)

birational geometry 🔥

[tidal torrent]

OHHH I get it

the field of reals extends the field of rationals and the field of reals is the extension field does that mean gal(R/Q) is the identity map

So it’s the trivial map basically

I’m not a true algebraist what does this mean

[tough raven]

From where did you get that R is a splitting field over Q?

[tall igloo]

yeah it isnt, there are many real numbers that arent roots of Q-coefficient polynomiais (see: pi, etc) and you're missing many roots (like i, etc). its some strnage field between Q and C

[earnest hatch]

Struggling with a proof: Prove that if N a submodule of M. And M/N and N are finitely generated, M is also finitely generated. I was thinking about defining the generating set of M/N to be some {x_1 , ... , x_n} and the generating set of N to be some {y_1, ... , y_i}. My intuition of the problem is that M would be finitely generated by the union of the two sets; however, I have no idea how to get there

[tardy hedge]

Any m in M is the preimage of something in M/N so you can probably just express that m in terms of those generators from that

[earnest hatch]

I will try thinking about this and working on it

Thanks

[tough raven]

(Swap domain and codomain in this sentence: anything in M/N is the image of something in M.)

[tidal torrent]

I meant extension field

Q(x)/f(x)) is the extension field over Q right

[tough raven]

Why don't you know what you meant?

[tough raven]

No, Q is a field and has no (non-trivial proper) quotients.

[tidal torrent]

Oh

Actually no it’s just gal(Q)

Trying to be an algebraist but saying fields are trivial is kind of vague with no context behind it

[tidal torrent]

I’m not too confident with the material I’ve learned in field theory yet

You know far more about it than I do, as you’re someone who’s probably had a solid background in representation theory

[tough raven]

That was a joke so it's kind of futile to try to take it seriously.

[tidal torrent]

Reason why I’m here to know why that is

[tidal torrent]

Actually this would be true for Q(sqrt2)) being a splitting field maybe an extension field

[tough raven]

It's both, but yes.

[tidal torrent]

Not for Q(x)/(f(x)) being an extension field

Aren’t groups also trivial as well

Specifically free and simple groups

[tough raven]

There is no Q/(f(x)). And I think neither fields nor groups (except when they have only one element) are trivial.

[tidal torrent]

Principal ideal generated by f(x) idk

Quotient ring

Oh I just realized I forgot the (x) for the integral domain Q(x)

I do that all the time

[rocky cloak]

The notation is Q[x]

[plucky coral]

Hi, I had asked a question related to field theory in #help-6 , was wondering if anyone could look over it and help, thanks

[scenic pendant]

hello i am having some trouble intuitively understanding something

i don’t know how to latex so lemme write it and send a pic

the orthogonal group is isomorphic to {-1,1} mixed with special orthogonal group?

[delicate orchid]

yes

[scenic pendant]

it doesn’t make sense in my brain at all, maybe the problem is that i don’t get what the semi direct product is doing

[delicate orchid]

with C_2 acting via the corresponding diagonal matrices

[scenic pendant]

i wish i had a more specific question to ask

[delicate orchid]

I should be more specific because this action actually depends on n mod 2

[scenic pendant]

oh right because of Z2

[delicate orchid]

not at all

consider the SES SO(n) -> O(n) -> C_2 with the second map being the determinant, if n is odd then -1 in O(n) has determinant -1, but if n is even then it lies in SO(n)

[scenic pendant]

i was typing up a storm, but i think my real question is that i need convincing, so im going to study the proof of this and pray that i am convinced somehow, idk i just feel like SO(N) is much smaller than O(N) so i dont get how this isomorphism is even onto

[scenic pendant]

oops

[quiet pelican]

There are at least ceil(n/2) such splittings that look like that

[delicate orchid]

in particular, in the first case then -1 commutes with all of SO(n) so this product actually becomes direct

[quiet pelican]

How do you define SO(n) and O(n)

[scenic pendant]

what is SES? what’s that stand for

[delicate orchid]

it's literally index 2

it's only half the size of O(n)

[scenic pendant]

and idk what index is

[delicate orchid]

why are you learning semi direct products when you don't know what the index of a subgroup is

[scenic pendant]

O(N) is matrices where A transpose = A inverse, SO(N) = things in O(N) with determinant 1

[scenic pendant]

that’s not what i was told

like i wasn’t taught that

[delicate orchid]

bad course lol

[scenic pendant]

i won’t know what i don’t know yk 😭

is index it’s dimension

[delicate orchid]

no it's the number of cosets

anyway

I guess I should ask what you actually want answered

is it the map defining the semidirect product you're confused about?

[scenic pendant]

yes

it says the map is from Z_2 to Aut(SO(N))

wait

[delicate orchid]

yeah that's not useful at all that's just restating what the presentation of the semidirect product looks like. I was going to say for convivence assume it's the map sending -1 to conjugation by whatever element you're sending -1 to in your section of the determinant map

[delicate orchid]

unfortunately this isn't unique lol but all choices generate an isomorphic group

[scenic pendant]

okay so the map let’s say f, is f(-1) = conjugate of -1

because it’s {-1,1} do we also have to define f(1) = conjugate of 1

[delicate orchid]

well yes otherwise it's not a group homomorphism. Also you can't just always pick f(-1) = the conjugation map by -1 in O(n). Writing c_x for the conjugation map induced by x, we have to pick f(-1) = c_x with det(x) = -1, as this x must be arising from a splitting of the determinant morphism O(n) -> Z/2Z. If n is even and we choose x = -1 in O(n) then det(x) = 1

sorry, awful phrasing

one moment

[scenic pendant]

okay that makes sense

[delicate orchid]

what you're saying works when n is odd without modification, but when x is even you'd need to pick something like, diag(-1, -1, ..., -1, 1) or something

[scenic pendant]

okok

sorry i didn’t realise i would have to consider even and odd n

[delicate orchid]

it's still a semidirect product it's just a fiddly detail with the precise action

in fact, when n is odd, because -1 is central this semidirect product becomes direct

which is fun

[scenic pendant]

yeah i have this theorem written down

this module is so hard to me lol, bcs it goes through groups of like order 4 and then suddenly i’m involved in matrix groups and geometry, i don’t think they’ve explained the underlying concepts well so it feels so out of the blue yk

like i didn’t know index was a thing

[delicate orchid]

are you a physicist by any chance

[scenic pendant]

and they don’t explain automorphisms well so i find myself reciting a definition without actually understanding anything

i’m not a physicist just a maths undergrad

[delicate orchid]

have you covered like, the first isomorphism theorem?

well no I guess not cause there's no way you've talked about quotient groups without index being mentioned

[scenic pendant]

yes i only understood today that it’s called that lol, we call it in the module, “the homomorphism theorem”

[scenic pendant]

u would think

i understand that a quotient group is like all of the left/right cosets

but never had index in the definition

[delicate orchid]

I guess it's a simple definition lol as I said it's just the number of cosets

[scenic pendant]

i remember they put index in the formal writing of Lagranges theorem, but it’s just “where i(H,G) is the index” and then no further elaboration ever

[scenic pendant]

this added a lot of understanding thank u for telling me

[elfin wraith]

Real, the way physics students seem to learn group/rep theory is wild

[delicate orchid]

I'm convinced half of the reason people think richard feymann was such a good expositor was because he's the only physicist to not be dog shit at pedagogy

[elfin wraith]

Nah I actually cant allow that, bar one, all of my physics profs were great. Im pretty sure physicts have been interesd (academically) in pedagogy for a good bit longer than mathematicans too. He may just be the one physcist who actually knows how to talk to people

Though not women...

[delicate orchid]

BLEEEEEEEEEEEEEEH

[thorn jay]

i mean he probably didnt consider them people

[kind temple]

and he didn’t brush his teeth or something

or wash his hands apparently (after #1)

[tough raven]

Semidirect products take time to get used to; if you want to get comfortable with them I would recommend trying to understand several different simple examples. I can suggest
(i) G = affine-linear transformations of the real line as the semidirect product of H = R^x (multiplicative group of dilations about the origin) and K = R (additive group of translations)
(ii) G = the dihedral group of automorphisms of a regular n-gon as the semidirect product of a cyclic group H = C2 of order two (capturing reflections) and a cyclic group K = Cn of order n (the rotations of the n-gon)
(iii) the group G = B2 of 2x2 invertible upper-triangular matrices (with entries in your favourite field - e.g. either real or complex entries) as the semidirect product of the group H = T2 of invertible diagonal matrices and the group K = U2 of invertible upper-triangular matrices with diagonal entries 1
(iv) your example of G = On(R), H = Z2 and K = SOn(R).

