groups-rings-fields

[thorn jay]

(because every normal subgroup is the kernel of a permutation representation; the natural action on G/N)

[lean sail]

quick question on what i think is supposed to be a relatively simple theorem and proof, but for some reason i am missing something. or, maybe there is a typo or mistake somewhere? this came from fraleigh. in the proof of the following theorem, it states that it is a proof by contradiction, but what are they actually negating?

[karmic moat]

m is squarefree

and d_1 is not 1

so d_1^2 divides m is a contradiction

[lean sail]

like, are we supposing that m IS divisible by n \ge 2?

what are we supposing?

[karmic moat]

no, you're supposing m is squarefree, i.e., you cannot divide it by the square of any integer

[lean sail]

ok but a proof by contradiction usually involves negating a hypothesis

[rustic crown]

we want to show k = 1, so we're assuming the negative of that.

[karmic moat]

^

[lean sail]

ok, is that where they say "each d_i \ge 2 divides d_{i + 1}"?

[rustic crown]

nu

[karmic moat]

"If k \ge 2, then d_1^2 divides m"

[rustic crown]

it starts at "if k >= 2... then something bad happens"

[lean sail]

this helps, i think i might be starting to see it.

[lean sail]

what's a little bit confusing might be the notation, they use $n$ in the theorem, but they when doing the proof, they use $k$, is $k$ supposed to represent $n$ in the proof?

[cloud walrus]

proofman

[lean sail]

like, i don't see an $n$ variable in the actual proof

[cloud walrus]

proofman

[rustic crown]

oh you might be mixing up two different thingies

d_1 is playing the role of n

and d_1 is >= 2 always.

[karmic moat]

the n in the statement of the theorem is also just them defining what a squarefree integer is

sometimes authors group in definitions in theorems

[lean sail]

this makes a lot of sense! so now i think i get it, if k = 2 then essentially d_1 will divide m twice

[rustic crown]

yeep

[lean sail]

or d_1^2 i should say

got it!

banging my head on the wall on that one for no reason.

[rustic crown]

(one pedantic thing which can ignore if you don't care: technically k = 0 is allowed in the theorem which happens when G is the trivial group with order m = 1. ofc the proof works, but people usually don't like to worry so much about edge cases, they can obviously be handled separately so no reason to worry about them in the meat of the proof)

[lean sail]

thanks for pointing that out!

[rustic crown]

yee

[lean sail]

yee yee

[twilit wraith]

"If A and B are ideals of a ring, the product of A and B is

AB = {a1b1 + a2b2 + . . . + anbn | n ≥ 1, ai ∈ A, bi ∈ B} .

Show that if A and B are ideals of a commutative ring and A+B = R, then A∩B = AB."

i have no clue frankly

[karmic moat]

do you already know one direction?

if not choose your favorite direction

and then we can work from that

[next obsidian]

Key thing to use in lots of ring theory:

write down 1 somehow

[twilit wraith]

what

theres only one direction

[karmic moat]

usually when you show two sets are equal you show each are contained in each other is what i mean

[twilit wraith]

oh i see

i dont know any direction

i have no clue where to go with this problem

[velvet hull]

Well, AB is always inside A∩B for ideals A,B

but the other direction isn't always true, so suppose it is true

and see what happens

[twilit wraith]

how can i be sure about that tho

maybe i just need to think about it a bit

[velvet hull]

ab is inside B

because B is an ideal

ab is inside A

[twilit wraith]

oh ofc

[velvet hull]

because A is an ideal

[karmic moat]

chmonkey gave an incredible hint actually

you know A + B = R

how can you write the element 1?

[twilit wraith]

im terrible with ideals so i gotta think about it

how do i know the ring has a multiplicative identity

[karmic moat]

you don't

wait

[velvet hull]

usually by assumption

[karmic moat]

okay i assume your textbook or whatever you're using assumes rings to have a multiplicative identity

lol i read it as multiplicative inverse

[velvet hull]

if they made you work with nonunital rings that's a bit cruel

[twilit wraith]

i gotta check if gallian does

it doesnt

i think i cant assume its unital

wait nvm the problem from gallian assume its unital

but the problem on my hw doesnt

so

ill just assume its unital

[karmic moat]

yeah

good rule of thumb

[next obsidian]

I’m pretty sure you need unital

[karmic moat]

whenever you see "ring" just assume unital

[next obsidian]

I can’t think of a counterexample cuz idk many rings without a 1

But this gives me large you need a 1 vibes

[twilit wraith]

i mean i can write 1 by taking a and a inverse and multiplying

for some a in A

[karmic moat]

a doesn't always have a multiplicative inverse

[twilit wraith]

right

[karmic moat]

don't overthink it

A + B = R

1 is in R

[twilit wraith]

ok well i know 1 is in some ideal

[karmic moat]

any ideal that contains 1 has to be the whole ring itself

[twilit wraith]

so i could take a1b1 - a1b1 + a2b2 - a2b2 ... + 1

[twilit wraith]

oh ofc

[karmic moat]

i'll give you a hint

[next obsidian]

Hiido

If R is Z

[karmic moat]

by assumption A and B are disjoint

[twilit wraith]

i missed the day we talked about this stuff

[next obsidian]

Pretend A = (a) and B = (b)

[karmic moat]

^

[next obsidian]

When is A + B = Z?

[twilit wraith]

only when either A or B is Z

[next obsidian]

No

[twilit wraith]

i mean if i have some two proper ideals of Z i cant generate all of Z

[next obsidian]

Yes you can

(2,3) = (1)

[twilit wraith]

oh right

bc all elements are 2x + 3y

man i really wish i didnt miss this day

[next obsidian]

So when is this true

[twilit wraith]

when a and b are coprime

given ur assumption

[next obsidian]

And if that’s the case is it true that AB = A\cap B?

That’s what you need to show

So assume again that this is Z

And work out why this is true

Then realize this argument works for any R

Except you don’t even need to say “coprime” because in Z “coprime” is literally just saying A + B = Z

[twilit wraith]

well all terms that define AB have to be divisible by both a and b

and so AB must be a subset of A \cap B because those elements also have to be divisible by a and b

[next obsidian]

Okay I guess it’s too simple here when you use divisibility

But try to prove this by using bezout’s lemma

And then manipulating an equation

[twilit wraith]

i see

well i know since A + B = R then ax + by = 1 for some x and y in R

but this is assuming that A and B are generated by only one element

[next obsidian]

Why do you even need x and y

No it isn’t

What’s the definition of A + B

[twilit wraith]

now that i think about it i have no idea

i assume its just like a + b for all a in A, b in B

[next obsidian]

It is

But then what does it mean for that set to be equal to R

also thing you may not realize

[twilit wraith]

for all r in R there is some a in A and b in B such that a + b = r

[next obsidian]

An ideal I is R if and only if a unit is in I, if and only if 1 is in R

So if you have I = R ever, you should write down 1 as something in I

[twilit wraith]

i see

so i know theres some a and b such that a + b is 1

then meaning that all elements of R are multiples of a + b

[next obsidian]

Yeah

But now use this to show A\cap B < AB

[twilit wraith]

well i know 1 is in A \cap B

actually wait do i

no i dont

ok so if i say that x is some element in A \cap B

then x(a + b)x is in AB

assuming a + b is 1

[queen quarry]

Say we have an epimorphism f: G->H and a free subgroup H_F of H, whose preimage under f is a free group of rank k. Can we say something about the rank of H_f, for example rank(H_f) ≤ k?

[quiet pelican]

You can say rank H_f \leq k, and that’s (clearly) all you can say
So WLOG we have an epi f: G -> H with G, H both free, by restricting to the preimage of H_f
Therefore H can be generated by k elements in H (the images of the generators of G)
But then we count the number of maps from H to Z/2Z
Since H is free on n generators, there are exactly 2^n such maps
If H can be generated by k elements, there are at most 2^k such maps
So n <= k

To show that any rank H_f \leq k is possible, you get maps F_k -> F_n for n \leq k by sending the first n generators to the n generators of F_n, and the remaining k - n to the identity

[queen quarry]

Oh perfect, thanks

[twilit wraith]

this seems like itd be easy to show that these two polynomials are equal as functions

idk if thats what its asking for tho

[wraith cargo]

err use wilson's theorem

[wraith cargo]

it is not

or actually

easier way to do it

use the freshman's dream

otherwise known as the frobenius

oh wait no that doesn't work here

okay actually I shouldn't think about this bye

[south patrol]

This is a good way to prove Wilsons theorem imo lol

[south patrol]

This leads to a proof. Consider the difference between the two; ||it is a polynomial of degree < p-1 with p-1 roots, so it is 0||

I guess this is ultimately what tterra is saying

[twilit wraith]

I see

And since I'd get p-1 roots from that i know that I've fully factored it

[static mauve]

There is a theorem I see that says that if we have a free abelian group G, then it is isomorphic to Z x Z x ... x Z (where the number of Zs is equal to the rank of G.)

Is this actually an if and only if condition? I was trying to consider if there are any groups that are isomorphic to Z x Z x ... x Z but are not free abelian... not but Im not coming up with any counter examples.

[karmic moat]

Z x ... x Z is a free abelian group

and thus any group isomorphic to Z x ... x Z must also be free abelian

also i think your theorem needs that G is finite rank

[kind temple]

if rank(G) = k, isn’t G = Z^k?

[karmic moat]

An infinite product of Z is not free

[kind temple]

mm true

[karmic moat]

Anyway isnt that also what this theorem is saying

If the rank of a group is k then it’s isomorphic to Z^k

Unless ur question also included k = infinity

[kind temple]

yes

[knotty badger]

Ok so for a local ring

The geometric idea is like

The germ of smooth functions defined at a point p

Where you’re allowed “singularities” like 1/(x - b), so long as its away from the point “a” that you’re focusing on

In this case something like (x - a) would be the unique maximal ideal

I guess. I’ve vaguely heard stuff about “generic points” in AG

I’m wondering if you could do like

Could you “localise at the circle x^2 + y^2 = 1” or something

So allow yourself to formally invert things that are nonzero everywhere on the circle

I guess this would be, like, functions defined on some open neighbourhood of the circle…?

