groups-rings-fields

[void cosmos]

so its like how bad Hom functor is not right exact

so like if P is projective then this would be trivial right

[rocky cloak]

A more illuminating description is that Ext is the set of short exact sequences up to isomorphism.

I.e. Ext(C, A) = {0 -> A -> B -> C -> 0}

(and you can give this a group structure by taking pullbacks and pushouts with direct sums)

[formal ermine]

haha

[formal ermine]

yeah

similarly Tor measures how a module fails to be flat

[rocky cloak]

You can also realize Ext^n as longer exact sequences, but then the equivalence relation is more complicated than just up to isomorphism

[rocky cloak]

You don't like yoneda-ext?

[formal ermine]

"illuminating"

[rocky cloak]

Well I certainly think it says more than "the derived functor of Hom"

It's a very concrete way to see what the elements are and how it acts as a functor

And it generalizes to settings where you don't have enough protectives/injectives

It also gives an immediate interpretation of the connecting homomorphism in LES

[elder wave]

he is talking about his name

[hidden haven]

But then it seems harder to motivate its purpose. Is there a way to understand why that definition would be useful?

[rocky cloak]

I'm an idiot

[hidden haven]

King

[rocky cloak]

Why short exact sequences are useful?

[hidden haven]

No, why that functor would be useful

[rocky cloak]

It describes short exact sequences

[hidden haven]

Lol

Maybe in a concrete setting?

Since this was the original definition of Ext I assume they needed this functor for something

Don't say to describe short exact sequences

[rocky cloak]

Is that not the answer though...?

[hidden haven]

Why would you care about the set of extensions of A by B?

[rocky cloak]

So you can understand how modules break into smaller modules

So you can understand complexes from their homology

[hidden haven]

Maybe I'll need an example lol

[rocky cloak]

I don't actually know the historical motivation, but these seem like natural questions to me:

Given a module B with submodule A and B/A = C, what can B be? When is B unique?

[hidden haven]

I agree with that and see how that is a question you would want to answer if you're working with homology or spectral sequences

[rocky cloak]

Though I guess for things like UCT and algebraic topology stuff it's more useful as a derived functor...

[hidden haven]

But never over the course of doing any of that have I ever felt the need to combine all extensions into a set and work with that set

So why did the people who came up with it feel the need to look at the set of extensions?

[rocky cloak]

Well, it's very useful to have a concept of Ext if you want to compute whether Ext is 0.

Things like LES it's an invaluable tool, I dare say I use it every day

[hidden haven]

Ye sure but working with them usually involves working with one extension at a time (the one you have in the LES) and not the set of all of them

[hidden haven]

This makes some sense yeah

[rocky cloak]

I'm talking about the LES in Ext groups

[hidden haven]

The LES you get from applying Hom to an SES?

For that, I feel like the other definition is more relevant? At least I don't see how this is any more illuminating than being a more concrete description

[rocky cloak]

Well the connecting homomorphism of s= A->B->C is literally given by s through yonedas lemma

[hidden haven]

The way I'm interpreting this is that the set of all extensions isn't too useful, it is the group of all extensions that becomes useful because then you have extra techniques to figure out whether the group is zero or not

[rocky cloak]

I don't see how some general snake lemma stuff is more illuminating than that

[hidden haven]

Ah fair

[hidden haven]

The reason I say this is that I don't see how it makes it easier to prove the uniqueness of something by saying the set of all possible things that do that something is a singleton

Whereas the group structure is actually extra info rather than just a translation of the problem

[rocky cloak]

Sure, but the group structure is there either way. You just take direct sum then push and pull the endpoints back.
I think that's still pretty natural, but here I guess I would understand if you didn't agree

[hidden haven]

Ye I agree with the group structure being natural. I was trying to come up with an explanation for why I would want to define the set of all extensions and I couldn't think of a scenario where it would help, but you have convinced me that at least defining the group of all extensions can be useful

[untold turret]

what's latex for this?

[hidden haven]

#latex-help and check on the detexify website

[untold turret]

detexify returned nothing :/ it is used for the cartesian product applied to self n times

will check out latex help, thanks

[hidden haven]

More standard notation for that would be \prod

[untold turret]

cool i didn't know of \prod, that works too. thanks

[rocky cloak]

I sometimes go with
$$\bigtimes^n M$$

[cloud walrus]

jagr2808

[wraith cargo]

tbh I always think of Tor and Ext as "filling in the rest of the Hom/tensor exact sequence"

[rocky cloak]

Guess I'm in the minority

[wraith cargo]

Also on another note is there some way to see why R^n lim is 0 for n=/= 0,1?

[hidden haven]

What do you mean by other lol what do you have in mind

[rocky cloak]

Hmm, so products are exact, so the limit is just a kernel. But I guess that just pushed the question to why the higher derived functors of kernel vanishes.

In the category of morphisms kernel is just the homfunctor from Z->0. Which has projective resolution (0->Z) -> (Z->Z)

Don't know if that is a very satisfying explanation...

[hidden haven]

That is indeed other

It is very satisfying actually

(for me)

The projective resolution map is supposed to go the other way right?

[rocky cloak]

No

[hidden haven]

Oh no ok this is like the truncated thing

Cool

[white oxide]

ye that makes sense to me, but could we also look at it as we can't permute the other two roots besides cbroot(2) since that wouldn't induce an automorphism fixing Q(isqrt(3))

[rocky cloak]

Well Q(a, b) is by definition generated by a and b, so any homomorphism is determined by where a and b is mapped. The key idea is that the other cube roots of 2 can be written in terms of cb(2) and isq(3)

[white oxide]

oh right

ahh

okay that definitely makes sense now

thanks!

[rocky cloak]

You might notice that G(Q(cb(2), isq(3)) / Q) = S3 is exactly described by permuting the roots of x^3 - 2

[white oxide]

right i think i made that observation earlier

[rocky cloak]

But fixing isqrt(3) fixes how the roots are related, so you can't permute one without the other so to speak

[white oxide]

it would be automatically isomorphic to Z3 because any group of prime order is cyclic right

[white oxide]

yea that makes sense

[chilly radish]

S3 is not of order 3

[white oxide]

Nah I was talking about the question I was working on

I know it’s of order 6

[chilly radish]

Oh

[white oxide]

S3

[chilly radish]

My bad didn't read context

[white oxide]

all g

[wraith cargo]

TIL

[rocky cloak]

Is this functors from I to an arbitrary abelian category or whats the statement here?

[wraith cargo]

any abelian category satisfying AB4*

[rocky cloak]

Hmmm, but if the category satisfies AB4^* shouldnt the argument I gave above show that R^nlim vanishes for n>=2...?

[wraith cargo]

o

[white oxide]

are the notions of a galois extension and a finite normal extension equivalent? fraleigh defines a finite normal extension as an extension E of F such that E is a separable splitting field over F, and a galois extension (according to google) is an algebraic extension E over F that is separable (finite implies algebraic)

[wraith cargo]

then it could be any abelian category?

I'll have to look up the reference

[rocky cloak]

galois extensions are seperable (which finite normal extensions need not be)

[white oxide]

huh my book defined them as separable

[rocky cloak]

They define normal extensions as seperable?

[white oxide]

they only defined finite normal extensions and not normal extensions if that matters

[rocky cloak]

thats weird, I would have thought finite normal just meant finite and normal

[white oxide]

yeah ig not

here's how they phrase the fundamental theorem of galois theory if that's relevant

well not in theorem format

but just a summary

[rocky cloak]

I would have replaced "finite normal extension" with "galois extension", but if thats what they wanna call it...

[white oxide]

yeah he does things a little differently...

galois extension sounds cooler too

finite normal extension is boring

[formal ermine]

what do you call $M^{\odot n} \coloneqq U(n^+) (v_M \tensor \cdots \tensor v_M) \subseteq M^{\tensor n}$

[formal ermine]

what @south patrol ?