For each example where I have given you groups G, H, K, I suggest the following exercise:

(a) Compute what conjugation of K by H looks like: for h in H and k in K, calculate h k h^{-1}. (It will come out to be in K; figure out which element of K.)

(b) Write an arbitrary element of G as a product of an element of H and an element of K. Convince yourself that there is exactly one way to do this. (Personally, I usually find th next step more intuitive if I use KH rather than HK here, but HK matches the convention you are being taught right now.)

(c) Express multiplication in G in terms of this "HK-representation": that is, for g1 = h1 k1 and g2 = h2 k2, consider g = g1 g2 and find out what the HK-representation g = h k is, expressing h and k in terms of h1, k1, h2, k2.

[languid trellis]

Let $M$ be an $A$-module. I would like to show that $A \otimes_A M \cong M$. I've constructed a bilinear map $f: A \times M \to M$ given by $f(a,m) = am$, hence this factors through the tensor product. So we have an $A$-module homomorphism $\overline{f} : A \otimes_A M \to M$. It is quite clear that this map is surjective, as any $m \in M$ is the image of $1 \otimes m$ in $\overline{f}$, but I am struggling to show injectivity. Does anyone have any advice?

[cloud walrus]

swifteeee

[tardy hedge]

its easier to write an inverse map instead

[languid trellis]

Wow

I am catastrophically stupid

[tardy hedge]

no i had the same exact issue first time i was trying that problem lol

[languid trellis]

lol, well that makes me feel a little better haha

[tardy hedge]

yeah i think with tensor products the more you can avoid dealing with the actual elements in M (x) N the better probably

[languid trellis]

i just universal propertied the other canonical isomorphisms but i couldnt do that here and i got lost

wow

[thorn jay]

||alternatively notice that a ⊗ m = (a * 1) ⊗ m = 1 ⊗ (am), so by bilinearity every element of A ⊗_A M is of the form 1 ⊗ m. Then of course f(1 ⊗ m) is zero iff m is zero iff 1 ⊗ m is zero so we are done||

[tardy hedge]

nice tensor symbol

[thorn jay]

with universal property its Yoneda lol:
||We have the adjunction
Hom(A ⊗_A M, N) ≈ Hom(A, Hom(M, N)) ≈ Hom(M, N)
So by the fully faithfulness of the Yoneda embedding we have that A ⊗_A M ≈ M||

[languid trellis]

oh yeah. neat

[tidal torrent]

How does one visualize the topology of a weyl group I don’t understand

This all seems trivial to me

[tough raven]

Do you mean that ax+b is not the same as a(x+b)?

[tough raven]

Why are you considering a topology on the Weyl group?

[prisma ibex]

if by Weyl group you mean the kind that shows up in Lie theory around root systems there is no topology here these are finite groups lol

[knotty badger]

None other than the discrete one

[supple ice]

it’s basically a discrete set

[marble hinge]

Finally got this big boy in print!

The paper is a little bit too smooth and thin as for my taste, but it will do. It’s 900+ pages after all, so I guess they needed some trade off in order not to make the book too thick.

[maiden crater]

NO WAY

I'm doing this right now lol

congrats !

in 13(b) I feel I'm missing something

[rocky cloak]

What do you feel is missing?

[maiden crater]

In my proof which I'm typing out now

Let $\phi^{-1} (M)$ not be a maximal ideal. Let There be a bigger ideal, call it $\Omega$. Then let $x \in \phi^{-1} (M)$ and $y \in \Omega \setminus \phi^{-1} (M)$. Then $\phi(xy) \in M$. This is as every ideal is closed under multiplication. So $xy \in \phi^{-1} (M)$. So $y \in \phi^{-1}(M)$

[cloud walrus]

wai

[maiden crater]

Like this doesn;t even use the fact that M is a maximal ideal, never mind the fact that phi is surjectuve

[noble nexus]

how do you conclude that y is in the preimage?

it doesn't follow

[maiden crater]

xy \in phi^{-1} (M). x \in phi^{-1} (M) so x^{-1} does too

[noble nexus]

what

if xy is in an ideal it doesn't imply x and y are in the ideal

[maiden crater]

I know but x^{-1} is in the ideal is it not, and xy is too?

[noble nexus]

and idk what x^{-1} is

how can you assume x is invertible

[maiden crater]

oh right

yea, that was my miskate

[noble nexus]

yeah if an ideal contains an invertible element it must be the entire ring

[maiden crater]

back to the drawingboard then I suppose

[maiden crater]

esentially what I want to do is show that if there's a larger ideal containing phi^{-1} M it is M , right

infact that is by defn

[noble nexus]

yeah

[maiden crater]

hmm

[noble nexus]

hint: correspondence theorem

[maiden crater]

yeah, just struk me

[maiden crater]

so I show S/M iso to phi^{-1} (M) do I not

[noble nexus]

uh not quite that

[maiden crater]

I think I skipped a bit of theory

nvm found it

[maiden crater]

ah yes, I skipped this as the proof used lattices which I hadn't done. Guess It's time to do it

[elfin wraith]

Honestly the better proof of that is just to use the correspondence theorem

Like its good to know multiple proofs, but to me this is like the clearest example of the power of the correspondence theorem and actually tells you why the result is true

[rocky cloak]

I mean "the proof used lattices" probably means "the proof is just the correspondence theorem"

[maiden crater]

Oh, I saw lattices as I was scrolling and got scared 😭

didn't even try to prove it by myself once I saw lattices"

[elfin wraith]

I guess, I just took it to be the very direct proof since (they had posted and now deleted) a elementary proof for R/P in the book

But yeah like if you mean correspondence theorem and youre scared of it, this is the perfect time to get comfortable with it

[maiden crater]

Here inclusion is of subsets, right

[noble nexus]

yeah

but yeah the fancy way of saying it is that the lattice of ideals of R/I is the same as the lattice is ideals of R containing I

[maiden crater]

I see. I suppose this is an excuse to read up on lattices aswell

thanks

[noble nexus]

its good to underestand lattices they come up quite a lot in math

basically they're posets with a notion of a "supremum" and "infimum"

(some only have it for finite # of elements, some can have arbitrary)

in the case of ideals, your "supremum" is the ideal generated by a bunch of ideals, and your "infimum" is the intersection

so essentailly a lattice isomorphism means a bijection that respects these two operations (which also implies that it respects the ordering if you think about it)

[maiden crater]

so join and meet in ordering theory

[noble nexus]

b/c a \leq b iff b = sup(a,b)

yeah join and meet are the abstract terms usually used though tbh I always say sup and inf b/c I forget which is join and which is meet half the time

[maiden crater]

fair enough. Okay, guess I'll spend an hour or two on this then

[noble nexus]

oh also I'm fairly sure an order isomorphism is automatically a lattice isomorphism, so really the only thing you care about is preserving the order

[maiden crater]

Eh, guess reading up on lattices will clear those questions

thanks

so essentially a lattice is a total ordering?

oh, right , two elements of P may be incomparable to each other, right

[noble nexus]

yeah in fact often lattices are very far from total orderings

the standard example of a lattice is the power set of a set under inclusion

where the meet and join are union and intersection

[maiden crater]

That does make sense

[tough raven]

partial ordering where you can do "unions" and "intersections"

[thorn jay]

except they may not distribute over eachother, unlike unions

better example may be the naturals under division, where meet and join are given by gcd and lcm

(this is in fact the opposite of the ideal lattice of Z)

lattice theory is cool its a shame that many concepts there are only implicitly used in modern day math

[maiden crater]

this is basically saying if an inf exists for. a set, then a sup exists?

oops this

[thorn jay]

yes

this is very nice in the case of so-called Moore families, or closure operators

[maiden crater]

wait it's saying if any set has a least element , that same set has a sup?

