groups-rings-fields

[final pasture]

(nvm I brainlagged for a small time )

yes

[chilly ocean]

Okay then. Thank you everyone for your help

I kinda missed this point

Thank you for being helped

[chilly ocean]

Lol

Don't make fun of me please.

[chilly ocean]

No no

I did not intend to make fun of you

[chilly ocean]

I thought you were being sarcastic

Hope you did not mind....

i forgive you this time

[warm holly]

Does the fact that if we take an application $\phi$, it's eigenvalues $\lambda,\lambda'$ and $v$ eigenvector associated to $\lambda$ and $v'$ associated to $\lambda'$ then $v$ and $v'$ are linearly independent follows from the fact that $V_{\phi,\lambda}\cap V_{\phi,\lambda'}={0}$ where $V_{\phi,\bullet}$ denotes a eigenspace associated to $\bullet$ eigenvalue?

[cloud walrus]

𝔻аniil

[warm holly]

(lambda' and lambda are distinct eigenvalues)

[vestal snow]

Lol i thought of proving that (which is a very simple proof), but I thought that it was too good to be true. Thanks for the help!

[flat spoke]

In the above example we have to check whether f(x) is irreducible over Q by applying the Mod p Irreducibility Test....

Can anyone explain me how they concluded that "h bar(x)" has no zeroes in Z2??

[carmine fossil]

If h(x) had a zero,it has to be 1

Clearly 1 doesn't satisfy

[flat spoke]

Can you pls elaborate this a bit...

[carmine fossil]

Suppose h(x) has a zero a,then h(x)=(x-a)g(x)

i.e.,constant term of g * a=1

i.e.,a is a unit

[flat spoke]

Ook...got it...Thanks a lot

[tulip crane]

how to prove that $\Pi iH_n(A{i})=H_n(\Pi_i A_i)$ with $A_i$ complex chains of $R$ modules?

[cloud walrus]

r2h

[limpid pilot]

Hi everyone

Can someone help me answering the following questions:

What are the concrete counterparts of a system, matrix, and eigenvalue?

Why do random matrices decode complex patterns so well?

<@&681260374879633482>

[old lava]

this seems more like a linear algebra question

but I suppose lin alg is technically part of abstract algebra

Matrices represent a lot of things well, since every linear transformation can be expressed as a matrix (once you fix a basis), and you can "zoom in" and approximate basically every function between vector spaces as linear locally, so matrices are a convenient, yet general representation of many complicated things

idk what you mean by concrete counter part to a matrix lol

it's like a generalized notion of a slope in some sense? (as least that's one way to think about it)

[vestal snow]

Can someone help me with this?

This is a very annoying exercise

I've shown that all elements a/1 of the localization with a irreducible in A and a not in the multiplicative set S are irreducible elements of S^-1(A)

I need to show that all irreducible elements of the localization are of the form a/s where a is irreducible in A and a is not contained in S

and then I need to show the uniqueness part

[gritty sparrow]

This seems to be untrue, if you localise Z at the set {6,6^2,..} 3/1 is actually a unit

[compact needle]

You want to be a little bit careful. In order for a/1 to be irreducible It is not enough that a is not in S, but rather that a does not divide any element of S. But the converse is also true, that if a divides an element of S, then a/1 is a unit. You should be able to use this to show that if a/s is irreducible, then you can write it in the form of a/s where a is irreducible and does not divide an element of S.

[vestal snow]

Yeah I sightly changed the strategy

First, notice that the localization at S is the same as the localization at the set of factors of the elements of S

At least in an integral domain

[compact needle]

I am confused about what exactly this means for an integral domain in general, but it makes sense for UFDs and should work for your purposes

[vestal snow]

Then prove that a/s is irreducible iff a is irreducible and a is not in S

After that it becomes pretty straightforward

I'm pretty sure that there is a way to prove this more directly

But it gets messy really fast

[gritty sparrow]

That is the standard proof i’ve read atleast, so I’m not so sure about that

[vestal snow]

Essentially the proof is careful bookkeeping

Which is why I described this as annoying

Also, I just realized your pfp is Rias Gremory lol

I always thought it was a leaf or something

[gritty sparrow]

I see you are a man of culture as well

[hidden haven]

A leaf?

[vestal snow]

I made two small mistakes which led me to the right conclusion

I need to prove that if a/s is irreducible, then a is also irreducible

Assuming that every factor of every element of the set S is contained in S

Let a=bc

Then a/1 = b/1 x c/1

Since a/1 is irreducible, we get WLOG b/1 is a unit

Using our choice of S, this implies that b is in S

However, we know that since a is not in S (as a/1 is not a unit), c is not in S

I don't know how to go from here

[vestal snow]

Do you remember the proof?

[gritty sparrow]

Well i’m pretty sure the claim isn’t true, for example if a is some irreducible, look at as/s=a/1 which is irreducible

But as is not irreducible

But you don’t need your claim exactly, with just the knowledge that a/1 is irred when a is irred we can do a factorisation in S^-1(A) and prove uniquesness

[vestal snow]

That's seems quite messy

Proving uniqueness that is

[gritty sparrow]

Well it isn’t so bad now that you know that S contains all divisors of its elements

It is a bit tedious, but it is not too hard

[vestal snow]

Hmmm okay. I'll try it in the morning.

Thanks for the help

[gritty sparrow]

Np

[glacial bough]

I'm checking if this set has structure of ring with the addition and product of complex numbers, do I just need to check if the addition is associative and the product is conmutative?

given that (Z,+,*) is a ring itself

[rustic crown]

One was would be to use the fact that C is a ring. So you would just need to check if this forms a subring. This saves you from verifying commutativity and associativity.

[glacial bough]

hmm I haven't learned anything about subrings

what do I need to check?

[prisma ibex]

you need to check that it's closed under these operations

you don't need to check associativity or commutativity or anything like that

you do need to show that 0 and 1 are in Z[i] (obvious) and that a and b in Z[i] implies a+b and ab in Z[i] (also obvious)

[glacial bough]

to check if its closed, it is just saying that (a+c) + (b+d)i is a complex number? because it has the form of a+bi?

same with the product?

[glacial bough]

also I dont understand what you mean in the second part

that the operations are closed?

[prisma ibex]

yes that's what it means for operations to be closed

[prisma ibex]

no

you're checking that sums and products of elements in Z[i] are elements in Z[i]

[glacial bough]

what does it means that an element is in Zi then?

I guess (a+c) is trivial because that is an integer, but I dont know what to do with (b+d)i

[prisma ibex]

uhhh a+bi is in Z[i] iff a is in Z and b is in Z

so both the statements reduce to statements about Z which are obvious

[glacial bough]

I also want to find which elements have inverse with the product

is that: a_1 * (a+bi) = 1 => a_1 = 1/(a+bi)

but i dont really know what to do from here

[prisma ibex]

Z[i] won't have inverses for the product generally

that's fine, it's a ring, not a field

[glacial bough]

and how to prove it? check if a_1 = 1/(a+bi) exists in Zi?

[delicate orchid]

1/(a+bi) = a/(a^2+b^2)-bi/(a^2+b^2), which means that a/(a^2+b^2) must be an integer which is only true if a is +/-1 and b is 0, same is true for b/(a^2+b^2) integer
=> a = 0, b = +/-1 so the units are {1, -1, i, -i} (omg isomorphic to Z/4Z)

I can explain why a/(a^2+b^2) is an integer => |a| = 1 b = 0 if you want but I think it’s clear

[sinful mirage]

can someone help me justify 22?

why can all the reps of the above given 5 dimensional algebra be obtained from the clifford algebra with dimension 4?

[hot lake]

"clifford algebra with dimension 4"

what does the ' mean in (20) ?

[sinful mirage]

B' is a priori not related to B. it's just a vector operator and mu_ are its components

A',A,C',C are numbers,B',B are vector operators

[hot lake]

so if you have 10 matrices satisfying (20) there exists 5 matrices gamma_alpha and beta satisfying (21) such that (22) holds ?

