groups-rings-fields

[chilly ocean]

but what is the question asking for

[ivory dust]

lol no need to be rude when you are asking others for help, no one owes you anything 😐

[dusty river]

Find all functions satisfying that equation

Functions meaning group homomorphisms here

[chilly ocean]

all functions

[dusty river]

I guess that works too actually

[chilly ocean]

i don't think you understand the question

[dusty river]

[gritty sparrow]

Why do you think that?

[chilly ocean]

fuck

<@&268886789983436800> profanity in this channel

[dusty river]

Deez nuts jokes are not against the rules it's ok

[stone fulcrum]

Yikes

[upper pivot]

They don’t just want group homos here i think.

[obsidian sleet]

as it stands profanity is not against the rules

[chilly ocean]

covering your ass i see

[upper pivot]

Someone post an appropriate copy pasta.

[ivory dust]

im getting unsolicited dms from them 😩 isnt that against the rules

[maiden ocean]

Who?

[ivory dust]

Queen's Purple Daisy

lol

[maiden ocean]

Dm modmail with a screenshot

[stone fulcrum]

Block if reasonable

[ivory dust]

keep bugging me to help them when I told them they were rude and should learn manners before asking for help

[maiden ocean]

Yeah send a screenshot

[stone fulcrum]

Did you message them first? Kek

Called it

[upper pivot]

This isn’t exactly unsolicited lol

Anyhow the drama s clogging up the channel prolly move somewhere else.

[ivory dust]

Yea guess not they added me i msged them asking why

[chilly ocean]

[wooden ember]

I’ve done all the parts of the hint except that psi is a bijection. I can show this for the infinite case but for the finite case I need to show (a,|K|)=1 and all I can show is (a,|phi_1(K)|)=(a,|phi_2(K)|)=1. I also see that I can pick any other a’ as long as it’s congruent to a mod |phi_1(K)|. I tried pigeonholing by generating successively new divisors of |K|/|phi_1(K)| by adding multiples of |phi_1(K)| and taking gcds with |K| but it doesn’t quite work. Tried other methods that don’t involve a 2-sided inverse and they don’t work either.

[warm saffron]

This question is asking you exactly what it states lol. We want to find all functions f that satisfy the given conditions. The solution simply makes use of what it means for h to be in G=<g> and what it means for the rest of the given conditions to be true.

@chilly ocean Please don't post anything if you're not going to be polite to people who are simply trying to help you.

[mild laurel]

@wooden ember yeah this is weird, cause its definitely not true that any choice of a will make psi bijective. But by using what you note about being able to choose any other a', you can always choose some other a' mod |phi_1(K)| so that gcd(a',|K|) = 1

[warm saffron]

word of advice....if their dms are annoying you can just block em especially since they weren't dming you with anything helpful and they were clearly being impolite beforehand.

[wooden ember]

But why? How do I show this

[mild laurel]

@wooden ember Convince yourself that you can reduce to the case where |phi_1(K)| and |K|/|phi_1(K)| are coprime. In that case, you can look at the system of equations x = a (mod |phi_1(K)|) , x = 1 (mod |K|/|phi_1(K)) and use the Chinese remainder theorem

[wooden ember]

Hmm alright I’ll think about how to reduce to that case

[blissful mauve]

hey guys Im trying to follow this from a YT video. Could anyone help me in solving this types of proofs?

for example Im not sure how to show that all the real numbers with addition are indeed a group

[nova plank]

Well, addition of real numbers is associative and commutative, and 0 is the identity element, and the inverse element of x is -x

So R is a commutative group under addition

[chilly radish]

Wth

They're defining a group as abelian

Why

[viscid pewter]

?

[chilly radish]

They're saying a set with a binary operation is called a group if it satisfies all group axioms + commutativity

[viscid pewter]

no

[chilly radish]

That's what the image is saying bro

[viscid pewter]

'the group is called a commutative group (Abelian group) if we have additionally that'

[chilly radish]

Oh

Shit

Didn't see that sentence

The fact that they're using additive notation for a grouo that's not necessarily abelian also threw me off

My bad

[viscid pewter]

yeah that's fair

[chilly ocean]

If R is a noetherian ring, and A an R-Algebra of finite type, why is then also A noetherian?

Is maybe A isomorphic to some polynomial Ring in finitely many variables over R, and then by Hilberts Basissatz A is also noetherian?

[wooden ember]

Do you think you could elaborate on “reduce”? I’m feeling pretty dumb…

[thorny flame]

what would be the extender version of this group?

[upper pivot]

Almost. A is iso to a quotient of some polynomial ring in R

And then as you said hilbert basis theorem.

[chilly ocean]

Thanks!

