groups-rings-fields

[final pasture]

What do you mean "making it from some ring" ? @trim grove

From Z[x], you can get F_p[X] from a quotient, and from F_p[X] you can get any finite field through another quotient in the usual way

Does that answer the question ?

[trim grove]

i know that form F_p[x] we find a irredcuiblle polynomial in that and can easly make field of order p^{order of that polynomial}, how to write F_p[x] form Z[x]?

[final pasture]

Consider the maximal ideal generated by the constant polynomial p and quotient

[trim grove]

okkk like Z[x]/<p>?

[final pasture]

Yes

[trim grove]

Thankyou

[ionic sparrow]

Hi, I’m working through aluffi’s Algebra: Chapter 0 and I’m stuck on an exercise.
“Prove that R[x] is an integral domain iff R is an integral domain”. I’ve managed to prove it for if R[x] is an integral domain but not the converse. Could someone give me a hint? Thanks. Any help is much appreciated, please ping me if you’re responding. I might take a while to answer any questions though

[weak oriole]

Think about the leading coefficient

[ionic sparrow]

Oh, I see now
Thanks

[misty harness]

I have a question on simple groups (groups where the only normal subgroups are the trivial group and the group itself): If G is a finite non-simple group and choose G_1 a non-trivial normal subgroup of G of maximum order. Then the quotient group G/G_1 is simple. How can you proof this? I've worked with contradiction. Suppose G/G_1 is not simple, then there exists a non-trivial normal subgroup, call this one

$nG_1$ with $n \in G, n \neq e_G$

[cloud walrus]

Luka835

[misty harness]

I'm not quite sure how to get a contradiction out of this. Is there someone who knows how one can proof this statement? 🙂

[sharp sonnet]

consider the preimage under G -> G/G_1

[misty harness]

Still using contradiction?

[proud bear]

Try to prove that if G' is a normal subgroup of G/G_1, then its preimage under that map is a normal subgroup that contains G_1

[misty harness]

I succeeded to prove that if $G'/G_1 \lhd G/G_1$ then $G' \lhd G$

[cloud walrus]

Luka835

[misty harness]

But not why $G_1 \subset G'$

[cloud walrus]

Luka835

[misty harness]

Any other approaches? Would be appreciated :))

[past temple]

G' doesnt contain G_1

stain is saying that the preimage of G' under the canonical projection contains G_1

[misty harness]

Yeah I used another notation: the preimage of G'/G_1 is G', that was easier to work with

[trim grove]

i got the fourth option , but i cant understand first 3 options, is there any direct results to do these ?

[trim grove]

but the only option true is 3 and 4

[proud bear]

Idk if you already got this by now, but G' contains the preimage of 1G_1 (thought of as an element of G'/G_1). And the preimage of 1G_1 is just G_1

@trim grove wait were you able to understand the solution I posted to that problem

[west violet]

What's the most beginner friendly book for intro to abstract algebra? or resource(it doesn't have to be a book)

hopefully self-contained too and makes very little assumptions

[trim grove]

which one?

[proud bear]

.

[trim grove]

i tried very hard , but i can't get that.

[proud bear]

which part didn't you get

[trim grove]

after fifth line

[proud bear]

this part? or the one before

[trim grove]

yesss after this part

[trim grove]

okk stain , let me try this one more time.

[proud bear]

wait. typing stuff up in LaTeX is slow

[trim grove]

Thanks a ton @proud bear ,

[proud bear]

@trim grove

oh wait there are typos lol

[proud bear]

should be this sorry

[next obsidian]

1. Isn’t true

Unless you aren’t requiring the 1 to be the same in which case wtf

3. Follows from 2. if you know that finite fields of every prime power order exist

[oblique river]

2 is false tho lmao

[next obsidian]

What? Did I misread it?

Wait what?

Isn’t this saying F_p < F_q iff p | q

[oblique river]

Yea

Which is false

[next obsidian]

Is the condition that q needs to be a power of p?

I thought divisibility was the condition

[hidden haven]

yes

[oblique river]

Yes

F4 is not a subfield of F8

[next obsidian]

Oof

[hidden haven]

divisibility of the exponent

[next obsidian]

Boy am I bad with finite fields

[trim grove]

yes this is a counter example

[oblique river]

Yea thats why i wrote it :P

3 and 4 are both true though

Which im sure chm knew as well. Right chm? ;P

[trim grove]

what about 3rd

[oblique river]

I just said it’s true haha

[trim grove]

is this any direct result

[oblique river]

Finite fields have order p^n

So if two fields have orders with gcd > 1 then they are powers of the same prime p

Say p^n and p^m

Both of which are subfields of p^mn

[trim grove]

Got it thankyou @oblique river

[next obsidian]

I did know 3 and 4, although the reason I knew 3 was a bit off

If I want to seriously get better at mixed characteristic stuff maybe I should first review finite fields

Lol

[trim grove]

and i want to leave this topic , Galois gives me nightmare

[white nymph]

if F is a field and F[x] polynomial ring, how to start showing that for phi in Aut(F[x]), phi(x) = ax+b, a and b in F with a =/= 0?

[hidden haven]

See what happens when the degree of phi(x) is not 1

||You will lose surjectivity||

[white nymph]

i see, if x maps to higher deg, nothing will map to x (as constants map to constant). and if x maps to constant, then all polys map to constant. merci

[plucky flicker]

What is the difference between free generating system and generating system. So G is a free group and S is a generating system of G, so every element of G can be written as a finite expression of elements (and their inverses) of S. But if S is a free generating system then this expression is unique up to simplification of terms like suu^-1t = st, where s is not t^-1. That's all?

[chilly ocean]

I had the same panic moment as you and realized I was thinking about divisors of the exponent of the prime, that is where it comes in. Ok thank god I'm not insane at least we were on the right track xD

[coarse storm]

I had not came across "generating systems" before. If it is the same as a generating set, then I do think the uniqueness hold. Or rather, if S is a generating set of a group G and it is free of relations, then the group is necessarily free.

[plucky flicker]

yes, I mean generating set

there is a theorem: if G is a free group of rank r, and X is a generating set with order r, then X is a free generating set

that's why I'm asking the difference

[plucky flicker]

hmm so basically that's why we call it a free group? cuz it is free of rrelations?

[coarse storm]

Yep. Free of relationships like in "Are you free?~ wink wink".

[plucky flicker]

I am sadge

[coarse storm]

A new year resolution... of length one.

Is that theorem in a book? How did they prove it?

[void knot]

Let $X$ be a set and $G$ be a group acting on $X$. So the isotropy group (also know as stabilizer) of $x\in X$ is defined to be [G_x:=\left{g\in G:: g\cdot x=x\right}.]
What does it means the next statement:
"
The set $\left{G_x::x\in X\right}$ of all isotropy groups is exactly one conjugacy class of subroup of $G$.
"
?

[cloud walrus]

RaD0N

[void knot]

conjugacy class of $x\in X$ is the set $\left{gxg^{-1}:: x\in X\right}$.

[cloud walrus]

RaD0N

[void knot]

ooups

so, the conjugacy class of $G_x$ is $\left{gG_xg^{-1}: g\in G\right}$.

