groups-rings-fields

[long narwhal]

yeah so i'm not sure how you're supposed to go from <3+2> = <2>

because since Z is a principal ideal domain that should be true

[delicate orchid]

by <3+2> do you mean <2, 3> or literally <5> - I'm kinda confused by this notation

[long narwhal]

sorry, i was using the wrong notation

i meant <2,3>

[elfin furnace]

so then <2, 3> = Z not <2>?

[long narwhal]

well <2,3> = 2Z + 3Z

but by the definition of a principal ideal domain <a,b> = <a> = <b>

[proud bear]

<2,3>=<1>=Z is principal

[long narwhal]

oh is it not <a1,a2>=<a1>=<a2>?

[elfin furnace]

no

[proud bear]

For Z, <a,b>=<gcd(a,b)>

[delicate orchid]

this is absolutely not the definition of a principle ideal domain

[elfin furnace]

pid just means that theres some p s.t. <a, b> = <p>

[delicate orchid]

PID => all ideals <S> for some set of elements S is equal to SOME <r> in the PID, not every element of S

yeah

for Z it's the gcd of S

[long narwhal]

does the one generator that's principal have to be equal to every other ideal?

[elfin furnace]

what do you mean by that

[long narwhal]

if a is a principal ideal of a PID,

does <a> = <a1, a2, ..., an> for any a1-an in the PID

i know <1> would work for all of Z i think but would that apply in general or just this case

[proud bear]

If a1,...,an are in <a> then yes. Otherwise no

[elfin furnace]

well 2 is in <1> but <2> is not equal to <1>

[proud bear]

Ah wait lmao

Nvm

[frozen quiver]

Is there a name for a module homomorphism that additionally is a homomorphism on the underlying ring?

[south patrol]

Do you mean a homomorphism M->N of R-modules where M, N are also rings?

That'd be a (homo)morphism of R-algebras

[frozen quiver]

I mean a homomorphism on a module that also maps the ring under the module.

The standard definition makes it an identity.

[chilly radish]

Well it's not gonna be a module homomorphism then

[frozen quiver]

Sure, I just wanted to know if that had a name.

[chilly radish]

Also you usually don't have an embedding of the base ring into the module

[prisma ibex]

sure why not

maybe the action isn't something as natural as say S_n acting on n elements by the natural permutation representation

[frozen quiver]

I think the homomorphism I am trying to get a name for is two separate morphisms, one on the ring part, and one on the abelian group part.

[deep sky]

Apparently it’s by still by permutation though, Is where my confusion comes from

do the higher numbers just not do anything?

We assume our three elements are just a b c starting in indices 1 2 and 3 right

[prisma ibex]

I mean any action by any group whatsoever on a set is by permutation, that's what an action is

[deep sky]

oh, because you number the positions the elements are originally in and then they have to be in positions formerly held by other set elements or else it’s not an action?

[prisma ibex]

no

an action of a group G on a set S is literally just a group homomorphism G->Aut(S), where Aut(S) is the group of permutations of S

so to have S_4 acting on a 3 element set you're just asking for a group homomorphism S_4->S_3

[deep sky]

Aut(s) means mapping each s to where another s formerly was, right?

Which is what a permutation is once we have an established initial order for S

[prisma ibex]

I mean it can fix elements but sure

we don't need an order on S

[deep sky]

Yeah ofc

yeah, but in first expressing the set you have some arbitrary first element in the first spot and so on

[prisma ibex]

you don't need this

the order doesn't matter

[deep sky]

I thought the whole point of a permutation is order does matter?

[prisma ibex]

I guess you need this if you want to write these things in the usual permutation notation

but this isn't strictly necessary

anyways

[deep sky]

Okay so technically you can say something bigger than S_n acts on n elements (call it S_m) but really it’s just whichever elements of S_n are found in S_m

Like anything you add is trivial

[prisma ibex]

I don't think this is the action you want

[deep sky]

wdym?

Like under what rule set of group action would (14) do something by permutation to a set of three elements

[prisma ibex]

sure so here's the natural action: consider the set of partitions of {1,2,3,4} into 2 sets of 2 elements

there are (4 choose 2)=6 such partitions

if we forget the order of these 2 sets, we have 3 such partitions

so let's say 12|34, 13|24, and 14|23 up to equivalence

so then (14) would for instance send 12|34 to 13|24

this action of S_4 on 3 elements is quite special, you don't really get this kind of thing for other symmetric groups

[delicate orchid]

no way S_4 acts on 3 elements that's cool

[pastel cliff]

should i do this by expanding every term of the jacobi or only using properties like anticommutativty and such

it looks like an easy computation but me is a lil stuck

[delicate orchid]

as V is a finite dimensional vector space what other form can you write the linear maps in

i.e. you can write them as functions and something else

[pastel cliff]

matrices...?

feel free to degrade me if that's dumb

though i do feel like \circ here is just matrix multiplication isnt it?

nvm the next question literally says "here is a matrix version gl(n, F)"

[delicate orchid]

yeah it's matricies

lol

so then you just show that matrix multiplication satisfies the Jacobi property

[pastel cliff]

[stone fulcrum]

This looks approachable. What's the book?

[pastel cliff]

https://link.springer.com/book/10.1007/1-84628-490-2

prof recommended it precisely for that reason :D

[stone fulcrum]

Thank you, stealing.

[pastel cliff]

hopefully it'll be the first book i can actually consume

slorp

[stone fulcrum]

c o n s u m e

[pastel cliff]

first chapter at least has been approachable

it helps that im also taking ann algebra course so i only recently learned about stuff like ideals and isomorphisms

i can slide pdf in dm's if you want @stone fulcrum

[delicate orchid]

crimes!

[pastel cliff]

we do a little trolling

[stone fulcrum]

Haha you're such a joker lol

We don't do that here haha

[pastel cliff]

i would never have such things

https://tenor.com/view/troll-pilled-gif-19289988

[delicate orchid]

need to learn lie algebras tbh

and then I can... do things... I guess

really my priority should be galois theory but I really can't be bothered

[lethal dune]

[[]]

[pastel cliff]

stupid question time

wtf is the adjoint

should i have learned this

[delicate orchid]

it's literally defined

[pastel cliff]

no i mean like

(ad x)

oh wait that's the map?

im confusing myself

[lethal dune]

ad wrt x

if that makes sense

[pastel cliff]

uhhhh

can u be a bit more verbose

[delicate orchid]

it's like

wait you haven't seen semi direct products

uhhh

it is what it is

it's just a map of maps so we tend to notate them like that

[pastel cliff]

in other news, water is wet

[delicate orchid]

it isn't though

[pastel cliff]

literally shut the fuck up

[pastel cliff]

wdym tho <3

[delicate orchid]

I really don't see why this is confusing you

[pastel cliff]

i am not a smart individual

i'll just read and be back later

[delicate orchid]

oh wait hold on they just use ad on it's own

wtf are they smoking

(ad x)y is, as ryu said, the adjunction of y wrt x

ad on it's own is just them doing the wacky tobbaccy

[pastel cliff]

ohhhhhh wait

so is ad x just the map being written weirdly (like it should be more like \phi(x))

it's really an element of gl(L)

[delicate orchid]

I'd write it as ad(x)(y) personally

you map x to a map and then put y through that map

[pastel cliff]

ok i think that makes sense

[delicate orchid]

the reason why I was thinking about semidirect products earlier is because you have a map f: H -> Aut(G) involved and the notation f(h)(g) or f_h(g) is used a lot

so it's a similar thing

(and gl(V) is just Aut(V) lol)

[pastel cliff]

i think im just a little stuck on what it means to map an element to a map

it kinda makes sense in the context of matrices but what about generally

[chilly radish]

Fraktur gl is End considered as a lie Algebra with the commutatoe

Not Aut

ad x may very well not be injective

e.g. if the Algebra is abelian

[delicate orchid]

I see

oh yes of course

I had gl = GL in my head

[barren sierra]

how is (c) not just dead obvious from the fact that zeta is literally already a root of f(x)????

and all the other roots are just powers of zeta

[pale jasper]

Hey, this is kinda basic but is this a sufficient proof that the intersection of two ideals is an ideal?

[south patrol]

Seems good

[pale jasper]

Thank you

[chilly ocean]

In field theory,
2 is defined as 1+1...
Is this correct?