[thorn jay]

not a least element

an inf

[maiden crater]

oops, yes

[thorn jay]

if every subset has a meet then every subset must have a join

however, kinda conventionally the meet of the empty set is the top element

[noble nexus]

that isn't really conventionally it basically must be true if you look at the def of the meet

[thorn jay]

yeah okay true

[noble nexus]

(which imples every complete lattice must have a top and bottom element)

[thorn jay]

peak

[noble nexus]

we love complete lattices

[thorn jay]

complete lattices? you mean closure operators?

[knotty badger]

I did this result in BY3

[thorn jay]

yes

[knotty badger]

The correct generalisation is “every total category is complete”

[thorn jay]

explicitly:
∨X = ∧{y | x ≤ y for all x ∈ X }

[knotty badger]

Yep

[maiden crater]

I read this as BYD and was like huh?

[knotty badger]

Nah it’s baby Yoneda 3

[maiden crater]

hmm, this is going to be fun to prove in the absence of additive inverses

I mean really all I need to do is each chain is bounded above for then zorn's lemma comes in clutch

[thorn jay]

its completely provable without zorn

[maiden crater]

huh, i see

lemme think ( It sounds simple , so I'm missing something obvious)

[knotty badger]

Now that I’ve written them I feel like it may have been better to swap the order of BY2 and BY3

[maiden crater]

For reals I'd basically take the negative seqeunce, I aim to do something like that I suppose?

[thorn jay]

well you only have the data of an ordering and a meet

[maiden crater]

is this an easy theorm to prove 😭

[thorn jay]

yeah

[maiden crater]

F should have had it by now

[thorn jay]

eh

consider the set { a in P | x ≤ a for all x ∈ X }

[maiden crater]

ah]

got it

thanks

Guess doing this for the frist time at 12 am is not the best idea

[elfin wraith]

This made me think, do any of the standard axioms of ZFC show up as theorems? (I.e. are there any “natural” theorems that turn out to be equivalent to one of the axioms)

I expect the answer is no, at least not in an at all interesting way

[thorn jay]

"every ring has a maximal ideal" is equivalent to a weaker form of choice

i believe

[elfin wraith]

Well yeah sure lol I mean there’s like 100 ways to do it for choice but I was wondering about any of the ZF axioms

I imagine the answer is no because they just don’t say “enough” but I hadn’t considered this before

[thorn jay]

nvm just choice

[thorn jay]

i don't know and dont think so yeah lol but im not an actual logician

[elfin wraith]

Yeah there’s about a million statements that turn out to be equivalent to choice

I may ask in foundations but that place scares me

[thorn jay]

lmao

that is valid

[thorn jay]

i love choice

[elfin wraith]

Same, honestly I’m at a point in my life where I am annoyed by notes or books etc when they say “caution, this requires choice”

Yeah, I use the axioms all the time, I’m not fussed

[thorn jay]

caution: you may need to make a based decision

[elfin wraith]

It’s honestly not even pointing out the use that irritates me, it’s specifically being like WARNING!!

My category theory notes do it and it just rubs me the wrong way lol

Like you’re a K theorist, why are you entertaining any skepticism of choice

[knotty badger]

i mean it's not really about scepticism of choice right?

it's more to do with constructive vs non-constructive

my favourite example of this is zermelo's theorem

https://en.wikipedia.org/wiki/Zermelo's_theorem_(game_theory)

[tough raven]

Maybe their target audience includes topos theorists

[noble nexus]

I think it's good to be aware of where choice is used and if it's necessary, even if I don't particularly care that much

[elfin wraith]

Yeah I mean this isn’t something I lose sleep over and I am joking in how I say it, but I do think displaying it as a warning is kinda silly. Mention it sure, but idk it feels silly to warn people about the use of one of the axioms from the commonly accepted foundations. Especially when most people in that room are topologists and algebraists

[knotty badger]

i don't see how it's silly

[elfin wraith]

Like I know Tom Leinster has a particular issue with this, his ETCS notes have a whole section dedicated to where our topology course uses choice unnecessarily and I just can’t help but feel that’s silly

[maiden crater]

What's going on here, oh NVM

[elfin wraith]

It’s the presentation as a warning that I think is odd, as though it’s something to be scared of or avoid

[maiden crater]

Not a thing i waana read at1 am😭

[elfin wraith]

I’m not saying that choice free maths is bad or uninteresting or anything

[noble nexus]

tbf there's at least one place in topology where using choice is imo quite bad form and that's proving net continuity implies continuity

[knotty badger]

it's just a note that the proof is nonconstructive

[noble nexus]

even though the choice proof is kinda the most obvious one

[mystic ether]

I am pretty sure there are quite a few things equivalent to foundation that show up as theorems

I don’t remember details tho

Also i think also replacement

Although maybe not for this last one cause any equivalent statements to a schema must also be schema

Anyway this question is a social one cause there are always SOME statements equivalent to all of our axioms they are just not always interesting enough to be theorems

[elfin wraith]

Yeah this is what I was getting at with “natural” which is of course vague and very open to interpretation. I’m sure there’s a bunch of not very interesting statements you can make to be equivalent to something or other

[tidal torrent]

by that means theres probably no topology on the weyl group nvm

isnt the weyl group one of the largest in size other than the monster group out of all the finite groups

its like over 600 million or something

[elfin wraith]

Infinity is significantly larger than any finite number you can think of

[karmic moat]

"the" weyl group?

are you thinking of a different group?

i can think of a pretty small weyl group

[elfin wraith]

There’s also arbitrarily large finite groups

The monster isn’t the biggest

[tidal torrent]

wait there are different types of weyl groups

[tough raven]

Equivalence is done in ZF

[elfin wraith]

It’s simply the largest sporadic group

[karmic moat]

you should probably read about the construction of a weyl group

[tidal torrent]

i knew it was a sporadic group but i didnt know of other groups larger than the monster group

[noble nexus]

S_100

[tulip otter]

so let $\alpha=\begin{pmatrix} a&b\c&d\end{pmatrix}, T=\begin{pmatrix}1&n\0&1\end{pmatrix}$ and $S=\begin{pmatrix}0&-1\1&0\end{pmatrix}$. WLOG assume $|d|\geq |c|$ and write $d=qc+r$ with $q,r\in\mathbb{Z}$ and $|r|<|c|$, then $r=d-qc$ and $\alpha T^{-q}=\begin{pmatrix}a&b_1\c&r\end{pmatrix}$ with $b_1\in\mathbb{Z}$. multiplying $\alpha T^{-q}$ by $S$ swaps $c$ and $r$ and then replaces $c$ by $-c$ (it doesnt matter what changes in the 1st row). Since $|c|>|r|$, we can write $c=q_1r+r_1$ and repeat a similar process to what was done above by first multiplying $\alpha T^{-q}S$ by $T^{q_1}$ and then multiplying by $S$. After repeating this process a sufficient number of times, the bottom left entry of the resulting matrix will be $0$. This matrix is of the form $\alpha\gamma=\begin{pmatrix}a_1&b'\0&d_1\end{pmatrix}$ with $\gamma\in\Gamma$. Now $\alpha\gamma\in SL_2(\mathbb{Z})$ so $\det(\alpha\gamma)=a_1d_1=1$, hence $d_1=c_1=\pm 1$. Using the remark that says $S^2=-I$, we can take $d_1=1$ and hence $c_1=1$ which means that $\alpha\gamma=\begin{pmatrix}1&b'\0&1\end{pmatrix}\in\Gamma$. Hence $\alpha=\alpha\gamma\gamma^{-1}\in\Gamma$ and $\Gamma=SL_2(\mathbb Z)$

[cloud walrus]

ali yassine

[tulip otter]

is this correct?

[vapid vale]

the finite simple groups have 18 infinite families and the 26 (or 27) sporadics, the infinite families have simple groups of arbitrarily large size given a parameter

and you can have nonsimple groups as well

[tulip otter]

Also is there a specific reason to why the hint asked for |d'|\leq |c|/2 rather than just |d'|<|c|

[tidal torrent]

lol ik what you mean when you said simply

was that intentional

[elfin wraith]

I would honestly love to say it was lol

[thorn jay]

Let n be the size of the monster group. Consider Z/n+1Z

[rocky cloak]

A Weyl group is a symmetry group of a root system. Different root systems have different symmetry groups, so they come in all different sizes.