[cloud walrus]

RaD0N

[hidden haven]

Yes, here they are saying that there is a subgroup H such that the set of conjugacy classes of H is exactly the set of all G_x

Equivalently every G_x is a conjugate of every other G_y and every conjugate of a G_x is G_x' for some x'

[void knot]

so, if the action is transitive (any orbit gives the whole $X$), then
[\left{G_x::x\in X\right}=\left{gG_yg^{-1}:: g\in G\right},]
for all the $y\in G$?

[cloud walrus]

RaD0N

[void knot]

yes

[hidden haven]

Yes that sounds right

[rustic crown]

no that isn't necessary. that thing could be generated by 1

like in Z, (7, 11) = (1)

yea sure, but you can't say anything certainly.

(2, 1+sqrt(-5)) isn't principal

but that doesn't mean (x, y) in general can't be principal

[hidden haven]

x = y

[rustic crown]

yep

[final oasis]

how do I show the multiplicative group $\mathbb{F}^*_q$ is cyclic

[cloud walrus]

bchaotic

[unique berry]

find a generator

[chilly ocean]

I dont get it whats F whats q

[unique berry]

F just means it's a field

[final oasis]

a Field of order q

[unique berry]

q is a prime number

[chilly ocean]

so q=p^n?

[unique berry]

or prime power

[chilly ocean]

ok

[final oasis]

yeah p^n

but can be p for now

[chilly ocean]

does F cyclic implies F* cyclic?

no

[final oasis]

what does a field of order p^n look like

I thought I had an intuition of it but it's wrong

would we take the fractions of $F^*_p[x]$ and quotient it by an irreducible polynomial to get a field of order $p^n$

[cloud walrus]

bchaotic

[final oasis]

I am not finding it in general so far it seems to be a number prime to p - 1 that's not the root of x^n = 1 mod p , n less than p

I am thinking maybe some restriction on n

like n | p-1 maybe but I am not sure

[final oasis]

original problem here

[cloud walrus]

causalloop

[final oasis]

Showing that the multiplicative group $\mathbb{F}^_$ is cyclic , for now q =p a prime ,
so far I have that if $a\in \mathbb{F}^_p$ ,is not a solution for $x^n = 1 $ , if $n | p-1 $ then a generates $\mathbb{F}^*_p$

[cloud walrus]

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

[final oasis]

but how can I show that such an a exist that is not the solution a any polynomial x^n = 1 with n dividing p-1

[desert marsh]

$f : R \to S$, $S = \text{img}(f)$. $A$ is the set of all ideals of $R$ containing $\text{ker}(f)$. $B$ is the set of all ideals of $S$. $g : A \to B$. Show that $g(I) = f(I)$ is a bijection.

Can someone help me do this?

[cloud walrus]

causalloop

[obsidian sleet]

uh

i don't know any ring theory yet

please avoid pinging people randomly for this

[amber stag]

Are you allowed to use the fact that F_q is the collection of the roots of the polynomial x^q-x?

[final oasis]

yup

I have some progress so for

[final oasis]

how would you use this

[amber stag]

I'm writing on a paper to make sure I'm not saying some bs

[final oasis]

this is Fermat's theorem yeah

[amber stag]

Ok got it. So we want to find a generator, i.e. an element g such that g^{q-1}=1 where q-1 cannot be replaced by any smaller number

[final oasis]

yeah

[amber stag]

Any nonzero element in F_q^* is a root of a polynomial in the form x^{p^n-1}-1. Since the number of roots of a polynomial cannot be larger than its degree, so there must be such a generator. Otherwise every non-zero element (there are q-1 of them) is a root of a polynomial of smaller degree.

[final oasis]

I am specifically looking at $F_p^*$ p a prime for now

[cloud walrus]

bchaotic

[final oasis]

this is what I want, keep in mind p-1 is the order of the multiplicative group of F_p

[final oasis]

every non zero element of F_p^n is a solution of this poly

x^n has a most n solution

[amber stag]

hmm. Seems what I said was indeed bs

[final oasis]

that's alright

What I need is a Theorem,
If G is finite abelian with x^n = e has at most n solution for each n , then G is cyclic. @amber stag

[amber stag]

talking to someone, will be distracted for a few min

[chilly ocean]

cringe

wait wut bchaotic

for every n, there exist x such that x^n = e?

wait wut

you just need to show the single generator

you are saying there exist a set of n elements of G for every natural number such that x^n=e

we are trying to prove that one of these elements generates the union of all of these sets

oh no nvm

you said has at most n solutions

i didnt see at most part

[amber stag]

Yes. Use the Euler totient function. Suppose there’s an element g of order r, then the subgroup generated by g is exactly the solution set of x^r-1. Especially all elements of order r are there. So the number of elements of order r is phi(r), where phi is the Euler phi function. Add all those numbers up for r and use the theorem, which I forget the name but it’s saying the number n is the sum of phi(r) where r divides n. Then we see there’s an element of order n.

[tribal moss]

Huh, is r a group element or an integer here?

[chilly ocean]

int right?

[tribal moss]

But "suppose there's an element r".

[chilly ocean]

he meant integer i think

only way this makes sense tbh

[amber stag]

An element g of order r

[chilly ocean]

Like let g be an element such that it satisfies g^r= e

[amber stag]

Thanks

[chilly ocean]

appreciate the proper use of totient func

[final oasis]

thanks I see it

[chilly ocean]

Aight who up happy new year it's time to classify groups of order 2022=2*3*337 in this channel now
how does wolfram know there's only 6?

n2 could be 1, 3, 337, 1011 since that still lets you have n2 = 1 mod 2
n3 could be 1, 337
n337 = 1

how to get some 6 for sure hmm idk

[vast quiver]

you also know that n2 divides 3*337, and n3 divides 2*337, right?

[chilly ocean]

wait why would n2 necessarily be 1

[vast quiver]

true

you are right

hmm so you have a normal Sylow 337 subgroup

call it P

[chilly ocean]

tru that is another thing, giga pog

[vast quiver]

and you have some Sylow 3 subgroup, call it Q

then because P is normal, PQ is a subgroup

[chilly ocean]

GIGA POG

[vast quiver]

and PQ has order 3*337

subgroup of index 2, so PQ gotta be normal

[chilly ocean]

same if we took a 2-subgroup right? yeah

yeee

awesome

[vast quiver]

so we got that G is the semidirect product of PQ and Z/2Z (the only group of order 2 up to isomorphism)

[vast quiver]

you could do that, but then you wouldn't have a subgroup of index 2

sorry index

index 2 subgroups are based

[chilly ocean]

oh I see what you did now, yeah since it has order 3*337 just lagrangang that mf it's index 2

[vast quiver]

mhm

[chilly ocean]

also wait PQ is guaranteed to actually be a subgroup since P and Q are normal and PQ=QP

[vast quiver]

we only needed one of the two to be normal

[chilly ocean]

ye that too

[vast quiver]

P is normal, Q might not be

[chilly ocean]

the Sylow 3 subgroup? o yeah since the self congugateness comes from there being only one of them rip

[vast quiver]

mhm

yeah n3 could be 337 and that would mean Q is not normal

but we know for sure that n337=1, so P is normal

[chilly ocean]

giga gaming

So
Order ------------ Potential # as far as we know
2 1, 3, 337, 1011
3 1, 337
337 1 (normal)
1011 1 (normal)

and quotients of order 6 and 2

[vast quiver]

well I'm not sure if there's only one subgroup of order 1011, but I do know that it is normal. It's kinda a different story for Sylow vs non Sylow subgroups I think

[chilly ocean]

yeah I put as far as we know this is just from Sylow info

[vast quiver]

Sylow subgroups are all conjugate, so there being only 1 is equivalent to normal

mhm

you know about semidirect products?