[delicate orchid]

consider F_2

[sharp sonnet]

yeah, more precisely its the image of 2 under the unique map from Z into your field

[sharp turret]

can anyone give a hint? just learned about finite fields and not sure what to use here

[tall jay]

If m and n are relatively prime, does that mean
$$[a]{mn} = [b]{mn} \rightarrow [a]_m = [b]_m$$

[cloud walrus]

beeswax

[languid walrus]

first statement means mn divides a-b, second m divides a-b

so this is always true, regardless of if m, n are rel prime

[tall jay]

If f is well-defined and injective, are we guaranteed that f-inverse exists?

[tribal moss]

No, it needs to be bijective.

[tribal moss]

Have you learned that the multiplicative group of a finite field is always cyclic?

[gusty cave]

lol this sentence is just so funny to read for the first time

[terse crystal]

F \ {0} must contain a cyclic group of order 364, so we need to find minimal p^r such that 364|p^r-1. the answer is 1093 I think

[terse crystal]

Sorry not 1093, should be 3^6=729. F_729

[prisma shuttle]

there are certain cases where being injective does imply surjectivity, which thus implies bijectivity and thus f-inerse is guaranteed to exist

the most basic case that comes to mind is the case where the codmain of the function is finite

u will often see this in proofs where its like take the map x-> ax and this map is injective and thus its surjective cuz everything is finite blahb lahb ah

|| ||

can someone expalin wut property 1 means here

i am not sure what $\gamma$ is suppposed to mean

[cloud walrus]

JustKeepRunning

[prisma shuttle]

cuz isn't this just llike the definitinn of an ideal

that it has to be closed under subtraction(which in this case is division)

[south temple]

it means that any element of a monomial ideal is a multiple of one of the generators

note that this is different from the usual definition of an element of an ideal

for example: if $I = \langle a, b\rangle$, then any element $x$ of $I$ $x = ra + sb$ for $r,s\in R$

[cloud walrus]

vov&sons

[south temple]

however, with a monomial ideal, we necessarily have either $x = ra$ or $x = rb$.

[cloud walrus]

vov&sons

[south temple]

Here's a formal lemma from Hassett's "Introduction to Algebraic Geometry"

[prisma shuttle]

can u explain why this is true??

[south temple]

I sent the proof of the lemma right above

[prisma shuttle]

oh oops ok i get it now

thx so much

[hallow kettle]

Just find the generator of Z36 and those elements have exactly the order of 36, is this what the question wants?

[thorn delta]

No. This group isn’t even isomorphic to Z36

It wants you to find all possible orders of elements of G

[hallow kettle]

GCD(4,9)=1 doesn't implies they are isomorphic to Z36?

[hallow kettle]

So do I list one by one like {(0,0),(1,0),.....(3,8)} and find their order one by one?

[next obsidian]

Pretty sure it is isomorphic to Z36…

[thorn delta]

Yea I’m wrong

[next obsidian]

Isn’t there a function which counts the number of each element of each order

Like the totient function?

[hallow kettle]

Haven't learn about that 😦

[next obsidian]

But it depends on two variables, and I think totient is only one

So that’s obviously fake lol

But this isn’t bad, you know everything is of the form x^n

Where x is a generator right?

You should be able to determine what the order of x^n is in terms of n and 36 pretty easily

[hallow kettle]

Yes, I have in my hand the generators are 1,5,7,11,13,17,19,23,25,29,31,35 so their order is exactly 36

[next obsidian]

No that’s not what I mean

Take 1 as the generator I guess

You should be able to determine what the order of n is in Z/36 easily via some sort of divisibility / gcd kinda thing

You can do it for an arbitrary n, you don’t need to just compute it for each number between 0 and 35

[tall jay]

Like if theres a finite domain and codomain. Then f injective iff surjective

[next obsidian]

You need them to be the same size too

[hallow kettle]

I think I am confused

[next obsidian]

Just compute what the order of n is in Z/36Z

You can do this just in terms of n and 36

You don’t have to compute what the order of 1 is, the order of 2, the order of 3,…, the order of 35, you can do them all at once

[hallow kettle]

How do I compute that?

I think I only know how to find the generators with gcd, could you give a bit more hint on how to get the order of n?

[thorn delta]

Do you know Lagrange’s theorem? This tells you something about the order of an element

[hallow kettle]

No... I just started taking this a week ago

[thorn delta]

So you don’t know that the order of a subgroup divides the order of the group?

[hallow kettle]

It says the order of group represents the number of elements?

[thorn delta]

Yea, order is number of elements of a group

[hallow kettle]

The order of elements and order of group are the same thing? Because I am asked to find order of element

[thorn delta]

The order of an element would be the order of the cyclic subgroup that the element generates

[next obsidian]

You’re looking for the smallest number k such that kn is a multiple of 36

There’s a nice way to describe this

This is (once you unravel them) by definition the order of n in Z/36

[thorn delta]

And in light of this fact, that number k should divide 36 for example

This is something that you can check for yourself too

I mean chm basically already said it

[hallow kettle]

This could be the Theorem that I need right?

[thorn delta]

Yeah, I mean this basically solves your exercise automatically

[hallow kettle]

Okay I don't know why my brain can't function now but I will try. Thanks to both of you! ❤️

[thorn delta]

Npnp

[prisma shuttle]

yes this is correct

how would u simplify the ideal $I=(x^6, x^2y^3, xy^5)$

[cloud walrus]

JustKeepRunning

[tribal moss]

Over Z[x,y]? It looks quite simple already.

[prisma shuttle]

is there any way to like make it simpler

cuz i need to draw a vector exponent graph of the x and y

and the only way i can see to do it right now is just trying to add and subtract elemenets until it "seems" that they are all accounted for

[tribal moss]

Hmm, I'm not 100% sure what a "vector exponent graph" means in this context -- but if it means what it sounds like it means, then I don't think adding and subtracting elements is even relevant.

[prisma shuttle]

part (a) of this

[tribal moss]

Yeah, that's what it sounded like.

So you get all of the multiples of x^6, which is the lattice point (6,0) and every point east, north, or northeast of it.

You get all of the multiples of x^2y^3, which is the point (2,3) and every point east, north, and northeast of that ...

Taking sums will not give you any new monomials, just polynomials with terms made from the monomials you already have.

[prisma shuttle]

oh sorry i meant products

would u just have to keep listing until it seems like you are done

[tribal moss]

No, I would do what I describe above.

Remember that for an ideal you're not looking for product of the generators.

You're looking for product of one generator with any element of the ring.

So each generator gives you a whole 90° wedge of the lattice.

[prisma shuttle]

ohhhhh i see now

thx u

[tall jay]

codomain and domain would have to be sets of equal cardinality for it to be true, right

[prisma shuttle]

yes but in some instances u can sort of "skip" this part

for example take some $a\in R$ for some ring $R$ and consider the map $x\to ax$ for all elements $x\in R$

[cloud walrus]

JustKeepRunning

[prisma shuttle]

suppose that this map is injective

then it has to be surjective if $R$ is finite

[cloud walrus]

JustKeepRunning

[prisma shuttle]

this is because the output is obviously a subset of $R,$ so it is surjective

[cloud walrus]

JustKeepRunning

[prisma shuttle]

and since $R$ is finite we can conclude that the mapping is bijective

[cloud walrus]

JustKeepRunning

[prisma shuttle]

this is jut like a specific case that came to mind

but ingeneral u would have to check for equal cardinalities

or use some other info that u are given

if u want an example of how the map $x\to ax$ can be used (trust me, its very common) refer here:

https://math.stackexchange.com/questions/60969/every-nonzero-element-in-a-finite-ring-is-either-a-unit-or-a-zero-divisor

[cloud walrus]

JustKeepRunning

[tall jay]

Ahh I see. Thank u

[cloud walrus]

NupurJ

[hard cairn]

tensor products* oops

[delicate orchid]

Ok here's my thoughts, Res(E) and W are both CH-modules so their tensor product is a representation of H x H, meaning the character of this tensor product is just the product of the character of W and the character of Res(E), similarly on the other side we have that Ind(W) is a CG-module blah blah so the character of the tensor product is just the product of the characters of Ind(W) and E

that's where I get a bit stuck - I do believe that frobenius reciprocity is the right way to go about this but I can't seem to think of how we map the above formulation into an inner product

we also have a formulation of the induced character using the centralisers of elements in G and H but I doubt that's useful here

[cloud walrus]

NupurJ

[delicate orchid]

that's basically exactly what I said good to know I wasn't talking nonsense

[hard cairn]

ah, yes, i see that now!

