groups-rings-fields

[pastel cliff]

baba is you

baba is e

[delicate orchid]

might try doing this with just tietze transforms lol

[pastel cliff]

what that

i once again forgor that bab = a^-1 holds in general for D_2n right

[delicate orchid]

anyway just double check the map x -> b y -> ab is an iso for me

wanna make sure my hunch is right

ok the tietze transforms give me that those two elements generate D_2n

so I think it's an iso

[pastel cliff]

what are tietze transforms

[delicate orchid]

stuff you can do a presentation that preserves the underlying group

we don't need to worry about even or odd cases btw
we have ((ab)b)^(n/2) = a^(n/2) for the even case

which is the only element that makes the baby jesus cry

[pastel cliff]

ok wait

what would f(xy) be

dont we have to specify that too

[delicate orchid]

what was the map lol

f(xy) = f(x) f(y) = a^2b

[pastel cliff]

x -> a y-> ab

i mean proving it to be a homo in the first place

[delicate orchid]

you're mapping generators to generators

I don't see why you think we need to prove it's a homomorphism

we can just assume that

[pastel cliff]

we can?

[delicate orchid]

as we're mapping into the group generated by f(x) and f(y)

[pastel cliff]

im rusty okay

[tight flare]

Hi rusty im Timo

[delicate orchid]

tietze transforms actually make this so much easier fuck it I'm just gonna write them down

[pastel cliff]

https://tenor.com/view/ja-gif-23934765

probably order

[delicate orchid]

<a, b | a^n = b^2 = 1, bab = a^-1>
introduce a generator c = ab
<a, b, c | c = ab, a^n = b^2 = 1, bab = a^-1>
substitute to get the relators in terms of b and c, noting that c and b have order two
<a, b, c | a^-1 = bc^{-1}, (cb^-1)^n = b^2 = 1, bc = bc^-1>
removing the generator a as it's clearly generated by the other two generators
<b, c | (cb)^n = b^2 = 1, cbc = b^-1>

hopefully this makes it clear
(formally we'd have to use something called the substitution test to extend the map a -> ab, b -> b into a group homomorphism but I don't care to do so as the relators are clearly the same)

very clearly order as neither x or y are numbers

[pastel cliff]

[delicate orchid]

now that I've written all that out

lets find the 1 liner on maths stack exchange

[signal rain]

If you wanted to prove that some H is a subgroup of G, would it suffice to find a homomorphism from G to H?

[delicate orchid]

just check the subgroup properties

[pastel cliff]

circling back a moment, mapping generators to generators guarantees a homo?

[signal rain]

Yes, that works, but would it also work to find a homomorphism?

[devout crow]

a hom is determined by where it sends generators

[signal rain]

Is it just the same thing with extra steps since you have to show that the identity gets mapped to the identity and stuff?

[delicate bloom]

I'm thinking you'd need an injective homo from H to G

[devout crow]

but ofc you can have maps that send generators to generators that aren't homs

[signal rain]

So an isomorphism?

[delicate orchid]

the fact that we were mapping directly into the generators (by definition of the group we're mapping into) allows it to be so

oh also we need a homomorphism so

if it's not a homomorphism then lol? what are we doing? why am I here? Lets just make it a homomorphism

[signal rain]

Ah, not surjective

[delicate bloom]

if we wanna be pedantic, this can't be right cause for all groups G, H there's a homomorphism from G to H that sends everything to the identity

[delicate orchid]

I presume they meant a surjective homomorphism

[signal rain]

Lol

That's my favorite homomorphism

[signal rain]

Going the other direction, from H to G, seems to have the same problem. So I suppose it seems like it probably wouldn't be a useful tool after all

How about a surjective, but not injective, homomorphism from G to H?

[delicate bloom]

what's wrong with this?

[signal rain]

What if you said your homomorphism was phi(h) = e for all h in H?

[delicate bloom]

I wouldn't cause it's not injective

[signal rain]

Oo

True

What if you just mapped H to itself?

[delicate bloom]

well a group is its own subgroup, seems fine to me

[signal rain]

Even if we didn't know that H was a group yet?

[delicate bloom]

well you should probably check something is a group before you start checking if it's a subgroup of something else

I suppose you could still go through with making an injective map between sets which obeys the f(xy)=f(x)f(y) property for their binary operations without knowing if anything is a group

[viscid pewter]

got curious and googled a thing https://math.stackexchange.com/questions/14588/does-the-order-lattice-of-subgroups-and-lattice-of-factor-groups-uniquely-det

i think this is saying even with quite strong details we can't get the group from the subgroups?

[delicate orchid]

that's very surprising

[viscid pewter]

idek if i understood what they're saying properly

[delicate orchid]

seems pretty cut and dry lol

[viscid pewter]

ok

[delicate orchid]

their counter example is so.... bruh

[viscid pewter]

yeah it's not incredible for intuition ngl

[delicate orchid]

wonder what was wrong with my proof

[hidden haven]

my
I think I found the problem

[delicate orchid]

moldi I'm actually crying

in real life

[hidden haven]

[viscid pewter]

moldi really just woke up and chose violence

[hidden haven]

Well I am about to go to sleep

[chilly radish]

The problem is you didn't study extension theory

[delicate orchid]

zero regrets

[chilly radish]

Don't come crying to me when suddenly you need to know about H^n(G,M)

[delicate orchid]

oh I will, and you'll jump on the very VERY rare opportunity to explain it

[devout crow]

mfw I need to know about the n-th cohomology of G with M coefficients

[dire bramble]

reviewing some basic group theory and I have a stupid question

I buy this for the set of cosets G/G_p for the isotropy subgroup at p, but does lagrange's theorem still apply even when the subgroup isn't normal?

[chilly ocean]

i've always seen lagrange's theorem stated with no assumptions on the group or the subgroup other than finiteness

(and a proper interpretation for infinite groups in terms of fancy stuff)

[dire bramble]

I thought the proof of lagrange implicitly assumes you have a well defined equivalence relation so the cosets are disjoint, which assumes then that the subgroup is normal

I guess in this case it's irrelevant since we still have an equivalence relation

[long nebula]

No it does not need to be normal

It works for any subgroup

[chilly ocean]

normality only comes in when you try to prove that the obvious operations on the coset space are well-defined group operations

[dire bramble]

makes sense, thanks!

[chilly radish]

Yea the coset relation is always an equivalence relation regardless of normality

[solar glacier]

whats a good teext to suppleement dummit and foote?

[chilly radish]

What are you looking to supplement

What do you feel is missing fron D&F

[chilly ocean]

there's nothing missing from d&f, but that's exactly its problem

[chilly radish]

Well "missing" doesn't just have to.be content

But if you're not satisfied with d&f maybe it'd just be best to study from another book?

[solar glacier]

I wanna know some other good recommendations for the graduate level. to aid dummit and foote, if they dont explain something well enough or

[grim wren]

Does anybody know where can I get more info about this?

[tidal crag]

Is z a complex nu...oh, it is. I remember something that was like this, but my head is somewhere else in terms of work.

[uncut girder]

@chilly radish do you know anything about prime factorization in hurwitz quaternions?

Also is there any analogous notion of "number rings" for finite dimensional skew-fields over Q?

[somber thorn]

can anyone help me with this?

i've tried 430

my current thought process was have $|S_6|=6!$ then subtract all the non-derangements

[cloud walrus]

sean

[somber thorn]

so i got $6!-1-\binom{6}{4}-2\binom{6}{3}-6\binom{6}{2}-24\binom{6}{1}=430$

[cloud walrus]

sean

[chilly radish]

No sorry, haven't really seen these.
As for the second thing, for division algebras over most of the theory of Vector spaces carries over pretty untouched so I presume you could talk about something like this, but the problem I see is idk if you could embed it into some 'algebraic closure'. I personally haven't seen something like this. It doesn't help that the brauer group of Q doesn't really have a nice structure (The less algebraically closed your field is generally the nastier the brauer group will be, since taking algebraic extensions kills division algebras as you probably know)

[smoky cypress]

The number of derangements on 5 elements is not 24, nor is the number of derangements on 4 elements 6

[wooden ember]

Am I being stupid or are the projective spécial linear group and projective linear group the same over C?

(Or over any algebraically closed field for that matter)

[upper pivot]

yeah they are

[wooden ember]

Good to know

[pulsar sorrel]

What section are you on?

[torn niche]

I want someone to verify this. I want to show that $C[0,1]$ is not Noetherian. For this, let $$A_n = { f \in C[0,1]| f(x)=0;\forall x\in [0, 1/n) }.$$
These $A_n$ are clearly form an ascending ideals. Now to show proper inclusion, we can use cutoff functions and that gives us an ascending ideals that do not stabilize. It this correct?