Could even be infinite depending on your definitions.

[mystic ether]

Algebraists seething over 4 year old chads

[elfin wraith]

Yeah? Well ill consider Z/n+2Z

[rocky cloak]

Z/0Z
Do I win?

[thorn jay]

finite group

well i mean if Z/nZ is finite for all n, Z/0Z must be finite too

[rocky cloak]

What's a little in-prefix between friends

[elfin wraith]

I must concede because 0 is in fact a natural number

[thorn jay]

unrefutable logic

[quiet pelican]

Z is finite because every proper quotient is finite

[thorn jay]

every residually finite group is finite

[cedar vault]

It looks fine, only I think you typed c1 instead of a1 in the 2nd and 3rd last lines.

[tulip otter]

ohh yea true, tysm for checking it and pointing this out

have a great day/night

[south patrol]

What if it's like mid afternoon

[tulip otter]

then its still day

[south patrol]

Valid

[maiden crater]

so I',m back

in 13b I simply need to show $R/\phi^{-1}(M)$ is a field, right

[cloud walrus]

wai

[noble nexus]

yeah that's one way, though I think the other way is a bit more direct

[maiden crater]

Showing if $R/phi^{-1}(M)$ is a field then $\phi^{-1} (M)$ is maximal?

[cloud walrus]

wai

[maiden crater]

oh right, I'll want the opposite direction

[maiden crater]

Very stupid question, but I'll first want to show $R/ phi^{-1}(M)$ forms a field right

[cloud walrus]

wai

[cursive spindle]

As a hint think of 2 maps starting from R and ending with some sort of field in such a way the kernel is exactly what you want

[maiden crater]

mhm

[maiden crater]

for this I'd need to know R has a maximal ideal

[cursive spindle]

Zorn

[maiden crater]

Ofcourse lol

[cursive spindle]

So your problem is you're not sure if R has a maximal ideal?

[noble nexus]

not sure if you need that

for this

[cursive spindle]

Fwiw every nonzero unital ring has a maximal ideal

[cursive spindle]

Yeah

just do the obvious thing

[noble nexus]

sure if you do that then phi^{-1}M is a maximal ideal

rather than showing its a field directly, its good to try and show its isomorphic to a field you already know about

given the information you have, there's only really one candidate

[cursive spindle]

hint: ||projection||

[maiden crater]

such as M/S

[cursive spindle]

hm

You mean S/M?

[maiden crater]

oops, yeah

S/M

[cursive spindle]

yup that's right

• my hint and you're basically done

[maiden crater]

which is direct from phi's surjectivity

[maiden crater]

The projection map now you said?

Isn't that overkill

[cursive spindle]

yes

huh

not really

[maiden crater]

hmm, lemme try both

I mean yeah, basically that

map a \in phi^{-1}(M) to phi(a)+M

[cursive spindle]

I'm sure there's an ugly way to do this but this is very neat and natural

[maiden crater]

This is yeah

Sad this didn't come to mind at once 😔

[maiden crater]

and now we're done

[cursive spindle]

It's okay that's one of the things you should be getting comfy with if you decide to do more algebra

Composing maps is helpful

[maiden crater]

I do want to do algebra so yeah

we're done now I think

[cursive spindle]

right yeah this is the homomorphism we got then the projection S -> S/M

[maiden crater]

awesome. Thanks so much!

[cursive spindle]

then you can check kernel is what you want

no worries

[maiden crater]

General advice: Other than practice(ofcourse) how to keep track of results

specifcially in algebra

[cursive spindle]

Apply them to what you care about

[maiden crater]

just to be clear, passge here basically refers to the homomorphism g-> g+ ((f(x)) right

[cursive spindle]

yes the bar is denoting that you're working modulo f(x)

[maiden crater]

the line" passage to the quotient" always cracks me up tbh

it sounds like a story

[tough raven]

I thought you were Ultra for a minute.

[maiden crater]

Not the first time

[tardy hedge]

Lol a story

[maiden crater]

the passage was treacherous, not all functions made it out , but all that entered were changed forever

[tardy hedge]

The Passage to the Quotient ™️ 🎥

[pulsar hound]

is there a difference between ker f and Ker f?

[maiden crater]

usually they both are just the kernel of a function

[delicate orchid]

the first letter is capitalised in the second one

[pulsar hound]

it was in the context of homomorphisms and i rememeber i saw both versions so i though they meant different things

[tardy hedge]

[maiden crater]

could you show us the context

and name the book you're using

[pulsar hound]

it s not in english though

[maiden crater]

Like if the book really wants to mess with you , it could do mean different things , but that's a horrible thing to do

like it's easy to write K instead of k

[delicate orchid]

they literally never mean different things. I will give chat 5 gifted if they're actually meaning different things

[pulsar hound]

this is me revisiting this class and i remember we wrote it with k

[maiden crater]

same thing, assuming 1' is the identity and this is a group

[pulsar hound]

yes

ty

[tardy hedge]

Lol

ker is baby and Ker is daddy

[karmic moat]

Can i have 5 gifted anyway

[rocky cloak]

I've seen people use one to denote the morphisms part of the kernel and the other the object. But I don't think that's a general trend

[maiden crater]

,w morphism

[cloud walrus]

Results provided by WolframAlpha

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

Requested by math_rocks

[maiden crater]

oh , cat theory

Can['[t wait to do this

[rocky cloak]

So basically the kernel (of a group homomorphism say) consists of a two things:

A group which is the kernel, and the inclusion map of said group into the domain.

Both of these things (the group and the inclusion map) are sometimes called the kernel of f.

[maiden crater]

Not OP, but never heard of kernel being used in teh 2nd conetxt

Also as an aside ideally how much algebra would I want to know before cat theory

Would groups , rings and fields upto an undergraduate level be enough

I suppose I should start with the appendix of D&F

And some places say toplogy , so I guess a good part of mukres

[delicate orchid]

no

[karmic moat]

Okay

[tardy hedge]

im so sorry amamono

[elfin wraith]

The difference for me is generally if I do \ker or \mathrm{Ker} because I have forgotten \ker works

[rocky cloak]

Not many hard prerequisites, so just depends on your motivations.

That's probably enough examples of categories to get started. But if you want to actually apply category theory to algebra you probably want to know a little about modules as well.

[maiden crater]

I see, thanks.

[elfin wraith]

The topology you want tends to be more algebraic, so probably most of munkres. The fundamental group is the prototypical example of a functor generally

But you don’t really need that

And then I guess chain complexes of modules are a very interesting category but not strictly necessary. At the end of the day it’s just a language to talk about stuff in, so the areas it touches are basically endless. You don’t need to follow every example

[maiden crater]

I see

Guess I'll have to wait atleast until next sem for cat theory then

[elfin wraith]

You don’t have to, category theory doesn’t really have any strict prerequisites beyond a basic familiarity with mathematical thinking. The thing is just that you want actual examples to get your hands dirty with, and they’re generally groups, rings, modules, chain complexes and topological spaces. But if you only know some of those you’ll be fine, if you want to learn it I think you probably know enough to survive you just won’t follow all the examples. That’s basically always true though

But I do personally think it’s better to know some topology first, just as motivation, but reasonable people disagree

[maiden crater]

I want to do it properly, not over and over again as I learn more

so I';ll wait

Maybe I'm misunderstanding this, but basically this is saying such groups have elements such that for each natutral number there exists an element with a bigger, yet finite order

[tardy hedge]

yeah the cat theory course im in rn isnt assuming much algebra background so the examples are like, pointed sets... partial functions ..

and some basic groups stuff

[rocky cloak]

Yes, every element has finite order, but the orders get arbitrarily large

[tardy hedge]

so i guess it is doable lol but yeah the examples arent as interesting maybe

[maiden crater]

and 3-generated means it's generated by a set of cardinality 3

[rocky cloak]

Fun fact for you: there exists infinite 2-generated groups such that every element has order p for a fixed prime p.