[south patrol]

Oh i'm just not sure what the notation means dw lol

[agile burrow]

Context might help

[formal ermine]

definition of a favorable module

CARTAN COMPONENT

that's what it's called

[south patrol]

Cartan component

ye

https://arxiv.org/pdf/1306.1292.pdf

[agile burrow]

The context helped

[formal ermine]

why are they making this so complicated lol

[south patrol]

Ngl idk this stuff lol

beyond basic lattice stuff

[formal ermine]

tl;dr of this is if nP cap Z = n * (P cap Z) then P is normal

[agile burrow]

Lattices seem neat

[next obsidian]

Walter seems neat

[formal ermine]

[south patrol]

Hm is there any pleasant description of the ith exterior power of the jth exterior power of a module

Like if our module M is a vector space of dim n then this new this has dimension ((n choose j) choose i) which is a bit gnarly

[delicate orchid]

I'm positively spooked solid

[south patrol]

Perhaps it is just too gross lol

I guess we get an interesting gnarly bigraded object lol

[delicate orchid]

you might be able to simplify something out of it using third iso maybe

not sure

[south patrol]

Yeah it's just cause the tensor power cause is more nicely behaved for obvious reasons

So I was curious about this

[delicate orchid]

((M^\otimes j/J)^\otimes i)/I lol this is a mess

[lime badge]

If we have a Galois extension $k \subseteq F$ and if E is an intermediate field, I just wanted to check the following argument for $E \subseteq F$ being Galois (seems too simple). Since $k \subseteq F$ is Galois, F is the splitting field of a separable polynomial $f(x) \in k[x] \subseteq E[x]$. So we can also view F as the splitting field of $f(x)$ viewed as a poly in $E[t]$. Doesn't this immediately imply $E \subseteq F$ is Galois?

[cloud walrus]

eternalway

[south patrol]

Yes that is correct

and is the proof I'd give too

[lime badge]

Thanks!

[frigid lark]

F/k Galois does not necessarily imply that F is the splitting field of a single seperable polynomial. F could the the splitting field of an infinite family of seperable polynomials, think Q^a / Q.

[lime badge]

We're only dealing with finite Galois extensions in my book. I should've specified that, my bad

[frigid lark]

Then you should be good

The argument in the infinite case isn't much harder, you just observe that an embedding of F into E^a, is also an embedding of F into k^a, hence the image of F under that embedding is F, giving normality, and if a in F is seperable over k, then Irr(k, a, X) has no multiple roots, but Irr(E, a, X) divides Irr(k, a, X), hence Irr(E, a, X) has no multiple roots hence a is seperable over E.

[lime badge]

What's a good resource to read about galois theory for infinite degree extensions?

[frigid lark]

I'm reading Lang, but for now it hasn't gone too far into infinite stuff

[lime badge]

I see, I'll check it out at some point

[lime badge]

I'm a bit stuck on the following problem. Let $k \subseteq F$ be a Galois closure of $k \subseteq E$. For $\alpha \in E$, prove that $E = k(\alpha)$ iff $\alpha$ is moved by all $\sigma \in \text{Aut}_k(F) \setminus \text{Aut}_E(F)$

[cloud walrus]

eternalway

[frigid lark]

What part do you need help with?

Also by Galois closure of E/k do you mean F is the smallest normal extension of k that contains E?

[lime badge]

Since $k \subset k(\alpha)$ is normal, there's a result I can use that for any algebraic extension $K$ of $k(\alpha)$ and every $\sigma \in \text{Aut}_k(K)$, $\sigma(k(\alpha)) = k(\alpha)$. So I think this gives me the first direction

[cloud walrus]

eternalway

[lime badge]

For a finite separable extension $k \subseteq E$, my book says the Galois closure is the smallest Galois extension $k \subseteq F$ of $k \subseteq E$

[cloud walrus]

eternalway

[frigid lark]

yep same thing

Have you tried applying the Galois correspondence?

[lime badge]

Right. So for the forward direction, since $k \subseteq F$ is a Galois extension, $E = k(\alpha)$ corresponds to the subgroup $\text{Aut}_E(F)$

[cloud walrus]

eternalway

[frigid lark]

For the backwards direction

[lime badge]

This basically gives the forward direction though, right?

[frigid lark]

yep

[lime badge]

As for the backwards direction, if $\alpha$ is moved by all $\sigma \in \text{Aut}_k(F) \setminus \text{Aut}_E(F)$, isn't it fixed by all $\sigma \in \text{Aut}_E(F)$? But as for using this to show E is simple, I was a bit confused

[cloud walrus]

eternalway

[frigid lark]

A bit more nuance there, you would have that it is fixed by only sigma in Aut_E(F)

Hence by the Galois correspondence, k(a) = E

A definition that E/k is simple is that there are only finitely many intermediate fields F/k such that F subset E

But that is simply true by the Galois correspondence

[lime badge]

I see. So if we assume that a \in E is moved around by all automorphisms in Aut_k(F) / Aut_E(F), is the idea that these automorphisms will basically create finitely many extensions of k in E?

[lime badge]

And yeah, I remember the book proving this result now

[frigid lark]

I haven't used the previous result to show that E/k is simple

[lime badge]

Wait, could you break down how you got this conclusion then if you're not using this other result about simple extensions?

EDIT: I think my book's wording is confusing. Are we proving E is simple in the first place or merely showing a generates it since we know E is simple from the fact that k --> E is a finite separable extension

[chilly ocean]

Does anyone know how one might go about selecting a basis for the subspaces corresponding to the irreducible representations? I already have a basis for each of the components of the canonical decomposition that I got using the standard projection operators.

[frigid lark]

So there is a bijection between the subgroups of G(F/k) and the field extensions of k contained in F. Specifically a subgroup H is sent to the biggest field that all elements of H fix, i.e if s is in H, then s restricted to that field is the identity. We observe that the automorphisms in G(F/k) that act trivially on a are also the automorphisms that fix k(a), hence by our assumption the same subgroup of G(F/k) that fixes E also fixes k(a). Hence by the Galois correspondence, E = k(a)

[frigid lark]

We might be trying to prove it by finding an a in E such that a is only fixed by the automorphisms in G(F/k) that fix E

But that proof seems inefficient

[lime badge]

Don't we merely know that there exists a sigma in Aut_E(F) that fixes a? We don't necessarily know that all automorphisms in Aut_E(F) fix a

[frigid lark]

Aut_E(F) means the set of automorphisms of F such that when restricted to E they are the identity

[lime badge]

I'm so sorry, its late here and my brain is fried

That was super obvious

Thanks so much for your help

[frigid lark]

That's alright

[spiral scarab]

So for a quotient ring R/I, I has to be an ideal so that the left ideal is equal to the right ideal, yeah?

Similar to a normal subgroup?

[coral spindle]

Left ideals are equal to right ideals anyway, this is what it means to be a subgroup of an Abelian group

While this is part of what's necessary, it's also important for it to be closed under the action of the ring

this is what allows multiplication to be well-defined

So the left ideal being equal to the right ideal is only half of the story.

[spiral scarab]

well for r in R and i in I r*i needs to also be in I

IIRC

[coral spindle]

yes I did indeed mention that it should be closed under the action of the ring.

[spiral scarab]

Im just trying to understand how for Z/nZ for say n=9, the quotient ring leaves us with {1,2...,8}

i understand that an integer modulo 9 isn one of the elements in that set

I just dont understand how the quotient gives us that set

[delicate bloom]

so you are thinking about the cosets, do you understand why the cosets 2+9Z and 11+9Z are equal for instance?

[coral spindle]

It doesn't give us that set, for two reasons!

Firstly It gives us the set {k + nZ | k in Z} rather than a set of integers. This is the definition of the quotient ring. We can choose particular values of k that represent every element of this set, giving us {0 + nZ, 1 + nZ, ..., (n-1) + nZ}. We could equally have chosen any of the infinitely many choices for values of k that produce all the elements of the set, but this is a nice choice.

We often just write 0, 1, etc instead of 0 + nZ, 1 + nZ for brevity. Note that you actually left out the set 0 + nZ = n + nZ in your list.

[spiral scarab]

Is it because we are partitioning the ring into ideals?

so like we are factoring a polynomial

in that sense?

[coral spindle]

We are not partitioning the ring into ideals. You are partitioning the ring into cosets of the ideal.

This part works exactly the same as in quotient groups.

Perhaps Mero knows what you mean by your comment about factoring polynomials, but I certainly have no idea.

[spiral scarab]

i mean in the same way integers can be factored into primes

[coral spindle]

I still have no idea how that's relevant.

I dare say it is not.

[summer path]

there is a related analogy with polynomials, but it is really not relevant in this context

[coral spindle]

Can I take this to mean that you do not see why 2 + 9Z = 11 + 9Z?

[spiral scarab]

It "rotates" the elements of Z9 so that we get the same set?

In a way

[coral spindle]

Sure

Here's a very useful fact.

[delicate bloom]

the polynomial thing sounds like you have a more complicated idea of what it "should be" in your mind luci, and that's crowding out understanding the concept as it really is, which is a different thing altogether

[coral spindle]

If J is an ideal, then a + J = b + J if and only if a - b is in J. You should try proving this.

[spiral scarab]

so we are using the ideal to split up the ring into cosets, and these cosets form a new ring?

[coral spindle]

Yes.

[tender wharf]

it's like a factor group lol

concept extended to rings

[spiral scarab]

is it because a-b congruent (mod J)?

[coral spindle]

congruent to what

[tender wharf]

J isnt a number lol

[coral spindle]

The definition of a being congruent to b mod J is a + J = b + J

[coral spindle]

We still talk about things being equal mod an ideal; this is fine.