[chilly ocean]

either it's been a long time or no LOL

[vast quiver]

cause at this point it mostly boils down to classifying all possible semidirect products of PQ and Z/2Z (for all the options that PQ can be)

aight

so first let's talk about all the options for PQ

P is a group of order 337, so it gotta be isomorphic to Z/337Z

Q is a group of order 3, so it gotta be isomorphic to Z/3Z

bc these are groups of prime order

[chilly ocean]

prime moment cyclic yeah

[vast quiver]

yuh

now we gotta talk about the possible ways of putting P and Q together

ie classifying the possibilities for PQ

[chilly ocean]

does Chinese remainder theorem not just give you Z/1011Z right away or nah wait what

[vast quiver]

so we could just start naming things if we wanted, but it would be hard to know if we exhausted everything

nope not quite

[chilly ocean]

wait whaa
OH its PQ not P x Q

[vast quiver]

yup

so some options we can just brainstorm right now for fun (before we start classifying more seriously)

include Z/3Z x Z/337Z

uhhh

hmm I'm struggling to think of anything else

well maybe that's it, let's talk about semidirect products

[chilly ocean]

wolfram thinks there is a second one for some reason

[vast quiver]

well we'll see

[chilly ocean]

okee

[vast quiver]

so there's a nice theorem that say that if a group G got two subgroups H,K such that for all h in H, k in K, hk=kh and if HK=G and if H cap K = {e}, then G=H x K

it's just a nice way of looking at some properties of G to recognize it as a direct product

there's a weaker thing where you drop some of those properties, and you don't quite get it to be a direct product, but you limit the possibilities

[chilly ocean]

I think you get that right away from looking at cardinalities if its finite

[vast quiver]

wdym?

[chilly ocean]

|HK| = |H||K|/|HcapK| I think so just do number theory pog moment if they're finite and of prime order

[vast quiver]

oh true in our case you get HK=G and H cap K = {e} for free

or something like that

the thing we don't have is that hk=kh

that elements commute

this is the thing that prevents us from deducing that PQ = P x Q

[chilly ocean]

unpog

[vast quiver]

aw yeah

[chilly ocean]

https://tenor.com/view/sueñitosgifs-traumatized-mr-incredible-los-increibles-meme-gif-24305569

[vast quiver]

so we gotta settle for less

so just in general there's a term for the situation we're in

if G got subgroups H,N, and N is normal, and H cap N = {e}, and HN=G, then we say G is the semidirect product of N and H

written as $G= N \rtimes H$

[cloud walrus]

Joseph

[vast quiver]

there are many possible semidirect products in general

for example, $D_{2\cdot 3}$ and $\mathbb Z/2\mathbb Z \times \mathbb Z/3\mathbb Z$ are both semidirect products of $\mathbb Z/2\mathbb Z$ and $\mathbb Z/3\mathbb Z$

[cloud walrus]

Joseph

[chilly ocean]

what being used to your own latex macros does to a mf

[vast quiver]

yeah lol

[chilly ocean]

nice

[vast quiver]

I think tex bot let's you define macros?

[chilly ocean]

oh cool

[vast quiver]

I'll have to check it out later

so now we want to figure out how to classify semidirect products, given a subgroup H and a normal subgroup N

the thing to do is to come up with a "external" semidirect product

or maybe that's a thing to do, that's how I think abotu it

uhhh

hmmm maybe I don't know semidirect products well enough to teach em

you can show that semidirect products are in correspondence with homomorphisms $\varphi: H \to \mathrm{Aut}(N)$

[cloud walrus]

Joseph

[vast quiver]

where the homomorphism says how H conjugates N

[chilly ocean]

wow sounds like a lot of stuff I did that was cool but forgot now lmao

[vast quiver]

yeah it do be like that

[chilly ocean]

that is cool there is something for this kind of scenario though

[vast quiver]

mhm

yeah we were lucky with 2022's factorization too

the strategy to keep going for 2022

would be recognize PQ as the semidirect product of P and Q---classify all the homs $\varphi: Z_3 \to \mathrm{Aut}(Z_{337})\cong Z_{336}$

then recognize G as the semidirect product of PQ and $Z_2$ (for each option for $PQ$). Then, for each option we came up with earlier, you gotta classify the homs $\varphi: Z_2 \to \mathrm{Aut} (PQ)$

[cloud walrus]

Joseph

Joseph

[vast quiver]

so there'll be some case work, but it's totally doable

other group cardinalities are bad

something with high powers of primes are in particular harder to classify because you won't be able to guarantee the existence of a normal Sylow subgroup

then you can't pull off all the cool semidirect product stuff

[chilly ocean]

that sounds not good

[vast quiver]

yeah it's cringe actually

[chilly ocean]

so the semidirect product stuff, does it guarantee or tell you something about what the group of order 2022 has to be? Like would one possibly end up with something like "o yeah we get 6 and only 6 groups like this" like how wolfram did

[vast quiver]

yup

[chilly ocean]

either way cool to squeeze more info out of this

oh COOL

where did you learn this stuff? class with a particularly good prof or a textbook

[vast quiver]

my group theory last quarter had a class project where we had to classify tons of different groups with cardinalities under 100

uhh kinda all over the place

[chilly ocean]

RIP
I have a degree and I never heard semidirect products til now lol

[vast quiver]

lol you're chilling, I didn't get them first pass either

semidirect products were an exercise in Lang, but I don't think Lang is a particularly good book, at least for group theory. I guess I mostly learned them from lectures

[chilly ocean]

noice

[vast quiver]

dummit and foote talks about em, I haven't read that though

[chilly ocean]

so basically 2022 has not terrible group theoretic properties

[vast quiver]

yeah it's a real good one when it comes to classification

[regal arrow]

What goes G \oplus H mean for groups

[vast quiver]

people write $\oplus$ only for abelian groups, and $G\oplus H \cong G\times H$

[cloud walrus]

Joseph

[vast quiver]

it's read as "direct sum"

[chilly ocean]

element-wise structure bois

[vast quiver]

and it's the same as direct product when doing finitely many abelian groups, but it's different form infinitely many

[regal arrow]

Why the convention of only using \oplus for Abelian groups

[vast quiver]

it has to do with the "universal property" of direct sums/coproducts

[regal arrow]

Fml

Thank you

[vast quiver]

lol it sounds worse than it is

[regal arrow]

I hate categories

[vast quiver]

nahhh they're not too bad

[regal arrow]

Only like 'em 'cause they make things work in ag and at

too abstract smh

[vast quiver]

it's just about saying that if you have a hom $f: G\to K$ and $g: H\to K$, there's a unique way you can put them together to get a hom $h: G\oplus H \to K$ such that $h$ restricted to $G\oplus {0}$ is $f$ and $h$ restricted to ${0}\oplus K$ is $g$

[regal arrow]

eck

[chilly ocean]

something like that must be unique? that's cool

[cloud walrus]

Joseph

[vast quiver]

it's the analogy of disjoint unions for groups

[hidden haven]

Why not just classify homomorphisms Z/6Z to Aut P and from S_3 to Aut P? I think those should be the only 2 order 6 groups

Because then you don't have to do it in 2 steps in which the second step depends on the first one

[vast quiver]

hmm

how do we know there is a subgroup of order 6?