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

[maiden crater]

wait, so every elment has prime order, or for every prime there exists an element of order p

[rocky cloak]

Every element has the same order

[knotty badger]

(Every nonidentity element)

[maiden crater]

what,every element has the same order, but is non-finite

a little mind boggling innit

this just means both the group and subgroup have a n-generator right

not the same one

Yea, that goes without saying

how would this work for the trivial group though

just {e}

I mean sure it's finitely generated, but it's not n-generated

[rocky cloak]

Typically people use n generated to mean having a generating set with cardinality<= n

I guess you can also think of it as allowing duplicates in your generating set

[maiden crater]

The second is much more useful for me, allows me to make sense of the 2nd statement more easily and have it be a direct result

well, ofcourse I have to show it has more than one generator

[candid patrol]

Give me a good exercice guys

Only finite group théory

[quiet pelican]

List all finite simple groups :3

[rocky cloak]

Prove burnsides basis theorem:

If S is a minimal generating set for a p-group G then there is an Fp vector space V and a group homomorphism G -> V sending S to a basis.

Or with a little more detail:

Let F(G) be the normal subgroup generated by commutators and pth powers in G (the Frattini subgroup). Then G/F(G) is an Fp vector space an a set is a minimal generating set iff it's image is a basis

[candid patrol]

yeah already done : (

A p-group is nilpotent, so there is a well defined homomorphism from G -> products of G/M where M describes the maximals subgroups,

and G/M ~ Cp

[maiden crater]

Let $A$ be generated by $(a_1,a_2 \dots , a_n)$, each being distinct. Then $A={a_1^{k_1} \dots a^{k_n} \mid k_i \in \Z , 1≤i≤n }$. Let $H≤A$. Then every element of $H$ must be of the form $a_i^{h_i} \dots a_k^{h_l}$. Suppose an element that's not of this form were in $H$, call it $t$. Then it follows $a_1^{b_1} \dots a_{i}^{b_i} \dots a_{l}^{b_l} \dots a_n^{b_n}=t$. We then multiply this by $a_i^{-(h_i+b_i)} \dots a_l^{-(h_l+b_l)}$. So $a_1b^{1}\dots a_{i-1}^{b_{i-1} a_{l+1}^{b_{l+1}} \dots a_n^{b_n} \in H$.

oops, wrong reply

[candid patrol]

np

[rocky cloak]

Groups of order p^2 being abelian?

Too easy maybe...

[maiden crater]

My idea is to ultimately show that then $H$ is the entire group( this is what you do for cyclic groups)

[cloud walrus]

wai

[candid patrol]

it is lol, G/Z(G) cyclic => G abelian

[rocky cloak]

Pick a random integers prove there is no simple group of the order / construct example

Or prove there is exactly one simple group of order 168

[maiden crater]

hmm

[cloud walrus]

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

[maiden crater]

would induction be a better idea, this is becoming a symbol soup

[candid patrol]

i'll try thx

[maiden crater]

NVM,got it!

[candid patrol]

I got a nice one too if you want : Determine the smallest group G such that G = Z(G) x D(G) where Z(G) and D(G) are proper and nontrivial. (D(G) = G')

[supple ice]

@candid patrol what are u doing here?

[candid patrol]

group theory of course

[maiden crater]

Let $A$ be generated by $(a_1,a_2 \dots , a_n)$, each being distinct and no power of any $a_i$ is any power of $a_j$. Then $A={a_1^{k_1} \dots a^{k_n} \mid k_i \in \Z , 1≤i≤n }$. Let $H≤A$. Then every element of $H$ must be of the form $a_i^{h_i} \dots a_k^{h_l}$. Suppose an element that's not of this form were in $H$, call it $t$. Then $a_1^{b_1} \dots a_{i}b^{i} \dots a_{l}^{b_l} \dots a_n^{b_n}=t$. Then $a_1^{b_1} \dots a_{i}^{-h_i-b_{i}} a_{l}^{-b_{l}-h_l} \dots a_{n}^{b_n} \in H$. But this is a contradiction to our assumption, we're done.

[cloud walrus]

wai

[maiden crater]

Pretty sure this is wrong though

Isn't it

[maiden crater]

If this is correct though probably a bit much of a symbol soup

eh, there are some small errors but is fine I think

[rocky cloak]

||C2 x A5||

||G' = Z' x D' = D', so D = D'. Smallest group where this happens would be A5 because all smaller groups are solvable.||

[maiden crater]

Let $A$ be generated by $(a_1,a_2 \dots , a_n)$, each being distinct and no power of any $a_i$ is any power of $a_j$.
This is as if it were then the group would be generated by a subset of $(a_1,\dots,a_n)$
Then $A={a_1^{k_1} \dots a^{k_n} \mid k_i \in \mathbb{Z} , 1≤i≤n }$. Let $H≤A$. Then every element of $H$ must be of the form $a_i^{h_i} \dots a_k^{h_l}$. This being a representative element of $H$, $h_j$ are fixed . Suppose an element that's not of this form were in $H$, call it $t$. Then $a_1^{b_1} \dots a_{I}^{b_{i}} \dots a_{l}^{b_l} \dots a_n^{b_n}=t$.
We now multiply across by $a_i^{-h_i} \dots a_{l}^{-h_l}$.
Then $a_1^{b_1} \dots a_{i}^{-h_i+b_{i}} a_{l}^{-b_{l}+h_l} \dots a_{n}^{b_n} \in H$.
But this is a contradiction to our assumption, we're done.

[cloud walrus]

wai

[maiden crater]

I feel I'm missing something

I got why this is wrong now

[delicate orchid]

Ok I choose 808017424794512875886459904961710757005754368000000000

[kind temple]

Z/nZ

[maiden crater]

Hi folxs

general advice wanted

So i just read this proof and it does make a lot of sense to me, however, I'm unsure of how I'd come up with this ( I suppose here we can just say it's a nutural extension of the case of n=1)?

And in general when is it a good idea to read the proof of a theom

I mean this a pretty simple proof 😭

[elfin wraith]

I often think that when reading proofs of theorems and honestly sometimes it is just hard if you’ve never seen that idea before and that was probably a great result when it was first published. I try to just give myself grace with that. Usually you’re coming into subjects that already have like 50-200 years of thinking done, it’s only natural there’s some stuff you’d never come up with on your own.
But the important thing is I suppose that you file the idea of the proof away, and maybe you can do something similar later.

For this specific proof, I would be thinking induction straight away because n=1 is pretty easy and I could probably just write that down straight away, so I’d want to see if I could extend it. Then as you say, the induction step is basically the same thing again.

I do get the feeling though, the one I really remember feeling recently was when I first saw the computation of the cohomology ring of RP^n, I followed the proof well enough but I just had no clue how anyone came up with that

[tough raven]

I wanted to say this so badly but I restrained myself.

[maiden crater]

tbh I spent ~1 hour trying to come up with a proof without this, in hindsight trying ot work from the general case to n=1 was a bad idea

[twilit wraith]

alright i have to show that the vector space of functions [a,b] to R is infinite dimensional

its a question im asked in abstract alg 2 which is why im asking here

my idea is that i assume theres a finite basis

f1, ..., fn

then for a linear combination a1f1 + ... + anfn = g for some g in the space where the image of g is at least size n, i can use these values to solve for each ai

[velvet hull]

Contradiction will be hard, it’s easier to find a big enough linearly independent set

[twilit wraith]

oh ur totally right

[acoustic igloo]

is there a way to show two groups are isomorphic by using a presentation of one of the groups?

[vapid vale]

this is generally an undecidable problem

but for specific groups there may be tricks

[acoustic igloo]

you could show some generators satisfy some relations, but i guess the problem is knowing whether they generate the "freest" group?

is this what you're referring to? https://en.wikipedia.org/wiki/Group_isomorphism_problem

[vapid vale]

yes

[acoustic igloo]

what if you have more information than just the presentations?

you can check two finite groups are isomorphic, but i'm wondering if there's a way to do it efficiently

[vapid vale]

i don't know anything about this problem beyond what i could read about it online, i think other people here may have more to say

[acoustic igloo]

ok thanks for your input 🙏

[acoustic igloo]

oh i found van Dyck's Theorem

[tough raven]

You can use that sometimes but there's no protocol you can run given the presentation that will definitely yield an answer.