[frank fiber]

you can say that the real numbers by definition is a field and so is an abelian group under +

@blissful mauve

[unreal portal]

If they're learning the definition of a group, I imagine they're not allowed to use the fact that R forms a field

[south patrol]

Ye just showing it directly is probs best

[vocal wolf]

Hi guys. If we let $\mathbb{R}$ be the additive group of real numbers and $\mathbb{A}$ be a subgroup generated by an irrational number, the second isomorphism theorem says(?) $\frac{\mathbb{A}+\mathbb{Z}}{\mathbb{Z}} \cong \frac{\mathbb{A}}{\mathbb{A}\cap\mathbb{Z}}$

[cloud walrus]

Brian485

[vocal wolf]

the second group is obviously isomorphic to Z but im not sure if the first one is because A+Z is dense in R and clusters around 0. Whats wrong with my reasoning?

[void cosmos]

what

[vocal wolf]

what do you mean?

[south patrol]

Yeah idk, is A+Z really dense in R?

If A is, say, generated by k then A = kZ, right?

[vocal wolf]

yep

yeah its here in my textbook A+Z is dense

[south patrol]

Yeah sure, I see that now ig

[urban acorn]

A + Z is absolutely dense

[south patrol]

Lmao yeah

I was dumb

[urban acorn]

it's a nice consequence of compactness of the circle

[south patrol]

how so?

ig makes sense intuitively as módulo one you'll cycle through the entirity of [0,1]

[urban acorn]

it's equivalent to saying that (the image under the projection of) A is dense in R/Z, which is topologically a circle

because x is irrational, all numbers of the form nx are distinct mod Z, and thus form an infinite set in the circle, so by compactness they must have a limit point, and then they get arbitrarily close to that limit points, so you can consider the subgroups generated by elements of the circle arbitrarily close to the identity, which get arbitrarily close to every element

[south patrol]

Oh sweet

Cheers

Do we not have that A+Z is isomorphic to Z^2 though, which would make sense of the above thing Brian shared?

Maybe im being thick though but that's clearly at least a bijection

[urban acorn]

yeah, I believe A + Z is isomorphic to Z^2

so, send (n, m) in Z^2 to nx + m

[south patrol]

yeah exactly

[urban acorn]

by definition A + Z is the image of this

and if this wasn't faithful, x would be rational

[vocal wolf]

mhm mhm

aah so then do we have A+Z/Z isomorphic to Z^2/Z = Z?

[urban acorn]

yeah, sure

[vocal wolf]

is that whats going on?

[urban acorn]

yes

side note

I like the fact that in abelian groups, the join of subgroups is precisely the element-wise sum

[oblique river]

Sorry to jump in but given that we’re treating A and Z as separate groups

I think it makes it more clear to say that (A+Z)/Z is iso to A

[urban acorn]

I sort of see what you mean

by they're trivially isomorphic

we only distinguish them with respect to their embedding in R

[oblique river]

Similarly, A/(A cap Z) is iso to A because A and Z are disjoint except for 0

[urban acorn]

even their embedding with R is equivalent up to automorphism of R, it's about the simultaneous choice of embedding of each

[oblique river]

Im just saying that in the context of this question, (A+Z)/Z is iso to A regardless of whether or not A is Z

[urban acorn]

but your point still stands, yeah

[oblique river]

Yes i like this way of phrasing

[vocal wolf]

thanks so much guys! what still bugs me is how A+Z is dense, won't that mean A+Z/Z cant have a generator? cause you can always find a smaller element

[oblique river]

No, as we’ve established, that group is cyclic and therefore has a generator

It’s just the group A again, so whatever the generator of A is

[urban acorn]

A + Z is dense in R, but that doesn't mean that you can "find a smaller element" within (A + Z)/Z

the quotient there is very important

[oblique river]

I think youre confused because youre trying to hold onto this “ordering”

But you cant order R/Z

[urban acorn]

yeah the ordering breaks down within the quotient

[vocal wolf]

right right, forgot things "wrap around" in A+Z/Z

[oblique river]

For example 0.25 < 0.5 < 1.25 even though 0.25 = 1.25 in R/Z

[urban acorn]

yes

[oblique river]

You cant talk about bigger or smaller

[vocal wolf]

thanks again guys

[urban acorn]

hmm

the people who asked for help wrt this question shouldn't worry about it

but

i was wondering

there's this concept of a "cyclic ordering"

[oblique river]

What is “cyclic ordering”

[urban acorn]

essentially it's a ternary operation where [a, b, c] means proceeding from a, we get to b before c

and I'm wondering how that would interact with quotients of - say abelian ordered groups by cyclic subgroups

probably very nicely

[oblique river]