[hidden haven]

The product of a sylow 2 and a sylow 3 subgroup?

Oh that may have larger order?

[vast quiver]

it might not be a subgroup

[chilly ocean]

or it might just not be a subgroup itself

yeah

[hidden haven]

Oh i thought we took closure under operations

[chilly ocean]

the set would be 6 elements for sure

[vast quiver]

oh sure then it might be bigger

I've taken the convention that HK is just the elements hk

[hidden haven]

Ideal convention lol

[vast quiver]

yeah lol

[chilly ocean]

oh by the way

Say I is an ideal

ive seen notation IR before

oh lol

i remember now

it was for hilbert basis theorem maybe?

yeah

let I be an ideal

IR[x] was the notation?

actually have no clue where i saw notation

but i think it might have to do with showing R[x]/I[x] iso R/I [x]

i forgot the proof

[hidden haven]

R[x] is an R module and you are taking the submodule IR[x]

[chilly ocean]

yes that was it

i think its supposed to be ideal with polynomials with coefficients in I

[hidden haven]

Yes it turns out to be that

[chilly ocean]

yeah we are given R is noetherian and we want to show R[x] is too

so we create this set A = ideal of R consisting of leading coefficients of polynomials in IR, unioned with 0?

nah nvm

i got to review it

i want to commit to memory because it was simple enough

[next obsidian]

,w factorize 2022

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Chmonkey 2.0#9767

[oblique river]

[next obsidian]

Well, to celebrate 2022 let’s show a group of order 2022 cannot be simple.

Suppose that it was, then by Sylow’s theorem it would have >= 338 Sylow-337 subgroups, and these intersect trivially leading to at least >= 338*336 elements which is 113,568 elements

So a group of order 2022 has a normal Sylow-337 subgroup, happy 2022 everyone!

Alternatively you could use that none of the numbers which are 1 mod 337 divide 6, other than 1

[hidden haven]

2022 has been celebrated 🥳

[sturdy marsh]

4n + 2

[next obsidian]

Be quiet

[sturdy marsh]

[next obsidian]

I solved it first

I just helped every person here who has an algebra qual this year

[sturdy marsh]

[trim grove]

Happy New Year To All 🎉

[trim grove]

Definition of irreducible element in integral domain is A non-zero non-unit element in an integral domain is said to be irreducible if it is not a product of two non-units, but what if the given domain is not an integral domain like Z_10, i want to check is 6 irreducible in it or not? any hint?

[lethal dune]

I'm not sure how to extend irreducibility to non-integral domains but with my best guess, you can write 6=1x6 or 2x3, either case one factor is unit

also 6=8x2

[trim grove]

then every element will be irreducible

[chilly ocean]

Wat
They just pointed out otherwise

[trim grove]

how to solve this, because i tried to find a isomorphic map with kernal <2x+4>, but get nothing, any hnit?

[wraith obsidian]

Well what do elements of this ring look like?

That's a reasonable thing to start with if you have no intuition at all about the structure at hand

[trim grove]

something like ploynomial with cofficients forn Z6 +<2x+4> type?

[wraith obsidian]

Okay. write down, say, five elements 🙃

[trim grove]

x+1, x+2, x+3 ,x ,1

[trim grove]

i am bad in solving these quotient rings

[hidden haven]

0,1,2,3,4,5 lol

[trim grove]

looks like 5 will again give 1?

[hidden haven]

Or rather their equivalence classes in the quotient

[hidden haven]

in what way?

do you mean 1=5 in the quotient?

hold on

shit we are not in an integral domain

😵‍💫

but is 1=5?

[chilly ocean]

How would 5 be 1 here? I don't see it

[hidden haven]

There might be some polynomial which when multiplied by 2x+4 gives a constant

Though it seems not possible because 4 and 2 have the order but I don't have a proof

[chilly ocean]

I can see all the polynomial terms with even coefficients yeeting down to constants

[hidden haven]

probably follows from 2x+4 being 2(x-1) and we can do the degree argument for monic polynomials

[trim grove]

is not there will be x+ a types polynomials?

[chilly ocean]

I can't read that sorry lol

[trim grove]

okk i'm thinking as if we divide with 2x+4, then variable part will get handelled by 2x , and constant part by 4, so posabillities of reminders are 0,1 for varaiable part , and 0,1,2,3 for constant part.

[hidden haven]

no you can't do long division nicely if the polynomial you are dividing by has a non unit leading coefficient

and even then the 0,1,2,3 constant part is wrong

because you won't be able to simultaneously kill both the variable part and the constant part

constant part is out of your control if you are trying to kill as many powers of x as possible

[trim grove]

okk i think i'm getting this now

[hidden haven]

maybe try doing long division with some random examples, this is something you will use a lot lol

should be comfortable with it

[chilly ocean]

Oh I just realized you can do something like 5x^k = 4x^k +x^k = 4+x^k when you have odd coefficients on powers of x
So all the elements will look like a power of x plus a constant I think

No clue how to proceed from there

Hard to believe that it ends up being a field according to the screenshot lmao

[trim grove]

answer given is 1,3

[hidden haven]

I think it is a true or false problem lol

definitely not a field

[chilly ocean]

Oh lol I thought it was like "Then: all of these are true"

[trim grove]

like if i want to solve 3x+1 +<2x+4>?

ok so can i write it like
2x+x+5+2 +<2x+4>
(2x+5)+x+2+<2x+4>
1+x+2
x+3
?

[hidden haven]

I have the advantage of having seen the problems TheStudent has posted before

[chilly ocean]

I think the next step would be to show those elements I claimed are a unique way of writing the elements of R or something like that

[coarse storm]

For part (a)?

Uh.. 1.

[chilly ocean]

x^k+a + I = x^t+b + I implies k=t, and a is b mod 6 or something like that

And then you'd be done

Boni

[coarse storm]

I think that works, yea. Uniqueness might be a little strong for proving non-finiteness though.

Hi!

[chilly ocean]

What could you do instead?

Cause idk how to show even my idea LOL

[coarse storm]

Proving uniqueness here would also mean showing that every element can take a given form, which you had explained you can do. But that is perhaps unnecessary.

[chilly ocean]

It just comes from 2x=2 and that odd* x^k = x^k + even*x^k

Every element is x^k+a

[coarse storm]

Wait... it does? How do you get x^2 + x to that form?

[chilly ocean]

Wait no lol rip I meant in a sum

F

Yeah like that just coefficient 1 or 0 on everything except the degree 0 term

My bad

[coarse storm]

Yea. That sounds right. You had phrased it correctly earlier.

It just feels a little fiddly to me to also show every element has that form.

[chilly ocean]

Yeah

[coarse storm]

What I had in mind was contradiction. Assume finite order n. Pick n + 1 elements of your form. Show they must be distinct, as you did. That is sufficient. I think.