[tender wharf]

oh, cool to know

[spiral scarab]

what about the chinese remainder theorem

i remember that one

[coral spindle]

My point is, luci, you should prove from the definitions of a + J and b + J that a + J = b + J iff a - b is in J.

This is a fact that only requires group theory, actually.

[spiral scarab]

I think I understand how the ring is being split up into cosets now. So going back to Z/9Z, the {1,2...,8} is just repeating for the numbers 1-9, then 10-19 and so on. And those are the cosets the ring is being split up into?

But we are just writing it once, since its just rotating?

[coral spindle]

Sure

You're still not writing 0 in that set, like I mentioned. It's {0, ..., 8}.

[spiral scarab]

My bad, forgot about the multiples of 9 lol.

So then we took the integers modulo 9, which is an ideal of Z, and then using that we split up Z into cosets which gave us the new ring Z/9Z?

[coral spindle]

The integers mod 9 is the ring Z/9Z

You're thinking of the multiples of 9, i.e., 9Z.

Apart from that slip of terminology, yes.

[spiral scarab]

Oh, sure, thanks. I think I got it now.

🙏

[untold turret]

what's the correct formal structure of a free Z module generated on an abelian additive group Z^n? it's not just strictly a free module because addition between elements of Z^n is already defined but the free Z module on the set L of n letters for instance doesn't have this

[rocky cloak]

$$A \mapsto R\otimes A$$

[cloud walrus]

jagr2808

[untold turret]

oh it's a tensor product, got it. thanks a lot!

[rocky cloak]

Wait, I think I misunderstood what you're asking. What are you asking? How addition is defined for free abelian groups?

In case that's what you're asking: the free abelian group on a set X is usually defined as the set of formal linear combinations of elements of x. And addition is defined as one would expect.

You can also define this as the set of functions X -> Z with finite support

Then addition is just pointwise addition

[south patrol]

I'm a bit confused what the question is too lol

[untold turret]

i'm confused too let me think in my corner thanks a lot for the response sorry for the muddled question

[south patrol]

wdym by "Generated on an abelian additive group"

Like you usually have the free Z-module on a set

[untold turret]

i didn't know how to ask the question but the solution is that "free Z-module on n letters" is a functor composition Set -> Ab -> Free Z-mod with the last one being an isomorphism

[coral spindle]

So you just had to recall that Z-modules are exactly the Abelian groups, is that it?

[south patrol]

Uh I disagree with that

What is your functor Ab -> Free Z-Mod

[rocky cloak]

There isn't really a natural functor Ab -> Free Z-Mod

[coral spindle]

I think this is just the inclusion Free Ab -> Free Z-mod

[south patrol]

the "free Z-module / abelian group on a set X" construction is just a functor Set -> Ab (or Z-Mod)

[untold turret]

right i'm hopelessly confused, i'm trying to figure out how it is possible that the quotient of free abelian groups of same rank generated on subsets of Z can be nontrivial

[rocky cloak]

Like how Z/2Z is nontrivial?

Even though 2Z is isomorphic to Z, so both free

[untold turret]

right, i'm trying to figure out what exactly is the difference that makes them be able to produce a nontrivial quotient

[rocky cloak]

Difference between what?

[coral spindle]

It's the way one is embedded in the other

[untold turret]

if the highest "equality" we can find between them is isomorphism, and yet they are isomorphic, where are we getting the nontrivial quotient?

[coral spindle]

The embedding of a submodule in a module determines the quotient

It is not enough to know what the modules are up to isomorphism; you must understand how they interact.

[rocky cloak]

They are isomorphic, but not equal

[coral spindle]

Of course this holds for groups, ideals, whatever you want to say.

I'm saying here that this is a general phenomenon and by no means is specific to free modules.

[untold turret]

am processing what you all have said, thanks a lot for the comments

[untold turret]

let $G$ be the infinite abelian group generated by the set $S={ (0,0), (2,3), (-2,-3)}$ with addition as in $\mathbb Z^2$, and consider the functors $F={f_i}_{i \in I}$ with domain powerset $P(\mathbb Z^2)$that have $G$ in their image. in some sense, there is a subset $F' \subset F$ of these functors that take an argument in $P(S)$ and sends them directly to $G$ in the most natural way, i.e., they generate $G$ by completing the generating set ${(0,0), (2,3), (-2,-3)}$ by adding the identity or the inverse if it is missing. how would you give a formal description of the functors in $F'$?

[cloud walrus]

nrs

[coral spindle]

Are you sure you mean for these to be functors?

It's not clear what these functors (functions?) are, even, but you say the functors yet you don't define them.

I personally have no clue what you mean.

[untold turret]

i want to get at "add additive identity and inverse if either one is missing"

[coral spindle]

I still don't know what you mean.

Add additive identity and inverse to what

[south patrol]

Yeah i don't udneratdn any of this either

[coral spindle]

Can you maybe put this in simpler terms.

[untold turret]

for instance (2,3) gets mapped to the abelian group who's elements are (0,0) , (2,3), (2,3)+(2,3), (2,3)^+3, ..., (-2,-3), (-2,-3)^+2, etc.

[south patrol]

Wait okay so like

What do you mean by functor nrs

[coral spindle]

Functor was completely the wrong name

This is just a function P(Z^2) -> {subgroups of Z^2}

Taking generators to the subgroup they generate

[south patrol]

So what does the notation {f_i} mean

[coral spindle]

I'm gonna ignore that part because it makes no sense lmao

OK, so what's your question then, what about it

There sure is a function sending subsets of a group to the subgroup it generates.

So what's the question

[untold turret]

well, isn't there more than one way to generate an abelian group from a set?

[coral spindle]

Yes.

The function I describe is not injective.

[coral spindle]

@coral shale The first ideal is zero iff the skew field is a field. The second ideal always contains a nonzero element, hence is the whole skew field.

[coral shale]

[coral spindle]

Remember: the only ideals of a skew field are 0 and the whole skew field. This doesn't depend on sidedness. Try proving this.

[coral shale]

Having not done non-comm alg, but anyways, I should've known this

[untold turret]

edit: i understood, thanks

[coral spindle]

Well if you have more questions, just ask.

Try to ask the actual question at hand instead of a fancy version of it. It's twice today that you've asked an unclear question, only to follow it up with a much simpler one!

[untold turret]

thanks a lot very appreciated

[real sparrow]

$$A\Bowtie B$$

Might be a good notation for Zappa–Szép product

[cloud walrus]

s5

[real sparrow]

Also im not sure if all of these products can be derived from an automorphism of AxB

$$\varphi:\text{Aut}\left(A\times B\right)$$
$$A\Bowtie_{\varphi} B$$

[cloud walrus]

s5

[real sparrow]

It probably more defines a specific type of Zappa–Szép product

$$\varphi:A\to\text{Aut}B$$
$$A\ltimes_{\varphi} B$$

[coral spindle]

Having fun there mate?

[real sparrow]

?

I'd like help understanding the Zappa–Szép product

[coral spindle]

Looked to me like you were just writing TeX

Why don't you ask your question explicitly and someone might be able to answer. It gets lost if you keep getting the bot to render your code.

[cloud walrus]

s5

[real sparrow]

@coral spindle what is the Zappa–Szép product

[coral spindle]

I don't know anything about the Zappa–Szép product.

I advise you not to ping specific people. Ask a question generally, and people who know will respond.

[real sparrow]

You brought it up so I thought I ping you.
And other people can see it as well.

[delicate orchid]

you brought it up

this is the only time boytjie has ever mentioned this product

[real sparrow]

You brought up me asking a question.

[summer path]

instead of having this meaningless conversation, why don't you just do as boytjie asked and ask whatever question you were going to ask originally, in one message

[rocky cloak]

If you have two subgroups H and K of group such that HK=G and they don't intersect. That's a Zappa-Szep product.

In particular this means that for k in K and h in H there are h' and k' such that kh = h'k'

So you can define it externaly by a function KxH -> HxK satisfying the necessary relations

[real sparrow]

I did

[real sparrow]

I mean exterior.

[real sparrow]

What relations.

Would it be equivalent to AxB homomorphism to Aut(AxB)

Or is it Aut AxAut B

wich would be less general.

[rocky cloak]

No, the group structure of AxB doesn't really relate to their Z product. So you shouldn't expect it to apear

Wikipedia lists the relevant reiterations

[real sparrow]

Ok

$$\left(\text{Aut},A\times\text{Aut},B\right)\hookrightarrow\text{Aut}\left(A\times B\right)$$

[cloud walrus]

s5

[lethal dune]

they got a point ngl

[real sparrow]

Me and my type theory mind sees this as a statement that the left is a subgroup of the right.

because spaces/sets are propositions.