[sinful mirage]

yes(this I should prove)

and i;m trying to justify why 22 holds and why can all the reps of the algebra 20 be obtained from the reps of 21

[hot lake]

is the B in (22) supposed to be a beta ?

[sinful mirage]

actually I think there should exist 4 gamma matrices instead of 5,right?

[hot lake]

uh yeah I was thinking 5 = 4+1 and messed up while writing it

[sinful mirage]

yeah I think it's a typo

[hot lake]

well it looks like solving for beta and gamma_alpha in terms of Bi, B'i is pretty easy so it should just be a matter of checking that everything works out fine

that plugging (22) into (20) gives equations that can be deduced from (21), and that plugging the inverse of (22) into (21) gives equations that can be deduced from (20)

[sinful mirage]

but what I don't see how can the dimensions match?

at one we have indices going from 1 to 5,at the other from 1 to 4

[hot lake]

what's puzzling to me is that they "compress" 10 matrices into 5 while keeping very similar-looking contraints

I don't see any problem with the indices

[sinful mirage]

i mean the dimensions is my issue

[hot lake]

B5 and B'5 give beta, and B'alpha and Balpha give gamma_alpha once you know beta

[sinful mirage]

one algebra is generated by 10 matrices,the other by 4

[hot lake]

5

but if it really does work, I don't see why that would be a problem

[sinful mirage]

i'll try doing this

[hot lake]

the equation B'iBi + B'iBi = -2 says that B'i is entirely determined by Bi, so it's not that surprising anymore that you can compress the data

[steady axle]

I have a real m x n matrix A. I want to prove that $N(A) \cap im(A^T) = {0}$
so here is a proof:
A has smith normal form $A=PD(a_1,\dots ,a_k,0,\dots , 0) Q $ and so $v\in N(A)$ means that $Qv$ has first k entries 0. let $A^Tu=v$ then $Q^T D()P^T u=v$. Also $D()P^T u$ is a vector whose only first k entries can be non zero so if we call it w then $v^TQ^Tw=0$ but this implies $v^Tv=0$ i.e. v=0.

[cloud walrus]

bert

[steady axle]

does this look good ?

[unique juniper]

Yes I think so

[vestal snow]

@gritty sparrow I found the correct statement for the irreducible thing: a/s is irreducible iff the unique decomposition of a contains exactly one irreducible not in S

[simple mulch]

Hey guys

Is this ok?

I wonder if I did the associative right

[devout crow]

yep

[simple mulch]

ok nice, ty

[tawny pine]

@simple mulch identity & inverse have dead weight. no need to show work to find the identity or inverse. just show that 2 serves as the identity and that 4/x serves as the inverse of each x

[simple mulch]

Oh ok, thats simpler

[gritty sparrow]

Yeah that sounds correct

[glacial bough]

how can I find all subgroups of (Z11,*)?

[sharp sonnet]

what have you tried?

[glacial bough]

I know that the identity (1) is one of them

[chilly ocean]

you mean (Z11)*?

[glacial bough]

yeah z11 and product operation

[chilly ocean]

there's another easy one

[chilly ocean]

all of Z11 won't be a group with product operation

[glacial bough]

what do you mean?

[chilly ocean]

what's the inverse of 0?

[glacial bough]

hmm I dont think all elements needs to have inverse in order for a set to be a group?

[chilly ocean]

then you don't know what a group is.

You might want to learn that before trying to prove something about subgroups.

[still stream]

I am confused in definition of group action. Is there any simple definition of this?

[hidden haven]

The definition is that 1•s = s and (gh)•s = g•(h•s), for all S in the set and all g,h in the group. If it's motivation that you want, then look at the example of the symmetric group on n elements acting on a set with n elements.
A group action is like a vector space, but instead of a field acting on an abelian group by scalar multiplication, it is a group acting on a set by permuting its elements

[sharp sonnet]

alternatively its a group homomorphism from some group G into the symmetric group of some set

[still stream]

That's easy one.

[sharp sonnet]

my favorite example is the group of symmetries of the cube acting on the edges/vertices/faces/etc of the cube

[magic owl]

the defn loch presents foramlizes what it means for a group action to tell you about symmetries of a space

you can think of an automorphism group as the collection of all acceptable symmetries and a group action is basically just a subgroup of that

for sets the automorphism group is the symmetric group

[lavish gale]

i am trying to show that the group of automorphism of Zn is isomorphic to U(n). i do know that each automorphism must map 1 to one of U(n). How do i make this statement concise?

[chilly ocean]

Note that elements of U(n) are coprime with n

Can you see what the order of them has to be?

[lavish gale]

ya i do know that any automorphism of Zn must map 1 to an element of U(n)

[chilly ocean]

i think that's not what i asked

[lavish gale]

every element of U(n) has order n

[chilly ocean]

can you use this to your advantage?

[lavish gale]

🤔

ahh

ty

[chilly ocean]

yo question, this problem seems too easy and I might be missing something. Given I ideal of R, show that if I has element e neutral wrt to multiplication, then I=eR and e is central idempotent of R. So I=eR because for any x in I x = ex. Central idempotent means ex=xe for any x in R. But if it's neutral then ex=x=xe. Is that all?

[delicate orchid]

yeah that works

if the multiplicative identity is in an ideal then the ideal is the full ring

cause 1x must be in the ideal if 1 is in the ideal for all x in the ring but 1x = x so for all x in the ring x must be in the ideal

[oblique river]

wew I think you misunderstood the question. It doesn't say that the the identity of R is in I, it says that I has an element which acts like the identity for all other elements of I, but not necessarily the identity of the ring

for example if you look at the ring R x R then R x {0} is an ideal and (1,0) is neutral w.r.t. multiplication in that ideal

but it's not the identity of the ring

[chilly ocean]

yeah, but I don't think I assumed it is an identity?

[hidden haven]

wew did

[chilly ocean]

ye ye ye

[oblique river]

Yeah sorry ledog i think youre good

Just talking to wew

[delicate orchid]

oh whoops

sorry my bad ledog

bit of a 4 times 8 moment if you know what I mean

[chilly ocean]

no I dont

[delicate orchid]

don't look at the pins

[chilly radish]

Trying to prove that that $(\sfrac{\Z}{p\Z})^{\times}$ is cyclic for a prime $p$, any leads (Please nothing too detailed, I wanna try breaking my head on this for a bit). I know up to and including sylow

[cloud walrus]

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

[chilly radish]

ffs

I mean the multiplicative group of the integers mod p

[chilly radish]

Please i'm dying

[delicate bloom]

it's a pretty hard proof; you don't construct a generator

[chilly radish]

Hmm

Damn

Ok what I'm trying to prove is that Aut(Z/pZ) has a unique subgroup of order 2

I thought since I know that it's isomorphic to the multiplicative group I could use that to prove it's cyclic then it has a unique subgroup of every order that divides it

But is there an easier way?

Cuz what I WANT to prove is that every nonabelian group of order 2p js isomorphic to the dihedral group, and i'm trying to do that through the semidirect product

But I need to prove that the automorphisms are the same (Namely the identity, and the inverse)

Where C2 acts on Cp

[sturdy marsh]

do you know the classification of finite abelian groups

[chilly radish]

No. I mean we're currently studying it but this problem set is due next week so I don't think that was the intent

[sturdy marsh]

do you know the sylow theorems

[chilly radish]

Yes

[sturdy marsh]

alright, then show that any finite abelian group is a product of its sylow subgroups

and then show that all the sylow subgroups are cyclic

[hot lake]

.... can't you just show that (Z/pZ)* has only 2 elements whose order divides 2 ?

[chilly radish]

You'd have to calculate the order of the elements for that tho

[hot lake]

one of them being 1

[sturdy marsh]

x^2 - 1 can have at most 2 solutions in Z/pZ

[chilly radish]

Hmm, that's definitely easier

[sturdy marsh]

yeah in the end you would almost certainly have to use a fact about # of solutions to a polynomial equation over a field

or a domain

[chilly radish]

I still wanna try showing this tho

Just for sport

[chilly radish]

In what way

[sturdy marsh]

you'll see if you try to write down a complete proof

[chilly radish]

For Z/pZ* having a unique subgroup.of order 2, or being cyclic?

