groups-rings-fields

[hot wadi]

For future reference, it suffices to show that each ideal contains the generators of the other

But this is still perfectly correct as written

[trail cave]

in part c here, how do I know that H is a subgroup of N(H)? I don't see how I can impose additional structure onto it

[toxic zephyr]

is H contained in N(H)?

[trail cave]

yes because cosets are partitions so aH=Ha for all a in H right?

I see now

ty

[trail cave]

what I said here made no sense but I figured it out

[proven mango]

Let $R_3$ be a field such as $\mathbb{Q}$ or $\mathbb{Z}_{17}$. Prove that $R_3$ is an integral domain.
If I want to prove that this is an integral domain, how would I show that for every $a,b \in R_3 \setminus {0_R}$, $ab \neq 0_R$.

[cloud walrus]

Devin ☆

[coral steeple]

I think this works, although I have been through several revisions of the proof.

The second line should have langle rangle instead of curly braces

[wispy light]

By definition, a field is a commutative ring whose units are its nonzero elements. The product of two units has to be a unit, but 0 isn't a unit, so...

[patent girder]

if G is a simple group of order 1/2 n!, must G be isomorphic to A_n?

[wispy light]

Sorry, I missed the word “simple”. But time to check the classification of finite simple groups, I guess.

[sonic coral]

no. there is A_8 and PSL(3,4)

both have orders 8!/2 and are not isomorphic

[patent girder]

ooh right

tyty

[fickle pagoda]

i want to prove that if G has odd order, x^2=a has a unique solution for all a in G

this is my process so far, I wanted to prove it with cosets
H={g^2|g in G}
[G:H]=2 because if a not in H, there is aH and H
so G has element of order 2 and that's bad

but is it true that [G:H]=2 or is the direction of my proof just bogus

[warm ember]

H is not a subgroup if G is not abelian

hint: use ||lagrange||

solution: ||say x^2=a=y^2 has order k, odd. then x^2k=e. because G has odd order, order of x divides k. if the order of x is l then a^l=e hence k=l. similarly y has order k. now suppose k+1=2m (remember k is odd). then raising each side to mth power we get x=a^m=y.||

[trail cave]

for my hw. I've done parts 1 and 2, I'm struggling with part 3 (both directions)

[restive birch]

well aHbH = bHaH iff abH = baH right

try right multiplying both sides by (baH)^-1

[trail cave]

so that I get (aba^-1b^-1)H = H?

am I able to conclude aba^-1b^-1 = 1 from that?

I got abH = baH but I wasn't sure about that move

[languid trellis]

How do we get to \sum x^i being the minimum polynomial?

oh its the cyclotomic polynomial

ups

[dim widget]

17 is prime

[opaque finch]

$xH = H$ iff $x \in H$ and in your case $x = [a,b]$

[cloud walrus]

mycroftholmes1703

[loud cipher]

My head hurts when I try to prove this. Let $R$ be a commutative ring with a $1$, then why $N(R) = \sqrt{0}$, where $N(R)$ denotes the intersection of all prime ideals of $R$. One inclusion is obvious, because $\sqrt{0} \subset P$ for any prime ideal $P$, but the other Inclusion I just can't prove it

[cloud walrus]

J_U_V3

[lapis latch]

It’s the more involved direction. You’d want to use Krull’s separation lemma to show that if an element isn’t nilpotent then there is a prime ideal not containing it

[minor fulcrum]

assuming choice, how many field maps are there R -> C

[tough raven]

The maps such that ℂ is algebraic over the image are in bijection with the left cosets of {id, conjugation} in Aut(ℂ), of which there are a lot - |ℝ|^|ℝ|. Since this is even equal to the cardinality of the set of functions from ℝ to ℂ, the answer is as many as there are functions from ℝ to ℂ.

[minor fulcrum]

oh wow, I knew there were a lot of wild automorphisms but I didn't know it was that many

[tough raven]

It's surprisingly straightforward to prove - just take a transcendence basis of ℂ and consider permuting its elements. I don't think even a single automorphism other than the obvious two can be constructed without any choice though 💀.

[grim hawk]

what group could correspond to switching of elements in a function? For instance switching $\phi_1$ with $phi_2$ in a function $$\partial_\mu \phi_1 ^\dagger \partial^\mu \phi_1 + \partial_\mu \phi_2^\dagger \partial^\mu \phi_2 - V[\phi_1^\dagger \phi_1 + \phi_2^\dagger \phi_2].$$. This is for some physics work but I am feeling very dumb trying to remember what this would correspond to

[cloud walrus]

PSCell

[tardy hedge]

Hmm uhh

I mean im sure other ppl could help u out better, im just an early grad student but like

You are wanting to define a group action im assuming?

[grim hawk]

Im trying to find the group that this would correspond to.

[tardy hedge]

Any group corresponds to some permutation

[wraith cargo]

wouldn't it be something like the symmetric group?

[grim hawk]

Wait maybe it is just S_2 since thats the permutation group

[wraith cargo]

but this is sort of a vague question

[grim hawk]

That makes a lot of sense

[tardy hedge]

Yeah i guess ur right if u just want a swapping action

[grim hawk]

Sorry for the dumb question!

[tardy hedge]

No dumb questions

Im curious how you implement that though

What is the purpose here? Just kind of curious what ur doing

Why do u need to involve groups to do that swapping thing

I dont study physics so im not really sure how groups are utilized precisely but group actions and modelling symmetries makes sense

[tardy hedge]

Oki

[tardy hedge]

Any nontrivial solution to x^p = e would be an element of order p right? Since p is prime the order of x cant be less than p

So when it says “the number of solutions to x^p = e is a multiple of p” that means there are a multiple of p elements that have order p?

Also, the proof for this was so cool. I hadn’t seen group actions in … “action” (lol) to prove things before

[grave sedge]

p or 1

[tardy hedge]

I dint get it

Oh nvm ur saying im correct right

Order 1 yes , ignoring the identity

[grave sedge]

The number of solutions is a multiple of p only if you include the identity

[tardy hedge]

so its incorrect to say there are a multiple of p elements that have order p

so are there a multiple of p - 1 elements of order p then? lol

[grave sedge]

No

There are a number which is congruent to -1 mod p

(which crucially implies it is a nonzero number, by the way)

[tardy hedge]

I dont really get it yet

if there are qp solutions to x^p = e, only one of them would be e, so every other element should have order p right

so why cant i say there are qp-1 elements of order p

[grave sedge]

Yes

Which is not a multiple of p-1

But it's a number congruent to -1 mod p

[tardy hedge]

oh i see what happened lol, when i said that originally i did mean qp-1 elements of order p

but it was ambiguous the way i said it

thanks

[grave sedge]

Ah yeah i see

[errant wedge]

@somber badge my friend did an example of the argument for n=m=3, it generalizes ez enough

[trail cave]

I finished problem 3 part i here but I'm struggling with part ii. I see that I can say either Z3 or Z5 are normal subgroups of Z15, hence is H is the inverse image of those normal subgroups, we find the quotient group G/H is isomorphic to both Z15/Z3 and Z15/Z5. Where do I go fromthere?

Can I conclude that the inverse images must have order 3 and 5 and so the indices are 5 and 3 respectively?

[wraith cargo]

you cannot

you can conclude their indexes are the same though

since G/H = G'/H'

[trail cave]

oh right, because the index is the number of distinct members of the quotient group

thank you

[grim hawk]

Was working on determining the symmetry group for a Lagrangian for two complex scalar fields. Very helpful in early quantum field theory

[tardy hedge]

Nice.

[golden turtle]

I have a few questions

this is the representation I was able to make

I am wondering, if it is a valid representation

and then secondly, I am confused about how to verify if it is, or is not, irreducible?

@digital seal hii

[digital seal]

presumably you wrote down the "obvious" D_8 action on R^2 right

[golden turtle]

yes

[digital seal]

oh bruh you're using D_8 for D_4

that's a sully

[golden turtle]

huh

my professor choice

I also disagre

[digital seal]

your professor is wrong

[golden turtle]

yes I know

[digital seal]

i'm just gonna assume your representation is correct

do you have any theorems about irreducible reps?

or do you just have to verify it manually

[golden turtle]

hm

Maschkes theorem I think

[digital seal]

that's probably not enough

[golden turtle]

would it be if I just let each element in D8 coorespond to th ematrix that does that symmetry

[digital seal]

yes

[golden turtle]

Do you have a theorem in mind

[digital seal]

i do

[golden turtle]

I can tell you if I've done it

Schurs lemma

we've done

[digital seal]

have you done any character theory

[golden turtle]

yes

[digital seal]

okay do you know how to compute the character for this representation

[golden turtle]

the character table?

don't I just take traces of all these matrices

[digital seal]

yes

there's a theorem that states the character has norm 1 iff the rep is irreducible

[golden turtle]

so

none of these matrices have trace 1

so none of them are irreducible?

[digital seal]

no that's uh

not a coherent statement

irreducibility is a property of the representation

[golden turtle]

I do have this theorem

[digital seal]

okay you can use that

you need to compute the character for the representation and sum it

[golden turtle]

so the representation

is

all of these matrices

together collectively

liek

[digital seal]

yes

[golden turtle]

ok

[digital seal]

it's a group homomorphism

[golden turtle]

and the character of that is the sum of their traces? or I guess I am probably not understanding the character

[digital seal]

the character is a function G -> C

it takes values on each element of the group

[golden turtle]

so

the rep is a function G->GL(V)

is the character

char(g)=trace(rep(g))

?

[digital seal]

yes

[golden turtle]

ok sick

yeah then I can find the character of each g

then what, sum them?

[digital seal]

well "sum"

look up the definition of the inner product and calculate the inner product of your character with itself

[golden turtle]

I know what an inner product is but what is the relevant inner product here?

[digital seal]

it'll be a sum of chi(g) chi(g^-1) divided by |G|

[golden turtle]

this

[digital seal]

yes

[golden turtle]

my freaking sir...