[delicate orchid]

is the restriction of the induced character just the original character, I should really know that off the top of my head lol

[hard cairn]

I'm not sure myself lol, let me try to look it up

[delicate orchid]

ah I don't think it is

[chilly ocean]

Why is this true.. plz help
:every finite field has characteristic p(prime)

[delicate orchid]

consider characteristic non-prime n

then n decomposes into the product of two numbers, pq = m

so in a field of characteristic m, pq = 0

so this "field" has zero divisors

which is a contradiction

[chilly ocean]

But a•b isnt equal to ab
3•a=a+a+a =/= 3a

[tribal moss]

Do 3·a and 3a mean different things to you? Which different things do they mean, then?

[main rover]

1. group theory
2. r means R120
3. f means to flip

I understand literally nothing except that the order of e is 1 and the inverse of e is e

[lavish nexus]

how does it work at i = 0
The H_i and C_i will all be 0 when i < 0 so can start considering at i = 0, which is rho_0 = r_0

we have $C_0 = B_0 \oplus H_0 \oplus (C_0/Z_0)$

[cloud walrus]

Iteribus

[lavish nexus]

$Z_0$ is the whole thing since $C$ is positive so just $C_0 = B_0 \oplus H_0$

[cloud walrus]

Iteribus

[lavish nexus]

But I don't see why $B_0 = 0$

[cloud walrus]

Iteribus

[hard cairn]

you can cut out an equilateral triangle from a piece of paper and try this out. basically, flipping twice is equivalent to doing nothing, so ff = e. Rotating thrice is also equivalent to doing nothing, so rrr = e. A flip followed by a rotation is the same as rotating twice and then flipping, so rf = frr. Similarly, fr = rrf. Keep doing this in the first problem to simplify the sequences

[main rover]

Ok perfect I will try now

[main rover]

so is rfrfrfrfrfrfrf = rf

ok so i got
rf
rf
e
rr

[main rover]

i have no idea about the order orbit or inverse

[stone fulcrum]

Useful relation on all dihedral groups:
rf = fr⁻¹

So you can swap an r and an f, but if you do, take the inverse of r

[main rover]

how exactly do i take an inverse

[stone fulcrum]

In this question, r means to rotate 120 degrees counter-clockwise

What does r⁻¹ mean?

[main rover]

rotate 240?

[stone fulcrum]

Yes that's true

So "take an inverse" means to do that instead

Well, let me put the equation again:
rf = fr⁻¹

[main rover]

ahh tyvm

[stone fulcrum]

That reduces them very fast

[acoustic fossil]

As part of a multiple step problem I'm tasked with finding U to be a normal subgroup of T and identifying the quotient T/U? Proving normal subgroup felt straightforward since ad != 0 it is invertible and can show tut^-1 in T. However, I don't understand how to take the quotient of these two, what would that look like?

Looking to find quotient T/U

[terse crystal]

the quotient is isomorphic to R^2. For any (a,b;0,d) from T, (a,b;0,d) is congruent to (a,0;0,d)

[chilly ocean]

Why the last sentence hold

[terse crystal]

Since when $a \neq b$, $$\begin{pmatrix}a&b\0&d\end{pmatrix}= \begin{pmatrix}1&\frac{b}{d-a}\0&1\end{pmatrix} \begin{pmatrix}a&0\0&d\end{pmatrix} \begin{pmatrix}1&\frac{-b}{d-a}\0&1\end{pmatrix}$$
When $a=d$,$$\begin{pmatrix}a&b\0&a\end{pmatrix}=\begin{pmatrix}a&0\0&a\end{pmatrix} \begin{pmatrix}1&\frac{b}{a}\0&a\end{pmatrix}$$

[cloud walrus]

Cogwheels of the mind

[terse crystal]

Your original picture shows the complete question can you send that again?

[chilly ocean]

@terse crystal

[terse crystal]

Because K=F(β), F(β) is isomorphic to F[x]/(irr(β,F)(x)). So n=[K:F]=[F(β):F]=[F[x]/(irr(β,F)(x)):F]=degree(irr(β,F))

[terse crystal]

a doesn’t equal d not a doesn’t equal b sorry

[desert dome]

Hi, I'm trying to prove that J = (x^2 - yz, xz - x) in k[x, y, z] is equal to the intersection of ideals (x, y), (z, x) and (x^2 - y, z - 1). It is easy to show that J is subset of the intersection but I don't know how to show the other direction. Could anyone give me a hint on that? Thanks!

[hard cairn]

Yes, and here r^-1 is r^2! That's definitely cleaner notation

[next obsidian]

I gave this some thought, and I couldn’t really come up with much. I think you can compute the intersection or (x,y) and (y,z) pretty easily, and then try to compute the intersection of that with (x^2 - y, z - 1)

The only other trick I can think of is, when you have something like “show that I = J” and one direction is easy, say J < I, you can try to show that I/J = 0

This has the benefit of, taking an element in I, once you go the quotient you can rewrite stuff using what’s in J

So as an example in this specific case, once you quotient by J, because you know xz - x = 0, you know xz = x, so you can replace any xz which shows up with an x

similarly you can replace x^2 with yz, so if you had eg x^3 this becomes xyz = xy

And stuff like that

This might make it easier to show that I/J = 0 (I in this case is the intersection)

Good luck

[chilly ocean]

could someone help me parse this? I dont understand what it is asking

[chilly ocean]

nvm I figured it out I think

[desert dome]

Tysm!! I'll think about it more

[strong yacht]

Is k an algebraically closed field? This looks like an algebraic geometry exercise

[prisma shuttle]

hey guys is something like $\mathbb{Z}[x,y]\cong \mathbb{Z}[x]\times \mathbb{Z}[y]$ true?

[cloud walrus]

JustKeepRunning

[prisma shuttle]

i feel like this is wrong but i am not sure how to show it

[chilly ocean]

zerodivisors

[prisma shuttle]

can u elaborate?

[chilly ocean]

one has zerodivisors and the other doesn't

[prisma shuttle]

i feel like ima missing smth obvious but it seems like both of them have no zero divisros?

[chilly ocean]

think about it a bit

[south patrol]

If you pick any two random finite field extensions they probably won't be isomorphic extensions

[pale jasper]

What does the set $2\mathbb{Z} +3\mathbb{Z}$ consist of?

[cloud walrus]

Scerball

[chilly ocean]

2(element of Z) + 3(another element of Z)

[pale jasper]

So 5 is in there?

For example

[elfin furnace]

every element of Z is

[south patrol]

Ah, sure. Well consider e.g. finite extensions of Q which have very different algebraic properties

For example Q(i) and Q(sqrt(2)) are p different

[delicate bloom]

$a\bZ+b\bZ=\gcd(a,b)\bZ$

[cloud walrus]

Merosity

[pale jasper]

Thanks

Is this because Z is a PID?

[delicate orchid]

Z is a PID because of this

it follows from bezout's lemma

[pale jasper]

Right

Is $\langle a, b \rangle = \gcd(a, b) \ \text{iff} \ R \ \text{is a PID?}$

[cloud walrus]

Scerball

[pale jasper]

Where $a, b \in R$

[cloud walrus]

Scerball

[pale jasper]

And $R$ is a ring

[cloud walrus]

Scerball

[delicate orchid]

good question

I can definitely conclude that $\langle a, b \rangle \subseteq \langle gcd(a, b) \rangle$ if $R$ is a Bezout domain, and all PIDs are Bezout domains

[cloud walrus]

Wew "Dithering" Tbh 📺

[devout crow]

I don't think so, iirc you can get rings where every finitely generated ideal is principal, but not all ideals are principal

[delicate orchid]

true, I did not consider the case of a gcd over an infinite set

just assume a noetherian ring and we're all good 😌

[pastel cliff]

for Z/mZ, there will always be m-1 elements right

[delicate orchid]

prove it

[pastel cliff]

well now i dont feel like it

[delicate orchid]

good cause it's wrong, there's m elements

[pastel cliff]

proof by stubbornness

[delicate orchid]

do not talk to me about stubbornness

[pastel cliff]

i forgor 0

https://tenor.com/view/the-j-shoebill-the-w-smg4discord-when-you-gif-23781183

[delicate orchid]

if m is prime then there's m-1 units I guess lol?