[cloud walrus]

বগলের চুল

[pulsar sorrel]

Generating functions equivalent to R[[x]] for any ring R.Are examples of using generating functions where R is ring of functions?
On a side note is there a structure where elements look similar to power series a0+a1x+a2x2+… but instead of x being an indeterminate x is a function f that you repeat compositions so you can rewrite as a0+f(a1)+f(f(a2))+…
So you can write as sum_0^infty f^n(a_n)?

[pulsar sorrel]

I think this is true but im not sure what cut off functions are supposed to be.

[broken stirrup]

How can i show that S_4 has no transitive subgroup of order 6?

using Galois theory

[rustic crown]

doesn't orbit stabilizer already rule that out?

the subgroup will transitively act on {1, 2, 3, 4} so will have size divisible by 4

[chilly radish]

What det said is true and elementary, but here is my significantly more convoluted solution using galois theory

Once my computer stops crapping itself

[wooden ember]

Who needs orbit stabilizer when you have field theory

[chilly radish]

Ok here goes:

1. ||Show that every subgroup of S_4 can be realised as the galois group of a normal extension of a char 0 field e.g. an extension E/F where F is finite over Q||
2. ||Suppose Gal(E/F) is a subgroup of S_4 that is transitive. Let f be the quartic polynomial such that E is its splitting field (Such an f indeed exists, why?), then f is irreducible iff Gal(E/F) is transitive (Why?)||
3. ||f is irreducible, so there is a degree 4 subextension of E/F (Why?), but this means that 4 divides [E:F] = |Gal(E/F)|, so it can't be of order 6||

Of course, this is just a convoluted reproof of orbit stabiliser using galois theory tools

idk if there's an easier way

[broken stirrup]

thank you all

[chilly radish]

Can someone help me find an exact definition of exactness in a general abelian category. Weibel defines exactness of $A\xrightarrow{f} B\xrightarrow{g}C$ as $\operatorname{ker}g = \operatorname{Im}f$, but i'm not sure what exactly this means. Do they have to be canonically isomorphic, if so, how is the canonical map obtained? Does he mean that they are equal as maps, or rather, as maps they both factor through each other?
\\
In general there are many choices for the image and kernel objects (Although of course they are all isomorphic via a unique isomorphism), but i'm not sure how to parse this to prove e.g. exercise 1.2.4.
\\
Before this he talks about how every map $f:B\to C$ has an epi-mono factorisation $B\to \operatorname{Im}f\to C$. I just wanna make sure i'm getting this part right, the maps here are given because $f:B\to C$ is a map that is $0$ composed with the cokernel map by definition, therefore it factors through the kernel of the cokernel (which is by definition the image). Here still the choice of the object Im f is not exactly canonical tho.

[cloud walrus]

ShiN

[chilly radish]

both factor through each other uniquely, i.e. they both satisfy the universal property, or in other words, ker g satisfies the universal property of the kernel of coker f

I wanna do this formally one time, but generally, would everything work the same if I thought of kernels and cokernels as in a concrete category, and carried out the arguments element-wise? Like, how can I guarantee that I can always abstract away element arguments for morphism-based ones

since generally proving exactness seems really tedious in a general abelian category

[chilly radish]

After looking more, I think equality here is like I said, which is the same as being equal as subobjects, i.e. the corresponding monos factor through eachother via an isomorphism (given any choice of kernels)

So a bit of abuse of notation ig

[agile burrow]

I think an equivalent notion is that g o f = 0 and coker f o ker g = 0

Ah nvm I think I'm saying the same thing you said, for some choice of isomorphism the kernel and image coincide, although the condition on morphisms can be checked regardless of which choice you make

[chilly radish]

I reckon you will need ker g to be universal wrt this property no still no? I don't think g o f = 0 guarantees this

[agile burrow]

I was thinking g o f = 0 implies a unique factorization of im f through ker g and the second condition guarantees a unique factorization of ker g through im f

I guess that's what I would take to imply equality

[chilly radish]

Ah yes I think you're right

[chilly radish]

So g o f = 0 but f factors as
Im f o e where e is epi (Where this factorisation is as before because coker f o f = 0 so by universality of Im f we have f = Im f o e with e an epi), so
g o Im f o e =0 but e epi implies g o im f =0, so by universality im f factors through ker g right.

The only thing here is thar weibel states there is such an epi-mono factorisation for f, but I can't seem to prove e:A -> Im f is epi

He states it in passing but it seems like a nontrivial fact

[south patrol]

Seems good to me, though just considering piecewise linear function ought to work

[chilly radish]

Ok i've been staring at this for 15 mins and i'm not sure how to prove it

f factors uniquely as f = Im f o e, then e is an epimorphism

Feels like it's the wrong direction but I know it's not

[agile burrow]

oops sorry i went to eat sandwich

so there's two things here, that e is an epi and that im f factors through ker g

[chilly radish]

That one i'm not concerned about

Like i've proven that given that e is an epi

Which should always hold regardless of any exactness

[agile burrow]

right

[chilly radish]

Do you know how to prove this

[agile burrow]

that e is an epi?

[chilly radish]

Ye

[agile burrow]

yeah, so let f: A -> B, we have f factors as A -> K -> B where K is the source of im f. Let e denote the map A -> K
Now consider the image of e, so we have the factorization A -> L -> K -> B where L is the source of im e

[chilly radish]

Ok

Ok I think I see it now

Wait

Nvm

[agile burrow]

then L -> K -> B is a mono through which f factors, hence by universal property of im f, it must be the case that im e is an iso

[chilly radish]

Wait why does this imply that im e an iso

Is an iso

[agile burrow]

since L -> B is mono and f factors through it, K -> B factors uniquely through L -> B

that unique factorization yields an inverse to L -> K

i think is how it goes

[chilly radish]

Hmmm

[agile burrow]

right because universal property of image is that im f is initial with respect to monos through which f factors

[chilly radish]

Right

That just falls out of the definition of im as ker coker f right

[agile burrow]

yeah basically

[chilly radish]

Ok so im e is an iso

[agile burrow]

so if im e is an iso, then 0 = coker(im e) = coker e implies e is an epi

[chilly radish]

Why does the last equality hold

[agile burrow]

oh yeah, because cokernels are epis and in abelian category every epi is the cokernel of its kernel

so i guess you'd write coker e = coker(ker(coker e)) = coker(im e)

[chilly radish]

Right yea

Alright cool

That's pretty nontrivial for smth remarked in passing

Oof

Like it's just a diagram chase but still

[agile burrow]

lmao yeah all these things tend to become kinda non-trivial when everything is universal properties

i don't think i would've thought to look at im e if i hadn't seen the proof before

[chilly radish]

Same

I didn't consider that

[agile burrow]

some might argue that proving things like epi mono factorization is overkill because elements make everything so much more intuitive

but i find it kinda fun

other people were like "can't you just use freyd-mitchell embedding and work with elements anyway"

how about you freyd-mitchell embed deez nuts in your mouth

[chilly radish]

I mean I prefer working with elements

But I appreciate the sentiment

I also think it's good to do at least once

[agile burrow]

i just wanted to make a deez nuts joke

[chilly radish]

Lmao

I also proved before that the kernels of the components of a chain map form a kernel

Which is nice

Just boils down to finding the chain maps

[agile burrow]

yeah it's kinda nice how everything comes together

cokernels too!

[chilly radish]

Yea ofc

It's just dual.tho

[agile burrow]

mfw appeal to opposite category

[chilly radish]

Reverse the arrows

Not gonna do it twice lmao

[agile burrow]

fair enough

[chilly radish]

Whatever, freyd mitchell to the rescue

[woeful flint]

I know that for a number field $\mathbb{Q} \subset K$ of signature $(r_1,r_2)$ we have that $r_1 + 2r_2$ is equal to the degree of the extension. I'm trying to wrap my head around the extension $\mathbb{Q} \subset \mathbb{Q}(\sqrt{2})$. To me it seems that there are two real embeddings the identity and the one interchaging $\sqrt(2) \mapsto -\sqrt(2)$ but aren't these complex embeddings too? This would give us that $r_1 = r_2 = 2$ but this is a contradiction.

[cloud walrus]

Kraft Macaroni

[cloud walrus]

Amorous aka Lucifer

[agile burrow]

We define a real embedding to be one contained in R and a complex embedding to be one contained in C but not contained in R

At least in this context, that's how I've always seen it just to avoid what you're saying

[woeful flint]

ah ok cheers

Also if $f\in\text{Emb}_\mathbb{Q}(\mathbb{R},\mathbb{C})$ and $f(\mathbb{R}) \subset{R}$ then $f$ is the identity?

[cloud walrus]

Kraft Macaroni

[woeful flint]

I'm tryinng to disprove this but I cant find a counterexample

[agile burrow]

You may not be able to disprove this

Remember that every real number can be realized as the limit of a sequence of rational numbers

Wait actually idk if this is true, I switched to topology mode mb

[woeful flint]

If it's true that these embeddings are continuous wrt the usual norms on R and C it would imply the result no?

