groups-rings-fields

[stoic rose]

After going through all these calculations you probably understand why I made that comment earlier

[dull root]

yea for sure, it was good practice tho. haven't done long divison in a while

[stoic rose]

Yeah it's still good to do some calculations at some point to get a feel for these finite fields, though I don't recommend going above F_4

[dull root]

So I have a related question. I know F_p^n : F_p is a galois extension, therefore a normal extension. Moreover

x^4 + x + 1 is irreduicble in F_2, so if some extension of F_2 contains some root of x^4 + x + 1, it must contain all roots, since any finite extension of F_2 has form F_2^n, ie it is a normal extension. Then is the splitting field of x^4 + x + 1 = F_2^4?

[stoic rose]

Yes exactly

[dull root]

Wait, x^4 has 4 roots. so why isn't F_2^3? enough. I have seen by the prior discuession that F_4 is not enough since it splits only into 2 deg 2 factors

[stoic rose]

x^4 splits in F_2

[dull root]

i meant x^4 + x + 1 srry

[stoic rose]

The extension obtained by adjoining ONE root of an irreducible degree d polynomial always has degree d. Now when the base field is finite that extension is automatically Galois so the splitting field itself has degree d.

[dull root]

Yes I recall that now. So since x^4 + x + 1 is irreducible, i take some root beta, and adjoin F_2(beta). This is a extension which has deg 4 over F_2, hence the field has size 2^4

[stoic rose]

[dull root]

So I also recall this consequence from the Galois correspodnce the subfields of F_2^n are exactly F_2^d where d divides n. Then if I know the splitting field of x^4 + x + 1 is F_2^4, and F_2^8 contains a copy of F_2^4 as a subfield, is it true that x^4 + x + 1 must split also in F_2^8?

[stoic rose]

Yes

[dull root]

I general, F_2^4n must split this poly then right?

[stoic rose]

Yes

[dull root]

So how we constructed F_2^4 was we took b, a root of x^4 + x + 1, and extended F_2(b), but how can I use this to find the other 3 roots other than beta.

1. So I know 1, b, b^2, b^3 form a basis over F_2. since b has deg 4

2. This means that the elements of F_2^4 = {0, 1, b, b^2, b^3 .... } and in general any F_2 linear combination of them is an eleemnt of the field right?

In the F_4 case, we knew the elemetns were {0,1,alpha, alpha^2} since I only had to list four, but what is the general way to find all elements of a given finite field. I know such fields are all simple exntesion of their prime field

[stoic rose]

For F_p^d you can pick an irrducible degree d polynomial f over F_p and take the extension F_p(b) where b is a root of f, and then a basis is given by 1, b, ..., b^(d-1)

So all elements are F_p-linear combinations of those basis elements

[dull root]

And to count the size of field it is p^n since we have n basis and p possible coefficients in F_p right?

[stoic rose]

Yep

[dull root]

I see, so I know what all the elements of F_2^4 look like. They are all F_2 linear combinations of {1, b, b^2, b^3} where b is a root of the irreducible poly in F_2: x^4 + x + 1. Then the other 3 roots must appear as some F_2 linear combination of {1,b,b^2 , b^3}, but how would I go about finding those other roots?

[stoic rose]

The simplest way to find the other roots is probably to apply the frobenius automorphism

[dull root]

The forbenius homomorphism is an injective ring hom on a field F to F that sends z to z^char(F). For finite fields, injectivity implies automorphism

Since F_2^4 is a galois extension of F_2, the forbenius automorph must map roots to roots, that is it must make b to another root, but can it map b to itself?

[stoic rose]

Yes, and the most important thing is that it generates the Galois group, so by applying it repeatedly to b you'll get all the roots

Because the Galois group acts transitively on the roots

[dull root]

Right, since the forbeniious auto generates the galois group, we also have the galois group is cyclic, of order 4 in this case

[stoic rose]

Exactly, so after applying it 4 times you should get back to b

[dull root]

But more generally, if I have a cyclic galois group, is it always true that the generator never fixes any roots

[stoic rose]

It doesn't mean anything for a single element to act transitively, but the Galois group always act transitively on the roots, yes

[dull root]

Why is it true? Say the galois group is C_n the cyclic group of order n, and let g be a generator. Since the galois group acts transtively on roots, it is possible to map any root to another. Then since g is a generator if g fixes some root, then what is the issue?

it means that every automorphism must fix this root as well

If a root is fixed by every automorphism, then the root must be in the base field right? By that I mean if L is a galois extension of K, then L is the splitting field over K of some polynomial p. Then we have a root in K

[stoic rose]

Yes

[dull root]

But why is that an issue, that a root can not be fixed by every automorhpism

of nvm, i see why

p was irreducible over K

[stoic rose]

Yes, the Galois group acts transitively on the roots of any irreducible polynomial

[dull root]

Yup I see. But back to the forbenius automorphism. Since char(F_2^4) = 2, the and b was a root of x^4 + x + 1 which was irreducble over F_2.

The Forbenius automorphism sends b - > b^2 - > b^4 = b + 1 - > b^2 + 1 - > b^4 + 1 = b so it indeed has order 4 and the four roots are exactly

b, b^2, b+ 1, b^2 + 1 which are all elements in F_(2^4)

[stoic rose]

Exactly

You could check explicitely that those are all indeed roots of x^4 + x + 1 if you want

[dull root]

Yea, I will multiple them to check it

Luckily for me, I dont have to distinguish between + and - here since we have char 2

[stoic rose]

Yeah that's convenient there's no sign error possible in characteristic 2

[dull root]

Yep it works :))

Next, if I move away from char 2 fields ):. Lets say I have the field F_7, so F_7 = {0,1,2,3,4,5,6}. To see x^2 + 1 is irreducible over F_7, it sufffices to just plug in all the values from 1 to 6 and see that none of them give 0 mod 7 right?

since if it has a root factor, it must have a root at one of the 6 pts from 1 - 6

Since F_7 is isom to Z/7Z

[stoic rose]

Indeed

[dull root]

Man I really hate finite fields, you would think they would be easier than extensions of Q since its finite, but I find it more confusing zzz

[stoic rose]

They have much nicer properties than finite extensions of Q and are easier in this regard, but they're indeed less intuitive (except for the prime ones).

One reason is that there's not really any canonical way to define F_p^n, you have to pick an irreducible polynomial of degree n over F_p and even though two different polynomials will give isomorphic fields there are many isomorphism and no canonical one.

So you have to make arbitrary choices to be able to talk about their elements, and even though they are finite computations quickly become to long to be done by hand, though computers are pretty good at handling them.

[dull root]

I actually have a question related to that. So for example F_p^2 is a simple extension of F_p some some irreducible degree 2 poylnomial. However, it is not the case that if I take another deg 2 irreducible polynomial, and extend by a root that those two extensions are the same, only isomorphic right

[stoic rose]

But the nice thing is that they have such a nice Galois theory that you often don't actually need to make explicit calculations and can get away with Galois theory arguments

[stoic rose]

Depends on what you mean by "the same" and "extending by a root". Usually the way we extend K by a root of f is by taking the quotient K[x]/f(x), and if you have different f then the elements of the quotient will be different equivalence classes, so the two fields won't be literally "the same" as sets.

[dull root]

They are certainly isomorhpic since they are simple extensions so we can just map one generator to the other, but not the same setwise

[stoic rose]

Well if you construct one using an irreducible polynomial f and the other using another g (of the same degree) then the isomorphism is not obtained by mapping the root of f you adjoined in the first field to the root of g you adjoined in the second field. It must be mapped to a root of f in the secodn field, which turns out to always exist (even though that might not seem obvious at first sight) but there are a several different roots that you can choose and they're not necessarily easy to find.

[dull root]

This is because the automorphisms act transtively on the roots of this irreducible poly, so it can not map to a root of g if they dont share roots. i see

yea, its not as simple as I thought it was then

[stoic rose]

Yeah the isomorphism must preserve the minimal polynomials

And the fact that there are a bunch of roots of f in the second field and no preferred one is what I meant when I said the isomorphism is not canonical

[stoic rose]

Coming back to this question, on the other hand if you fix an algebraic closure of F_p, and consider only subextensions of this algebraic closure, then there is actually exactly one extension of each degree, so adjoining a root of f or a root of g give you literally the same subextension, not just isomorphic ones.

The problem of course is that you have to make infinitely many arbitrary choices to define that algebraic closure, so it's definitely not better in terms of giving explicit descriptions of the elements

[dull root]

Then I don't see why this following question I have should be true. I know x^2 + 1 is irreducible in F_7, and of course i is a root of it. Then for F_7(i), it is also true that any quadratic (not even just monic irreducible ones) must split in F_7(i). Since this is a normal extension, It means that F_7(i) is the unique degree 2 extension of F_7 right?

[stoic rose]

Right

Up to isomorphism

[dull root]

This is more than isomorphism right? If F_7(i) splits every quadratic in F_7, then any deg 2 extension of F_7 must be = right?