[sturdy marsh]

being cyclic

[chilly radish]

Ah gotcha

[sturdy marsh]

you might be able to get away by clever counting arguments

in the Z/p * case

idk

[chilly radish]

I'll give it a try

When I have time that is

[hot lake]

I'm not sure what you want him to do for example if p=17 that sylow theorem doesn't tell you anything

[sturdy marsh]

if you write it as a product of sylows

[hot lake]

G = G

done

[sturdy marsh]

not Z/17Z

Z/17Z *

[hot lake]

yeah that has order 16

which is a prime power

[sturdy marsh]

okay, now count solutions to a polynomial equation

if it wasnt cyclic

everything would satisfy x^8 - 1

which has at most 8 roots

in Z/17

this is the proof that you do while showing that a finite multiplicative subgroup of a field is cyclic

[hot lake]

yeah it's pretty much about solutions to polynomial equations in a field

[sturdy marsh]

so in general, write it as a product of sylows H_1 \times ... \times H_K, show that all the sylows are cyclic, and then (h_1, ..., h_k) would be the generator of the product

where each h_i generates H_i

[chilly radish]

isn't the product cyclic only if the order is coprime

[sturdy marsh]

the orders are coprime

[chilly radish]

or are we taking a representative for each sylow subgroup

for each p-sylow subgroup I mean

[sturdy marsh]

there is a unique p-sylow

the group is abelian

all the p-sylows are conjugate \implies only 1 p-sylow

[chilly radish]

oh right

I forgot about the abelian part haha

btw I finished the problem using the fact that there's a unique element of order 2 in Z/pZ* so thanks!

Totally slipped my mind, went straight to trying to prove it's cyclic

[vestal snow]

What exactly does agree on overlaps mean here? Let $\pi_i: U_i \longrightarrow Y$ be a map of ringed spaces (i.e. there is a sheaf map $\varphi_i: O_Y \longrightarrow \pi_{i*}O_X|_{U_i}$. How exactly do I restrict the $\varphi$ to the overlaps?

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

The only way I see to define it is by $\varphi_i|{U_i\cap U_j}:O_Y\longrightarrow \pi_i|{U_i\cap U_j*}O_X|_{U_i}$

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

Does this seem reasonable?

[compact needle]

This should be correct

[vestal snow]

Thanks

I figured I'd verify this before trying to prove this

[compact needle]

Literally this should be saying that the compositions $(U_i \cap U_j,\mathcal{O}X\vert{U_i\cap U_j}) \rightarrow (U_i,\mathcal{O}X\vert{U_i})\rightarrow(Y,\mathcal{O}_Y)$ and $(U_i \cap U_j,\mathcal{O}X\vert{U_i\cap U_j}) \rightarrow (U_j,\mathcal{O}X\vert{U_j})\rightarrow(Y,\mathcal{O}_Y)$ are the same maps

[cloud walrus]

Turgul

[compact needle]

But that's what $\pi_i\vert_{U_i \cap U_j}$ should mean

[cloud walrus]

Turgul

[cloud walrus]

Yes ツ

[dusty river]

@unique juniper e_1^3 - 3e_1e_2 + 3e_3, where e_i is the elementary symmetric polynomial of degree i. See https://en.wikipedia.org/wiki/Newton's_identities.

[unique juniper]

thank you so much

[deft oasis]

if algebra is already an abstract mathematical object then what is the abstract algebra?

[junior moth]

#math-discussion

[delicate orchid]

algebra has numbers in it

[carmine fossil]

ever heard of convolutions?

[deft oasis]

here and there, what of it? 🤠

[carmine fossil]

abstract algebra is an abstraction of algebras

And abstraction is a term that will be defined soon

[tough raven]

What if I want to abstract out "abstract" and consider other adjectives?

[carmine fossil]

Ergo adjective theory is born

[tough raven]

Ahh

I see

Enlightening

Should I start enlightened algebra

Thanks

[quaint tree]

Does anyone know any good online lectures for learning galois theory? I'm planning on using Stewart's Galois Theory for textbook and exercises, but lectures would be useful too.

[final pasture]

careful, that looks a bit like a categorical idea

[tough raven]

Save me

[simple mulch]

Can anyone give an hint to solve 6?

[hot lake]

solve 5

[simple mulch]

I did

its 1 or a word with length n with 1s

(?)

[hot lake]

???

[simple mulch]

isn't?

[hot lake]

say n=7, what's the inverse of 1110100 ?

[simple mulch]

0001011 ?

[hot lake]

what's 1110100 + 0001011 ?

[simple mulch]

a word with full 1?

[hot lake]

what did you get for question 4 ?

[simple mulch]

0?

[hot lake]

is 1111111 = 0 ?

[quaint tree]

Shouldn't every element be its own inverse?

[simple mulch]

but 4 asks for the identity element?

[hot lake]

yes

[simple mulch]

1 + 0 = 1

0 + 0 = 0

[hot lake]

what's the definition of inverse

[simple mulch]

x*x' = e

[hot lake]

besides, 0 isn't even in B^7

so is it true that 1110100 + 0001011 = e ?

[simple mulch]

good point

[quaint tree]

Would it be easier to do the exercise for $\mathbb{B}^{1}$ first, then try $\mathbb{B}^{2}$ to see what carries over?

[cloud walrus]

Bannanachair Monarch

[hot lake]

maybe

[quaint tree]

And then after doing that try $\mathbb{B}^{7}$

[cloud walrus]

Bannanachair Monarch

[simple mulch]

I didn't used the definitions to solve the exercise

[quaint tree]

Well that's an issue.

[simple mulch]

That's why this happened

[quaint tree]

Use the definitions next time

[simple mulch]

overconfidence

Yeah.. now I know lol

[quaint tree]

Like, how do you prove anything without using the definitions of the objects you're proving things about? Unless you're building off of other theorems that have used the definitions, but what theorems would be helpful for this?

[simple mulch]

I don't know of any.. thanks anyway

[simple mulch]

Is it ok?

[quaint tree]

You're assuming $a+b=a-b$ and showing that $a=a$ in that proof. Try going the other way.

[cloud walrus]

Bannanachair Monarch

[west violet]

any good intro to crypto books? ik crypto requires understanding of Elementary NT as well as basic abstract algebra so I was wondering if any book covered that as well

[ivory dust]

if we try to factor a polynomial over Q[x] and we factor it as far as we can then one of the factors is irreducible we must stop there correct?

No otherway to factor

[uncut girder]

You need all the factors to be irreducible, not just one

[devout crow]

given that $\text{GL}(2,2) \cong S_3$ and $\text{GL}(2,3) \cong Q_8 \rtimes D_6$, is there a similar expression for $\text{GL}(3,2)$?

[cloud walrus]

∧res

[compact needle]

$\text{GL}(3,2) \cong \text{PSL}(2,7)$, if that's what you are looking for. It is the second smallest nonabelian simple group, after $A_5 \cong \text{PSL}(2,5)$.

[cloud walrus]

Turgul

[vestal snow]

Do we need locally ringed morphism for the red part?

$p\in \pi^{-1}(D(b)) \iff b \notin \pi(p) \iff b\notin Spec(\pi^#)(p) \iff b \notin (\pi^#)^{-1}(p) \iff \pi^#(b)\notin p \iff p \in D(\pi^#(b))$

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

Am I implicitly using something about pi being a locally ringed morphism here?

[gritty sparrow]

What do you mean by $spec(\pi^#)$ exactly? In any case I’m pretty sure that the equality between the second and fourth expressions needs the morphism to be of locally ringed spaces

[cloud walrus]

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

[gritty sparrow]

The proof is to localise spec(A) at p, and spec(B) at pi(p) and because it is a morphism of locally ringed spaces the whole diagram commutes, with the preimage of the maximal ideal being the maximal ideal in O(Spec(B))_pi(p)

[vestal snow]

Can you clarify which expressions you are referring to?