[untold turret]

would i be correct in saying that the general form of this problem is solved by quotients of free groups?

[real sparrow]

$$A\hookrightarrow B$$
$$A\twoheadrightarrow B$$

[cloud walrus]

s5

[real sparrow]

subgroup and quotient group

A is a subgroup of B
B is a quotient group of A

could also say A is a multiple of B

[celest furnace]

You trolling?

[real sparrow]

No why

[coral spindle]

I don't see how that's relevant really, neither do I really see what the 'problem' is. Every group is a quotient of a free group, and the fact that there is not a unique generating set for a group is not a problem at all.

[untold turret]

take a set S and add additive identity and inverse if either one is missing is the same as taking the relevant quotient on the free group on S, no?

[coral spindle]

No not at all

I cannot see why you'd think this is the same thing

The group generated by S will be, but this has nothing to do with including the identity and inverses

You don't need to include these things in a generating set; you get them for free when you see what they generate.

[untold turret]

but isn't "generate the group" formally given by taking the quotient on the free group on S?

[coral spindle]

No, it's not

We can only generate a subgroup of a group.

[cloud walrus]

boytjie

[coral spindle]

Note that this comes with information about how the subgroup embeds in G!

[cloud walrus]

boytjie

[coral spindle]

But this doesn't have the information about how <S> embeds in G without this.

This is a different thing for that reason!

[untold turret]

i see, thanks a lot

[coral spindle]

So generating the group can be seen as taking the image of this map

and this, being a subgroup of G, will come with information about how it embeds

[untold turret]

i see i see

[glossy crag]

Crossed products

[formal ermine]

what is the difference between a free vector space and a vector space?

[coral spindle]

All vector spaces are free, so as it happens there is none.

A free module is a module that has a basis. There are rings, believe it or not, for which not all modules are free (shock horror I know) so it does require some work to show that there is this correspondence.

But indeed, every vector space has a basis.

[formal ermine]

yeah I know that

like you said every vector space is free

so I was confused if there is a difference

[coral spindle]

OK

[tribal moss]

Indeed, all vector spaces are free, so saying "this is a free vector space" doesn't convey any more information than "this is a vector space".
But "this is the free vector space on such-and-such set" is meaningful.
(And omitting "free" in the last phrasing would yield word salad; it's the word that describes the relation between the vector space and the indicated set).

[formal ermine]

oh, what does the latter sentence mean?

[tribal moss]

The free vector space on a set A is the set of formal linear combinations of elements of A.
Or (equivalently up to isomorphism) the set of functions f: A -> R such that f(a)=0 for all but finitely many a.
Or (equivalently up to isomorphism) any vector space with a chosen basis and a chosen bijection between that basis and A.

[coral spindle]

You could see it as saying that the dimension is the cardinality of the chosen set.

[formal ermine]

ah I see

my ta said something like "a vs V is reflexive if V** = V (dual of dual is itself)" but I can only find func anal definitions of reflexive online

did he perhaps mean something else?

[coral spindle]

No I reckon he probably meant that

That isn't standard terminology though, not for vector spaces in the abstract

He will likely soon prove that a space is reflexive iff it is finite-dimensional

[formal ermine]

he only said it to prove something else

this is a class on geometry

[formal ermine]

right

[hidden haven]

That's only the standard definition when V* refers to the vector space of continuous functionals on V

V needs to be an inner product space in order to take this kind of dual twice

Or at least normed ig

With this definition you can have infinite dimensional reflexive spaces (eg Hilbert spaces should be reflexive because of Riesz)

That's why you can only see funky anal definitions. If you remove the continuity requirement, this isn't very useful as a definition because of what bowtie said

[coral spindle]

bowtie

Yeah so as I thought, context was being missed.

It wasn't the dual but the continuous dual that mattered.

[hidden haven]

I have never properly read your name and I shall maintain this trend

[coral spindle]

Hey I wasn't complaining!

[hidden haven]

[rocky cloak]

You don't need more than a topological vector space, no?

Seems, you do. The topology on the dual space usually used in this context (the strong topology) requires more data than just the topology of V it seems.

[coral spindle]

Right, you'd have to have a topology on k too

Or right, you're saying we need to give a topology to the dual itself. I see.

[untold turret]

i finally fully understood, i think, what you were saying: taking the quotient of a free group "generates" a group only up to isomorphism

[coral spindle]

Yes.

[untold turret]

ok i understand now, thanks a lot for your time

[coral spindle]

I believe that Clifford's theorem excludes a normal counterexample.

Briefly: when we restrict a kG-module to a kH-module, pick some simple kH submodule. Then by normality, the distinct translates of this simple submodule will be simple kH-modules, and the sum of all of these will be the whole of the module. A sum of simple modules refines to a direct one always.

[coral spindle]

Oh ofc to be clear, we're restricting a kG-module which is a direct sum of simple modules, so we can restrict our attention just to simple kG-modules.

[untold turret]

we consider subgroups of $\mathbb Z^2$. consider $G_1$ generated on $(2,0), (0,3)$ and $G_2$ generated on $(2,0), (0,1)$. we consider their Z-module structure to represent the group homomorphism $G_1 \to G_2$ by a matrix. clearly $G_1 \to G_2$ is represented by a matrix with integer entries, but shouldn't there be a matrix with integer entries representing $G_2 \to G_1$? the matrix for $G_1 \to G_2$ is $\begin{bmatrix} 1 & 0 \ 0 & 3 \end{bmatrix}$ but its inverse has rational entries

[cloud walrus]

nrs

[coral spindle]

So you say the group homomorphism as if there is only one

but there could be many. It seems you've chosen one, which is fine.

[untold turret]

oh! the matrix has an inverse if it is an isomorphism

[coral spindle]

Anyway, in general there will not be an inverse for a matrix of integers, even if the determinant is nonzero, as you have discovered. For that reason typically there will not be a matrix with integer entries which is inverse.

Yes.

[untold turret]

i see! thanks a lot

[wooden ember]

if you want a matrix inverse with integer entries, a necessary and sufficient condition is for your determinant to be +-1

[balmy belfry]

Matrix in abs alg chat?

[wooden ember]

(this is in general true over a PID, unit determinant corresponds to invertible matrices over your ring)

[coral spindle]

Ooh linear algebra is a subfield of abstract algebra so spooooky 👻

[untold turret]

woah strong theorem, will work on that, thanks

[wooden ember]

can occasionally come in handy

[rocky cloak]

true over any ring right? at least commutative

[south patrol]

Adjugate matrix :)

A adj(A) = det(A) I, so if det(A) is a unit A is invertible; converse is trivial

[rocky cloak]

Seems you can generalize it even to noncommutative rings, but it gets hairy: https://ncatlab.org/nlab/show/quasideterminant

[wooden ember]

hurb yeah not sure why i was thinking of PID's

[chilly ocean]

maybe you were thinking of smith normal form

[wooden ember]

i guess i always assume you need to be over a PID for matrix stuff to work nicely idk, got confused

[untold turret]

let $\phi$ be a map from free $\mathbb Z$-module on cartesian product $\prod_{i \in \mathbb N}\mathbb Z/i\mathbb Z$ to $\mathbb Z$-module direct external sum $\bigoplus_{i \in \mathbb N} \mathbb Z/i \mathbb Z$ defined by $z_0(z_1, z_2, \dots) \mapsto (z_0z_1 \mod 1, z_0z_2\mod 2, \dots)$, that's an isomorphism right?

[cloud walrus]

nrs

[south patrol]

Not sure what your notation z0(z1,z2,...) means there like

Wait "free Z-module on the product" must be a typo

[untold turret]

it's the free Z-module on the words generated by the cartesian product on Z/iZ with i in naturals N

[south patrol]

Why use that notation then? The free module on the set just ignores all the group structure

Unless you're using a non-standard meaning of free Z-module on a group or smth

[untold turret]

cartesian product has no structure no?

[south patrol]

Okay you mean the product of sets then sure

[untold turret]

so its just a set of words like any other but i might be confused

[south patrol]

Well it's not a well-defined map because like

e.g. (1,1,1,...) isn't an element of that direct sum

[untold turret]

but (1,0,...) is not in that cartesian sum

i think

[south patrol]

If you mean (1,0,0,0,...) then that is in the cartesian product

[untold turret]

1 is not in the set of the group quotient Z/Z, no? or am i doing a full derp

[south patrol]

I assumed n started at 0 lol

But also like 1 = 0 in Z/Z

so this is tangential

Replace it by (0,1,1,1,...) if you wish

[untold turret]

ok but do i have an isomorphism then? max confusion

[south patrol]

No, because your map isn't even well defined

Also, more seriously, $\bigoplus_{i \in N} \mathbb Z/i\mathbb Z$ is countably infinite, whilst $\prod_{i \in N} \mathbb Z/i\mathbb Z$ is uncountably infinite, even before you take the free $\mathbb Z$-module on it

[cloud walrus]

potato

[formal ermine]

posting screenshots because I'm too lazy to tex

nu : C[x]\0 -> Z

let this be a valuation

define these

then isn't C[x]_m/C[x]_>m just (C \ 0)x^m

[untold turret]

isn't 0, 0, 1, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 5, ... a sequence bijective on \prod Z/iZ? but thanks for the well-definedness comment, i think a slight tweak can fix it

[formal ermine]

and thus the associated algebra is C[x] \ 0 again?