[agile burrow]

Yeah

[woeful flint]

So we just need to show that these are bounded since both R and C are banach

then they're continuous and hence the result

[agile burrow]

Idk anything about Banach spaces but you sound confident so I agree

[woeful flint]

hahaah I might be chatting shit lol

[hidden haven]

You are taking an embedding as a Q vector space?

Then you can take a basis of R over Q and get a non identity embedding by exchanging some basis elements

[agile burrow]

😮

I was thinking embedding as Q algebra

[hidden haven]

Ah

Then this is true

R only has the trivial ring endomorphism

[agile burrow]

Oh right, because any map is gonna be order preserving

[hidden haven]

Yep

[upper pivot]

all banach spaces are C vector-spaces so R isnt one

[lethal dune]

banana space

[chilly ocean]

I actually find it that people in the Banach space theory usually prefer to talk about real Banach spaces, and only then translate to the complex case
For example, there are theorems about separating convex sets using linear functionals, which for complex case we need to introduce functional Re(f) instead for every complex linear functional f

[upper pivot]

you find wrong.

[chilly ocean]

nice argument

Anyway, what I was going to say. For working with C*-algebras and such, I can see why people would rather work with Banach spaces over C, but clearly both approaches have their merits

[upper pivot]

thats probably fair, just for the things I do which is indeed operator theory

you never hear R lol

[warm wyvern]

how're normal subgroups special?

or are they just a common algebraic special case?

[south patrol]

do you mean normal subgroups or smth else?

[warm wyvern]

oh, yeah, sry

[chilly ocean]

normal subroups are the ones whose quotients are groups, seems pretty special to me

[south patrol]

Normal subgroups of a group G are precisely the kernels of homomorphisms out of G, which also means they often occur (though ultimately this is equivalent to the above via first iso)

[warm wyvern]

I see

quotient groups are two sections away tho, so ig I'll wait to get to that

[chilly ocean]

i don't think you should wait

[warm wyvern]

should I dig in straight away?

[chilly ocean]

[warm wyvern]

[wooden ember]

What’re you reading from @warm wyvern

[warm wyvern]

artin

why?

[wooden ember]

Just wondering, I hear artin is one of the best ones to learn algebra

Wish you a lot of fun, i learned group theory last summer by reading through a few chapters of d&f and it was a ton of fun 👌

[warm wyvern]

thanks

[chilly radish]

Define the obvious group structure on the quotient G/H for some subgroup H. Check what is a necessary and sufficient condition for this action to be well-defined on representatives and hence give a group structure

bam you have normal subgroups

they fall out of the definition basically

[chilly ocean]

[rustic crown]

there's another perspective which is pretty useful to keep in mind. instead of thinking of normal subgroups as anything "special" think of them as just another type of "sub-structure"

If you're doing set theory, then one such is a subset and other one is called an equivalence relation. One way to define an equivalence relation on X is by starting with a function f : X --> Y and saying a ~ b in X if and only if f(a) = f(b).
The first isomorphism says that all equivalence relations arise in this fashion, one can take Y to be the quotient X/~. The second isomorphism theorem says an elementary statement about how these two types of structures interact with each other and the third isomorphism theorem says something when you have two things you can quotient with, one being finer than the other.
There's also a correspondence theorem for both these sub-structures.

You can transport these theorems to other algebraic structures, like groups, rings, modules, ...
The equivalent notion of "subset" for groups is nothing but a subgroup. This is when the inclusion map H --> G becomes a map of groups.
You get normal subgroups when you try to quotient by an equivalence relation G --> G/~ and if you want this to be a map of groups, then a little work shows that one is forced define a ~ b iff aN = bN for some normal subgroup N.
For the case of ring theory, you have subrings and ideals and all the above theorems generalize nicely. For sets and rings, the second structure is not really a special case of the first one, it came out of a different construction entirely.

i'll sleep now, good night

[hidden haven]

cosubstructure

[devout crow]

to what extend does commalg generalise ideas from galois theory? my prof mentioned that integrality and integral ring extensions generalise algebraic field extensions, but he also said there are "many" other examples of this kind of thing and I was wondering what those are

am asking because I'm planning on taking commalg next year without galois

[opal osprey]

If $G,H$ are finitely generated abelian groups, then the existence of a monomorphism $f : G \rightarrow H$ and a epimorphism $g : G \rightarrow H$ implies that $G$ and $H$ are isomorphic?

[agile burrow]

certainly the free parts of G and H are isomorphic, i'm not sure about the torsion off the top of my head

certainly f and g both send torsion elements to torsion elements, hence the torsion subgroups of G and H have the same order

as for whether they are isomorphic

well I guess the generators for the torsion subgroups are sent to elements with the same torsion since f is a monomorphism, which implies that the torsion subgroups are isomorphic?

[agile burrow]

so yes, methinks

[torn niche]

why there's no gcd? since t⁵|t⁵ and t⁵|t⁶ and for any other r, r|t⁵?

[proud bear]

i guess t^5 does not divide t^6 in R because t is not in R

[torn niche]

i see

[noble dust]

That's weird

What does that even mean

Because for example t^3 is a divisor of both in R

I don't get what they mean by there is no gcd

[rustic crown]

gcd is a common divisor such that any common divisor divides the gcd.

t^3 is not a gcd because t^2 is also a common divisor but t^2 doesn't divide t^3.

[iron vessel]

What are the best books for learning Cluster Algebras? Also what prerequisites are necessary?

[upbeat fulcrum]

I find this really hard to understand but interesting, is there any other way to make this more simple?

https://mathoverflow.net/questions/416436/is-there-a-formula-i-can-use-to-count-the-number-of-k-potent-elements-over-gauss

[winter yoke]

I've been stuck on the following problem, any ideas? $f \in S = \mathbb{R}[x,y]$ is a 1234 polynomial if we can write $f(x,y)=g(x^{12},y^{34})$ for some $g$ in $S,$ and we define $T(f)$ as the corresponding $g.$ If $f, g, h$ are 1234 polynomials and $f=rg+sh$ for some $r, s \in S,$ prove $T(f)=uT(g)+vT(h)$ for some $u, v \in S.$

[cloud walrus]

aadfg

[devout crow]

is R a general comm ring? or the reals? I think it makes a difference

[agile burrow]

There's a book by Fomin, Williams, and Zelevinsky that I've heard good things about

As for prereqs, I'm not too sure. I imagine some familiarity with quiver representation theory would be helpful, but someone could maybe get by knowing Lie algebra representations?

I haven't actually learned cluster algebras myself so it's hard to say, but that's what I'd guess based on what little I've seen

[agile burrow]

Seems like there are also a number of lecture notes available online

[lethal dune]

the 1234 one again

[agile burrow]

bring back weed polynomial 😔 ✊

[hidden haven]

https://tenor.com/view/walter-bonk-walter-bonk-nelson-dog-gif-15721111

[winter yoke]

The reals

[winter yoke]

People didn't treat the problem seriously. It's basically the same now but hopefully that no longer happens.

[warm wyvern]

what're representatives?

[warm wyvern]

that does sound special, now that I know what it means

[chilly ocean]

elements of G/H are certain subsets of G. a representative of an element of G/H is an element of that set

what shin meant (and what is an elaboration of what i wrote) is the following: the elements of G/H are of the form gH for g in G. the "obvious group structure" is (gH)(g'H) = (gg')H. now check precisely what it means for this to be well-defined in terms of H

you will see that what i wrote falls out of it

[warm wyvern]

that means gH=Hg for all g which implies H is normal

(also, please don't hesitate to ping me when you reply)

[south patrol]

Perhaps I'm being dense, but in this exercise, what stops me from taking k a field of characteristic p - let's say Fp for definiteness - and n=p; doesn't (1,...,1) in (F_p)^p generate a proper, non-trivial A-submodule of M?

[Essentially I've just chosen this since the action of S_n just becomes trivial]

If the characteristic is zero there's an easy proof though

[chilly ocean]

It looks the same way to me

[south patrol]

Sure yeah

[warm wyvern]

for people that went through artin, I'm assuming the starred exercises are hard?

also, if I feel like I understood a chapter's content quite well, can I just do the miscellaneous problems and move on?

the section problems don't seem particularly challenging

[delicate bloom]

depends, drilling through some problems can help make it really stick, and won't hurt you. And you can try to generalize on the questions or use them as a spingboard to ask more interesting questions

but you don't need to do all the problems either, just gotta figure out what balance works best for you

[dire python]

where's this from?

[fallow plume]

If I'm proving that there's infinite extensions with an Abelian Galois group, G, then it suffices to show that any abelian group has an extension (which is a pretty straightforward proof) and then just use G^n right?
I think each composite group of G^n would correspond to a different subfield right?

or am I too tired to think straight

[tribal moss]

When you say "there's infinite extensions", do you mean that there are infinitely many different (field) extensions (of Q?) that all have abelian Galois groups, or that there is one extension of infinite degree whose Galois group is abelian?

