groups-rings-fields

[arctic dune]

Is the dimension of a lie algebra representation equal to its rank as a linear transformation?

[chilly ocean]

An element always commutes with its multiplicative inverse. What it means is that every left (or right) inverse of an element commutes with it. Which is the same as saying that being a left (or right) inverse implies it to be an inverse

[glad osprey]

Yeah, I know, I was just trying to explain to OHHELLNAH why it doesn't imply commutativity in general

[wary sorrel]

I'm probably missing something, but in Griffiths-Harris they use

Let $L$ be a characteristic $0$ field, $K \le L$ a subfield. Suppose for all $x \in L$, $x$ is algebraic over $K$ with $\deg_K(x) \le d$. Then by the primitive element theorem, $L/K$ is finite and of degree $\le d$.

[cloud walrus]

shingtaklam1324

[wary sorrel]

I'm not sure how the primitive element theorem applies in this case? cuz that's about finite extensions

so the proof feels circular

[chilly ocean]

Maybe they're using it on K(x)/K?

[wary sorrel]

right

so [K(x) : K] <= d for all x

but to go from this to [L : K] <= d is non-trivial as far as I can tell

https://math.stackexchange.com/a/677047 there's a few more steps involved than what G-H claim

[chilly ocean]

Yeah, I was going to suggest something similar to that answer

[rain grove]

In a ring with divisors of zero show that there exists x,y != 0 that xy = yx = 0

How to do this?

[coral spindle]

What's the definition of divisors of zero that you know?

[rain grove]

x!=0 is divisor of zero if there exists some y!=0,that xy = 0 or yx = 0

[coral spindle]

Very nice

Hmm

Right OK

Here's a thought.

[chilly ocean]

Suppose x!=0 and y!=0 but xy=0. If yx=0 you are done. Else can you think of somehing nonzero z such that zyx=0 and yxz=0?

sorry if i interrupted your thought

[coral spindle]

No I was going to say something similar

It's worth noting that this argument ^ is "without loss of generality," so we can assume that xy = 0 rather than yx = 0, since if that didn't hold, a totally symmetric argument would prove the same thing.

[rain grove]

oh cool for z = x

[chilly ocean]

why is yx² = 0?

[rain grove]

yeah not necessarily, so then for z = yx

[chilly ocean]

That should work

This more strongly proves that a ring with divisors of zero also has a nilpotent of order 2 element

[rain grove]

Hm yep, might be useful later

[chilly ocean]

err

[rain grove]

there are so many little tricks with these starting excercises... I feel like I should know them by now but they always get me

[chilly ocean]

I was careless saying this, maybe its actually false sorry

Exercise: Can you find a ring with zero divisors that has no nilpotent elements?

Its an Extra exercise for you if you want

why did my color change

[rain grove]

lol i noticed that

lvl up

[rain grove]

idk maybe M_2(R)

[chilly ocean]

$\begin{bmatrix} 0 & 1 \ 0 & 0\end{bmatrix}$

[coral spindle]

You get a new colour if you talk a whole lot

[cloud walrus]

borel equivalence relation

[chilly ocean]

this is nilpotent there

There's a very simple commutative example

[rain grove]

shi yeah ofc...

Z_15

[chilly ocean]

mmhm

[rain grove]

3*5 = 0

[chilly ocean]

there are smaller ones

Z/6Z is the smallest

[rain grove]

Im not at quotients yet, that is Z_6?

yeah ig

[coral spindle]

It is yeah

[coral spindle]

Well done for spotting this

[chilly ocean]

I wrote it like Z/6Z because the notation Z_6 is a bit ambiguous, some people use it to mean Z/6Z, others to mean 6-adic numbers and yet others to mean localization at a prime ideal (altho (6) is not a prime ideal so that wouldn't make sense for 6)

[rocky cloak]

I guess Z/2 x Z/2 is even smaller

[chilly ocean]

That one has a nilpotent

[delicate orchid]

everything is that ring is an idempotent I don't see how anything can be nilpotent

[rocky cloak]

0^2 = 0

[delicate orchid]

ok FINE

[rocky cloak]

😉

[chilly ocean]

Oh I'm blind I misread it as Z/4Z for some reason lol

[chilly ocean]

This one is actually the smallest yes, sorry

[urban geyser]

I got a bit confused again—GLn(R) is non-abelian for n>1 and R* is abelian, so how can the determinant from GLn(R) to R* be a homomorphism? Would the multiplication table not be entirely different? How can two groups not be homomorphic, but a function between them (like Det) be a homomorphism?

Is commutativity not a group structure we care about being preserved?

[coral spindle]

Homomorphisms don't preserve commutativity.

Isomorphisms do

Groups being "homomorphic" isn't saying a huge amount.

N.b. people don't tend to use that term

[urban geyser]

Ohh.. i see, so two groups can have different multiplication tables but still be homomorphic?

[coral spindle]

OK I want to ask you what homomorphic means

What does it mean to you

Bc like I said, that's not a term that people use

[urban geyser]

For an f:G-->H, f(xy)=f(x)f(y) for all x,y in G

[coral spindle]

That's a homomorphism between G and H, sure

Or I should really say, from G to H

So what does it mean for groups to be homomorphic?

[urban geyser]

There's a structure preserving morphism from one to the other?

[coral spindle]

So you mean there is a homomorphism f : G → H?

Then every group is 'homomorphic' to every other group.

Via the homomorphism f(x) = e.

[urban geyser]

huh.. so even if it preserves group structure it doesn't say much?

[coral spindle]

Not necessarily

[urban geyser]

I'm confused entirely then..

[coral spindle]

It is less that a homomorphism "preserves" structure. No, it can simply collapse everything down.

It's more like it "respects" structure

A typical function f : G → H just doesn't care about the groups at all

But homomorphism actually let us say something, typically.

The structure is not preserved, but respected. I tend to say that if we didn't look at group homomorphisms, the maps would just be bullshit that we can say nothing about.

[urban geyser]

I see..

Okay back to the drawing board I suppose...

[coral spindle]

Do you have a question?

[chilly ocean]

I think a nice way to think about of homomorphisms is analogous to implications, while isomorphisms are analogous to equivalences

[tribal moss]

And you practically never care about just having some homomorphism. It's generally always either a case of:

• Such-and-such particular map, which we already care about for other reasons, is a homomorphism, and that lets us conclude interesting things about it.
• Or, we're looking for a map with particular properties, one of which is that it must be a homomorphism. But then that's almost always just one among several demands we have for the map we need.

[chilly ocean]

And indeed if you have an equation that holds for some elements in your domain, after applying an homomorphism to every element, you get an equation that holds in the codomain too!

(in particular if you send everything to 0, you always get 0=0 which is trivially true)

[urban geyser]

I see—if isomorphisms are bijective homomorphisms, and homomorphisms dont preserve commutativity but isomorphisms do, how is "respecting" of commutativity implied in bijection? I mean, there exists mapping from GLn(R) to R* that are bijective for n>1, but also bijective for n=1—yet one is noncommutative and the latter is

[knotty badger]

technically isomorphisms are a little more than just bijective homomorphisms

[coral spindle]

It isn't implied in bijection! You need the fact it's a homomorphism for that

[knotty badger]

but being a bijective homomorphism implies that you get an isomorphism here

[chilly ocean]

The image of an homomorphism of a commutative group is commutative

[coral spindle]

Like I said up here, if you don't look only at homomorphisms, you're gonna get random nonsense, as you've realised!

[urban geyser]

Im also quite confused because in Harvard's online video course for abstract algebra, he says that the determinant is not an isomorphism because "GLn(R) is nonabelian and R* is abelian"

well i get it now

[coral spindle]

Well of course, because if groups are isomorphic then every group-theoretic property is preserved

One might even call that the definition of such a property...

[urban geyser]

okay I need to sit down, Im walking as I write and I cant concentrate well

[tribal moss]

Note that the determinant is not bijective: there are different matrices that have the same determinant.

[coral spindle]

And tbh, observing that GL_n(R) is not Abelian but R^x is is sufficient to prove that

[urban geyser]

Yes... but there do exist arbitrary mappings from GLn(R) to R* that are bijections

[coral spindle]

Yes, as we discussed just above.

[tribal moss]

Those are not homomorphisms, though.

[urban geyser]

yes

Alrighty lemme get back to my desk and actually focus to clear this up

[tribal moss]

You need both "homomorphism" and "bijective" at the same time about the same map before you can conclude that the groups behave similarly.

The special thing about this combination is that a bijective map has an inverse map, and the inverse of a homomorphism (when it exists) is itself a homomorphism.
So an isomorphism means that you now have homomorphisms both G -> H and H -> G which are inverse to each other. That means that you can move between the groups in both direcetions, which is a lot stronger than being limited to one direction.

[chilly ocean]

This. The fact that they are an homomorphism with an inverse that is also an homomorphism is the important part. There are other areas of math where a morphism can be bijective and its inverse not be a morphism, and when that happens not all structure is preserved.

[lone niche]

Homomorphisms preserve commutativity. They don't preserve non-commutativity I guess.

[chilly ocean]

They preserve commutativity in the image

not in the whole codomain

[urban geyser]

I see; so even if there exists bijection from one group to another, the determinant simply isnt one of them since it takes an infinite amount of matrices to the same value in the codomain and "information is basically lost" is how I heard it expressed; however, the determinant IS a homomorphism even if it doesn't preserve commutativity. So an isomorphism is a "bit more than just a bijective homomorphism" in the sense that it also implicates commutativity, even if commutativity isnt implied in bijection nor homomorphism.

[mighty kiln]

Is this a good notion of "preserve"

[coral spindle]

Well

[mighty kiln]

Almost nothing is preserved since there is a homomorphism between any two groups

[knotty badger]

ye Grp is connected

[chilly ocean]

bijection doesn't preserve anything except number of elements

[urban geyser]

I see, so if I start with a noncommutative domain, I might very well have a commutative codomain. But if I have a commutative domain, I will have at least a commutative image in the codomain

[knotty badger]

the main thing is just this

image of abelian is abelian

[coral spindle]

That's right!