[chilly ocean]

Right ok

More like contraduction amirite 🥁 💥

I kid

Still how tf would you do that

Definitely sufficient

[wraith obsidian]

No time to respond in more detail, but I think about „modding out 2x+4“ as „adding the equation 2[x]+4=[0], i.e. 2[x]=[2] (where [x] means the coset of x)
So you have [0] up to [5] and then [x] (in combination: [x]+[0]…[5]), multiples [kx] reduce to ([x] if k odd)+[constant rest] or because the even part becomes a constant due to the new equation

CRT would also help but is less informative for you I guess

[hidden haven]

problem here is that [0] up to [5] need not be distinct

at least not immediately

since not domain

and we have all the powers of x

I don't think those reduce

[wraith obsidian]

Oh true, I'm stupid

I shouldn't answer when I'm in the midst of traffic

[hidden haven]

dw I made the same exact mistake 🫂

[trim grove]

Drive safe

[weary bane]

Hello i am trying to understand this proof in Linear algebraic groups by humphrey and i understand evry step exept the last line where he says that (G,G) and G'_u are connected thus G_u is conmected

Can anyone help me please understand how he came to this conclusion?

[wooden ember]

I know nothing about this but $G_u$ is the complete preimage of $G_u'$ in the quotient by $(G,G)$ and both of these are connected so maybe there's some theorem youve seen previously that allows you to conclude?

[cloud walrus]

𝓛ittle ℕarwhal ✓

[weary bane]

G_u is the unipotent elememt of a connected solvable algebraic group

[wooden ember]

this wont help me help you i have zero familiarity with any of this stuff i just tried to relate to the common mistakes i usually make when i dont understand a proof

best to wait for someone qualified

[weary bane]

Okay thank you for the help <3

[chilly ocean]

With the information you would usually get from looking at sylow 2-subgroups of groups of order 26, why aren't the reflections {1, s} in D26 normal? They obviously are not since rsr^{-1} = r^2s

What "goes wrong" with Sylow here, since you know n2 must be 1 right (since 13 is prime and no other subgroups of order 2 of D26, not the info from Sylow)?

[trim grove]

there can be 13 sylow 2 subgroups

[chilly ocean]

Yes
But in D26 there are not

[trim grove]

https://math.stackexchange.com/questions/1410582/normalizer-of-a-sylow-2-subgroups-of-dihedral-groups

does this will help?

[chilly ocean]

Wait lmao it's just because there are literally 13 sylow 2 subgroups 💀
sr^ksr^k= 1 yeah

[broken stirrup]

With Sylow theorems it's so easy to tackle, but how can I prove it without Sylow Theorems?

it smells like induction but I don't know how to use induction on this one

[cyan marten]

If I remember correctly, you use regular permutation representation of G. I saw it somewhere in Dummit and Foote.

[white nymph]

Can someone help me understand this problem?
I got stuck on part b.) (I tried using tower theorem)... and I'm not even sure I did part a.) correctly

[woven delta]

If $m_{F,\alpha}\in F[X^n]$, then $m_{F, \alpha^n}$ is going to be exactly $m_{F,\alpha}$ with $X^n$ replaced with $X$

[cloud walrus]

'quid

[woven delta]

The degree of the field extension $F(a)$ is the degree of the minimal polynomial of $a$ over $F$, and it's easy to see from what I said above that $\deg(m_{F,\alpha})=n\deg(m_{F,\alpha^n})$

@white nymph

[cloud walrus]

'quid

[white nymph]

thanks @woven delta

[oblique river]

not necessarily

f(x) = x^4 + 1 is irreducible over Q

but its galois group is (Z/2Z)^2

which doesnt have an element of order 4

so in particular it doesnt have a 4-cycle in it

[hidden haven]

All you can say is that it acts transitively on the roots but that doesn't imply that there is an n-cycle

[hidden haven]

does tensor distribute over direct product for R-algebras

It is not obvious from the adjunction and I don't wanna check if the isomorphism for the module case respects multiplication

actually I guess there is no tensor hom adjunction in this case

only tensor diagonal adjunction since tensor is the left adjoint to the diagonal functor C → C x C

is that right?

because I am trying to solve this

first part is done, need to do the L=k^a case

wait is k^a direct product of k or alg closure of k

it is algebraic closure

I spent so much time stressing about this problem and the only clue to k^a being the alg closure was that the a is not in the standard math mode font

[next obsidian]

k^a isn’t a field (as a direct product)

Oops

[next obsidian]

Verifying it is works and is basically trivial. It sends (a,b) (x) c to (a (x) c, b (x) c) and this is just obviously multiplicative

[hidden haven]

Nice very cool

[hidden haven]

Ye I thought we are just embedding k into it with the diagonal map and then tensoring

But Lang has to use epic notation

[next obsidian]

Yeah but when L is an extension of K

You expect L to be a field

And a direct product of rings isn’t ever an integral domain even

[hidden haven]

I thought that part was separate from the extension part

[next obsidian]

[hidden haven]

[next obsidian]

Anyway this is an interesting problem

Oftentimes you can actually test separability using an explicit basis

I forget what condition you need, it might be a characteristic assumption?

But you look at the trace of the extension

[hidden haven]

I haven't seen this

[next obsidian]

You take a basis of L over K

Number it x_1 through x_j

Then multiplication by x_i is a K-linear map on L

So you can write it as a matrix (with entries in K)

Then take the trace

So what you do is look at
det(trace(x_ix_j))

This being 0 or not is independent of choice of basis

Then it’s separable iff this is non-zero

[hidden haven]

I'd have to think about why that is true

[next obsidian]

This is in I think section 24 of Matsumura

https://stacks.math.columbia.edu/tag/0BIE

And also in here probably

[hidden haven]

will take a look

after this assignment

apparently (iii) obviously implies (iv) and I don't see it

[next obsidian]

I wonder if this lets you show the “suffices to take L = k^a” part

[hidden haven]

somehow Hom(P,A) (x) E encodes pairs of maps from P to a free module and free module to E I suppose?

[next obsidian]

Hmm

[hidden haven]

I showed it by using the fact that all k-modules are flat because it is a field and then for an inseparable E I took k[b] for some inseparable element b and then showed that this tensored with k^a has a nilpotent

and then this embeds into E (x) k^a so that has a nilpotent

[next obsidian]

For iv WLOG the map f is surjective

And this E is also fg

[hidden haven]

I don't see how

[next obsidian]

Just take the image

You only need to take maps into the image of f to make the diagram commute

[hidden haven]

The natural map is f (x) e maps to (p maps to f(p)e) btw

I think

[next obsidian]

Yeah

[hidden haven]

but does the property given in (iii) restrict to submodules

[next obsidian]

Huh?

E is arbitrary in iii

[hidden haven]

oh no

E is a fixed module

apparently this is all equivalent to E being flat

over commutative ring A

[next obsidian]

Hurbed

Okay

[hidden haven]

sorry I didn't bother stating the set up because I thought you were good enough with commalg to deduce that yourself

I shall lower my expectations of you

[next obsidian]

Is your ring Noetherian

No

They wouldn’t specify finitely presented them

Can we assume the ring is coherent?