[digital seal]

what are these notes

[golden turtle]

they're always like this

you know

[digital seal]

how are you surviving in this class

[golden turtle]

you

I'm not really

[digital seal]

💀

[golden turtle]

the final is 50%

so

I am just coasting until then

[digital seal]

[golden turtle]

I think I have like 90% on the HW?

which is the other 50%

and then

I might be fucked

but I am trying

you see

[digital seal]

[golden turtle]

it is just really hard to learn from my sir

and his notes

and his speed

[digital seal]

is this the same class as the one you were talking about before

where you did the Ext and Tor functors

[golden turtle]

so I do this, and if the sum is 1 then its irreducible. Say I do this, and the sum isn't 1, then my representation was bad (which could easily be the case since I just picked the most basic one that came to mind) what then should I do?

[golden turtle]

Yes, same class

all of everything I post is the same class

[digital seal]

bruh momento

[golden turtle]

we did module theory category theory now rep theory

I think we are going through all of d&F in like 1 quarter

[digital seal]

skull

[digital seal]

uh it should be fine

[golden turtle]

why?

[digital seal]

because i'm pretty sure this representation is irreducible

[golden turtle]

and what if I am unlucky in the future

just fucked?

[digital seal]

yes

[golden turtle]

great

[digital seal]

there are a number of ways to find all the representations

you should refer to your notes for those

[golden turtle]

💀

snow

[digital seal]

read uh

[digital seal]

fulton and harris

[golden turtle]

more generally if the subring A[b] sits inside a finitely generated A-submodule C that is also a subgring of B then b is integral over A, for then if c1, ... , cn generate C over A, then multiplication by b, regarded as an A-module map from C to itself, admits a matrix M with respect to the c_i

[digital seal]

[golden turtle]

this matrix satisfies its characteristic polynomial, which is monic, by the cayley hamilton theorem, whence multiplication by b and b itself satisfy the same polynomial

whence

by b and b itself

[digital seal]

what even is the main point of this slide

that the integral elements form a ring?

[golden turtle]

you think I fuckign know the main point of any of these slides?

[digital seal]

[golden turtle]

wait until the one about how he privately calls the clifford groups

[digital seal]

[golden turtle]

he gets freaky in this slide

[golden turtle]

the main point of this slide is much clearer

when bro admits in the slide the explanation is rough you know it is bad because rough is generous even for the explanations that he thinks are goood

[digital seal]

oh my god

[upbeat dirge]

god damn

those slides look more painful than decrypting burned up papyrus from pompei

[golden turtle]

im so fucked u have NO ideea

[upbeat dirge]

what’s the topic (if you know lol 😂)

[golden turtle]

of that slide???

[upbeat dirge]

of the course

[golden turtle]

the composition problem

whatever the fuck that means

he said "i mention the composition problem" so I belive him..

[upbeat dirge]

what’s the course

group theory ?

[golden turtle]

but idk what that is outside of this "rough" explanation

[golden turtle]

abstract algebra

we jump around

[upbeat dirge]

lol

[golden turtle]

so far module, category, and rep theory

[upbeat dirge]

maybe try picking up the reference book

[digital seal]

this is shocking

[upbeat dirge]

or just any book

[digital seal]

[golden turtle]

i mostly read d&f

but it doesnt help because i cant read it all in 1 day

like he expects

[digital seal]

nobody can 💀

[upbeat dirge]

can you do the exercises well?

like do you usually manage to finish them

[golden turtle]

for the homeworks?

I usually spend all week trying to do the 3 he assigns and get like 1 or 2 done

[upbeat dirge]

yikes

my group theory teacher did MIT and his slides are fucking divine I swear

[golden turtle]

I hate you

[upbeat dirge]

compared to my fields medalist last semester who probably put at most 5 minutes into writing each lecture note page

[digital seal]

imagine getting slides

[vast stump]

so true

[upbeat dirge]

lol that’s true

[vast stump]

notes? what are those

[golden turtle]

my prof went to MIT too

unforuntately they arent all good

apparently

[upbeat dirge]

damn

[digital seal]

i only know to copy down from the blackboard

[vast stump]

indeed

[upbeat dirge]

copy.. you mean understand

[digital seal]

no i mean copy

[upbeat dirge]

sad

I hate copying everything the teacher says

usually I try to active recall at home

[digital seal]

understanding comes naturally

[golden turtle]

yeah same

I mean like

I just look at these notes

and

it is like

what do they call it

osmosis

it seeps into me

[upbeat dirge]

so you mean the acid trip seeps into you ?

[digital seal]

if i don't copy, im not gonna have a copy

[golden turtle]

this course is y i wont get into grad school

but it is ok

because it is very funny

to online strangers to see my lecture notes

#worth it

[digital seal]

you can return to your linalg study group once you're done

and view it in a new light

[golden turtle]

I am takign linalg next quarter actually

thats exciting

with a non insane profe

so it will be good

b4 we get too far gone

this sum is called the character of the rep?

and the overline(d(g)) is the character of g^(-1)?

oh god snow is gone

[digital seal]

no

(c, d) is the inner product of c and d

[digital seal]

overline d(g) is the complex conjugate of d(g)

you can prove that it's equal to d(g^-1)

[golden turtle]

oh but we are taking (c,c)

[digital seal]

yes

[golden turtle]

where c is the rep

and ya

[digital seal]

so it will be the sum over |c(g)|^2

[golden turtle]

conjugate of d(g) is d(g^(-1))

it does sum to 1 btw

Letsgooo

[digital seal]

[long swan]

interesting font lmao

[hidden wind]

> 6x + 3
> reducible in Z[x] as it factors as 3(2x+1)
> irreducible in Q[x], as 3 is a unit

sometimes i feel too many words

[slim kayak]

Somewhat decent mnemonic for gauss's lemma for polynomials tho

Irreducibility in Z[X] is stronger than irr. in Q[X]. The missing piece is being primitive

[tardy hedge]

Having trouble finding where the left S-module structure on S (x) N needed that “compatibility” relation

To me it looked like that compatibility relation was only needed so that a homomorphism from N into S (x) N worked, not that the compatibility relation was needed to define a left S action on it

The context in that paragraph was that they started the tensor product with a special case in that S is a ring, and R is a subring of S

The “extension of scalars” tensor product case

[grave sedge]

You define s'(s⊗n)=(s's)⊗n, but you then need [s'(sr)⊗n] = s'[(sr)⊗n] = s'[s⊗(rn)] = (s's)⊗(rn)=[(s's)r]⊗n

Where s,s' ∈ S, r ∈ R and n ∈ N

[grave sedge]

(here the first and third equality follow from the way you defined the S-module structure, the other two are from the R-tensor product structure)

[tardy hedge]

I think i see it better now thanks

It has been very tricky for me to understand exactly where each part of tensor product structure is used where and why it is defined that way

[tardy hedge]

Ok so you need that compatibility so that the S-action on S x N is well defined

[elfin wraith]

I think I’m just being kinda dumb here, but why is J(R/J(R))=0 for any ring R?

It’s the intersection of all maximal (left or right) ideals of R/J(R) which by the correspondence theorem are maximal ideals of R which contain J(R) but I don’t see why this is generally just 0?

I think I’m just getting a bit lost in all the inclusions there but I’m not seeing it

[vast quiver]

the correspondence theorem is definitely the thing to do.

You want to know J(R/J(R))? Intersect all maximal ideals of R/J(R), which corresponds to intersecting all maximal ideals of R containing J(R), which is just intersecting all maximal ideals again (since every maximal ideal contains J(R)), so this intersection recovers J(R) inside R, which corresponds to 0 in R/J(R)

[elfin wraith]

Oh god yeah of course, it’s just the fact that we’re quotienting out by J(R)

Thank you! Been staring at that for far too long

[vast quiver]

haha yeah no problem!

[elfin wraith]

Trying to prove the Hopkins-Levitsky theorem and getting stuck worryingly early there

[tardy hedge]

Got confused by this before. Im not sure why this is not a group homomorphism.

i((m1,n1)+(m2,n2)) doesnt equal m1 (x) n1 + m2 (x) n2?

[vast quiver]

correct, they are not equal

[tardy hedge]

why?

What does i((m1,n1)+(m2,n2)) even map to?

i cant figure out what it maps to just based on the formula for i given

[vast quiver]

being linear in each argument should have you do something like “foiling” from high school, or distributing

so i((m1,n1) + (m2,n2)) = i((m1+m2,n1+n2))

since that’s the group structure on the product M x N

[tardy hedge]

really?? I thought (m1,n1) + (m2,n2) cant be simplified further. I thought M x N was a free Z-module with basis (mi,ni)

[vast quiver]

ohh yeah I think they have separate notation for that free group

M x N should denote the product of groups, no free group shenanigans

I’ve seen people write something like F(M x N) to be the free group on this set

I’m not sure what notation your book uses

[rocky cloak]

In your text they say that they map from MxN to the free group then to the tensor product.

So i is the composition
MxN -> F(MxN) -> M(x)N

[tardy hedge]

oh so the i: M x N -> M tensor N is not even considering M x N as a group, just a set?

[rocky cloak]

Yeah

[tardy hedge]

so i( (m1,n1)+(m2,n2)) wouldnt even make sense to write?

[rocky cloak]

Well MxN does have a group structure, hence why they're saying that i isn't linear

But yeah, best to just think of MxN as a set

[tardy hedge]

saying i isnt linear is referring to a group structure on M x N that is NOT the free Z-module structure on M x N

right?

If you are considering the map from the free Z-module to the quotient, that SHOULD be a group homomorphism, shouldnt it? isnt that just the projection map?

[vast quiver]

To be clear, M x N doesn’t have a “free Z-module structure”. The free Z-module on M x N is an entirely different (and very huge) abelian group. For example, for m,0 in M x N, we have also 2m,0 in M x N. Then, 2(m,0) and (2m,0) give different elements in the free Z-module, which I’ll denote F(M x N). As a set, it is much larger than the cartesian product M x N.

But to your other point, yes, i is not linear when we equip the cartesian product M x N with the usual group structure.

On the other hand, the map from F(M x N) -> M (x) N is linear, and it is the projection map as you said.

So, the problem here is exactly the M x N -> F(M x N) is not linear (otherwise i would be a composition of linear maps).

[tardy hedge]

yea that makes more sense, thx

[vast quiver]

yeah no problem!