[urban geyser]

To recap: bijection preserves "size" (maybe an informal term), and homomorphisms preserve group structure (associative binary operation with an inverse and identity element), and an isomorphism preserves size, associativity, invertibility, identity, and ALSO commutativity given that we have a homomorphism to and from, in both directions

[coral spindle]

Isomorphisms preserve absolutely everything about a group, provided it's actually about the group itself and not the underlying set.

[knotty badger]

welllllll

ya gotta be careful if the group has a topology or smth

[coral spindle]

So e.g. the statement "this group is Abelian" is a property that is preserved by isomorphism, but for example "this group's elements are all subsets of N" is not

[knotty badger]

(R, +) is iso to (R^2, +) as groups after all lol

[coral spindle]

(Because the latter property is not about the group, it's about the set)

[urban geyser]

So for example the set of actions you can perform on n elements?

[urban geyser]

Very interesting

[coral spindle]

I'm not entirely sure what that means

[urban geyser]

Sorry, the set of bijections from a set of n elements to itself

Like doing nothing, cycling everything one position over, flipping the first and second elements and fixing everything else, etc...

Or i might be saying distracted malarkey

[coral spindle]

I'm still not sure what that means

So you're talking about the symmetric group S_n

[urban geyser]

Yes

[coral spindle]

And the question is?

[urban geyser]

the underlying set of n elements is not what we care about, but the group itself

[knotty badger]

ahh

[coral spindle]

Yes

[knotty badger]

i think they mean the underlying set of the group

[coral spindle]

But this is not the same underlying set

[knotty badger]

which in this case has n! elements

forgetful functor and all

[coral spindle]

The underlying set of S_n is not {1, ..., n}, it is the set of bijections from {1, ..., n} to {1, ..., n}

[urban geyser]

Oh i see nvm youre right

ok gtg

[coral spindle]

This is a fairly common misunderstanding, it's good to clear it up

o/

[urban geyser]

ill brb tho!

[surreal dagger]

5 b) is ideals generated by cosets b+(a) such that b divides a in R?

[coral spindle]

Yes, they are ideals (b)/(a) where b divides a, in other words

[surreal dagger]

Thank you

[rapid junco]

here, how does one prove that this map is well defined?

if f = g, then for u arbitrary we have Tf(u) = f(u, \xi) = g(u, \xi) = Tg(u)

just wanted to make sure im not making a stupid mistake.

[delicate orchid]

I don’t understand how a function has a coordinate for one

But this is just an evaluation map so yeah it’s well defined? I don’t see where well definedness could even fail.

[delicate orchid]

This works btw

[arctic trail]

oh yeah

[arctic trail]

yes

$\mathbb F_2[x,y]/(xy)$

[cloud walrus]

Trivial Lemma

[surreal dagger]

In practice, how do I think about the disjoint union in this case?
In the case of two Rings A and B, elements are of the form (p_a, 0) or (p_b, 1) for prime ideals p_a and p_b that live in A or B?
Moreover, is it easy to appreciate why we consider the disjoint union and not the union at this point in time? In this short chapter only 1 page introduced spec

[grave sedge]

I'm not entirely sure about your characterization of prime ideals of AxB, but once you do characterize what those primes look like you should have a feel for why it's the disjoint union (try to think about the inclusion lattice if you havent seen the zariski topology yet)

And it's disjoint union just because if you have something like AxA you distinguish between primes from "the first A" and "the other A"

[surreal dagger]

They should be of the form p_A x B or A x p_B, so I can clearly identify them with either an element of Spec A or Spec B.
p_A x p_B does not work, because the product is prime
Thanks, btw what do you mean by inclusion lattice?

[grave sedge]

Something that looks like this (in this case you're drawing all the subsets of {x,y,z}, in your case it would be the prime ideals of your ring)

(btw i've discovered this thing is called hasse diagram)

[surreal dagger]

I see, so they divide nicely on two sides if I where to write it out

[grave sedge]

Indeed

[grave sedge]

By the way this is not the usual way to represent spec R, at some point you'll see you can put a topology on spec R, and the topology of spec RxS is going to be the disjoint union topology of spec R and spec S

Where disjoint union topology means having the two spaces "next to each other"

[surreal dagger]

Cool, tbh the author gave the definition but I cant really appreciate it yet

[barren sierra]

How do I prove that every finite group where every maximal subgroup is conjugate to each other is of prime power order?

[quiet pelican]

Do you have the sylow theorems?

If so, then consider whether you can use them to construct maximal subgroups of different orders

[urban geyser]

Just to make sure I’ve got it before I go to bed—you can have a homomorphism from a non-abelian to an abelian group, because being noncommutative is not a group structure, it’s the absence of one; meanwhile, you can’t have a homomorphism from an abelian group to a non-abelian one, since, there, a group structure is actually being disrespected. The reason why isomorphisms must respect commutativity is because, between a group that is abelian and one that is not, there must be homomorphisms in both directions. Sure, from non-abel. to abel. is fine, but it won’t work the other way around, and therefore can’t be an isomorphism. What I’m not sure I get is that I was told that an isomorphism between two groups means those two groups have the same multiplication tables—and in reality they’re just relabellings of one another. But commutativity is shown by symmetry along the main diagonal of a table; so how can a mapping from a non-abelian (not-diagonally-symmetric) to an abelian (diagonally-symmetric) group be a homomorphism? (let alone an isomorphism, i.e. having the same multiplication table?)

Would it be because we can rearrange one of the tables to get the other?

Sorry for the long message, I just feel this is elementary enough to make sure I have a grasp on it

[knotty badger]

You can definitely have a homomorphism from an abelian group to a non-abelian group

[urban geyser]

[knotty badger]

The result is that image of abelian is abelian, so the image will just be some abelian subgroup of the codomain

Even non-abelian groups can have abelian subgroups

[urban geyser]

was this incorrect?

[knotty badger]

The trivial subgroup is always abelian, for example

That’s correct

This doesn’t mean that you can’t have a homomorphism from abelian to nonabelian

You may be conflating image with codomain here

[urban geyser]

I think I see the confusion

A homomorphism only need to respect the group structures of the domain for its image in the codomain, not necessarily the entire codomain

[knotty badger]

Yes

So you cannot have a surjective group homomorphism from abelian to nonabelian

[urban geyser]

so for example, the determinant—maps from non-abelian to abelian, but there’s no requirement that this direction be non-surjective. So the entirety of the codomain could very well be abelian, given that the image of the morphism is the entire codomain, even if the domain itself is not. However, the inverse, abelian to nonabelian, can never be bijective since the image of the morphism is not the entire codomain, but only that subset of the codomain which is also abelian

[knotty badger]

Yep

[urban geyser]

Alrighty—and how could I understand why people say thay isomorphisms involve identical multiplication tables despite commutativity being shown in such a table?

[knotty badger]

I mean isomorphisms do you give you identical multiplication tables

Any group isomorphic to an abelian group is abelian

[tribal moss]

It sounds like you're still being confused between isomorphisms and homomorphisms here. Homomorphisms do not mean just a relabeling of the group or its multiplication table.

[urban geyser]

Yes… homomorphisms if I understand correctly just need to preserve that group structure/multiplication table for the image, but there could be a bunch of other elements and entries that just arent included in the original group’s table?

[knotty badger]

I think the first iso theorem can help clarify things here too

every group homomorphism takes the form of a quotient map, followed by an isomorphism, followed by an inclusion map

[urban geyser]

So I can have a homomorphism between two groups with completely different multiplication tables, but since abelian to non-abelian cannot be surjective, it cant be bijective, and this morphism cant be an isomorphism.

[urban geyser]

Ah, havent gotten there yet so hopefully that helps

[knotty badger]

yep!

[knotty badger]

yes, the first iso theorem can be thought of as a “classification” of all group homomorphisms

quotient maps encode the idea of “forgetting details” about the group multiplication, so abstraction

isomorphisms encode the idea of “relabelling” in a bijective way, which respects the group operation

and inclusion maps encode the idea of “modelling” a smaller group within a larger group

[tribal moss]

Hmmm, I think it still sounds like you're expecting too much of a homomorphism when you want it to "preserve the multiplication table".
The only thing a homomorphism needs to do is to make sure that it maps x and y to things that multiply to the same thing it maps xy to.

[hidden wind]

i like the terminology “respect the group structure” mwahaha

[urban geyser]

And these multiplication tables become identical once we set the constraint that the homomorphism is bijective, since now it cant be going from abelian to non-abelian (which would make it non-surjective, only the image in the codomain is abelian), and it cant go from non-abelian to abelian (which could make it non-injective(?), such as in the determinant which takes many/infinite elements in GLn(R) and crams them into a single element in R*). For the homomorphism, which may very well show two completely different multiplication tables, to be bijective, it must end up showing identical tables

[urban geyser]

Making note of this for when I encounter it, thank u

[rain grove]

in a ring K we have an idempotent element e^2 = e. Show that for every x, e+ex(1-e) is idempotent. Now find some idempotent element from M_3(R) that is not a diagonal matrix and is not rank 1.

I showed the first part and the second I was just guessing for some time and came up with [110 // 000 // 011]. But I feel like this problem was trying to give me a way to find it without guessing and idk how

[knotty badger]

Yes, so you can prove as a theorem that every bijective homomorphism must be an isomorphism

[urban geyser]

Yes… I watched a video on the definition f(x+y)=f(x)*f(y), but admittedly was hard for me to think of what that formalization means except for it producing the same table (up to the elements of the first group G) since I’m still not accustomed to “mathematical maturity.”

[tribal moss]

You should really try to stick to f(xy) = f(x)f(y), because it looks like you're extrapolating wildly too much from the softer descriptions in words you have seen.

[urban geyser]

yiss

[wary sorrel]

(aluffi calls this the "canonical decomposition")

[knotty badger]

nice!

[wary sorrel]

well I mean M_3(R) has plenty of idempotents (e.g. diagonal ones, or rank 1 ones), so you just need to choose a "nice" x and use the result you showed

[tribal moss]

I think what you were expected to do in the second part was to come up with an easily idempotent matrix such as Diag(1,0,1), and then stick it into the e+ex(1-e) formula together with an arbitrary 3×3 matrix as x, and see what it gives.