[stoic rose]

If you mean literally equal as sets then clearly no

[chilly ocean]

How would you go about showing that certain polynomials form maximal ideals in fields?

That would be the best way to show that R[x]/(x^2 + x + 1) is a field, correct?

[stoic rose]

There are no nontrivial ideals in fields

Oh do you mean polynomial rings?

[prisma shuttle]

yes this is the most common way to show that something like this is a field

also could someone help me with this question

Find all zero divisors and nilpotent elements of $\mathbb{Z}_n[x]/I,$ where I the ideal of vanishing polynomials over $\mathbb{Z}_n$.

[cloud walrus]

JustKeepRunning

[stoic rose]

You can use the fact that K[X] is PID so the maximal ideals are those generated by irreducible elements

[stoic rose]

Note that this quotient is isomorphic to the ring of functions Z/nZ -> Z/nZ

[prisma shuttle]

i thought of that

but i am not sure how to proceed from there

[dull root]

So the Forbenious automorphism on F_7(i) just gives the conjugation action. How canI use this to show that F_7(i) splits any quadratic in F_7? A quadratic in F_7 either has 2 real roots of 2 complex roots, but it is not even true that the real part of the roots need to be in F_7

[prisma shuttle]

plus what would a zero divisr/nilpotent even mean/look like in this isomoprhic ring?

[stoic rose]

Write down what's the condition for the product of two functions to give 0

And its shouldn't be too hard to identify from there the zero divisors

[prisma shuttle]

ok i got it reduced down to the set of all polynomials functions which have a nonzero element in the tuple (P(0), P(1), \cdots, P(n-1)) and which have a polynomial corresponding to these output values

but i only know that lagrnage interpolation resolves this problem over division rings; is there an analog of lagrange interpolation over non divisoin rings?

[stoic rose]

Hum right, I didn't thought about that

I think that every function over Z/nZ is still represented by a polynomial even though usual Lagrange interpolation doesn't work, but I'm not sure

By CRT it would suffice to prove it for Z/p^nZ

[stoic rose]

Hum nevermind, that's actually false

For example in Z/4Z if f is represented by a polynomial then f(0)+f(1)+f(2)+f(3) is always even, so you cannot represent all functions by polynomials

[umbral arrow]

Anyone good at abstract algebra? Need help with a few problems

Dm me

[south patrol]

Kinda vague lol why not ask here

[prisma shuttle]

can someone explain what this theorem is saying

its on page 3 in this paper

https://core.ac.uk/download/pdf/82496173.pdf

idg what it means by F_n is uniquely determined by F, along with the iff statement and all the other uniquely determined stuff

shouldn't the F_n uniquely determined F intuitvely (like to construct F as a function that vanishes everywhere)

[vestal snow]

Let $I$ be an ideal of a noetherian ring $R$ such and $M$ be a finite $R$-module such that $IM=M$. Is it true that depth$(M,I)$ is finite?

[cloud walrus]

Finitely Many Bananas

[vestal snow]

This is essentially the same as asking "can Ext_R^n(R/I,M) be 0 for all n"

[barren sierra]

Can someone check a proof for me?
Here's the problem:

Here's an extremely relevant proposition I used

Here's my proof

[prisma shuttle]

I think its correct

[barren sierra]

[oblique leaf]

What does it mean for $$M\bigotimes_F V^\vee$$ to be an R-module but tensored over F? In this context, R is is an F-algebra, M an R-module and V dual an F-vector space

[cloud walrus]

alyosha

[oblique leaf]

I have to show that this is a short exact sequence given 0->U->V->W->0 is exact

I'm trying to show injectivity and surjectivity of the most natural possible maps

does anyone have any hints?

[next obsidian]

Is F a field? If so, you want to first show the dual sequence is exact. From there writing a basis for M replace M with F^(+)alpha for some alpha. Then show that tensor product distributes over direct sum. The result is that you basically just multiplied your exact sequence

[next obsidian]

You can multiply by R-coefficients on the left using the fact that M is an R-module

[oblique leaf]

yes

[oblique leaf]

but then what does it mean to tensor over F?

[next obsidian]

You just form the normal tensor product

If it helps, M is also an F-vector space

You can multiply by F-coefficients by sending the elements in F into R via the algebra map

Then you define your action of F on M by going through R first

[oblique leaf]

so does it mean that ax\otimes b = x\otimes ab if a is in F but not when a is in R

[next obsidian]

Yup exactly

[oblique leaf]

thanks, then i was also wondering what this means?

does it mean this for all v in V?

[next obsidian]

I have no idea what this means lol

[oblique leaf]

f_w and f_w' are elements of W dual and psi: V to W

im trying to show injectivity

but im not sure what it means for two tensors to equal

[next obsidian]

Yeah idk sorry

I don’t think about this concretely like this

I am using more general stuff

I would first show that the diagram consisting of the dual VSes is exact

Then you don’t need to worry about the dual

You just have to show that tensoring an exact sequence remains exact

[oblique leaf]

are you suggesting that I first show that 0->W dual ->V dual -> Udual -> is exact?

[next obsidian]

Yes exactly

Then you can simply show that given an exact sequence, tensoring with M will keep it exact

[oblique leaf]

doesn't the tensor functor only preserve exact sequences when M is flat?

[next obsidian]

Everything is flat over a field

Because everything is free

[oblique leaf]

but M is an R-module right

not necessarily free?

[next obsidian]

But you’re tensoring over F

So over F it’s a vector space

Aka has a basis over F

Yeah?

[oblique leaf]

ok i think i can go from here

thank you so much!

[next obsidian]

Actually wait

I think I have another proof

This one is gigabrain though

Every exact sequence of vector spaces is split exact

This is preserved under applying functors

Oh wait

Nvm I think this won’t work you need to show exactness in the middle

Hehe whoops

Well wait

The dual is left exact and this argument shows it’s actually exact as well

Similar for tensor products

¯_(ツ)_/¯

[iron veldt]

hey i was wondering if anyone could point me in the direction of getting started understanding symmetry groups? (ie U(1) and SU(2) type stuff) from a more intuitive understanding? I feel like i can't find anything on how the naming scheme works or even anything on the specific symmetry groups in general

[oblique leaf]

Actually i now see how you can argue that the ses is exact, but that's when you see them as F-vector spaces, however it seems that I need to prove exactness while considering all those tensor products as R-modules, does your process still hold?

[next obsidian]

Exactness is measured via image and kernel

So it doesn’t matter what you view them as modules over

In fact, you can just consider exactness as abelian groups

[oblique leaf]

but don't image and kernel depend on the homomorphism?

[next obsidian]

Yes, but it won’t matter

The maps still exist, like umm

If the map was R-linear

It’s also F-linear

[oblique leaf]

I think we start out with F-linear

how does that extend to R-linear?

[next obsidian]

I think you should spend some more time thinking about it, but what it’s linear over doesn’t matter

[oblique leaf]

okok

thank you!

[next obsidian]

It’s because of how the R-module structure is defined

Basically like

R only acts on M

And it’ll be like rm (x) n maps to rm (x) phi(n)

When you had a map phi:N -> L or whatever

[stoic rose]

Could you be a little more precise about what kind of thing you're interested about these groups? Are you looking for ressources about the general theory of Lie groups, or about their use in physics, or you just want to understand the naming scheme?

If it's just the naming scheme then it's not particularly difficult : O(n) is the group of n by n real orthogonal matrices while U(n) is the group of n by n complex unitary matrices, and adding S means you add the condition of having determinant 1

So O(n) correspond to linear transformations of an n-dimensional real vector space preserving the inner product, or equivalently symmetries of the (n-1)-sphere (in dimension 2 and 3 that's just the reflections and rotations which are easy to visualize)

And SO(n) just adds the condition that the orientation is preserved, so you keep rotations but not reflections

For U(n)/SU(n) it's similar but now you're looking at transformations of an n-dimensional complex space which preserve the complex inner product. It's harder to visualize for n>=2, but it's really analogous the the real case.

If you want to learn more about Lie groups in general Bocherds has a nice lecture series about them

[mint seal]

Borcherds is king

[frail zealot]

true

[oblique leaf]

ok I thought about it a bit more, and I just don't understand how M is also an F-vector space. I know that R is an F-algebra, but I don't know how that is relevant?

[next obsidian]

If R is an F-algebra we have F < R as F is a field

[oblique leaf]

yes

[next obsidian]

From there, given x in M and a in F, we can define a•x just pretending a is in R

As we know how to make sense of a•x when a is in R because M is an R-module

This doesn’t actually even require F < R, if F wasn’t a field so R being an F-algebra just meant we had a map phi:F -> R, we could define a•x to be phi(a)•x

This is something called “restriction of scalars”

Because in the case F < R we’re sort of restricting our scalars from everything in R to just those in F

If you look up restriction of scalars you’ll see it’s an adjoint to something called extension of scalars which is given by tensor product :)

[oblique leaf]

Ok yeah sorry, that was dumb

thank you so much for your explanation!