[velvet steeple]

Let $f: G \to G'$ be a group homomorphism, let $H'$ be a normal subgroup of $G'$ and let $H = f^{-1}(H')$.
[\begin{tikzcd}
G & {G'} \
{f^{-1}(H')} & {H'}
\arrow[from=1-1, to=1-2]
\arrow[from=2-1, to=1-1]
\arrow[from=2-1, to=2-2]
\arrow[from=2-2, to=1-2]
\end{tikzcd}] Then $f^{-1}(H')$ is normal in $G$ [proof cut out]. We then obtain a homomorphism [G \to G' \to G'/H'] composing $f$ with the canonical map of $G'$ onto $G'/H'$ and the kernel of this composite is $H$. Hence we get an injective homomorphism [\overline f: G/H \to G'/H'] \hr \ My question is: why do we need to include that $f^{-1}(H')$ is normal in $G$? It doesn't seem to be used at all.

[surreal dagger]

I think to write G/H you need H to be normal in G, as only quotienting by a normal subgroup gives a groupstructure on the quotient.
But since H was identified as the kernel of the map, you automatically get for free that H is normal in G

[velvet steeple]

Ah, I was thinking it's enough for H to be a subgroup. Do you know off the top of your head which group axiom fails when making it a not-normal subgroup? And yeah, from H = f^(-1)(H') it should right away follow that the kernel of G' -> G'/H' is H, so the proof is kinda unnecessary

[surreal dagger]

The proof is only unecessary if you know that kernels are normal subgroups!
The condition that the subgroup is normal is exactly whats required to make the group operation in the quotient well defined.
As in: If you think about what is required that the group operation in that quotient is well defined it pops out that your group has to be normal

[velvet steeple]

The proof is only unecessary if you know that kernels are normal subgroups!

Yeah, I do

As in: If you think about what is required that the group operation in that quotient is well defined it pops out that your group has to be normal

Ah

Yeah that's reasonable, since for *: (G/H, G/H) -> G/H we want xHxH to be in G/H and that is fulfilled when xHx = H

Thanks

[elfin wraith]

Realsing theres something im a bit unsure of, Im working on a proof of the Hopkins-Levistky theorem and ive just shown that J(R)^n/J(R)^{n+1} is noetherian, but im actually quite unsure of what J(R)^0 is, would it be all of R? From the definition of the product ideal I really cant think of a reasonable definition of what this should be

[velvet steeple]

Sorry, I think the last thing I said here is incorrect

[elfin wraith]

The main reason I think it could potentially just be R is that knowing R/J is noetherian would be somewhat helpful for me right now

[tough raven]

I think I^0 = R for an ideal I is reasonable.

[velvet steeple]

you think about what is required that the group operation in that quotient is well defined it pops out that your group has to be normal

Can you elaborate please?

[elfin wraith]

Im dyslexic

[tough raven]

Chill

[elfin wraith]

Is there any reason for I^0=R? I cant really think of a great reason for that beyond the fact that it makes this chain I have rn work

[rocky cloak]

I mean, I^n is supposed to be the n-fold product of ideals, so you want
I^n I^m = I^n+m

[tough raven]

Let G be a group and H a subgroup. There is a multiplication on G/H such that the quotient map G → G/H respects multiplication and sends the identity to an identity iff H is normal (in fact, just sending the identity to an identity forces H to be normal).

[elfin wraith]

Ah yeah of course

[tough raven]

If I is generated by a set S, then I^n is generated by all n-fold products of elements of S. For n = 0, this means that I^0 should be generated by the empty product, which should be 1.

[tough raven]

And this.

[elfin wraith]

Sick, massivley helpful for what im trying to do, proving this theorem has taken me far longer than it reasonably should have

[tough raven]

Although, if you have proved J(R)^n/J(R)^{n+1} is Noetherian, you should look at that proof when n = 0 and see what it assumes J(R)^n to be in that case.

[elfin wraith]

Im thinking it should be R, im trying to show that R is Noetherian, so im guessing that having R/J is Noetherian is probably helpful

my proof that J^n/J^n+1 also doesnt really make use of the powers of n anyway, just that its an R/J module

[tough raven]

Well, R/J is an R/J-module

So if you define J^0 = R, J^0/J^1 = R/J is an R/J-module

(This is related to wanting J^m J^n ⊆ J^{m+n} for all m, n; taking m = 1, it gives us J J^n ⊆ J^{n+1}, so that J^n/J^{n+1} is an R/J-module. So you want J J^n ⊆ J^{n+1} for n = 0 as well.)

[elfin wraith]

But i get the feeling that I should be using the fact that the jacobson radical of a artinian ring is nilpotent somewhere so im slightly unsure of my argument that J^n/J^n+1 is noetherian tbh

[velvet steeple]

just sending the identity to an identity forces H to be normal
With the canonical map, e gets sent to eH and we want e to be in H if that is supposed to be the identity. So we want eH = H and He = H, thus eHe = H.

Does that look reasonable?

[tough raven]

Is e the identity or an arbitrary element?

[velvet steeple]

e is the identity on G

Ok, this still didn't prove that H is normal though

[elfin wraith]

(mainly because when its proven in the notes it mentions its a helpful lemma for hopkins levitsky and im yet to make use of it, only thought is that we now know J^{m-1} is Noetherian for some m but im not sure how this helps)

[tough raven]

OK nvm it has to preserve multiplication as well

[velvet steeple]

I think we can leave out the identity mapping to identity then? That just makes it a homomorphism

[tough raven]

Yeah, that follows from preserving multiplication (since the map is surjective).

[tough raven]

The Jacobson radical of a semisimple ring is 0 soooo

[velvet steeple]

The map being surjective is irrelevant, no? It preserving multiplication makes it a group homomorphism and implies f(e) = e

[elfin wraith]

yeah oops, added semisimple for no reason there, 0 is certainly nilpotent though so I guess what is said isnt wrong

[tough raven]

If you know that the codomain is a group, yes. If the codomain is a monoid, not necessarily. (Consider n ↦ n+1 on ℕ with the max as the binary operation.)

[velvet steeple]

Yeah

So, we have f(xy) = f(x)f(y). First, f(xy) = xyH = f(x)f(y) = xHyH. Thus yH = HyH = Hy. and so yHy^(-1) = H and this implies H is normal, since y was arbitrary.

[velvet steeple]

Does my last justification look reasonable?

Thanks

[elfin wraith]

The more I think about it, I dont think R/J(R) is actually interesting here, im not sure that in general J(R) should be Noetherian but im not seeing how I go from J^n/J^n+1 is Noetherian to R is noetherian

[surreal dagger]

If I localize a integral domain at an element and then take the fraction field, it should be isomorphic to just taking the fraction field in the first place, right?

[south patrol]

Yes exactly, in fact this is a general fact about localisations which is a good exercise from the definitions

(The general fact being that if you have two subsets S \subset T then A[S^-1][T^-1] = A[T^-1])

[velvet steeple]

Regarding the last part, I agree that we can get an injective homomorphism $f: G/H \to G'$, but why can $f:G/H \to G'/H'$ be injective?

[surreal dagger]

Okay thanks, will do. From the definitions meaning they both satisfy the same universal property?

[surreal dagger]

Since H is the kernel of f and f: G/H -> G´/H´ maps g+H to f(g)+H´ then f(g)+H´=0 if f(g) in H´ if g in H

[tough raven]

What is G'?

[velvet steeple]

Ok that sounds good

Thanks

[surreal dagger]

No problem, you exactly quotiened out the kernel of f i.e. all elements that got mapped to zero are already zero in the quotient hence the induced map G/H -> G´/H´ is injective

[elfin wraith]

Could anyone point me in the right direction wrt Hopkins-Levistky, im really struggling to see how to complete the proof. Ive got that $J^n/J^{n+1}$ is Noetherian for all $n \in \mathbb{N}$. I know that $J^m = 0$ for some $m\in \mathbb{N}$ since $R$ is Artinian, so we have the chain $0 \subseteq J^{m-1} \subseteq J^{m-2}/J^{m-1} \subseteq \ldots \subset J/J^2 \subseteq R/J$ but im unsure how to show this implies $R$ is Noetherian

Of course if we knew J(R) was Noetherian we'd be done but I dont think thats going to be true in general, and im not having many other ideas

[cloud walrus]

Nope

[tough raven]

Do you know that if N ⊆ M and N, M/N are Noetherian then M is Noetherian?

[elfin wraith]

Yeah

Well, not subset, submodule but sure

Im not sure how to make use of that though, I thought maybe some sort of inductive argument to show that J is Noetherian, but I couldnt quite see how to do it

Oh wait no maybe I do see it, J^{m-1} is noetherian, as is J^{m-2}/J^{m-1} hence so is J^{m-2}

Inductivly then J^0 is noetherian thus R is

I think that works?

[tough raven]

Yep, that's exactly it.

[elfin wraith]

jesus christ that took me far too long

Ill take it though, it did make me think much more carefully about submodules of quotients so thats good I suppose

[tough raven]

@elfin wraith How did you show that J^n/J^{n+1} is Noetherian BTW?

I'd like to learn this.

[elfin wraith]

Each J^n/J^n+1 is semi simple and artinian so it’s a finite direct sum of simple modules, and simple modules are trivially Noetherian

Showing they’re semi simple and artinian just comes from the fact that R is artinian and they’re annihilated by J

[tough raven]

Why is it semisimple?

[elfin wraith]

An artinian module is semi simple iff its Jacobson radical is 0

Actually is that true, or is it only true of the ring

[tough raven]

Hmm. IDK enough about radicals of modules; I should learn more IG.

Can we say that J is finitely generated?

I think I can see the rest from there.

[elfin wraith]

Yeah I think I might need to rework my proof there slightly I think it’s only true that a ring is semi simple if it’s Jacobson radical is 0, but like the quotient definitely is semisimple, my reasoning might just be a bit off

[tough raven]

I think R/J Artinian, trivial Jacobson radical => it is semisimple and then any module over it is semisimple.

IDR how to prove that either though.

[elfin wraith]

Yeah honestly I’ll think about it more in the morning I’m tired now but my original argument was that they’re R/J modules which are precisely R modules which are annihilated by J, but then since R is Artinian J(J^n/j^n+1)=0 implies J^n/J^n+1 is semisimple

I’m pretty sure this doesn’t quite work as is though, I don’t think that last implication is true

[chilly ocean]

why is $gp({[n]}) \subset \mathbb{Z}/d\mathbb{Z}$ is cyclic of order m by construction?

[cloud walrus]

whattodowithanything

[chilly ocean]

can you give more context on gp( { [n] } )?