[urban geyser]

Maybe—I’m just trying to see where that formula came from; as I imagined it, it’s basically saying that f(x+y) ends up occupying the same position as f(x)f(y), and doesnt just run off wildly somewhere else. I thought I understood how this demonstrated respect of group structure, since you can see the structure on a table, but if i should just throw this table idea out the window, I’m not quite sure how to think of f(x+y)=f(x)*f(y) for x,y in G.

[knotty badger]

i suppose one perspective is that it comes from a commutative diagram…

[tribal moss]

It definitely sounds like you're thinking too much in terms of multiplication tables. That is rarely a productive way to think about groups.

[knotty badger]

yes tables are sometimes a nice perspective for things like cosets, but otherwise not often very useful

[urban geyser]

this was what i was shown

and i think it’s tripping me up a lot

I think I just want to “get it” and then only ever use the short formula

[knotty badger]

interesting, so this question boils down to “why are group homomorphisms the right notion of map between groups”

[tribal moss]

I’m not quite sure how to think of f(x+y)=f(x)*f(y) for x,y in G.
It quite literally says: If you have two elements of G, you can either combine them in G and send the result through f, or you can send your elements through f separately and combine them over in the other group. In other words, you don't need to keep track of whether the move from G to H happens before or after you multiply.

[urban geyser]

Could I equally say it doesn’t matter whether I do the operation in G or in H?

[knotty badger]

yes!

[urban geyser]

Oh my god it just clicked

[tribal moss]

Yes, exactly.

[knotty badger]

so i think one thing to stress here is

the result you get is the same either way

that’s the definition of homomorphisms

[urban geyser]

aHHH god it was right there

[knotty badger]

but, the operations you perform are different

computing f(xy) is a different operation to computing f(x)f(y)

[urban geyser]

Im sorry yall that took me a second, its crazy because sometimes you just need a small sentence to make things click

[knotty badger]

indeed, it may be the case that one of these is easier to compute than the other

[urban geyser]

yisssss

[knotty badger]

so the homomorphism gives you the choice of which operation to perform

knowing that they give the same result

and that’s pretty useful

[urban geyser]

it could be f(x+y)=f(x)#f(y) where f is a homomorphism from (G,+) to (H,#)

[knotty badger]

as a concrete example, if you want to compute the sign of a composite permutation

it’s usually a lot easier to compute the signs individually, and then multiply them

than it is to compose the permutations, and then take the sign

[urban geyser]

Thank you, that last small phrase is what made it click for me—I was overthinking it

[knotty badger]

we know these give the same result

[urban geyser]

yess

[knotty badger]

but they are genuinely different operations

[urban geyser]

Oh goodness

[knotty badger]

and in particular we get to choose whichever is more convenient

[urban geyser]

Now it also makes sense how you can have homomorphisms from abelian to non-abelian groups and vice versa

[urban geyser]

I can see how wildly useful that is

[knotty badger]

in general, this kind of thing is called like

[urban geyser]

math

[knotty badger]

“denotational” semantics vs “operational” semantics

[urban geyser]

[knotty badger]

the former is what the result is

the latter is how you get to it

like to small children, 1 + 9 is operationally different to 9 + 1

because for 1 + 9, you have to start at 1 and count up by 9, and that’s hard

whereas for 9 + 1, you start at 9 and count up by 1, and that’s easy

[urban geyser]

Beautiful, I love that there’s even a term for it—this clears up a lot

[knotty badger]

these give the same result, because addition is commutative

but it’s still true that one is easier to compute than the other

so they’re not operationally the same

but knowing they give the same result is useful, cause then you can choose whichever is more operationally convenient

[urban geyser]

would the proper way to phrase this be that homomorphisms preserves denotational semantics but with different operational semantics?

[knotty badger]

so it’s like

f(xy) and f(x)f(y) are denotationally the same, but not operationally the same

[urban geyser]

Im not sure if “preserves” is the right word here

yes yes i see

Ahhh thats so nice

Im happy I finally got it

[knotty badger]

yeah, so it’s often useful to consider a sense in which things are the same, and a sense in which they’re not

[tribal moss]

(For what it's worth, I don't think this usage of "denotational semantics" and "operational semantics" is standard -- and I'm speaking as someone who did a PhD in a relevant field for those two terms).

[urban geyser]

me all day because i didnt have a chance to actually sit down and focus until now, so ive just been stressing about morphisms

[knotty badger]

often what equations are doing is showing things that are operationally different are denotationally the same

[knotty badger]

ah i see, i just picked this up from some type theory friends of mine

im by no means an expert in it

[urban geyser]

I imagine it’s just kinda a cool aside, and I admit it does actually help me think of this a lot better

Humour me if I’m misphrasing this entirely, but I’m trying to find the words to express myself—I just noticed a bad tendency I inherited from elementary algebra is that operation and result are the same, that you should try to care as little as possible about the difference in the steps since everything is secretely the same and youre just unearthing the result, and it feels like it took me a second to realize that the operation is as real a part as the result now

[knotty badger]

yeah, i think knowing the separation is useful

[urban geyser]

Not sure if that makes sense

[knotty badger]

between operation and result

[tribal moss]

In theoretical computer science, "denotational" and "operational" are two styles of defining the intended meaning of a programming language. In denotational semantics you carefully describe a way to define a single mathematical object that represents everything there is to say about a piece of program text, and its entire behavior is encapsulated in that one object. (For denotational semantics after roughly the 1990s, this single object is often a morphism in an appropriately intricate category). On the other hand in "operational semantics" you instead describe an idealized model of a machine that executes the program, often going back to the same piece of text several separate times during the execution.

[knotty badger]

hmm, so tropo, what would you say the concept im referring to should be called?

im perfectly willing to change the terminology I use here

if you have a suggestion in mind

[urban geyser]

corny side-note, this is why I want to learn math even if I’m never going to make a dime, nor be famous, nor be even all that good at it—it might sound silly for someone who’s done a PhD, but the rush of satisfaction when something makes sense is great. You guys make math worth learning

okay corny aside over resume activities

[tribal moss]

Hmm, good question. I think it's primarily the word "semantics" that makes me think my field's terminology was being appropriated.

[knotty badger]

i see - so then, just “denotational” vs “operational” would be fine?

[tribal moss]

Yeah, they're fairly neutral.

[knotty badger]

ok!

[urban geyser]

Ah before I go, what pronouns may I use for u pseudonium?

[knotty badger]

hmm, im feeling she/her atm!

[urban geyser]

awesome uwu thank u u twooo

its sleepy time 😻

time for more math tomorrow

[chilly ocean]

sleep well

[tribal moss]

Interesting. So often we see struggling students who have fallen into the other extreme with elementary algebra -- they seem to view expressions just as a meaningless symbolic game and insist of asking "is this an allowed rewriting" rather than "would this give the same result".
It's useful to be reminded that you can also get too far in the other direction.

[knotty badger]

Yes i think knowing that both are important is useful

There’s a sense in which things are the same and a sense in which they’re not

[chilly ocean]

Not sure I understand. "allowed rewritting" is equivalent to "give the same result".. as long as "give the same result" is provable, that is

[quiet pelican]

“Allowed rewriting” implies you’re thinking about it as an arbitrary thing

[tribal moss]

Yes, you know that, and I know that.

[quiet pelican]

And not something that logically follows from axioms

[knotty badger]

perhaps unsurprisingly, im inspired by the way Eugenia cheng phrases these things

[tribal moss]

But the students I'm talking about seem to have only very hazy ideas that expressions even mean something.

[knotty badger]

what kinds of expressions do you mean, tropo?

[tribal moss]

Plain old high-school algebra expressions.

[knotty badger]

ah, i see

interesting, so students really just see them as collections of symbols?

[tribal moss]

Some students, at least.

[knotty badger]

right, yea

and - how about the students that don’t?

what kind of meaning do they attach?

[tribal moss]

I would hope the meaning of "here's a recipe for a computation you can do once you have decided which value you will let the letter x stand for".

Or something in that vein.

[knotty badger]

sounds like a function!

[chilly ocean]

I'm a formalist so I also only see them as collections of symbols too. But I have intuition about them which helps guide what manipulations might be useful, so maybe in practice that intuition is equivalent to what you mean by meaning

[knotty badger]

indeed I’ve occasionally tried out “view algebraic expressions as functions” on this server, and it sometimes turns out well

very occasionally though, it’s certainly not the standard way to view them

[chilly ocean]

If you have a term with a free variable in some language L and an L-structure, then indeed it gives a function from that L-structure to itself

[knotty badger]

mhm mhm

[tribal moss]

E.g. with some helpees, if they rewrite (x+5)² to x²+5² and you explain that doesn't work because if we plug in x=1 we get 36 on one side and 26 on the other, then you can almost feel the blank stare through the screen, and then perhaps they'll collect themselves and reply. "Okay. So it's not allowed to do that?"

[knotty badger]

in general I’ve found that expressing things in a “functional” way is surprisingly helpful for understanding

cat theory gives me the confidence to do this, but I don’t usually need to introduce it to phrase things in such a way

[knotty badger]

e.g. it gives you the benefit of type-checking

[chilly ocean]

This sounds like they think math is a bunch of complicated rules someone made up, rather than a few simple fundamental rules, and the more complicated ones are consequence of the simples ones

[chilly ocean]

So I don't think the issue is necessarily whether they view it as a meaningless game or not. The "would this give the same result" is a very simple rule, while the "is this an allowed rewritting" is a complex one

And usually in math we tend to prove the complex ones from simpler, fundamental ones (axioms)

[knotty badger]

I guess if I was helping that student, I would first try to see if I could convince them that one can view algebraic expressions as functions

in the sense of an input-output machine, not as a special relation between sets

[tribal moss]

But in order to use that simple rule, you need to have an idea of an expression giving a result when you evaluate it.
My hypothesis about these students is that if you say "try to plug in x=1", then at best what they will hear is "write a new expression where you have textually replaced every x with 1, and then follow the approved PEMDAS rewriting rules until you end up with a single number after the equals sign".
Whereas ideally they would hear "carry out the calculation this expression specifies, in the case where the input called x is the number 1".
(This is the same type of students who tend to abuse the = sign to mean "... and here is the outcome of the next step in the procedure", so if they're differentiating they'll write "x²+5x+3 = 2x+5" without a blink).