[pale jasper]

$\mathbb{Z}/m\mathbb{Z} \cong \mathbb{Z}_m$

[cloud walrus]

Scerball

[delicate orchid]

isn't this true by definition

[pastel cliff]

yeah

[pale jasper]

Is it?

[devout crow]

I think the only way it's not is if you define Z_m just as the cyclic group with m things

rather than a quotient of Z

[pastel cliff]

If $R$ is a domain, then there is a unique field containing $R$.

[cloud walrus]

μ₂

[pastel cliff]

this is false right

[devout crow]

yes

[pastel cliff]

the uniqueness thing right

like it's not necessarily unique

[devout crow]

yeah

[pastel cliff]

kk just wanted to make sure

[chilly ocean]

the first example of "domain" you should always think of produces a counterexample

[pastel cliff]

Z

contained in both Q and C right

[devout crow]

are you asking or telling

[pastel cliff]

telling with smidgeon of asking

[chilly ocean]

you know this

[delicate orchid]

really asking if Z is a subset of Q...

[pastel cliff]

lmao

[delicate orchid]

of course it isn't! Z is a set of numbers and Q is a set of equivalence classes!

[pastel cliff]

[devout crow]

Z is equivalence classes in N x N 🙂

[delicate orchid]

ok Q is a set of equivalence classes in Z^2

if we're going to be that pedantic

[pastel cliff]

the voices

[delicate bloom]

believe in urself

[pastel cliff]

what's special about commutative rings

[delicate orchid]

they commute

[pastel cliff]

https://tenor.com/view/patrick-star-spongebob-squarepants-wow-mindblown-gif-13169052

Every commutative ring is a subring of some field.

[delicate orchid]

I mean you probably aren't aware of how powerful a property that is

[pastel cliff]

im being asked if this is t/f

probably not

[delicate orchid]

every domain is I believe

[pastel cliff]

i know a domain is

[devout crow]

fields don't have nonzero zero divisors

[pastel cliff]

ooo so a commutative ring that isnt an integral domain would disprove

[delicate orchid]

every domain R can be embedded into Frac(R) if you want an explicit formulation

[pastel cliff]

shoulda thought of that ty

[hidden haven]

Imagine saying non zero zero divisors and thinking that you're sane

[next obsidian]

Integral domains are exactly the subrings of fields

[pastel cliff]

every sixty seconds, a minute passes in africa

[desert dome]

We just assume k is a field, not necessarily algebraically closed.

[next obsidian]

I don’t think it helps regardless

I think this is writing a primary decomposition, but knowing that doesn’t really help much regardless

[main rover]

Ok so inverse maeks sense but how would one take an orbit

[desert dome]

Hi, I'm trying to show that k[x, y]/(1-xy) is isomorphic to k[t, 1/t]. I see one way to solve this is to parametrize x = t, and y = 1/t. I'm not sure why could we just do that?

[next obsidian]

Use the first isomorphism theorem

your parameterization is describing a map k[x,y] -> k[t,1/t]

and then you just have to show that the kernel is (1-xy)

[desert dome]

Ah I see.. Thanks!

[upper cape]

Could anyone tell me if this is fine? I have a matrix A over a field k, and I am considering the algebra k[A]. Would it be correct to say that k[A] is isomorphic to some quotient of the polynomial algebra k[X]?

[next obsidian]

what is k[A]?

[upper cape]

just elements of the form f(A), where f is a polynomial

[next obsidian]

well...

I am assuming A is a square matrix?

[upper cape]

yes

nice

[next obsidian]

or else I think you have issues

[upper cape]

so it all works

[next obsidian]

yeah then this is exactly right

What you've done is

you have a map k[X] -> n-by-n matrices over k

given by sending X to A

and the image is exactly k[A]

so you can apply the first isomorphism theorem

[upper cape]

ahhhh

I should have seen that

thanks!!

[next obsidian]

no worries!

This is very useful, good job noticing that it's a quotient though!

[stone fulcrum]

@main rover
Still looking for it?

[stone fulcrum]

Wait, I agree I don't know what "orbit" means in this context

[main rover]

My bad, I figured it out

[lilac trench]

What's the point of the last statement? Is E not already suitable enough for a splitting field?

[tardy yacht]

The definition of splitting field in Dummit requires that f(x) does not factor completly in any proper subfield, which we ensure by taking the intersection of all such E

[desert dome]

Hi, I'm trying to work on this problem. I'm a bit confused on what this question means. Do we need to show for example, B can be embedded into A, i.e., there is an injective homomorphism from B to A, and B is a k-algebra? Or we need to show something more?

[steady axle]

If a map A-> B is flat epiimorphism and A is reduced, does that imply that B is reduced?

[broken stirrup]

I didn't get underlined part

[proud bear]

i think this is true because if I and J are ideals, then IJ is contained in I intersect J

[broken stirrup]

oh yeah you are right

I didn't know that

well it's easy to see though

thanks

[proud bear]

no problem 🐋

[next obsidian]

That’s what it’s asking, yeah

[broken stirrup]

bro this is way more abstract for this channel, try asking this in preal-and-algebra channel

[broken stirrup]

i don't mean to be rude just wanted you to know that questions must be posted in appropriate channels

[chilly ocean]

i see

i under

stand

[iron vessel]

Hey guys, I need to find wether there exists a quadratic extension of $L = \mathbb{Q}(\sqrt{17})$ such that it is cyclic over $\mathbb{Q}$. I was thinking of using cyclotomic extension, more precisely, since 17 is congruent to 1 mod 4, we have that $\sqrt{17} \in \mathbb{Q}(\zeta_{17}) = K$, and then we obtain that $Gal(K/\mathbb{Q}) \cong (\mathbb{Z}/17\mathbb{Z})^{\times}$, which is an isomorphism of cyclic groups of order 16. Then, we observe that there exist unique subfields of $K$ for every divisor of 16, i.e., 2,4,8. I am not sure how to continue however, so any tip is welcome.

[cloud walrus]

Évariste Galois

[iron vessel]

Naturally, there is $\eta_2$, the gaussian period of degree 2 in $K$ that has precisely $\dfrac{17-1}{2} = 8$ number of terms.

[cloud walrus]

Évariste Galois

[iron vessel]

In particular we have $\mathbb{Q}(\eta_2) = \mathbb{Q}(\sqrt{17})$.

[cloud walrus]

Évariste Galois

[iron vessel]

with $f_{\mathbb{Q}}^{\eta_2} = X^2 + X - 4$.

[cloud walrus]

Évariste Galois

[strong yacht]

There was some discussion about subfields of cyclotomic extensions in this chat a week or so ago? An automorphism that generates a cyclic subgroup of order 4 is $\sigma: \zeta \mapsto \zeta^4$ which permutes the set ${\zeta, \zeta^4, \zeta^{13}, \zeta^{16}}$ so i think a good candidate for the field you are looking for is $\mathbb{Q}(\zeta+\zeta^4+\zeta^{13}+\zeta^{16})$.

[cloud walrus]

Greenman

[iron vessel]

Yep, i figured it out at the end. More precisely i got that the unique cyclic quadratic extension of $\mathbb{Q}(\sqrt{17})$ is equal to $\mathbb{Q}(\sqrt{\dfrac{17+\sqrt{17}}{2}})$

[cloud walrus]

Évariste Galois

[iron vessel]

By using exactly the trick you advised me to use

[strong yacht]

Nicely done

[thorn delta]

Galois doing Galois theory

[lapis trail]

Ikr

Galois just let her go man

Not worth it

Me with a time machine

Go back and stop the duel that killed Galois

Anywayyyy

[chilly ocean]

Why does this hold

[lapis trail]

Is there a way to check if an irreducible of GF(32) is primitive other than brute force?