[chilly ocean]

the subgroup generated by [n]

[chilly ocean]

since this subgroup generated by an element [n], can you compute the order of an element [n] in Z/dZ?

$ kn = 0 (mod ,d) \Rightarrow k = m $ since $d = mn$

[cloud walrus]

whattodowithanything

[chilly ocean]

so order is m

is this kinda correct?

[chilly ocean]

i think so, in general in a cyclic group Z/nZ of order n, the order of an element m is n/gcd( m, n )

so your subgroup has an order m and it is generated by an single element therefore it is cyclic subgroup

[chilly ocean]

what do you mean by"order of an element m"

by element m you mean [m]?

[chilly ocean]

yes

i see, so the order of [m] being n/gcd(m,n) is a general property

[chilly ocean]

Yes in Z/nZ

[spare orbit]

what’s a non example of a homomorphism?

[serene dune]

map identity to a non-identity element

the easiest one i think

[spare orbit]

why exactly is it not a homomorphism

[serene dune]

take G, (S_n \ AutG) as bijections are all not homomorphism

[serene dune]

recall what is homom i.e. the definition

[spare orbit]

I don’t quite understand this

[serene dune]

focus on the former for now

$f: G \to H$ such that f is a homomorphism,
$\forall g \in G$
$$f(e_G)f(g) = f(e_Gg)$$

[cloud walrus]

nastasya

[serene dune]

try to think now ?

[toxic zephyr]

can someone give me a hint for part a of 7 here? I'm supposing ax=0 for a,x nonzero but I can't find a contradiction

also, the idea is going to be that b is the inverse of a right?

[serene dune]

a comment, there can be element which is not zero divisor still it does not have inverse

[toxic zephyr]

in a division ring?

[serene dune]

well i made no restriction on the structure

[toxic zephyr]

part d says it's going to be a division ring so that's why I'm asking that

like if we are going to show it's a division ring, then b is going to be the inverse right?

[serene dune]

ohh thats interesting now im gonna solve it, haha

[spare orbit]

then $f(e_G*g)=a\bullet f(g)$?

[cloud walrus]

sushi

[serene dune]

yeah see how a has to be e_H

strech the equation to both directions and find the click

[spare orbit]

I think I get it

thank you

[serene dune]

for finite groups there is another way
but this one works in general

i always had the finite argument in my head

[toxic zephyr]

oh how about this:

• ax=0, a,x nonzero, then axa=0
• a=aba=aba+axa=a(b+x)a
• b+x≠b contradicting the uniqueness of b

[serene dune]

that works

[toxic zephyr]

sweet ty

[serene dune]

i didnt expect b to be the inverse, but anyways

[next obsidian]

I kind of find these weaker algebraic structures fascinating

You can say a lot about when silly slightly weaker axioms imply the full set of them

Or when finiteness plus X actually gets you Y

[toxic zephyr]

my thought was aba=a. if there was an a inverse, we could apply it and then ab=1.

[next obsidian]

But I don’t think they’re much worth studying other than for funsies

[toxic zephyr]

idk if my solution for c was very good. i basically used that ab acts as an identity

[next obsidian]

With any x?

On both sides?

[toxic zephyr]

yeah

[next obsidian]

In that case it should be fine

I don’t really think you have any other canonical way to produce an identity

[serene dune]

i was thinking if we can find some structure which creates equivalance with some know structure (division ring here)

that would be fascinating

[toxic zephyr]

for all x, ax=abax implies a(bax-x)=0 so bax=x. similarly xab=x. so we just need ab=ba.
a(ab-ba)a=a^2-a^2=0 so ab-ba=0 qed

[next obsidian]

In the case of like, a finite set with a regular multiplication (multiplication is injective let’s say) you pigeonhole to find a single thing where e•x = x and then show e works as identity everywhere

But in this case there’s really nothing, it has to pop out of the hypothesis somewhere and the only thing you’re really given is that for each a there’s a b where…

[toxic zephyr]

this is a pretty cool exercise tho yeah

[serene dune]

so its a field ?

[toxic zephyr]

but it's interesting that it sort of relies on the uniqueness of b. idk if R would necessarily be a division ring if the b is not unique

[next obsidian]

Only a and b commute

[serene dune]

oh right!

[next obsidian]

Nah no way, I can’t give you an example

But there’s no shot

[toxic zephyr]

yeah without uniqueness you get zero divisors

[next obsidian]

I think any division ring has this property, just set b to be the inverse, so a non-field division ring gives you an example

[serene dune]

yeah the structure is kinda nice, leads to some truth then blocks it

[toxic zephyr]

if aba=aca=a, and R has no zero divisors then b=c

[serene dune]

so division ring implies the hypothesis of the problem ?

[next obsidian]

I mean doesn’t it? Just take b = a^-1

[serene dune]

thats strong if true

[toxic zephyr]

yeah definitely

[serene dune]

oh yeah

[toxic zephyr]

haha

[next obsidian]

This is basically showing that a stupid notion of invertible is invertible

I think this has a name, without uniqueness

[toxic zephyr]

that's kinda what I meant by that beings the whole point of the problem: if aba=a and you can invert then ofc b will be the inverse

[toxic zephyr]

i like "stupid notion of invertible"

[next obsidian]

Ah

https://en.m.wikipedia.org/wiki/Regular_semigroup

I knew I had seen it before

What’s interesting is that a regular semigroup is not a group

Note what you did here, consider R\{0}, at best this is normally a semigroup

It has an associative multiplication

But when you considered the fact that R also has a + which worked well with x you got to upgrade this to being a full-on group. You got 1 and inverses

Now consider the case of a regular semigroup, where you have this weird aba = a thing, but no notion of +

You can’t show that this is a group, no 1 for you, and no inverses

https://en.m.wikipedia.org/wiki/Von_Neumann_regular_ring

[next obsidian]

Okay wait, maybe the problem is also uniqueness idk

[next obsidian]

Anyway, when you drop uniqueness these are Von Neumann regular rings

[serene dune]

i appreciate this raw input

[next obsidian]

https://en.m.wikipedia.org/wiki/Weak_inverse

also maybe of interest

Also who tf are you nastasya r u new

[serene dune]

idk how to reply to this

[next obsidian]

That’s kind of an unanswerable question

It’s just rare that the algebra channel gets someone new in it

Idk if this is a sign that i haven’t been tending to the channel enough recently

🚬

[serene dune]

u can find any previous conversations bw us, if it exists

[next obsidian]

Bad memory IG

[serene dune]

search box: from, mention

[serene dune]

same from this side

[next obsidian]

I searched and there wasn’t any from u to me

Whatever, the age that I knew everyone who used this channel is long past IG. It’s not 2020 anymore

🚬

[serene dune]

no, i started doing maths this summer

but pls come to this channel to vomit some of those raw chain of thoughts, i find those more valuable than most texts

[next obsidian]

maths

[serene dune]

blud is writing gigan tabs and calling it algebra

https://cdn.discordapp.com/emojis/1246899065615159296.webp?size=44

[next obsidian]

Idek what means

[serene dune]

kinda insider joke, but gigan is a band and someone told me this when the new album came out

[next obsidian]

[serene dune]

do u suggest reading sir Rotman ?

[next obsidian]

I didn’t read it

[elfin wraith]

Can’t speak for his intro to algebra but his algtop and homological algebra books are decent

Not read either cover to cover or anything, but I’ve used them both as secondary sources (to Hatcher and Weibel) so he is capable of writing a good book if that means anything

Although he defines the direct sum in an incredibly strange way that I don’t really understand in his homological algebra book

[next obsidian]

Lol how

And don’t say it’s via functions with finite support

Cuz that is the proper definition

Unless you want to try to argue infinity-tuples are a rigorous thing to make a definition of

[elfin wraith]

He first defines an external direct sum, which just seems to be the same as the normal direct sum, then says the normal (internal) direct sum is the same as the external one when it exists

I don’t know when it doesn’t exist or what the difference is but like it’s the choice he made

[next obsidian]

Lol

I guess he just wanted to say when you get a canonical decomposition

Via disjoint, jointly generating submodules

Weirdge

[elfin wraith]

Yeah I think you could be right but it just feels like an odd choice that I’m not sure he justifies particularly well, I was mainly just confused what squareplus could possibly mean when I first opened the book

[next obsidian]

Lmfao

That’s bad

Really occasionally I open a ring theory thing and see “quasilocal”

[serene dune]

i was asking for the group theory
and 2vol abs aleg

[next obsidian]

Which means local but not Noetherian

Cuz local is meaning Noetherian

[elfin wraith]

That feels like the French are to blame

Don’t love that

[next obsidian]

Not really

I mean, maybe originally

But it’s kinda extant only from Hochster I think

Or something

Idk

[elfin wraith]

Idk just adding an extra property to a definition then calling the usual thing quasi is very French

[next obsidian]

Lmao

[mighty kiln]

Je shall mange a quasibaguette oui oui

[next obsidian]

Arki red fox

[serene dune]

merci

[chilly ocean]

i think i did this one, which book? Jacobson?

i want to find the kernel of Ring homomorphism Z[x] -> R, f(x) -> f( 1 + \root2 ).

is there any thought process to find a function f such that f( 1 +\root2) without hit and trial?

[lusty marlin]

If f(1+√2)=0 for a polynomial f with integer coefficients, then f(1-√2) is also 0
One therefore obtains that (x-1-√2)×(x-1+√2)=x²-2x-1 divides f

Hence the kernel is the ideal generated by x²-2x-1

[dull tiger]

Good morning everybody. I'm currently learning about the Burnside Lemma and I found this online. Could maybe someone help me out here?
Like I do know the symmetry groups, but i dont really get how they do it here.

Why is it 4 over 2 for the identity? The group action is permutation, so isn't this like a graph without edges? As in nothing gets permutated.

For the second point: we choose 2 vertices to connect out of four, which will automatically give two two-cycles. But whats happening with the invariants?

and do i understand it right, that a fixed point in this case is just the same graph after permutation? So if i have the graph with 1 connected to 2 and 3 connected to 4, this would be a fixed point of the second case? Cause permutating 2 gives 1 and permutating 4 gives 3 which leaves the graph as it is?