[knotty badger]

like, I think I’ve probably brought up in #math-pedagogy before about how “substitution” as a concept can be surprisingly hard to teach

and I honestly don’t think we got a good answer to how to teach it

[dull ginkgo]

not sure how to do this off the bat

$f_i$ being a basis of a free submodule $K$ is equivalent to $A*$ being an injective endomorphism. Assuming that's the case, then $A^*$ induces the determinant endomorphism on $\wedge^n R^n \cong R$.

[cloud walrus]

The Library of Babble

[dull ginkgo]

actually lmao

I have a really stupid way to do the one direction

Because being injective on End_R(\wedge^n R^n) is basically NOT being a zero divisor

[chilly radish]

You already did this exercise

Did you not

Oh maybe it was someone else

[dull ginkgo]

I did a tiny bit of this one before but never completed it

Mainly because of the issue that some matricies have adjugates that are the zero matrix

which kind of throws out my idea of using the adjugate

[dull ginkgo]

Eh i just used the induced maps on the exterior powers :3

[woeful sage]

Neam's algebra arc (insert sotrue)

speedrunning dummit and foote in this thread

I'm at section 1.7 rn, gonna start chapter 2 soon

[ripe glacier]

How's speed running going

[rain grove]

For a matrix $A \in M_n(\mathbb{C})$, we say that it is diagonalizable if there exists an invertible matrix $P \in M_n(\mathbb{C})$ such that $P^{-1}AP$ is a diagonal matrix. Show that such a matrix can be written as $A = \sum_{i=1}^{r}\lambda_i E_i$ for some $\lambda_i \in \mathbb{C}$ and idempotent matrices $E_1, \dots, E_r \in M_n(\mathbb{C})$, such that $E_iE_j = 0$ for all $i \neq j$.

[cloud walrus]

OHHELLNAH

[rain grove]

Idk what to do here

Are E_i some already defined matrices or in this case we only know they are idempotent and E_iE_j = 0

[woeful sage]

going great

[rain grove]

Ig I can do $P^{-1}AP = d_1E_1+d_2E_2+\dots + d_nE_n$ where $E_i$ is a matrix with all 0 except a 1 in its $i$-th diagonal place

[cloud walrus]

OHHELLNAH

[rain grove]

And now I need to show that $P^{-1}E_iP$ is a idempotent and $P^{-1}E_iP \cdot P^{-1}E_jP= 0$

[cloud walrus]

OHHELLNAH

[rain grove]

oh and that is obviously true! lol

[rain grove]

In the Rings chapter I have this fraction... for some element $u\neq 0,1$ and $u^2 = 1$ and an invertible element $(1+1)$: $(\frac{1-u}{2})^2 = \frac{1-u}{2}$ and it says they denoted $\frac{1-u}{2}$ as $(1+1)^{-1}(1-u)$.

[cloud walrus]

OHHELLNAH

[rain grove]

But If I try to multiply $((1+1)^{-1}(1-u))^2 = (1+1)^{-1}(1-u)(1+1)^{-1}(1-u)$ I can't get to $(1+1)^{-1}(1-u)$ because it's not necessarily commutative right?

[cloud walrus]

OHHELLNAH

[coral spindle]

But 1 commutes with everything, so 1+1 does too, so its inverse does too.

[junior badge]

Is the image of a primary ideal under a surjection primary?

[rocky cloak]

The image of a general primary ideal need not be primary. but in the correspondence theorem, primary ideals correspond to primary ideals

[junior badge]

Okay, I see. Is there some nice counterexample where Q is primary and f(Q) not?

[rocky cloak]

Take C[x, y] ->> C[x, y]/(xy)
And your ideal to be (0)

[junior badge]

Ah, nice. Thanks!

[crystal vale]

Let R be a commutative ring, A in M_n(R) then for every y(column vector in R^n) there exists x in R^n such that Ax=y.

Is A invertible? If yes then any hint? Why R to be commutative here?

[tribal moss]

Hint: In succession, let y be the columns of the identity matrix.

[crystal vale]

And why R to be commutative? To define the determinant concept?

[tribal moss]

I don't think there's necessarily any need for R to be commutative, other than to free you from spending extra effort figuring out whether the answer works for noncommutative rings too.

[crystal vale]

Actually the question is let R be commutative and if f is a surjective endomorphism of R^n then f is bijective.

Here Endomorphism as module homomorphism between R^n.

So if A is a matrix associated with f, A follows the above condition.

Yes in linear Algebra we used rank nullity argument

[tribal moss]

So we get a right inverse easily, but we need a left one.
If only R was a domain, we could switch to its field of fractions to argue that a right inverse is also a left inverse ...

[dull ginkgo]

I think as long as the ring has the property that non-zero divisors are units then it works out

By passing to the n-th wedge power and observing the determinant, and how “injectivity and surjectivity” are actually right/left cancellativity in an endomorphism ring

That and Cayley Hamilton ironically

Endomorphism rings are so fucking cool

You can characterize the R-endomorphism ring via the centralizer of the embedding of R into End_Z(M) by left multiplication (or R^op into End_Z(M) via right multiplication)

And then if M is simple we have cool properties in the double centralizer, consisting of additive maps that commute with any R-linear endomorphism

[rocky cloak]

I wonder if there's a very good characterization of when any surjective endomorphism on R^n is an isomorphism.

It holds if R is commutative, and if R is Noetherian.

But surely it holds for other rings as well

[dull ginkgo]

I think that’s true if R is commutative in general due to Cayley Hamilton

The converse I think holds iff non-zero divisors are units

Applying it to R^1 = R

Because being a unit is basically being an “isomorphism”

[rocky cloak]

I'm not sure what you mean here, non-zerodivisors being units seems way too strong for doing anything

[dull ginkgo]

And being cancellative in either direction is injectivity or surjectivity

[dull ginkgo]

By converse i mean

Injective => surjective

[rocky cloak]

Oh, I see

[dull ginkgo]

I mean that’s just what came to me immediately, at least it implies one direction

[rocky cloak]

Anyway, is it equivalent to IBN maybe?

[dull ginkgo]

And you can use the functorial map into the n-th wedge power and using $\mathrm{End}_R(\wedge^n R^n) \cong R$

[rocky cloak]

Hmmm, seems weaker on the face of it, but maybe not

[cloud walrus]

The Library of Babble

[dull ginkgo]

My brain keeps going “DON’T FORGET IT’S AN OPPOSITE RING” but it’s commutative

This happened like 3 times during this convo

[rocky cloak]

Hmmm, so a surjective map would split. Making R^n a direct summand of itself. But the other summand doesn't necessarily have a full copy of R.

Still, seems pretty close to IBN...

[dull ginkgo]

IBN?

[dull ginkgo]

I’m unfamiliar with IBN (once again most of this is me deducing stuff from intro ring theory and module theory stuff in Jacobson)

[south patrol]

Invariant basis number

[dull ginkgo]

Oh i see

[south patrol]

I.e. R^n iso to R^m iff n = m

[dull ginkgo]

Is that not held by any commutative ring

[south patrol]

It is yes

[dull ginkgo]

Me using wedge powers implies I’m using commutative

[south patrol]

I didn't read context aha

[dull ginkgo]

I know you can’t easily make a tensor product of sided modules but

[south patrol]

Just explaining what was said

[dull ginkgo]

Can you still construct a tensor algebra

Mainly to be adjoint to the forgetful of R-alg to R-mod

Oh yeah no shit gets fucky fast

[rocky cloak]

It holds for any commutative ring and any Noetherian ring yes. But also many other rings

[south patrol]

Tbh I've not heard of such a construction

[next obsidian]

Sorry how is the original question true without a Noetherian hypothesis

With only commutativity

[south patrol]

Do you mean this chmonkey?

[next obsidian]

Yeah

Like why is it true for a commutative ring?

[rocky cloak]

Determinants for example

[next obsidian]

Thonking

Never mind I do believe it

[dull ginkgo]

Actually it breaks down!

Consider au (x) bv = a(u (x) bv) = ab(u (x) v) = a(bu (x) v) = (bu (x) av)

[next obsidian]

I wanted to Nakayama the kernel

But that needs Noetherian

[dull ginkgo]

Do we necessarily need Noetherian-ness for endomorphisms of R^n to be integral in End_R(R^n) over R? (Assuming commutativity)

[next obsidian]

[dull ginkgo]

I thought it depended upon viewing det(M) as a polynomial valuation map

[south patrol]

No this is just generalised Cayley Hamilton

[next obsidian]

You just need fg I think

This is the determinant trick

[south patrol]

Rephrased in a slightly unfamiliar way (to me) lol

[dull ginkgo]

Oh that’s the only way i am aware how to prove it

[next obsidian]

Yeah haha

That’s why I responded

Then I thought more

[dull ginkgo]

The ramblings of a psycho because I don’t know much abs algebra and the terminology used

Frankly a lot of it is me going “ooo this looks cool” and fucking around with it for a while

tldr endomorphism rings are cool

I am not doing math for college or any other reason outside of being a hobby so i do what i want :3

[crystal vale]

But if R is commutative then if AB = 1 then BA = 1, right?

[south patrol]

Pog

This is more fun tbh

[knotty badger]

Sounds fun yeah

Don’t think I’ve ever done that..

[tribal moss]

I'm not sure I can immediately argue for that, and that seems to be the point of the long subsequent conversation too.

[south patrol]

Are A,B elements of R

Or matrices over R

[tribal moss]

Matrices.

[rocky cloak]

Hmmm, now I'm wondering. Is IBN equivalent to having a homomorphism to a division ring?

One direction is true, but I can't find anything about the other direction

[rocky cloak]

If A and B are square matrices, then yes

[south patrol]

Say AB = 1. Then A: R^n -> R^n is surjective so its determinant is a unit so it is invertible and by uniqueness of inverses BA = 1

[tribal moss]

That seems to assume what we were trying to prove in the first place.

[dull ginkgo]

Are Dividion rings IBN?

[south patrol]

Yes

[dull ginkgo]

Hm i might try to prove that

[south patrol]

Standard proofs (without exterior powers) that dimension is well defined for f.g. vector spaces still go through for division rings

In fact it is true for arbitrary dimensions again with the same proof as for vector spaces

[rocky cloak]

They are Noetherian for starters

[dull ginkgo]

You can’t use exterior powers with a noncomm ring unless the annihilator contains the commutator subring

i thought

[dull ginkgo]

oh

[south patrol]

That's why I said without

[dull ginkgo]

Oh yeah

[south patrol]

Nws

[tribal moss]

Oh, I see it now, this goes via adjugate matrices, right?