(x-a)^2+b^2

[delicate bloom]

since we're in characteristic 2, that's equal to (x-a+b)^2

[lapis trail]

Oh my b

I was responding to @chilly ocean

[delicate bloom]

oh didn't read

haha

[lapis trail]

Thanks though

I just found the 6 irreducible polynomials of degree 5 over F_2 now I have to find a primitive one

*F_2[x]

[fallow plume]

so I've already done this proof:

but I wanted to ask where the assumption for the characteristic of the field came in

I think i did it implicitly without realizing it

oh wait is it just polynomial long division fails

[cloud walrus]

Kraft Macaroni

[fallow plume]

it's def true if it's finitely generated PID-modules from CoFGM Thm

but I can't think of anything else just yet

@woeful flint got it, third iso thm:
(M/J')/(J/J') ~ M/J

so J/J' is never trivial

[woeful flint]

Ah that makes sense

Thanks so much?

!

[fallow plume]

np

[woeful flint]

that wasnt a question lol

[fallow plume]

lmao it's chill

[fallow plume]

well this is awkward. i should've just read the sketch of the proof for 5 and it mentions it

[lapis trail]

A polynomial p(x) of a given degree d is primitive over F_2[x] if F_2[x]/p(x) is isomorphic to GF(2^d), is that right?

Nevermind

[lethal dune]

is it true for non commutative ring that the nil radical is unique?

[south patrol]

Well if you're defining it as the set of nilpotent elements then it won't be an ideal necessary; are you viewing it as the intersection of all prime ideals?

[prisma shuttle]

i think its guaranteed to be an ideal only for commutative rings

[lethal dune]

that's not how nil radicals are defined on a non commutative ring

the def goes like this. The nil radical is the maximal ideal of R wrt the given property that for every element a of N, the principal ideal (a) is nil ideal.

unlike in commutative case, you can't say it's the intersection of all prime ideals

[next obsidian]

I don’t see how this isn’t well-defined?

Let S be set of all a such that (a) is a nil ideal

Claim: (S) = nilradical

[lethal dune]

No that's not what the definition is

It also has be an ideal

let me quote

[next obsidian]

I am showing that this (S) is the nilradical

Oh Hurb

Nvm my proof uses commutativity

TFW binomial theorem is fake

[lethal dune]

The nil radical of a ring R is defined to be the radical ideal with respect to the property that "A 2-sided ideal is nil". and denoted by N(R)

Radical: A two sided ideal I in a rign R with 1 called a radical ideal wrt a specified property P if

1. the ideal I possesses the property P
2. the ideal I is maximal for the property P

[next obsidian]

[lethal dune]

Theorem: For any ring R the nil racial N(R) exists and it's characterized by
N(R) = {a | the principal 2-sided ideal (a) is a nil ideal}

I hate non commutative stuffs

[next obsidian]

Yeah I guess I don’t see why that’s closed under addition

Without commutativity

ChmonkaS

[lethal dune]

given here, sum of nils are nil

[next obsidian]

Oh huh I guess I do see a proof

So wait, in this case it is well-defined right?

Like it definitely is unique

[lethal dune]

definitely an ideal because it's defined to be

[next obsidian]

I just mean like this is by definition going to be the largest one with this property

And you can show that this is actually an ideal, because a priori it’s just defined as a set

[lapis trail]

To find the subfields of GF(64), they would be generated by modding out the irreducible polynomials of degree 6 from F_2[x], is that right?

Nevermind

[lethal dune]

how do I conclude the uniqueness then?

[next obsidian]

It’s literally defined as a specific set of things

There’s only one such set

[pastel cliff]

do algebras not require an identity element

[delicate orchid]

evidentially not

[pastel cliff]

Apart from Lie algebras, most algebras one meets tend to be both associative and unital.

[delicate orchid]

what's the problem here

[pastel cliff]

nothing lol just hadnt notice the lack of a unit in a lie bruh

lie algebra -> lie algebruh -> lie bruh -> libra

[delicate orchid]

I can attest to this btw

every algebra I've worked with thus far has been both those things

[pastel cliff]

is there a name for this relationship

[chilly radish]

This is a dirty lie

The unital part

It's true associativity is mostly assumed outside of lie algebras

[pastel cliff]

welll uhhhh

is there notation for an ideal of a ring

like how the triangleeq one exists for normal subgroups

[delicate orchid]

leq

I use triangleeq because I'm weird

[pastel cliff]

can someone explain what a commutator is pls

this doesnt really help

[delicate orchid]

hm I wonder

[chilly ocean]

it's a thing that measures commutativity

[delicate orchid]

in group theory the commutator is (usually) defined as [x, y] = x^-1y^-1xy or [x, y] = xyx^-1y^-1 etc.

[chilly ocean]

is this in a lie groups context? you can give a geometric interpretation to the lie bracket using flows

[delicate orchid]

general lie algebras

[lavish nexus]

xy-yx

[delicate orchid]

^ basically lol

[pastel cliff]

yeah it's ch1 of a UG book on lie algebras

[delicate orchid]

you can kinda see why it's called the commutator in that image as well

[stone fulcrum]

The commutator of x and y is [x,y]

[chilly ocean]

i'd be careful writing this, since lie brackets need not be of the form xy - yx. but in many contexts (e.g. matrix groups) this is fine

[stone fulcrum]

Yeah there's other contexts for the word "commutator" but that's not what the book is using atm

[pastel cliff]

hm

i shall not overthink and just move on then

[delicate orchid]

I mean there's not much to think about lol it's a definition

[pastel cliff]

https://math.stackexchange.com/questions/3415698/prove-that-mathbbqi-is-a-subfield-of-mathbbc

is the proof 1) there really enough

it feels like they just shows closure and kinda inherits the rest from Q being a subfield of C, is that enough

ok i guess it obviously is, but should they have been more explicit about it then?

[chilly ocean]

see "subring test"

their proof is fine

[upper cape]

If I have a left linear map which is a left isomorphism, does that mean that its inverse is also a left isomorphism?

[lavish nexus]

f(ax)=af(x)
$f^{-1}(ax)=f^{-1}(af(y))=f^{-1}(f(ay))=ay=af^{-1}(x)$

[cloud walrus]

Iteribus

[lavish nexus]

first =: f is onto
second =: left linear

[upper cape]

Thanks!

[barren sierra]

kinda crazy how normal subgroups are called normal cause of Galois

[marsh goblet]

I'm trying to show that Q/Z has only one subgroup of order n
now the cyclic subgroup generated by 1/n is of order n, and so that works

however, the cyclic subgroups generated by 2/n,3/n etc also have order n

Now I'm assuming those are all going to be equivalent to 1/n, but I can't really put into words as to why that is. I'm certain it's because we are quotienting out Z but I can't formalise that properly bc smol brain

[delicate orchid]

see if you can construct an isomorphism between these different groups

[lapis trail]

Explain

[barren sierra]

My prof said that Normal subgroups are called such because they correspond to normal extensions of fields when talking about the Galois Correspondence

[lapis trail]

Idk what that means 😭

If I had a time machine I'd save galois' life

[south patrol]

Yeah it's kinda cool

[barren sierra]

Tho tbh idk why normal extensions are called normal

[south patrol]

Although ultimately ig with the Galois correspondence you can just say they correspond to other Galois extensions right lol

True hm

[barren sierra]

Yea

[south patrol]

I wonder if it's just to mean like it's a well behaved extension lol, hm

Normal seems to be used weirdly in maths lol

[fallow plume]

am I misunderstanding here or is this contradictory

suppose E/K is a finite extension of say, char 0 that isn't simple. then by the above we have its seperable. totally seems reasonable to me esp given the def of a seperable field

but then Wikipedia says that implies its simple? it's clearly not right or am I misunderstanding

[lapis trail]

Mathematicians suck at naming things

Learning about reproducing kernel spaces, I ask the professor in what sense is it a kernel, and it has nothing to do with algebra

[barren sierra]

lmao

[lapis trail]

Polynomials are wassup though

[next obsidian]

Every finite extension in char 0 is simple

The primitive element theorem tells you that any separable extension is simple, and any extension in char 0 is separable

That’s why your counterexample doesn’t work, no such thing exists

[chilly ocean]

I know this is a dumb question and the answers don’t really matter, but I’m curious. Which do you all think is a better notation to denote the integers modulo n, Z/nZ or Z_n? If either, why?

[full panther]

does the alpha in the evaluation homomorphism $\phi_{\alpha}: R[x] \to R$, $f(x) \mapsto f(\alpha)$ have to come from any particular set?