[gritty sparrow]

I mean in that chain of equivalences you wrote here

[vestal snow]

Did you mean second and third expressions?

[gritty sparrow]

Well i didn’t really get what the 3rd expression was (are you applying the spec functor on pi^#?). I meant the second and 4th expressions

[vestal snow]

Spec(\pi^#) is the map induced by \pi^# on the spectrum of the ring

Oh I think I see your point

I proved that pi = spec(pi^#) earlier

and I had used local there

Thanks. That cleared it up

[gritty sparrow]

No problem

[fickle pewter]

Hii, I wanna ask some question if you guys don't mind.

[dusty river]

go ahead

[fickle pewter]

Lately, I study about r-ideals in a commutative rings.
Based on the journal, it define r-ideals as a part of ideals.

A proper ideal I in a ring R is called an r-ideal, if ab is an element of I with ann(a)=0 implies that b is an element of I, for every a,b are element of R.

Do you think that r-ideals can be find on a ring? I mean, can I generate it on a ring which is not commutative? or is it only for commutative ring?

I've learned it for a month and can't find the answer :((

[dusty river]

I think you can define it the same way for non commutative rings? but then they may not be very useful

[fickle pewter]

I think the same way but ideals in a rings divided into right and left.-.

[dusty river]

even then the definition would work, you just won't have as many nice properties. You can take ann(a) to be the 2 sided annihilators of a or something like that

I haven't seen this definition before though, so I'm just guessing

[fickle pewter]

ahhh I see

[fickle pewter]

my teacher gave me this topic to see if it can be defined in rings and integral domain

anyway, thank you for sharing your thought

it helps me a lot

[dusty river]

np

[frank fiber]

how to prove that direct sum conmutes with homology in R-mod?

[vestal snow]

IIRC you have to use the fact that direct sum acts as both the product and the coproduct in R-mod

[ancient night]

Is there a name for a special case of Euclidean domain where there exists Euclidean division function, i.e. where q and r are unique quotient and remainder.

[final pasture]

that doesn't give you a name but:
https://mathoverflow.net/questions/109785/what-structure-supports-division-to-a-unique-quotient-and-remainder

[quaint ivy]

well, the citation in the top answer shows that any Euclidean domain with that property must be either a field or a poly. ring over a field, which is probably there isn't a name for it

[ancient night]

@quaint ivy I didn't see a proof that there is no single alternative. Anyway, thank you!

@final pasture And you too! I don't know why Google didn't suggest me a link to this question on MO.

[quaint ivy]

often it's best to just search MO and Math StackExchange directly for these questions

[ancient night]

n-cat lab has been a great aid for me as well. In addition, I could run into an article about it on arxiv.

Yet your point does make sense if Google search lets me down like this

[quaint ivy]

I use DuckDuckGo, which has https://duckduckgo.com/bang

so you can e.g. type "!mo [query]" to search [query] on Math Overflow

and same for many other websites, you can even submit your own

[ancient night]

Oh, that's awesome! Maybe there is a plugin for it in Google Chrome too, but I didn't use one before.

Hmmm... Thank you once again!

Ah, I see. It is like doing ``https://mathoverflow.net/" + tab + search term

[quaint ivy]

yeah but shorter, I just set DDG as my default search engine

[quaint ivy]

anyway I might read the paper cited here later I wasn't aware of this result actually

[final pasture]

me neither

[quaint ivy]

related MSE question https://math.stackexchange.com/questions/2171776/uniqueness-of-the-remainder-and-quotient-in-an-euclidean-domain

[kindred mist]

This is correct, right? (I know how to prove it unless I am missing something silly)

[kindred mist]

[gritty sparrow]

Yes this is correct

In fact i think we only need R to be an integral domain for this theorem

[kindred mist]

thanks

[vestal snow]

The map on the topology would be $\oplus D_i \mapsto D_0$ and the map on the distinguished basis would be natural map $A_f\mapsto ((S_{\bullet})_f)_0$ right?

[cloud walrus]

Have a Banana, Bitch

[vestal snow]

To see that this map is a map of locally ringed spaces, notice that if $Q_0$ is a prime ideal of A and g is caontained in $Q_0$, then g is contained in $\oplus Q_i$ (the construction of $\oplus Q_i$ is a bit weird, but all we're using here is that it contains $Q_0$.)

[cloud walrus]

Have a Banana, Bitch

[gritty sparrow]

Yeah this looks correct, I don’t really get your argument for why it is a locally ringed space though. I would just say that “gluing” locally ringed morphisms on a base will give a locally ringed morphism on the whole space

[vestal snow]

That's what I did. I probably should have clarified, but the idea was:

1. Construct the morphism from using the distinguished base
2. The stalk at each point is the natural map $A_{Q_0} \longrightarrow (S_{\bullet})_{\oplus Q_i}$ and we can check that this is local using my argument

[cloud walrus]

Have a Banana, Bitch

[gritty sparrow]

Oh, I see what you mean now. Yeah that looks correct

[vestal snow]

By no zeroes, I'm guessing that it means that there is no point p such that all f_i become zero in the local field of p

Is this what the exercise is trying to say?

[gritty sparrow]

Yes

[vestal snow]

The obvious way to do this is to send each point p to the value point [f_0(p),...,f_n(p)] where f_i(p) represents the image of f_i at the local ring at p

But the image of f_i at the local ring at p might not be an element of B

[oblique river]

I interpreted "functions on X" to mean "functions X to B"

maybe that doesnt make sense though

[vestal snow]

Like set-theoretic functions?

[oblique river]

no, scheme-theoretic functions

oh wait sorry

X to Spec (B)

wait nvm that makes no sense

because there's one map from X to Spec(B)

nvm me lol

[vestal snow]

If it helps, this example was right under it

[oblique river]

wait

no this is all fine

because the points of P^n(B) are not just

(n+1)-tuples of elements of B

for example, think of A^n

even A^1(Z) = Spec Z[x]

it's not just Z pointwise, because there are nonlinear prime ideals, like (x^2 + 1)

I think you're trying to think of P^n(B) as if B were algebraically closed

where you could just think of it as tuples of elements of B

[vestal snow]

Let's think of it as Proj B[x_0,...,x_n]

Which is the right way to think of this right?

How do I construct a map from X to this using f_i

[oblique river]

I don't think that's the right way to think about it

Proj behaves really badly

[vestal snow]

Affine glues?

[oblique river]

and I'm not even sure the mapping property you want is true in this case

yeah I think you want affine glues

if you could do the Proj thing, basically you would be looking for a map B[x_0, ..., x_n] into the sections on X

[vestal snow]

Smh this is supposed to be an easy exercise

[oblique river]

and you would basically want x_k --> f_k

so just... affinize that

[gritty sparrow]

Well how i would do it is that for each fi we have a map X_fi to a standard open Ui =spec(B[x0,..xn] by thinking of the map from that ring to the section sending xk to fk/fi

Does that make sense? Intuitively this is very similar to your coordinates thing

(Also X_fi cover X obviously)

[oblique river]

something like that should work, yes

[vestal snow]

Thanks for the help

I think I'll just skip this one

[gritty sparrow]

Damn, that is a little sad. It shouldn’t be too bad now with this map, but i’m not sure

[vestal snow]

I might come back to it later, but I'm kinda busy speedrunning AG to do time consuming applications (that are not relevant to function fields)

Also, Buncho you're into arithmetic geometry right?

[gritty sparrow]

I see, interested in some number theory stuff?

[vestal snow]

Yeah! I'm actually gonna be continuing a research project this fall with a professor

I somehow got away with not knowing any AG till now, but I need to know some fancy stuff like differentials, line bundles, stacks to continue

@oblique river Is it fine if I DM you for some advice?

[gritty sparrow]

Damn that is pretty impressive, best of luck

[oblique river]

what kind of advice?