[echo orbit]

this is an example from lang's algebra i came across

OHHHH!!!!

im wrong

sometimes you just gotta rubber duck debug

[gritty sparrow]

No, you are right. There is no such canonical map, for example when G=Z, H=2Z and K=0, we can’t get a map from Z/2Z to Z by mapping cosets to cosets, the only map is the trivial one

I assume lang just mixed up H and K

[echo orbit]

he didn't

i did

😛

[gritty sparrow]

Oh lol

[echo orbit]

i was subbing in H and K for an indexed infinite seq

this is from the groups chapter section on inverse limits

[gritty sparrow]

I see

[echo orbit]

that was my only misconception with it

poor reading 😛

[timid plover]

in interpolating the following data points using the quadratic Newton/Lagrange polynomial: (3, 7), (4, 2), (5,6), how do i find the correct Newton/Lagrange interpolating polynomial for the data set?

(these are two separate questions difference being one considering Newton and other considering Lagrange)

[delicate orchid]

Not abstract algebra my friend, but if you’ve got the formula for the general n-point Lagrange polynomial you can just substitute in and get your answer lol

[broken stirrup]

so, to prove (a), I found a middle linear map in between R/I \otimes B and B/IB. Now I'm considering a homomorphism g and want to show that they are mutual inverses. Kernel of g should be IB but i can't proceed to show that

I kinda see that why it's true but can't put into math properly

[thorn delta]

you can show that IB \subset ker g, and this is good enough to get a well defined map g : B/IB --> R/I \otimes B

[broken stirrup]

well i've already shown this inclusion, the other way around seemed harder to me

you are right, that's indeed enough

[thorn delta]

ye, all you need is a map B/IB --> R/I \otimes B. showing that this is actually the inverse of you other map is equivalent to ker g = IB

[broken stirrup]

thanks

[thorn delta]

np. There's another way to do this too

The idea is you tensor the SES
$$ 0 \to I \to R \to R/I \to 0$$ with $B$

[broken stirrup]

i guess it has something to do with short exact sequnces

[cloud walrus]

kxrider

[broken stirrup]

oh yes

[timid plover]

oops sorry what would be the appropriate channel for this type of problem then?

[delicate orchid]

Maybe #computing-software ? This server is kinda lacking in numerical methods channels

[prisma shuttle]

can someone explain what this theorem is saying

its on page 3 in this paper

https://core.ac.uk/download/pdf/82496173.pdf

idg what it means by F_n is uniquely determined by F, along with the iff statement and all the other uniquely determined stuff

shouldn't the F_n uniquely determined F intuitvely (like to construct F as a function that vanishes everywhere)

[tribal moss]

There must be a typo somewhere. The lower-case f that is mentioned before the "iff" does not appear anywhere after the "iff".

[prisma shuttle]

no this f just refers to the polynomial function determined by the polynomial f (like the mapping f:Z_m\to Z_m, giving the values of f(0), f(1), f(2), \cdots)

so its basically just saying a condition for $F$ to vanish over $\mathbb{Z}_m[x]$

[cloud walrus]

JustKeepRunning

[prisma shuttle]

idk the part i don't understand is the iff a uniqueness condition like shouldn't it be the other way around?

that F is uniquely determined by arbitrary polynomails F, and a_k?

but then how do I use the uniqueness condition?

[tribal moss]

My first guess would be that it's sloppy phrasing. I think it's trying to say that

(f = 0) iff (exists Fn, ...,ak)(F = FnSn + ....)
oh and by the way then you're going from left to right, Fn and ak are uniquely determined up to such-and-such

It also looks like the reference to "Definition 1" should really be "Definition 2" which suggests there are limits to the care that was taken with the phrasing.

[prisma shuttle]

do u know if there's another way to phrase the theorem thats easier to understand?

i am still having a lot of trouble understanding wut the theorem is tryign to say

[tribal moss]

I'll admit it's not intuitively clear to me how it works.

[broken stirrup]

how do you intuitively understand tensor product of modules?

[next obsidian]

Experience

[delicate orchid]

universal ploperty

[next obsidian]

This mf said plopertyxD

[delicate orchid]

it may not seem like it at first but that is actually factual info

honestly when it comes to tensoring modules I do tend to go off of either the quotient construction or the universal property (even though they might be the same thing ;) )

[edgy glacier]

Hi, i hope its okay to ask book recs even though it appears the book recs have dissapeared,

does anybody have an abstract algebra book with lots of worked theorems?

Because I'm trying to self teach i'd need to see the theorems properly proved to know im not making mistakes

[proud bear]

(btw, i think book recs has disappeared because you have given yourself the studying role)

[chilly ocean]

dummit & foote, they describe the proofs quite wordy as well

[frank fiber]

if i have an homomorphism in gap, how can i get the image of the homomorphism under a given element?

[chilly ocean]

gap?

[frank fiber]

yes, the software gap

[frank fiber]

i didnt find it in the manual

[stoic rose]

You mean just just evaluating f(x) given an homomorphism f and an element x?

[frank fiber]

yes

[bleak abyss]

So

[cloud walrus]

Several Sloths

Several Sloths

[bleak abyss]

Well ignoring mod center stuff

We can spit out a matrix

[cloud walrus]

Several Sloths

[bleak abyss]

Oh I assumed a few minutes passed so you guys were done

I'll migrate

[stoic rose]

I don't know gap but that should definitely be in the first examples in the documentation about homomorphisms, that's like the simplest thing you can do with an homomorphism 🤔

[vivid onyx]

I have a question about convex optimization and linear programming. Where is the best place for me to post my problem?

[hidden haven]

#numerical-analysis perhaps

[edgy glacier]

peabrain abstract question: I'm trying to define the vector space of vectors that are linear maps over the rationals and the field that is the rationals

When you're trying to prove that the linear map

f(x1,x2,...,xn) =c1x1 + c2x2+...+cnxn

Has the additive inverse

-(f)(x1,x2,...,xn) = -c1x1 + -c2x2 + ... +-cnxn

(where all of the c1x1 are individually rational numbers, and the additive inverse exists by abelian group)
how do you prove that f and -f sum to the zero vector?

The zero vector being a map from n rationals to (0, 0, ..., 0)

I'd say by induction, then in the induction step, n iterations of associativity in (moving f_{n} to the end)

But then I'm not sure if that's a "good" proof method 🥴

I know it's trivial but i don't want to mess up on trivialities

[delicate orchid]

am I misunderstanding this or
f(x1,x2,...,xn)-(f)(x1,x2,...,xn) =c1x1 + c2x2+...+cnxn-c1x1 + -c2x2 + ... +-cnxn = c1x1-c1x1+c2x2-c2x2...+cnxn-cnxn = 0

works for all finite n

[edgy glacier]

right it works

but if you're proving it works where c1x1 c2x2 etc are elements of a group

then don't you have to do iterated associativity to prove you can rearrange to that?

It's hairsplitting but I'm trying to self learn 😅

[delicate orchid]

of course they're elements of the vector space, the vector space is closed under scalar multiplication

and addition in a vector space is abelian by definition

[edgy glacier]

Yes of course they're elements of the vector space my concern is about iterating the associativity in the proof

I think it's all good tho

[delicate orchid]

I fail to see how associativity really matters here, apologies

the fact that you even write "c1x1 + c2x2+...+cnxn" without parentheses implies associativity of addition

[edgy glacier]

yeah i mean because you have to move c1x1 + -c1x1 and so on

that's n times

n associativities for moving one element so that addition is applied to it and it's additive inverse

Then n more associativities

sorry 🥴 it's NBD, the prof is just incredibly precise

So if i elaborate on my thought process he'd probably expect to see that

And i just don't want to miss basics

[delicate orchid]

that's commutativity not associativity

but don't apologise it's good when you're starting to really think through this stuff

[wild solar]

Hi, does anyone mind helping me with part 2? I guess I'm supposed to prove 3x^2y+3xy^2 is not divisible by 5 or something but I just couldn't get why the number 5 would appear anywhere.

[next obsidian]

I think you should just look at like, what happens to 1 + 1 or something, probably

[wild solar]

okay thanks

🙂

[next obsidian]

[edgy glacier]

you can make a group by pointwise extension of a group right 🤣

Like for instance the function S->G is a group under pointwise addition and pointwise zero where pointwise addition is defined
+(f1,f2)(x) = f1(x)+f2(x)
pointwise zero is
0(x) = 0
And pointwise inverse is
-(f1)(x)=-(f1(x))

I can't see what's wrong with this but when i look up pointwise extension of a group i see nothing

[vast quiver]

your group structure on {f: S -> G} is pretty much how we define G^|S|, i.e., the direct product of G with itself |S| times

so there's nothing wrong with what you're doing at all, people definitely study direct products

e.g., G^2 can be identified with the set of functions f: {1,2} -> G, where (g1,g2) maps to the function f, where f(1)=g1, g(2)=g2

[wild solar]

I don't understand why he wrote x^2,y^2 and xy are irreducible in this ring but he said they are reducible in Q[x]. Sorry I'm a little unfamiliar with these concepts but can anyone help me out?