[rocky cloak]

Yes

[south patrol]

That is one way to do it

[rocky cloak]

Do you know another way to do it?

[dull ginkgo]

Or End_R(R^n) ~= M_n(End_R(R)) ~= M_n(R^op) and considering matrices of division rings yeah

[south patrol]

Personally I thought in terms of exterior powers (exterior powers take surjections to surjections, so det(A)R = R, so in particular det(A)a = 1 for some a, so det A is a unit)

[dull ginkgo]

I sorta see it

[south patrol]

And then A is invertible eh okay sure

There you use adjugates

Lol

[south patrol]

Actually this is silly. You can phrase this as multiplicativity of determinants

Lol

[dull ginkgo]

Functor :3

[south patrol]

Tbf I guess if A is not injective neither is multiplication by det A

So you don't need adjugates for that

[tribal moss]

Is there a neat direct argument for that?

[south patrol]

Well it was that if wlog Ae1= 0 then det(A)e1 ^ ... ^ en = Ae1 ^ ... ^ Ae_n = 0

Or again adjugates

[dull ginkgo]

If R is commutative, then End_R(\wedge^n R^n) ~= R. Thus because the functorial / multiplicative determinant map from End_R(R^n) preserves surjectivity, we have left-cancellativity in R which then becomes right cancellativity too because R is commutative :3

[tribal moss]

Perhaps it's just me being too cautious about doing stuff over general rings rather than fields, but is it obvious that we can extend e1 to a basis in the ring case?

[crystal vale]

Yes

[south patrol]

True that WLOG was bad I think. It should be that det A is a zero divisor

I think

[crystal vale]

I want to show that in R, the commutative ring if AB = 1 in M_n(R) then BA=1.

Here I used that if R is a commutative ring, a matrix A in M_n(R) is invertible if and only if its determinant is invertible in R.

Since AB is invertible so det(A)det(B) is invertible now since R is commutative therefore detA and det B is invertible in R. Therefore A and B are invertible.

Let CA=1=AC. So C = C•1 = C(AB)=1(B) = B.

Is it correct?

[crystal vale]

A, B are in M_n(R)

[south patrol]

For example consider A given by multiplication by 2 in Z/4 which has determinant 2 != 0 and isn't injective

[dull ginkgo]

Let x be in End_R(K):

• x is surjective iff x is left cancellative
• x is injective iff x is right cancellative
• x is a left unit iff it admits a endomorphic section
• x is a right unit iff it admits an endomorphic retract
• x is a unit iff it is an automorphism

[tribal moss]

What is K now?

[dull ginkgo]

Some R-module

[south patrol]

To prove my altered claim, note that if Av = 0 then you can write v = sum lambda_i e_i and note that 0 = Av ^ Ae2 ^ ...^ Aen = lambda_1 (Ae1 ^ .... ^ Aen) = lambda_1 det(A) (e1 ^ ...^ en)

So det(A) lambda_1 = 0

[dull ginkgo]

This is just rephrasing types of morphisms in the context of multiplication in the endo ring :3

[south patrol]

Here I assume lambda_1 nonzero lol

[dull ginkgo]

The silly funny part is that if you have more than one retract, then you have infinitely many

[tribal moss]

I'm ... afraid I've lost the thread now.

[dull ginkgo]

FUNCTOR FTW

[tribal moss]

But I'm trying to make sense of both Potato and Mizalign at the same time, and I think they're presenting competing proofs for the same original claim ...

[dull ginkgo]

No i was just writing stuff out for my own retention lol

Sorry

[rain grove]

Problem: Show that in a group where for all elements x,y: (xy)^2 = (yx)^2 holds then its center has squares of all elements.

Proof in the book: xyxy = yxyx, if we write y = yx^{-1} then it follows xy^2 = y^2x so y^2 is element of the center.

When they say y =yx^{-1}, doesn't that mean y is now an element of the group that can be written as some element y an inverse of another element (so in my mind not necessarily any element). So now I would have to show that any element can be written like a = yx^{-1} for any two elements x,y. But that is obviously true for y = a and x = 1, but then that means x = 1 and that (xy)^2 = (yx)^2 equations doesn't really mean much if x = 1.
... hopefully I explained what I don't understand lol

[dull ginkgo]

Center is the squares or contains the squares?

[rain grove]

contains

[rocky cloak]

It's not y = yx^-1, but rather replace y by yx^-1

[tribal moss]

Okay, so what is it you have proved here? I.e. what was the "altered claim"?

[south patrol]

Det A is a zero divisor if A is not injective. It isn't true det A js necessarily zero

[delicate orchid]

cool if true

[south patrol]

Actually this is in my dissertation, lol

The sort of funny fact which is elementary but I never stumbled upon because it is only for matrix rings over commutative rings w zero divisors, which isn't a common situation for me at least aha

[delicate orchid]

I can see it for n = 1 so I believe it for n = a billion

[south patrol]

Taking top exterior powers reduces to n = 1

[tribal moss]

Okay, and that proves that if A not injective then multiplication by det A is not injective either.
Wait, does that help us with avoiding adjugate matrices when proving right inverses are left inverses?

[rocky cloak]

A better phrasing might be take x and z arbitrary and set y = zx^-1, then you get that x and z^2 commute.

[dull ginkgo]

If you wanted you could compute the centralizer of two squares :3

not centralizer

Commutator

[x^2,y^2]

And use your relations to annihilate it

Or if you want to do it in a cool way

x^2 = (xy^-1 y)^2 = (y x y^-1)^2 = yx^2y^-1 so squares are invariant under conjugation

[south patrol]

Uhh so say AB = 1 so det(A) is a unit. In particular A must be injective by what I just said, and clearly A is surjective, so it has an inverse

I mistakenly said if A is not injective det A is zero, when really det A is only necessaeily a zero divisor, but all we needed was det A isnt a unit

[rain grove]

Ok I see, in the book it aslo says replace and not y = yx^{-1}. But still if I want that square of any element is in the centre then y would have to be any element. But y = zx^{-1} for any z and x. So I still have to prove that any element in the group can be written as zx^{-1} for some other elements z and x? But when I tried to prove that I can just say y = y . 1^{-1} = y but then x = 1 and... you see what I mean? Im sure im doing smth wrong tho

[tribal moss]

Oh right.

[dull ginkgo]

Okay this is probably my favorite proof of this lol

[rain grove]

Ok I like this a lot

[dull ginkgo]

Let (xy)^n = (yx)^n for any x,y
Then for any x, y
x^n = (xy^-1 y)^n = (y xy^-1)^n = yx^ny^-1 thus x^n y = yx^n so n-th powers are central :3

[dull ginkgo]

This is fucking ridiculous because i had a problem like this in Jacobson and it took me like half a page and then me just sitting on the fucking toilet i find a one line proof

Brain is stored in the colon

[rain grove]

Ok so now this is also true for rings. Groups with this proprety can be non-commutative (Quaternion group), I need to show that a Division ring with this proprety is commutative.

[dull ginkgo]

Same proof

Everything commutes with 0, and everything outside of 0 forms a group

[rain grove]

How is this same proof.. The only difference between groups and rings in this context is that rings have distributive law. So I need to use that right? Or no?

[dull ginkgo]

A better question would be to see if it holds for general monoids :3

[dull ginkgo]

The set of units of a ring form a group

In a division ring, every nonzero element is a unit

So we have that squared units are central with other units

The only thing to check too is 0

But 0 always commutes :3

[rain grove]

Set of units? There is one unit for + and one unit for *, so set of units is just {0,1} ?

[dull ginkgo]

In a division ring, the set of units is the set of invertible elements

[coral spindle]

(In fact in any ring)

[dull ginkgo]

In a division ring, every nonzero element has an inverse

[coral spindle]

The clash of terminology is a little unfortunate, but it's important to keep in mind that this is what people mean when referring to "units of a ring."

[mighty kiln]

Surely additive / multiplicative identity

[coral spindle]

Neutral element, identity, unit – there are several terms that people use

[dull ginkgo]

@rain grove in a ring, the multiplicative identity and additive identity aren’t called “units”, the units refer to elements with an inverse :3

[rain grove]

Yeah ok I understand, on top of that my book is in a different language too so im trying my best haha... and yeah from that just follows that squares of all elements commute with everything. I need to show that all elements commute with everything. So then it would be a field right?

[dull ginkgo]

Yes! A field is a commutative division algebra!

By definition. Some older literature calls division rings fields, and “modern” fields commutative fields

Same for vector spaces. Modules over division algebras are often called vector spaces anyway

[south patrol]

Particularly french

[dull ginkgo]

]:3[

[rain grove]

Ok I had a hunch to use distributive proprety and got $x(x+y) + y(x+y) = (x+y)^2 = x^2+xy+yx+y^2$. Now because $a^2$ for any $a$ commutes with everything also $a^2 + b^2$ for any $a$ and $b$ commutes with everything. So $(x+y)^2-x^2-y^2 = xy+yx$ commutes with everything. But now im stuck.

[cloud walrus]

OHHELLNAH

[dull ginkgo]

Once again

x and y commute for any nonzero element

and 0 commutes with everything

[rain grove]

Im sorry, why x and y commute for any non zero element?

I mean this is what im trying to prove

[tribal moss]

If you're in characteristic not 2, then set y = 1/2 in "xy+yx commutes with everything".

[rocky cloak]

Okay, so two things.

It's not necessary to prove that any element can be written as zx^-1. The proof establishes that xz^2 = z^2x for any x and z, hence any square is in the center. There is no need for y to take up every possible value.

Secondly, it's should be very straightforward to see that any element y can be written as zx^-1 (just pick ||z = yx||)

[dull ginkgo]

Also

I wonder if it holds for cancellative monoids

Probably not

[rain grove]

Is there always such element y = 1/2?

[tribal moss]

You're in a division ring, right?

[rain grove]

Yes

ohh so that is the inverse of (1+1)

[tribal moss]

Yes.