[vestal snow]

The stuff in AG that I can skip (for now) for the research thing I wanna do

[oblique river]

oh I see -- I can try to answer that, yeah, but I might not be able to give a great answer

if I'm not super familiar with the kind of problem you're doing

but yeah, go ahead

w/ number fields?

[rugged sage]

Hello, sorry if this is the wrong place for this but I was wondering how I could get started with Abstract Algebra?

(I'm mostly interested in Group Theory and Category Theory, but from what I've been told starting with AA is a good place to start)

[dusty river]

#book-recommendations has a bunch of AA books listed in a pinned message

[thorn delta]

How are we considering these objects in the isomorphism below? AM only ever talks about tensor products of A modules as A modules, but I think this is at best an isomorphism of abelian groups, right?

[hidden haven]

They are simultaneously A and B modules so I imagine the isomorphism is as both A and B modules

Which I'm guessing is the definition of a bimodule isomorphism

Wait nvm only N is the bimodule

[final pasture]

Aren't the resulting tensor products both A and B modules ?

so like, this iso should be a bimodule iso ?

[thorn delta]

well hmm, maybe something like a((m\otimes_A n) \otimes_B p) := (am \otimes_A n) \otimes_B p is well-defined for a similar reason as M \otimes_A N

[final pasture]

actually nvm idk

[hidden haven]

oh yeah A can act on one entry and B on the other

[thorn delta]

yea okay, pretty sure that would work now

[unique juniper]

for part b

a primitive root may not equal the same (primitive root)^i

?

[dusty river]

i will depend on what b you pick

if you pick a itself then i=1

[unique juniper]

oh right

[gray comet]

Yes, without more assumptions on M and P this would only have the structure of an abelian group

[gray comet]

Not in general, no. You would need M to be an A-bimodule, for M \otimes_A N to be a (left) A-module

Since $N$ is an $(A,B)$-bimodule, however, $M \otimes_A N$ is a right $B$-module, which is why the tensor product $(M \otimes_A N) \otimes_B P$ makes sense in the first place

[cloud walrus]

Tormeson

[young frigate]

I need to prove that with a finite group G and a subgroup U if this is true, U is a normal subgroup of G and I have no idea how to start

[chilly ocean]

lagrange

[hidden haven]

How Lagrange?

[chilly ocean]

look at cosets

not lagrange I guess lol

I just remember the proof of index 2 implies normal which is the same thing

[hidden haven]

Right

[young frigate]

Sorry for the late answer, lagrange just says that the order of a subgroup must be a divisor of the order of the group

Or am I missing something?

And in addition to that that the order of G/U must be the order of G divided by the order of U

So I guess the quotient group G/U has the order 2, but what do I do with that?

[hidden haven]

How many groups of order 2 are there?

Actually you can't quotient

[young frigate]

Z_2

[hidden haven]

Because you don't know if it's normal yet

But as ledog said, look at the cosets

[young frigate]

isn't this lagranges theorem?

[hidden haven]

Yeah, but that's not useful here

Look at ledog's later message

[young frigate]

what's a coset?

[hidden haven]

Well he only says look at coset

oh ok wait

[young frigate]

wait a sec

oh

nvm

I know what it is

[hidden haven]

So you need to show that for any g, gUg' is U, where g' is g inverse

[young frigate]

english isn't the language I'm studying maths in

[hidden haven]

Ah I see

[hidden haven]

So try to do those 2 multiplicative one at a time and see where you land

[young frigate]

But I don't know what G is

I mean I guess it's a finite group, and thus its kind of cyclic

[hidden haven]

Yes but you can still do something from the fact that a left coset of U is either U itself or disjoint from U

Kind of cyclic?

[young frigate]

that's not a proper property, but each element has an order, so you can make a group from any element

generate would be the better term I think

[hidden haven]

Ahh right

[hidden haven]

But you can use this

[young frigate]

a left coset, so gU

[chilly ocean]

there are only 2 cosets, do you see this?

[young frigate]

yes, since we already have all but two elements in U

so we only have a left and a right coset, since one of the elements must be the neutral element

[chilly ocean]

yeah, the cosets are therefore U and G\U.

[young frigate]

ok, and then?

[chilly ocean]

you wanna show taht xU=Ux

[young frigate]

the element we used for the coset must be self-inverse, right? so we get xU=U

[chilly ocean]

no

xU=U for x in U

[young frigate]

right

[chilly ocean]

so xU=U=Ux for x in U trivially, but you want to show that's true for any x in G (this is the definition of a normal subgroup)

[young frigate]

for x in G it only needs to be xU=Ux, right?

[chilly ocean]

yes

[young frigate]

Give me a minute, I can't concentrate that well anymore

[chilly ocean]

I leave the rest to you, take it easy and think it over.

[young frigate]

okay, thanks

[young frigate]

I think I got it: for any g in U, gU=U=Ug, and for any g not in U, gU=G\U=Ug

[hidden haven]

Yep

[young frigate]

whoooo

[hidden haven]

[frank fiber]

in R-mod, the kernel of $f:A \rightarrow B$ is the object $ker f$ with the inclusion $f:ker f \rightarrow A$ right?

[cloud walrus]

Or x1

[frank fiber]

so, is every monomorfism a kernel inclusion?

[gritty sparrow]

Yes let N be a submodule of M, N is the kernel of the standard map from M to M/N

@frank fiber

[frank fiber]

thanks

[thorn delta]

I think you might be missing some context here: this is a commutative algebra text, so they are using "ring" to mean a commutative ring w/ identity. hence the left/right module distinction is not important.

[gray comet]

Ah

Then yeah, that's a bimodule, and it should be a bimodule isomorphism

[unique juniper]

what does it mean for Cubics to be an X-dimensional family

[oblique river]

Can you give some more context?

[unique juniper]

sure

[oblique river]

I would say that it means that if you look at the space of all cubic curves parametrized by their coefficients, that space has some dimension

[unique juniper]

wdym parametrized?

[oblique river]

Depending on some parameter

For example, if i said “consider the family of curves y = x^3 + a”

That would be a one-dimensional family

And a is the parameter

For each value of a you get a different curve

[unique juniper]

ok yes

ty

[oblique river]

👍

[ivory dust]

How does one show a field of 4 elements is not isomorphic to a subfield of a field with 8 elements

without using the fact that if lFl = p^n then its subfields have p^m elements where m divides n

(not allowed to use that 😐 )

Is showing multiplicative subgroups of groups of order 8-1 = 7 (i.e., F* = F - {0}) not being isomorphic mult. subgroup of order 4-1 = 3 sufficient?

i feel like thats a disconnect and goes from proving things about fields into group theory

[void cosmos]

think of vec spaces

or yea u had a good idea ig

u can look at the multiplicative gropus

and then like compare elements orderwise and stuff

@ivory dust

[void cosmos]

so yea

the orders of elements in the multiplicative group are all of order 3 (nonidentity)

for the field of 8 elements the order of elements are all of order 7 (nonide)

so if the one field is a subfield of the other

[ivory dust]

but can we extend back into field theory from grp theory?

[void cosmos]

the multiplicative group of this would be asubgropu of that

[void cosmos]

wdym

[ivory dust]

All non-id elements of subfields of a field of 8 elements have order 7 w.r.t multiplication, all non-id elements of field with 4 elements have order 3 w.r.t multiplication because of their multiplicative subgroups. end of proof?

[unique juniper]

check out galois theory

links fields and groups nicely

[ivory dust]

The textbook? im using it

[unique juniper]

no

[ivory dust]

idk why this theorem:

[ivory dust]

. isnt allowed on my problem set 😭

[unique juniper]

oh

lol

[thorn delta]

this works. There is no "disconnect" going from groups to fields here. Every field has an associated multiplicative subgroup. If K is a subfield of F, then the multiplicative subgroup K* of K is necessarily a subgroup of F*.

[ivory dust]

is that iff statement?

[thorn delta]

i doubt it

[frank fiber]

how can i prove that every epimorfism is a cokernel in R-mod?