[next obsidian]

The point is inside of R you don't have access to x

so you can't say that x^2 is reducible because x^2 = x * x, since x isn't in R, so you can't talk about that right hand side inside of R

inside of Q[x,y] though you can use x, so this is a valid expression showing that x^2 is reducible

[wild solar]

oh okay that makes sense... ty so much 🐶

[red imp]

if I've done something and gotten $\bZ\oplus\bZ\cong A\oplus(\bZ/2\bZ)$ then have I messed up somewhere? It doesn't feel right for such an $A$ to exist but I'm not sure

[red imp]

abelian groups btw

[chilly ocean]

compare order of elements on each side

[red imp]

ah okay

ty

I thought there might be some tricky isomorphism if A had two generators

[pastel cliff]

can i show an isomorphism between fields by showing isomorphisms between the additive and multiplicative groups respectively

actually is that even a smart thing to do

[red imp]

yeah field isomorphisms are exactly ring isomorphisms

so it works

[chilly ocean]

hmmm

[pastel cliff]

hmmmm

[red imp]

hmmmmm

[pastel cliff]

im thinking about this bc i need to prove that any two four-elt fields are isomorphic

but idk if that's the right approach

it can probably work but might be unnecessarily lengthy

[next obsidian]

Write both of them as an extension of F_2

This is sort of how the general classification of finite fields go, but to get the F_p^2 fields is easy to do ad hoc by considering how they are extensions

This is particularly easy for F_2 because there’s so few polynomial in each degree

[pastel cliff]

im gonna assume an extension has a formal definition, and we unfortunately didnt learn that :(

[next obsidian]

Your class sucks

Your prof needs to be replaced

[pastel cliff]

lol yeah

[red imp]

you could cheat and use the fact that any four-element ring is either cyclic or the klein 4-group

only one of these is a field, qed

[next obsidian]

That doesn’t do anything?

[pastel cliff]

we also stopped short of the group actions chapter and instead jumped to a whirlwind on basic stuff with rings

im taking a better algebra class next semester lol

[red imp]

why not? a 4-element field is a 4-element ring, so it either has to be Z_4 (a field) or the klein 4-group (not a field)

[next obsidian]

You have to classify 4 element rings now???

You’re trying to solve a strictly harder problem

[red imp]

yeah that's why I said it was cheating

[next obsidian]

And I really think calling Z/2Z X Z/2Z a klein 4 group is silly because a group isn’t a ring lol

[red imp]

-_-

[pastel cliff]

https://mathhelpboards.com/threads/any-two-fields-with-4-elements-are-isomorphic.11035/ i caved

i thought about this too but i like my original idea better so i'll try that lol

[warm holly]

Could someone clarify please why the algebraic multiplicity of eigenvalue gives the size of Jordan block of the matrix with that eigenvalue on diagonal?

[red imp]

so from your answer I'm guessing that if $A\oplus B\cong C\oplus D$ then $A$ has the same order as either $C$ or $D$?

[next obsidian]

He isn’t talking about the order of the groups, he’s talking about the order of elements inside the group

This isn’t true because you could compare like, uh
Z/pZ (+) Z/qZ and Z/pqZ (+) {e}

[red imp]

ah yeah

so looking at the order of the generators should be enough

[pastel cliff]

Hint: First prove that 1+1=0, and then show that the nonzero elements form a cyclic group of order 3 under multiplication

i found these hints for the four elt field problem

but how does the second one help

[next obsidian]

I’m gonna be honest

Write down the elements of a field with four elements

0,1,x,y

Figure out what relations are required, like 1+1 = 0

What x•y is forced to be

Shit like this

Once you do this you can take another field {0,1,c,d}

And you can just define a map sending 0 to 0, 1 to 1, x to c, y to d

And just show it’s a homomorphism

Like when there’s four elements

You can literally bash out the addition and multiplication tables

There’s 16 entries to fill in, and a ton of them are already fixed

Like 1• anything else = the thing

0• anything = 0

[next obsidian]

Also Z_4 isn’t a field it has zero divisors

There’s at least 3 rings with 4 elements

Z/4Z, (Z/2Z)^2 and F_4

[pastel cliff]

but that's so sillyyyyyy and uglyyyyyyy

you're probably right

but still

[next obsidian]

It should be possible to write it out in like 3 minutes

Not even kidding

Because of commutativity there’s only 10 things to compute

For multiplication 7 of those are enforced by ring axioms

For addition 4 of them

[pastel cliff]

thank you mr chmonkey

[red imp]

oh true I was thinking of groups again lul

I've been working with groups all day

the real way to cheat would be to use the fact that all fields of a given finite order are isomorphic if they exist

[pastel cliff]

could i even stop short of showing the tables

and just say "this is a unique construction of a four element field"

[north field]

Yo

Let f : G to G’ be surj homo

Let H’ normal subg of G

Let H = f^{-1}(H’)

Then H is normal and the map x |-> f(x)H’ is a homo of G to G’/H’ whose kernel is H

How do i prove this

[north field]

Wot

???

[red imp]

he's saying to use the first isomorphism theorem

[north field]

Bruh

On what page in Lang Algebra does he teach first iso theorem

Did i miss it

Ok can you just

Prove it along with the first iso theorem

Then ill catch up

[thorn delta]

I don’t think he names it, but iirc it should be one of the first few pages of the groups chapter

[north field]

Ok well im on like page 40 something

[wild solar]

Sorry to disturb again but could anyone help me with part 3? I actually don't understand how the hint is used...

[north field]

Can some1 just do a deep proof

[thorn delta]

you don't really even need first iso for anything. Do you know what you have to show to prove x |-> f(x)H' is a homomorphsim?

[cursive temple]

Note that if 5 = xy for x and y in Z+Zsqrt(2), then N(x)|N(5) and the same for y

[wild solar]

okay let me try

sorry I don't know how to proceed

so a^2-2b^2=1, right?

[pastel cliff]

am a lil stuck

can't get a contradiction from a^2 = b

[thorn delta]

if a has order 4, then a generates F

[pastel cliff]

uhhhh

how is that no bueno

im missing soemthing dumb

[thorn delta]

err hm actually nvm that there's something weird you've done here (F, *) is not a group. However (F - {0}, *) is a group of order 3

[pastel cliff]

ooh shoot

does removing 0 still work for the whole field tho

[thorn delta]

yea, basically the "group of units" of any ring is a group

[pastel cliff]

ohhhhh right right

[thorn delta]

in fact for finite fields, F - {0} is always cyclic

[wild solar]

okay I'm still stuck.. can anyone help me PLS

[pastel cliff]

sorry quick doubt, but what i meant is that like, proving a claim about this gives us something about the whole field right?

idk if that makes sense

[chilly ocean]

What does "any monomorphism in R-Mod is a kernel" mean

[chilly ocean]

We need to assume N(x) = N(y) = 5 and show it's not possible

If 5 = xy

Since N(x) = 1 iff x is a unit

So we have a^2-2b^2 = 5

And you need to show this has no solutions

Squares mod 5 are 0, 1 and 4

So which of those satisfy a^2 = 2b^2 (mod 5)?

And then you get only one possibility, from which you get a contradiction

So there's no element of Z[sqrt(2)] with norm 5

Anything with norm 25 must be prime in this ring

[pale jasper]

Could anyone explain the valuation function on $\mathbb{Z}_{13}$?

[cloud walrus]

Scerball

[wild solar]

okay wow, thank you so much!!!! I guess I wasn't very aware of how to use the fact that squares mod 5 =0,1,4, but thats super cool! Also just a little question, your last line, do you mean if N(x)=25 then N(y) must be 1 and therefore y is a unit, proving 5 is prime?

[chilly ocean]

Yeah, either N(x) = 1 or N(y) = 1

[wild solar]

Also why is the case a mod 5=0 and b mod 5 =0 cause contradiction?

I got confused again

[chilly ocean]

Because then both a, b will be divisible by 5

[wild solar]

yeah

[chilly ocean]

So a^2, b^2 will be divisible by 25

But then 5 will be divisible by 25

[wild solar]

im sorry but why

[chilly ocean]

a^2 -2b^2 = 5

[wild solar]

ohhhh

okay im dumb

thankssssss

[chilly ocean]

Np

[hidden haven]

There exists map with that monomorphism as its kernel

[chilly ocean]

Where we treat kernels as the whole inclusions of a submodule?

I mean, more abstractly here

Monomorphism satisfies the definition of a kernel basically?

[hidden haven]

Ye

[hidden haven]

For some map

[next meadow]

How do you find the maximum ideals of (R)? I think (<2>) is the only maximum ideal of (\mathbb{Z}{8}) and the only maximum ideals of (\mathbb{Z}{30}) are (<2>, \ <3>) and (<5>). My guess is that the maximum ideals of (R) are just the direct products, but how do you prove this? \
There's also a theorem in the book that the factor ring (R / I) is a field iff (I) is a maximum ideal. Since (R / I) is finite, I can show that (R / I) is a field by showing that it's an integral domain, so maybe that gives some constraint on what the maximum ideals are. But it seems complicated thinking of the problem that way.