[south patrol]

What do you mean a bijective sequence

lol

I am confusion

[untold turret]

a bijection N -> prod Z/iZ

[south patrol]

Such a bijection again can't exist by cardinality reasons

[untold turret]

why not? the index i is in N/{0}, so what element does the above sequence not cover?

OH

i understood

i think

big mistake of mine

[south patrol]

I'm confused what you mean by 0,0,1,0,1,2,... being a bijection

[untold turret]

right, it doesn't make sense

thanks a lot for the help

yeah you are super right, thanks again

[formal ermine]

bump

[real sparrow]

$$\langle a,,b,|,a^2=e,,b^3=e,,ab=b^-a\rangle$$

[cloud walrus]

s5

[real sparrow]

S3

Smallest non abelian group

e, a, ab, abb, b, bb

[empty rose]

do you have a question about it...?

[real sparrow]

No

[untold turret]

it's abstract algebra so i hereby formally allow it

[real sparrow]

$$\langle a,,b,|,a^3=e,,b^7=e,,ab=b^2a\rangle$$

[cloud walrus]

s5

[real sparrow]

Frobenius group 21

Smallest odd order non abelian group.

[untold turret]

indubitably

[real sparrow]

I think it's neat

Because order 15 doesn't work even tho it's 3x5

You need 3x7

Due to C7 having a automorphism group of C6, C3 as a subgroup

S3 being C2xC3 semidirect product for a similar reason.

Automorphism group of C3 being C2

[real sparrow]

factoring a finite group can be shown with a graph, where each point is a finite simple group, where arrows show where each group maps to what automorphism of the other.

no arrows just say that the relation between them is a direct product in a sense.

[untold turret]

is there a useful structure theorem for modules over dedekind domains?

[maiden ocean]

https://en.wikipedia.org/wiki/Dedekind_domain#Finitely_generated_modules_over_a_Dedekind_domain

[untold turret]

thanks a lot!

[maiden ocean]

np

[frigid lark]

Excercise 11-13 from chapter 3 of Lang deals with projective modules over Dedekind rings if you are interested

You can show that a projective finite module over a Dedekind ring is isomorphic to the direct sum of ideals of that ring

[untold turret]

very cool stuff, thanks a lot!

[rocky cloak]

Yes, so the associated graded algebra is C[x] again. (It's no longer an algebra if you remove 0).

[frigid lark]

I am still unsure of what A_N is defined as here?

Is it the subgroup of A generated by elements of period N?

[rocky cloak]

Yeah, that seems to be what it is, if you look at the example

[frigid lark]

ah, so it's not common notation

[rocky cloak]

I don't think its necessarily uncommon, but the notation A_N is used for quite a few different things

[frigid lark]

Alright

[formal ermine]

ah right I forgot that in the direct sum we can choose all elements to be zero

thanks

[glossy crag]

I know that the Brauer group of a finite field is trivial, but I also just read that the same is true for any algebraic extension of a finite field. Is there a simple way of deriving this from Wedderburn's theorem or does one need something more?

[formal ermine]

why eisenstein

it just needs to be the root of a polynomial

you can trivially find a polynomial which has that number as a root

then look at its factorization

[tribal moss]

The algebraic numbers are closed under addition and taking roots.

[formal ermine]

what polynomial did you get

you could try reduction mod p

[tribal moss]

Hmm, this might lead nowhere, but: Q[11^(1/7)] obviously has degree 7 over Q, so if you can show that Q[your number] has degree 11 over Q[11^(1/7)], then no integer polynomial of lower degree than 77 can have it as root.

[void cosmos]

yo

are bimodules important

[formal ermine]

what is a bimodule

[void cosmos]

let R and S be rings

A is a bimodule given its a left R module and a right S module

[elder wave]

Only seen them in the context of TP over noncomm rings so far

[delicate orchid]

they're analogous to bisets for group actions, which are very useful

[void cosmos]

fuckem

[delicate orchid]

so I imagine they're quite important

[void cosmos]

fuckj

[delicate orchid]

or maybe they're just useful for me dunno how many people work with biset functors

[lethal dune]

an integral unimodular even quadratic form has index divisible by 8, what's a short proof of this?

[delicate orchid]

wtf

[rocky cloak]

Any R module M is naturally an End(M)-R bimodule. This comes up a lot for me as a tool to understand End(M).

For example tilting modules, which characterizes derived equivalences, are bimodules in this way.

Bimodules also apear in Moritas theorem, about how every Morita equivalence is given by Hom/tensor product with a specific type of bimodule.

Hochschild cohomology is the derived functor of the functor sending a bimodule to the subspace where rm = mr. So bimodules appear there too.

[rocky cloak]

A Z/5[x] module is exactly a vector space over Z/5 with an endomorphism describing the action of x. So you can describe it by a square matrix over Z/5. Two such are isomorphic if the matrices are similar (do you see why?).

So the question is can you find two 2x2 matrices that are not similar

[feral agate]

I am trying to understand the proof of that the product of disjoint cycles being equal to a permutation is unique up to the number of disjoint cycles. I don't understand why the claim in blue is true. In particular, why $c_i(x) = c_j(x)$ implies $c_i^m(x) = c_j^m(x)$

[cloud walrus]

FrankF

[tribal moss]

One has the property that X³·v = 0 for every v in the module, the other doesn't.

Wait, as Z-modules? That just means their additive groups are isomorphic, which they must be because they are vector spaces of the same dimension over F_5.

[rocky cloak]

You can use linear algebra as I first suggested, or note that both are generated by 1, so you can use that to construct all homomorphisms between them

[formal ermine]

do you know what we call the quotient in english?

in german it's sheet/leaf

[rocky cloak]

Idk

[formal ermine]

also what is the total order on Z^n

is it just by sum?

[elder wave]

Lexicographical probably

[formal ermine]

oh lmao I'm stupid

it says

"let blabla be the minimum valuation wrt a total ordering on Z^n"

then the first exercise is "show that its a valuation if we choose the hom lg ordering"

[void cosmos]

just wanna saay

Hom(Sum Ai , B) ~= Product Hom(A_i,B)

is one of the coolest yet easiest shit ever

[slim kayak]

Someone is gonna love category theory

[void cosmos]

yea idk but this is just weird and magic af

how simple the proof is and like yea obviously this follows from the universal properties

but at the same time the first time i saw this i was just okay this is going to be rough

[formal ermine]

what's the proof

and what structure

[void cosmos]

over modules

[slim kayak]

I heard it described once as:
So abstract that there is only one real thing to try

[void cosmos]

the isomorphism is that you look at the tuple of g_is , those are maps that are being sent to B

for each i

so you the universal property gives you a map from the direct sum to B

which is an element of the left hand side

so thats ur isomorphism

you send each element in that product to the map it induces from the universal property

[formal ermine]

which uni prop

[void cosmos]

thats it lmfao

of the direct sum

[delicate orchid]

of the direct sum

[formal ermine]

what's it

[delicate orchid]

just google the diagram

[void cosmos]

the direct sum is like the product

[delicate orchid]

it's the coproduct so yeah they're dual

[formal ermine]

too lazy I have a slow phone

[void cosmos]

it comes equiped with inclusions such that any fake maps trying to be inclusions u cna get there the same way

"thats how i remember it "

[delicate orchid]

what

[void cosmos]

forget it its just my intuition

im sure u have ur fucked up intution for other things as well

right 😄

[slim kayak]

Thingy goes in vs thingy goes out is better

[delicate orchid]

I don't have intuition for this stuff I bash definitions together in my head

[void cosmos]

wow

so good

i wanna learn about other things likke Hom

[delicate orchid]

functors?

[void cosmos]

and what properties they have that would be cool

[slim kayak]

The functor?

[void cosmos]

but i dont see any motivation

or like what problems does this solve

yea ig functors

[slim kayak]

Homological algebra is pretty useful

[delicate orchid]

^

[void cosmos]

yea but honestly i see no fun in the problems

they are just boring details for me

the theorems and the definitinos are the cool part

weirdly

[delicate orchid]

also if you have any object that's like, not 1 dimensional (like a 1-category), any morphism between these objects will be analgous to a functor

[slim kayak]

Uh, then idk. You wanna learn category theory for its own sake?