And then you have, like, ideals that correspond to points on the circle

Those generated by (x - a) and (y - b)

But then you also have an ideal generated by (x^2 + y^2 - 1)

Would that describe, like, a generic point on the circle…?

[karmic moat]

I can only speak in the context of irreducible varieties over algebraically closed fields (as in Hartshorne chapter 1)

[karmic moat]

Sort of, given a variety X and a point P, you can define a “local ring of P on X” which is actually the ring of germs of regular functions near P

So it’s a ring whose elements are germs

The unique maximal ideal is given by germs of regular functions which vanish at P

And so in this sense you can think of local rings as looking “locally” at germs around some point

I don’t think there’s anything about singularities in just that, but maybe I misunderstood what you’re saying? But we can say a variety is nonsingular at a point P if the local rings of P on X is a regular local ring

[knotty badger]

Yes actually I was just thinking about this

Is there a way to describe the ring of germs in terms of the ring of global functions?

[karmic moat]

In the case of affine varieties, to each point P you can map it to the ideal of functions vanishing at P

This gives a bijection between points of your variety X and maximal ideals of the affine coordinate ring A(X)

[knotty badger]

If you have some subset S, could you have an ideal of functions that vanish somewhere on S?

[karmic moat]

And also you get that the local rings of P on X is isomorphic to the localization of A(X) at the ideal of functions vanishing at P

[knotty badger]

And then localize at that ideal instead

[karmic moat]

I believe so but I’ll have to think about it

[storm kiln]

Isn't that the coordinate ring? Intersection of all local rings on a variety is its coordinate ring

[knotty badger]

For context I have 0 AG knowledge

[karmic moat]

This is certainly true

[storm kiln]

i.e. k[x, y] / (x^2 + y^2 - 1)

[knotty badger]

Hmmm I’m not entirely sure if this is what I mean

What I mean is something like

k[x, y], but you formally invert anything that never vanishes on the circle

[storm kiln]

where gamma(V) is defined to be k[x, y] / (x^2 + y^2 - 1) in this case

[karmic moat]

There’s a map which sends a subset A of affine (resp projective) space to an ideal of functions (resp homogeneous functions) which vanish at P

[knotty badger]

Yeah I think maybe the difference here

I’m not considering functions which vanish on all of A

But functions which vanish somewhere on A

So $\exists$ instead of $\forall$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[storm kiln]

so functions that are defined on the entire variety right?

that would be the intersection of local rings

[knotty badger]

Idk what a variety is but

Hm

[karmic moat]

Yeah you can take a subset A’ of A and then you have that I(A) is a subset of I(A’) where I(-) is the ideal of functions vanishing at -

[knotty badger]

Hmmm

I think I see what you mean

[storm kiln]

irreducible algebraic set

[knotty badger]

So for the circle

If I have a germ for each point x_0 on the circle

Taking the “intersection” is essentially asking these germs to be compatible, right?

And so it should glue to a continuous function defined in some neighbourhood of the circle

[storm kiln]

all of them are elements of the field of rational functions on your variety

[knotty badger]

Sure sure, rational then

[storm kiln]

basically k(x, y), but quotiented out by the ideal that makes it have different elements represent different functions

maybe ask someone is #algebraic-geometry as well, I'm also quite new to this

[knotty badger]

Ah is this all algebraic geometry

[storm kiln]

the basics, I usually don't understand a thing being said in that channel 😂

[tall karma]

someone help how do cosets work

[elfin wraith]

Could you be any more specific about your question?

Broadly speaking cosets are a partition of your space into distinct equivalence classes. Say for example you take Z and quotient out by 5Z, giving the space Z/5Z.

The elements of this space are cosets, and they look like 0+5Z, 1+5Z,…,4+5Z. Essentially you have grouped up all of the numbers which are multiples of 5, one more than a multiple of 5, etc

The idea is the same regardless of what you’re quotienting, wether that is say R[x]/<x^2+1> or S4/A4, you essentially look at the “remainder” modulo what you’ve quotiented out by

[slim wagon]

not really

you kind of jump around a bit.

there is no element in <x> not in <x^a>
this is the first problem, because it's not true, in general, for any a - you're kinda trying to prove it for a, n coprime

The second paragraph is a little bit meaningless as is, but i might be misunderstanding your reasoning

I don't know if that's necessary. since if <x> and <x^a> have n distinct elements (and trivially <x^a> is a subset of <x>) then you get the result almost immediately that <x> = <x^a>

OK so the first paragraph is just outlining what you're doing, if im understanding correctly.
The second paragraph still doesn't really make sense

Ok think i get what you've written.

ord(x^a) = n/gcd(n, a)
if you're allowed to use this result that's great and you're idea is fine although i would make it read slightly better by changing things like x^a, x^(2a),... x^n to end with x^(an) (x^n = x^(an) has not been made clear) and some parts of the statement aren't contributing anything

And i would rewrite the first paragraph im not sure if your phrasing provides enough justification mainly because its a bit ambiguous

[vagrant zinc]

Sometimes I wonder where the fuck did all this come from hahahaha.

[tardy hedge]

Would it be correct to say homomorphisms are determined by kernels up to automorphisms onto the image?

(I originally just said “up to automorphism” but saying up to automorphisms onto the image would be more accurate i guess?)

[south patrol]

Idk what you mean by automorphisms onto the image

I guess you mean up to automorphisms of the image ye

Like this is just the statement of the first iso theorem ig

[tardy hedge]

Yeah i just didnt really think of it in that way i guess

In terms of i guess the kernel tells you everything besides exactly where the individual elements are going

[delicate orchid]

You can say even more: Every group homomorphism factors as a canonical projection onto a quotient followed by an isomorphism, followed by an inclusion. I can prove it in numerous unhinged manners

[elfin wraith]

What’s your most unhinged/a particularly fun way?

[tardy hedge]

ya lets hear it

[coral spindle]

My money is on biset functors

[elfin wraith]

It would be very unlike wew to bring those up

[thorn jay]

I'm sure it's going to be some obscure algebraic logic stuff

[delicate orchid]

oh yeah I meant to reply to this like 2 hours ago

[delicate orchid]

"n... no! that's not true!" Said the suspicious "Remark 3.1.3 in Bouc's "biset functors for finite groups""-shaped 'jak

[coral spindle]

Pay up bitches

Oh btw Wew

Me and my collaborator proved this cool result involving a particular biset functor

If/when it comes out I’ll send it to you

[delicate orchid]

which one huh it better not be some kinda generaised burnside ring bullshit

not generalised I've forgotten what they're called

ts pmo I can't remember it

all I can remember is that the generalised burnside ring corresponding to the representation ring functor is called the global representation ring

[coral spindle]

Oh no it’s extremely niche darling

But it involves a very cute classification result

[delicate orchid]

the augmentation map 🥀

[coral spindle]

If I say too much here I will become extremely googlable

Oh nah wew it’s about rationality you know me

[delicate orchid]

87.154.45.96

[coral spindle]

Yippee

[delicate orchid]

if it's the kernel of the brauer map (brauer relations functor) I'm going to do something comedically exaggerated

that's the only thing in my mind that lies in the intersection of "niche", "rational", "biset functor", and "classification result"

[elfin wraith]

What is a biset functor anyway

[coral spindle]

Take it away Wew

[delicate orchid]

representation of the biset category

[elfin wraith]

Ahhhh I see now

[delicate orchid]

so you know what a bimodule is

[elfin wraith]

I do

[delicate orchid]

a biset is the same thing but not woke (no ring actions just group actions)

so a (G, H)-biset is a left G set and a right H set such that (gx)h = g(xh)

[elfin wraith]

Ah ok so you’re just imitating good algebra

(Non com ring theory)

[delicate orchid]

you can do this with bimodules over rings too it's just nowhere near as nice

like the bicategory construction goes through just fine but that's a seperate discussion

[elfin wraith]

So is the biset category just the category with these things as objects? What would be the morphisms?

[coral spindle]

Show em the tensor product of bisets wew… it’s such a sexy definition

[delicate orchid]

no these are the morphisms

[elfin wraith]

Oh yeah of course

[rocky cloak]

Is every biset a union of double cosets just like normal G-sets?

[elfin wraith]

Nicer than the regular tensor product?

[delicate orchid]

point is, if $B(G, H)$ is the set (abelian group, ring) of $(G,H)$-bisets then there's an associative, unital composition map $\circ \colon B(G,H) \times B(H, K) \rightarrow B(G, K)$

[cloud walrus]

☻ Wew Lads Tbh ☻

[delicate orchid]

this composition is the tensor product boytjie was referring to

[coral spindle]

It linearises to the ordinary one

[delicate orchid]

every (G,H)-biset can be identified with a G x H^op-set, so yes

[elfin wraith]

I’m not sure I know what you mean by that

[rocky cloak]

I guess, it's not quite then, since there are subgroups of GxH^op not of the form AxB

But close enough or whatever

[delicate orchid]

yeah you need to Goursat's it, but it still behaves nicely

hold on I'm trying to type like 8 lemmas at once

[delicate orchid]

specifically, if $X \in B(G, H)$ and $Y \in B(H, K)$, $X \circ Y = \frac{X \times Y}{(xh, y) \sim (x, hy)}$

I'll fail your render in a fuckin minute

[coral spindle]

From a (G,H) biset you get a (kG, kH) bimodule by taking the free vector space. This is what I mean by linearising. The linearisation of the tensor is the tensor of the linearisations

[delicate orchid]

7

[cloud walrus]

☻ Wew Lads Tbh ☻

[delicate orchid]

this is identical to how in the tensor product of bimodules we have $xa \otimes y = x \otimes ay$

[cloud walrus]

☻ Wew Lads Tbh ☻

[delicate orchid]

I joke about how this is the tensor product of F_1 modules because I am mentally ill

anyway questions questions any questions

[rocky cloak]

It's clearly the tensor product of F1[G]-modules

[delicate orchid]

(F1[G], F1[H])-bimodules 😭

[rocky cloak]

Well, can you really take tensor product of things that aren't bimodules

[coral spindle]

Isn’t it glorious how it’s defined precisely because of the group structure? I love it

[elfin wraith]

What does all of this buy us? Like it’s a pretty nice analogue with bimodules and the tensor product, but what can we now do group theory wise that we couldn’t before? What problems does this solve?