[hidden haven]

In particular, for any inclusion M \↪ N, the quotient map N → N/M has the inclusion as a kernel

[cloud walrus]

gnosis

[iron vessel]

Let $L = \Omega_{\mathbb{Q}}^{x^5 - 2}$ be a splitting field of $x^5 -2$ over $\mathbb{Q}$. Prove that there is an exact sequence of groups:

$0 \xrightarrow{} \mathbb{Z}/5\mathbb{Z} \xrightarrow{} Gal(L/\mathbb{Q}) \xrightarrow{} (\mathbb{Z}/5\mathbb{Z})^{\times} \xrightarrow{} 1$.

[cloud walrus]

Évariste Galois

[iron vessel]

I'm not sure where to start. I haven't seen many exact sequences before.

[strong yacht]

By Sylow's Theorem, any group of order 20 (which is the order of your splitting field) must have a normal subgroup of order 5 (a group acts on its set of Sylow p-subgroups by conjugation, but group of order 20 has only one Sylow 5 subgroup [check], so it must be normal).

So there is an inclusion of Z/5Z as a normal subgroup of your Galois group

You just need to check that when you quotient your Galois group by Z/5Z, you get (Z/5Z)*, and this will give your exact sequence

[edgy glacier]

when you're solving a system of linear eqns through Gauss Jordan, is this the sort of vector space you solve over?

because you're solving multi linear maps

I'm a bit stuck on that tbh

my brain is melting, a vector space seems so general, why are you solving over linear eqns? Because you have no multiplication axioms for the vectors? 🥴

[hidden haven]

Your guess is good, depending on what you mean by direct products. For a starting point, try proving that any ideal J of R x S looks like (J cap R) x (J cap S).

But just to be sure, maximal ideals of the product ring won't quite be the direct products of the maximal ideals of R and S

[pastel cliff]

any nudges on the second one i dont see it

i didnt write that proof of i) i know it could be prettier

[chilly ocean]

try thinking of complex numbers as scalings and rotations

polar form

[pastel cliff]

https://tenor.com/view/hmmm-thinking-batman-gif-14744673

oooooo i think i see it

but would that suffice as an argument that they're isomorphic

could i like consider a "basis" of each one and work from that

[chilly ocean]

what argument?

you haven't written anything down

[thorn delta]

an explicit isomorphism would suffice as an argument

[chilly ocean]

and it's the only argument you should give in this case

[pastel cliff]

i guess i played myself there

generators to generators is what i meant!

[barren sierra]

potentially

just show that all the operations work

[chilly ocean]

please just write down an explicit isomorphism

[barren sierra]

or well that addition and multiplication are preserved

(and to do so yes you need an explicit isomorphism)

[edgy glacier]

What happens when you exchange the equalities in axioms like group axioms with equivalence classes

[woven delta]

We do the opposite when we take quotient groups

[marsh goblet]

So apparently all elements in a ring that have left inverses do not necessarily have right inverse
I've seen some counterexamples, but I've seen a proof for groups that left invertible implies right invertible -- so what about rings changes that?

as in, what about this breaks?

[next obsidian]

You’re assuming that everything has a left inverse

The issue with rings is some elements may have a left inverse while others don’t, so you can’t show that everything has a right inverse

Crucially, we start with a’s left inverse b, then use b’s left inverse

In a ring there’s no way to guarantee that b has a left inverse

[marsh goblet]

right and b's left inverse might not exist

[next obsidian]

This proof does work if you assume everything non-zero has a left inverse though

[marsh goblet]

ahh i see
tysm

[next obsidian]

[prisma shuttle]

hey guys do any of u understand how to answer this question

https://mathoverflow.net/questions/421794/generalization-of-lagrange-interpolation-over-non-division-rings

[chilly ocean]

|lambda(E)|=[K:E]
Why is this true

I kno.now

[daring hinge]

What's an example of a ring without identity that does not contain maximal ideal?

[leaden scaffold]

The even integers mod 4

[frail perch]

Suppose I have a complex representation whose character is real (takes a real value for each element in the group), is it true that the representation itself comes from a real rep? (I.e equals the tensor product of a real representation with the complex numbers)

[wraith ether]

Hi, for my abstract algebra course my prof has sent us assignments. One of these assignments is about irreducability of polynomials. I've spent some time to learn about irreducible polynomials, but I'm still confused by the question as it does not specify whether they should be irreducible over a specific field. To quote the question:

(1) x^4 + x + 1 - is the polynomial irreducible? What about primitive?

Everywhere I look to learn about irreducibility, it is specified over what field

Should I be able to answer that question without knowing the specific field?

[prisma shuttle]

no you need to know it

[wraith ether]

In that case I'll email the prof, thank you

[prisma shuttle]

did your prof specify it in class?

yea u should

[wraith ether]

No they did not

Maybe there's some general assumption (my guess would be Q, rational numbers) but I'm unsure

[prisma shuttle]

but if it was this would be relatively trivial

just plug in +=1 using RRT

ima guessing its more thanthat

my guess its probably over $\mathbb{Q}[\sqrt{n}]$ for some integer $n$

[cloud walrus]

JustKeepRunning

[prisma shuttle]

but yea u shoudl email the prof

[wraith ether]

rrt = rational root theorem?

[prisma shuttle]

yea

[wraith ether]

It is the first assignment though, out of 8 so could be a warming up

[prisma shuttle]

oh maybe yes

idk just ask to clarify

clarifcaiton never hurts 🙂

[wraith ether]

I will, thank you!

[chilly ocean]

I'm not sure how to see that that the last condition is indeed equivalent to saying \mu and \eta are coalgebra maps

does a 2 out of 3 property hold for (co)algebra maps

[frail perch]

Anyone has smth about this?

[chilly radish]

I have a biquadratic $x^4 + 2bx^2 + c$ which is irreducible over a field of char $\neq 2$. If $E$ is its splitting field with roots $\pm \alpha,\pm \beta$, I want to show that $[E:F]=8$ iff $c,c(b^2-c)\notin F^2$. The hint I was given is to show that $F(\alpha^2)=F(\beta^2)=F(\sqrt{b^2-c})$ which I did, but i'm pretty lost

i'd like just a small hint

[cloud walrus]

ShiN

[next obsidian]

https://groupprops.subwiki.org/wiki/Faithful_irreducible_representation_of_quaternion_group

Look at “summary” and then “type of representation”

It has real characters but not all reps come from real reps

[frail perch]

Interesting, thank you!

[strong yacht]

Maybe F(alpha.beta) and F(alpha.beta, sqrt(b^2-c))

[chilly radish]

I don't see how this helps exactly

[strong yacht]

Containment of fields F < F(alpha.beta) < F(alpha.beta, sqrt(b^2-c)) < F(alpha, beta) = E

[chilly radish]

Well yea, I understand that, I don't see how you relate this to the condition c(b^2-c) \notin F^2

the condition that c is not a square can be implied I think since alpha*beta=sqrt c

[strong yacht]

Yes, that's right, so that's some progress towards E being degree 8

Want each of those extensions to be degree 2

[chilly radish]

yes, that was my train of thought too, but i'm still not sure how to relate the condition c(b^2-c)\notin F^2 to this

[strong yacht]

Ok, let's prove the next field containment is strict

I.e. can't have F(alpha.beta) = F(alpha.beta, sqrt(b^2-c))

You mentioned that alpha.beta = sqrt(c), so want to prove: F(sqrt(c))=F(sqrt(c), sqrt(b^2-c)) is false

[chilly radish]

ok yea I see that now

I think

wait nvm

[strong yacht]

Well suppose the containment wasn't strict. Then it must be that sqrt(b^2-c) = q * sqrt(c)+r for some rationals q and r

[chilly radish]

yup

well, the base field isn't necessarily Q but that doesn't matter

just that it's not char 2

[strong yacht]

Yeah sorry, I'm so used to Q

I don't think it matters

[chilly radish]

it doesn't

[strong yacht]

Re-arrange the equation: [sqrt(b^2-c)-q*sqrt(c)]^2 = r^2

Expand: b^2-c-2sqrt(c(b^2-c))+q^2c = r

Here is one place we're using Char != 2

[chilly radish]

ok nice

yea

[strong yacht]

Ok, now E is at least degree 4

One more to go

[latent vector]

ayo

can anyone help me?

i have to show that GL(n, Zp) has a cyclic subgroup of order (p^n)-1

for every prime p and positive integer n

the hint says to consider the degree n extension of Zp

the degree n extension of Zp is Zp^n.

this group is cyclic, so let a generate it.

thus |<a>| = p^n -1

but what does this have to do with GL(n,Zp)?

[chilly radish]

the degree n extension of Z/p as a field is absolutely not (Z/p)^n. The latter isn't a field with the direct product structure so unless you mean something else when you say that you're mistaken

[latent vector]

no

not (Zp)^n

Z_(p^n)

[chilly radish]

ah ok

[latent vector]

sorry for miscommunication

[chilly radish]

np

[latent vector]

i'm guessing i have to show that this isomorphic to a subgroup of GL(n,Zp)?