[void cosmos]

idk

[slim kayak]

Or do you wanna use functors and limits, all that stuff

[void cosmos]

the only math that interests me now is number theory honestly

[delicate orchid]

chain complexes come to mind, not quite a functor but you have a similar commutative square thing going on

[void cosmos]

but im studying hungerford for an exam

[slim kayak]

Algebra book, right?

[void cosmos]

yes

and it had this out of the blue sction on Hom

and its cool

proving left exactness and this property

[slim kayak]

That's like a free demo of homological algebra

[void cosmos]

also it was funny for me proving that given f: A--> B induces a surjective map Hom_R(P,A) --> Hom_R(P,B) is qequivalent to P being projective

[slim kayak]

Do you like going through diagrams too? Element wise?

[void cosmos]

i thought this was going to be hard but then its literally the definitoin

[void cosmos]

when its easy

like i did the five lemma it was cool

first experience of me doing diagram chasing

but then it got harder and just weirder

i hate it when im doing problems and im just mindlessly unfolding definitions but i have no plan

and at the same time i get there

[slim kayak]

That's all math tbh

[void cosmos]

it feels like trash

[slim kayak]

First five messages*

[void cosmos]

yea

i think analysis was the math where i can have an actual plan

and it can actually work

[slim kayak]

Maybe geometry?

Algebraic topology ans algebraic geometry are drowning in quick and easy but stylish category theory stuff

[void cosmos]

yea i looked up itno AG and for me it was just so much theory

and that was good but then i asked what problems couldnt be solved without this new theory and now it could be

and it was just " some number theory problem " ( weil )

XD

i was like wheres the geometry in this

[delicate orchid]

yeah the geometry is in the wacky Zariski topology and the funny schemes

did you get to varities

[void cosmos]

yea thats the intro

and it was actually ssuper cool

but then it got to like nakayama';s lemma and commie algebra basically

andd i was just like yeah i saw this movie b4

and i ignored it

[delicate orchid]

yeah skip that nonsense

[void cosmos]

nah i meant i ignored the whole thing

i cant do AG without commie algebra

ig

[slim kayak]

Communist algebra

[void cosmos]

is it normal for you guys

that some problems u just mindlessly unfold the definitions

and it suddenly solves itself out

but u had no plan ? u were just "investigation"

[delicate orchid]

those are the best problems

[slim kayak]

Me trying to understand the proj construction:

[void cosmos]

what proj construction?

projective modules?

or whats that

[slim kayak]

Making projective schemes out of a graded ring

[delicate orchid]

wtf

[slim kayak]

?

[formal ermine]

ong

@void cosmos how is the learning going

[slim kayak]

Bro, you gotta work on your communication skills. Why you so confusing 😭

[formal ermine]

wym

[slim kayak]

Your construction

[delicate orchid]

what

[formal ermine]

oh lmaoooo

[formal ermine]

unaware

[delicate orchid]

aware

[formal ermine]

I don't like projective schemes either

[void cosmos]

im in a league game now but its going great but slow

[slim kayak]

Apparently I will see them a lot moving on. But I don't think gluing affine schemes will get me far

[void cosmos]

i will never know what a scheme is..

its so advanced

its literally 80s math

so its must be op

[formal ermine]

it's like a manifold

[slim kayak]

Honestly word problems in combinatorial group theory have been worse

Or well, at least better at making me feel like I suck

[void cosmos]

any problems in combinatorics are just a remainder

that im nto good enoug

lmao

[formal ermine]

toric geometry is worse

and newton okounv theory

that's stuff from 2013

[vital forge]

hello

?

anybody there

?

[slim kayak]

Like, toric varieties? I think one of our introductory courses introduced them before schemes even, why are they that bad?

[formal ermine]

the geometry side is annoying

so many defns

[slim kayak]

Classic alg geo moment

[coral spindle]

Oh no so many definitions!

Anyway a solvable group is one whose derived series reaches the trivial group after a finite number of steps, where the derived series is def—

[formal ermine]

so true king

[fast stratus]

i'm confused, how is this a isomorphism? don't we need to show $\phi(ab) = \phi(a) \circ \phi(b)$

Here, they showed $\phi(ba) = \phi(a) \circ \phi(b)$

[cloud walrus]

Helium

[fast stratus]

it makes sense if x maps to x^-1... i guess, it's a mistake

[agile burrow]

You're correct, the map should send x to x^-1

[real sparrow]

Group theory should be called Automorphism theory.

[agile burrow]

What makes you say that

[real sparrow]

Small joke that's slightly serious

[agile burrow]

Good one

(I don't get it)

[summer path]

(I also don't get it)

[next obsidian]

Hi Walter

[slim kayak]

(The mathematical community wasn't ready for such humor)

[agile burrow]

Hi chmonkey

I'm still learning rep theory of sl(3, C)

[next obsidian]

Swag

Idk how hard that is

Give it ur best

[agile burrow]

Thanks. It's not too bad since everything's relatively concrete, but it introduces most of the machinery to do semisimple Lie algebras in general

[south patrol]

A group is just a groupoid w one object is the meme i assume idk

[next obsidian]

Just remember to try ur best and have fun :)

[frigid lark]

Cause all finite groups can be represented as the automorphisms of a field extension of Q that fix Q

[white oxide]

Why are Lie algebras cool and why do ppl study them

[tribal moss]

Chiefly in order to study Lie groups, which are groups that have a topology making them differentiable manifolds.
The associated Lie algebra is the tangent space at the identity, with a Lie bracket that encodes enough information about the group operation to reconstruct almost the entire thing. Since a Liee algebra is in particular a vector space, it allows using linear algebra to express information about the original group.

(TL;DR) an element of a Lie algebra represents the derivative of a curve that takes values in a group.

[white oxide]

oh that sounds like it has something to do with representation theory

in particular the part about using lin alg to understand the original group

[chilly ocean]

"If f is a homomorphism from a ring A to a ring B, not necessarily onto, the range of f is a subring of B."
What does "onto" mean when used like this? That the function is not necessarily surjective?

[tribal moss]

Yes.

[chilly ocean]

thanks

[coral steeple]

Am I allowed to use inverses to prove this? It seems like it might be circular, since inverses are defined in terms of the identity

[cloud walrus]

person2709505

[open sluice]

what is your definition of subgroup, incidentally

[coral steeple]

A subset that is closed under the parent operation containing the parent identity and every element's inverse

The problem I have is that if you remove the identity requirement then the inverse criterion needs some interpretation

[open sluice]

I had that problem before

I think of it like this: your elements of H all live in G, so they will obey the laws of G
they just have some additional neighborhood rules on top of that

so every element does indeed have an inverse, or else G would not be a group

[cloud walrus]

person2709505

[open sluice]

think about what it means for H to be closed under multiplication
now think about how you'd prove uniqueness of the identity in your typical group setting

(just throwing ideas out there, I haven't completely thought this through)

[coral steeple]

I guess since we want H to be closed, and we don't know a priori 1_G is in H, $aa^{-1_H}=1_H$

[frigid lark]

Could someone help me with the surjectivity of the map in part c

[open sluice]

I thought about it more and I think I may better understand the point of confusion now

do you think the inverse operation of H is the same as that as in G

or would you allow it to be different

well is there an inverse operation here

[coral steeple]

It would be nice if it were the same but it would make more sense if it were different lol

[white oxide]

How do y’all find interesting topics of research in algebra? Do u just read random papers, or talk to professors, etc

[summer path]

#1100503863586455632

[summer path]

Talk to grad students, or go to office hours and ask your professors

Former if latter seems too hard

[rocky cloak]

You certainly don't need inverses to prove this.

In fact the identity has a unique identifying property that doesn't include any other element.

[hidden haven]

Not sure what you mean but I think you do need inverses because this is not true for monoids. ℤ/6ℤ is a monoid under multiplication and has a subsemigroup {0,2,4} in which 4 is an identity.

You'd need a cancellative monoid at least

[rocky cloak]

The exercise says that they usually defined a subgroup to have the same identity, show that it's enough to assume they have an identity at all.

So you don't need to assume anything about how inverses behave in the subgroup. It's even true when H is a submonoid of a group

[hidden haven]

Oh ok I thought you meant that G doesn't need to be a group

[rocky cloak]

Yeah, G should be a group. But the only assumptions you need on H is that it is a group and the operation coincides. That identity and inverses coincides will follow

[hidden haven]

Ye ye

[ember field]

Is category of finitely genreated R algebras abelian?