[cloud walrus]

beachcow

[barren sierra]

I think as long as alpha is in R it's chill

seeing as the codomain is R

[fallow plume]

oo gotcha. i guess I was always thinking that like, Q(\sqrt{2}, u_3) where u_3 is a nontrivial third root couldn't be written as a simple extension but I guess I never thought of a single element which could satisfy both.
even though now in retrospect there's obviously one for this case

[barren sierra]

the former since it's more evocative of quotients

[full panther]

hmm im a bit conflicted then because you can consider roots of polynomials in R but also those that are outside of R, and my book uses the evaluation homomorphism for the definition of roots of polynomials

roots being the alpha such that f(alpha) = 0

i presume you can only consider such homomorphisms when R is contained in the set alpha comes from

[fallow plume]

I think I agree with @barren sierra
I think by former he meant Z/nZ and not "integers modulo n" though
I personally swear by it enough that I go through the pain of having to do \Z{}[x] to do my polynomial rings since I have \Z[n] be Z/nZ

[chilly ocean]

That's why I use the former more frequently as well. Additionally, Z_n can be potentially confusing with the notation for the p-adic integers, although context certainly allows one to differentiate between the two.

[fallow plume]

I thought people used \Q_p for p-adics

[cloud walrus]

Arr0w_04

[fallow plume]

ah gotcha

if i wanted to learn more about p-adics, what would i go into? I heard analytical number theory is one but I'm not actually sure

[chilly ocean]

I think Dummit and Foote has some exercises discussing them, however there are some other books on them as well.

https://www.amazon.com/p-adic-Numbers-Introduction-Fernando-Gouvêa/dp/3030472949

https://www.amazon.com/Numbers-Analysis-Zeta-Functions-Graduate-Mathematics/dp/0387960171

I haven't read either of those, but those are two examples.

[fallow plume]

oo thanks 🧡

also getting recced grad level texts

[grand sigil]

I liked this one

Very quick and dirty

[chilly ocean]

Why is a factor ring formed by a maximal ideal always a simple ring?

[lethal dune]

by factor ring, did you mean quotient?

if yes, use correspondence theorem

[chilly ocean]

I can never remember my isomorphism theorems

Okay it follows directly from the correspondence theorem

[lethal dune]

there's only one theorem

[chilly ocean]

? There are four

[tall jay]

Is this true, or should m be in M and n in N?

[lethal dune]

yes

[tall jay]

So m in M and n in N need not be true?

[lethal dune]

What I meant is "m ∈M and n ∈N"

[lethal dune]

is this definition alright for tensor products? I feel something off since we are taking a map to an abelian group so kinda feels like we are loosing the module structure

A bit doubtful

[south patrol]

Lol idk why they use a group instead of another R-module

[lethal dune]

ikr

there's something seriously wrong with my prof

[south patrol]

Like what they have works for tensor product of abelian groups

So ig u could say oh and its compatible w the scalar multiplication etc

Seems slightly odd to me though

[lethal dune]

it does

because the right side has nothing to do with R

[south patrol]

Also feels slightly funny as like

They seem to say M \tensor_R N is the tensor of M and _R N

But the subscript is part of the tensor product not N

[lethal dune]

Lol

nah it's part of the tensor product

[chilly radish]

Because the Tensor product doesn't have an R-module structure necessarily if R is not commutative. The module structure of M,N is preserved by the fact we are looking at R-biadditive maps (It shouldn't be bilinear maps because the Tensor product isn't necessarily a module)

You either need commutativity or one of the modules to actually be a bimodule to get back a module structure

[south patrol]

Fair enough lol I assumed that R would be commutative

[lethal dune]

you mean even if M is R right module and N is left R module, M \tensor N is not necessarily a R module?

[chilly radish]

Correct

[lethal dune]

but it's a Z-module

[chilly radish]

Aka abelian group

[lethal dune]

yes

[chilly radish]

The scalar multiplication might not be compatible with the scalar multiplication of both M and N

Associativity can break down

When R is not commutative or one of the modules is not a bimodule

[lethal dune]

kinda confused ngl

[chilly radish]

If you try and give it a module structure, then on one hand you have
r(s(m x n))= rs(m x n)
You want this to be compatible with the scalar multiplication in M and N so you want
r(mxn)= mr x n= m x rn
So on one hand
r(s(m x n))= r(ms x n)= msr x n
But on the other hand
r(s(m x n))= rs(m x n)= mrs x n
And you want these to be equal, which is usually a problem. Same thing happens if you try and endow it with a right structure

It is a Z(R)-Module tho, but looking at Z(R)-Biadditive maps instead of R-Biadditive maps mans you're forgetting the R-Module structure somewhat unless R.is commutative, so it won't be a tensor product of R modules anymore

[lethal dune]

i see

[cloud walrus]

cgadski

[cloud walrus]

Arr0w_04

[lethal dune]

No

[proud bear]

@chilly ocean no. Let G=1 be the trivial group and H any group, for example. It would be true if phi is surjective

[lethal dune]

true when φ is surjective

[chilly ocean]

Right, that is what I thought but I wanted to make sure. Thanks ryu sama and stain!

[broken stirrup]

Hi

Im studying representation theory. What does it mean by F-basis? M might not be a free module how do we know that it has a basis in the first place

[chilly ocean]

every vector space has a basis

[upper pivot]

It’s an a module but a is a F algebra, so M is also a F vector space right

[broken stirrup]

oh yes thanks

[round jay]

Let G be a group, and let [G,G] be the subgroup of G generated by all
elements of the form aba−1b−1. (This is the commutator subgroup of G; we will
return to it in §IV.3.3.) Prove that [G,G] is normal in G.

Any tips for this? I tried showing that [G, G] is kernel of some homomorphism with domain G but I'm not sure how to find such a homomorphism.

[brittle quiver]

I think directly via definition is easier. For any $g \in G$,
$gx^{-1}y^{-1}xyg^{-1} = (gx^{-1}g^{-1})(gy^{-1}g^{-1})(gxg^{-1})(gyg^{-1})$

[chilly ocean]

Like what 1345631 said, you can just use the definition of a normal subgroup and things should work nicely.

[cloud walrus]

1345631

[round jay]

Why should the right hand side be in [G, G]?

[brittle quiver]

It's of the form x-1y-1xy. Try to verify this

[late thicket]

I'm having difficulty with abstract algebra tooic sub group

[chilly ocean]

we might be able to help, if you ask a question

[broken stirrup]

also, what does tha matrix of a_M look like?

[late thicket]

Can you tell me how to understand in easy way about groups, subgroups, permutation groups, quotioent groups

Is there any online teacher or videos which is helpful in it?

[grand sigil]

The concepts are really easy to get when citnextualized

You should read any introduction on groups

[frail zealot]

groups represent symmetries of some object(s), for example the cyclic group represents rotational symmetries of a flat object. an element of the group is essentially an action that applies your symmetry like a rotation or reflection or what ever else there is. the group operation is just a way to compose these actions like first rotate and then reflect.

subgroups are subsets that are also a group. they can be things like smaller sets of symmetries of an object like just the rotations for example but there are many more types of subgroups too.

a permutation group is just the set of all permutations (bijective functions from one set to itself) via cayleys theorem all groups a a subgroup of the permutation group. so you can think the permutation group as all symmetries on a set of a certain size.

qoutient groups are easier to understand if you understand if you understand integers mod n cause it's the same concept

[chilly ocean]

groups are group actions

[frail zealot]

a group acts on itself so yea

[delicate orchid]

specifically group actions on vector spaces

[tranquil parcel]

It seems the galois group of a polynomial is independent of my field as long as its coefficients are in that field. Am I understanding this correctly?

[south patrol]

Well galois group of x^2 +1 regarded as a poly C[x] is trivial since it already splits but galois group of x^2 +1 in Q[x] isn't trivial

Unless you mean smth else

[wraith obsidian]

Hey, sorry, I've kept this message as unread for over a month in hopes that I would have time to come back to it, but yeah that didn't happen and I still know virtually nothing about group topology…
very curious fact though.

[tranquil parcel]

See that makes sense to me. What am I supposed to think when a question just asks "Find the galois group of the polynomial x^3 - x + 1"

[south patrol]

I would assume they mean as a poly in Q[x]

[wraith obsidian]

Isn't that clear tho? The cohomology H is functorial in the group G in the first argument, and H is isomorphic to any of its conjugates

[wind sphinx]

Okay I know that b and c are reducible, but how do I reduce them?