[delicate orchid]

oh you want a representation of F_(p^n) over F_p

[latent vector]

what does that mean?

wait yes

i do

is GL(n, Zp) isomorphic to Z_(p^n)?

[delicate orchid]

I'll have to think about this - we're taking these as groups under multiplication right?

[latent vector]

yeah

no way right?

GL(n, Zp) has p^n entries minus those that are not invertibnle

but |Z_(p^n)*| = p^n-1

[delicate orchid]

nah, the former has order (p^n-1)(p^n-p)...(p^n-p^(n-1)) iirc

wrong reply

[latent vector]

oh yeah

[delicate orchid]

meant to reply to the message below

[latent vector]

that's correct

[delicate orchid]

so they're def not iso

[latent vector]

i think i just do have to show it is isomorphic to a subgroup

i.e. show that it is a valid representation

is there a natural injection?

[delicate orchid]

I'm trying to think of one

I need some kind of faithful group action of F_p^n on a set of size n and then I think I can get you a representation of degree n

[latent vector]

what about the action of the generator on the set F_p^n

that's bigegr

does that induce an injection

[delicate orchid]

I think the way to go might to just find an element of order (p^n)-1 and take the cyclic group generated by that element

cause I am writing a lot of stuff here that might be way too in depth

[latent vector]

https://math.stackexchange.com/questions/2828925/prove-that-gl-n-mathbbf-p-contains-an-element-of-order-pn-1

i just found this

this does not help me though....

[delicate orchid]

my mad ramblingsthoughts:

F_p^n \cong F_p[x]/(q) where q is degree n, these elements all look like some degree r < n polynomials with coefficients in F_p
is the action of F_p^n on the set {1, x, x^2, x^3, .., x^n} faithful - that might be something
so if we chose the set {1, x, ... , x^n} as our basis vectors of a vector space over F_p (which is valid as the polynomials already have coefficients in F_p, so taking coefficients these as our scalars is alright), then the action of elements of F_p^n by regular multiplication in F_p^n on this set of basis vectors is faithful, I think
so we map each element x of F^p_n to the corresponding permutation matrix generated by how x permutes the basis vectors - this is an injective map as the action is faithful, hence the kernel is 0, so the image of the generator (with order p^n-1) under this map also has order p^n-1.
As for what this map is I have absolutely no clue but it exists

[latent vector]

what IS the group action on the set {1, x, ..., x^n} given by F_p^n?

don't we just care about the generating element's action?

[delicate orchid]

ah no, this doesn't work

jesus this is difficult

ok new plan

[latent vector]

wait!

i found a stackexchange post

that is related

"It is easy to show existence of element of order 𝑞𝑛−1 (by embedding 𝔽∗𝑞𝑛 in this group)."

https://math.stackexchange.com/questions/3319875/maximum-order-of-element-of-finite-general-linear-group

let's embed this motherfucker

[delicate orchid]

oh yeah lemme just "embed it"

yeah I'll embed it alright

watch this shit

[latent vector]

ahaha

[delicate orchid]

hold on I need paper but I think I've got it

[latent vector]

kk

[delicate orchid]

nope sorry another false alarm

I was trying to write the generator a as a polynomial and then have map it to a matrix such that the product with the vector (1, x, x^2, ..., x^n) gave me a but it doesn't preserve multiplication so it's useless

[latent vector]

i think constructing it is harder than just showing that it exists

[delicate orchid]

this is a disgustingly hard problem

[latent vector]

this isn't even the last problem on the pset 😭

[delicate orchid]

ok time for my last ditch attempt, I'm gonna try and use character theory here

[latent vector]

tf is character theory

is a character related to free groups?\

oh nah

[delicate orchid]

I just cannot for the life of me get an injective representation

there's some diagonal matrix that works

that's pretty much the best that I've got

[latent vector]

fuck

[delicate orchid]

check the answers and tell me what they say lol

[chilly radish]

I think I got it. Thanks a lot!

[delicate orchid]

and if I see the word "galois" anywhere I'm actually quitting the server

[delicate orchid]

At least my construction worked lol

[chilly radish]

I'm just not 100% sure where I used the hypothesis that f is irreducible.

hmm maybe it's in the converse

I'm not sure if the conditions imposed are sufficient without the hypothesis that f is irreducible or not

[latent vector]

bruh

[strong yacht]

That's a good question

Is irreducibility perhaps used to show that the final field containment is strict? I haven't checked

[latent vector]

@delicate orchid thanks for ur help 🤝

[delicate orchid]

I tried

[strong yacht]

Wys about Galois bro

[delicate orchid]

because I don't know galois theory

[latent vector]

this is my galois theory class @vast copper

@delicate orchid

[delicate orchid]

for gods sake I knew it

funny field extenstion 🤪

[chilly radish]

the way I did it it reduced to the fact that the other 2 containments are strict

i'm doing the converse rn

hmm wait I might have a mistake

ok yea I think my approach doesn't work actually

ping me if you have an idea. I'm a bit stuck rn on the last inclusion but i'll try again tmrw

[strong yacht]

Yeah, i got it

The last field containment is F(alpha, beta) > F(alpha.beta, alpha^2)

Why is this strict? Because the right field doesn't have alpha. alpha satisfies the following poly with coefs in F(alpha.beta, alpha^2): X^2-alpha^2

X^2-alpha^2 is irreducible because if it wasn't, then the quartic we started out with would be reducible since it can be expressed as (X^2-alpha^2)(X^2-beta^2) by assumption

[chilly radish]

not sure I follow. It would be reducible over F(alpha*beta,alpha^2), not over F

or uh, over F(alpha^2) even

[strong yacht]

Sorry, I think I might have written some nonsense

Hmm

[chilly radish]

I tried going in this direction too haha

[strong yacht]

Oh wait no, I think this is right: I've shown that alpha is not in F(alpha.beta, alpha^2), since its min poly in this field is degree 2

Oh wait yeah, it is wrong lol

Complete nonsense, I don't see it right now, sorry

[chilly radish]

no problem

lmk if u have an idea

I think this containment has to hold whenever f is irredducible fwiw

like the strict containment

since the other two may fail because of the other conditions but the degree of E/F cannot go below 4

[latent vector]

what is the difference between $End_R(M)$ and $End(M)$?

[cloud walrus]

Monku

[chilly radish]

End(M) refers to just abelian group endomorphisms

Subscript R means only R-linear maps

[latent vector]

wdym by R-linear map

[chilly radish]

f(ra)=rf(a) for r in R

[latent vector]

ah

i see

[chilly radish]

In addition to respecting the group structure

So in particular End_R(M)\subset End(M)

[latent vector]

yea ofc

okay

ty

this makes sense actually right

if we say R is not only commutative, but a field

and we consider it as a vector space

all homomorphisms between vector spaces are linear maps, no?

[south patrol]

Homomorphisms between R-modules are precisely the R-linear maps, yes

[chilly radish]

Well yea, but you can have maps between vector spaces that only respect the group structure but aren't necessarily linear

[latent vector]

huh

not bijectively?

or injectively*

[chilly radish]

There are non-linear maps R->R (even bijections iirc) that satisfy f(a+b)=f(a)+f(b)

[latent vector]

ur spitting fax rn

i got my linearities confused.

okay

so

sweet i got it

[latent vector]

fuck i need more help

i was not in class-- how does this induce a Z2[x] module

[delicate orchid]

Looks like the definition of scalar multiplication to me

[strong yacht]

@chilly radish I just realised why we ran into a wall earlier, we didn't actually use any Galois theory

[chilly radish]

I didn't see a way to really use galois theory here, but a friend gave me a different approach

[strong yacht]

What did your friend say

[chilly radish]

you can use the fact that for a field F, F(sqrt a, sqrt b)/F is degree 4 iff a,b,ab are not squares in F, and from this you can conclude everything

also suggested just like, making size arguments about the galois group using the fact that certain elements are/aren't squares to disqualify certain automorphisms

[chilly radish]

morally this is similar to what you suggested, it's just an easier and less direct appraoch

so like first you apply this to F(sqrt(b^2-c)) with alpha and beta

and then to F(\sqrt(b^2-c),\sqrt c)

[tribal moss]

If M is an R-module and f is an R-linear map M->M, then we can make M into an R[x]-module by defining

(a+bx+cx²+dx³+...)·m = a·m + b·f(m) + c·f(f(m)) + d·f(f(f(m))) + ...

In this case this just works out to p·v = p(alpha)v when p in Z2[x] and v in F, and the multiplcation on the right happens in F.

[latent vector]

gotcha

thank you g

[strong yacht]

To get to the final part in that chain I gave, recall alpha*beta = sqrt(c), alpha^2 = -b+sqrt(b^2-c). The field F(sqrt(c), sqrt(b^2-c)) is a Galois extension of F which has Klein-4 Galois group and generated by the automorphisms A "fix sqrt(c) and permute +/- sqrt(b^2-c)" and automorphism B "permute +/- sqrt(c) and fix sqrt(b^2-c)". Suppose for contradiction that F(alpha, beta) = F(sqrt(c), sqrt(b^2-c)). Then from the automorphisms just found, AB(alpha) = beta or -beta, and AB(beta) = -alpha or alpha, but this would give an element of order 4 in the Klein-4 group which is a contradiction.