Hmm it seems reasonable but im still maybe a little suspicious

[urban acorn]

i.e. intuitively

it should even generate the circle topologically

you can do that anyway by gluing the cosets of Z

supporting the intuition that the algebraic structure naturally induces these quotient objects

[urban acorn]

but yeah i understand your suspicion, i personally didn't bother with verifying the formal construction of the cyclic ordering, I was just like "yeah seems reasonable"

i just happened to encounter it because of some guy on reddit

and i thought "huh, I wonder if that applies here"

if anyone wants to, you can google the definition, try to verify that these quotients do naturally have a cyclic orderings, and that it respects the group operation

[chilly radish]

Hi i'm back again to check that i'm not dumb. Suppose you have a short exact sequence of chain complexes of free abelian groups
$$0 \rightarrow S' \stackrel{i}{\rightarrow} S \stackrel{p}{\rightarrow} S'' \rightarrow 0$$
Such that $S',S''$ are acyclic, and I wanna show that $S$ is acyclic.
So we know from the exact triangle that this sequence induces an exact sequence of homology groups
$$\begin{tikzcd}
\ldots & {H_n(S')} & {H_n(S)} & {H_n(S'')} & {H_{n-1}(S')} & \ldots
\arrow[from=1-1, to=1-2]
\arrow["i", from=1-2, to=1-3]
\arrow["p", from=1-3, to=1-4]
\arrow["d", from=1-4, to=1-5]
\arrow[from=1-5, to=1-6]
\end{tikzcd}$$
Where $i,p$ are the induced maps and $d$ is the connecting homomorphism. Note that since $S',S''$ are acyclic, for every $n \geq 0$ we infact have the exact sequence
$$\begin{tikzcd}
0 & {H_n(S)} & 0
\arrow["i", from=1-1, to=1-2]
\arrow["p", from=1-2, to=1-3]
\end{tikzcd}$$
So it follows that $H_n(S)=0$ and so $S$ is acyclic as required

[cloud walrus]

ShiN

[fossil shuttle]

yep!

[chilly radish]

yay

[frank fiber]

what is a pollack group?

[mild laurel]

If |phi_1(K)| and |K|/|phi_1(K)| share a common prime factor p, then since a and |phi_1(K)| are coprime, you know that p doesn't divide a, a + n|phi_1(K)| will always not be divisible by p. Thus, you can look at |phi_1(K)| and |K|/p instead

[cursive temple]

for some reason i just cant wrap my head around why s/1 should be a unit

can i just pick 1/s to be its inverse

seems too easy

also cant i run into some zero divisor shenanigans

[mild laurel]

yes you can pick 1/s to be the inverse, but you need to confirm that s/s is equivalent to 1/1

I'm not sure what you mean by zero divisor shenanigans, the problem discusses that by saying that its a monomorphism iff no element of S is a zero divisor

[cursive temple]

So by the defn of equivalence i need to show u(s1-1s) = 0 for some u

Why doesnt this show that every a/1 is a unit in RS^-1

[oblique river]

if a isn't in S then 1/a doesn't exist in RS^-1

[cursive temple]

Good lord

Yeah im going to sleep

Thank you

[oblique river]

sleep well

[chilly ocean]

Don't let the bed bugs bite!

[crimson falcon]

Is the a general "strategy" to find stuff like maximal ideals in particularly common rings? For example, on polynomial quotient rings R[x]/f(x)? Or everything is case by case basis?

[urban acorn]

there's gotta be something about rings of that particular form because they're incredibly important

especially when R is a field, and if it's algebraically closed on something

and they should be very well behaved

One strategy I can see off the top of my head when F is a field is to consider the natural projection $\pi : F[x] \to F[x]/(p)$, then for every ideal $I \subset F[x]/(p)$, we have that $\pi^{-1}(I)$ is an ideal of $F[x]$, and the last is a PID.

I think that would mean that all ideals are given by a polynomial that divides p, or something like that, I'm too tired to make sure.

Another strategy is for a maximal ideal M, consider the field that is the quotient by it. What could it be?

[cloud walrus]

All groups are abelian

[crimson falcon]

ohhh yeah that should do nicely when you can factorize the polynomial on something nice and simple

[urban acorn]

funny enough, like how I initially thought this would be nicely behaved on algebraically closed fields, their polynomials factor into c(x1-k1)...(xn-kn)

okay, so I've been thinking about this a little, and you asked in particular about R[x]/(p), but let's think about R[x]/I when R is an integral domain

there two important kinds of maximal ideals: those that contain a non-zero constant, and those that don't

I think that the first of these precisely consists of the ideals of the form (J, x) when J is a maximal ideal of R

And I think the second of these has something to do with fields containing R and solutions to all polynomials in R

[urban acorn]

oh, I forgot something, I was talking about ideals of R[x], what I didn't mention is that maximal ideals of R[x]/I are precisely the images under the projection of maximal ideals of R[x]

[crimson falcon]

And if K is an integral domain, and we have p irreductible, then the ideal generated by p is maximal if i undertood correctly

[urban acorn]

Maybe you're right, I know this in the case k is a field, then it's because k[x] is a PID

[crimson falcon]

ah yeah, in my syllabus says for integral domains "maximal among principal ideals"

but if we have PID that's all ideals

[urban acorn]

I think "maximal among principle ideals" means there's no principle ideal strictly between it and the whole ring.