[fallow plume]

infinitely many different field extensions of Q with the same galois group isomorphic to G

[tribal moss]

For every abelian G, or just for one G that you can choose?

[fallow plume]

any abelian G

[wooden ember]

I definitely second what merosity said, especially when it comes to finding the right balance: I’ve been on both ends of the spectrum, doing all it no exercises and neither work. Unless you really like the book you’ll lose motivation with the first, and it won’t help you learn much better (I don’t remember half the exercises I did in d&f), and if you do none you’ll feel like you understood but let misconceptions settle in which will make later sections a lot harder

[fallow plume]

(except trivial*)

[tribal moss]

When you say "any abelian G", do you mean for every abelian G, or for some abelian G that you get to choose?

[fallow plume]

basically fix any possible abelian group G, with the one exception of the trivial group. I'm proving there's infinitely many distinct extensions of Q such that the Galois group is isomorphic to G

so for every

[tribal moss]

When you say "any possible ablelian group G", do you mean that you get to choose which group it is, or that you need a constuction that works for every abelian group G?

[fallow plume]

yea still for every, sorry if it was frustrating if i was wording that ambiguously 🤣

[tribal moss]

Okay thanks. Then I've got nothing :-)

[fallow plume]

ah okay. I'mma keep thinking about the G^n thing

like,,, it really obviously seems like it should work but

it seems trivial

cause one of our books is D&F. the first part of the problem was already proven in D&F (corollary 28 of 14.5 which proves that there's a field extension for any abelian G), and then the second part is this question and that's it

it seems suspiciously easy

ig it's right

weird problem ¯_(ツ)_/¯

[fallow plume]

just a general question, are all cyclic Galois extensions of Q cyclotomic?

i can't seem to word it well enough for Google to give me anything

[tribal moss]

Q[sqrt(2)] ?

[delicate bloom]

tangentially related, you might be interested in the kronecker-weber theorem

[fallow plume]

mmm yea Ig that does mean I need to word it more carefully if this is even worthwhile

[south patrol]

How is this meant to make sense - $\phi_1(g),\phi_2(g)$ aren't even in the same group, let alone words in $X_1 \cup X_2$.
I assume what's meant is if $w$ is a word in $X_1$ and $v$ a word in $X_2$, then $wv^{-1}$ is a relation if there exists $g \in G_0$ such that the image of $w$ in $G_1$ is $\phi_1(g)$ and the image of $v$ in $G_2$ is $\phi_2(g)$?

[cloud walrus]

potato

[tribal moss]

Yeah, that must be the idea. Perhaps it would be clearer to say that the pushout is a quotient of G1 × G2 by the normal closure of (phi1(x), phi2(x)^-1).

[south patrol]

Or should it be the free product? (That's what I seem to see online I suppose).

[south patrol]

But thanks. I decided to take this idea and it does work to give you a pushout

[tribal moss]

Oh yes, free product, sorry.

[south patrol]

No worries, thanks

[tribal moss]

I'm too used to thinking about abelian groups, apparently.

[south patrol]

So am I :/

I guess another way to get around the notation be just to take the canonical presentations (as in X1 = G1 etc)

I reckon that may be what they intended given how the notes continue

Ah, yes they said let them be the canonical presentations - I can't read (though notation doesn't emphasise this ig)

[upbeat fulcrum]

so im struggling to do this one using hensels lemma can someone help me to begin with this?

[wild solar]

hi, I'm confused about the part I circled in red. From what I understand a-1 does not need to exist as rings don't necessarily have multiplicative inverses, and this R is clearly not a division ring(1/f(X) is not well-defined). So how can we justify something like this?

I could be missing something trivial, but this is really confusing

[chilly ocean]

alpha is a non zero constant, you're allowed to divide by it

[wild solar]

but why can we say this element belong to the ring M as well

sorry, U

[chilly ocean]

by the argument written before your red box

[wild solar]

why?

I thought that justifies a not 1/a

[chilly ocean]

or are you confused about how a constant could be an element of U, a set of functions?

[wild solar]

no I understand that

I'm just confused about why 1/a belongs to U

[chilly ocean]

it doesn't need to, U is an ideal

a(1/a) is in U because a is in U and U is an ideal

(a meaning alpha, of course)

[wild solar]

OMgggggg I forgot M is the ring of all continuous functions

sorry

I understand now

thank you thank you

[chilly ocean]

[jovial flint]

$\mathbb{Z}_5\oplus\mathbb{Z}_5$ is a field no?

[cloud walrus]

Bilboswaggins

[rustic crown]

nope, (1, 0) doesn't have a multiplicative inverse.

[jovial flint]

ah yes missed that

thx

[hidden haven]

It's not even a ring

[jovial flint]

why not?

[hidden haven]

For rings, you take direct product rather than direct sum. I mean you can define the ring operation on the direct sum too but the direct sum notation probably shouldn't be used

Direct sum usually means coproduct/biproduct, and coproduct of rings is tensor product instead

Tldr better to use \times instead of \oplus for rings

[jovial flint]

I have no idea what this sentence means lol

but thanks

[hidden haven]

ye mb lol

It just suggests a different kind of property

than that of products

[tribal moss]

Finite direct sums or rings are the same as finite products, so it's really only a relevant distinction once you generalize to more advanced properties.

[hidden haven]

Direct sums of abelian groups do have that property so there it is fine

[hidden haven]

How do you define direct sums of rings?

[tribal moss]

Ah no, you right, direct sums don't usually even come with a multiplication.

[jovial flint]

herstein uses this def

but these are abelian wrt to .

[jovial flint]

is this what you meant here?

[hidden haven]

Yeah but here the book is also defining multiplication on it

Weird

[tribal moss]

So it's not that it doesn't work, it's just unusual terminology.

[hidden haven]

Some authors will speak of the direct sum $R\oplus S$ of two rings when they mean the direct product $R \times S$, but this should be avoided since $R \times S$ does not receive natural ring homomorphisms from $R$ and $S$

from wikipedia

[cloud walrus]

Moldilocks1337 ✓

[hidden haven]

Basically direct sum implies the existence of maps R → R + S and S → R + S, and these don't necessarily exist for rings so not good to use this notation (they do exist for abelian groups)

[tribal moss]

It's not even a proper coproduct for non-unitary rings (where plausible-looking injections do exist).

[hidden haven]

very sad

[jovial flint]

what does natural homomorphism mean here?

[hidden haven]

It's category theory talk, but you can think of it as a homomorphism which doesn't depend on the rings R and S, but is given by the same "formula" for all rings

If you are really curious you can look up natural transformations but they can be very confusing in the beginning and the above intuition for them works for almost all cases

[jovial flint]

the intuition being this?
wouldnt something like phi(r)=(r,0) or phi(s)=(0,s) work?

[hidden haven]

Yes.
That doesn't work because it doesn't carry identity to identity

Hold on does Herstein define ring homomorphisms to have this property in the first place?

[jovial flint]

it does refine ring homomorphisms

[hidden haven]

Interesting

[hidden haven]

Then the issue is this lmao

[tribal moss]

As I was pointing out, even if the ring homomorphisms are not required to preserve 1, the direct sum still doesn't have the universal property of a coproduct.

[hidden haven]

You do have natural homorphisms but those need to be "universal"

Which is another category theory thing

[jovial flint]

[hidden haven]

[rustic crown]

Rings are better than Rngs >.<

[hidden haven]

Lol just do what the book says I was just trying to be annoying with that one message

I have never encountered a rng in the wild I think

How do those even occur

Please don't say direct sum of rings

[rustic crown]

is every Rng a subrng of a ring?

[tribal moss]

Hmm, ideals are subrngs.

[tribal moss]

Yes, there's a universal way to embed a rng in ring.

[hidden haven]

Neat

[rustic crown]

nice, never thinking about rngs then

is the construction nice?

[hidden haven]

But rings arise as endomorphisms of abelian groups or abelian groups with extra structure, but where do rngs come from

[tribal moss]

If the rng happens to have an identity, that still doesn't map to 1 under the undersal embedding.

[hidden haven]

lmao

Not neat

[hidden haven]

Freyd adjoint functor theorem should work very nice

Because product of rings coincides with their product as rngs

Surely there's a simpler construction 🙈

[south coral]

I'm trying to understand an old paper and I've come across a seemingly easy algebra problem. P is the symmetric group S5 with 120 elements, and G is the group of the trefoil knot in the picture, that is, it is presented as <a,x|axa=xax>. The author claims that there are two non-trivial presentations (presumably homomorphisms) of G in P that map x to a 5-cycle and gives them. However, I don't see how to get that result without doing a lot of very tedious calculations. Does anyone have any tips?