[chilly radish]

nice

that's pretty clean

[north field]

Let G be a group and K a normal subgroup. Assume that K and G/K are solvable. Then G is solvable.

How

[lavish nexus]

Just put them together?

There’s no problem with quotient groups being abelian

(Use 3rd iso if necessary )

Then normality
If G1/K normal to G/K what about G1 to G

[north field]

what

so we have

G = H0 > H1 > ... > Hn = K

and if we say G'=G/K

wee have G' = H'0 > H'1 > ... > H'm = {e}

but where does the first isomorphism theorem com ein

[lavish nexus]

First H0/H1=(H0/K)/(H1/K)

by 3rd iso
the latter is abelian because G/K is solvable

Normality we have H1/K normal to H0/K
take any h0K then h0h1h0^-1K is in H1/K
so any h0h1h0^(-1) =h1’k for some h1’ in H1 and k in K

but K is contained in H1

so H1 is normal in H0

@north field

[upper cape]

Is it true that M_n(R)/M_n(I)=M_n(R/I)? Where R is a not necessarily commutative ring, and I a two sided ideal.

[north field]

But it says to use 1st iso

[wooden ember]

a hint to prove that K\sub L and L\sub F algebraic extensions => K\sub F algebraic extension?

[chilly ocean]

look at the degree of some a in F over K, try to conclude its finite. Use polynomial over L

[wooden ember]

that's what ive mainly been trying 😭

tried getting rid of the coefficients in L by taking polynomials in K but im struggling

[chilly ocean]

there is f(x) such that f(a) = 0 and f(x) = sum a_i x^i where a_i are in L.

Maybe look at the extension K(\sum a_i)

[wooden ember]

hmm okay havent looked at that extension

god im a moron

of course

lmao thanks for the confirmation

[chilly ocean]

whats your hollow knight completion?

[wooden ember]

112% in regular, 110% in steel soul

[chilly ocean]

damnn

[wooden ember]

havent gotten around to actually doing p5 in steel soul, havent been bothered

it's literally the same as in regular

[broken stirrup]

(b) and (c) ?

[delicate orchid]

Pretty sure this works for b: Take any two vector spaces A and B both with dimension greater than 1 over a field F. Let a_1, a_2 be distinct basis vectors of A and b_1, b_2 be distinct basis vectors of B, then a_1 \otimes b_1 + a_2 \otimes b_2 cannot be written as a single a' \otimes b'

as for c) I'm thinking we take $\bQ \otimes \bQ$ as a $\bQ$-vector space and consider $1/2 \otimes 2 = 1/2 (1 \otimes 2) = 1/2(1 \otimes 1\times 2) = (1/2\times2)(1 \otimes 1) = 1(1 \otimes 1) = (1 \otimes 1)$ Think this works.

[cloud walrus]

Wew "Planks" Tbh

[wooden ember]

just to make sure im not being stupid, if i find a field L in which a polynomial over K splits, then i can directly conclude that the splitting field is K(a_1,...,a_n) where a_1,...,a_n are its distinct roots in L right?

[delicate orchid]

is that not the definition of a splitting field

[wooden ember]

it pretty much follows directly yeah i just dont wanna be dumb

[delicate orchid]

fairs

[chilly radish]

Anyone know where I can find a list of subgroups of $\sfrac \mathbb{Z}/4 \times \sfrac \mathbb{Z}/4$. I looked in groupprops but the page for it didn't list subgroups

[broken stirrup]

have you checked GAP?

[cloud walrus]

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

[broken stirrup]

i believe it can list subgroups

[chilly radish]

i'm not aware of GAP

what is it

[delicate orchid]

computational group theory software

[chilly radish]

oh

is it accessible, cuz I don't really wanna learn a new language just for this one problem

[delicate orchid]

fairly accessible

but your group is small so finding all the subgroups by hand - while tedious - is probably doable

actually since it's abelian I can find all the subgroups (up to isomorphism ) from the character table

[chilly radish]

if u wanna do that I won't stop you

I only need the order 8 subgroups thb

tbh

[broken stirrup]

Imma get done with this assignment then I'll learn representation theory

I'm taking it actually

[delicate orchid]

all my representation theory is self taught lol

[broken stirrup]

where did you learn it?

[delicate orchid]

I did my master's dissertation on it so I learnt during that

[broken stirrup]

I'll write thesis on random walks on graphs

but first i have to study amenable groups

[delicate orchid]

never heard of them

[broken stirrup]

in order to determine whether a group is amenable or not, you link it to a probability problem of random walks on graphs

[delicate orchid]

ah, that'll be why I haven't heard of them

fascinating that groups come up in a probability problem though

[delicate orchid]

I'm pretty sure {(i, j) : i in {0, 1, 2, 3}, j in {0, 2}} and {(i, j) : i in {0, 2}, j in {0, 1, 2, 3}} are the only subgroups of order 8

[chilly radish]

alright sweet ty

[broken stirrup]

lol isn't it too much for a hint?

what else am i supposed to do? am i missing something?

like by hypotesis an ideal must also be a free-module and then you show that every two elements are linearly dependent so there must be only one element that generates it, therefore every ideal is principal and by definition R is PID

[tough raven]

Yes, and the roots don't need to be distinct thought it said the polynomial had distinct roots, sorry.

[chilly radish]

yea, if the roots aren't distinct your notation is just redundant but it doesn't change the field

if you want the minimal generating set you need distinct roots tho

[tough raven]

Oh, then it should be Z/2 × Z/4 and Z/4 × Z/2, since such a subgroup is the same as surjective hom to Z/2 is the same as two homs from Z/4 to Z/2 which are not both trivial.

[chilly radish]

true

nice thanks

[tough raven]

Given an ideal $I$ in a lattice $L$, I'm pretty sure there's a lattice homomorphism $L \to L^\prime$ which is universal wrt making every element of $I$ minimum in the codomain, by taking the quotient of $L$ by the smallest congruence under which any two elements of $I$ are equivalent.
Is there a good concrete description of this congruence? If $L$ is distributive, I think the following works:
$$x \sim y \iff \exists i, j \in I \colon x \vee i = y \vee j.$$

[cloud walrus]

Raghuram

[warm holly]

Could someone clarify please why the algebraic multiplicity of eigenvalue gives the size of Jordan block of the matrix with that eigenvalue on diagonal?

[tough raven]

The Jordan canonical form has the same characteristic polynomial—in particular the same algebraic multiplicities—as the original matrix, as it is similar to it.
For an upper triangular matrix the eigenvalues are the diagonal entries, with algebraic multiplicity of an eigenvalue the number of times it occurs on the diagonal.
Thus the algebraic multiplicity of an eigenvalue of the original matrix is the number of times it occurs on the diagonal of the Jordan canonical form i.e. the total size of all Jordan blocks for that eigenvalue.

[broken stirrup]

that's the homomorphism that is mentioned in Corollary 5.3

[warm holly]

alright, thank you

[tough raven]

One is a homomorphism, the other an element of a tensor product of abgroups of homomorphisms.
However, observing that Corollary 5.3 is bilinear as a function of f and g, there is an induced map from the tensor product to the Hom group which sends the second f (×) g to the first.

[tough raven]

So perhaps the difference is pedantry.

(Cue lots of embarassment if there actually was a significant difference.)

[broken stirrup]

i actually wrote something similar but then i deleted it

[pale jasper]

Could somebody please explain why $\mathbb{Z}_4$ is not a PID?

[cloud walrus]

Scerball

[broken stirrup]

so yeah, in the former one, there is a corresponding balanced function while in the later one not necessarily

[pale jasper]

Or even a hint

[chilly ocean]

what's Z_4?

[pale jasper]

Z mod 4

Sorry

[chilly ocean]

so just 0,1,2,3?

[pale jasper]

Yes

I think I have shown that it is a PID but the answer sheet I have tells me it's not

[broken stirrup]

it's not a field

maybe it meant field?

[chilly ocean]

yeah it is a PID, so reread the sheet maybe

[pale jasper]

Right

[tough raven]

The ideals are principal but it's not an integral domain.

[pale jasper]

How can I prove this is a PID?

[tough raven]

Am I wrong?

[tough raven]

It's not because it's not an integral domain
Which is part of the definition of a PID.