Its from this website:
https://artofproblemsolving.com/wiki/index.php/Burnside's_Lemma

[chilly ocean]

ah great thank you

[elfin wraith]

I worked out what’s actually happening here, you show R/J is Artinian semi simple by first showing R/J is Artinian, which is straight forward because R is and J is an R submodule, then J(R/J)=0 by the correspondence theorem

J^n/J^n+1 is Artinian because it’s just a quotient of submodules, and then it’s semi simple because a left Artinian ring R is semi simple iff every left R module is semi simple, the proof of that is showing that submodules of semisimple modules are direct summands, and these are all finite by Artinian so no worries there

I noticed that this is actually given as a theorem in my notes but it’s kinda buried because we only give an outline of a proof (lecturer skips any proof that uses zorns lemma for some reason)

The notes for my course say there’s a proof given in Lams book “First course in non commutative rings” section 1.2 but for some reason springer isn’t letting me log in to look at it, I think the vague idea would that some of the simple submodules are contained in the submodule, then you can write the original semi simple module as like the submodule plus the simple submodules that don’t contain it, something along those lines is my feeling but idk I’ve not worked out the proof

[somber sleet]

HEy guys, why are we interested in groupautomorphisms

[chilly ocean]

I'm watching a youtube video of a lecture I missed and my professor blasts by this proposition while I'm sitting here trying to understand what being minimal wrt inclusions means. Could someone explain this for me?

[spice whale]

the only subfield is itself

[chilly ocean]

Oh I see so it's "prime" in some sense

That makes sense

Thanks

[spice whale]

generally, in a collection of sets S, A is minimal with respect to inclusion if the only subset of A in S is A itself

[glad osprey]

How do I find the Smith normal form of a matrix? For example, the SNF of [2 0; 0 5] should be [1 0; 0 10], but I'm struggling to understand the steps

[tough raven]

One way is to use that if SNF(A) = diag(a1, a2, ...), then ak = gcd of determinants of all k⨯k submatrices of A. (This won't give you the matrices you multiply A by on the left and right to get SNF(A) and for larger matrices can get unwieldy since if A is m⨯n, it has mCk ⨯ nCk such matrices, and that's (m+n)Cmin(m,n) in total.)

Another way is to run the algorithm for the SNF found in proofs.

IDR it but in this case I think you can do it ad-hoc. We "want" to get 2 and 5 in the same row or column so we can take their GCD, so start with C1 := C1 + C2. Then get the GCD into the (1,1) entry by R2 := R2 - 2 R1 (Euclidean algorithm for GCDs which happens to terminate after one step here), R1 := R1 - R2. Then clear everything in the first row/column with R2 := R2 - R1, C2 := C2 + 5 C1. You now have [1 0; 0 10], so you are done.

[keen terrace]

Guys quick question.. if a group G is generated by one of his elements, does that also mean that G is ciclic?

[void cosmos]

yeah

[glad osprey]

That makes sense, thanks I tried to extract an algorithm from the proof given in Basic Abstract Algebra, but it was somewhat hard to understand. Do you know of any other textbooks talking about the Smith normal form?

[keen terrace]

Perfect, thx

[tough raven]

I thnk I ended up learning it by reading multiple online lecture notes until I felt comfortable with it (but note that I don't remember the algorithm, and I've never had to practise running it myself).

I think the key step is to somehow get the (1,1) entry to be the gcd of all the entries of the matrix. Once you do that, you can clear the first row and column as above and recursively work on the rest of the matrix.

[glad osprey]

I see, thank you

[elfin wraith]

Is it only true that maximal ideals are prime if the ideal is 2 sided? I think even using the non commutative definition of a prime ideal we still need that M is 2 sided

I wrote up a quick proof so I know it is true if M is 2 sided, but it kinda hinges on that, I don’t see a way to remove that condition

[void cosmos]

where

wdym only

its true in noncommrings aswell

[chilly ocean]

i have to determine the structure of Z[x]/ (x^2+3, 3), i know how the structure looks of Z[x]/(x^2 + 3 ) and Z[x]/ ( 3 ) but i am not sure about original question

in these types of problems i am facing problems

[void cosmos]

ur modding out the 3s

right

like (3) is contained in (x^2+3,3) right

so all ur coefficints in these polynomials cant have 3s

so for starters this is Z_3[x]/(x^2+3)

right

[elfin wraith]

My proof was basically let M be maximal, IJ subset M, assume neither in M, then M+I=M+J=R, so 1 =(m+i)(m’+j) but to conclude 1 is in M for a contradiction id need to know that both m’j and im are in M, which only works if M is 2 sided no?

It could just be that this isn’t the best proof, but im not seeing another one rn

[chilly ocean]

yes

is it something ax+b, where a and b are in Z/3Z?

[void cosmos]

no continue

[elfin wraith]

Wait actually im dumb, the addition commutes so I can just swap it there

[void cosmos]

yes

no not that ur dumb

but yeah you just continue

[void cosmos]

yes

[chilly ocean]

is there any more description? what are the better ways to deal with it?

[void cosmos]

just understand that modding out is just remainders ig

like if ur modding out something by x^2

[chilly ocean]

modding out?

[void cosmos]

quotient

[chilly ocean]

ah

[void cosmos]

/

then all ur going to get are linear polynomials

etc

[chilly ocean]

yes

[void cosmos]

also usually with these problems the correspondcone theorem helps

[chilly ocean]

how?

[void cosmos]

you know what the ideals of a quotient look like by looking at the ideals that contain the thing your quotienting out by

this is the (third?) iso theorem

this is kinda why when you divide out by a maximal ideal you get a field

or a divsion ring atleast

[chilly ocean]

yes

[chilly ocean]

yes

thank you

[void cosmos]

yw

[proper tide]

How to approach to this question..please explain it stepwise

[hot wadi]

What is L800?

[proper tide]

Z_800

That is Z

[barren sierra]

Can you find a subgroup of Z_800 which is isomorphic to Z_2? Can you find a subgroup of Z_200 which is isomorphic to Z_4?

And then do you see why finding those groups suffices?

[tardy hedge]

I do not understand the argument about uniqueness

At. The very end

[bitter rover]

A homomorphism is uniquely determined by its values on the generators, in general

If an element g can be written as some product of generators, say g=abcd, and f is a homomorphism then f(g)=f(abcd)=f(a)f(b)f(c)f(d)

So if you know the values on the generators then you know the value for any element g — write g in terms of the generators and use the fact that f is a homormophism.

[tardy hedge]

Yes, I guess I wasn’t sure why Psi cant be defined some other way though. We want psi(m x n) to be equal to phi(m,n) (the middle linear map)

Ok nvm i know this is obvious im just being weird

Is there necessarily one unique one because if we want psi(m x n) = phi(m,n) then once we find one psi that satisfies that, oh it determines the rest of the map because m x n generates M tensor N

[proper tide]

@barren sierra yeah I find subgroup of Z_800 which is isomorphic to Z_2 and similarly for another..then what to do

[barren sierra]

What subgroups did you find. And then notice that the big group in question is a direct sum of the two groups I mentioned

[proper tide]

{0,400} and {0,50,100,150}

Yes then By their direct product got a set..Is that the required subgroup of Z_800 ×Z_200 which is isomorphic to Z_2× Z_4???

[barren sierra]

Nice

[proper tide]

Is that the required subgroup of Z_800 ×Z_200 which is isomorphic to Z_2× Z_4??

[tardy hedge]

Wew where are u 😢

[rotund aurora]

is "uponthewitnessing" a meme? I know Wew says it, but what does it mean

and isn't the guy of the emoji gothamchess?

[tardy hedge]

I got no idea lmao, ive only seen it in this server. i find it funny tho

[bitter rover]

Yes

[tardy hedge]

thanks

[dull tiger]

I have a group of order 90. If it has a subgroup isomorphic to $C_3 \times C_3$, why can't it have a subgroup isomorphic to $C_9$?

[cloud walrus]

mathrie

[next obsidian]

Idk how much group theory you know, but both of these are Sylow-3 subgroups and those are conjugate and hence isomorphic. Thus you’d conclude C_3 x C_3 ≈ C_9 which isn’t true

If you don’t have access to Sylow, you can show that the set (C_3 x C_3)(C_9) has size either 9,27, or 81. 9 corresponds to the two sets being the same which isn’t true cuz again C_3 x C_3 isn’t isomorphic to C_9. The middle is when they intersect at a subgroup of order 3, and the last one when they intersect trivially

If you can then show one of them is contained in the other’s normalizer then you get that this set is a subgroup of G, and hence its order has to divide 90 which they don’t, but idk how you can show the “contained in the other’s normalizer” very easily

[dull tiger]

great thanks! I do know sylow, so that explanation is very nice :). Why is it a subgroup of G if one is contained in the normalizer of the other?

[next obsidian]

Theorem

Usually it’s proven as a lemme in third iso theorem

It follows from the statement that it K < G is normal then KH is a subgroup

You just apply this where G = N(K) cuz if H < N(K), then well K is normal in N(K)

[next obsidian]

So you apply this

[dull tiger]

ah okay, many thanks!

[next obsidian]

[chilly ocean]

Prove that if F is a field then F[x] is isomorphic to F[t].

f(x) -> f(t), works? If yes then why do we need a ring to be a field if we take the ring commutative then also work?

[south patrol]

How are you defining these objects lol

To me this just seems like a silly question

[opaque finch]

Let $\alpha$ be a root of $x^2 + ax + b$ and $\beta$ be a root of $x^3 + px + q$. How do I construct a polynomial whose root is $\alpha + \beta$ and coefficients in $\mathbb{Q}(a,b,p,q)$ ?

[cloud walrus]

mycroftholmes1703

[chilly ocean]

F[x] the set of all polynomials over F

[tawny dune]

Does the problem ask you to construct it or are you just trying to prove the sum of two algebraic numbers is algebraic? Because there is an easier way to do that, namely ||showing a number is algebraic if and only if it is contained in a finite extension of Q||

[chilly ocean]

For me also

[opaque finch]

problem asks me to explicitly construct. Sum of two algebraic numbers is algebraic, that is a fact can be easily proved by extension of fields, so I don't want that

[opaque finch]

If I get the sense correctly, then I believe you want $x$ to be an indeterminate and $t$ to be fixed constant of $\mathbb{C}$. Now if $t$ is algebraic over $\mathbb{F}$ then such an isomorphism doesn't work, since then if you let $m_{t}(x)$ be the minimal polynomial then you have $\mathbb{F}[t] \cong \mathbb{F}[x]/<m_t(x)>$. To prove this isomorphism, you just have to use your map, which in this case will not be an isomorphism, but the kernel will be indeed be the denominator. But if $t$ is transcendental over $\mathbb{F}$, then it is indeed an isomorphism and exactly your isomorphism works well. And you are right, similar arguments holds for any commutative ring

[cloud walrus]

mycroftholmes1703

[chilly ocean]

But in question there is no given condition on t, so i think both x and t are indeterminate

[opaque finch]

But if $t$ is indeterminate then the question makes absolutely no sense. Cause by changing indeterminate notation keeps the ring "same" which is actually much stronger than isomorphism.