[gritty sparrow]

If f is an epimorphism f:M->N, it is the cokernel of the map i:ker(f)->M

[void knot]

,tex How to prove that if the only factor congruences of an algebra $\bold{A}$ are the trivial ones, then the algebra is directly indecomposable?

[cloud walrus]

DoJ

[gritty sparrow]

Otherwise the congruences given by the projection maps will give rise to a non trivial factor congruence

[frank fiber]

the cokernel of $i:ker(f) \rightarrow M$ is the proyeccion $p: M \rightarrow M/im(f)$, why is $p=f$?

[cloud walrus]

Or x1

[gritty sparrow]

M/im(f) is isomorphic to N (because f is an epimorphism)

[frank fiber]

thanks

[void knot]

,tex So if we suppose that $\bold{A}\overset{\Psi}{\cong}\bold{B}\times \bold{C}$ and choose $\phi_1=\pi_1\circ \Psi$, $\phi_2=\pi_2\circ \Psi$, what we will get is $\bold{B}\cong \frac{\bold{A}}{\text{ker }\phi_1}$ and $\bold{C}\cong \frac{\bold{A}}{\text{ker }\phi_2}$, right?

[cloud walrus]

DoJ

[void knot]

,tex In our case the $(\text{ker }\phi_1,\text{ker }\phi_2)$ is a pair of factor congruence on $\bold{A}$ and I know that $\bold{A}\cong \frac{\bold{A}}{\text{ker }\phi_1}\times \frac{\bold{A}}{\text{ker }\phi_2}$

[cloud walrus]

DoJ

[gritty sparrow]

Yes, that is the nontrivial factor congruence that gives us a contradiction

I mean, does that seem ok to you?

[void knot]

sure, thx

I wrote it down and it works perfectly

[gritty sparrow]

Nice

[thorn delta]

i have a sort of general question about proofs of statements like this

The basic idea for i) is that you can take a SES, tensor with M, and then tensor with N, and that is the same as tensoring with M \otimes N.

However, the maps you get by applying the functor (-\otimes M)\otimes N are different than the maps you get by applying -\otimes (M\otimes N). Therefore, to show that M\otimes N is actually flat, it suffices to show that the diagram below, where the phi's are isomorphisms and the top row is exact, commutes:

this isn't too bad to show, but then for ii) the idea is that for any $A$-module $K$, $$K\otimes_A N \cong K\otimes_A (B \otimes_B N) \cong (K \otimes_A B) \otimes_B N$$ as $A$-modules

[cloud walrus]

kxrider

[thorn delta]

and so the "phi" in this scenario would be this composition of isomorphisms. You could do the same thing I did for i) to complete the argument, but this seems super cumbersome, and I was wondering if there was a better/simpler way to think about this?

are all these isomorphisms natural in the categorical since? That would make things easier, right?

[gritty sparrow]

Well, there is a conceptually clearer way of thinking about this, but the amount of work is the same: In both cases the isomorphisms are natural in K

[gritty sparrow]

Yes exactly, unfortunately you still pretty much have to show that all the squares commute

[thorn delta]

i see, but I guess once you know that each step is natural, then you can argue that a composition of these is natural.

[gritty sparrow]

Yes

[ivory dust]

Hey guys so I know cyclotomic polynomial phi p (x) = x^p-1 + x^p-2 + ... + x + 1 is irreducible over Q and Z for p prime

im asked to show f(x) = x^p-1 - x^p-2 + ... - x + 1 is irreducible over Q

and i realized f(-x) = phi p (x) when p is odd prime

then since f(-x) = phi p (x) is irreducible this implies f(x) is irreducible

when p odd prime

but what about when p = 2

also is my logic correct above^

cuz f(x) = x-1 is case when p = 2

isnt that reducible since f(1) = 0

[hidden haven]

That seems correct. x ↦ -x is an automorphism of Q[x] so it must map irreducibles to irreducibles and reducibles to reducibles

[hidden haven]

Don't get what this means

oh I see what you're saying

Try to think about why having a root usually means that the polynomial is reducible

And why that doesn't work here

[ivory dust]

well im assuming it means it can be factored into two polynomials

but my previous thought was that there exists a such that f(a) = 0

so im guessing degree 1 polynomials arent reducible

by euclidean algorithim essentially(?)

[hidden haven]

Usually this follows from the factor theorem

Which says that a is a root of a polynomial iff x-a divides it

But here, you get that 1 is a root, so x-1 divides x-1

That doesn't tell you anything

So finding roots to prove reducibility only works for degree greater than one

Because you get a non unit factor

Here the claim is: All linear polynomials over a field are irreducible.

[hidden haven]

I don't see how

[ivory dust]

Oh I see

[hidden haven]

You can prove this directly, using the definition of irreducibility

[ivory dust]

eucl alg means f(x) = a(x)b(x)+r(x) but r would be 0

if im thinking correctly?

idk maybe im just saying nonsense ☠️

[hidden haven]

That's not Euclid's algorithm, that algorithm finds GCDs

It's just long division

But what are you trying to do with it?

[ivory dust]

oh nothing i just thought it could serve as justification

since deg(r(x)) = 0

might be a condition for irreduciblility

[hidden haven]

Nope, just a condition for divisibility by b(x)

[hidden haven]

Use the definition of irreducibility itself, you can prove this from first principles

[ivory dust]

okk yea i had to review the defn

my proff said it like this

non zero p(x) e F[x] is irred if its deg >= 1 and no factorization p(x) = f(x)g(x) such that df>=1, and dg > dp

and its monic and unique

[hidden haven]

Hmm that seems weird

[ivory dust]

thats irrelevant tho

[hidden haven]

Both df and dg should be ≥ 1

And ≤ dp

Either of these conditions gives the other

dg>dp doesn't make sense

[gritty sparrow]

The opposite does though

[hidden haven]

Because when you multiply polynomials their degrees add

[gritty sparrow]

Ie dg<dp

[hidden haven]

Yeah but with the opposite it's still not the correct def

[ivory dust]

o dg<dp*

i wrotei t wrong

my bad

[hidden haven]

Becuase then f = p and g = 1 makes every p reducible

[ivory dust]

and yea deg p >= 1 implies df, dg >= 1

i think

[hidden haven]

Right

[ivory dust]

dg<dp was how i copied the note

[hidden haven]

So do you see that linear polynomials will always be irreducible?

[ivory dust]

yess thank you 😩

[hidden haven]

[ivory dust]

is my logic for this correct?

cuz f(ax+b) irreducible implies f(x) irreducible?

im not sure how to prove this^

[hidden haven]

Yes, look at this

[ivory dust]

actually ik how to prove f(x+a) irred implies f(x) irred

[hidden haven]

Do you know what isomorphisms are?

[ivory dust]

Yea for the most part

more familiar w them in group theory

[hidden haven]

Ah

So in rings you can define isomorphisms too

[ivory dust]

ring iso is grp iso but also f(xy) = f(x)f(y)?

and f(1) = 1?

[hidden haven]

Yep

Exactly

[ivory dust]

so x --> ax+b is automorphic and that concludes proof?

[hidden haven]

Yep, for non 0 a

This works in Q, not in Z

Try to prove that

[ivory dust]

but Q to Z is an iff statement

[hidden haven]

In what sense?

[ivory dust]

from some dudes thm?

right

oh idk

[hidden haven]

Gauss lemma?

[ivory dust]

yes!

[hidden haven]

Ah it has extra condition

[ivory dust]

eisensteins criterion

• guass lemma

gauss*

[hidden haven]

Irreducible and primitive in Q iff irreducible in Z

The primitive part is what makes that thing work

Otherwise working in Z and Q are quite different. For example in Z[x], x ↦ 2x is not an automorphism

Also in Z[x], the definition of irreducibility is different from the one you gave

[ivory dust]

Hmm i copied my proffs notes and it says given f(x) e F[x] f(x) factors in Z[x] iff f factors in Q[x]

or f irred in Z[x] iff irred in Q[x]

didnt learn abt primitive

yet

[hidden haven]

That is wrong. For example, 2x factors in Z[x], as 2 times x

[ivory dust]

We learned about Eisenstein criterion too

[hidden haven]

But in Q[x] this isn't considered a factorization

Because 2 is invertible

So notice that in a ring, if c is invertible, you can write any r as r times 1 = r times c times c inv

So we don't consider these factorizations

We only consider proper factorizations, ie r = ab is a factorization if neither a nor b is invertible

In Q[x], this definition matches the definition you wrote, because deg f ≥ 1 is saying exactly that f is not invertible, non zero

Because if deg f = 0, and f is non zero, then f is a constant rational number

Which is invertible

[hidden haven]

Perhaps you are missing 'monic' here?