[acoustic igloo]

the theorem gives you a homomorphism, and you can map onto a generating set so you know it's onto, but the problem is if it's not one-to-one. then you get an isomorphism with a quotient group i guess (rather than the whole group)

[twilit wraith]

ok so i have to show that for all endomorphisms phi: V to V with V a finite dimensional vector space, there is an integer m such that the intersection of phi^m and the kernel of phi^m is 0

im pretty sure i have it i just wanna make sure my argument is rigorous

if phi is injective we are done as ker phi = 0

thus suppose not and repeat

then im phi has dimension less than V

for some k <= m, phi^k is either an injection from im phi^(k-1) to itself or it maps to a space with smaller dimension

but if we keep coming across spaces with smaller dimension, the process has to terminate since the space itself is finite dimensional

so either ker phi^k = 0 if theres an injection or im phi^m = 0 if there never is an injection before m

[whole basalt]

Is there a way to take a physical object and assign two rigid rotations such that they act like the quaternions i and j? I want to teach my class about quaternions but I'd like to have a physical way of thinking of them.

Given that as a complex number, i is a 90 degree rotation, I'm wondering if something similar can be done in 3D, since that's what Hamilton was trying to do with them anyway.

[tough raven]

I think any two rotations by 90° about perpendicular axes work?

[whole basalt]

Not unless you're identifying some of the faces of, say, a cube

And I'm having a hard time visualizing it

If it turns out it's not possible, I'd be open to any "hands-on" way of visualizing quaternions

[tough raven]

No, I'm wrong.

The quaternionic group has only 1-dimensional and 4-dimensional irreducible representations over R, and the former all have -1 act trivially. So the smallest dimension in which you can faithfully represent them by rotations/reflections (or equivalently linear maps) is 4.

[whole basalt]

That's annoying. 😛

Oh well, thank you!

It's good to know such a thing doesn't exist in terms of just rotations

Though it means I'm back to the drawing board on how to teach it more concretely

[glass sinew]

saw a d&f exercise answer proving two groups isomorphic based on same generators/relations, given the two groups are of the same order (which was already explicitly given)

wondering about this myself now

[tough raven]

OK one thing you could do is let i, j, k be rotations by 180° about x, y, z axes. This is not faithful but it is faithful except for -1. So you can remember the extra thing as some "twist" (I think this is probably related to spinor stuff).

Like draw a ribbon at the point acted on such that it gets twisted by 180° on two i's if you can come up with such a way of drawing it.

[maiden crater]

Kind of confused here. This is what I did.

Let $A \subseteq I \subset \R$ where $A$ is a prime ideal. Then Let $a \in A,b \in I \setminus A$. Let $a,b$ both be non-zero divisors. Let $n=o(a)≠o(b)$. Then $a^{n}b \in A \implies b\in A$. This is a contradiction. Thus $A$ must be a maximal ideal

[cloud walrus]

wai

[maiden crater]

The question ofcourse is why doe such a,b need to exist

[rocky cloak]

What is o(a) here? And how are you concluding that b is in A?

[maiden crater]

order of A , so a^n=1. And I'm assuming ab \in A

[rocky cloak]

If a is a unit then A is not a prime ideal

[maiden crater]

That makes sense. Should have not missed that

[quiet pelican]

The way I’d do it is via the “I maximal <=> R/I is a field” and “I prime <=> R/I is an integral domain”

[maiden crater]

yea, I thought of that at first too

shame I didn't prove it using that

showing R/A is a field is fairly easy too 😭

[maiden crater]

this is key to said proof ofcourse

[rocky cloak]

I guess if you want to follow this approach consider m and n such that
b^n = b^m
with n and m different (must exist because R is finite).

Then
b^n-m (b^m - 1) = 0 is in A so either b is in A or (b)+A = R

[maiden crater]

How does this prove the second proposition

[rocky cloak]

The meet and join are only defined in terms of the order.

E.g. the meet is the largest thing smaller than all the Hi.

[maiden crater]

yea, but the meet and join need not belong in the set right

[rocky cloak]

Well they do actually, but that's not relevant

[knotty badger]

if you really want you can use the defining (universal) property of meets and joins

[maiden crater]

which is

[knotty badger]

https://pseudonium.github.io/2026/01/27/Baby_Yoneda_3_Know_Your_Limits.html

[rocky cloak]

Just try to write down what it would mean for

$$\left[\bigcap H_i\right]^a$$

to not be the meet of $H_i^a$

[cloud walrus]

jagr2808

[knotty badger]

essentially, $x \leq \bigwedge_{s \in S} s \iff \forall s \in S, x \leq s$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

and $\bigvee_{s \in S} s \leq x \iff \forall s \in S, s \leq x$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

these are the universal properties of meet and join

[noble nexus]

the intersection of subgroups is a subgroup, the subgroup generated by any number of subgroups is a subgroup

but yeah the fact that is being used here is that any order isomorphism is necessarily a lattice isomorphism

and you can get fancy with it but really it's just a consequence of the fact that the meet and join are the greater lower bound and least upper bound

[maiden crater]

I got that bit, I just didn't see how it's immediate

[maiden crater]

there's some element between those two elements greater than H_i^a

[noble nexus]

If you remember from real analysis, if a is an upper bound for a set S that is \leq all other upper bounds, then a must equal the supremum

The same idea is true here

(and yeah you can see this using universal properties and category theory but if you aren't comfortable with that you can also just prove it directly)

[maiden crater]

I know no cat theory :(

[noble nexus]

this follows immediately because you get sup S \leq a but also a \leq sup S

[maiden crater]

okay, so I just replicate that

[noble nexus]

right because sup S is a least upper bound but also we assumed a is a least upper bound

[maiden crater]

right

[noble nexus]

hence they're the same and we can say "the" least upper bound because it's unique

[maiden crater]

makes sense

got it

thanks

really sorry for such an obvious question 😭

[thorn jay]

a lattice is completely decided by its order so an order isomorphism induces a lattice isomorphism

[noble nexus]

All good it's an important point

Because for example a monotone function is not necessarily a lattice morphism

[thorn jay]

i really wish they'd teach like general lattice/order theory more

[noble nexus]

I do too but tbh it doesn't really fit into any class

It would be nice to maybe cover lattices in a proof class I think

just because they come up everywhere

[thorn jay]

yeah

[noble nexus]

I remember doing posets but not much more than that

[maiden crater]

mhm, I've only done very little order theory , but it's so fun

Like a few pages ( so basically nothing)

[thorn jay]

like the first couple sections of the first chapter in Burris and Sankappanavar is super informative

the stuff about algebraic lattices, compact elements, relations with closure operators

[noble nexus]

and having lattice theory lets you make a lot of statements more precise

mainly a lot of the time in math you have two natural lattices and prove they are order isomorphic, and sometimes you then also go through the proof that they preserve the other stuff (even though it follows)

[maiden crater]

I;m using roman rn 😔 , shoudl get hugerford instead , but too late for that [ because this is a reading project]

[thorn jay]

and Galois connections of course

[maiden crater]

Maybe I'm just forgetting the notation here , but isn't HvK the join of H and K

so is this just saying the join of HK=HvK

[solar shore]

yes

[thorn jay]

yippee permutability mentioned

[acoustic igloo]

https://www.youtube.com/watch?v=ilVS-ZY5QP4

this video shows how to do it in 4d

more here at 5:50 https://www.youtube.com/watch?v=RhuaPhahHbU

[tough raven]

Right, but I think they are looking for a 3D visualisation so they can show it to people.

[maiden crater]

On a tangent ( as this came up in my group theory book), would cardinal arithmatic have any pre-requitsites

[tough raven]

Still, a video for 4D is nicer than just "there exists a 4D faithful real representation".

[maiden crater]

context

[tough raven]

If you just want the answers, not really.

[tough raven]

This doesn't even need cardinal arithmetic.

[maiden crater]

eh, it doesn't, I know

but the book mentions it like so

And it has a section about cardinals , so I was wondering if there's any good reason to do it right now

[vapid vale]

Don’t talk about my size

[tough raven]

What's in that section? Is there a table of contents you can share?