[dull ginkgo]

characteristic 2

Anticommutivity :3

[tribal moss]

Wait, what exactly are we assuming here? I lost the beginning of the conversation, I think.

At least I can't find anything that speaks about anticommutativity.

[rain grove]

original problem is:

Show that the center of a group, in which $(xy)^2 = (yx)^2$ holds for all elements, contains the squares of all elements. Such a group can be non-commutative. (find an example). Prove that, however, a division ring in which $(xy)^2 = (yx)^2$ holds for all elements is commutative

[cloud walrus]

OHHELLNAH

[tribal moss]

Thanks.

[rain grove]

Huh ok so characteristic is order of additive identity in the ring... what is wrong if characteristic is 2?

[tribal moss]

When the characteristic is 2, there's no "1/2" element, because 2=0.

[rain grove]

Ohh I see

so if characteristic is 2 then if xy + yx commutes also xy + yx +0= xy + yx -2yx = xy-yx commutes

[coral spindle]

Nicely done

[rain grove]

and then (xy-yx)^2 = ... = 0 (because squares commute) and I still need to show there are no divisors of 0

[tribal moss]

A division ring can never have (nonzero) zero divisors, since zero divisors cannot be invertible.

[tribal moss]

That's neat!

[rain grove]

yess nice

[tribal moss]

I wonder if there's a way to avoid needing a special case for char 2.

[tribal moss]

We have in general
(xy+yx)(xy-yx) = (xy)²-(yx)² + yxxy - xyyx = 0+0 = 0
so any two elements must either commute or anticommute.
Suppose a and b are nonzero elements that only anticommute. Then
(1+a)b = b + ab whereas b(1+a) = b + ba.
By assumption b+ab != b+ba, but b+ab + b+ba = 2b, so 1+a and b don't anticommute either, which is a contradiction ... unless 2=0; dang!

[coral spindle]

Well I mean, saying that two elements must commute or anticommute is sufficient for char 2 anyway

[tribal moss]

Sure sure, but that's still a case split.

[coral spindle]

So I guess while the proof isn't perfectly the same, at least you just have to do a little extra rather than a completely different approach

[arctic trail]

I mean isn't that the definition of a unit in any ring

[arctic trail]

yeah

easy to check invertibility if and only if det A is invertible in the ring of entries

This just goes to show how important the adjugate matrix is

[lean sail]

reading through Gallian chapter 1 - introduction to groups - quick question, what exactly does the author mean here "distance from the image of p to the image of q"

like, a literal image? or image as in when you execute a function on a variable?

[barren sierra]

well considering that this is talking about an object in the plane

it's both

So this is a hexagon in the plane

one plane symmetry of this hexagon is rotation about the center of the hexagon by 2pi/6 = pi/3 radians counter clockwise

this moves F into A, A onto B, B onto C, etc etc

so this is a plane symmetry of the hexagon (it maps the hexagon onto itself and preserves distances)

hence this rotation is an element of D12, the symmetry group of a regular 6-gon (hexagon)

@lean sail

so not only is it talking about the image of the function described by this rotation

but here we are talking about a literal figure in the plane (a literal image of sorts)

[hidden wind]

i have four days until my abstract algebra exam, ehehe, i’m so not ready

[barren sierra]

same!

well, 2 of them this week for me

we'll get through it

[still dew]

I was never ready for all the exams I have given so far

Just do it 👍

[hidden wind]

i have not even gone through half the curriculum

why am i like this

[lean sail]

is it saying kind of: consider points p, q... they will be the same distance from one another before and after the "image" is computed?

[hidden wind]

oh well i’ll pull through

[barren sierra]

yea that's the definition of a plane symmetry

think about the example I wrote out of rotation

[lean sail]

it's like... the image isn't stretching, it's either just rotating or reflecting

[barren sierra]

yup

[lean sail]

just a weird way to phrase it i think.

[barren sierra]

In fact, every element of D12 can be written as some sequence of rotations and reflections, which you'll probably see soon

[lean sail]

just $D_{12}$?

[cloud walrus]

proofman

[barren sierra]

no, D_2n for any n

I was just talking about D_12 since that's the example I looked at above

[lean sail]

oh gotcha, thank you

[surreal dagger]

I want to show that non units are closed under addition if 1+b is a unit for non unit b, assuming to the contrary that they are not:
Then (a+b)c=ac+bc=1 hence ac=1-bc. This should give me a condratiction but I dont see it, could someone give me a hint to see why 1-bc is a unit?
I know non units are closed under multiplication by any ring element.
If 1-bc is a unit then ac is a unit then a is a unit and we have a contradiction.

1+b being a unit means (1+b)d=d+bd=1 hence d=1-bd which kind of looks like 1-bc

[quiet pelican]

c is a unit and b is not, so bc is a non-unit, so -bc is a non-unit

[tribal moss]

It doesn't even matter that c is a unit: anything times a nonunit must be a nonunit.

[surreal dagger]

I just write it as 1+(-bc)?

[quiet pelican]

Yes

[surreal dagger]

Thanks

[rocky cloak]

Assuming we're in a commutative ring at least. But what they're trying to prove is true in the noncommutative case as well, so perhaps best to not assume.

[tribal moss]

Oh.

[surreal dagger]

Ah right thanks for that aswell, I missed that too

[rocky cloak]

I guess the proof requires a little more work than the argument above in the noncommutative case though

Or, no I guess it's fine since we're using that c is a unit

[barren sierra]

this implies noncommutative localization is a thing

I think

since 1 + a is a unit for all non-units a in R if and only if R is local

[rocky cloak]

I'm not sure it implies that, but it is true

[barren sierra]

it doesn't lol

[rocky cloak]

It implies local rings exist

[barren sierra]

just a funny thought tho

[rocky cloak]

A ring is local if
It has a unique maximal left ideal
iff
It has a unique maximal right ideal
Iff
It has 1+x a unit for any nonunit x

Iff the sum of nonunits is a nonunit
Iff
The nonunits forms an ideal

[tribal moss]

(But the equivalence of two of those conditions doesn't imply that there's something that satisfies them).
((We know there is for other reasons, ofc)).

[rocky cloak]

Yeah, fair

[arctic trail]

that's false

3+2 in Z/6Z

[quiet pelican]

They want to prove “b is not a unit => 1 + b a unit” => “non-units are closed under addition”

Which is true

[barren sierra]

Why is what is written here enough to show that the Galois Group is S3?

Why do we not have to rule out Z/6Z?

of course if I start writing out automorphisms I can find ones that don't commute

[quiet pelican]

We get that it is a subgroup of S_3 via its action on those roots

[barren sierra]

ah I see

so if the Galois group was Z/6Z, there would necessarily need to be 6 roots available to permute?

[quiet pelican]

There could be 5

Because Z/6Z embeds in S_5

[barren sierra]

oh yea

Ok then there would necessarily need to be 5

[quiet pelican]

I’m not sure if that’s impossible for other reasons though

But it does prove “at least 5”

[barren sierra]

yea hence why I said necessarily

[rocky cloak]

Has to be a transitive subgroup as well.

So you'd need 6 roots

[quiet pelican]

Oh yeah ofc

[rocky cloak]

Or I guess the polynomial doesn't have to be irreducible

In which case 5 is fine

[quiet pelican]

(x^2 - 2)(x^3 - 9) I suspect is fine for that but I can’t be bothered to prove it so 🙃

[grave sedge]

x^3-9 doesnt work

[rocky cloak]

Well I think that would give you C2 x S3

[quiet pelican]

Oh yeah I’m stupid (again)

x^3 - 1 then?

[rocky cloak]

But something like the minimal polynomial of cos(2pi/7) should do it

(times x^2 - 2 or whatever)

[rocky cloak]

That has Galois group C2

[barren sierra]

x^3 - 1 = (x^2 + x + 1)(x - 1)

[rocky cloak]

So (x^3 - x^2 - 2x - 1)(x^2 - 2)

[lone niche]

I'll give you a general statement: Let f(x) be a separable and irreducible polynomial of degree n over a field K. Then the galois group of f over K is a transitive subgroup of S_n.

[hidden wind]

whew how does that work

S_5 is much bigger so it's not too surprising i guess but i don't see it :S

[quiet pelican]

(123)(45)

[hidden wind]

right

[coral spindle]

Fun(?) fact: given some n, the smallest m for which Z/nZ embeds into S_m is given by the sum of the prime power constituents of n. So e.g., for n=6 we have m = 2 + 3 = 5. For n = 12 we have m = 3 + 4 = 7.

Getting from some number m to the largest n is harder!

[delicate orchid]

simply compute the lcm of the parts of every partition of m

[next obsidian]

Bonus: find a closed form for this number

[errant wedge]

I have a bit of a silly question. I see the ring hom properties listed as
f(x+y) = f(x)+f(y)
f(xy) = f(x)f(y)
f(1) = 1
a lot. f(0) = 0 isn't listed bc it follows, since f(x) = f(x+0) = f(x) + f(0) for all f(x) => f(0) = 0
We have f(x) = f(x1) = f(x)f(1) = f(x1) = f(1x) = f(1)f(x) for all f(x) => f(1) = 1, unless f is the 0 map. are there any other rng homs that would be miscounted as ring homs if we drop f(1) = 1 explicitly besides the 0 map?

Im guessing not

Yeah wait no

cuz f(1) = 0 forces everything to be mapped to 0, nvm

[lapis latch]

You can have non-zero maps where f(1) \neq 1. A good example is if you have a commutative ring with an idempotent element a then f(x)=ax satisfies the first 2 properties. The image is a “ring” in its own right just not with the same multiplicative identity

[next obsidian]

No this does not imply that f(1) = 1

Consider the map R -> R x S sending x to (x,0)

[hidden wind]

uwu

[next obsidian]

You’ll have f(1) = f(x) but f(1) is not 1

The issue is that inverses aren’t guaranteed to exist (or that zero divisors can exist)

[errant wedge]

right ok ty

[crystal vale]

Let R be a commutative ring such that f: R^n -> R^n is an injective module homomorphism, is f bijective?

Since { f(e_1), f(e_2),.., f(e_n) } spans im f, also { f(e_1),...,f_(e_n)) set of linear independent elements, so { f(e_1),...,f(e_n) } is a base for im f

And if R is commutative, and if R^m is isomorphic to R^n then m = n.