[cloud walrus]

mycroftholmes1703

[chilly ocean]

Yes

What do I now?

[opaque finch]

can you tell me the context for the question ? By that I mean, like if it's an exercise then for which section this exercise is in.

[chilly ocean]

It is an exercise from Herstein's Ring last supplementary exercise

[opaque finch]

Can you tell me the book please ?

😅

[chilly ocean]

Topics in algebra by Herstein

[toxic zephyr]

hungerford

[elfin wraith]

Could anyone look over this proof for me, im slightly unsure about my justification that we can just pull back but I think the idea is there, Im trying to show that if R is left Noetherian, and S is a multiplicitive set then R<S^-1> is also Noetherian

We know that $R$ is left localisable so $R \subseteq R\langle S^{-1}\rangle$. Then if $I$ is a left ideal of $R\langle S^{-1}\rangle$, $I \cap R$ must be an ideal of $R$, as this is simply the restriction of $I$ to a subring.
Clearly we have that $(R\cap I)\cdot(R\langle S^{-1}\rangle) \subseteq I$ since $I$ is an ideal, hence closed under multiplication. The other direction is similarly clear to see, $I \subseteq R\langle S^{-1}\rangle$ by the definition of an ideal, and $R\cap I \subseteq R\langle S^{-1}\rangle$, so this is still just $R\langle S^{-1}\rangle$.
Thus, every ideal of $R\langle S^{-1}\rangle$ is just the projection of an ideal $I$ of $R$ into $R\langle S^{-1}\rangle$ via the canonical homomorphism. So consider a chain of ideals in $R\langle S^{-1}\rangle$,
[I_1 \subseteq I_2 \subseteq\cdots.]
Each of these has the form,
[J_1R\langle S^{-1}\rangle \subseteq J_2R\langle S^{-1}\rangle\subseteq\cdots,]
for ideals $J_i$ of $R$. Now pulling back along $\varphi^{-1}:R\langle S^{-1}\rangle \to R$, the inverse of the canonical homomorphism, we see,
[J_1 \subseteq J_2 \subseteq\cdots,]
however this chain stabilises since $R$ is Noetherian, thus so must the chain $I_1 \subseteq \cdots$, hence $R\langle S^{-1}\rangle$ is also Noetherian.

[cloud walrus]

Nope

[toxic zephyr]

it looks like my prof won't be doing nilpotent or solvable groups. does anyone have any recommended texts or readings that have a good treatment? hungerford's approach is kind of dry and they say specifically it's a "group theoretic approach" but "historically they're talked about in terms of polynomials".

[toxic zephyr]

my understanding is that nilpotent groups are groups such that if you take commutators with commutators with etc. then eventually, the result is in the center (it commutes with everything). sort of makes sense since i'm used to nilpotent meaning A^k=0, and here we have G^k={e} (for G^k=[G^(k-1),G])

but solvable i'm still not quite understanding. it seems that to be solvable you have that commutators of commutators of... eventually commute with each other. but what does that mean? why do we care? what makes that "solvable"?
so not being solvable means that if you keep taking commutators, eventually things don't get "simpler" or smaller and the k-commutator subgroup is non-commutative and fixed by the commutator action (idk if that makes sense or if the commutator of a subgroup is an action)?

[grave sedge]

Solvable groups are those that have a composition series with abelian quotients

[tough raven]

Write successive powers of a+b as linear combinations of 1, a, b, ab, b^2, ab^2, writing the coefficients in columns next to each other (start with (a+b)^0 = (1)1 + (0)a + ...). Eventually the columns will be linearly dependent, and that linear dependence means a polynomial relation satisfied by (a+b).

Specifically, (a+b)^0, ..., (a+b)^6 gives you a 6⨯7 matrix, so there must be a relation.

Another way to do it is using resultants; this is given as an application on https://en.wikipedia.org/wiki/Resultant.

[next obsidian]

I would go even further to say you have one with simple quotients

Which is as far as you can possibly hope to push it

[tough raven]

Simple abelian*

[next obsidian]

Err

I meant to say cyclic

And yeah simple abelian I guess

Basically I just wanted a bunch of Z/pZ

[rocky cloak]

So you don't consider for example Q to be a solvable group?

[next obsidian]

All groups r finite

[rocky cloak]

Ah of course. My mistake

How could I forget such an obvious fact

[next obsidian]

I gVe the most ass lecture today

At least for the like first 40 mins

How to give the most unhelpful explanations of Aut(G) and Aut(Z/nZ) ≈ (Z/nZ)^x

[tardy hedge]

😂

thats so funny dude

[next obsidian]

Were u in the crowd

[tardy hedge]

Indeed

[gilded finch]

https://arxiv.org/pdf/2410.20392

damn i cant believe i missed this

[hot wadi]

I didn’t realize pure group theory like this was still an active area of research

Very cool

[chilly ocean]

ah

[chilly ocean]

what the exact structure of ring Z/12Z [x]/( 2x - 1 )?

the polynomial 2x-1 is not monic so we have not a basis for given ring

how do i compute residue of x^2 in given ring?

[rocky cloak]

Notice that modding out 2x-1 is exactly adjoining a multiplicative inverse of 2.

So x^2 will be an inverse of 4. That means also that things y such 4y = 0 will now become 0, since you can divide by 4.

So this is the same as Z/3Z [x] / (2x-1)

[chilly ocean]

got it, is it final descripition or are there any more clarification left?

[rocky cloak]

Well you can simplify Z/3Z [x] / (2x-1) quite a bit more (remember Z/3Z is a field)

[chilly ocean]

so Z/3Z[x] is Euclidean ring so Z/3Z [x]/ (2x-1) is isomorphic to Z/3Z, correct?

[rocky cloak]

Yes, that's right

[chilly ocean]

thank you

[opaque finch]

So actually solvable groups arise the from Galois' original motiavtion for developing his theory, when is a polynomial equation solvable by radicals. The motivation for a subnormal series comes from the pretty notion of Galois correspondence, all normal subgroups actuallu correspond to a normal extension. The quotient abelian condition is related to some parts of Kummer theory and the fact that abelian groups can be direct product of cyclic groups, which makes some properties of extension easier to analyse.

However your concern about the commutator of commutator of...... and why do we care has an interesting notion. We know how much we like, (I definitely do like I dunno about you) abelian groups and particularly a specific class of abelian groups called cyclic groups. Our fact is that we want to find the notion between such a definition of solvable group and investigate its relation to high commutator subgroups. Just like we have field extension we wants to have some notion Group Extensions and how do we build up the group from it's successive extensions. Solvable groups has a finite solvable series so it can be constructed back to original by finitely many extensions.

@toxic zephyr sorry for not replying to your message, but instead I am mentioning you.Hope this answer helps

[opaque finch]

Try Peter Cameron's notes and blog posts

[crystal vale]

if G is a group of odd order and x not equal to e then x^-1 is not in cl(x) because | cl(x) | is odd and if x^-1 in cl(x) then | cl(x) | will be even.

if y in cl(x), then y^-1 will be in cl(x) because x^-1 in cl(x). SInce y^-1 cannot be y, so they are different, hence |cl(x) | will be even.
is it correct?

[tough raven]

You could elaborate more on why y^{-1} in cl(x). Otherwise, yes, this is correct.

[crystal vale]

let y in cl(x) so y = gxg^-1 for some g.
Now since x^-1 in cl(x) so x^-1 = hxh^-1 for some h. Now y^-1 = gx^-1g-1 = ghxh^-1g^-1, thus y^-1 in cl(x)

[crystal vale]

if G = HK, where H and K are normal and H and K are abelian subgroup, then G is abelian, correct?

because H, K are normal so they commute each other, h1k1h2k2 = h1h2k1k2 = h2h1k1k2 = h2h1k2k1 = h2k2h1k1

i used H and K are abelian subgroup

[serene dune]

i think so coz only the trivial semi direct product is allowed now

i did something funny ig

[crystal vale]

I have to determine the number of Sylow 2-subgroups of D_2m, where m is an odd integer at least 3. I think it is exactly m

[serene dune]

each reflection is an involution

[crystal vale]

yes

[serene dune]

they all will generate order 2 subgroups

[crystal vale]

yes

[serene dune]

things get more interested when m is even,

u get more involution

[crystal vale]

yes

[tough raven]

If H ∩ K is trivial, then yes (hkh^{-1}k^{-1} has to lie in H ∩ K for any h ∈ H, k ∈ K). Otherwise, not necessarily: in the dihedral group of order 8 <r, s ∣ r^4 = s^2 = srs^{-1}r = 1>, consider H = <r> and K = <s, r^2>, which are both abelian and normal (index 2), but don't commute with each other because srs^{-1}r^{-1} = r^{-2} = r^2 ∈ H ∩ K.

[elfin wraith]

Wondering if anyone can point me in the right direction for this problem. Let $A$ be a ring and let $P \triangleleft A$ be such that $A/P$ is a domain. Let $S:= A\setminus P$ and observe that $S$ is multiplicatively closed. Suppose that $S$ is left localisable. Prove that,
[S^{-1}P:= {s^{-1}p: s\in S, p \in P}]
Is the jacobson radical of $A \langle S^{-1}\rangle$.

[cloud walrus]

Nope

[elfin wraith]

I want to prove that this is a left ideal before anything else, its closed under addition, so I just need to show its closed under A<S^-1> multiplication, but im really unsure how to do this

Like elements of A<S^-1> are just words, and we cant generally simplifiy them. S is left localisable, so we have an overring of A/P, B where we can simplify everything to the form s^-1a, but im not sure where this gets me

Weve got the universal property of A<s^-1> and quotient ring floating around, weve also got that P is prime (Since A/P is a domain) but again not quite seeing where this takes me

[tardy hedge]

A map being R-bilinear does NOT a priori mean that phi(mr, n) = phi(m, rn) right

From the definition of R bilinear it does not seem that way

From the definition anyway it only says M and N are left R mods anyway too

[lone niche]

If R is commutative then yes it follows. If R is not commutative then the definition gets a bit funky. You can check the property on wikipedia. You mention M and N are left R modules but you have R acting on the right on M, is that a typo?