[maiden crater]

sure

[maiden crater]

see section 1

[tough raven]

Sure if it's 3 pages just read it

All of those preliminaries (except misc, ofc I don't know what that is about) are probably going to be useful to be familiar with sooner or later (or sooner).

[maiden crater]

I'm reading 'em as and when they first pop up in theory, I suppose that's good?

[tough raven]

Also works

[maiden crater]

There are |HK| blocks are the map ,f, is surjective right

[solar shore]

wdym

f is defined to be surjective ye

[maiden crater]

and hence there are |HK| pre-image sets

[solar shore]

yes

[maiden crater]

cool, was obvious but for some reason took 5 minutes to click 😭

[solar shore]

np

surj just means if your codomain has x elements then there are exactly x fibers

does it show u how you can get theformula for |HK|?

and where |H \cap K| comes from

[maiden crater]

it does, but I'lll do that bit tomorrow once I read the bit on cardinalities ( as it does that using the same method)

[solar shore]

good luck!

[maiden crater]

TYSM!

[solar shore]

idk how your book does it but tl;dr, basically another pair of h, k in the same fiber can be written as a product of only elements from H or only elements from K, hence an element in H \cap K

[maiden crater]

yup, I think that's what this book does too, and I was thinking of trying something similar anyways

[solar shore]

then rewriting into unique elements you end up getting |HK| * |H \cap K|

division badaboom

very cool, gl friend

[solar shore]

so peak

[maiden crater]

This book is SOO GOOD

[solar shore]

currently struggling

[maiden crater]

with D&F?

[solar shore]

yeah, just so many exercises

[maiden crater]

D&F is a slog , yeah. But been very rewarding for me so far

[solar shore]

id like to get through a bunch of 4.5 today

but there's 56 exercises

[maiden crater]

[solar shore]

like come on

[maiden crater]

aim for like 20 at most

maybe skip the repititive ones

I didn't use D&F for groups, so can't comment

[solar shore]

oh im doing all of them

just for shits and giggles

[maiden crater]

Have fun /srs

[solar shore]

if you ever wanna contribute

or at least, solution check

[maiden crater]

will have a look soon, I'm on a bit of a time cruch rn as I have a reading project

So need to finish the readings and exercies planned for this friday :)

[solar shore]

npnp

[maiden crater]

The last bit doesn't seem to be clicking

,ti

[cloud walrus]

The current time for math_rocks is 12:00 AM (IST) on Wed, 04/02/2026.

[maiden crater]

checks out

[solar shore]

which part?

[maiden crater]

got it lol, just so late I'm slow now

got it for naturals atelast

cadinals , tomorrow :)

The way of calcultaing |f^{-1}(x)| si beautiful thouhh

I'd never have come up with that way myself tbh, would have thought of a much more complicated way, which is sad

[solar shore]

yea get some sleep

good night friend

[maiden crater]

Yea this is probably why, good night

[novel star]

can you check my proof for the following exercise please

Let $G$ be a finite group acting on a finite set $S$. Writing $[w]$ for $1\cdot w$, we have the direct sum
[
\mathbb Z\langle S\rangle = \sum_{w\in S}\mathbb Z[w].
]
Define an action of $G$ on $\mathbb Z\langle s\rangle$ by defining $\sigma [w]=[\sigma w]$ (for $w\in S$), and extending $\sigma$ to $\mathbb Z\langle s\rangle$ by linearity. Let $M$ be a subgroup of $\mathbb Z\langle S\rangle S$ of rank $#(S)$. Prove that $M$ has a $\mathbb Z$-basis ${ y_w}{w\ins}$ s.t. $\sigma y_w=y{\sigma w}$ for all $w\in S$.

[cloud walrus]

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

[novel star]

proof:

It suffices to prove the transitive case (just take direct sum). Let $w$ be in $s$ and consider the set $S_w$ defined by
[
S_w = \bigcap {\sigma\in G_w}\text{fix }\sigma,
]
where $\text{fix }\sigma$ is the set of fixed points of $\sigma$ (I don't know the proper notation for this). Observe that there is a basis for $M$ that consists of ${d_w\mathbb Z[w]}{w\in S}$, where $d_w$ is some nonzero integer. Let $g\in G$ be in the pointwise stabilizer of the image of this set under some $\sigma '\in G\G_w$. Then $\sigma '^{-1}g\in G_w$ and so $G_{\sigma 'w} = \sigma 'G_w.$ We see that we can define the desired basis ${y_w}$ by setting $y_w = \text{lcm}(d_1, ..., d_w)\sum_{s\in S_w}[s].

nvm this is wrong

[cloud walrus]

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

[novel star]

this is wrong because it assumes that there is overlap between the $S_w$, which might not be the case

[cloud walrus]

qchs

[rocky cloak]

I don't see why M would have a basis consisting of multiplies of the [w]s

[novel star]

free abelian group of same dimension as group containing it

[rocky cloak]

Say G=S=C2={1, g} and let M be generated by 1+g and 1-g

Then M does not have such a basis

[novel star]

yes i know

i need to figure out how to make a basis to play nice with the d_k

[novel star]

{1, g} forms a z-basis though

[rocky cloak]

Not for M

Since M doesn't contain either one

[novel star]

that doesn't have rnak 2 though

[rocky cloak]

It does

[novel star]

o wait

[rocky cloak]

Like there exists bases for ZS and M with this property, but the basis given by S doesn't work

[novel star]

hmm

[rocky cloak]

I'm also sceptical of your original claim

Like I don't think M is just a permutation module

[novel star]

yeah the basis thing screws it up

[rocky cloak]

So yeah maybe there is some mistake in your exercise or maybe you wrote it down wrong

[novel star]

a mistake in the exercise is somewhat likely

since its lang

here is langs statement

and the theorem he cites

[rocky cloak]

Yeah I just found it myself, so then it clearly is a mistake on langs part

(as the cited theorem is different)

[novel star]

is that why it took way longer than normal to solve (incorrectly)

[rocky cloak]

Your solution more or less solves the correct statement though I guess

Or I mean, once you have the basis consisting of multiplies of things in S you're done

[novel star]

yay

ty

[twilit wraith]

this question is confusing me

how is it not immediate that phi restricted to W is a linear transformation?

if phi behaves nicely with V elements then it does with W ones as well

is it asking to show that phi restricted to W is an endomorphism?

and same with phi bar

[rocky cloak]

It is immediate yeah. I guess phi-bar very slightly less so.

Main part of the problem is the second part I guess

[white oxide]

Could I have a hint for 5 please?

[twilit wraith]

Oh

Idek why theyd ask this then

The first part i mean

[rocky cloak]

Hint 1: ||For any polynomial f in Q[x] there is a rational number you can multiply it by to make it into a primitive polynomial in Z[x].||

Hint 2: ||if p(alpha) = 0, then f divides p||

[white oxide]

Multiplying by the lcm of the denominators gives a primitive polynomial in Z[x] right?

Nvm

[rocky cloak]

At least since f is monic

[white oxide]

So we know that we can multiply f by some coefficient B such that Bf is in Z and is primitive

Bf(a) = 0 and so f divides Bf but I'm kind of stuck here..

Maybe the contradiction is to show that Bf is the product of two polynomials, one of which isn't primitive?

The problem is that if we're assuming that f is in Q for contradiction this won't work

Oh wait hold up

Alpha is an algebraic integer

[acoustic igloo]

a group G with a normal subgroup N doesn't necessarily have a subgroup isomorphic to G/N, right?

[alpine plank]

G = Quaternion group, N = {±1}

[acoustic igloo]

thanks 🙏

[tulip otter]

Tho if G is abelian then such a subgroup exists

Well at least if G is a finite abelian group and N is a subgroup of G such that the orders of N and G/N are relatively prime, then there exists a subgroup M of G isomorphic to G/N. Moreover, G=N\oplus M

[dim widget]

not true (e.g. Z = G, pZ = N) but it is true for finite abelian groups!