It shows every base of R^n has cardinality n, right?

So here we can show that { f(e_1),..., f(e_n) } is a base for R^n.

But I am not sure

[lone niche]

Let f:Z->Z by f(1)=2.

[crystal vale]

I see

Thank you

[mystic ether]

Every base of R^n is a LI subset of cardinality n but that doesn’t imply the converse.

[lavish nexus]

Calculate the number of homomorphisms from F_p^n to F_p^m

if m < n just 1

If m >= n do I just find the number of order p^n-1 elements in F_p^m*

Because F_p^n = F_p(α)

α has such order

[crystal vale]

Okay

[lone niche]

I think there's two cases here (for m>n)

When n|m, and when n does not divide m.

Any homomorphism is an embedding

so if n does not divide m, by a degree argument F_p^n does not embed in F_p^m

So only one homomorphism in this case too.

If n|m, then with the same idea of counting how many ways F_p^n embeds in F_p^m

Suppose we have a fixed embedding F_p^n->F_p^m

then composing it with an automorphism of F_p^m gives us another embedding.

And there are m automorphisms of F_p^m

[lavish nexus]

right this way it’s clearer

[lone niche]

By Galois theory, there is a unique subfield of size p^n.

So we are looking for automorphisms of that subfield only.

Which are n, so I think there's n homomorphisms.

[arctic trail]

Ohhhh

So 1+(a+b) is a unit?

[arctic trail]

isn't 1+b being a unit for all non units equivalent to the ring being local

[quiet pelican]

Yes

[lavish nexus]

The problem with just sending the generator to the elements that have the same order is that it might not be Frobenius

It’s only a multiplicative group iso

like α to α^11 in Z_2^4

has to do α to α^p

[lone niche]

I am back to thinking there's m homomorphisms, because I think any homomorphism F_p^n->F_p^n->F_p^m, where the second arrow is our fixed embedding, the left arrow an automorphism of F_p^n extends in m/n possible ways to F_p^m.

ping me if you get the answer some day please

[urban geyser]

Is $\Phi : G \to \text{Aut} (G)$, such that $\Phi : g \mapsto \phi(g)$ a homomorphism for $g \in G, \phi(g) \in \text{Aut}(G)$? Likewise, is $\phi(g) : G \to G$, where $\phi(g) : x \mapsto gxg^{-1}$ for $g,x \in G$, an automorphism?”

In other words, can I associate to every $g$ an automorphism of $G$; more specifically, is there a homomorphism which maps every element of $g$ to an automorphism of $G$ such that the centre of $G$ is the $\ker (\Phi) = { g \in G : \phi(g)(x)=x \mathrm{; for ; all ;} x \in G }$, i.e. all $g$ in $G$ for which conjugating by them returns the trivial automorphism. This can only be if $g$ commutes with $x$ for all $x \in G$, i.e. $gx=xg$, so that $gxg^{-1} = gg^{-1}x = x$. The identity is trivially in the kernel, but any other commutative element in $Z(G)$ is also in the kernel. The first part of this problem would include showing that $\phi(g)$ is an automorphism by showing that (a) it's a homomorphism such that $\phi(g)(h_{1}h_{2}) = \phi(g)(h_{1}) \phi(g)(h_{2})$ and admits an inverse $\phi(g^{-1})$. Then (b) we'd need to prove that $\Phi$ is a homomorphism by verifying that $\Phi(g_{1}g_{2})(h) = \Phi(g_{1}) \Phi(g_{2})(h)$. There's no requirement that $\Phi$ be surjective nor injective. Are there any other mappings aside from conjugation (by g) which could work here, since conjugations by $g$ are only a subset of the automorphisms of $G$?

[cloud walrus]

thimg

[urban geyser]

Be gentle

[knotty badger]

So it’s a little difficult to interpret your first paragraph

But it is indeed true that there is a group homomorphism G -> Aut(G)

Which sends a group element g to the automorphism “conjugation by g”

These are called the inner automorphisms

You can show that these always form a normal subgroup of Aut(G)

[knotty badger]

And indeed, I think the kernel of this homomorphism is the center of G

[dull ginkgo]

Does anyone have a good intuition of the crossed product for rings

Maybe as a generalization of semidirect products of groups

[urban geyser]

To make sure, the inner automorphism is φ(g) : x ↦ gxg⁻¹, which forms a normal subgroup of Aut(G)—every element g can be paired with an inner automorphism. The image of Φ in Aut(G) is the subgroup of inner automorphisms Inn(G), and the kernel of Φ is the centre of G (since any g that commutes will be invariant under conjugation)

Out of curiosity, is conjugation the only possible automorphism we can do this with? Someone tried left multiplication by g but that obviously couldn’t work, it wouldn’t exhibit a homomorphism; if we let φ(g)(h)=gh, then φ(g)(hh’) =/= φ(g)(h)φ(g)(h’)

But surely there’s more than just conjugation by g which could serve as the automorphism unto which g is mapped to—or maybe I’m missing a key detail that for any arbitrary group G, the elements of g can only be paired with specifically inner automorphisms.

“Normal subgroups are important because they (and only they) can be used to construct quotient groups of the given group. Furthermore, the normal subgroups of G are precisely the kernels of group homomorphisms with domain G, which means that they can be used to internally classify those homomorphisms”

Ah nvm I see

But I’d still like confirmation of what I said before to make sure I’m good before moving on

[grave sedge]

Is your question "Is the homomorphism G→aut(G) given by sending g to conjugation by g the only homomorphism G→aut(G)?"

[mighty kiln]

The explicit construction looks the same

[urban geyser]

I don't wanna ask stupid questions but I imagine we'd have the same result if we sent g to conjugation by e

Im in the car right now but if Im reading correctly then yes thats my question

Be gentle if it's silly

[grave sedge]

Then the answer is no

[urban geyser]

yiss

Good i have hope

[grave sedge]

As you said, there's the homomorphism given by sending everything to conjugation by e (which is equivalent to sending everything to the identity of aut(G))

[urban geyser]

Yiss

[grave sedge]

But you can also have things which are not given by conjugation

[urban geyser]

Are there any notable examples aside from identity and conjugation?

[grave sedge]

If G=Z/4Z, then aut(G) is cyclic of order 2, and conjugations are all trivial

[grave sedge]

You can have a nontrivial homomorphism from Z/4Z to a cyclic order 2 group though

[urban geyser]

Ah ah i see

[grave sedge]

Like in general you can have lots of "noncanonical" homomorphisms from G to aut(G) (if you consider them as two random groups and forget that one is the automorphism group of the other)

[urban geyser]

Alrighty good, i didnt have a chance to think of a specific concrete example but it would've been odd that only conjugation and identity do this

thank yew

[grave sedge]

Np

[dull ginkgo]

There’s the inclusion of regular elements that satisfy what appear to be weird and unintuitive relations

[mighty kiln]

?

The construction is to take the tensor product of the R-modules A and R[G] then define multiplication by (a⊗g)(b⊗h) = (aφ(g,b))⊗(gh) right

[dull ginkgo]

Which construction are you looking at

https://encyclopediaofmath.org/wiki/Cross_product

[mighty kiln]

Hmmmm

I guess I assumed ρ = 1

[dull ginkgo]

horror

*HORROR*

[delicate orchid]

I don't get it what's horrifying

calling a vector space P does make me sick actually

[dull ginkgo]

Cross product definition

[delicate orchid]

? ok?

[delicate orchid]

this is exactly analogous to a semidirect product

there's even a 2-cocycle I can see in there dawg

[dull ginkgo]

What the fuck is that

[delicate orchid]

who are you?

[dull ginkgo]

What is a 2-bicycle

Debilitated jacobi identity

[delicate orchid]

lemme think about this one

yeah that's right

Although I've only seen it for R[G] acting on another group algebra

[dull ginkgo]

How is it a direct generalization of semidirect with the 2-cocycle

[delicate orchid]

this case is literally the group algebra of the semidirect product

I don't know how much more direct u want your generalisation to be

you have a group G acting on a module A, and u form a new module using that action in the way darny described

[dull ginkgo]

Oh I thought you meant the generalization included the 2-cocycle part

[delicate orchid]

yeah the trivial 2-cocycle

[delicate orchid]

pattern match the first line of this to the first line of #groups-rings-fields message

[dull ginkgo]

What’s the deal with 2-cocycles anyway

Do they only really get motivated in homological contexts

[delicate orchid]

nah

but yes

[dull ginkgo]

I also just realized you can do semidirect products for monoids if you’re lazy and consider monoidal actions via maps into the endomorphism ring

But that sounds ill behaved as fuck

Cancellativity going bye bye probably makes it miserable

[dull ginkgo]

Fair enough

[delicate orchid]

u can think of them as "twists" on multiplication defined so that associativity is still ok. Like for instance projective representations can be thought of as SET maps f: G -> GL(n, C) such that f(x)f(y) = a(x, y)f(xy), where a(x,y) is some 2-cocycle. This becomes an actual homomorphism when we pass to PGL(n, C) because all the scalars get squished

that's the main motivation for me outside of cohomology

[chilly ocean]

It is, so much that people have defined a notion of a special kind of split extensions of monoids that behaves better and more similar to the group case called Schreier split extensions of monoids.

[dull ginkgo]

I need an excuse to look into homology, I just get rabbitholed massively by complexes and sequences

[delicate orchid]

draw a graph. Like a bunch of nodes and shit. Then work out it's graph homology

ok maybe don't make it too many nodes

[dull ginkgo]

I don’t know graph homology

[delicate orchid]

do you know regular homology

[dull ginkgo]

Not much, I know the def of a (co)homology for a complex

And have done some silly shit

[delicate orchid]

cause graph homology is just homology on a 2-skeletal (maybe 1-skeletal? I forget if it's a < or <= kinda deal) space

so ur complex is just like

C_2 -> C_1 -> C_0 lol

and ur boundary maps are just literally "what are the end points of this line"

so it's a great toy example to play around with

[dull ginkgo]

The fuck is a skeletal space bro horror

Eh I’ll get to it in Jacobson 2

[rocky cloak]

Does it actually contain any useful information as well?