[crimson falcon]

yeah that's my take too

[urban acorn]

so we probably don't have that (p) is maximal

that felt wrong anyway

okay, so

[crimson falcon]

Unless we're in a field, in which case all ideals are principal no?

[urban acorn]

yes, but it's not generally true for integral domains

[crimson falcon]

yup

[urban acorn]

let R = Z and p = x^2 + 1, p is clearly irreducible

but Z[x]/(p) is the Gaussian integers, which are very cool, but certainly not a field

and thus (p) isn't maximal

okay, so, let I be a maximal ideal of R[x] and let k in I be a non-zero constant. Then let J = (k, x). I think we can prove (k) is a maximal ideal in R and I = J.

let's first prove J is proper and contains I. then since I is maximal this gives I = J. We can show J is proper because 1 isn't in J. It can't be expressed as kp + xq.

[crimson falcon]

J = (k, x) not familiar with that notation sorry :p but : https://math.stackexchange.com/questions/504551/image-of-a-maximal-ideal/504604
Images of maximal ideals under surjective morphism (and between commutative rings) are either maximal or the entire ring

And the canonical projection to the quotient is surjective

[urban acorn]

(k, x) is the ideal generated by k and x

[crimson falcon]

ah gotcha

[urban acorn]

and I mean in R[x] with k considered as a constant

[urban acorn]

I'm trying to figure out how to prove J contains I

cause if it's not true, my theory about what the maximal ideals look like will be wrong

okay, it's not true, but in a way such that the overall proof is salvageable

[next obsidian]

Are you trying to classify primes of Z[x]?

[crimson falcon]

(imma be honest, following at 80% here). But, for fields at least, gathering all i get that the ideals generated by irreducible polynomials are maximal, and their image under the canonical projection to any quotient ring is either the entire quotient ring or maximal.

[urban acorn]

not primes, maximal ideals

[next obsidian]

Ah

That takes you most of the way to classifying prime ideals

This is kinda hard

[crimson falcon]

(psss: I'm just nodding my head along)

[next obsidian]

At the very least they can’t be principal

I think you’re trying to figure it out yourself so I won’t spoil it

[urban acorn]

yeah

well, two things (1) we're actually thinking of a general integral domain, (2) I'm not looking for a complete classification per se, I just have a theory that I want to prove that classifies them into two types, shows that one is the ones of the form (k, x) in R[x] where (k) is maximal in R, and then just give a certain condition on the ideals of the other type

[crimson falcon]

Honestly, I was just doing some very basic exercises about finding maximal ideals on some simple rings, and i realized that i didn't had any "main ideas". You know, like 'check this, and check this, and then do this" and had solved the few i had done because there was some easy characteristic

But this turned out to make me go ahead and review a lot so, very productive discussion xD

[urban acorn]

yeah then they asked here for general strategies for finding maximal ideals of rings of nice forms, and I started thinking about it, which lead me to what I'm doing now

[next obsidian]

This will be false

[urban acorn]

It's too vague for it to be false.

[next obsidian]

Mainly because of dimension

I mean the maximal ideals being (k,x)

[urban acorn]

I didn't say that.

I just have a theory that I want to prove that classifies them into two types, shows that one is the ones of the form (k, x) in R[x] where (k) is maximal in R, and then just give a certain condition on the ideals of the other type

[next obsidian]

Oh, I see

[urban acorn]

I think that condition will have to do with fields that contain R

[next obsidian]

I don’t think you’ll get a satisfying condition looking at trying to classify maximal ideals of R[x] over general integral domains

They could have arbitrarily many generators

[urban acorn]

Yeah, what I'm looking for is less of a classification and more of a partial description

what I think though, and don't spoil for me whether this is true

is that the ideals of the form (k, x) will be precisely those that contain any non-zero constant

[next obsidian]

So

A maximal ideal which contains a non-zero constant

Has a decomposition like (k,x)?

[urban acorn]

I think it's probably true, yes, but don't spoil it for me yet.