[gritty sparrow]

no

[rocky cloak]

They are not even additive

[gritty sparrow]

for example: the initial and final objects are not the same (0 ring vs R)

[inland gull]

What are the c_0 and c_{00} = R^\inf_00 = { r is inside R^\inf; r_i = 0 except for finite number of i}, called? Not the functions which are n times derivatable.

[tribal moss]

https://en.wikipedia.org/wiki/Sequence_space just calls them "the space of null sequences" and "the space of eventually zero sequences", which is not particularly fancy names, but ought to work as descriptions in prose.

[ember field]

Why is zero not initial in Rings. Why would we want both identities to be preserved?

[coral spindle]

Because that's the definition of a ring homomorphism lol

If you look at non-unital rings, wherein the rings don't necessarily have units and the homomorphisms therefore are not forced to preserve them, the zero ring is then initial.

Having a unit is a very nice property for a ring to have and so we tend to look at those.

[south patrol]

... and if we're looking at rings with identity, then the identity is a key part of the structure of the ring and hence something we'd like to preserve

[rustic crown]

yee identity is a 0-ary operation, and you're asking it to be preserved :3

[ember field]

What is 0-ary

[rustic crown]

like for a group, you have a 0-ary operation e : * --> G
a unary operation, i : G --> G
and a binary operation m : G x G --> G

a map of groups G --> H would then be asked to preserve all these 3 operations, but we only write it for m, because other two follow from it >.<

[frigid lark]

I've gotten to the point where every $\phi \in$ Hom $(H_\Gamma(N), \mathbb{\mu}N)$ is of the form $ \phi(\sigma) = \sigma(y)/y$ where $y = \sum{\tau \in H_\Gamma(N)} w_\tau \tau(a)$ for some $a \in \Gamma^{1/N}$, where $w_\tau = \phi(\tau)$, but I can't quite figure out how to show $y \in \Gamma^{1/N}$?

[cloud walrus]

parrottea

[wraith cargo]

can you show that $\Gamma \to \mathrm{Hom}(H_\Gamma(N), \mu_N)$ is surjective with kernel $\Gamma^N$?

[cloud walrus]

ironyincarnate

[frigid lark]

That's what I'm trying to do

I've got the kernel part

I just need the surjective part

[wraith cargo]

ah

sorry but what's the mapping here exactly

[wraith cargo]

ah cool

[frigid lark]

And $H_\Gamma(N)$ = Gal$(K(\mu_N, \Gamma^{1/N}) / K(\mu_N))$

[cloud walrus]

parrottea

[frigid lark]

I think (wasn't properly defined)

[wraith cargo]

show that an arbitrary element of H_Gamma(N) doesn't fix the y defined in your message above is what I'd try

[frigid lark]

That's not necessarily true, as what if that element is in the kernel of phi?

[wraith cargo]

how does that impact anything?

the way you defined it it has to be an element of K(mu_n, Gamma^1/N)

[frigid lark]

Well sigma(y) = phi(sigma)y by construction

[wraith cargo]

OK but what does sigma(y) = 0 imply?

remembering that sigma fixes K

[frigid lark]

It's a multiplicative map

So if sigma is in the kernel, phi(sigma) = 1

[wraith cargo]

ah okay that makes more sense lol

hmmm okay then

lol you can prove this with cohomology

[frigid lark]

How?

[wraith cargo]

there's some weird SES that you take cohomology on and one term goes to zero

[wraith cargo]

bit of a dumb thought, but since the map Gamma -> Hom(H_Gamma(N), mu_N) is a group hom can you just show that the identity on Gamma is in the kernel of Phi, so that you can assume wlog that y isn't in the kernel?

or wait I think I might be thinking of the map wrong

[frigid lark]

H_Gamma(N) is a Galois group afaic

[wraith cargo]

god this problem is fucked

[frigid lark]

Since I assume everything is finite could a counting argument work for surjectivity?

[wraith cargo]

what book is the problem from btw

[frigid lark]

Lang

Chapter 6

[slim kayak]

Can someone recommend a source that covers the reidemeister-schreier algorithm? The one where you construct covering graph schreiergraph corresponding to some subgroup and then find a maximal subtree in the graph to then determine a presentation of the subgroup.

[frigid lark]

Wait, this is wrong, it's meant to be $\phi(\sigma) = \sigma(y^{-1})/y^{-1}$

[cloud walrus]

parrottea

[lethal dune]

I don’t get the definition

Is it a polynomial or a function

[delicate orchid]

both lol

I get it though, it's not clear enough for some

it's the image of a polynomial under an evaluation map

[lethal dune]

How is it taking value in R then?

[delicate orchid]

judging by the line below, it's taking each entry in an element of gl(n, R) as a variable

so f would be an element of R[x_1^1, x_1^2, x_2^1, x_2^2] when n = 2

[lethal dune]

So a polynomial in n^ 2 variable assembled in a way that it takes values from gl_n?

Ok cool tks

[delicate orchid]

yus

np

[rocky cloak]

Guess in general you might call a function V -> R polynomial if it is a finite sum of products of linear functions V -> R.

So if you choose a basis for V* this just the polynomial ring with indeterminants given by the basis.

[lethal dune]

doesn't sound right

[rocky cloak]

How so?

[delicate orchid]

sounds off cause you're implictly dodging the issue of irreducible polynomials

take a Q-vector space V, I wouldn't describe the function x^2-2 : V -> Q via the product (x-sqrt(2))(x+sqrt(2))

[rocky cloak]

?? You describe it as the sum of x*x and -2

[delicate orchid]

is x*x linear?

ohhh I see now yesss yess

I misread what you said

[rocky cloak]

It is the product of two linear functions

I should probably replace sum with linear combination, since we only get the constant function 1 from the empty product

[night onyx]

I imagine it means a map f : gl(n,R) -> R such that f(A) = p(a_11, a_12, ...., a_nn) where p is a polynomial in n^2 variables. The determinant is an example of such a map.

[lethal dune]

yeah, already been answered, tks

[wraith cargo]

How does the existence of such a resolution follow from the hypothesis of R being Noetherian

[delicate orchid]

syzygy spotted

[rocky cloak]

R is noetherian if submodules of finitely generated modules are finitely generated. Therefore finitely generated is the same as finitely presented is the same as finitely resolved

[glossy crag]

I have a question about crossed products (the setup is kind of long):

The context is $A$ is a $K$-CSA, $L/K$ is a finite Galois extension with group $G$ and $L$ a splitting field of $A$, $V$ an $n$-dimensional $L$-VS, $h:A\otimes_KL\to\operatorname{End}L(V)\subset\operatorname{End}K(V)$ an isomorphism of $L$-algebras. Next, for every $\sigma\in G$ there is a $K$-automorphism $h^{-1}\sigma h$ of $\operatorname{End}L(V)$, where $\sigma:A\otimes_KL\to A\otimes_KL$ is $a\otimes\lambda\mapsto a\otimes\lambda^\sigma$ (composition is left to right and maps are written exponentially), so by Skolem-Noether there are units (defined up to $L^\times$) such that $x^{h^{-1}\sigma h}=u\sigma^{-1}xu\sigma$ for all $x\in\operatorname{End}L(V)$. Since $u\sigma u\tau$ and $u_{\sigma\tau}$ induce the same automorphism, there are $c_{\sigma,\tau}\in L^\times$ with $u_\sigma u_\tau=u_{\sigma\tau}c_{\sigma,\tau}$.

Now for the question: since the $u_\sigma$ are defined up to factors of $L^\times$, the coefficients $c_{\sigma,\tau}$ change if we change $u_\sigma$ according to the rule $c'{\sigma,\tau}=c{\sigma,\tau}(a^\tau_\sigma a_\tau a_{\sigma\tau}^{-1})$, where $u'\sigma=u\sigma a_\sigma$. The text claims that if we change the isomorphism $h$ or the $L$-VS $V$, the coefficients will change according to the same rule. Can anyone help me work this out?

[cloud walrus]

leave_no_norm

[barren sierra]

for anyone still wondering about the Spamakin's weird symbol saga

I have figured it out

after obtaining a WIP english translation of a different text

it is indeed central product, confirmed

[formal ermine]

was

[topaz heart]

i've been staring at this for a while and i can't see how to get the forward implication

the lhs seems so much weaker than the rhs

[formal ermine]

what is a block and what does Delta^x mean

[coral spindle]

A block is a subset of a G-set whose translates are either the block itself or disjoint from the block.

Delta^x probably just means the action of x on Delta, but I was also wondering.

I'm confused as to why they add this "In the case that G is finite...". Isn't that exactly the same statement?

[hidden haven]

I think the second one is stronger because it implies that you can separate either of the two

Whereas the first one only allowed separating out one or the other

[coral spindle]

Oh I see what you're saying

[topaz heart]

yeah Delta^x is x acting on Delta

[coral spindle]

I really want to impose some minimality condition but I can't quite see how it would work

[topaz heart]

on the size of Delta?