[shrewd flame]

Sorry to cut the convo but any idea how in Hamermesh he's going from the first to the second line in both equations? Looks simple and I had some ideas related to symmetry, antisymmetry n dummy indices but it's wrong and I don't see it

[tribal moss]

It wouldn't be universal if it always preserved 1, because the inclusion of {0,3} into Z/6Z doesn't preserve 1, so it wouldn't factor through an embedding that does preserve 1.

[rustic crown]

internet says look at Z + eR where we define e to be an idempotent.

[shrewd flame]

terrible pic hahah i'll try to get a better one

[rustic crown]

(n + er)(m + es) = nm + e(ns + mr + rs)

[hidden haven]

lol you thought I cared

[agile burrow]

Is coproduct of noncommutative rings also tensor algebra?

[rustic crown]

idts

[agile burrow]

Sad

[tribal moss]

No, it's more akin to a free product of groups.

(which is to say, horribly complicated)

[pulsar sorrel]

Why is free product of groups horribly complicated?

Because there are a lot of elements?

[tribal moss]

In the same way that free groups are.

[pulsar sorrel]

The only time I had free groups introduced were in a coding section of algebra 2 and algebraic topology

I havent seen them in other contexts

What are some ways in which they are horribly complicated?

Answering questions about subgroups maybe?

[hidden haven]

Free groups are being used whenever you describe a group using generators and relations

[agile burrow]

Pushouts of groups keep me up at night

[pulsar sorrel]

Yeah but they are without the relations

[tribal moss]

IIRC, every finite group is a (sub?)quotient of the free group on 2 generators, so they have all the complexity.

[hidden haven]

I mean that the free groups without relations are used whenever you have a group defined by generators and relations

Generators and relations is the same as saying quotient of free group on those generators by the normal subgroup generated by those relations

[pulsar sorrel]

I know the definition of a pushout group but I dont know where they are used

I have a book called examples of groups

and it talked about the construction

[agile burrow]

Van Kampen theorem is the first that comes to mind

[hidden haven]

That's the only one I know yeah

[pulsar sorrel]

That does use a pushout diagram lol

[hidden haven]

It was why people defined free products in the first place

[pulsar sorrel]

But thats only example I know

[agile burrow]

Groupoid one is better anyway 🙃

[pulsar sorrel]

I have a question that is off-topic. Does anyone have any cool animal/dinosaur/math posters?

[hidden haven]

2 questions got buried in this conversation

[south coral]

I realized that since (xa)x(xa)⁻¹=axa(xa)⁻¹=a, we know that a is also a 5-cycle and so there are only 24 options to check. I suppose that may be it? still a lot of calculations but doable by hand

[south coral]

I am confused again. For x=(12345), there exist more values of a than the given two that make axa=xax, e.g. a=(24153), so it seems I have misunderstood what is meant by "up to inner automorphisms of P" or perhaps "non-trivial representations". Does anyone have any thoughts on this?

[hot lake]

your a'=(24153) and their a=(13542) are related by conjugation by x

and x conjugated by x is still x

so the pairs (x,a) and (x,a') are related by conjugation by x

so those are the same representation up to a inner automorphism

[south coral]

ah right, thank you! and the only way to get this result is still via just a lot of calculations?

[hot lake]

idk I would have to think about it to know if there is a smart way lol

[south coral]

right, sorry. I suppose these kinds of papers really do get away with skipping over a lot of details huh

definitely a humbling reading experience

[delicate bloom]

what conditions are necessary for a ring to have zero divisors but no nontrivial idempotent elements? I tried making a few examples but didn't get anywhere

I just sort of expected that since it's easy to take a ring with idempotents and make zero divisors that the converse would be equally easy but seems not the case

[tribal moss]

How about the dual numbers?

[agile burrow]

any local ring with zero divisors should work too

i'll have to think about necessary conditions though, that sounds much harder

[delicate bloom]

cool thanks that sounds like a good example

[tribal moss]

The minimal example would be Z[x]/<2, x²> -- that is, the dual numbers over F_2 -- with just 4 elements.

Um, and also Z/4Z :-)

[delicate bloom]

nice I'll think about this more

[delicate bloom]

haha I think I just barely missed these nice examples you gave, I looked at the ring with j^2=1 and knew it had zero divisors and then thought about the Z/pqZ cases and that was a bust too and came here haha

[tribal moss]

In fact, the integers modulo p^n for n>=2.

[delicate bloom]

my original thinking was sort of tarnished cause I was thinking I could create a circle of implications that 'has idempotent => has nilpotent => has zero divisors => has idempotent"

[agile burrow]

i'm wondering if maybe local ring with zero divisors DOES characterize them

but idk how to prove this

or at least, i don't have a counter-example off the top of my head

[hidden haven]

I would imagine not because local rings are a strong version of decomposability as a direct product

[agile burrow]

right, yeah it's a strong condition

[hidden haven]

Existence of non trivial idempotents is equivalent to existence of a decomposition as a direct product

Yeah and if you look at Jacobson he calls a module M decomposable if End(M) is a direct product and strongly indecomposable if End(M) is local

And these coincide for nice modules, I think for finite length

[agile burrow]

semi related, End(M) is local iff M is simple, right?

[hidden haven]

So it's a close approximation still

[hidden haven]

For finite length yeah by what I said above

Wait

Ah yes

[agile burrow]

or am i thinking division ring?

[hidden haven]

Not by what I said above but I think that's a result in Jacobson? Let me think

Yeah division ring lol

Schur's lemma

[agile burrow]

oops, right

[hidden haven]

Division rings are local though

Converse doesn't hold

Sad

[agile burrow]

:(

but yeah, methinks I should be able to find an example of such a ring that isn't local

[tribal moss]

Z[x]/<x²> isn't local, is it?

[hidden haven]

Right, (x, 2) and (x,3) are maximal and not the same

[delicate orchid]

Z[x] saves the day once again

[chilly radish]

Not even a PID smh smh

[south patrol]

at least it's a ufd

[chilly radish]

It's not even Dedekind

[rustic crown]

dimension 2 is okie

[winter yoke]

Same problem but with 4 and 20 instead, and it's from my abstract algebra study. I changed the numbers because people weren't treating it seriously.

[warm wyvern]

Artin book club

[glossy widget]

if I want to create a group out of a set, can I just impose a cyclic structure on it to achieve that goal?

[chilly ocean]

what does "impose a cyclic structure on it" mean

[glossy widget]

make a cyclic group out of the set

like this for example

[chilly ocean]

if the set is uncountably infinite, you're not going to be able to make it a cyclic group

[glossy widget]

yeah true

[chilly ocean]

if your set is in bijection with a known group, you can use that bijection to transfer over the group structure

something something, the unique group structure making your bijection an isomorphism

[chilly ocean]

i still don't really get it lol

hmm

[glossy widget]

it's an analogy to this

[chilly ocean]

ah

[chilly ocean]

you'd basically be doing this

lemme draw something

[glossy widget]

alright

[chilly ocean]

if your set is {paper, magnet, corndog, garlic} you can assign to it (one of a few possible) group structures by identifying it with the cyclic group of order 4 as such

"identifying it with..." is the same as picking a bijection with {paper, magnet, corndog, garlic} and {0, 1, 2, 3}

i see what you mean now

[glossy widget]

and I see what you meant with the bijection thing

[chilly ocean]

the one in my picture is
paper -> 0
magnet -> 1
corndog -> 2
garlic -> 3

if you have a set and you specifically want a cyclic group structure, then you can pick a bijection with a cyclic group and do this kind of thing

(unless the set is uncountable, in which case you're not going to get a cyclic group structure)

[glossy widget]

hmm could I also get a different group structure from a finite set?

[chilly ocean]

what do you mean?

[glossy widget]

for example a symmetric group structure (the one which describes permutations)

[chilly ocean]

let me write formally what i'm talking about. let's say you have a set $S$, a group $G$, and a bijection $f\colon S \to G$. for $s_1, s_2 \in S$ i define $$s_1 \cdot s_2 := f^{-1}(f(s_1) \cdot f(s_2)).$$ then this gives a group structure on $S$ (in fancy terms, it's the unique one making $f$ an isomorphism)

[cloud walrus]

TTerra

[glossy widget]

hmm yeah this makes sense

btw is this what free groups are? are they more general? or are they entirely different things?