[next obsidian]

Ok

[urban acorn]

oh wait, not quiet

I can see that this is wrong but I know how to fix it

replace k, an element, with a maximal ideal of R

[next obsidian]

Yeah

This sounds potentially true now

[urban acorn]

so it reduces to (k, x) when R is a PID

[next obsidian]

I’m not sure

But your initial conjecture was wrong

Dunno about the revised one tho

[urban acorn]

Yeah, I still have it all improperly organized in my head.

[urban acorn]

I think I thought of something like this earlier then lost the part where it's any maximal ideal

[next obsidian]

I am mainly thinking about polynomial rings over Z because those are easy ways to get examples

[urban acorn]

okay, so let I be a maximal ideal of R[x]. let phi : R[x] -> x be the evaluation at 0 homomorphism

[next obsidian]

But those are also maybe too well-behaved to have the results there generalize for arbitrary integral domains

[urban acorn]

We wanna see if either phi(I) = 0 or phi(I) is a maximal ideal

if there was an ideal J properly inbetween phi(I) and R, phi^-1(J) should be properly inbetween I and R[x], no?

yes okay so phi(I) is indeed a maximal ideal, but I now think that's not quiet what I should've been looking for

[next obsidian]

Set theoretically there’s no reason it should be proper in R[x]

You’d have to argue using the properties of phi

[urban acorn]

alright

[next obsidian]

I and phi^-1(J) I mean

Errr

Actually nvm

[urban acorn]

yeah, just let y be outside of J

[next obsidian]

I think you’re fine because phi is surjevtive

[urban acorn]

okay yeah that's not what I want to think about right now, I think this is what I need:

The intersection of I with the constants.

If I'm correct, either it's 0, or it's a maximal ideal.

Indeed it has to be an ideal, because it's the pullback of I under the inclusion of R in R[x] as constants

okay, I think this is the idea that I'll need:

Let I be a maximal ideal of R[x] that contains a non-zero unit, and let J be a maximal ideal of R containing the preimage of I under the inclusion of R in R[x].

Can we then show that (J, x) contains I?

It contains everything with constant term in J, and R only has constant terms in J.

and it's proper since J is maximal and thus proper in R

so since I is maximal, I = J

Am I missing something?

[urban acorn]

oh no, this probably isn't true

every constant polynomial is in J, but that doesn't mean that every polynomial's constant term is in J

[urban acorn]

okay, maybe I was wrong with this statement

[crimson falcon]

(i tapped out, way ahead of me)

[urban acorn]

yeah lol

I myself am struggling with trying to figure out what's happening, and apparently this is a very hard problem, I'm largely thinking out loud

if I do get a result I think you'll find interesting I'll @ you in a format that's more easily digestible than hearing me think out loud before I got it myself, if I even will figure this out

right now I'm thinking of (4, x+2) in Z[x]

[crimson falcon]

I did find however that you were right on the thing right at the beginning about the ideals being by the factors of the polynomial

[urban acorn]

Oh, yeah.

In the case you're looking for ideals of F[x]/(p) with F a field

[crimson falcon]

https://tenor.com/view/caddyshack-i-got-that-going-for-me-bill-murray-nice-its-something-gif-5467327

[chilly ocean]

@next obsidian can you describe in what ways you are so mommy?

[urban acorn]

@next obsidian I think I was wrong, I actually think (2, x+1) in Z[x] is a counterexample.

I think I can show (2, x+1) in Z[x] is maximal with an argument involving (x+1) in (Z/2Z)[x] being maximal

[next obsidian]

I got assigned this by moth

[urban acorn]

but even without that

[next obsidian]

After I got “mommy moth” in some buzzfeed ass “what moth are you?” Quiz we all took

Yeah

@urban acorn

[urban acorn]

even if it wasn't maximal, there's a maximal ideal that contains it

[next obsidian]

It is maximal

You end up with F_2

[urban acorn]

yeah

[next obsidian]

By exactly the calculation you did

[urban acorn]

the picture is indeed much more complicated than I expected

[next obsidian]

Yeah

You can replace x with f(x) though

[urban acorn]

maybe it would be fun to try to classify the maximal ideals of Z[x] containing a non-zero constant.

[next obsidian]

Do you want a spoiler?

[urban acorn]

alright, hit me

[next obsidian]

They literally all do

[urban acorn]

huuuh

as in, all the maximal ideals contain a non-zero constant?

[next obsidian]

Yes

A prime number

[urban acorn]

wow

[next obsidian]

Proving this is kind of a pain

[urban acorn]

even more wow

[next obsidian]

This is the hardest part of classifying prime ideals of Z[x]

[limpid edge]

what's so special about vector spaces over fields with nonzero characteristic

apparently shit is supposed to break?