[hidden haven]

bruh I don't even know what it means

[next obsidian]

That the ring is a coherent module over itself

I KNEW THIS WAS LAZARDS THEOREM

[hidden haven]

I shall raise my expectations of you again

There is a hint

but it says

(iii) implies (iv) is easy from hypothesis

what kind of hint is this

is this some encrypted hint

[next obsidian]

Hmmmm

NGL I don’t see it lol

This is in uhhhh

That Atiyah Macdonald alternative

[hidden haven]

Damn

btw do you wanna see the endsem for this course

it was shit hard

all problems were like this

I was so depressed when I had only solved 2 problems fully 3 hours in

thanfully we got 5 hours in the end

[next obsidian]

Maybe try inducting on the number of generators of P?

iii gives it to you when P is cyclic

[hidden haven]

So I was trying to do this by first taking an element of Hom(P,E) whose preimage is f (x) e, just a pure tensor

and I still couldn't see it

I will try inducting on generators of P

[next obsidian]

Lot there’s a typo in 3

When E is finitely presented it becomes an isomorphism

[hidden haven]

lol ye

[next obsidian]

You know that characterizes finitely presented modules?

[hidden haven]

they spotted that and sent an email annoying

[next obsidian]

If this is true for all direvt systems you’re finitely presented

[hidden haven]

oh I didn't

Nice

[next obsidian]

I don’t know how to prove that direction

Kekw

[hidden haven]

lol

[next obsidian]

Anyway the proof is a simple application of 5-lemma

[hidden haven]

ye

This was the first problem I figured out

Then I did 6

[next obsidian]

Just like all these functorial isos for finitely presented things

[hidden haven]

Then I stared at the paper for hours

and finally got all 4 parts of 1

and then 2,4,5 all submitted partially

[next obsidian]

Lmfao problem 4

That’s just annoying

The separable field tensor product is a field hs to be false right?

[hidden haven]

yes

[next obsidian]

Okay

[hidden haven]

C tensor C

over R

[next obsidian]

Oh yeh

[hidden haven]

can't be a field because 4 dim alg

[next obsidian]

…

Just compute it manually lol

[hidden haven]

lmao why would I do that

[next obsidian]

Have you proved that in clas??

Lol

[hidden haven]

nope

[next obsidian]

TFW

[hidden haven]

wait what are you talking about

[next obsidian]

IIRc it’s literally just an application of CRT

[hidden haven]

proved what

[next obsidian]

The dim 4 thing over R

Can’t be a field

[hidden haven]

isn't that just C being alg closed

[next obsidian]

Oh yeh

True

[hidden haven]

nice

don't scare me, every point is precious in this one

[next obsidian]

Anyway it just becomes C/(x^2 + 1)

And then you use CRT lol

[hidden haven]

Ye lol I proved that in this assignment today

[next obsidian]

For 5

[hidden haven]

1,2,3 were all there

[next obsidian]

If you own Algebra 1 by Bourbaki it’s in there

[hidden haven]

Why would I buy fr*nch books

[next obsidian]

Lmfao for 6

[hidden haven]

lol even saketh found this exam really hard

[next obsidian]

(i) (ii) and (iii)

Then (a) (b) (c)

[hidden haven]

I got so lucky, I have solved all the exercises in AM ch1,2,3,

[hidden haven]

ye lmao

[hidden haven]

so I remembered the solution

[next obsidian]

So for 1

True

False

I believe true, I think I did this before

[hidden haven]

Yes

[next obsidian]

In fact I think Q/Z is super special

Then for iv it’s obviously false lol

[hidden haven]

Without having done Baer's criterion in class btw

lol

[next obsidian]

Like take a flat thing that’s not free then an infinite direct sum of it

[hidden haven]

I gave Q over Z as an example

[wooden ember]

gives perspective to see a graduate exam when you think your exams are gonna be hard

[next obsidian]

2 is just Galois theory gross

[hidden haven]

yes it is very gross

[next obsidian]

3 is easy as we said

[hidden haven]

is infinite direct sum of flat flat

oh tensor commutes with infinite direct sum

damn

how do you find a not free flat thing though

saketh mentioned an example to me after the exam

something like Z[sqrt -5] or some shit

[next obsidian]

There’s a lot lol, take something projective that isn’t free

[hidden haven]

but idk why that's flat

[next obsidian]

Okay but

[hidden haven]

Idk examples of those either lmao

[next obsidian]

An example

Umm let me think

Oh

Polynomial algwbra

Lol

[hidden haven]

I gave Q as example, that is neither, but is localisation of Z and localisation of Z will be flat over itself

[hidden haven]

algwbra

isn't that free

[next obsidian]

No

Oh uh

I guess so huh

[hidden haven]

lol

[next obsidian]

Over a Noetherian ring take a power series

[hidden haven]

damn

[next obsidian]

This requires a different problem tho

[hidden haven]

how would I even show that such a thing is flat

[next obsidian]

That’s in there

You know the finitely presented thing

That you proved

You use that

[hidden haven]

ye

bruh

[next obsidian]

You can test flatness on fg ideals

Which are fp in Noeth

[hidden haven]

nice

[next obsidian]

Then you get that infinite products of flat stuff is flat

Oh I guess it’s slightly different but

The proof that the thing commutes with infinite products

For finitely presented is exactly the same

[hidden haven]

I see

[next obsidian]

Anyway that alone gives you a flat thing that isn’t free

[hidden haven]

wait

infinite product is just a limit

and hom is continuous

[next obsidian]

It’s on the wrong side

You have to show that tensor and infinite products commute

[hidden haven]

hom is continuous on second factor

oh

[next obsidian]

When the thing you tensor by is fp

Yeah

[hidden haven]

I see

ok so I have one more problem to do (other than the one I asked before

[next obsidian]

Ugh

[hidden haven]

So in b part

[next obsidian]

I don’t really know the norm stuff

[hidden haven]

I have no clue what tensor means there

is that supposed to be just multiplication in those rings

[next obsidian]

Where does N(x) go

[hidden haven]

to A

[next obsidian]

A -> C?

Or is it an element?

[hidden haven]

N: A (x) B → A

and this tensor is over smaller ring R

so x (x)_B y makes no sense to me

unless that is the multiplication of A (x) B

but I don't see why that would be lol

[next obsidian]

Oh lmfao

x is itself an element in A (x) B

[hidden haven]

ye

[next obsidian]

We’re looking at a map from

(A (x) B) (x)_B (A (x) B)

Except I suspect that really should be over R

[hidden haven]

bruh

[next obsidian]

NGL

[hidden haven]

That is so annoying

why can't he just write that

[hidden haven]

nvm 3 to 4 actually was pretty easy

These obviously compose to f lol

[next obsidian]

Oh lmfao right it can be a sum of simple tensors

Hurbung my life

[cloud walrus]

mns

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

[simple mulch]

Can anyone check?

pls

[next obsidian]

It doesn’t make sense to write phi^-1(yxy^-1)

phi^-1 is not a function since phi isn’t assumed to have an inverse (okay technically it’s a function on power sets but whatever)

[simple mulch]

good point

[next obsidian]

n’ also totally disappeared

But I see

This is because you tried to swap to G’ using phi^-1 which doesn’t exist

Your strategy is pretty close to correct

You want to show that gn’g^-1 is in phi^-1(N’)

This is saying that phi(gn’g^-1) is in N’

From what you’ve written down so far, you should be able to prove that final statement that I wrote

[cloud walrus]

mns

[simple mulch]

one thing, should I write another n'' when using the fact that N' normal G' such that I can swap them?