[elfin wraith]

As a sort of aside to this question, we never really spoke about why we actually care about localisation or rings of fractions in the non commutative case, like sure it’s nice that we get inverses, but we can’t do it for all rings, and isn’t the usual motivation for localisation in the commutative case algebraic geometry which is pretty commutative?

Mainly interested in a hint for how to show it’s an ideal, but any uses for the left ring of fractions would also be appreciated

[rocky cloak]

R-bilinear would be
phi(rm, n) = phi(m, rn) = rphi(m, n)

The condition
phi(mr, n) = phi(m, rn)
is called being R-balanced.

As you might notice, R-bilinear is something about a pair of left modules (or possibly pair of right modules) while R-balanced is a statement about a right and left module.

Really R-balanced is the thing related to the tensor product, but in the connotative case it's equivalent

[tardy hedge]

Ok yea I was kinda thinking something like that thanks!

I am spending so much time trying to understand all the details of tensor product

[elfin wraith]

There’s a really nice explanation in chapter 12 of “a primer of algebraic D modules”

[rocky cloak]

So the localization of a ring always exist, to invert r you can just take the coproduct of R and Z[x] modulo the ideal (rx-1, xr-1). But you usually impose restrictions to get a nicer construction (coproducts of rings are a bit unwieldy).

I'm no expert, but I think the main motivation for localization of noncommutative rings is for noncommutative geometry. Basically you have techniques that "deform" schemes and other geometric objects and when you translate that to rings you sometimes get noncommutative rings. So you kind of need to reconstruction algebraic geometry in the noncommutative setting.

[elfin wraith]

Provided you’re comfortable enough with modules anyway, I found the explanation via bimodules to be the most sensible explanation I had seen anyway

[elfin wraith]

Ah now that you say that I do remember the lecturer saying there’s a more general notion of localisation in the non commutative case when your ring isn’t a domain which we don’t cover because it’s just a bit of a mess, I’m guessing this is what she was meaning

That does sound quite interesting, I wasn’t aware there was much to non commutative geometry, but equally I don’t know much about regular algebraic geometry. Is that at all related to like deformation of rings and deformation theory? My comalg lecturer has mentioned it before but never in much detail

[crystal vale]

yes my bad their intersection is trivial

[rocky cloak]

Yeah, don't know much about it myself. But I believe that's related yeah.

I guess this would be a good thing to ask your lecturer about, they can maybe answer more specifically about what your learning.

[rocky cloak]

Anyway, to your actual question.

Notice x (s^- p) = (x s^-)p = (t^- y) p = t^- (yp)
for some t in S and y in R, by the definition of being left localizable.

[elfin wraith]

Ahhh yeah I was getting caught up in thinking of the words like letter wise if you will, just taking it as a single element makes the Ore condition much more clear

[nimble folio]

i dont really like how he presents modules

id rather read d&f and once you get to tensor products switch to rotman

[elfin wraith]

Someone in advanced algebra was incredibly confused about direct sums of modules this morning and I instantly knew it was rotmanns homological algebra lol

It’s such an odd presentation

[nimble folio]

Yea

he put his introduction to modules in the section on categories and functors

i thought that was really silly

[elfin wraith]

In fairness I would’ve assumed modules were just a prereq to homological algebra but

The main reason I’ve even looked it is just to see if I can reasonably define Tor without mentioning derived functors

And I guess I can just define directly for the tensor product but I’m still unsure if I want to do that, but even rotman goes for derived functors

[elfin wraith]

So ive worked through the rest of this problem (took a break for a while), and ive shown that it has a unique maximal left ideal and that this is the Jacobson radical of A<S^-1> (I know this would mean that the ring is local in the commutative case but im not sure if that generalises) but I cannot for the life of me see where ive used that P is a prime ideal of A

Im guessing ive implicitly assumed I dont have zero divisors somewhere but im not sure where, any chance youd be able to take a quick look? Mainly just curious because im pretty sure my proof works, but im failing to see the relevance of A/P being a domain

We know that $S^{-1}P$ is a subgroup of $A\langle S^{-1}\rangle$, so we need only show it's closed under left $A\langle S^{-1}\rangle$ multiplication.
Let $x \in A\langle S^{-1}\rangle$. We see,
[xs^{-1}p = (xs^{-1})p = (t^{-1}y)p = t^{-1}(yp),]
for some $t \in S$ and $y \in A$ since $S$ is left localisable. Notice however that $yp \in P$ since $P$ is a left ideal of $A$, so we have that $S^{-1}P$ is a left ideal of $A\langle S^{-1}\rangle$.
\\
Suppose $I$ is an ideal of $A\langle S^{-1}\rangle$ such that $S^{-1}P \subsetneq I$. Then there must exist an element $s^{-1}a \in I$ such that $s^{-1},a \in S$. Then, by definition, there exists an element $a^{-1}s \in A\langle S^{-1}\rangle$.
However, since $I$ is an ideal we see that,
[s^{-1}aa^{-1}s = 1 \in I,]
giving us $I = A\langle S^{-1}\rangle$, so we must have that $S^{-1}P$ is maximal.
\\
To see that this is unique, suppose $M$ is a maximal ideal of $A\langle S^{-1}\rangle$. It cannot contain a unit, as then it would be all of $A\langle S^{-1}\rangle$, contradicting it being a proper ideal. So $M$ only contains non-units, thus it must be a subset of $S^{-1}P$. However, as $M$ is maximal, $M \subseteq S^{-1}P \subseteq A\langle S^{-1}\rangle$ implies that $M = S^{-1}P$, so $S^{-1}P$ is unique.
\\
The Jacobson radical is the intersection of all left ideals, and the only maximal left ideal of $A\langle S^{-1}\rangle$ is $S^{-1}P$, so $J(A\langle S^{-1}\rangle) = S^{-1}P$.

[cloud walrus]

Nope

[elfin wraith]

Is it just in the step where I say s^-1aa^-1s = 1? Since A/P is a domain, S must have no zero divisors? I dont think thats it though because they should be units

[cloud solar]

If in a finite unit ring we have the element 2 a nilpotent one then the order of the ring is a power of 2?

[rocky cloak]

I mean, that A\P is multiplicatively closed is just saying that A/P is a domain.

[cloud solar]

??

[rocky cloak]

Yes that's correct

[peak root]

Suppose $x,y$ in $G$ are elements of finite order in $G$. p is a prime number which is coprime to the order of x and y. We know that $x^{p^n}$ is conjugate to $y^{p^n}$ for some n. Prove that x is conjugate to y. I know that we could deduce that x and y has the same order but that does not necessarily mean $x$ is conjugate to $y$. I feel like it should be a straightforward argument, but I could not get it right. Can someone give me a hint?

[cloud walrus]

Dong_Valentino

[spice whale]

think about bezout's lemma and the product of the orders of x and y

[fluid kelp]

I am reading the book Finite fields and there is an exercise that I have spent a while trying to prove. However, it has not come to success. This is exercise 3.97 by the way.

Any help would be greatly appreciated!

[peak root]

I see that now. Thanks a lot.

[fluid kelp]

I don't know if this correct etiquette or not but could you please help me with the question I posed 2 message above this? If it's wrong to do this I apologize

[spice whale]

in general don't ping random people

[fluid kelp]

I didn't ping

[spice whale]

i don't know how to do this

[fluid kelp]

I responded

[spice whale]

you did ping me

[fluid kelp]

Replied

[spice whale]

yes it was a ping reply

[fluid kelp]

Oh I didn't know

Apologies

[spice whale]

apology accepted

[rose axle]

What have you tried

[fluid kelp]

Well for one, I tried using cyclotomic polynomials. This obviously wouldn't work. Then I tried using orders and primitive roots and that didn't work too well unfortunately.

The previous exercise was easily done using cyclotomic because it was x^2n+x^n+1 and it is basically x^3n-1/x-1 so that worked out well. This one is a lot harder and I have struggled with it for a while.

Not sure if I am headed in the right or wrong direction.

[stark gull]

How do you prove that $x^3-3x+1$ is irreducible over $Q$?

[cloud walrus]

lebzul

[rocky cloak]

Rational root theorem oughta do it

[stark gull]

it doesn't work

[serene dune]

i forgot the theorem completely, should revise

p|a_0, q|a_n

[lone niche]

Why not?

[sonic coral]

it’s a cubic, so if it were reducible, it would have a linear factor

a neat thing about that polynomial is that given any root alpha, alpha^2 - 2 is another root. one root is -2cos(pi/9)

[rocky cloak]

Well, it does work. But where are you running into trouble?

[stark gull]

i'm srry I confused the Rational Root Theorem with Gauss's Lemma.

[rocky cloak]

Well, same thing in this case

Rational root theorem is just a special case of Gauss' lemma

[stark gull]

So I use the theorem 'if p(x) is quadratic or cubic with no roots in K, then p(x) is irreducible over K,' and to verify this, I use the Rational Root Theorem by assuming it has a root in Q and arriving at a contradiction, right?"

[stark gull]

yep i forgot this

[tough raven]

Let a be a root of the polynomial x^4 + x + 1 = 0 over F_2. Then your desired conclusion is equivalent to the combination of (i) a has degree 4 over F_2 (i.e., X^4 + X + 1 is irreducible over F_2) (ii) the polynomial X^n - a is irreducible over F_16 iff n = 3^k 5^m for some k, n.

I assume (i) can be done by hand somehow. For (ii), a hint is ||look at the multiplicative orders of a and the root of X^n - a||, although I haven't figured out fully how to solve it.

[rocky cloak]

. @fluid kelp

Should work similarly for 5

[tough raven]

Oh, oops.

Wait, these were asked by different people, right?

[rocky cloak]

Yeah, similar question asked by someone else some time ago

[tough raven]

I think it'll need to add a lemma to show that ||y is not a 5^th power => X^5 - y is irreducible|| at the inductive step.