[next obsidian]

If you have char p

And you add something to itself p times it dies

It can make things kind of awkward

[limpid edge]

oh that is pretty awk

norm doesn't make sense too i think

[next obsidian]

I’m sure there’s other stuff

[urban acorn]

whenever you have a vector space over a field of characteristic 0, you really at the very least have a vector space over Q

so a lot of structure which feels general is really just Q

so when you lose Q, you lose a lot

I think there was something with representations I don't quite remember, but I remember where to find it

[next obsidian]

There is

You don’t want to be divisible by the characteristic in order to get like… Artin-Wedderburn?

Altho I guess that’s group rep theory

It’s something about p not dividing |G| then some result holds

This is what makes rep theory over C for example so nice, but conversely it means rep theory is only intersetinf in positive characteristic

[urban acorn]

oh yes

that is familiar

representation theory over C is also nice because it's algebraically closed I think

[next obsidian]

Yeah

[limpid edge]

look how based axler is

[next obsidian]

I mean it’s C

[limpid edge]

casually throwing this in there at the end of chapter 1

because students will definitely understand this

[urban acorn]

I mean, it's a good pedagogical approach to just work over R and C when teaching linear algebra to undergraduates

[limpid edge]

i guess since he brought up fields he kind of had to include that

but it still could have been brought up later

or he could've developed fields like a normal person

[urban acorn]

and they are told "later you will learn that R and C are these things called 'fields', and you can in general define a vector space over any field. most of the work you'll do over R and C holds in general, except some fields that have this property that 1 + ... + 1 = 0."

[limpid edge]

big picture moment

[urban acorn]

you know R and R^2 are isomorphic as additive groups?

[limpid edge]

no i did not know that

[urban acorn]

They are vector spaces over Q.

And a cardinality argument shows that they have the same dimension.

[limpid edge]

wait uh

these vector spaces are inf dimensional?

[urban acorn]

yes

[limpid edge]

oh ok

[urban acorn]

but they still have bases, so you can say their "dimension" is a transfinite cardinal encoding how big their basis is

if these cardinals are the same, you have a bijection between these bases, and that extends to a linear isomorphism between them

[limpid edge]

ok so the general definition of isomorphic vector spaces is dimension having the same cardinal

well equivalent to the map preserving stuff

but yea

[urban acorn]

well, the definition would be the existence of a linear map that is a bijection

but that is equivalent to requiring the existence of a bijection between the bases

and by definition of cardinality, that is equivalent to saying that the cardinalities of the bases are identical

[limpid edge]

i see

where do people pick up set theoretic stuff like this at uni

because most (first year) linear alg courses aren't talking about cardinality?

because no way everyone takes a set theory course

[kind temple]

not to butt in, but i very much dislike this fact

axiom of choice fuckery

[urban acorn]

cringe

axiom of choice is clearly the right thing to do

you don't even need axiom of choice to make them vector spaces over Q, you just need it to produce bases

but here's the thing

the reason this is icky for you is that there's more to R and R^2 than their Q-space structures

so it's morally wrong to consider R and R^2 as mere Q-spaces

for example, this additive isomorphism between R and R^2 is clearly not continuous

yo BanAnaMan is here

[oblique river]

yeah

my "bad takes about aoc" sense was tingling

[limpid edge]

why do you need aoc again

[urban acorn]

LMAO

[limpid edge]

when are you taking an infinite product of sets

[urban acorn]

always

[oblique river]

what do you mean

[limpid edge]

in the construction of the basis

like they're all just there

why do you gotta cartesian product them

[oblique river]

where are you taking an infinite product when talking about R and Q?

[urban acorn]

it's just absurd to say that you can make a choice of element for each of a collection of set, but there is no simultaneous choice of elements out of each.

[limpid edge]

oh uh

i thought c was ~c

[urban acorn]

lol

[limpid edge]

classic mistake

(not classic im just an idiot)

[urban acorn]

lmao

[limpid edge]

i read some thing nami recently posted about if you accept ~c then there should exist an infinite product of sets which is empty

and that fucked me up

now im good tho

[oblique river]

yes, axiom of choice is equivalent to "the cartesian product of every family of nonempty sets is nonempty"

[crimson falcon]

For half a second i was like "Why is someone talking about ocasio cortez in abstract algebra"

[kind temple]

if G is a group and H is a normal subgroup of G, is the direct sum of G/H and H isomorphic to G?

[urban acorn]

No. For example, take S3 and the subgroup <(1 2 3)>.

Or D_n and the subgroup <r>.

[proud bear]

Z/4Z also works

[urban acorn]

which this is actually a special case of.

[urban acorn]

oh, that's right

I don't know how I missed that simple example.

[kind temple]

ok. thanks guys. i couldn't think of a quick one of the top of my head

[urban acorn]

Z4 is definitely the nicest example

[kind temple]

hmm. why does Z/4Z work as a counter example?