[delicate orchid]

typing up a fuckin massive paragraph that explains exactly that

[elfin wraith]

Fair play lol

I went to someone’s masters defence about homotopy for graphs and asked basically that question to which they had no good answer and I felt really quite bad

They eventually said it just gives you fun topos stuff to play with

[coral spindle]

It’s really stunning actually I was super impressed by this stuff

I’m sure wew will explain

But fr this gets us some amazing theory down the line

[delicate orchid]

The point is is that there are certain "elementary" bisets from which all others are constructed - Given an iso phi and G, H, N being what you think they are these are $\text{Def}{G/N}^N \in B(G/N, G)$, $\text{Inf}{G/N}^N \in B(G, G/N)$, $\text{Res}_H^G \in B(H, G)$, $\text{Ind}_H^G \in B(G, H)$, and $\text{Iso}(\phi) \in B(\phi(G), G)$.

These names are rather telling. One of the main selling points is that these restriction and induction bisets obey mackey's theorem under composition. So any functor from this biset category to some module category is automatically a global mackey functor (think representation ring, burnside ring, cohomology etc. anything with a notion of restriction and induction). So now your question might be ``ok so why not just study mackey functors u nerd" and the point is (at least for my research personally) is that this notion also encodes precisely how inflation and deflation should behave w.r.t restriction and induction, so you can have extra structure on your mackey functor if you want.

oh for FUCKS sake who fucked up the fackin latex

[cloud walrus]

☻ Wew Lads Tbh ☻

[delicate orchid]

@coral spindle do u want to talk about semisimplicity and simple biset functors or should I or like what

I'm very much rambling at this point and can tell it will become incoherent

[coral spindle]

You are far more knowledgeable than me about these things Wew, and also I’m on my phone

[delicate orchid]

for example, inflation and restriction as maps of group representations behave exactly like their bisets do

[coral spindle]

I will say I think the theory of biset functors generated by p-groups is so hot

[delicate orchid]

yeah you get different theory by restriction to different full subcategories

I mean there's SO much shit here. Like the fact that the restriction biset is the "opposite" biset of the induction biset gets us not only Frobenius Reciprocity, but Shapiro's Lemma

just because you swap the actions around! Opposite bisets (can, under the appropriate mapping) induce adjoint functors

such fun and games

[elfin wraith]

I feel like I do not at all have the background in representation theory to understand any of that but it certainly sounds powerful

[delicate orchid]

that's the thing - Shapiro's lemma is a result about cohomology

it's really general

and if you take your biset functors mapping into a nice enough module category (over a ring characteristic 0, I think?), the biset category's category algebra is semisimple

so the funny thing is, in Serge Bouc's bouc, he decomposes the representation ring functor itself into "irreducible representations"

which I find VERY funny

[elfin wraith]

I also only know about regular homology

This does all seem very cool and very powerful though, I do at least know what a biset is now

[delicate orchid]

yur. I've refrained from actually generalising it as far as I can

because these are still groups! they're one object categories

why only one? but that's a discussion for later

[elfin wraith]

It does sound sick though, I really do fuck with algebra, I really can’t wait to start my masters and get back to this nonsense

I do hope one day I can say more about it than “sounds sick”

[delicate orchid]

I cannot recommed Serge Bouc's "Biset functors for finite groups" enough. He basically invented all of this

[coral spindle]

It’s a really good book

I got jumpscared bc in the early chapters it cites one of the academics at my uni

[keen badge]

Let $q\in\bQ$. Why $\bZ[q]$ is a Noetherian ring?

[cloud walrus]

𝒢𝒾𝓃𝑔𝑒𝓇 𝑀𝒶𝑔𝓂𝒶

[thorn jay]

A quotient of a Noetherian ring is Noetherian

And Z[x] is Noetherian, as Z is

[keen badge]

ok

[keen badge]

I don't see how to finish

[elfin wraith]

If you see how Z[q] is a quotient of Z[x] then you’re done, so where are you stuck?

[keen badge]

I don't see why it is a quotient

[delicate orchid]

q = a/b implies Z[q] = Z[x]/(bx-a)

[thorn jay]

Local bars hate this one trick!

[ivory talon]

Hi, I know this isn't exactly the proper channel for it however I'm doing a course on groups and rings and I really need help with some set of problems
if anyone got time to help me i'd be very grateful, it should be pretty elementary

[white oxide]

Can somebody help me see why this is true

[crystal vale]

so when i say A is an algebra that means A is ring and also vector space such that a(xy) = (ax)y = y(ax), right?

[mighty kiln]

4 ⊗ x = 2 ⊗ 2x = 0
2 ⊗ x can't be written as 1 ⊗ 2x tho

[jade mason]

I think you can consider the bilinear map 2Z x Z/2Z -> Z/2Z which sends (2n,y) to n*y

[jade mason]

This sends (2,1) to 1, so by univ property the induced linear map on the tensor product sends 2 ⊗ 1 to 1

And thus 2 ⊗ 1 must be nonzero

[ivory talon]

Okay, I'm with some sub group of $S_4$, Let $a = \begin{pmatrix}
1 & 2 & 3 & 4\
2 & 3 & 4 & 1
\end{pmatrix} & b = \begin{pmatrix}
1 & 2 & 3 & 4\
3 & 2 & 1 & 4
\end{pmatrix}$ and so the sub group is defined as: $K = {a^{i} \cdot b^{j} | i \in {0,1,2,3}, j \in {0,1}$ and so I'm trying to find the Conjugacy classes of $K$ but i'm not sure exactly how they are defined as the lecturer never really explained how the conjugancy classes of a group are defined... I do know how conjugacy between elements is defined. Any help? Thanks in advance

[cloud walrus]

Yuval

[ivory talon]

If it's any help which I think it is the sub group is isomorphic to $Dih(8)$

[cloud walrus]

Yuval

[delicate orchid]

I agree it's D_8

so if g is in some group G, the conjugacy class of g is just the set {xgx^-1 for x in G}

[ivory talon]

so say I found that $a^(-1) = a^3$ then I can say that $Cl(a) = {a^3}$ and I can say vise versa that $Cl(a^3) = {a}$?

[cloud walrus]

Yuval

[ivory talon]

oopsies $a^{-1}$

[cloud walrus]

Yuval

[delicate orchid]

no, not given that information

g and h are conjugate if and only if there exists some x such that xgx^-1 = h

[ivory talon]

sorry what I withheld some informatio not on purpose, I know that $(a \cdot b)^{-1) = (a \cdot b )$

$(a \cdot b)^{-1} = (a \cdot b) $

[delicate orchid]

so $b^{-1}a^{-1} = ab \iff a^{-1} = bab$, now using the fact that $b$ is order $2$ we have $b = b^{-1}$ so yes. But I wouldn't do it like this. Just compute $xyx^{-1}$ for all $x, y$ in your group, making use of the fact that conjugacy is an equivalence relation to speed up your calculation

[cloud walrus]

☻ Wew Lads Tbh ☻

[delicate orchid]

what I mean by the last part is the following:

if x and y are conjugate, then their conjugacy classes are the same set, so you only need to work out the conjugacy class for either x or y

you don't need to do both

you want me to do a simple example?

[ivory talon]

so that would mean that $Cl(a) = Cl(a^3)$?

[cloud walrus]

Yuval

[delicate orchid]

yes it would!

[ivory talon]

but wouldn't that mean that the elements inside are the same which they are not?

[delicate orchid]

but they are the same

cl(a) = {a, a^3} = cl(a^3)

[ivory talon]

huh but that would mean b a b^-1 = a no? which is incorrect?

[delicate orchid]

why would it mean that

from the top

[ivory talon]

because by the definition of the Cl set the elements in Cl(a) are elements which are equal to b a b^-1 no?

[delicate orchid]

no

cl$(a) = {g \in G \colon \exists h \in G \text{ such that } hgh^{-1} = a}$

[cloud walrus]

☻ Wew Lads Tbh ☻

[delicate orchid]

you're fixing b for no reason

b varies over the entire group

[ivory talon]

but by that definition wouldn't that mean that only $g$ that there exists such an h for him is in the conjugacy class? 😦 why would $a$ also be in the set?

[cloud walrus]

Yuval

[delicate orchid]

because $1a1^{-1} = a$

[cloud walrus]

☻ Wew Lads Tbh ☻

[ivory talon]

oh that's right.......

[delicate orchid]

also aaa^-1 = a

[ivory talon]

I get it now

okayy

yesss

thank you

[delicate orchid]

no worries

[tardy hedge]

Bro said Dih

I didnt kno math could get so freaky 😜

[coral spindle]

Wait until you hear about the dicyclic groups

[tardy hedge]

Is it really just as simple as there is a group map that sends 2 (x) 1 to something nonzero, so 2 (x) 1 cant be 0 in tensor product?