[chilly ocean]

free groups are a different thing

free groups are a way of building a (much larger) group out of a given set

what you're doing here is making a given set into a group

[glossy widget]

ah I see

thanks for the help and clearing things out

[warm wyvern]

sure

as long as the arrow represents multiplication by "onion" or smth since groups require a binary operation

from there you can extrapolate how multiplication works for other elements of your set and verify that it satisfies the groups axioms

[chilly ocean]

[warm wyvern]

your group would behave the same way as a cyclic group ofc

the problem is, since you're the one imposing a structure similar to a cyclic group on this collection of objects you won't really gain much insight about these objects

and I'm under the impression that's why we study groups in the first place

[chilly ocean]

you might as well just study the group whose structure you're imposing on your set

[glossy widget]

yeah true, that's a cool insight

[oblique river]

https://math.stackexchange.com/questions/105433/does-every-set-have-a-group-structure

@glossy widget maybe relevant to your original question

[glossy widget]

thanks for the link

[oblique river]

(summary: it is equivalent to the axiom of choice)

[warm wyvern]

[chilly ocean]

everything is

[brittle minnow]

So I'm looking at the Lorentz group and I see a section on the Surfaces of transivity. Could someone help me understand what this represents?

[cloud walrus]

potato

potato

[south patrol]

*as subspaces of V tensor V

Oh okay I'm silly

Ignore lol

[wispy urchin]

tensor spaces themselves can be considered as subsets of formal products or quotients so yea i guess

[south patrol]

Ok i'm back - I'm a bit confused about the construction of the symmetric power here; the other constructions I've seen seem to define S^n V as the image, not the cokernel, of this map. What am I overlooking?

[cloud walrus]

potato

[south patrol]

also, this nlab page references stacks project where they give a construction equivalent to what I originally wrote, which is weird

[oblique river]

@south patrol here's what wikipedia has to say https://en.wikipedia.org/wiki/Symmetric_algebra#Relationship_with_symmetric_tensors

[south patrol]

Yeah that seems to agree with it being the image of p_A (in nlab notation)

Odd, might ask about it on nlab if it could be a mistake

[oblique river]

ive noticed some inconsistencies within nlab before too

[south patrol]

Oh sure

(also in what I wrote above I forget to say take the subspace generated by those sets)

[chilly ocean]

"abelian" means "for all a, b in G, ab = ba", so "non-abelian" means "there exist a, b in G such that ab != ba"

not "for all a, b in G, ab != ba"

it's perfectly fine to have some elements that commute with eachother (such as any element and its inverse)

it's only when we can find some that don't commute that the group is non-abelian

[chilly radish]

A lot of nonabelian groups do have abelian subgroups

In fact if you have a nontrivial subgroup you have an abelian subgroup or are yourself abelian (namely a cyclic subgroup)

Either an element generates a cyclic subgroup or it generates the group in which case it is abelian

[chilly radish]

Lmao @walter weibel proves snake lemma in abelian categories by defining the smallest abelian subcategory containing all relevant objects and using freyd-mitchell

Based

I wonder if he expects us to prove 5-lemma like this or in an element-free way

Not sure I understand the construction here, idg what he means by thinking of f as a chain complex, as in, the complex Im(f)?
Also what the obvious way is for thinking of C, B[-1] as double complexes. I'm assuming just setting the 0th row to be C and then all the other rows 0?

,rotate

[chilly radish]

Actually idg what he means by using the sign trick either. The sign trick is used to identify double complexes with complexes in Ch(Ch(A))

So what's the double complex here to begin with

[hidden haven]

The general idea for abelian category diagram chases is to use subobjects

Instead of elements

More generally you consider subquotients

And you can do a lot of set theoretic stuff with these. For example you can intersect 2 subobjects by taking their pullback

So if you want to prove the snake lemma for abelian categories, then you can start by looking at the filtration induced on each object in the diagrams by the kernels and images of all involved maps

Ravi vakil has a cool post about this on 3b1b's blog where he does this filtration stuff by drawing Venn diagrams and it works really nicely

[south patrol]

Oh nice ig we're both reading through weibel then shin

I found these things proofs awkward as he's unclear if we can assume we can just work w elements but when I checked solutions online everyone did use elements

[chilly radish]

I just meant because weibel proves snake lemma using freyd-mitchell, does he expect us to do thr diagram chase in an element-free way or use elements

[hidden haven]

If you're asking which one helps more later then it is neither

[agile burrow]

If I recall, Aluffi does something similar to what Moldi mentions, but I found it pretty instructive. If you construct an appropriate pullback using the top row and a pushout using the bottom row, you can view the right-most kernel as a cokernel and the left-most cokernel as a kernel, which is how you construct maps from a kernel to a cokernel (since the universal properties usually suggest the other direction)

I believe the four/five lemma have similar proofs, I remember doing them for an exercise

[chilly radish]

I se

see

[agile burrow]

Wait does Weibel do chains or cochains

Indices messing up my answers :(

[south patrol]

mostly chains nearer the start but mentions cochains too not long after

[lethal dune]

why is it called short five lemma tho

most books only have 3 lemmas

[agile burrow]

I love when my rows are short exact sequences

[agile burrow]

I haven't learned double complexes yet, but if I understand the construction correctly, the idea is to look at two rows, where the top row is B[-1] and the bottom row is C. The differential in the top row is just the differential of B with the sign change while the differential in the bottom row becomes a combination of the chain map and the differential in C

To make it into a complex, you just take direct sum of the two rows

I'll think about this a bit more because I need to learn total complexes eventually so if the mapping cone is a baby case then that's good for me :)

[south patrol]

Shin I remember finding that q weird, iirc when i googled it people thought it might have mistakes

I'll need to look at what I did to remember why though lol

[chilly radish]

I think I will bring it up to my prof

And skip for now

[chilly radish]

Are you looking at 127 or 128

[agile burrow]

128

[chilly radish]

So you're saying taking the 2 rows and letting all the other rows to give a double complex?

I don't understand tho how this is a double complex formed 'from f'

[agile burrow]

the vertical maps are induced by f, no?

or wait, maybe i'm not thinking of it correctly

i'm trying to reconcile viewing the mapping cone as a direct sum of B[-1] and C with a weird differential and the perspective of double complexes

[agile burrow]

ok yeah after looking at the text myself, I don't think i understand what he's doing

weird indexing stuff gets in the way, but this is the picture I have

but realistically I missed some sign changes to the f

[celest cairn]

Is this computation even possible?

[chilly ocean]

of course it's possible

you can always take the tensor product of two rings (implicitly, here, over Z), and generally of two R-modules over R

Z_n \otimes Z_m is naturally isomorphic to Z_gcd(n, m)

[celest cairn]

Ok

Ok, so how would I take the tensor product of it? I’m currently in precalc but learning abstract algebra for fun.

[chilly ocean]

are you asking for the definition of the tensor product?

[celest cairn]

Sure I guess. Don’t know much about them.

[chilly ocean]

i won't rewrite it here, but ill suggest you find it in an abstract algebra textbook

[delicate orchid]

I forgot how the proof of this went and I think I've managed to re-do it but I just wanted to check that the map ||1 (x) a -> a+(d) where d = gcd(n, m)|| is the right morphism for the job

[agile burrow]

this exercise must have a typo, right? the only map between these complexes is the zero map, and the zero map will always induce zero in the derived category

[delicate orchid]

lol

[agile burrow]

this is correct

[agile burrow]

fwiw, Exercise 6.9 was showing that a non-zero morphism in the derived category could still induce the zero map in cohomology

[waxen hedge]

You need to use that $A/I \otimes_A B = B/I$ (if $A$ is a ring, $B$ is a $A$-algebra and $I$ is an ideal of $A$) and that $n\mathbb{Z}+m\mathbb{Z}=\mathrm{gcd}(n,m)\mathbb{Z}$

[cloud walrus]

Adrien

[delicate orchid]

nice

[waxen hedge]

( it's essentially the same than using your morphism)

[delicate orchid]

I just manually reasoned that every element in that tensor product had to be of the form 1 (x) m and d(a (x) b) = 0 seperately

[agile burrow]

yeah, i just assumed you brushed some stuff under the rug when you only define it on 1 \otimes a

[delicate orchid]

yur

[tribal moss]

You can even get away with only defining it for 1 otimes 1.

[delicate orchid]

True

Map generator to generator

[agile burrow]

nvm, my mistake was thinking that the functor from chain complexes to the derived category was full

derived categories are making me become the joker

[warm wyvern]

(b) relates to a proper subgroup, right?

[oblique river]

not necessarily

it doesn't say that H is a proper subgroup

for example, let n = |G|. then G has exactly one subgroup of order n, namely itself, and therefore that subgroup is normal

(which is true -- any group is a normal subgroup of itself)

[chilly radish]

||So true||

[warm wyvern]

OH

I basically just skipped that "of order r" part lel

thanks for the clarification :D

[south patrol]

I feel like the quantification in the question is almost ambiguous tbf (although ig then you'd add a comma if it truly meant there were just one subgroup?)

[tribal moss]

If you truly meant there were just one subgroup, then "of order r" would not contribute any meaning to the claim at all.

[vestal snow]

Is the completion (at the maximal ideal (x,y)) of $k[x,y]/(xy)\cong k[[x,y]]/(xy)$?