(like about the graph or whatever)

[delicate orchid]

your highest simplices are 2-dimensional (triangles (the faces of the graph))

[dull ginkgo]

Brother I barely know what a simplex is

or graph theory

[delicate orchid]

U can get euler's formula from it iirc

[dull ginkgo]

I know the basics of algebraic homology

[delicate orchid]

for the plane, of course

[delicate orchid]

u don't need to know any graph theory u just need to know what a line is

[delicate orchid]

and the number of connected components of course

[dull ginkgo]

I’ll come back to it eventually

Is this mapping just a set mapping or is it Group-homomorphic

[surreal dagger]

What are the underlying reasons that every subring of Q is a localization of Z and do you know of other Rings that interact in interesting ways with each other regarding localization?

[dull ginkgo]

Because then wouldn’t the above sigma be sigma(g_2 g_1)

Giving a sort of conjugation - commutivity relation

[delicate orchid]

but instead it's a conjugation anti-commutivity kinda deal

I'll be honest I'm not sure why it's flipped. Possibly due to g acting on a Hom space (if that's the construction you're still working with)

[dull ginkgo]

h

oh cool

Okay this makes a bit more sense now

[barren sierra]

is there anything useful we can say about the multiplicity of roots of p_red when we aren't looking at an algebraically closed field?

[topaz solar]

well if you still are a field so as to allow funny division stuff, it'd still be 1 no?

(with a silly root-free part)

[barren sierra]

oh hm

yea idk what I was thinking

I thought there were issues with (x^2 + 1)^2 over R

[rotund aurora]

That breaks down in char p tho

[barren sierra]

but I did my math wrong

yea I should have specified

char p is weird

[topaz solar]

yeah but also many things involving p' shenanigans

[barren sierra]

I've seen some things about dealing with derivatives in char p

(useful for algebraic complexity)

[dull ginkgo]

So if $\rho(a,b)$ is the cocycle map into the ring, then the crossed product is the group ring $R[X]$ quotiented by the relation generated by relators of the form $(\alpha - \sigma_g(\alpha)\rho(g,h))gh$?

[cloud walrus]

The Library of Babble

[delicate orchid]

I am positively plussed parsing this purely preposterous proposition

there's no quotient at all I don't get it

[dull ginkgo]

Sorry let G be the group

[delicate orchid]

it's just this but u put the cocycle where u put it

I think

[dull ginkgo]

In $R[G], g(\alpha h) = \alpha gh$. In the crossed algebra: $g(\alpha h) = \sigma_g(\alpha) \rho(g, h) gh$

[cloud walrus]

The Library of Babble

[delicate orchid]

I haven't actually thought about this with a non-trivial cocycle

no scalar multiplication should work the same, it's still an R-module

[delicate orchid]

I don't see how that's equivalent to what u said but I believe this

[dull ginkgo]

I realized it wasn’t

Because the scalars don’t commute with the coefficients

In the crossed algebra

[delicate orchid]

burrrrpppppp

[dull ginkgo]

Okay this relation makes it a bit easier for me to understand

[delicate orchid]

wdym u have the definition right there

I'm presuming u know what all the letters mean

[dull ginkgo]

Yes

I can’t prove associativity with their def

[dull ginkgo]

If $\sigma(ab) = \sigma(a) \circ \sigma(b)$ like standard actions then it’s fine, in which this is equivalent to commutivity, not anticommutivity

[cloud walrus]

The Library of Babble

[dull ginkgo]

$(ag * bh)ck = (a \sigma_g(b) \rho(g,h)gh)ck = a \sigma_g(b) \rho(g,h) \sigma_{gh}(c) \rho(gh,k) ghk = a \sigma_g(b) \sigma_{gh}(c) \rho(g,h) \rho(gh,k) ghk = a \sigma_g(b) \sigma_{gh}(c) \sigma_g(\rho(h,k)) \rho(g,hk) = a \sigma_g( b \sigma_h(c) \rho(h,k)) \rho(g,hk) = ag( b \sigma_h(c) \rho(h,k)hk) = ag(bh * ck)$

[cloud walrus]

The Library of Babble

[dull ginkgo]

Actually that’s sorta interesting

[delicate orchid]

what you checking boss? associativity?

[dull ginkgo]

Let $A, B$ be monoids with $\sigma: A \rightarrow \End(B)$ being a monoid morphism. Also allow $f : A^2 \rightarrow B$ to be some map.

Equip $B \times A$ with binary operation $* : (x, a)(y, b) = (x \sigma_a(y) f(a,b), ab)$. This operation is associative iff f satisfies the same relations as rho

[cloud walrus]

The Library of Babble
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[dull ginkgo]

Just proved this

[delicate orchid]

rho being

[delicate orchid]

oh a 2-cocycle

yeah

[dull ginkgo]

Just instead of conjugation at the bottom

[dull ginkgo]

Yeah it forced f to be a cocycle with the commutativity relation

[delicate orchid]

now do u get what I meant by "twisting the multiplication in a way that ensures it's still associative"

[dull ginkgo]

Yes

:3

Thank you

[delicate orchid]

from here understanding their connection to group extenstions is just a hop skip and perhaps even a jump away

[dull ginkgo]

I didn’t expect an iff relation even for monoids lol

You can verify it by doing (1,g)(1,h)(1,k) and then checking (1,g)(x,h)(1,k)

(1,g)(1,h)(1,k) = (f(g,h),gh)(1,k) = (f(g,h)f(gh,k),ghk) = (1,g)(f(h,k),hk) = (\sigma_g(f(h,k))f(g,hk),ghk)
Thus: f(g,h)f(gh,k) = \sigma_g(f(h,k))f(g,hk)

[mighty kiln]

Is this true for all integral domains

As in, a subring of k(R) containing R is a localization of R

[dull ginkgo]

Consider the group of units of this subring

The group of units between {-1,1} and Q^times each uniquely determines a subring. It’s a localization because of it’s complement

What can you say about the complement of the group of units?

[chilly ocean]

I think so: Suppose T is a ring between R and its field of fractions. Let S be the set of elements of R that become invertible in T. Then S is a multiplicative subset and the localization of R at S should equal T.

[long obsidian]

Is anybody here familiar with the notion of a semifield P=(P,+,x) where (P,x) is an abelian group and (P,+) is a commutative semi group where x distributes over +. In fact most fields are not semi fields because the distributive law of fields is + distributes of x ( in this notation)

Apparently it is a basic fact that if a semifield P has a neutral element 0 over addition that is multiplicatively absorbing then P consists of a single element.

Can anybody see how to this is the case? I tried to do stuff like if a is in P then ax1=ax(1+0)=ax1+ax0 but didn't get anywhere

[coral spindle]

Well then (P, x) is a group with an absorbing element, so it must be trivial. Just look at its inverse and conclude that it must be the identity, but then by definition everything is the identity.

[candid dove]

Why is M a direct sum in 121

Shouldn't it be N

Considering that we are supposed to use 1st isomorphism theorem with the projection map?

To be clear i actually don't understand the hint in 121

[grave sedge]

The (a) in the direct sum should be the submodule generated by the preimage of a generator of (a)

(i.e. an element of N whose first coordinate is a)

Btw the direct sum decomposition should be of N, not M

[candid dove]

Yeah that's what I was saying

It should be N = sth + sth?

Ok actually I'm kinda stuck could you pls help me poiny out the key steps in 121

[delicate orchid]

if we're using strong induction it should be of M, but pi should be M -> R

[candid dove]

Oh ok

Wait im more confused

How exactly are we doing induction

[delicate orchid]

I think the idea is that you're assuming it's true for all k < n, and then showing that M splits as a direct sum R^(n-1) (+) R, which we're assuming are both free

and is thus itself free

oh wait I'm going the wrong way do everything I said backwards

[candid dove]

Sooo

M splits..

Then what

[hidden wind]

i know there is the classificiation of finite fields, but is there any sort of general classification of characteristic 0 fields?

[candid dove]

I have actually proved an intermediate result that free module with a finite basis is R iso to R^n

So I think we can directly assume M is R^n for some n

[grave sedge]

But M doesnt split that way

[delicate orchid]

Use the classification of F.G. modules over PIDs. Failing that: N < M = R^n, and let pi be the projection M -> R sending stuff to the first coordinate. By induction, we know that all submodules of R^k are free for all k < n, hence ker(pi) is free. The image is isomorphic to R and is therefore obviously free, if N is completely contained in either the image or the kernel we're done by induction. So assume N \cap ker(pi) and N \cap img(pi) are non-zero. Then the restiction pi|_N of pi to N gives us a decomposition N = ker(pi|_N) (+) img(pi|_N), these are submodules of ker(pi) and img(pi) respectively and are thus also free by induction, thus N is a direct sum of free modules and is therefore free

[grave sedge]

You want to show N splits so it's R ⊕ something that by induction is free

[delicate orchid]

I misread the question. Sue me

[delicate orchid]

I don't know what the hint is talking about

(a) need not be free

[grave sedge]

(a) is free because R is a pid

[candid dove]

No classification problems of the first statement before this and this is a problem course

Wait lemme show u the previous problems

[delicate orchid]

are you actually going to ignore the next four lines based on a joke?

[candid dove]

Me?

What joke?

[delicate orchid]

I'm not actually using the god damn classification

[candid dove]

[delicate orchid]

I've definitely heard this before and I don't buy it. U can show me a proof, an elegant simple beautiful proof. And I will not buy it

I will remain stuck in my ways

[grave sedge]

π: N→R projection on first coordinate, then image of π is (a) for some a. Take x in N such that π(x)=a, and show that N=(x)⊕ker(π). Now since ker(π) is a submodule of R^(n-1) you know it's free with ≤n-1 generators, and since (x) is free with ≤one generator you get N is free with ≤n generators. (If a=0 actually you need to do something slightly different, i.e. you must take x=0 or else the direct sum thing won't work)

[candid dove]

🔥

Aight thanks guys ❤️

[dull ginkgo]

Turns out it’s more general, you need the following: $\rho(g,h) \sigma_{gh}(x) \rho(gh,k) = \sigma_{gh}(x) \sigma_g(\rho(h,k))\rho(g,hk)$

[cloud walrus]

The Library of Babble

[dull ginkgo]

Which is frankly, fucking ugly

[delicate orchid]

I don't know what any of those letters mean dawg