[delicate orchid]

Q_2n

[coral spindle]

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

[delicate orchid]

4n
woke alert

[coral spindle]

🤭

[glad osprey]

A being a ring implies that it is associative and (usually) unital, but in some contexts we don't require algebras to be associative or unital

[crystal vale]

Oh

[glad osprey]

Like Lie algebras are neither unital nor associative unless I'm mistaken

[south patrol]

Well that is a very different story ig lol

Just "algebra" is used in various different ways

[glad osprey]

Btw, is (ax)y = y(ax) true in general? Aren't you assuming commutativity?

Wouldn't you say a Lie algebra is a kind of algebra over a field? Or are we talking about different kinds of algebras?

[south patrol]

Imo it would be misleading to say that. Algebra over a field already has a meaning

[glad osprey]

But I don't see why a Lie algebra isn't an algebra over a field? It's a vector space with a bilinear product? Or wait, is it bilinear?

[south patrol]

Eh okay maybe this is a thing where conventions vary a lot lol

[tough raven]

The Catalan numbers count ____ words.

[south patrol]

I guess ye with the most general definitions it is an algebra over a field

In my experience when someone says algebra over a field it is usually more specific (an ring with a central embedding of the field)

[glad osprey]

I see

[south patrol]

Then in AG and similar things it would often be assumed commutative

[crystal vale]

Yes

[tough raven]

Yeah. I think it's because it's pretty rare to study general algebras over a field. If someone is talking about Lie algebras they might want algebra to mean "just a bilinear map", but they don't actually need to mention general algebras much anyway, just Lie algebras in particular.

[thorn jay]

Some algebras aren't even over a ring but that's just too woke for some people

[pliant rivet]

How do I prove that f(z,w) = w^2 - h(z) is irreducible if h(z) isn't a perfect square? I'm not really familiar with C[z,w]

[storm kiln]

If it's reducible, then it would also be reducible over C(z)[w]

[pliant rivet]

ah, I suppose that does it

thanks

[white oxide]

Oh nice thanks

Also this is the example they were talking about, do you know how what you proved would follow from this example?

[tardy hedge]

y are u writing 4

[tardy hedge]

yea tbh im trying to figure out what this example is even illustrating too

[jade mason]

I dont think this is supposed to be a proof of the earlier claim

[jade mason]

Another way to see this btw is that tensoring by Z doesnt do anything

In fact, the isomorphism N ≅ Z ⊗ N sends n to 1 ⊗ n, and via the isomorphism Z ≅ 2Z we identity this with 2 ⊗ n ∈ 2Z ⊗ N

So in this case the element is 0 iff n is already 0

[tardy hedge]

Can i get a hint for why 1 (x) n = 0 means n = 0?

@jade mason

[twilit wraith]

"Show that for every positive integer n there are infinitely many polynomials of degree
n in Z[x] that are irreducible over Q."

is this just showing that there are infinitely many (n+1)-tuples that satisfies eisensteins criterion

or more specifically i just let the nth coefficient be odd and the rest be even with the constant being 2 mod 4

[kind temple]

i think its simpler than this

my hint is to look at ||x^n - 2, x^n - 3, x^n - 5||

[twilit wraith]

Oh ofc

I already used the eisenstein thing tho

[kind temple]

yea, this one uses eisenstein too.
more generally, x^n - p^m is irreducible over Q whenever p is prime and n and m are coprime

[rocky cloak]

So you're looking at Z(x)N and say n is nonzero. Then you want to show that 1(x)n is also nonzero.

Use the universal property of the tensor product and construct a bilinear map ZxN -> N

[ivory talon]

Hey peeps I have a quick question, say I have the permutation group $S_4$ and a subgroup of $S_4$, $K$ what would be the meaning of the characteristic permutation?

[cloud walrus]

Yuval

[coral spindle]

It would be good to share the context you’ve seen this in

[long geyser]

for a book at D&F's level, that seems fine to skip

[elfin wraith]

I mean the proof is incredibly straight forward and I think something they presume you know. I would recommend thinking about why it’s the case

[long geyser]

yeah to me it's pretty clear what the partition corresponding to an equivalence relation should be and vice versa, and verifying these things won't take long at all

[elfin wraith]

I also take slight issue with saying they’re the same, they have different definitions. Any partition induces an equivalence relation and any equivalence relation partitions a set, so the notions are in bijection sure, but idk if I’d say they’re the same, certainly isomorphic as concepts anyway

That’s just pedantry though

[long geyser]

I'm very fine with saying they are the same thing, to be honest

[thorn jay]

Yeah

[long geyser]

as long as it's not in a foundations context maybe

[thorn jay]

You'll always treat them as the same thing anyhow

[elfin wraith]

Yeah I mean I’m just being pedantic, they’re equivalent but the same is slightly stronger. For all intents and purposes it doesn’t matter

[thorn jay]

Definitions are often given as a formality / refresher

[elfin wraith]

You should try to do it yourself, it’s not hard and it should be clear why it’s true if you understand the definitions

[long geyser]

books give a lot of definitions that they assume the reader knows (or at least has seen before)

just to set up expectations/notations if nothing else

[thorn jay]

In my brain the corresponding partition of an equivalence relation is part of the equivalence relation itself lmao

[long geyser]

or a refresher

[thorn jay]

Relations come up in introductory math courses I presume?

[long geyser]

no, but a class like an intro proofs class probably

[elfin wraith]

I learned about them in my first ever class at uni, which included like an intro to sets

[long geyser]

sometimes it can also be referred to as discrete math

but theres always a class that covers like

[thorn jay]

Surely you don't start your math bachelor with abstract algebra right

Lol

[long geyser]

basic set theory, basic logic, functions relations

basic cardinalities

etc.

[jade mason]

Same up to a canonical isomorphism

[long geyser]

yeah please dont let d and f be your first math book thats insane

[thorn jay]

[long geyser]

https://richardhammack.github.io/BookOfProof/

free, popular, great book on this kind of stuff

you can skim it, or read the parts about relations if youre confused abt them specifically

[kind temple]

a little silly, but given a set X, you can form the category of equivalence classes on X with arrows being drawn if one equivalence class contains another.

you can also form the category of partitions of X, arrows being drawn if one is a refinement of another.

these categories are isomorphic.

[thorn jay]

Why use a category for that

We have a very nice thing called lattices

Algebraic lattices, even

I will not stand for lattice erasure

[kind temple]

can't lattices be viewed as categories?

[thorn jay]

They can

Lattices are posetal categories with finite products and coproducts (i.e. finite limits)

[kind temple]

idk, i don't use lattices enough to think of using them lol

[thorn jay]

Algebraic lattices are apparently locally finitely presentable posetal categories

But like

At that point the order theoretic definition is nicer

[thorn jay]

All the fault of those silly category theorists

[long geyser]

how do you even learn about lattices anyway, all I've learned about them has been through osmosis

[thorn jay]

I'm active in universal algebra

And lattices are basically the language in which a lot of universal algebra is spoken

(esp when working with congruences and such)

[kind temple]

i feel like they come up naturally quite a bit, but are sparsely discussed

[long geyser]

oh yeah I feel like lattices come up quite a bit actually

when doing analysis

[thorn jay]

They do

[long geyser]

or algebra

[kind temple]

yea, they are kinda ignored over more standard things

[thorn jay]

As I said

Lattice erasure

[kind temple]

can't erase something that was never there lol

[thorn jay]

That's just not true

Lattice theory was very popular a while ago

[kind temple]

how long ago? what made it die down?

[delicate orchid]

everything started becoming NP-hard

[thorn jay]

Category Theory

[long geyser]

categories I'd assume

[thorn jay]

Lattice theory was trying to be a grand foundation for math

[kind temple]

so, it was the hot thing before 1940?

[thorn jay]

Smt like that yeah

[thorn jay]

And don't get me wrong, it is very powerful

Just not enough

[kind temple]

where does it kinda start to break down?

[thorn jay]

For example, studying simple groups

Lattice theory won't really help you there

[kind temple]

but category theory will?

[thorn jay]

The way of thinking, I'm pretty sure

[kind temple]

i see

[elfin wraith]

Fully aware, they’re certainly equivalent, the same just feels very strong to me.

I also take issue when people define the nth cyclic group to be Z/nZ and similar things so maybe I’m just needlessly pedantic about that

It doesn’t actually matter though in any way shape or form

[thorn jay]

But, regardless or not if CT was useful there, lattice theory was decidedly not useful, which meant it could not be a grand foundation / unifying theory

It has its place in math though, usually in logic related areas

[kind temple]

sounds like some interesting math history

[thorn jay]

I read a very long stack exchange post on it

[south patrol]

Dawg

[thorn jay]

What :3c

[south patrol]

lol just a funny sentence

[thorn jay]

This is why lattice theory is better

[delicate orchid]

"posetal categories" 🥀 or as I like to call them, posets 🕊️

[thorn jay]

https://mathoverflow.net/questions/150544/was-lattice-theory-central-to-mid-20th-century-mathematics

Found it

[thorn jay]

No that's not category theory enough

[delicate orchid]

ok fine (0,1)-categories then

[thorn jay]

❤️

[woeful sage]

I feel the same about order theory in general

They come up a lot in algebra and topology but not enough to justify making every student take a whole course in order theory

[white oxide]

Let M be an A-module. Is S^{-1}M an S^{-1}A module in the sense that (a/s) * (m/t) = (am/st)?

[mighty kiln]

Yes

[thorn jay]

Sadly qwq

[keen badge]

If P=<f1,f2,f3> subset C[x1,dots,xn] is there an algorithm/nice way to check if P is prime

[elfin wraith]

Something to do with Gröbner bases iirc

I kinda erased them from my mind after my comalg class but yes I’m pretty sure using Gröbner bases you can just keep eliminating variables and you end up with something easier to check.