[cloud walrus]

Finitely Many Bananas

[vestal snow]

Here k is a field

I don't think k[x,y]/(xy) is flat as a k[x,y] module, so the short exact sequence thing won't work

Wait

My bad. The SES thing does work

[vestal snow]

I forgot that polynomial rings over fields are noetherian

[dapper nebula]

typical scheme theory enjoyer

[west sinew]

How does one compute the intersection of ideals in a polynomial ring? For example, I want to show that, in $k[x, y]$ (where $k$ is a field), $(x^2, xy)=(x)\cap (x, y)^2 = (x) \cap (x^2,y)$.

[cloud walrus]

adi

[sharp sonnet]

why not just do what you would do for sets?

[dapper nebula]

That would be ideal

[delicate orchid]

ba dum tsh

[prisma shuttle]

wut does the notation (x,y)^2 mean

[sharp sonnet]

its the ideal multiplied by itself

[west sinew]

Yeah

[delicate orchid]

yeah

[cunning dust]

yeah

[west sinew]

😔

[west sinew]

What do you mean?

[chilly radish]

No

[warm wyvern]

is there a surjective homomorphism $\varphi\colon \bR^\times\longrightarrow \bQ^\times$?

[cloud walrus]

DarQ

[warm wyvern]

how about if we changed the operation?

also, is an injective endomorphism of a vector space an automorphism?

[chilly radish]

Only in the finite dimensional case

Necessarily

[warm wyvern]

I see

[chilly radish]

If you want a counterexample in infinite dimensions, consider the space of bounded sequences l^infty and the right shift operator (a_1,a_2 ,a_3,...)->(0,a_1,a_2,...)

There's also corresponding left shift operator which is surjective but not injective

[warm wyvern]

what's l^infty?

[chilly radish]

Space of bounded sequences

i.e. sequences (say complex) which are bounded as sets in C

[delicate bloom]

I'm thinking we know phi(-1)=-1, so we can focus on the positive numbers. If we look at the logarithm as a homomorphism of the groups from * to +, we can think of them as vector spaces over Q and mapping surjectively with some axiom of choice nonsense probably

[oblique river]

If you just want a simple counterexample you dont even need bounded sequences

Just look at the space of all sequences

Also i think the answer to the previous question about R* and Q* is “no”. R* is a divisible group but Q* is not.

Eg if phi(2) = x then phi(2^(1/n)) = x^(1/n) which implies that x = 1

@warm wyvern

I think the issue with mero’s argument is that while positive R* and R are isomorphic via the logarithm, positive Q* is not isomorphic to Q.

[warm wyvern]

I'm here, I'm just looking at the definition of a divisible group

[chilly radish]

I'm still in functional analysis mode so I went for a banach space & bounded operator lol

[oblique river]

You dont need the full definition, i just proved it for you

But i was originally just appealing to the general fact that a quotient of a divisible group is divisible

[warm wyvern]

ok

how about if we change the operation to (+)

[oblique river]

Well, proved via example

Which operation?

Also what is this for, why are you looking for examples like this

Like if i know what you want i can maybe give a better example of it

[warm wyvern]

i.e. what if we have $\varphi\colon \bR^+\longrightarrow \bQ^+$ instead

[cloud walrus]

DarQ

[warm wyvern]

I don't remember, I wrote them down yesterday when I was sleep deprived

I think I was trying to prove the dim is well defined or smth

[oblique river]

Then yes, and mero already gave a proof

Or an outline of one

[warm wyvern]

oh

I'll need a second to parse that

[oblique river]

Just ignore the log stuff

That was for when you had the times instead of plus

R is a vector space over Q

Choose a basis and send one basis element to 1 and send the rest to 0

[warm wyvern]

bruh

[delicate bloom]

oh nice, divisible group argument is clean and nice

[warm wyvern]

is "choose a basis" the axioms of choice nonsense part?

[oblique river]

Yeah

Also what’s wrong with saying R is a vector space over Q

Sorry i thinj maybe im assuming you know more linalg than you do

[warm wyvern]

I've never seen it before and it reminded me of some trail of thought when I was trying to prove a function with the property f(a+b)=f(a)+f(b) needn't be continuous

[warm wyvern]

no need to apologize lel

you guys just gave me a lot to think about and research lol

[oblique river]

Oh, when you were talking about dimension i assumed it was of vector spaces

[warm wyvern]

yeah

it was

[oblique river]

Oh haha my b

[warm wyvern]

I'm the one pestering you with my indulgences

it's my b

[delicate bloom]

indulging in mathematics? criminal!

[tall mountain]

@warm wyvern you can also derive a contradiction from the fact that every element of R has a square root in R, and a surjective homomorphism would imply that the same is true of Q

[warm wyvern]

yeah, that's like a special case of buncho's argument

[agile burrow]

I woke up feeling like a Lie algebra

[lethal dune]

do additive functors make sense outside modules?

[oblique river]

sure, they make sense between any two additive categories

[lethal dune]

i see

[lethal dune]

how xu=yu=0?

[cloud walrus]

det

$x(x \otimes y) = x \otimes xy = xy \otimes x = x(y \otimes x)$

[rustic crown]

[cloud walrus]

det

[lethal dune]

but like why not x ⊗ y ≠ y ⊗ x? What am I missing?

oh I see why that is

1 ∉ I

[rustic crown]

yee

[opaque pilot]

Does [G:H] tell us the number of distinct cosets?

got it

[warm wyvern]

[G:H] is the number of distinct cosets of H in G

[warm wyvern]

so all n dimensional vector spaces over a field F are classified since they're isomorphic to F^n

is there any nice classification for infinite dimensional vector spaces?

how about just real vector spaces?

[oblique river]

All vector spaces have a basis

I.e. they’re isomorphic to F^S for some set S

(Or think of S as a cardinal; that’s the dimension)

This is assuming axiom of choice

[warm wyvern]

F^S is F acting on S?

[oblique river]

Think of it just as a vector space with S as a basis

Like F^n has a basis of n elements

every vector space has a basis, and two vector spaces are isomorphic iff their bases have the same cardinality

[warm wyvern]

ah

[lethal dune]

In Aluffi I saw a definition of IM where I is an ideal of R (commutative with 1) and module M, and
IM := {rm | r ∈ I, m ∈ M}
and claims it's a submodule, I don't understand why

clearly r1m1+r2m2 has to be of the form rm but r1, r2 may not have a gcd so

Is the def supposed to be { finite sun of rm | r ∈ I, m ∈ M}?

[oblique river]

yes

you are correct

IM is generated by elements of the form rm; maybe that's what they were trying to say

[lethal dune]

doesn't look like it

also "submodule of R"

[oblique river]

hmm, yeah guess it's a typo

another typo haha

[lethal dune]

cool tks

[agile burrow]

@lethal dune aluffi is semi-notorious for errata, much of which is documented online here: https://www.math.fsu.edu/~aluffi/algebraerrata.2009/Errata.html

[lethal dune]

[agile burrow]

I have consulted this page several times while reading through the book lmao

[lethal dune]

they have mentioned it there

[lethal dune]

the book is good tho

[agile burrow]

oh yeah, i love it

[rustic crown]

also there are two printings of the book. second one is soooo much better than the first

[terse crystal]

@winter hawk Sure, if you have further question about it then DM me or ask me in this channel, it can be classified as an algebra question

[arctic storm]

Q1. Why can't exp(x) be a vector of F[x]? After all F[x] is infinite dimensional so, it is possible for a vector in F[x] to utilize all the basis(which is also infinitedimensional). No?

And again, why is it necessary for all vector in F[x] to be composed of only finite number of monomials? (after all you have infinite many monomials)

[arctic storm]

Q2. If n is fixed, then why is S an infinite sequence? Shouldn't it be finite because you have finite basis vectors cardinality n? FS would then be a space of finite sequences with infinite number of non zero elements?(assuming F is infinite dimensional)

Thanks!

[inland otter]

F[x] is defined as the set of all polynomials over F which by definition always are finite sums.
You can look at F[[x]], ie. The ring of power series, if you want to take infinite sums but this has the drawback that you cannot evaluate power series at a random point, as on general fields you do not have any notion of convergence

Also not sure how you would define n! over a field of finite characteristic

[inland otter]

n isnt fixed here, n indexes your sequence.

[tribal moss]

It doesn't look like n is fixed anywhere in that screenshot. And S is not an infinite sequence -- it is a set whose elements are infinite sequences.

I'm not sure of the exact definition of the notation FS in use there; but it looks like it might be the span of S as an F-vector space. Since each element is an infinite sequence, linear combinations of those elements are of course also infinite sequences.

(assuming F is infinite dimensional)
That doesn't make sense -- F must be a field here, and a field always has dimension 1 as a vector space over itself.