[solar shore]

hey group theory geniuses

so Bungo was helping me with this earlier, and i eventually got it, but i was curious

i was doing this problem showing that these two matrices generated the unique sylow 2-subgroup of SL_2(F_3), and i solved it with the technique of using the projection homomorphism

from G -> G/Z(G)

i was wondering why it was important to factor out G by {I, -I}

[vapid vale]

what does why mean? beyond that the quotient is easier to work with

[solar shore]

oh

i guess thats my answer

was just curious as to why we chose Z(SL_2(F_3))

easier to work with

good enough for me

[vapid vale]

that subgroup PSL_2(F_3) is isomorphic to A_4

[solar shore]

yeah, but i guess it was just a curiosity as to why the center

[vapid vale]

i mean quotient group

well the center is conained in every sylow 2-subgroup

[solar shore]

i see

[alpine island]

what is "a minimal prime ideal" (of a commutative ring)? Surely the only prime ideal that is minimal under inclusion is the nilradical?

[wicked patio]

The nilradical isn't necessarily prime though

[elfin wraith]

I think you need to associate your minimal prime to another ideal I, so it’s the largest prime extension of I in a sense

[wicked patio]

For example in Z/6Z

[elfin wraith]

I guess not in a sense, that just is what the smallest prime ideal containing I is lol

[alpine island]

oh good point

[alpine island]

it doesn't mention any other ideal though

where Sigma is the collection of multiplicative subsets of R (which do not contain 0)

[elfin wraith]

I think the criterion here is actually the nilradical is prime iff there’s only one minimal prime but my comalg is mega rusty so I could be slightly mistaken there

[wicked patio]

I heard somewhere that the nilradical is the intersection of all minimal primes

So I think that's true

[alpine island]

nilradical is the intersection of all primes in general I know

[wicked patio]

I guess that's sufficient

[alpine island]

that sounds correct, because it would itself be the only minimal prime ideal.

[wicked patio]

For any collection of sets, there is a unique minimal element in the collection iff the intersection of all sets in the collection is in the collection

[alpine island]

I feel like I don't need unique here

[alpine island]

it's clear minimal elements do exist because Zorn's lemma

chains of multiplicative subsets are fine

[thorn jay]

be careful, Zorn has a condition which is not always satisfied

oh huh, apparently you can

that's interesting

[rocky cloak]

Union and intersection of a chain of prime ideals is prime!

[thorn jay]

oh right of course lol

[tulip otter]

Ohh right, Tysm for the correction.

[tulip otter]

Actually I checked an exercise which I remembered to check if I was correct and that's how I wrote this after the false claim, but I didn't delete the false one because I wasn't sure whether it's true or no and I said that someone would probably correct me hahaha

Anyways tysm, have a great day/night

[maiden crater]

is this too concise/ missing anything

[tardy hedge]

Oh is it because intersection of two minimal primes cant be prime cause it would contradict the fact that the starting two were minimal

[maiden crater]

just to be clear, here the primes are arbitary, right

[maiden crater]

I suppose I'll need to use symmertic groups here

[knotty badger]

hang on

.

ah right i see

a and b can't commute, makes sense

[maiden crater]

got it I think

a=(1 2 3 4 5), b=(4 5 6)

[twilit wraith]

ab = (1234)(56)

[maiden crater]

right, and that had order 6

[quiet pelican]

It has order 4
Which is not prime

[twilit wraith]

6 is not prime

(If you wanna use symmetric groups, because of how disjoint elements of the group behave when composed, think of a prime r such that there are primes p,q such that p + q = r)

Thats a big hint

[tulip otter]

the way you proved ab\notin Z is weird, as in the phrasing looks weird to me, also how does it follow that a^k=e from (ab)^k=e and ab=ba?

[maiden crater]

That;s what I was going for 😭

I has this exact thought in mind

[maiden crater]

also how does it follow that a^k=e from (ab)^k=e and ab=ba?
a and b can't be inverses as they don't belong to the same subgroup, so the only option is both a^k and b^k=e

[twilit wraith]

You actually end up getting only a single triplet of primes that work

Note that since r must be greater than p and q, r must be at least as big as the third prime number

And so r is odd

[maiden crater]

mhm

and so one must be 2

[twilit wraith]

Yes

[maiden crater]

p=2 and q is say 5

[twilit wraith]

Oh I suppose you dont get a single pair

Oopsies

[twilit wraith]

That works

So then r is 7

What can we do with those primes now?

[maiden crater]

then I have to work with S_7 and do something with that

[twilit wraith]

Good idea

[maiden crater]

(12) and (12345) is the first thing that comes to mind

[twilit wraith]

Hmm

Whats the issue when you compose these permutations?

We're trying to get a permutation of length 7 right

[maiden crater]

Well this will reduce the length of the perm

[twilit wraith]

True

Is there a different set of permutations we can use?

[maiden crater]

not if they're disjoint

then the order will be the product of their individual orders

[twilit wraith]

Exactly

Ohhh wait I mightve goofed hold on

Yeah oops

I blanked and thought the order is the sum

I think we gotta do a little more heavy work then

[maiden crater]

So what instead I want is to reduce the order

[twilit wraith]

Yes

[maiden crater]

also congrats on green

[twilit wraith]

So like i would work with 2,3,5 here personally

[twilit wraith]

Thank you!

[maiden crater]

oh I think I got it

(12)(123)=(32)(1)=(32)

[lament flame]

I'm trying to prep for a tutoring session with someone struggling with a rings & fields class

Does anyone have a simple example of a non-injective, non-surjective ring homomorphism?

I'm trying to help them understand images and kernels

[frigid epoch]

Z -> Z/nZ[x] or something

[vapid vale]

You can also just take a really silly map of direct products

[lament flame]

Hmmm... yeah polynomials over Z_n should be good enough, ty

Probably any homomorphism that's the kinda "interesting" I'm looking for would be too much for them ^^;

[frigid epoch]

you could also do something like Z/9Z -> Z/12Z

[tardy hedge]

yea i was gonna say^

something like that

[lament flame]

...Is that allowed?

[frigid epoch]

the nonzero one

[vapid vale]

you can talk about matrices

[tardy hedge]

there are maps between them if they not coprime

[vapid vale]

That’s a place where they’ve seen images and kernels

[lament flame]

Hmm okay yeah, I could make a matrix example I think

[frigid epoch]

wonder if non-in/surjective maps of topological spaces would induce such maps of C*-algebras by gelfand-naimark 🤔

[twilit wraith]

Primes gotta be distinct

[maiden crater]

Oops🥀

[noble nexus]

for Gelfand duality a surjective map between top spaces induces an injective map of C* algebras and vice versa I believe

you have to be a bit careful if they're only locally compact and not compact

[exotic verge]

idk if this is the right place to ask... i know eventually areas of analysis, especially functional analysis, become algebraic. i'm reveiwing undergrad real analysis content right now and trying to pick up more algebra, and i was wondering if anyone has a resource for how algebraic ideas can be applied to prove standard undergrad analysis theorems

ik this isn't the point of undergrad analysis, i already went through the standard proofs, so i'd wanna try something like this for fun

[karmic moat]

i suppose one example is the Fundamental Theorem of Algebra

in the sense that you can prove it analytically (and i think this is done in most undergrad analysis courses)

and you can also prove it algebraically

[exotic verge]

anything more interesting? haha

[karmic moat]

there are certainly a lot of connections between analysis and algebra

but other than the fundamental theorem of algebra, i cant think of any at the undergrad level lol

[proud vigil]

i didnt know there was an algebraic proof of the fundamental theorem of algebra, does it not use any results that rely on analysis?

[karmic moat]

oh i guess it uses intermediate value theorem lol

bummer

https://mathoverflow.net/a/10626

this answer gives one via galois theory but does use the intermediate value theorem lol

[quiet pelican]

You fundamentally need analysis, because C is an analytic object
You either need some form of ordering or topology to form the reals
I think the IVT is the least analysis you’re gonna need for a proof
In the sense that like, it’s a “minimal” topological requirement for the reals

[proud vigil]

right that makes sense

[quiet pelican]

What analysis does it use?

[quiet pelican]

That’s the same proof we were talking about before

[elfin wraith]

That is a sick proof, I hadnt seen that before

[karmic moat]

Yeah the thread is full of fun proofs lol