Looking in IVA or some other book about Gröbner bases is probably helpful. I think @barren sierra knows a good amount about them too (sorry for the ping if not)

[barren sierra]

Checking if an ideal is prime via Gröbner bases is quite hard iirc

I've seen it in papers but not textbooks, or at least neither of the Cox, Little, and O'Shea texts

[elfin wraith]

Obviously there’s the usual method of taking quotients and stuff, but yeah I’m not sure if there’s a generic algorithm in that case

But also uh probably ask the alggeo people, they probably know every prime ideal of C[x…]

[barren sierra]

https://www.sciencedirect.com/science/article/pii/S0747717188800403

This exists

And these algorithms are implemented in computer algebra systems such as Macaulay2 and perhaps SageMath

So if you just need this for a specific example or a few for your own curiosity then use those perhaps

If you need this for homework then the answer highly depends on the polynomials in particular

If you need a general algorithm uhhhh get reading

Perhaps better expositions exist lemme try to find

[jade mason]

Are there any resources for learning / practicing algebraic manipulation of rings and modules

[jade mason]

Like computing tensor products or quotients via the isomorphism tricks

[barren sierra]

Here are some citations from Ideals, Varieties, and Algorithms by Cox, Little, and OShea @keen badge with probably better exposition than what I linked

[barren sierra]

Probably just practice problems in textbooks

[jade mason]

Do you know any textbook with a good collection of such problems

[barren sierra]

I liked Rotman’s Advanced Modern Algebra

[ivory grotto]

could someone with enough patience explain to me why a ring has such a deceiving name?

[elfin wraith]

Deceiving?

[ivory grotto]

yup, like, why a ring? i read the definition, and ring is probably in the top worst names to categorize it at

[long geyser]

The term "Zahlring" (number ring) was coined by David Hilbert in 1892 and published in 1897.[13] In 19th century German, the word "Ring" could mean "association", which is still used today in English in a limited sense (for example, spy ring),[citation needed] so if that were the etymology then it would be similar to the way "group" entered mathematics by being a non-technical word for "collection of related things". According to Harvey Cohn, Hilbert used the term for a ring that had the property of "circling directly back" to an element of itself (in the sense of an equivalence).[14] Specifically, in a ring of algebraic integers, all high powers of an algebraic integer can be written as an integral combination of a fixed set of lower powers, and thus the powers "cycle back".

[elfin wraith]

I mean you’re reading into it too much I think, names are just names

Afaik there’s 2 main theories on why it’s called a ring. 1) is that it’s the same idea as a group, it’s a collection of people who share some common theme (like a spy ring, or not so nicely, a grooming ring)

[long geyser]

https://en.m.wikipedia.org/wiki/Ring_(mathematics)

[elfin wraith]

The other idea is that it’s because you sort of circle around and stay within the ring, like a boxing ring. Hilbert (who defined them) spoke about how you stay within the object so that’s the other theory

Ultimately no one’s quite sure, Hilbert never explained his thought process, but I don’t think it really matters what it’s called

I don’t think the name field is any more descriptive than ring. Arguably it’s worse because it breaks with the whole group, ring, category type theme

[ivory grotto]

truetruetrue

alright thank yall, ill go read the wikipedia def also

[karmic moat]

Atiyah-macdonald

[vivid kestrel]

In a division algebra D, does any maximal subfield of D contain the center of D?

[karmic moat]

Hochster huneke have a book called “integral closure” or smth like thay

[barren heath]

Applying this algorithm from wikipedia (image 1) to my rubiks cube (image 2) results in an orientation of images 3 and 4. I don't know how to prove it but I don't think the corner algorithm given along with the edge algorithm under it are the generators for the orientations of the rubiks cube because this orientation, (image 5) which IS a legal orientation of the cube (image 5 is from algorithm 3: F U F D F' U' F D' R2 U' L2 U R2 U' L2 U F2), isn't possible (or at least I can't make a combination of the corner algorithm from the wikipedia page in image 1 (shown in 2 and 3) that has the effect where two non-diagonal corners get turned). Is there a way to get the effect in image 5 from the wikipedia algorithm and I just haven't found it yet, or is the wikipedia algorithm not one of the generators of the rubiks cube group?
(sorry for long msg)

[vivid kestrel]

nevermind i see why

[rocky cloak]

I can agree that it's not necessarily an insightful name, but is there anything deceiving about it?

It doesn't signal any other meaning to me other than just being a word. Especially not compared to things like "normal" or "regular" which are both overused and usually signal the opposite of its English meaning.

[kind temple]

it kinda signals circular, or round, no?

[rocky cloak]

Which is very fitting right?

Since Z/n is kinda the canonical example of a ring

[kind temple]

right, but if you’re studying rings for the first time and you encounter simple examples like R or Z or polynomial rings

the looping behavior isn’t really present

[rocky cloak]

I guess I don't really see what kind of behavior one would expect

[kind temple]

i guess i’m against seeing it as just a word. even as you just pointed out, along with lems, there is some cyclic behavior associated with a ring

idk

maybe i’m trying to force meaning where there isn’t any

[livid mirage]

hi

i need help or im gonna fail my class

[elfin wraith]

There is also the other interpretation though, ring means more than one thing in English

[kind temple]

true

[long geyser]

to be clear I don't think it's a good name lol, I just copy pasted why it is (plausibly) named that way

[livid mirage]

can anyone help

[elfin wraith]

But again, I do think it’s largely irrelevant, things are called stuff. Somethings helpful smart stuff, other times just stuff. It don’t matter

[elfin wraith]

Not unless you ask a question

[livid mirage]

oh um

i dont know algebra.

[long geyser]

eh, I agree for sth as ubiquituous as a ring, but how we name things affects how we interact with them

how we memorize them eg

[kind temple]

torsion elements play a large role in ring/module theory, no?

[livid mirage]

can u help me?

[kind temple]

im like, fighting at the lunch table for why its a good name lmao

[elfin wraith]

Again, not unless you ask a question lol, what specifically are you stuck on

[livid mirage]

idk just idk algebra

[rocky cloak]

A better question might be, what would make a good name?

[kind temple]

hmm

[rocky cloak]

The nice thing about ring I guess is that
set, group, ring and field all mean the same thing

(or maybe that's a bad thing)

[kind temple]

ha

[sturdy spear]

Hi jagr

[elfin wraith]

For sure we should name stuff helpfully when we can, morphism etc is good, but at the end of the day I’m not sure it matters. Group, set, category, ring, there’s a clear theme there but they all just mean “a bunch of stuff” it doesn’t tell you much, and I don’t think it really has to. Fork doesn’t mean much in isolation

[south patrol]

Ring

[kind temple]

a bunch of stuff

[elfin wraith]

There isn’t really much anyone can do about that. If you can post a question you’re stuck on and explain where you’re lost we might be able to help

[long geyser]

yeah I agree regarding those terms, my gripe is moreso with how more specific things can be named like theorems

[elfin wraith]

What specifically are you studying? Intro group theory? Representations of Lie algebras? Localisation in rings? We need a bit more than “I can’t do algebra” if we are to help (which I’m sure we’d like to!)

[long geyser]

aesthetically (I think) it's pretty cool that theorems get named after mathematicians, but I think it would synergize more with how our brain works if what we named things related more to their use

[elfin wraith]

Ah yes, like the Baire category theorem

[long geyser]

yeah

[kind temple]

what should it be named?

[elfin wraith]

I see your point, I don’t really have any feelings about it tbh, things are just called what they’re called

[rocky cloak]

Often it's hard to make a good name though.

And then a deceiving name might be worse than a generic one

[long geyser]

even a name like dense open set theorem would work way better, but I'm sure we could think of better ones

or G delta theorem

which you might say, isn't G delta kind of a dumb name just like baire, answer is yes but by repeating a term already used you'd be strengthening past memories instead of having to remember new ones

[elfin wraith]

I mean there are some theorems name like that, nullstellensatz, the height one which I’m blanking on rn

[long geyser]

to be clear I'm not saying all theorem names suck

[elfin wraith]

I actually think in the specific example of baire, the name is almost so bad that it makes it more memorable

[rocky cloak]

Krull's height theorem?

[kind temple]

is it pronounced bare or bye-air?

[elfin wraith]

That’s the one, it’s usually in German though I’m sure, I forgot who’s theorem it was though

[long geyser]

I pronounce bare

[elfin wraith]

I’ve always said it like bear, no idea if that’s correct.

[kind temple]

same

[rocky cloak]

René-Louis Baire (French: [bɛʁ];

[long geyser]

essentially bare

[rocky cloak]

So like bear, but you choke on your R

[kind temple]

behrghri

[elfin wraith]

Who’s the other person with the controversial name, I remember they’re Norwegian, but for the life of me I can’t remember who it is (my brain isn’t working so well today)

[rocky cloak]

Sylow I guess

[elfin wraith]

That’s the one

[rocky cloak]

Not really controversial, just has a nonEnglish sound

[surreal magnet]

Is the following statement true?

Let $G$ be an abelian group, and $G > H_1 > H_2$ subgroups. If $G / H_1 \cong G / H_2$ then $H_1 = H_2$.

[elfin wraith]

Well controversial in so far as no one I’ve spoken to, here or in person, had any clue. Never got to ask a Norwegian about it though

[cloud walrus]

jp

[kind temple]

i remember the name gelfand-mazur but never the theorem

[elfin wraith]

I say it like Sil-ov, unsure if that’s correct

[rocky cloak]

Like See-lahv or Soo-lahv

Really a sound between ee and oo

[tough raven]

If they're just isomorphic, then no. If the canonical quotient map is an isomorphism, then yes.

[kind temple]

do you speak languages other than english?

[tough raven]

G/H1 = (G/H2)/(H1/H2) by the third(?) isomorphism theorem. So if you have a group G with a proper quotient G/H isomorphic to G (e.g.: G = ℤ^ℕ and H = ℤ ⨯ 0 ⨯ 0 ⨯ ...), then you can take G = G, H2 = 0, H1 = H.