[fluid kelp]

Oh wow

Very similar question

[fluid kelp]

Thanks for the help. I also have another proof that I am not sure works. Do you think I can send it in dms and you could tell me if it is viable or not?

[crystal vale]

let f: G -> G/N, where N is a normal subgroup of order 2 of G. Now let H is a normal subgroup of G/N of order 3, i know f^-1(H) is a normal subgroup of G but how can i say it has order 6? yes it contains N

[next obsidian]

Given a group homomorphism f:G -> H, let y in im(f). Let x in f^-1(y) be arbitrary, then show that f^-1(y) = xker(f)

What I’m saying is, every fiber is a coset of the kernel, so they each have size the kernel

[crystal vale]

yes

[next obsidian]

Apply that to your situation, the kernel is N and you have 3 fibers

So 3•2 = 6

[fluid kelp]

If you have the generators for subgroup of index 2 in a group, how do you prove it is the only index 2 subgroup?

[crystal vale]

got it thank you

[next obsidian]

This isn’t an answerable question

Like, it’s unique to the situation, because this statement isn’t true in general

C_2 x C_2, a generator of C_2 x e is just (1,0)

But it isn’t the only index 2 subgroup

[fluid kelp]

What about like C_3 x C_2

What would happen there

Or maybe C_5 x C_3

[next obsidian]

You don’t have an index 2 subgroup of the latter one lol

And for the former there’s only 1

But like, these are all different situations so idk what you’re really even trying to ask

[fluid kelp]

How would you prove that

[crystal vale]

Let G be a group of order 60 and it has normal subgroup N of order 2, how can i say Sylow 3-subgroup is normal subgroup in G?

[next obsidian]

I looked at it

An order 3 subgroup is cyclic so it’s generated by an order 3 thing and there’s only 2 of those

And those generate the same subgroup

Or you can say C_3 x C_2 = C_6 by the Chinese Remainder Theorem and cyclic groups have unique subgroups of any given order

Idk

[fluid kelp]

Same

I'm just asking for the sake of asking

Lol

[hidden wind]

what’s some nice topic to look at to get used to thinking with exact sequences

prof working some prev exam problems and he just pulls exact sequences everywhere

[elfin wraith]

Exact sequences are everywhere in homological algebra so maybe take a look at some books on that? There isn’t really a huge amount to exact sequences though tbf, it’s just telling you that the image of one map is the kernel of the next

[hidden wind]

hmm i’ll get to alg top later, unfortunately i don’t have time to delve into that rn

but thanks it’s good to know these’ll show up there

and yeah i get that they don’t have all that much content, what i don’t get is why that content is so important

something something group quotient i guess but i don’t quite understand the something something ehehe

[elfin wraith]

Im not sure exactly what to say about it, maybe someone with more experience can provide a better answer but they’re just kinda nice to work with

For example if you have 0->A->B then you know A->B must be injective

You then get like SES and LES and these are just kinda very nicely behaved, and you’ve got stuff like the snake lemma that tells you how to make these LES

Maybe the thing that’s nice is that exact sequences can just be generally phrased in terms of abelian categories, so you can say stuff about abelian groups, modules, vector spaces etc

[south rain]

So in a proof where we show that every non-abelian group G of order 8 we have to have an element of order 4 it was argued that if we would have no element of order 4 we must have an element of order 8 or all elements have order two, in both cases this would mean G is abelian, which is a contradiction. I understand that any group of order 8 has normal subgroups of order 2,4,8. But why must we have elements of order 8 (or the other case) if we have no element of order 4?

[slim kayak]

Really just trying to do the diagram chase proofs yourself. Short exact sequences should be straightforward to understand, and long ones are mostly a cohomology thing

[grave sedge]

If all elements have order 2 then the group is abelian: ab=a(ab)²b=a²bab²=ba

[south rain]

Yes, I'm aware, but still, why is it that if no element has order 4, either one element has 8 or all have order 2?

[grave sedge]

What are the possible orders?

[fluid kelp]

Hi, when is x^4n + x^n + 1 divisible by x^4 + x + 1 versus x^4 + x^3 + 1 when n is not divisible by 3 or 5. this is in F2

[vast stump]

2 and 8 are the only possibilities, if none are order 8, then all are order 2

[chilly ocean]

I'm having trouble understanding this explanation for why Q⊂R is not finite. Why does the fact that pi is transcendental imply that Q(x) is a subfield of R?

In fact, it's actually kind of unintuitive to me that Q(x) would be a subfield of R

[grave sedge]

Isomorphic to a subfield of R

[chilly ocean]

What is this subfield? It is not clear to me what sending x to pi really looks like as a field of rational functions over Q

[grave sedge]

You send f(x) to f(π)

[fluid kelp]

Matteddy could you help me with my question please

[south rain]

Alright, then I'm clearly missing something. We have to have a order of an element which divides the group order?

It's probably a basic theorem which follows from Lagranges theorem right?

Like, it can't be that there is no element with order 2,4 or 8

[lusty marlin]

In a finite group, the order of each element divides the order of the group.

[south rain]

Alright, which follows from Lagrange by considering $g^{|G|}$ where $|G|$ is then somemultiple of the order of g?

[cloud walrus]

dellinger

[grave sedge]

A subgroup of index 2 is normal, in general

So there's a surjective homomorphism from your group into C2

The C3 generator must be in the kernel because it has order 3, if the generator of C2 was also in the kernel then the homomorphism would be trivial

So the kernel must be the subgroup generated by y and xyx^-1 (i'm calling x and y the generators of C2 and C3 respectively)

[grave sedge]

It's the kernel of the unique surjective homomorphism from the free product to C2

And any index 2 subgroup must be the kernel of such a homomorphism

[fluid kelp]

How do you know it is a unique surjective homomorphism

[fluid kelp]

Is that by definition

[grave sedge]

This

[grave sedge]

No, because as i said above a subgroup of index 2 must be normal, and the quotient is the unique group of order 2 (i.e. C2)

[fluid kelp]

Can it alternatively be represented as <xy^2x, yxy^2>

[grave sedge]

No, because yxy² is not in the kernel

[fluid kelp]

Why is that

It's to free product to C2 so I thought yxy^2 would be in it

[grave sedge]

Ok let me state the homomorphism explicitly

you send x to 1 and y to 0

This is the only surjective homomorphism from the free product to C2

yxy² then must go to 1

So it cannot be in the kernel

[fluid kelp]

That makes sense.

[fluid kelp]

I am not entirely sure how it easily follows that the group is generated by those 2

I have been trying to figure it out for sometime but I cannot make the connection unfortunately

[grave sedge]

The subgroup is "the words with an even number of x's"

Ig if you want to do it formally you can argue by induction on the length of the word (written using only x's and y's, no inverses)

If the leftmost thing is a y then you have y times something with an even number of x's and less characters so by induction you can write that as a product of y and xyx

If the leftmost thing is an x then there must also be a y (else the xx simplifies) and notice that xy=(xyx)x so again you get xyx times something shorter

[elfin wraith]

Let $R$ be a ring. Suppose $M = N \oplus P$ is a direct sum of right submodules. Suppose that the only $R-$module homomorphisms $N \to P$ or $P \to N$ are zero.
Let $f \in \text{End}_R(M)$. Show that $f(N)\leq N$ and $f(P)\leq P$. Deduce that there are surjective ring homomorphisms,
[\phi_N: \text{End}_R(M) \to \text{End}_R(N), \enspace \phi_p: \text{End}_R(M) \to \text{End}_r(P)]
whose kernels intersect trivially and sum to $\text{End}_R(M)$. Hence, describe $\text{End}_R(M)$ as a direct sum of rings.

[cloud walrus]

Nope

[elfin wraith]

Anyone able to point me in the right direction for the first part in showing that f(N)\leq N at least? I kinda get the vibe here but im struggling with how to actually show it

[toxic zephyr]

the set S of 0 and all zero divisors of a ring R is just the torsion of R as an R module right? and to show it always contains a prime ideal, can we just pick the annihilator of any element a of S?

[rocky cloak]

N is the kernel of the projection M -> P

[rocky cloak]

It might depend how you define torsion for non-domains, but it is the set of elements with nontrivial annihilator yes.

The annihilator of an element isn't necessarily prime though. What you can do though is notice that the complement of S is multiplicatively closed, then think about prime ideals in a localization.

[tardy hedge]

A middle linear map to L gives u a group hom from the tensor product to L, if the tensor product is also an R-module, does this middle linear map extend to a bilinear map?

[fluid kelp]

jagr2808, does the following idea work to show that 3^a5^b is irreducible in the x^4n + x^n + 1 in mod 2 question?

Let f(x) = x^4+x+1 in mod 2
f(x^(3^m5^n)) = you can just plug in
We have that totient function of 3^m+1 is 2 x 3^m and similarly for 5^n+1 we have 4 x 5^n
Taking order of everything we have 4 x 3^m x 5^n and this is also the degree of f(x^(3^m5^n)). thus it is irreducible.

[rocky cloak]

Middle linear meaning f(mr, n) = f(m, rn)?

If so, then not necessarily. I mean it's not even given that L is an R-module. But even if it was you would need something relating the map to the R-module structure

[rocky cloak]

Maybe, but I'm not sure I'm following.

Like you take phi(3^m 5^n) and then how does that connect with something being irreducible?

[fluid kelp]

That's like a hole in my argument. I am not sure. I assumed that if order of a polynomial = the degree, it is irreducible. This is in mod 2

Do you know if that implies it is irreducible?

[rocky cloak]

I'm not sure what "the degree in mod 2" means

[fluid kelp]

the phrase mod 2 is poorly placed

[rocky cloak]

Well then I'm even more confused.

What does "the order of a polynomial = the degree" mean?

[fluid kelp]

Um

[fluid kelp]

When you take the order in mod 3^m5^n

I have the order as: 4 x 3^m x 5^n
I have the degree of the polynomial as: 4 x 3^m x 5^n
Does this mean anything with regards to irreducibility?

[rocky cloak]

I'm still not sure what it means at all.

You take the order of what mod 3^m 5^n ?

[fluid kelp]

Using the totient funtion we have phi(3^m+1) = 2 x 3^m and phi(5^n+1) = 4 x 5^n. specifically, the order of the roots of f(x^(3^m5^n)) is 15 x 3^m x 5^n.

so then