[next obsidian]

No

You can’t just swap the elements like that

Rewrite the first thing in that chain of equalities using x and y like in your original proof

It should then be obvious that this lies in N’ because N’ is normal

You’ve also never defined phi_1

[simple mulch]

that was the element from the fact that N' is normal with G'

[next obsidian]

That doesn’t make sense

Unless

Are you trying to use that gN = Ng??

Rather than the fact that gNg^-1 = N?

[simple mulch]

well yes g'N' = N'g' for all g' in G'

[next obsidian]

If so, you shouldn’t use phi_1(n’), rather you’d need to declare that phi(g)phi(n’) = n’’ or something like that, for n’’ in N’

[simple mulch]

yeah

exaclty

[next obsidian]

Well that’s I guess a way to do it but

That seems a little bit…

Inefficient

[simple mulch]

why so

[next obsidian]

Why not directly use that g’N’g’^-1 = N’?

Then you’d get that phi(g)phi(n’)phi(g^-1) is in N’

[simple mulch]

that would follow directly

[next obsidian]

Just by noting that phi(g^-1) = phi(g)^-1

[simple mulch]

oh I see

yeah

[next obsidian]

Anyway, yeah this is all there is to it

[simple mulch]

one more question

If I may

[next obsidian]

Sure

[simple mulch]

then I have phi as an epimorphism and N normal with G. I need to show phi(N) normal G'.
Can I show that for any g' in G' and n in N we have phi^{-1}(g'ng'^(-1)) in N? Since now we have an inverse for phi

[next obsidian]

I believe so, but you want to be careful since you only get an inverse on one side

You also might need to show that the inverse is actually multiplicative

I would avoid using an inverse

Instead just say something to the effect of “there exists x in G such that phi(x) = n” or something to that effect

[simple mulch]

and y in G such that phi(y) = g'

[next obsidian]

Yeah

Using one-sided inverses are just a recipe for disaster

It makes things unclear, it’s sort of harder to follow

And sometimes you’re not sure if what you’re doing is kosher

[simple mulch]

interesting

thank you very much!

[next obsidian]

No problem

Also here’s a specific case where things might turn bad

Suppose you have a map f: G -> {e}

Then your “inverse” really is just a choice of where you send e since every element of G maps to e

But there’s no reason to expect f^-1(e•e) = f^-1(e)•f^-1(e)

Since this is also just equal to f^-1(e)

So if this were multiplicative, since f^-1(e) can be anything in G, you’d have that g^2 = g for all G in G, but this is preposterous

(Or I guess just note that f^-1 doesn’t have to send the identity to the identity so it isn’t necessarily a group homomorphism)

[simple mulch]

indeed

[next obsidian]

Anyway the point here is just, if you decided to try and use an inverse, you might accidentally try to use that f^-1 is a group homomorphism

But this isn’t true

[simple mulch]

we must make sure f is an epimorphism first to use f^-1

[next obsidian]

Yes, but in the case I outline f is an epimorphism

But f^-1 doesn’t have to be a homomorphism

[simple mulch]

how is f an epimorphism if it doesn't satisfy injectivity?

[next obsidian]

Epimorphism is surjectivity

[simple mulch]

oh lol

endomorphism is the surjective one

[next obsidian]

No endomorphism is a map G-> G

Epimorphism is surjective

[simple mulch]

oh shit

[next obsidian]

And monomorphism is injective

[simple mulch]

isomorphism is the one

[next obsidian]

Uhhh

I mean an isomorphism (as a result) is just a bijective homomorphism

A priori it’s a homomorphism with a two-sided inverse (which is also a homomorphism)

[simple mulch]

I see, more pratice to do! grazie

[next obsidian]

Np

[hidden haven]

@next obsidian I think this doesn't make sense if you view N as a map from (A (x) B) (x) (A (x) B) because then the N(x) and N(y) on the other side are in A so you are viewing the other side as A (x) A but then is the first thing a free A (x) A module? Because the middle tensor is over B and I feel like that spoils freeness

I asked a couple classmates and they said they interpreted the tensors as just multiplication

So that seems like a safe thing to do lol

But such a bad way to phrase a question

[hidden haven]

It seems that it is free with basis (1 (x) 1) (x) (1 (x) e_i)

[pastel cliff]

im trying to prove that gcd(a,m) = 1 iff a is a unit in Z_m

can anyone gimme a pointer pls

[chilly ocean]

bezout

[pastel cliff]

im following a book and that gets mentioned 200 pages from now

though i do see how that makes it kinda easy

[pastel cliff]

what is the relationship between isometries and metric spaces

my book is going into analysis

[chilly ocean]

#point-set-topology

[pastel cliff]

oh really? i just learned what a metric space is yesterday and the definition of an isometry i just came across reminded of it so idk

[hidden haven]

metric spaces are analysis/topology

isometries are certain kinds of maps between metric spaces

[pastel cliff]

ahhh

i just saw that they both had to do with the idea of "distance" between points so i thought they might be related

[primal beacon]

Hi I have this problem : Let $n\geq 1$ , show that if the unit group $\frac{\Z}{n\Z}^*$ isn't cyclic, the nth cyclotomic polynomial $\phi_n$ is reducible in $\frac{\Z}{p\Z}[X]$ for all p prime

[cloud walrus]

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

[primal beacon]

Hi I have this problem : Let $n\geq 1$ , show that if the unit group $\frac{Z}{nZ}^*$ isn't cyclic, the nth cyclotomic polynomial $\phi_n$ is reducible in $\frac{Z}{pZ}[X]$ for all p prime

[cloud walrus]

rayane

[primal beacon]

I wanted to know if someone has any idea where to start without fielld extensions ? and if not, with field extensions ?

[white nymph]

sorry to post on top of rayane's question... but I need to leave the house now. any hints on how to begin this problem?

[amber stag]

Just take the derivative?

[hidden haven]

Also need to show that it splits

[hidden haven]

If v is a root of that polynomial, then so is v+1

[upper pivot]

Im reading a book on differential galois theory, im stuck on smth. Let $E\supset F$ be an integral $F$ algebra, and $T=E \otimes_F E$, then the element $d= a \otimes 1 - 1 \otimes a$ is not nilpotent as long as $a\nin F$

[cloud walrus]

Kanga gang enforcer John

[upper pivot]

The book actually talks about differential field F and differential algebras, but i suspect that information is not necessary for this part

Nvm just realize its an integral domain lmfao

[south storm]

Exposed

[hidden haven]

Is it always? Tensor of C with itself over R is not a domain

[cloud walrus]

Hausdorff

[hidden haven]

Prove both inclusions

[cloud walrus]

Hausdorff

Hausdorff

[hidden haven]

If you can write 2 as Z linear combination of 4 and 6, you're done. Don't do it with an arbitrary element

Showing that the generators are there is enough

[median pawn]

There's a small glitch here but let's overlook that. I meant x = 2c, and x = 2c_1 + 3c_2

LOL 2 = 6 - 4

Okay this was trivial, thanks

I believe it's true that the ideal generated by a_1, ..., a_n is basically the principal ideal generated by their gcd?