[rocky cloak]

Norwegian and Esperanto, and I guess some sentences in a handful of others

[tough raven]

On the other hand, if the canonical map G/H2 → G/H1 is an isomorphism, then in particular it's injective, so its kernel H1/H2 is trivial - which means that H1 = H2.

[elfin wraith]

I’m having a hard time working out what that would be, more similar to ü or to ö in German?

[kind temple]

oh, i didn’t know esperanto had speakers

cool

[south patrol]

It also has headphones

[surreal magnet]

@tough raven a that makes sense, thanks for the counter example

[elfin wraith]

You underestimate the nerdy-ness of linguists, pick a language and someone will speak it. The number of conlangs out there is absurd

[rocky cloak]

More like ü.

Basically the German Norwegian vowel comparison is
o - å
u - o
ü - u
? - y

[kind temple]

ig i do lol

[surreal magnet]

A professor at my uni speaks it fluently!

[rocky cloak]

Wouldn't surprise me if Esperanto has more speakers than Norwegian 😛

[elfin wraith]

Norwegian is really cool, I started trying to learn a few years ago but ended up giving up on account of laziness and also the fact there’s about as many Norwegian speakers as there are Scots

[rocky cloak]

If you throw Swedish and Danish into the mix it becomes a bit bigger though

[surreal magnet]

There are also native esperanto speakers i think, a couple raised their child with it

[elfin wraith]

That is true, I imagine the conversion between the 3 languages is considerably harder if you’re non native though

[rocky cloak]

Sure, but the same can be said about talking to someone from western / northern or southern Norway (assuming you are learning eastern Norwegian, as is usually taught)

[elfin wraith]

That’s fair, I’m aware there are like “2 Norwegians” is that the split?

[rocky cloak]

Not really,

There's two written standards (bokmål & nynorsk), and there is of course some correlation between your choice of written standard and your spoken dialect. But Bokmål is used by like 90% of people, so it's not really a determining factor

[surreal magnet]

So if we know that $H_1$ and $H_2$ are of finite index (I do in the argument I was trying to make) we are done , since the canonical map $G / H_2 \to G / H_1$ is always epic and those are finite sets of same cardinality, it is an isomorphism.

[cloud walrus]

jp

[ashen heron]

intersection of subgroups is a subgroup and subgroups have 1 at least

does the definition say it has to?

"belongs to" as opposed to "subset of"

H is the subgroup there. The intersection is taken over all subgroups containing A.

[kind temple]

groups are just sets with additional structure

so anything that applies to sets will apply to groups

the empty set is a subset of every set

so it’s a subset of every group

all subgroups of a given group

yea

nah, i just saw what u were getting tripped up on

never too late

[cloud walrus]

UGOBEL

[rocky cloak]

So, have you already shown that G = ZQ and you just need to show that the product is direct and Z = UxV, or are you missing more things?

[vivid kestrel]

so given a simple k-subalgebra B of M_n(k), we know that B = M_l(D) for some D and a skew field D by wedderburns theorem. then Z = Z(B) the center of B is isomorphic to Z(D) in the obvious way i suppose. how does one see that C(Z) (the centralizer of Z in M_n(k) is isomorphic to M_k(Z) for some natural number k? such is stated in lecture notes without any further elaboration

[candid patrol]

The product isn't direct at all because if Q = <a,b>, then a^2 € Z cap Q

[rocky cloak]

Well, can be chosen to direct* then

[candid patrol]

I don't get you, Z = C_G(Q) btw mb

[rocky cloak]

I thought Z was just an abelian group.

If it's something specific, then you can't choose it

[rocky cloak]

Anyhoot, you can use the classification of finite abelian groups to write Z as a product of cyclic groups.

If Z has an element of order 4 you can construct a non-normal subgroup, so Z = {±1}xUxV (where ±1 is in Q)

[vivid kestrel]

the context is a version of the double centralizer theorem if that helps

[cloud walrus]

UGOBEL

[candid patrol]

< a^2 >Q = Q of course but... Is that all I need ?

[rocky cloak]

So
ZQ = (CxH)<a^2>Q = (CxH)Q
so what you need to show is that CxH and Q don't intersect.

Then it follows that ZQ = (CxH)xQ since both groups are normal.

Or more explicitly you can show that they commute using that their commutators are in their intersection

[candid patrol]

Thanks, yes, I understand better now, but how do you go from $(\langle a^2\rangle \times C \times H)Q$ to $\langle a^2\rangle,(C \times H),Q$? It’s simply because $\langle a^2\rangle$ has trivial intersection with $C \times H$ and commutes with $C \times H$, right?

[cloud walrus]

UGOBEL

[candid patrol]

And are both normals of course *

[rocky cloak]

I mean if you have a group
AxB, then this is also equal to AB

[candid patrol]

Right yes, thanks a lot ! 👑

[kind temple]

say G is a group with subgroups H and N. is {h in H : hn = nh for all n in N} a thing? like centralizer of N over H or something like that?

[keen badge]

C_H (N)=C_G(N) ח H

[kind temple]

thanks, knew it was something like that

[keen badge]

Let K/F be a field extenstion with a,b in K, algebraic over F, with the same minimal polynomial.
Does it mean that F(a)=F(b)?

[glad osprey]

F(a) and F(b) will be isomorphic, but not necessarily literally equal. Consider for example a = cbrt(2) and b = one of the other roots of x^3 - 2

[keen badge]

so I can't assume that b in F(a)

great, thx

[tough raven]

Is any discrete subgroup of the group GL_2(ℚ) (with the subspace topology from GL_2(ℝ)) conjugate to a subgroup of GL_2(ℤ) (GL here meaning determinant ±1)?

[cloud walrus]

c squared

c squared

[kind temple]

like, i guess in the free group on two elements, every word can be written uniquely as products of powers of a and b

so when you have a^2n = (aba^-1b)^k for non-zero k, then one has a non-trivial power of b and the other doesn't

so these can't be equal for non-zero k

so a^2n has to be trivial, in which case n = 0

does that sound right?

[tough raven]

Observe that in ℤ ⨯ ℤ/2ℤ with a := (1, 0) and b := (0, 1) the relations are satisfied. Thus there is a homomorphism G → this: a ↦ a, b ↦ b. However, for any m, n ∈ ℤ, a^m = b^n ⇔ (m, 0) = (0, n) ⇔ m = 0, 2 ∣ n in this group. So in particular a^m = b^n in G cannot happen unless m = 0, in which case a^m = 1 and b^n = a^m = 1. Thus any element in H and in K must be 1.

[kind temple]

ill try this out later tonight. does mine sound right tho?

[tough raven]

This seems wrong, because words which become trivial in G can be products of conjugates of the relations, not just products of the relations.

But I might be missing something.

[kind temple]

mhm yea, you're right

it seems arguing that way is difficult...

[tough raven]

My idea is basically yours but counting the powers of a rather than the powers of b, and this will be zero for any product of conjugates of the relation.

[kind temple]

i like this idea

thanks for the suggestion

i guess thats what i should have been doing in the first place

you naturally want to study maps out of this thing anyways

[tough raven]

You can think of this as a neat way of packaging invariants-based argument.

[kind temple]

sick

[tough raven]

You can define an invariant of words called "net power of a" and show that the relations don't change this invariant.

But this is so common a sleeker way is to define "net power of a" as a group homomorphism from the free group to ℤ, and the fact that it's invariant translates to this homomorphism descending to G.

[kind temple]

yea, i was just about to say something about this

i think a natural way to come up with your map is to look at what happens when you send each generator to 1 in Z in each component of Z x ... x Z (one for each generator)

[tough raven]

I came up with my map by taking the abelianisation of G.

[kind temple]

yea

that too

[tough raven]

Because fg abelian groups are really easy to understand.

[kind temple]

true

[tough raven]

Relations become just counting powers

[kind temple]

mhm. this trick gonna stick with me

[tough raven]

This helps with any argument based on counting powers of generators

Since the power counting amounts to the obvious map from Free(n) to ℤ^n, which is exactly the abelianisation of the former.

[kind temple]

alr neat. gonna work through this in more detail later. gtg for a bit tho

[tough raven]

This is actually why I said ℤ ⨯ ℤ/2ℤ, to make the abelianisation clearer. Since for this question I only counted powers of a, I could have just used the first factor of ℤ.

Question: Let G be a discrete subgroup of SL_2(ℝ) such that phi(ℝ) ∩ G is non-trivial, say phi(ℤa) for some positive real number a, where phi: ℝ → SL_2(ℝ): x ↦ [1, x; 0, 1].

Is it true that there must exist a positive real number R such that for z, w ∈ ℂ with im(z), im(w) ≥ R, z and w are conjugate by G (under the action on the upper half-plane by Möbius transformtions) iff they are conjugate by phi(ℤa), i.e., w - z is an integer multiple of a?

I think that it suffices that for g = [a, b; c, d] ∈ G and im(z) ≥ R, |cz+d| ≥ 1 with equality iff g ∈ phi(ℤa) Stab(z).

(This uses the formula im(g⋅z) = det(g)/|cz+d|^2 im(z).)

[kind temple]

this is nicely captured in a naturality square coming from the the adjoint pair U : Ab —> Grp and Ab : Grp —> Ab

[rocky cloak]

Something like the group generated by
diag(2, 1/2)
cannot be in GL(Z), because the trace is not an integer.

[candid patrol]

Are there any interesting applications to be drawn from this classification

[vivid kestrel]

nvm got it, hit me like a truck

[barren sierra]

I don't see where I use non-commutativity here.

So $I^2 = I$ trivially. 
Thus, there exists $a \in I$ such that $am = m$ for all $m \in I$.
In particular, $a^2 = a$ so that $a(a - 1) = 0$.
Since there are no non-zero divisors in $A$, either $a = 0$ or $a = 1$ which contradict $I$ non-zero and $I$ proper respectively.

[cloud walrus]

Spamakin🎷

[rocky cloak]

What do you mean?