Because monic polynomials are always primitive

So Gauss lemma applies

[ivory dust]

Ohh no i copied it correctly but his notes must be wrong but i think it doesnt matter for my current problem set cuz im trying to prove things are irred over Q[x]

so then that implies its irred in Z[x]

and also an extra condition that its primitve (didnt learn yet)

but the if holds

implication* holds

cuz im not going the other way

[hidden haven]

Yeah just replace primitive with monic for now

Right

[ivory dust]

so primative is when the

polyn is monic or gcd of its coeff is 1?

gcd pairwise?

[hidden haven]

Yes, you can remove the monic thing because it's redundant

Not pairwise

GCD of all of them

[ivory dust]

like gcd(a,b,c,d...)

oh ok

whats c(f(x)) (content) of f

[hidden haven]

Never heard of this gimme a sec

Oh it's just GCD of coefficients

[ivory dust]

oh so if you factorize a polynomial in Q[x]

as f = c(f) * g then g is primitive

im just reading my textbook

kinda get it

didnt cover in lect prob need it for my next problem set tho

so eisenstein holds --> irred in Q[x] ---> by gauss irred in Z[x] (also primative)

[gritty sparrow]

The last step doesn’t need gauss, it will always be true. The other way around needs gauss

[hidden haven]

Yeah I stated gauss as the equivalence

Just to make things simpler

[safe citrus]

how do i describe H?

[hidden haven]

What can you say about the order of H?

[safe citrus]

um

its same as G?

[hidden haven]

Why?

[safe citrus]

it has all g^n

is that right? idk

[hidden haven]

The hint is given in the problem

Lagrange's theorem

[safe citrus]

ohh

cuz only p or 1 divide p, and g^n not = to 1?

[hidden haven]

Yep

Well I guess you should say g²≠g

[safe citrus]

oh

[hidden haven]

Or like g^n = 1 ≠ g for some n

[safe citrus]

oh so that also proves G = H right?

[hidden haven]

Yes, you get that H has 1 or p elements, but we've found 2 elements already, 1 and g

So H and G have the same no of elements and since they're finite they must be equal

[safe citrus]

ok thankss

[safe citrus]

for describe all possible groups G is it just this

[hidden haven]

Well you get that G just consists of powers of a single element

[safe citrus]

o

[hidden haven]

So you can say exactly what G is upto isomorphism

Have you seen any such groups?

[safe citrus]

uh

integers?

but a prime number of integers

wait nvm

uhhh

[hidden haven]

Integers certainly yes

Only problem is they are infinite

[safe citrus]

yeah..

[hidden haven]

Have you seen quotient groups?

[safe citrus]

like modular?

oh like mod 7 would work?

since 7 is prime?

[hidden haven]

Yep

Z/pZ

[safe citrus]

i c

[hidden haven]

Is finite, has p elements

And all elements are just 1 element's multiples

You can show an isomorphism from G to this

[safe citrus]

right

ah ty for help

:)

[hidden haven]

[frank fiber]

if f is a homomorfismo and f(a)=f(b), then a+ker(f)=b+ker(f)?

[hidden haven]

Yes

[steady axle]

I have written a detailed solution for this problem, because I was not very confident. Can someone take a look?

Notice that S_6 contains such a subgroup namely the subgroup generated by {(1 2),(3 4), (5 6)}.
Note that in S_4 and S_5 the group of order 8 is a sylow 2 subgroup. But the sylow 2 sugroups in both these groups contain element of of order 4 of the form (a b c d) so this subgroup cannot be the one in problem.

[dusty river]

That is correct

I think someone asked the same problem a couple weeks ago

[steady axle]

[frank fiber]

is every free module a vector space?

[dusty river]

no. Z is a free module over Z but not a vector space over Z because Z is not a field

[frank fiber]

and if two free modules have basis with same cardinality, then are isomorfic?

[chilly ocean]

ye

[gray comet]

Try proving why

[chilly ocean]

can someone help me with this I want to show that for the root system of a lie algebra, there is an element in the wely group which sends all the positive roots to negative roots

suppose that there are n positive roots, i can see that its enough to find an element in the weyl group with height (or i think this is more often called "word length") equal to n

but im not sure why such an element must exist

[chilly ocean]

okay so

if w was the longest word but had length less than n then it does not send some alpha to a negative root

but then setting w' to be w followed by applying the reflection generated by alpha will send alpha to -alpha

[hot lake]

but what if it changes the others ?

[chilly ocean]

sorry yes thats what i can't figure out

[chilly ocean]

seemingly this follows from that the wely group acts transitively but I don't see why that is the case

[vestal snow]

Can someone ELI5 group schemes?

[brazen pumice]

well if you know groups and schemes

then a group scheme is a scheme that also has a group structure

[vestal snow]

So like a topological group?

[brazen pumice]

yeah but all the maps need to be scheme morphisms

[vestal snow]

Okay

Part of my question also was why should I care about group schemes?

[ivory dust]

is Q(pi) = a+b*pi

🤔

[void cosmos]

yea where a and b are in Q

[vestal snow]

Wait no

Q(pi) is all rational functions with pi

Which is isomorphic to Q(x)

[ivory dust]

why is Q(sqrt2) = a+bsqrt2 but

Q(pi) something completely] diff

[vestal snow]

Q(something) is basically attach (something) to Q and then fieldify it

[ivory dust]

my question is to describe the elements of Q(pi)

[vestal snow]

This is the answer (unless you're using non-standard notation)

[ivory dust]

would it be a0 + a1pi + ... + anpi^n n e N?

[brazen pumice]

no, it's (a0 + a1pi + ... + anpi^n)/(b0 + bmpi + ... + bmpi^m)

where m>0 and bm != 0

[ivory dust]

wtff

😩

[brazen pumice]

you need to close it under division

[void cosmos]

ohh yeaa

[brazen pumice]

the reason why Q(sqrt 2) can be just polynomials is because sqrt(2) is algebraic

[void cosmos]

lmao yea mb

didnt think

Q(pi) is the field of fractions of the one is aid

i said*

[ivory dust]

so would F = Q(pi^3) = (a0 + a1pi^3 + ... + anpi^3n)/(b0 + bmpi^3 + ... + bmpi^3m)

m>0, bm =/= 0?

then what would F(pi) equal?

[void cosmos]

pi is algebraic over Q(pi)

[brazen pumice]

pi is algebraic over F = Q(pi^3)

because it satisfies X^3 - pi^3 = 0

so F(pi) would be a + b pi + c pi^2 where a,b,c in F

[ivory dust]

these notes claim F(pi) is finite dimension vect space over F w basis {1,pi,pi^2}

oh i see

hmm

ok i need to digest this

[brazen pumice]

you need to understand Q(cbrt(2)) first

[ivory dust]

cbrt2 is not transcendental right?

but its not algebraic cuz its not constructible?

[brazen pumice]

what does algebraic mean?

[vestal snow]

not every algebraic is constructible

[ivory dust]

solution to integer polynomials right?

[brazen pumice]

that's algebraic integer, but cbrt(2) satisfies that too

so let x = cbrt(2) and try combining powers of x to get an integer

[ivory dust]

x^3 = 2

[brazen pumice]

there you go

[ivory dust]

oh so cbrt2 is algebraic

[vestal snow]

What does it mean when someone says "an integral affine k-variety is rational"?