[coral spindle]

I would like to have the intersection between \Delta and \Delta^x to be minimal, but nvm

I'm not sure this is true. Maybe you can help me check if I'm not way off with this counterexample

[topaz heart]

yeah i spent some time looking at this

[coral spindle]

I think the action of Z on R by addition and the set \Delta = [-0.1, 1.1] is a counterexample with the points \alpha = 0.4 and \beta = 0.5

[hidden haven]

That's not a transitive action

[coral spindle]

Oh of bloody course thank you

[hidden haven]

You can probably look at the action of alpha inverse beta, this takes beta to alpha

And maybe see where alpha goes? Idk

[coral spindle]

Yeah I think so but we also need that to be something that breaks \Delta being a block right

Here's my idea so far.

Since $\Delta$ is not a block, there are elements $p, q \in \Delta$ and $g \in G$ such that $g\cdot p \in \Delta$ and (now this is the annoyance) the element $g \cdot q$ is either in $\Delta \setminus \Delta^x$ or $\Delta^x \setminus \Delta$.

[cloud walrus]

Bowtie

[coral spindle]

Let's just deal with the second case I guess

Now I'm trying to think of a way to use a translation from alpha to beta here

but I can't quite think of it

[hidden haven]

Maybe look at the set of all x such that both α and β are in Delta^x. Take the intersection of Delta^x indexed by these x. This contains α and β and I would hope that this is a block

Not sure what to do with this even if it is a block though

[topaz heart]

annoying cause u have the other way

intersection of blocks is a block

but i dont think the intersection of two subsets being a block tells u much

[hidden haven]

Ye I'm not saying that

[coral spindle]

Perhaps it's simpler to do this contrapositively. If there were a pair for which this didn't hold, I think we can show that Delta is a disjoint union of blocks

Well ok, that doesn't help does it

[barren sierra]

(is there an online sagemath forum that's somewhat active?)

I don't see how (iii) is possible either

(i) and (ii) make sense but like

a single clean presentation for arbitrary central products seems unlikely

unless the answer for (iii) is just "here is a unified presentation for nonabelian groups of order p^3, now just take central products"

[delicate orchid]

yeah i and ii are basically identical, you have the two isomorphism classes
So. Is there a nice interaction between the central product and presentations?

let us google

[barren sierra]

Have been googling 💀

[delicate orchid]

ok well we already know the generators, as it's a quotient of a product

might not be a minimal set of generators but I don't recall that being my problem

[delicate orchid]

but what is the quotient exactly

ok it's this thing

so the extra relators are of the form \theta(h)k = 1

So, G = <X|R>, H = <Y|S>, G central H = <X, Y | R, S, [blah, blah], \theta(g)h = 1 for all blah blah blah>

[barren sierra]

ok so nothing cleaner

I was wondering I was missing that something cleaner existed other than just the general thing of products of presentations

[delicate orchid]

this is just me bashing stuff together there's probably a better way to do it that gives you a nicer presentation

but this'll work

[barren sierra]

it will work indeed

Anyone know how to do central products in SageMath? Or GAP but I'd prefer SageMath
I can't find any official functions but maybe some library or something exists that I cannot find immediately

Or is there a better forum to ask on

[summer path]

Possibly #computing-software ? Although maybe not

[glossy crag]

Anyone?

[white oxide]

i'm struggling to see how Coim g is isomorphic to C; i know that it follows from the first iso theorem and that g is an R-module epimorphism, but i can't see how ker g = {0}

maybe i'm blind or smt

[open sluice]

coker

that autocorrects to Coker

[white oxide]

coker and meth

[rocky cloak]

Why would ker g equal 0?

[white oxide]

oh wait oops i totally forgot the first isomorphism theorem lol

my fault thanks

[white oxide]

yas

[formal ermine]

@white oxide still, very based name

[white oxide]

ikr

im keeping this forever

@ashen heron don't sully my ass go do epsilon delta proofs

[formal ermine]

let epsilon <

[ashen heron]

these algebra people sound nice

[formal ermine]

ok time to do geometry

[open sluice]

where would you even talk about geometry, like hilbert axioms
or is that not a thing anymore and everyone just does alg/diff/idk geometry

[white oxide]

to show that this diagram is commutative do i just have to show that each of the colored arrow pairs are equivalent lol

[formal ermine]

doing toric geometry rn

[jovial tundra]

what's the meaning of the Z and U

[formal ermine]

you look at fans and the toric varieties they induce n stuff

[formal ermine]

unitary group and cyclic group

[coral spindle]

Yeah, it's a funny metatheorem but you can show that if the smallest cycles in a diagram commute then the whole diagram commutes

For some definition of 'smallest cycles'

[white oxide]

who's bowtie

imposter

[coral spindle]

That's what I'm sayin'

[white oxide]

impostoor

impostor

[formal ermine]

AMOGUS

WHEN THE IMPOSTER IS SUS

[coral spindle]

But yeah maybe you should take some time to convince yourself that you can just show that the smaller cycles commute

[white oxide]

oh damn ok

[coral spindle]

Try it :)

Just for this one

[jovial tundra]

ah thank you, I missed the definition they gave 🤦‍♂️

[paper aurora]

say L = [x,y] is a lattice of C, (the additive subgroup of complex numbers ax+by with a,b, running over the integers, also a,b form an R-basis for C). if L' has index p in L, why must L' contain pL = [px,py]? i am stuck on this fact reading a proof. might be relevant is the fact that pL has index p^2 which i can show

[balmy belfry]

Hhfvj

[rocky cloak]

So if G is a group H a normal subgroup of index n and g is in G then g^n is in H. Applying this to your example we see that both px and py are in L'.

[paper aurora]

why must pL be a subgroup of L'?

sorry i was misinterpreting things, i think i see now lol

[real sparrow]

The Klein bottle is the semidirect product of 2 circles.

It's fundamental group is the semidirect product of Z and Z

[wraith cargo]

I don't believe that's true?

[delicate orchid]

the fundemental group is <a, b | aba^-1 = b^-1> right?

in which case it is a semidirect product of Z with Z

but my question is why post random non-sequiturs

[south patrol]

Yes

[delicate orchid]

thank you for the peer review fellow scholar "potato"

[south patrol]

Lol

Nonymous referee

[formal ermine]

potentially a stupid question

[rocky cloak]

I think it's some general construction where you can build the Eilenberg-Maclane space of a semidirect product out of the the EM spaces of the two other groups. Maybe a question for #point-set-topology

[formal ermine]

let v_1, v_2 be valuations, is v(a) := min(v_1(a), v_2(a)) a valuation again? I already proved the same thing for quasi-valuations

I'm not sure how to show equality for v(xy) = v(x) + v(y) nor can I come up with a counterexample

[rocky cloak]

If v1 is the 2-adic valuation, then sand v2 is the 3-adic, then v(2) = v(3) = 0, but v(6)=1.

[wraith cargo]

isn't the fundamental group Z x Z_2?

hm
maybe they're isomorphic

[delicate orchid]

can't be isomorphic, Z x Z_2 is abelian

[wraith cargo]

OK maybe I have it weird in my head lol

fuck it's H_1 I was thinking of

confirmed idiot

[delicate orchid]

not so fast, the image of the associated map with the semidirect product Z \ltimes Z is C_2
so it's not completely unrelated

[south patrol]

Ye

[rocky cloak]

Also H1 is the abelianization of the fundamental group so should be pretty similar

[delicate orchid]

yeah you just take the uhhhhh bar resolution of the ermmmmm the uhhh

[south patrol]

Yeah abelianisation gives Z x Z/2 easily

[wraith cargo]

clearly then applying the forgetful functor gives us back Z \ltimes Z because strains brain adjoint something dies

[south patrol]

One cool thing I looked at today was A-linearisation of a space

But now this chat is topology

Gross

[delicate orchid]

actually have you heard of this potato

[south patrol]

Yes

[delicate orchid]

it's staying topology

[south patrol]

Need to learn about it tho lol

[delicate orchid]

you can make a bunch of simplitical sets from a group operation and it agrees with the bar resolution in some way I'm forgetting

[south patrol]

I just know it comes up in some constructions yes

[delicate orchid]

jagr can probably inform me in 5-15 minutes

[south patrol]

Ig this is one way to produce K(G,n)s and stuff

[delicate orchid]

dunno what that is

ok I learnt about this in a very weird way

[agile burrow]

lol simplicial K(G, 1) for discrete groups

[delicate orchid]

I first learnt about it when reading about partial groups

so groups is a special case of that

[south patrol]

Lol okay sure

[real sparrow]

why not

[delicate orchid]

I dunno much about it but it's cool

[real sparrow]

the wikipedia page literaly says it is