You didn't use anything you just stated the result. Which doesn't hold.

[barren sierra]

Agh I forgot the start "Suppose towards contradiction Nakayama v3 holds for M = I"

[rocky cloak]

Okay, I see what you're saying.

I guess it's a little weirdly formulated. Where you would use commutativity is in such an ideal not existing

[barren sierra]

yea wording may not be the best, I've got a headache 😵‍💫

And ok that makes sense

cause any ideal satisfying I^2 = I in a commutative ring would have to be zero right

[rocky cloak]

Well any finitely generated proper ideal in a commutative ring without zero divisors would be

[barren sierra]

?

take any finitely generated ideal in k[x], k a field?

or you mean such ideals satisfying I^2 = I

yea yea

[rocky cloak]

You need all the assumptions is what I'm saying

[rocky cloak]

I'm not even sure there are noncommutative examples of this though.

Either way this exercise is very far from proving the failure of Nakayama v3.

And for that I think you can get away with much simpler things, constructing M yourself

[barren sierra]

The sheet gives an example (I do not know any lie algebra stuff so I cannot confirm this)

I am going to try to construct an example myself though

[rocky cloak]

I'm surprised the enveloping algebra doesn't have zero divisors, but it appears to be the case

[barren sierra]

idek what an enveloping algebra is (I really need to sit down and learn lie theory at some point)

[rocky cloak]

You just take the free algebra generated by your lie algebra modulo the relations that
xy - yx = [x, y]

It's the adjoint of the forgetful functor from associative algebras to lie algebras

[barren sierra]

yea I looked up the definition and I see there's some universal property (largest such algebra etc etc)

ah

Yea I guess specifically I just don't know it's usefulness in Lie Theory

[rocky cloak]

Or from a practical perspective
A lie algebra g-module is exactly a U(g)-module

[thorn jay]

We mathematicians like to turn every kind of representation we have into some representation of algebras lmao

[rocky cloak]

Yup
Quivers -> path algebra
Lie algebra -> enveloping algebra
Bimodules -> universal enveloping algebra
Groups -> group algebra

Everybody loves a good algebra

[south patrol]

I wonder how hard it is to show directly that they are categories of modules lol

[thorn jay]

Lmao even quandles have
Quandles -> adjoint group -> group algebra

[thorn jay]

Can't you show it's abelian and then use some embedding theorem?

[south patrol]

Like I assume you can exhibit a compact generator but idk how to do that nicely without just writing down the ring lol

Sure but that is weaker

[thorn jay]

Right

[thorn jay]

Center of the category

Wait that only works if it's over a commutative ring

[rocky cloak]

Yeah finding a progenerator feels like it would be the same as just constructing the algebra.

And I guess showing that the forgetful functor has an adjoint would not count as "directly showing it"

[rocky cloak]

Like embedding it into Mod Z(C)?

Think that should work for small categories, but it doesn't really show it's equal to Mod ZC

[thorn jay]

I meant that, if C is equivalent to some category of modules over a commutative ring, then it must be abelian and equivalent to the category of modules over its center

[rocky cloak]

Right, but I guess that's a different question.

Once you know it is a module category how to identify the ring, as opposed to figuring out if it is a module category in the first place

[languid trellis]

I'm having an absolute brainfart trying to show that in a PID, prime implies irreducible. Any tips?

Like I know I need to use the fact I'm in a PID at some point because this doesn't hold for more general rings. Also the definitions I'm using are p prime if p | ab implies p|a or p|b, and p irred if a|p implies a is a unit or associate to p

Wait I've figured it out

I think this holds for more general rings actually, and we need PID for irred implies prime

Suppose p is prime. Consider a|p. Then p = ab. By primality, either p|a or p|b. If p|a then a = up, where u is a unit. If p|b then b = cp, which implies ab = acp which implies ac =1, so a is a unit. So a is either a unit, or a unit multiplies by p, so p is irreducible.

[tardy hedge]

pretty sure prime implies irreducible in any domain

[languid trellis]

ya that's what i've concluded

that was somewhat silly

[vast verge]

Okay I jumped ship to Fraleigh's book, let's hope I struggle less on this one

[delicate orchid]

I’ve been trying to do this for my representations for 3 years

[thorn jay]

You'll get it sometime

I promise

I talked with the math gods and they said it should be possible

[tough raven]

I think this follows from the associated graded algebra not having zero divisors (use leading terms).

[tough raven]

That's how you know it deserves to be called a representation.

[vast verge]

Could someone explain why this isvwrong?

[velvet hull]

1 goes to 2 (on the right), then 2 goes to 1 (on the left)

I don't know why you sent it to the 2 on the bottom

[vast verge]

The left permutation is inverted (it has -2 power on the top)

[velvet hull]

oh, I missed that got it

[vast verge]

The answers said I should be getting 6 instead of 4

[tardy hedge]

In defining the equivalence relation for when a/b = s/t, we have u(at-sb) = 0 for u in multiplicatively closed subset. Why do we need u to be in S?

[velvet hull]

because u appears in the bottom

u(at - sb) = 0 is basically telling you that s = au and t = bu

and you can "cancel"' any u you want from a common fraction

[tardy hedge]

Sorry i dont really get this

[tardy hedge]

Why?

[tardy hedge]

Like if a/ut = b/ut then a/t = b/t?

[velvet hull]

no, we put u there essentially to allow us simplify fractions

i.e. au/bu is equal to a/b

which is not possible if you remove the u from the definition

and u has to be in the multiplicatively closed set for dividing by bu to even be defined (set b=1)

[tardy hedge]

So i think if we make the relation u(at-bs) = 0 with u in the ring then its still an equivalence relation on pairs (a,b) but we cant make the operations work?

[velvet hull]

what do you mean

[tardy hedge]

Yea sorry, one sec

[velvet hull]

actually, now that I think about it again, I was mistaken

the cancelling already happens if you remove u

[tardy hedge]

If u remove u where? In u(at-bs)=0?

Ik that we just need the u there so that we can define localization for non integral domains

But i was wondering what is different if we had u in the ring and not in S. Because i think that way still at least defined an equivalence relation

[velvet hull]

the immediate problem is that if you set u = 0 the localisation collapses into being the trivial ring

I would guess that similar behaviour happens even if you exclude 0

[tardy hedge]

similar collapsing behaviour?

i was thinking maybe the operations of addition and multiplication may not be well defined anymore this way or something

[rocky cloak]

So you want
a/b to equal ua/ub.

So if
a/b = ua/ub = s/t
you want
a/b = s/t

And then the second equality would still be
u(at - sb) = 0

And as for u should be in S as opposed to something else, you only divide by things in S.

[tardy hedge]

i dont really undersand the point here tbh

[rocky cloak]

You agree that
a/b should equal ua / ub right?

[tardy hedge]

I guess that would be nice

[velvet hull]

by 1(at-sb) = 0 alone we have that au/bu = a/b.
But then since 1 is in S, set b = 1 such that you get u is in S as well

[rocky cloak]

So if ua / ub = s/t, then it must also be the case that
a/b = s/t

So a/b = s/t if
ua t = s ub
which we can rewrite as
u(at - sb) = 0

A (imo) less enlightening explanation is also just that it's needed to prove transitivity of the equivalence relation

[tardy hedge]

yea the transivitity is still satisfied if u just say u(at-sb) = 0 for u in ring

so i was wondering what goes wrong if thats what u say

[rocky cloak]

Oh, that's what you're asking.
Well then everything just becomes equivalent cause you can just set u=0

[tardy hedge]

What if you exclude u = 0

[rocky cloak]

So in this localized ring the point is you can divide by stuff in S.

So the equation
u(at - sb) = 0
you divide by u to get
at = sb
and then
a/b = s/t.

So for this to work you need u to be something you can cancel

That might work for some u outside S, it depends a lot on the ring and S

[ivory talon]

Hi peeps I got a question, I have this $Q_8$ group and i've been asked to find it's automorphism group and so I can find some automorphisms of $Q_8$ as well as inner automorphisms however how do I make sure I can actually "guess" all of them? I guess my question is, is there some theorem or some rule which I just haven't been taught that will help me effictively find all automorphisms of $Q_8$?

[cloud walrus]

Yuval

[ivory talon]

I should mention $Q_8$ isn't a random group, its the quaternion group

[cloud walrus]

Yuval

[velvet hull]

any homomorphism originating from a group is uniquely determined by the choice of image of any fixed generating set

[ivory talon]

basically what you're saying is that if $Q_8$ is generated by ${i,j}$ then I can use different permutations of some element $q \in Q_8, \phi(q) = i \cdot q$ or replace $i$ by $j$ or replace $q$ by $i$ or $j$?

[cloud walrus]

Yuval

[ivory talon]

Did I get that right?

I meant different permutations of $i,j$ in the automorphism

[cloud walrus]

Yuval

[ivory talon]

also if that is the case how can I make sure that the homomorphism actually is bijective

[ivory talon]

and like that all the permutations combined are the order of $Aut(Q_8)$

[cloud walrus]

Yuval

[rocky cloak]

Basically just checking.

Like i would need to be sendt to some element of order 4, and j would need to be sendt to a different element of order 4. Then most of the cases are the same so it's not so much to check what gives an automorphism

[ivory talon]

so basically any automorphism defined will have to map $i \rightarrow j/k/i, j , j \rightarrow j/k/i, j , k \rightarrow j/k/i, j $ and $1 \rightarrow 1/(-1), -1 \rightarrow 1/(-1) $

[rocky cloak]

An automorphism will send 1 to 1, but yeah i/j will map to ±i/j/k

[ivory talon]

yeah forgot the minus sides

okay.. I think I get it

but

Wouldn't defining so many homomorphisms and proving each of them is actually an automorphism take hours and hours?

there are so many combinations of automorphism which will bring me said mapping...

and the question explicitly states to find all automorphisms of $Q_8$

[cloud walrus]

Yuval

[ivory talon]

if I'm just being lazy tell me lol

[rocky cloak]

Well if you squint your eyes there's really only like 2, since they all look basically the same

[ivory talon]

2 automorphisms and the rest of the will basically be of similar form? Wdym?