[gritty sparrow]

Z/2Z can be embedded as a subgroup, and the quotient will be isom to Z/2Z. But Z/4Z is not the direct sum of Z/2Z and Z/2Z

For completeness, if {1,x,x^2,x^3} is Z/4Z then {1,x^2} is isomorphic to Z/2Z

[kind temple]

oh. thought that Stain meant Z and 4Z, so that Z/4Z + 4Z was not isomorphic to Z

but i guess both interpretations work, since Z/4Z + 4Z is also not isomorphic to Z, right?

[gritty sparrow]

Yes

[kind temple]

the reason why Z/4Z + 4Z is not isomorphic to Z is not obvious to me : (

it just doesnt feel like they should be isomorphic

[gritty sparrow]

(1,0) should be an element of order 4, this wouldn’t exist in Z

[kind temple]

whoops i need to brush up on my group theory lol

[uncut girder]

I need help with part a

How do you show A is closed under inverses ?

[upper pivot]

So if you have shown the identity is in A

Then it is also in xA

[uncut girder]

I have not shown the identity is in A

Oh I have to use the fact that A is a subset of a group

[carmine fossil]

You can choose an x in A and find an a such that xa=x

Now see x and a as elements of G and you get a=1

[next obsidian]

I don’t get your question tbh

[stone fulcrum]

Painful sigma notation is the way to go

[next obsidian]

Just take two arbitrary elements, it has n terms

Add them and you’re fine

If you multiply you introduce higher degree stuff but you can replace that with lower degree terms using the fact that p_alpha(alpha) = 0

[fickle tapir]

i dont follow this part

[next obsidian]

You can write alpha^n

In lower degree terms of alpha

Via that polynomial

Something like

Alpha^3 = 2alpha^2 + 45alpha -7 or something

This lets you just inductively take any monomial alpha^k for k >=n and turn it into some polynomial in alpha of degree < n

[fickle tapir]

via p_alpha(x) right?

[next obsidian]

Yes

[fickle tapir]

i guess the problem is im not fully wrapping my head around that polynomial

[next obsidian]

It just is a degree n polynomial which alph satisfies

Plug in alpha and you have like
alpha^n + lower degree terms = 0

Just move half of it to one side

[fickle tapir]

hm yeah

Lets say x, y in Q(alpha)
Then x = b_1 + b_2a + ... + b_{n-1}a^{n-1}

[next obsidian]

Yeah

And y is similar but like with c_i

Just bash out what their product is then reduce all the alpha terms with too large a power

[fickle tapir]

$x = b_1 + b_2\alpha + ... + b_{n-1}\alpha^{n-1}$

[cloud walrus]

add

[fickle tapir]

$y= c_1 + c_2\alpha + ... + c_{n-1}\alpha^{n-1}$

[cloud walrus]

add

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

[fickle tapir]

for representing something like this wouldnt sigma notation be useful?

i may be overthinking this but this could be represented as some double sum i think

[next obsidian]

It’s a double sum

Just describe the coefficients on alpha^i

I’ll leave you to the rest, you should be able to finish

Honestly

You don’t even need to write down the product

Just say

xy =
d_0 + … + d_{2n-2}alpha^{2n-2}

Just introduce new variables, the coefficients literally don’t matter

Then using an inductive lemma prove this can be rewritten only using alpha^0,…, alpha^n-1

The exact coefficients don’t matter since they can just be whatever

[fickle tapir]

oh ok

would the multiplicative inverse be the hardest part to show here

[next obsidian]

Probably

that one involves bashing it out

That’s the only way I can think to do it at least

[fickle tapir]

what does bashing it out mean specifically?

[next obsidian]

Manually come up with the inverse

In terms of the coefficients of x

For that you’ll probably have to manually write out the product because now the coefficients matter

[fickle tapir]

oh ok

[wooden ember]

I ended up doing it differently but thanks

[crimson falcon]

da moment when you see see "uh, this should be easy" and turns out you're trying to rediscover the Chinese reminder theorem (but you're a bad student and dont remember bezout )

[crimson falcon]

So, is K[X]/fg isomorphic to K[x]/f x K[x]/g? if i understand things correctly, the above theorem says that's the case, no? Except maybe if fg are not coprimes?

[delicate orchid]

yes that's the case

one of the conditions of the Chinese remainder theorem is that f+g must be the entire ring, which is only true if a unit is in f+g so they must be coprime

[crimson falcon]

ah yes, thanks, last part was escaping my mind

[fickle tapir]

Also I have shown that If the roots of the cubic equation x3 + bx2 + cx + d = 0 are
r1, r2, and r3, then b = −(r1 + r2 + r3). where these roots may be the same as each other

since the equation has a rational root r1 = m/n and has r2, r3 real roots
then b = -(m/n + r2+ r3) is also constructible. Then must r2 and r3 be necessarily constructible since the constructible numbers are in a field? I'm not sure if this suffices

another suggestion i was given is to factor out the rational solution, but im not exactly sure what to do with that

[oblique river]

I would give the same suggestion.