[grand sigil]

Algebra rules

Starting with b for example

assume x^3+2=(ax^2+bx+c)(dx+e) for a,b,c,d,e in Z_3.

Solve for a through e.

If we expand rhs we get
(ad)x^3+(ae+bd)x^2+(be+cd)x+(ce).

We need ce=2 and ad= 1.

ae+bd=0 and be+cd=0

[chilly ocean]

x^3 + y^3 = (x + y)(x^2 - xy + y^2)

[grand sigil]

That is quickest way.

Then you need to find y^3=2 in Z_3

[strong yacht]

Quadratics and cubics are very special cases. If they are reducible, then they must have a linear factor in their factorisation. So e.g. sufficient to show a poly is irreducible by testing X=0, 1, 2 for these exercises and confirming none of them are roots (this does not work in general).

So e.g. X^3+2 must have X+1 as a factor, since (1)^3+2=3=0 mod 3.

And continue applying this idea essentially

[grand sigil]

must have x-1 =x+2 as factor?

[strong yacht]

Indeed, my bad I miswrote

[grand sigil]

Only love

[wind sphinx]

omg I'm so dumb!

lmao I didn't think about how mod behaves with negative numbers

that helped a lot Cole Lee

Thanks everyone else

[prisma shuttle]

can someone explain wut a left polynomial is

vs a right polynomial

like wut is the distinction

is it just that in the general form of a polynomial in a polynomial ring $R[x],$ a left one is expressed as $\sum_{i=1}^n a_ix^i$ while a right one is defined as $\sum_{i=1}^n x^ia_i$?

[cloud walrus]

JustKeepRunning

[lavish nexus]

probably

Your R might not be commutative

[prisma shuttle]

yea it seems to be in the paper ima reading

but i am a bit confused about this part of the paper

can someone explain the red part

if u want link to the full paper it is here: https://libgen.ggfwzs.net/book/81145072/cc70ba

[chilly ocean]

left/right just seems to distinguish where the coefficients are

also, "left polynomial" in the first line of the second page has a footnote that might answer your question

[prisma shuttle]

yea it does confirm thx

but do u get my question about the red highlighted part

idk wut the paper is trying to say

[chilly ocean]

idk didn't read

[desert dome]

Hi, I'm trying to prove that J = (x^2 - yz, xz - x) is a radical ideal in k[x,y,z]. That is, suppose there exists n such that p^n in J for p in k[x, y, z], p in J. Could anyone give me a hint on that? Thanks

[prisma shuttle]

just wondering is k here commutative or no

[desert dome]

k is a commutative field

[prisma shuttle]

ok ok nice

[next obsidian]

The solution I outlined before does work, I worked it out.

Compute the minimal primes over J, then look at their intersection [ which is sqrt(J)] mod J. Use relations in J to simplify the ideals in the intersection until the intersection becomes easy to compute, then conclude that sqrt(J) mod J is 0.

There ends up being very little computation

[prisma shuttle]

@next obsidian could u elaborate on how u can compute the minimal primes over J?

[next obsidian]

Suppose p is a prime containing J. Use the primal of it to conclude certain combinations of elements must be in it, and no matter how you do it, the ideal those elements generate is prime. They also contain J, so if p was minimal then it equals them

One example is, you conclude that one possibility for p is (x,y)

[desert dome]

Thanks! But this problem asks us to prove it is radical first, then find the intersection. I'm not sure if I can switch the order tho...

[next obsidian]

Of course you can

It doesn’t matter in what order you do it, computing it as an intersection shows that its equal to its radical, so it constitutes a valid proof of the first point

Proving that its radical without computing the radical as an intersection means you’d be showing if f^n is in J then f was in J and this is hardly realistic, it will be absurdly hard to do so

[chilly ocean]

nullstellensatz moment

[desert dome]

I see... Thanks!

[next obsidian]

It really won’t help, because showing that this is the ideal to the associated algebraic set is basically gonna just be showing its radical lmao

[chilly ocean]

i know

giving unhelpful answers is my forte

[next obsidian]

Swag

[torn frigate]

let E be a vector space over K : dim_K(E) = n
and F a generating family of E with n element

prove that F is basis of E

[next obsidian]

#linear-algebra

[chilly ocean]

you could start by noticing that every generating set contains a basis

[desert dome]

Sorry another question. Currently I find out these three ideals and the intersection of them = J. How could we conclude they are the only minimal primes over J so the intersection = sqrt(J)?

[next obsidian]

Well if you’ve found 3 primes containing J whose intersection is J, call that ideal I

Then you know I = J < sqrt(J) < I, the last equality because you’re taking an intersection over fewer ideals

But this implies all of these < are equality

[torn frigate]

Why

any proof for that?

[chilly ocean]

no

[next obsidian]

no proof exists for it 😧

[torn frigate]

r u srs?

[chilly ocean]

yeah :(

[torn frigate]

i feel like u can extract a basis by removing dependent elements contained in the generator @chilly ocean

[chilly ocean]

two pings too many pings

[torn frigate]

sowwy

am i correct tho?

[next obsidian]

why don't you write down a proof, then you don't need to ping someone to know if you're right

[torn frigate]

still wondering how to do it

thats why idk what proof should i write kek

[next obsidian]

do you believe in doing any work on your own?
I'm telling you to figure out a proof on your own, this isn't hard

[torn frigate]

if it wasn't hard i wouldn't ask u

[next obsidian]

you haven't even worked on this for ten minutes

[torn frigate]

how'd u know?

[next obsidian]

because TTerra gave you an idea and you immediately asked how to do it

You have an idea

"remove dpenednet elements"

[torn frigate]

you could start by noticing that every generating set contains a basis

[next obsidian]

just write that down

[torn frigate]

This?

[next obsidian]

figure out why that idea proves it

[torn frigate]

i tried that i want something else

[next obsidian]

how long did you try to prove that?

[torn frigate]

using the method to extract basis from the generator?

[next obsidian]

Yes

[torn frigate]

well i just used induction on it

[next obsidian]

then why are you asking if it's right?

you literally proved it

[torn frigate]

no no im asking for a more straightforward method

or maybe contradiction

[next obsidian]

this is the straightforward method

[torn frigate]

not really

[next obsidian]

show that bases are the same size

it literally is

[torn frigate]

idk

well i gave this problem too much attention

cba

moving on

[desert dome]

Ahh Thanks! _(:_」∠)_

[lavish nexus]

f: C. -> D. is a quasi-isomorphism if H_n(C.) = H_n(D.)

this definition seems completely independent of f

like if H_n(C.) = H_n(D.) which only has to do with the d's then I can shove in any f to be a quasi-isomorphism

or I guess as long as f makes the diagram commute which still isn't really any condition on f since any morphism f has to satisfy that

[late thicket]

I need to know about books which is helpful for begginer and youtube too please duggest me sir@grand sigil

[lavish nexus]

dummit & foote

[next obsidian]

Because this isn’t the definition

f is a quasi isomorphism if the map it induces on homology is an isomorphism

[grand sigil]

Im not exactly sure what is good. I personally learned most my algebra from a textbook by Fraleigh. If you want to use youtube you should check out Richard Boechards or Socratica.

[modern hawk]

Hi, could someone help me understand the Number Theoretic Transform

(I assume it's abstract algebra, though redirect me if I'm wrong) I don't quite understand what's happening in this example

is the weird letter next to X just a coefficient

and why is it to the power of 128 too suddenly

https://dsprenkels.com/ntt.html it is taken from here

[prisma shuttle]

you mean the $\mathbb{Z}_q$ or the $\zeta$?

[cloud walrus]

JustKeepRunning

[lethal cipher]

I'm not sure if this is an easy question. Say Gal(L/Q) is isomorphic to A_n. Is there a way to find a polynomial s.t. L is the splitting field?

[eternal furnace]

should 2 not be the other way around?

[modern hawk]

the zeta, sorry, had to run some errands, I got some rundown from someone about it

[odd flame]

Hey guys this is probably a really straightforward question but i dont understand how to go about it

This is asking how to prove that the quarternions are a subring of M(2,R)?