[hidden haven]

In PIDs yes

Because bezout's lemma applies

Not in UFDs and such

Example is the ideal generated by x and y in R[x,y]

This is a proper ideal, but if R is a UFD then the gcd is 1

[median pawn]

Ahh perfect

[next obsidian]

Wut, in a UFD (x,y) = (gcd(x,y))

fg ideals are principal in a UFD for this reason

I think

except I doubt that fg ideal => principal thing

but I thought that gcds always exist in a UFD

maybe the issue is that (a,b) is properly contained in (gcd(a,b))?

hurb

[hidden haven]

ye I just gave a counterexample lol

[hidden haven]

this one

what you saying would make every noetherian UFD a PID

[next obsidian]

is a UFD always Noetherian?

why can't you have a non fg ideal even if you always had (a,b) = (gcd(a,b))?

I don't think a UFD is Noetherian

I think k[x1,x2,...] is probably a counterexample

[hidden haven]

not in general

[next obsidian]

But maybe there's something else going on

[hidden haven]

im saying every noetherian one would be pid

[next obsidian]

oh

I accidentally a word

[hidden haven]

[next obsidian]

ffs

[iron vessel]

I may have asked this in the past, but does anyone have a cool visual representation of a finitely generated module over a pid. I am writing a report on it and would like to add a cool image in the beginning.

[primal beacon]

hey I think its probably impossible to represent due to torsion in general you can try to represent $Z\bigoplus Z/2Z$ I think it would make no sense since it's a math report pt

[cloud walrus]

rayane

[primal beacon]

put key exemples ? it could be good

[hidden haven]

make the torsion part into a circle and you just have higher dimensional cylinders

[primal beacon]

easy ahah

[primal beacon]

$V\cong\bigoplus_{i=1}^rK[X]/(di(X))$ and $G{ab}\cong\bigoplusZ^{\bigoplus r}\bigoplus_{i=1}^{k}Z/p_i^{k_i}Z$

[cloud walrus]

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

[primal beacon]

I can't seem to be able to write latex but the first one shows vector spaces as $K[X]$ modules and the second one classifies abelian groups (add a Z under the +r) you can put this somewhere

[cloud walrus]

rayane

[iron vessel]

Ah ok, thanks guys !

[untold cloud]

Hi, guys, i am confused about graded algebra, for example $A=\bigoplus_{i=1}^\infty A^i$, does it mean every element a in A is in $A^0\oplus...\oplus A^n$ for finite n rather than $a\in A^0\oplus ...\oplus A^n\oplus...$?

[cloud walrus]

cuiyuze0728

[rustic crown]

they're more or less the same thing, assuming you identify that finite direct sum as a submodule of A

elements of an infinite direct sum are already finitely supported

(just one minor point, that big oplus should start indexing at i = 0, by looking at your description later on)

[untold cloud]

oh yeah, sorry

so there is no situation such that $a= a_0\oplus a_1\oplus...$ for infinite sum, where $a_i\in A^i$

[cloud walrus]

cuiyuze0728

[untold cloud]

where only finitely many a_i=0

[rustic crown]

yep

[primal beacon]

but it doesn't have to start at 0 right ?

[rustic crown]

one construction of infinite direct sum is given by looking at the infinite direct product and the submodule in which an element is 0 at all except finitely many places in the tuple.

[untold cloud]

thanks

[rustic crown]

also this is just a little bit weird to say.

[untold cloud]

finitely many nonzero

[rustic crown]

wait that makes it wrong then

an element of the direct sum is non-zero at finitely many places

so situations where infinitely many are non-zero don't happen

[untold cloud]

thanks

[rustic crown]

we don't need to specifically say that as a_i are still allowed to be 0, we're just saying finitely many are non-zero.

[primal beacon]

yeah yeah I meant since he wrote $\bigoplus_{i=0}^{n}A^i$ when it can be in any finite sum

[cloud walrus]

rayane

[misty harness]

Hi, how can I show this ringisomorphism is surjective?

[hidden haven]

what is R

[misty harness]

Oh right

The subset of polynomials over Q without a linear term

[hidden haven]

The subring of that generated by Q, z^2 and z^3 is all of R

And the image of a ring isomorphism is a subring

In this case containing all of those things

[misty harness]

So if I get it right, it is obvious that $\text{Im}(\psi) \subset R$, and to show that $R \subset \text{Im}(\psi)$, we only need to show that $\mathbb{Q} \subset \text{Im}(\psi)$ and $z^2,z^3 \in \text{Im}(\psi)$ ?

[cloud walrus]

Luka835

[hidden haven]

yes

[misty harness]

In our algebra course, we didn't really see the notion of a 'ring generated by ...', what notation is often used for this?

[willow mason]

R = (a,b,c,...)

[hidden haven]

That's for ideal though

[misty harness]

Oh, right it's an ideal!

[willow mason]

just uhhh abuse the notation or smthn

[hidden haven]

If you haven't seen subrings

you can say that if you have an element of R

[willow mason]

idk [] maybe if not ideal

[hidden haven]

write it as a polynomial in z^2 and z^3

like these 2 variables

which you'd have to prove

[next obsidian]

You use []

For a ring generated by things

[hidden haven]

and then write the preimage of this in terms of the preimages of z^2 and z^3

[misty harness]

Okay, I see now. Thank you

[white nymph]

looking for help/guidance with this problem. in part a.) i showed that p had no roots in F and was thus irred since deg(p) = 2 (i think this is the correct strategy?). part b.) i was not quite sure how to handle. im assuming we want to use characteristic 2 somehow (ie that u = -u), but im not sure how to proceed

[next obsidian]

You’re exactly correct

I think, oh fugg

Oh right

[iron vessel]

Correct me if I’m wrong but can’t you use that if u^2 - t = 0 then u=+-sqrt(t)

[next obsidian]

So use the fact that in char p, (x + y)^p = x^p + y^p

And yeah

You know if it has a root in K then it splits

But you can just write it as

(x - sqrt(t))^2

Because this is then x^2 - t = x^2 + t

[terse crystal]

In a field of characteristic 2 any element x we have that x=-x

(2x=0)

[trim grove]

i discarded the option 3rd , because that will give x^4+x =x(x^3+1) , which is factrozation in non units ,
but facing trouble in 1,2,4 , the given polynomial is of unit content, so there is if and only if condition over irreducibility over Q and Z, Now for option 1, this polynomial reduces to x^4+x^3+x^2+x+1 , which is a cyclotomic polynomial and i know that it is irreducible over Q , but here we are doing this over Z2, is this irreducible over Z2 or not? please give me a hint.

[iron vessel]

Have you tried using the rational root theorem for 2?

Similarly for 4.

[trim grove]

sorry for that i edited the question

[iron vessel]

So one of the 4 is wrong?

[trim grove]

you have to find which of these four is/are true, it can be all , or less then four

[iron vessel]

So have you tried using the rational root theorem for question 2 and 4? That is to show it has no linear root. Then since it is of degree 4 you can see if f(x) is the product of two degree 2 polys

[trim grove]

is it worth it to use rational root theorem here?

[iron vessel]

Do you have any other idea in mind?

I usually try to check if I can apply Eisenstein but in this case you cannot.