[lethal dune]

I wonder which one I have then

[gilded gull]

I'm doing a problem where I need to examine the relation fTg where fTg iff f(3) = g(3) and check if it is an equivalence relation and then describe the set of equivalence classes

Obviously it is an equivalence relation

each equivalence class is then the set of functions that map to a specific y in the cartesian plane

how am I supposed to write this ^ description in set notation

[devout crow]

do you know the definition of an equivalence relation?

[gilded gull]

Yes?

So in this case T has to be reflexive symmetric and transitive

f(3) = f(3)

f(3) = g(3) implies g(3) = f(3)

[devout crow]

ok yeah so what's the question

[gilded gull]

f(3) = g(3) and g(3) = h (3) implies f(3) = h(3)

yeah

How do i write the equivalence classes in set notation

[gilded gull]

^^

[maiden ocean]

People use brackets sometimes

as in [f] for the equivalence class of f

[gilded gull]

is that it

do I have to further elaborate more than: each equivalence class is then the set of functions that map to a specific y in the cartesian plane

[devout crow]

[f] := {g : f(3)=g(3)}

[gilded gull]

o wait so i literally just write

the relation

oml I'm stupid LMFAO

tysm

[devout crow]

np but for future reference this is not really an abstract algebra question

[gilded gull]

o

I wasn't sure just since the basis for it is abs alg but

[delicate orchid]

I mean if it’s part of a question about a quotient of like, continuous functions on R or something, then it’s abs alg

[devout crow]

true I suppose

[gilded gull]

idk its in an abs alg textbook

ik its a really surface level thing i just wasn't sure how to write the set ahaha

[gilded gull]

but also I do have to do this :p

[delicate orchid]

Yeah in general [x] = {x : xRy} for some equiv relation R

[dull ginkgo]

Let A and B be commutative rings
Assume there exists a monomorphism f (injective ring homomorphism) from A to B, and there exists an element x in B that is outside the image of f

Let Hom(B,A) be the set of homomorphisms from B to A, and allow two homomorphisms be equivalent if they map x to the same element of A. If Hom(B,A)/~ [set of equivalence classes] is bijective to the set of elements of A, then does B have to be isomorphic to the polynomial ring of A?

i know jack shit about abs algebra. I thought of this 2 days ago while working.

[warm wyvern]

I know jack shit about abs algebra

[devout crow]

I might be reading this wrong, but what's wrong with taking A=B?

[rustic crown]

if B = Z[x] and image of A doesn't contain x, then Hom(B, A)/~ is always bijective to A

[dull ginkgo]

Injective.

& x outside of the image

[devout crow]

oh there exists, sure

[rustic crown]

oh forgot about injective mb

[dull ginkgo]

I haven't read any book about it lol

I mainly just know basically what a homomorphism is

and fuck around with that

[proud bear]

what about $B=(A\times A)[x]$ or something

[proud bear]

does that work

[dull ginkgo]

no

[rustic crown]

also btw what do you mean by "the" polynomial ring? if A = Z and B = Z[x, y] then also it works right? Hom(B, A) = A^2 after quotienting does become A

[proud bear]

there is an injection A->B. and there always exists a homomorphism mapping x to any element of A so Hom(B,A)/~ should be in bijection with A no?

[dull ginkgo]

ig

Well, the quotient Hom(B,a)/~ is bijective to A^2 instead

which isn't bijective to A i t h i n k

w a it

[rustic crown]

well if A is infinite then A and A^2 are bijective, but here it is A i think

[dull ginkgo]

Assume N and A are bijective

use this bullshit

not explaining this fuckery

Recognize f(x)'s is the sum of n from 0 to N, g(x) is a branch of it's inverse.

god here we go.

You start with asking how many points are drawn using the diagonal argument for every point on the y axis

and every diagonal line has as many points as the y coordinate of it's intercept + 1

(number of elements from 0 to the intercept lol)

fuck it, moving to discussion

[delicate orchid]

King shit

[dull ginkgo]

I did this while at work waiting for an avionics panel I spray painted to dry

[south patrol]

Yh i don't see why just using like R[x,y] over R as a counterexample doesn't settle it

or am i misinterpreting the discussion

lol

[devout crow]

this seems like a "free object" type thing

[south patrol]

Is this question motivated by Z[X] representing the forgetful functor on Ring

[devout crow]

like, you're really asking if we can hit everything in A

[delicate orchid]

R[x] is a pretty free object ngl

[devout crow]

from x

[dull ginkgo]

"forgetful functor" or some dogshit

[south patrol]

yh

Natural bijection between hom(Z[X], A) and the underlying set U(A) of A

But yes

[dull ginkgo]

Aren't rings just a set with two binary operations satisfying axioms, and homomorphisms are SPECIAL functions between the sets.

So why the fuck do we need forgetful functors to make the ring a set.

[delicate orchid]

This is a fairly interesting construction actually

[south patrol]

We don't?

Well

[dull ginkgo]

someone before got pissy at me

[south patrol]

I mean okay depends on what your view is lol

[rustic crown]

we we don't "need" it, but if you wanna talk about function from a set to a ring, it's nicer to put everything in the same "place" and then talk about the functions

[south patrol]

I think the point is more that it allows us to be more clear over which category we are working and things like that

[rustic crown]

also forgetful functors are often not very interesting, what's more interesting is their left/right adjoints...

[dull ginkgo]

Aren't categories just like, an alternative notion of "sets" with a slightly modified inclusion to say that they contain elements that have axioms over it's elements

[delicate orchid]

A category is a graph

[dull ginkgo]

not inclusion

[south patrol]

this also seems to extend to say 'aren't forgetful functors useless' but ye

[south patrol]

I'm not really sure what the second half of this sentence says / intends to say

[dull ginkgo]

nvm

I had a question that I answered myself in my head that I was originally going to say afterward

ignore that

[south patrol]

But yes informally they are collections of objects with maps/arrows between them

[dull ginkgo]

I did some bullshit on my own before which was sorta like

[rustic crown]

bigggg directed graphs >.<

[dull ginkgo]

ZFC y'know

[south patrol]

Sure

[dull ginkgo]

with their two relations between variables, equation and containment

with axioms

[south patrol]

ok

[dull ginkgo]

but I added a third, that's LIKE containment but is slightly different

not reflexive, some other shit that helps allow containment of all forms of sets with structure but without self refernece

til I caused self reference lol

[south patrol]

oop

[devout crow]

so just to kind of formalise this problem: given a commutative ring $B$ and a subring $A \leq B$ and an element $x \in B \setminus A$. the ring $\Hom(B,A)$ has an ideal $I = {f : f(x) = 0}$. then we are asking whether the map $\Hom(B,A)/I \to A$ given by
[f + I \to f(x)]
is surjective

[cloud walrus]

Average J∘du=du∘j enjoyer

[devout crow]

right?

[dull ginkgo]

i gues

This started with essentially this

[devout crow]

so I guess we can extend this to a map Hom(B,A) --> A so we can forget about the ideal

[rustic crown]

Hom(B, A) may not be a ring

[dull ginkgo]

I was working with fields a little bit, and I wondered why so many of their properties related to it's polynomial ring are important especially for shit like automorphism groups

and it's stabilizers when considered as actions on elements in the form of galois groups

then I wanted to consider more shit about polynomial rings,

had an idea while at work

then came across that question I had

[devout crow]

oh because of multiplicative identity issues?

[dull ginkgo]

... what

what the fuck it has structure besides being a set

composition makes NO sense

[rustic crown]

yea and multiplication in general

[dull ginkgo]

crying

[devout crow]

what's wrong with (f.g)(b) := f(b)g(b)

[rustic crown]

no idts

oh i meant f+g may not preserve multiplication

[dull ginkgo]

you can't compose them because the domains and images doesn't line up

[proud bear]

is there even a 0 element if you require that 1->1 for ring homs?

[rustic crown]

(f+g)(ab) = f(ab) + g(ab) = f(a)f(b) + g(a)g(b) and this isn't same as (f(a) + g(a))(f(b) + g(b))

yee Hom_ring is kinda boring

[devout crow]

oh rip

[dull ginkgo]

H O W

What is the operation

can't be composition

[rustic crown]

no 0 😦

and f+g won't be a ring map

[south patrol]

Oh okay fair uh

wait what did i mean

[rustic crown]

Hom_set(a set, a ring) is a ring

[dull ginkgo]

h o w

[south patrol]

Sorry yes confused it w something else that was dumb lol

[dull ginkgo]

Is the operation not composition

[rustic crown]

nope, it's pointwise addition and multiplication

[dull ginkgo]

o

[south patrol]

I was thinking of modules over a given ring lol

oops

[delicate orchid]

How can composition when S -> A not A = S so chain ->->->->... break?! ☝️ 🤨

[south patrol]

indeed

[dull ginkgo]

yeah

[south patrol]

using S and A in the same sentence when S isn't a multiplicative subset of A should be banned

[dull ginkgo]

[proud bear]

lol

[dull ginkgo]

If F is a field or commutative integral domain or whatever semantic bullshit