[round light]

How do I show that ( 2Z, +, . ) is not a ring with unity
is it sufficient to just say that there exists no element 'a' in 2Z that follows a.b = b.a = b

[chilly ocean]

yes

[round light]

Ok thank you

[round light]

How do I show that phi is onto

I know that I need to show every element of Real number set R has a pre-image in Domain R, but not sure how to approach it

[delicate bloom]

are constant functions continuous?

[round light]

Yes !

[round light]

Ok so do you mean we can always define a continuous function f(x) =c, for all x belongs to [0,1] and for any real constant c ?

[strong yacht]

I'm not sure if this is what you're precisely asking, but this got me wondering if every finite group appears as the Galois group of some polynomial with coefs in Q.

After Googling, it turns out this is an open problem, but S_n and A_n are proven (by Hilbert) to always be the Galois group of some polynomial with coefs in Q

Constructions are given here: https://en.wikipedia.org/wiki/Inverse_Galois_problem

[delicate orchid]

yeah, ||and then since c is any real number the image of all constant functions at 1/2 is the real line||

[prisma shuttle]

I think $\zeta$ is a primitve 128th root of unity

[cloud walrus]

JustKeepRunning

[chilly ocean]

Does anyone know of where I can get my hands on a complete solutions manual to Dummit and Foote?

[prisma shuttle]

i saw there was one on quizlet

but u need quizlet plus

for me it was worth it

[next obsidian]

Doesn’t exist

[chilly ocean]

Ugh that sucks. I tried looking it up on the web and I found partial solutions to the exercises, but it was mostly for group theory. It had less and less solutions for ring theory and field/galois theory, much less the later chapters...

[prisma shuttle]

oh yea i saw this one too

it wasn't complete though

[chilly ocean]

Yeah they would skip certain exercises, and sometimes whole chapters towards the end. It's a great book, super unfortunate I can't check my answers.

[prisma shuttle]

no there is one on quizlet

thats complete

i'll send it here

https://quizlet.com/explanations/textbook-solutions/abstract-algebra-3rd-edition-9780471433347

you need quizlet plus though

[chilly ocean]

Sweet, I'll check it out.

[next obsidian]

I am very suspicious that someone has solved and collected every single exercise

At the very least that they’re correct

[chilly ocean]

It says "verified" but I am not sure what that means lol

[prisma shuttle]

it seems to be done though?

[next obsidian]

I doubt it

[prisma shuttle]

at least i've been using it for like the first 5 chapters

and it sbeen pretty good

[next obsidian]

Yeah that isn’t deep enough to where I think it would become dicey

There’s a number of people who’ve done all of the groups stuff

Someone in this server did it even

[prisma shuttle]

yea but i think its still avaluable resource

[next obsidian]

Sure

[prisma shuttle]

if u have any questions u wanna check

u can always just ask in this server

[chilly ocean]

I could use that one too, do you know where to find that?

[next obsidian]

I just am doubtful it actually has all of them

Idk I haven’t seen that user in ages

They didn’t publish them

[chilly ocean]

Also, what are your thoughts on using Artin instead of Dummit and Foote? I have never read Artin, but I heard that it was comparable in terms of quality...

[next obsidian]

¯_(ツ)_/¯

I think honestly you should pick whatever you like the best

What you enjoy reading most, what exercises you find most engaging

The treatments of all the standard texts are roughly the same honestly

[chilly ocean]

hmmmm ok thanks

[next obsidian]

Ppl will debate the nitty gritty and split hairs that X does this, Y doesn’t!

But they’re so similar I think just using the one that you’ll actually enjoy using is most important

A textbook you don’t want to read isn’t gonna teach you as much as one you will read haha

[chilly ocean]

Yeah, true. I just had a weird experience with calculus and linear algebra where there are books that are genuinely better than the others. The ones I took for my classes were fine, but then I figured out that Spivak existed, for example. I just don't want to make the same mistake with abstract algebra where I put a bunch of time into it and there was a significantly better book out there.

[next obsidian]

Yeah that’s fair. I think just stay away from Lang is all I can say

Lang is too hard for a first pass

[chilly ocean]

LOL yeah lang is really really terse

[next obsidian]

The differences between the books are impossible to really judge for yourself until you already are familiar with algebra

There’s definitely differences, but none is really “better” than the other in total

Some just treat X better but Y worse

And if you’re new, you can’t judge which between X and Y will be more important for you

I’m sure whatever books you’re looking at, Artin, D&F, etc are all roughly equal

[chilly ocean]

OK, thanks for the advice @next obsidian! Appreciate it.

[brittle quiver]

Let G,H be groups and phi: G->H a group homomorphism. Suppose H is abelian and phi is injective. Is G abelian?

[south patrol]

first iso

[brittle quiver]

Nvm I got it

[barren sierra]

how do I draw these diagrams in LaTeX

nvm I found an online tool

[cursive temple]

quiver is pretty good

https://q.uiver.app/

[barren sierra]

yea that's what I found

it's quite nice

[tranquil parcel]

Is there a way I can use fixed fields to answer c or do i do something by hand

[spiral wolf]

Honestly, I don't see how c) is true, but I could also just be being dumb. Like, sigma_2(z) = z^2*z = z^3 =/= z

[tranquil parcel]

how do you know that's what sigma_2 of z is?

it only says that it does that for the root a of f

[spiral wolf]

Oh, of course, I'm dumb

[tranquil parcel]

i know sigma1 and sigma5 are inverses and i tried to use that somehow

same w 2,4 and 3 is its own and id is its own

wait doesnt s3 have like 4 self inverses tho

[spiral wolf]

Just thinking out loud, so sigma_i(z) = sigma_i(1/2 + a^3/2) = sigma_i(1/2) + sigma_i(a^3/2) = 1/2 + 1/2 sigma_i(a^3) (since sigma is an automorphism)

Assuming what I've said is true, I think that should give you the right direction to go

[barren sierra]

Ngl I got no idea how to do this.

omega could map to itself or omega^2 right?

[tranquil parcel]

Depending on what the discriminant of your polynomial is.

Have you covered discriminants?

Oh that makes perfect sense @spiral wolf ty ty idk how i missed that

[barren sierra]

no

I'm trying to map it to this example

but it's hard since I don't know what L is

[tranquil parcel]

Ok so L|Q is degree 3 (given) and adjoinint w is a degree 2 extension on L (since w isn't already in l and its minimal polynomial over Q is the degree 2 one dividing x^3 - 1)

[barren sierra]

I mean there is only one field of order 3

does that mean Gal(L/Q) must be Z/3?

[tranquil parcel]

Yes, since it is a finite galois extension of that degree, it must have a group of that order. We really only have one order 3 group

[barren sierra]

oh group my bad that's what I meant

[tranquil parcel]

So what do we know about the galois group of L(w) over L?

[barren sierra]

it must be degree 6 by tower law

I got that much

oh wait no

over L

it must be degree 2

[tranquil parcel]

ye i was just gonna use that to derive that the extension totals out to degree 6

[barren sierra]

yea I got that immediately it's just determining automorphisms

is the only valid automorphism omega -> omega^-1

and thus this generates a group of order 2, Z/2

[spiral wolf]

I don't want to interrupt your conversation, but I'm forgetting the wording for something. If I have an irreducible quadratic polynomial and I want to declare alpha to be a root of this polynomial, what do I call alpha exactly? Not just a root, right, since it's irreducible and I'm creating a field extension?

"Suppose f(x) has an irreducible quadratic factor in Q[x]. Let alpha be a ??? quadratic root ??? of that factor"

[tranquil parcel]

I always just call it a root, but the extension itself is called quadratic

[spiral wolf]

Fair enough, wasn't sure if I was allowed to just say "root" without some qualifier explaining that it's part of the extension

Another question, in this problem I'm to show that x^5-x-1 is irreducible mod 5, which I have done. I am then supposed to conclude it is irreducible in Q[x]. What theorem am I supposed to use here?

[south patrol]

As this is primitive, it's irreducible over Q[x] iff irreducible over Z[x]. And what would happen if it factored properly over Z[x]?

[round light]

Thanks 👍

[barren sierra]

Is it enough to state that Galois Group of order 3 => Z/3 and then we have Q(omega) has two automorphisms: the identity and the map omega -> omega^-1 which forms a cyclic group of order 2? Then since omega is not in L we have these are separate fixed fields and so we have Z/3 x Z/2

I am unsure how to show this isn't S_3