I know what an integral affine k-variety is

and I know what rational maps are

I'm guessing it means that the map from the scheme to Spec k is dominant?

[sturdy marsh]

rational = birational to projective space

@vestal snow

[vestal snow]

Oh okay that makes sense

[vestal snow]

Let C and D be equivalent categories and A and B be objects of C such that their "images" in D are isomorphic, then A and B are isomorphic

Is this true?

I don't know the precise definition of equivalent categories, but the definition would be pretty sucky if it didn't imply this

[sturdy marsh]

yup

[thorn delta]

the construction of the colimit involves taking the direct sum of of the M_i and quotienting out by the submodule generated by elements of the form x_i - mu_ij(x_i) for x_i \in M_i. Since the mu_ij are just inclusions here, then isn't this submodule 0?

but that would imply the colimit is their direct sum, not their sum

[vestal snow]

mu_ij are not necessarily inclusions

[thorn delta]

in this case they are

according to the problem

[vestal snow]

Oh my bad

[gritty sparrow]

Well, more precisely, xi is not the same as mu_ij(xi) because we are taking the abstract direct sum here

As in, not the direct sum inside M, but we are defining the sum to be finite sequences of elements from the Mi’s. Does that make sense to you?

[thorn delta]

yeah, it does make sense.

so we don't have x_i - mu_ij(x_i) = 0 in then direct sum but in the quotient we do have this, which looks a lot like the regular sum of submodules.

[gritty sparrow]

Yes exactly

[thorn delta]

also, yea i can see how union-ing might be easier here. Given f_j : Mj \to N such that f_j = mu_ij f_i its not hard to see how u could define a map f : \bigcup M_i \to N satisfying the universal property we want

[tough raven]

(assuming mu_ij is from M_i to M_j)
You are quotienting out by x_i - mu_ij(x_i) in the direct sum. The x_i is an element of M_j as a submodule of the direct sum. The mu_ij(x_j) is in M_i as a submodule of the direct sum. In particular, because it's the external direct sum, M_i and M_j are disjoint (except 0) in this direct sum.
So for non-zero x_i, x_i - mu_ij(x_i) is not zero in the direct sum.

For example, let I be finite, so all the M_i are submodules of one of them, say M_1. Then the colimit should be M_1, right? Not the direct sum which is bigger.

[tough raven]

Oh, you said this already
Oops

[vestal snow]

What does A^1 (without subscript) mean?

[gritty sparrow]

I think it is just A^1_Z.

[brazen pumice]

which is Spec Z[x]

[surreal basalt]

I have a question regarding the first (supposedly trivial) part of this mathoverflow question

https://mathoverflow.net/questions/272550/a-simple-colimit-in-the-derived-category

Where do we need the complex to be bounded below?

It would seem to me as if any complex would work, although I have seen several people claiming that so surely I’m missing something

[shut halo]

I think #category-theory is more appropriate for your question

[surreal basalt]

Ty, I’ll repost there

[shut halo]

Hey, I'm confused by this definition:

I don't quite get what the definition means by * agreeing on R inA

[hidden haven]

A contains a copy of R. There's an involution * on A, and an involution # on R (both have the same name in that statement). That is saying that # = * restricted to R

Wait the copy of R thing is only true if R is a vector space I think

But the usual thing is that (rx)* = r#x*

Equivalently r* = r#

Viewing r ∈ R as the element r times identity in A

[steady axle]

can someone take a look at my solution to this problem?

(a) From the given equation (a,b,c)=(1). If $(a^{15},b^{16},c^{17}) \neq (1)$ then $(a^{15},b^{16},c^{17})$ is contained in some maximal ideal. So in particular $a^{15}\in $ this maximal ideal so $a\in$ this maximal ideal. Similarly b and c. But then this maximal ideal contains $(a,b,c)$, a contradiction. \
(b) Let P be a prime ideal. let $a \not \in P$. $a^n-a=0 \in P$. So $a(a^{n-1} -1) \in P$ so $a^{n-1}-1 \in P$. this tells us that every non zero element of $R/P$ is a unit.

[cloud walrus]

bert

[hidden haven]

Looks good

[ivory dust]

@brazen pumice for Q(pi) you said its elements of form (a0 + a1pi + ... + anpi^n)/(b0 + b1pi + ... + bmpi^m) where m>0 and bm != 0
why is it bm=/=0? shouldnt it be all the denominator together cant be 0?

and is there any restriction for if n>m?

or can they be differing degrees

sorry to bother you again

[brazen pumice]

no

[brazen pumice]

well they're equivalent

[ivory dust]

oh ok thank you sm

[vestal snow]

How is G_m over an arbitrary scheme defined?

Seems easy when the scheme is affine as maps of schemes correspond to maps from the global ring of the affine scheme to the other scheme

[marsh fractal]

basic ring theory question here: how would i deduce from the ring axioms that a(-b) = -(ab)

not too sure where to start

[gritty sparrow]

It will just be Sx spec(Z[x,x^-1])

[vestal snow]

We haven't covered products of schemes yet

So I'm guessing I should come back to it later

Thanks tho

[gritty sparrow]

ab+a(-b)=a(b-b) = a(0) =0. (For the last part remember that a(0)=a(0+0)=a(0)+a(0), now cancel an a(0) on both sides)

[gritty sparrow]

Once you know the product of schemes, there is a nice way to define group schemes, it will be the group object in the category of schemes. A group object of a category is an object G equipped with a morphism m:GxG->G satisfying some diagrams. m is supposed to be like the multiplication function in normal groups. I feel like this is the “correct” way to define things

I forgot to mention the identity and inverse morphisms as well, sorry

What definition are you using for group schemes?

[marsh fractal]

thanks

[vestal snow]

Is this how you multiply in Mor(X,G). Take two morphisms. They induce a unique map from X to G x G. Compose this with m

[gritty sparrow]

Yes exactly

[vestal snow]

Verifying the group axioms with this multiplication is PAIN

[gritty sparrow]

Yeah it does seem pretty annoying, have fun

[vestal snow]

If anyone has done this verification, is there any thing insightful about it or can I just skip it?

[gritty sparrow]

I don’t know if i’m remembering this correctly, but I remember this just being a specific verification of the fact that Hom functors preserve limits. I haven’t done this recently though so i’m not really sure about that

[spice lance]

hi! can I ask for help for this problem on the dihedral group D8? :))

I expressed s as a^5b using

and since (a^5b)(a^5b) = e, then the the order of s is 2

<s> then is equal to {e, a^5b}

so the factor group D_8/<s> has elements D8 and (a^5b)D8

then using the cayley table, D_8/<s> is isomorphic to Z_2

is what I did correct? sorry for the bother

[quaint tree]

Dihedral group notation can be confusing; some people use $|D_{n}|=n$, some use $|D_{n}|=2n$.

[cloud walrus]

Cool I can change my nickname

[quaint tree]

It looks like your book is using $|D_{8}|=16$, because $\text{ord}(a)=8$.

[cloud walrus]

Cool I can change my nickname

[quaint tree]

So, by a theorem (I think a corollary of the first isomorphism theorem but honestly I'm drunk), $|D_{8}/\langle s\rangle|=|D_{8}|/|\langle s\rangle|\ne2$

[cloud walrus]

Cool I can change my nickname

[dusty river]

nice nickname

[quaint tree]

Thanks

[ivory dust]

is the splitting field of p(x) = (x^2+1)^2 over Z3

Z3[x] / (x^2+1)?

or i can say Z3(a) where a is root of x^2+1

but that root doesnt exist 🤔 in Z3

confusion

[dusty river]

yes

[dusty river]

"Z_3(a) where a is a root of that polynomial" is defined as that quotient

[ivory dust]

and this is the case

because (x^2+1) is irred in Z3?

otherwise it wouldnt be the case

?

[dusty river]

yes

[ivory dust]

Why dont we square it

in the ideal under Z3[x]

[dusty river]

dont get what you mean

[ivory dust]

actually dumb question

i was asking cuz p(x) = (x^2+1)(x^2+1)