[pale jasper]

Wait so it isn't

Omg

Zero divisors

Thanks

[prisma shuttle]

usually to to prove/disprove things are PIDs u want to look at the chain of types of rings

the chain goes from like most general to most specific

usually the important ones are as follows:

[pale jasper]

I need a diagram of this lol

[prisma shuttle]

rings > commutative rings > integral domains > UFDS > PIDS > Euclidean Domains > Fields

there are some other ones that aren't as well known that you can see here:

https://en.wikipedia.org/wiki/Unique_factorization_domain

but the ones I wrote above are the ones u should remember

becuase they all come up really often

so for example if u want to prove that something is not a PID and u show its not an integra; domain then that is sufficient

[glacial prism]

Its a PIR though. PIRs are just PIDs that aren't D

(Quotients of PIRs are PIRs; Z is a PID so a PIR)

[fickle brook]

if one were to ignore the restriction that p be prime in the construction of the prüfer p-group and set p to 10 (obtaining a group that's isomorphic to the additive group of terminating decimals mod 1), would the result be iso to Z(2^infty) oplus Z(5^infty)?

[wooden ember]

What would something like (e^(2pi*i/2^n), e^(2pi*i/5^m)) with m=/=n be identified with?

[long obsidian]

If I have a field extension whose degree is a composite number n , does the Galois correspondence and sylow theory imply the existence of an intermediate fields whose degree are the primes that divide n

[next obsidian]

No, first you would need to assume your extension is Galois to do that, and you would need subgroups of index p for a prime dividing n which Sylow doesn’t afford you

[south patrol]

This is quite likely a silly question but is there a nice interpretation of commutators and derived series in terms of field extensions via the Galois correspondence?

The 'other' way of viewing solvable groups in terms of composition series w prime factors obviously seems far more natural, what with radical extensions

[chilly ocean]

Does every symmetric function have same coefficients?

[chilly ocean]

It does as far as i understood ....

[chilly ocean]

Polynomial?

Then no

you may want to rephrase your question

[chilly ocean]

Please let me know any symmetric function in F(x,y,z) over F=Q

[prisma shuttle]

$xyz$ works?

[cloud walrus]

JustKeepRunning

[prisma shuttle]

i mean as long as permuting teh variables doesn't chagne anything its symmetric

assuming u are using the usual defintion of symmetric?

[tribal moss]

$\frac{22}{7}xyz^2 + \frac{22}{7} xy^2z + \frac{22}{7}x^2yz$

[cloud walrus]

Troposphere

[chilly ocean]

Please let me know any symmetric function in Q(x,y,z) that doesn't have same coefficients..

[tribal moss]

The sum of mine and JustKeepRunning's.

If it's really Q(x,y,z) and not Q[x,y,z], you'll also get functions like $$\frac{22xy + 22xz + 22yz}{1+2x+2y+2z}$$.

[cloud walrus]

Troposphere

[prisma shuttle]

hey guys can someone help with this problem

Find the zero divisors of $\mathbb{Z}_m[x]/I,$ where I is the ideal of vanishing polynomials over $\mathbb{Z}_m[x]$

[cloud walrus]

JustKeepRunning

[prisma shuttle]

From first iso thm, it follows that \mathbb{Z}_m[x]/I is isomoprhic to the ring of all polynomial functions

And it is well known that every polynomial function uniquely corresponds to a representation $F=\sum_{k=0}^{n-1}b_kX^k,$ where $n$ is the smallest integer satisfy $m|n!$ and $0\leq b_k < \frac{m}{(k!,m)}.$

[cloud walrus]

JustKeepRunning

[prisma shuttle]

so basically i am just trying to find all zero divisors over the set of polynomilas $F=\sum_{k=0}^{n-1}b_kX^k$ with $0\leq b_k < \frac{m}{(k!,m)}$ (i.e. all polynmials in this set for which there exists another polynomila in the set such that their product is a polynomila that vanishes over all elements of $\mathbb{Z}_m$)

[cloud walrus]

JustKeepRunning

[prisma shuttle]

does anybody know how to approach this problem

[fickle brook]

i am tempted to say their product but im not sure tbh

[long obsidian]

If I have a tower of field extensions F/E/K could I get away with calling the dimension of F as an E vector space the codegree of E? I.e. codef(E)=[F:E]

[north field]

is this just first iso theorem

it seems more generalized

[next obsidian]

This is a combination of the first iso theorem, third iso

And also the fourth iso / lattice iso, or whatever you call the theorem that tells you subgroups of G/N are the same as subgroups of G containing N

[north field]

bruh

i hate how lang explains it i dont understnd

[next obsidian]

Apply the first iso so you replace G’ with G/K for K the kernel

Take H’ a normal subgroup of G/K, by that fourth iso or whatever you know H’ is of the form H/K for H a normal subgroup of G containing K

The inverse image of H/K is just H

Then G’/H’ = (G/K)/(H/K) ≈ G/H by the third iso

[north field]

Why cant he

explain that shit

Before

he says this

and when he uses them

why cant he say that

is this 2nd iso

[next obsidian]

Yeah

[north field]

ffs

[vestal snow]

Does anyone have the full version of MAGMA? I have the student version (which has a memory limit) and I want to run some computations for research

I have the code, all you would need to do is plug the code in and send me what it returns

[north field]

when one writes $HN$ for two groups do they mean ${ hn : \forall h\in H, \forall n \in N }$ if $H<G$ and $N<G$

[cloud walrus]

nonagonclass

[north field]

@next obsidian

[broken stirrup]

yes

but note that this is not always a subgroup

[north field]

yes ofc

[broken stirrup]

if one of them is a normal subgroup then it's a subgroup and equivalent to H V N which is a subgroup generated by H and N

[north field]

ok so $HN={e, h_1,\cdots, h_{|H|-1}, n_1,\cdots n_{|N|-1}, h_1n_1, h_2n_1, \cdots, h_{|H|-1}n_{1},\cdots, h_1n_{|N|-1},\cdots h_{|H|-1}n_{|N|-1} }$

oops

[cloud walrus]

nonagonclass

[north field]

@broken stirrup

[broken stirrup]

Basically, you have identity, everything in H, everything in N and every possible product of elements so yes

[north field]

so there might not be inverses for the $hn$'s?

[cloud walrus]

nonagonclass

[north field]

where $h,n\neq e$

[cloud walrus]

nonagonclass

[north field]

since

ig

$(h'g')(hg)$

[cloud walrus]

nonagonclass

[north field]

uh

g' could be the inverse for g

hmm

[broken stirrup]

generally no, but if either one of them is normal then NH=HN

[upbeat fulcrum]

can anyone help me with this proof? kinda stuck here

[south patrol]

Are you allowed to use that F^x is cyclic? (just checking in case this leads to that or smth)

[upbeat fulcrum]

yes

[south patrol]

Sure. Well this reduces to asking - in Z/nZ (since any finite cyclic group order n isomorphic to Z/nZ) , how many x (mod n) are are there such that xm = 0 mod n

That should give you gcd(m, n)

[upbeat fulcrum]

can you help me start the proof so i know what should I do next?

[south patrol]

Hm well with the reduction to Z/nZ I'd say that if g is a generator of F^x then (g^a)^m = 1 iff am = 0 mod n where n = |F^x|

Now from there it is a little calculation and I can't really give a hint without just giving the answer

[upbeat fulcrum]

wdym? i cant tell kinda lost here

@south patrol im really having a hard time to prove this one ngl this is a pain in my ass

[south patrol]

Hm well consider pairs (x, y) such that xm + yn = 0

Idk if you know the general solution to that

[upbeat fulcrum]

can you show me the proof so I can study it?

[simple mulch]

So, there exists mZ = {..., -m, 0, m, ...} but is there something like mZ/nZ ?

So if m = 2 and n = 8, 2Z/8Z = {0,2,4,6}

is this ok?

[vast quiver]

yep! but you’ll need m to divide n for this to make sense. this is actually a special case of the 4th/lattice isomorphism theorem. Not only is mZ/nZ is a subgroup of Z/nZ when nZ is a subgroup of mZ (ie m divides n), but these are the only subgroups.

[kind temple]

5Z/3Z = {0, 1, 2} doesn’t make sense because?

[delicate orchid]

3Z is not a subset of 5Z let alone a subgroup

how could quotienting possibly make sense

[vast quiver]

well I think it’s a matter of convention/notation at that point

[kind temple]

why can i not just quotient by the equivalence a ~ b iff a mod n = b mod n?

[vast quiver]

i think that you make a good point c squared

but in the more abstract setting people will write NH/N rather than H/N when H doesn’t contain N

or you could call pi the projection map and say pi(H)

[delicate orchid]

2nd iso theorem moment

[vast quiver]

or shoot

er no i’m good?

sorry I haven’t done groops in a bit

[kind temple]

i see. so this would be nZmZ/nZ?

[vast quiver]

well Z is an additive group

so you would write (nZ + mZ)/nZ

and nZ+mZ is just gcd(n,m)Z

[kind temple]

thanks!

[vast quiver]

yup!

[delicate orchid]

I'd like to clarify that no, it is not a matter of convention/notation - you must have a normal subgroup for a quotient to make sense, there's a correspondence between equivalence relations that maintain the group structure and normal subgroups

you can do it as sets, sure

but the result will not be a group

[sly crescent]

When are two Lie algebras considered to be isomorphic?

[chilly ocean]

when there's an isomorphism of the underlying vector spaces that preserves the brackets