[rocky cloak]

Like
i goes to ±i/j/k, so that seems like 8 choices, but clearly it doesn't matter which one you send i to since it's all symmetric.

So let's just say i goes to i.

Then j can't go to i or -i. j going to j is the identity you know that one, to -j that one's inner you know that one too, so just check k and -k

[ivory talon]

umm sorry if it's a stupid question I just feel as if I'm missing something,
why can't i go to -j or -k?

[rocky cloak]

It can. I'm just saying the calculation would be identical so no need to redo them

There's 8 possible choices for what you can do with i, so just pick one of them, count, then multiply by 8

6*

Not 8 sorry

[ivory talon]

okayyyy

now I'm with you...

so

I will get one identity automorphism and another automorphism of a different form which since the group is pretty much symmetric will also cover the rest of the cases?

and obviously I get the trivial automorphism from i -> i so I should just search for the second automorphism prove it is indeed an automorphism and say that because of the group's symmetry it behaves this way for the rest of the automorphisms ?

and for example I'll take i and map it to j and do the calculations for that one and then generalize

[minor lance]

Sorry for necroreposting but can you elaborate on how absurd it was

[twilit wraith]

ive come to learn since that pretty much every sporadic group is pretty crazy

it was just constructed from a very particular graph with a lot of vertices

[minor lance]

Santa Claus told me to research Janko groups

Well not exactly research

He just said look them up

[grand yarrow]

I was reading the definition of the tensor algebra and they mention the multiplication is just "this extended by linearity".

Do they really mean that if we take any two elements of the direct sum
i.e. functions alpha from N to the disjoint union of T_i

The product of any two, say alpha and beta, of these is just

Alpha . Beta(n)= sum_{(i+j=n)} A(i)B(j)

[tribal moss]

They mean something like an element $\mathbf{v}\in T^kV$ has the form
$$ \mathbf{v} = \sum_{i=1}^n a_i(v_{i1}\otimes v_{i2} \otimes \cdots \otimes v_{ik}) $$
and similarly $\mathbf{w}\in T^\ell V$ has the form
$$ \mathbf{w} = \sum_{j=1}^m b_j(w_{j1}\otimes w_{j2} \otimes \cdots \otimes w_{j\ell}) $$
Then define the product $\mathbf{vw}\in T^{k+\ell}V$ as
$$ \mathbf{vw} = \sum_{i=1}^n \sum_{j=1}^m a_i b_j (v_{i1}\otimes\cdots\otimes v_{ik} \otimes w_{j1} \otimes\cdots\otimes w_{j\ell})$$

And then, of course, one needs to check that this is well-defined, i.e. that if v or w can be written in those forms in several different ways, then the various expressions for vw actually evaluate to the same element of the larger grade.

[cloud walrus]

Troposphere

[grand yarrow]

i understand this part but I wanted to know how they extend this multiplication to all of the tensor algebra

[grand yarrow]

is it in this way that I wrote?

[tribal moss]

Ooh, yes, now I can see what your question actually was. Yes, that's how you extend it to non-pure elements.

Sorry for answering the wrong question :-)

[grand yarrow]

Thanks!

[near tapir]

If f: R —> S is a ring homomorphism, and a in R being a zero divisor implies f(a) is one aswell, does this mean that f has to be injective?

[tardy hedge]

Why do u think f has to be injective

Shouldnt that property just be true for any map

[tribal moss]

It might be that the element that kills a maps to 0 in S, such that f(a) is not a zero divisor.

[tardy hedge]

O yea

Well is that condition enough to say f is injective

[tribal moss]

No. If R is a integral domain, then the condition about zero divisors is vacuously satisfied, but not all homomorphisms out of a domain are injective.

[tardy hedge]

Non domain case

[near tapir]

Let $a \in R$ be a zero-divisor. Then there exists a nonzero $b \in R$ such that $ab = 0$ or $ba = 0$. Then $f(ab) = f(a) * f(b) = 0$. Not sure how to guarantee $f(b) \neq 0$ here.

[cloud walrus]

hzh

[tardy hedge]

I dont think u can guarantee anything

Can someone think of a counter example lol

[near tapir]

Yeah that makes sense, the exercise was to find out if a being a zero divisor implies f(a) is one aswell.

What about mapping Z to any trivial ring as a counter example

[rocky cloak]

Z -> Z/n
Works as a counterexample

[glad osprey]

I don't think that works as a counter example, because Z does not have any zero divisors. To find a counter example, see if you can find a map from a ring with zero divisors to one without

[near tapir]

What about 0?

[glad osprey]

0 is not a zero divisor AFAIK

[near tapir]

The book I’m using has 0 as a zero divisor in every nontrivial ring

[glad osprey]

Hmm, but not in the trivial ring? I guess that makes sense, but then the counter example is boring

I was thinking of Z/6 -> Z/3 as counter example

[tardy hedge]

Counter example to what

Oh nvm

Of just a zero divisor not mapping to zero divisor ?

[glad osprey]

Yep, 2 is a zero divisor in Z/6 but not in Z/3

[tribal moss]

If you don't like the condition being vacuously satisfied in Z, then how about Z/4Z -> Z/2Z?

The zero divisors in Z/4Z are 0 and 2; both map to 0 in Z/2Z, but the map is still not injective.

[near tapir]

How does this work?

[tribal moss]

If you want at least one zero divisor to map to a nonzero zero divisor, then Z/8Z -> Z/4Z.

[tribal moss]

Every zero divisor in Z (that is, just 0) maps to a zero divisor in Z/n, but the homomorphism is not injective.

[rocky cloak]

0 maps to 0, and that's the only zero-divisor in Z. So all zero divisors map to zero divisors

[near tapir]

Oh okay, thanks guys

[tardy hedge]

I usually excluded 0 as a zero divisor

[rocky cloak]

I usually include 0 as a zero divisor, but then I say things like "Z has no zero divisors"

#Living on the edge

[glad osprey]

Hmm, I don't think this example works, since 0 is not a zero divisor in Z/2, so it doesn't map zero divisors to zero divisors

(using hzh's definition of zero divisors)

[tribal moss]

By hzh's definition 0 is a zero divisor in Z/2Z.

There's a nonzero element a such that a·0=0, namely a=1.

[glad osprey]

Oh, I got it mixed up, 0 is a zero divisor except in the trivial ring

I really don't like including 0 as a zero divisor

[rocky cloak]

Why not?

Seems to me it mostly simplifies things

[glad osprey]

Dunno, just feels wrong. Integral domains shouldn't have zero divisors!

[tribal moss]

And fields shouldn't have non-units.

If 0 isn't a zero divisor, then it gets trickier. Suppose f is not injective, so f(x)=0 with x!=0, and further suppose y is a nonzero zero divisor in R. Then f(xy)=0 which is not a zero divisor, but xy is zero divisor unless it is 0 itself.

[rocky cloak]

I think it's pretty natural to surpress the word nontrivial/nonzero
e.g
In a field every (nonzero) element is a unit
In an integral domain there are no (non-zero) zero divisors
A simple ring has no (nonzero) proper ideals

[rocky cloak]

Z -> Z/n still works though

[glad osprey]

Yeah, that's a good point

[tribal moss]

Yeah, so it's only if 0 is not zero divisor and we hate vacuous truths.

[tribal moss]

Oh, but if xy=0, then x was a zero divisor, and by assumption it maps to 0.

So in conclusion, if

• f: R -> S,
• R has at least one nonzero zero divisor, and
• f maps every nonzero zero divisor in R to a nonzero zero divisor something nonzero in S,
  then f is injective!

[south patrol]

This is cute

[tardy hedge]

I dont really get what troposphere was talking about near the end

[tribal moss]

I was trying to find a way to modify the original claim so it would become true. But it got somewhat mutilated in the process ...

[glad osprey]

I think the idea is that you can check if a map is injective just by looking at where non-zero zero divisors are mapped?

[south patrol]

Provided they exist

[tardy hedge]

If R->S and R has at least one nonzero zero divisor and all nonzero zero divisors map to something nonzero then f is injective?

That was tropo statement?

[south patrol]

Ye

[tardy hedge]

Whats the proof? Its annoying me that i cant really see it rn

[south patrol]

Here and following discussion

[tardy hedge]

What ive been getting hung up on was he said f(x) = 0 with x nonzero, supposed y is a zero divisor then said xy = 0

But how do u know that x is the pair that makes xy = 0

And not some other element

[south patrol]

There's a bit more after

So it's like: say r is a nonzero zero divisor and that f(x) = 0 for some nonzero x. Then f(rx) = 0. But rx is a zero-divisor, so by our assumption on f, we must have rx = 0. But then x is a non-zero zero-divisor, so f(x)=0 contradicts our hypothesis on f

[tough raven]

Re the last, why not suppress proper too?

[rocky cloak]

It would feel improper for me to do so xD

[tardy hedge]

thanks, thats kinda weird i guess

[tribal moss]

We can state the same proof more directly.
Don't assume anything about injectivity of f, just that rs=0 in R for some fixed r,s both nonzero, and that f maps nonzero zero divisors to nonzero.
Let x be any nonzero element of R, and I will prove that f(x) is nonzero too.
Now, either xr = 0 or xr != 0.
If xr=0, then x is a nonzero zero divisor, so f(x) != 0 by assumption.
Otherwise, since (xr)s = x(rs) = 0, xr is a nonzero zero divisor, so f(x)·f(r) = f(xr) != 0 by assumption. This requires f(x) to be nonzero too.
In either case f(x) != 0.