Can you show that the coefficients of the remaining quadratic are also constructible?

[cloud walrus]

add

[oblique river]

no

you shouldn't have an x in the denominator...

you're supposed to be factoring out (x-r) where r is the rational root

not just x

[fickle tapir]

oh i see. x^3 + bx^2 + cx + d = 0, in factored form is 0 = (x - r1)(x-r2)(x-r3)
and lets say r1 is the rational root

then factoring it out gives us 0 = (x - r2)(x- r3)

[oblique river]

that's right, but now i'm suggesting that you try to express the coefficients of that remaining quadratic in terms of b, c, d, and r1

[fickle tapir]

$0 = x^2-r_2x - r_3x + r_2r_3$

[oblique river]

so that you can prove that they're also constructible

[cloud walrus]

add

[fickle tapir]

oh

[dusty river]

Why the "=0"?

[fickle tapir]

i know that b = -(r1 + r2 + r3)

[oblique river]

so -(r2 + r3) = b + r1

and that's the x-coefficient in our quadratic

that's a good start

what about the constant term r2r3

[fickle tapir]

b = -r1 - r2 - r3
so
r2 = -b -r1 - r3 and r3 = -b -r1 - r2

[oblique river]

that's true but not really useful here

can you express d in terms of r1, r2, and r3

[fickle tapir]

im unsure

[oblique river]

how did you prove that b = -(r1 + r2 + r3)

[fickle tapir]

i expanded the factorization and looked at the coefficient of x^2

[oblique river]

so why not just do that again?

[fickle tapir]

i thought i had to work with the new equation whoops

-(r1r2r3)

[oblique river]

okay, so now can you express the constant term of the new equation in terms of d and r1?

[fickle tapir]

d = -r1r2r3, so then r2r3 = -d/r1

[oblique river]

yes

so now can you write the new quadratic in terms of b, c, d, and r1?

[fickle tapir]

what i have so far is $x^2 + (b + r_1)x - \frac{d}{r_1}$

[cloud walrus]

add

[oblique river]

okay, now what adjective can you apply to the coefficients of that quadratic equation

(hint: it's really the only adjective we've been throwing around so far)

[fickle tapir]

what does adjective mean

[oblique river]

oh sorry, my bad

basically like

descriptor

[fickle tapir]

rational?

[oblique river]

so rational is an adjective but not correct

all we know is that r1 is rational, we don't know anything about b + r1

but we do know that it's constructible

because both b and r1 are constructible

similarly for d/r1

[fickle tapir]

constructible yeah

[oblique river]

but now we are all set because

roots of quadratics with constructible coefficients are themselves constructible

[fickle tapir]

and we could we show this using quadratic formula right?

[oblique river]

that's right

the important thing is that square roots of constructible numbers are constructible

[fickle tapir]

what about in the case when the square root in the quadratic formula turns out negative?

[oblique river]

it can't

the problem says that all three roots of the cubic are real

[fickle tapir]

ohhh

so essentially each arithmetic operation on each term in the quadratic root expression yields a constructible root as well in the end

[oblique river]

yep

the two roots of the quadratic are the two missing roots of the cubic

and they are the two roots you get when you apply the quadratic formula

and since the coefficients are all constructible, and all the manipulation you do in the quadratic formula (multiplying, adding, square root, etc) all preserve constructibility

these two new roots must be constructible as well

so yeah that is the proof :)
... or is it?????

(there is one small detail we are missing right now. one step that we did that might be illegal in certain cases. can you find it?)

[fickle tapir]

since r1 is rational, it could be equal to 0?

im concerned about the -d/r_1

[oblique river]

that's right

(sorry for my delay)

so if r1 isn't 0, our current proof works

but what if r1 is 0? then d = 0 as well

so then what does our quadratic look like now after we factor out x?

[oblique river]

yes, but even easier

is 0 = x^3 + bx^2 + cx = x(x^2 + bx + c)

since d = 0

and now your quadratic is just x^2 + bx + c and both b and c are constructible

so the rest of the proof still works out

[fickle tapir]

oh yeah

and we can use the similar reasoning with the quadratic formula

as a justification

[oblique river]

yep

[fickle tapir]

thanks 🙂

[oblique river]

np

[cursive temple]

how would i show that the x_i are in the center of R[x_1, \ldots, x_r]

for r = 1 you can just use the formal definition i think

im not sure on how to multiply in R[x_1, \ldots, x_r] lol

this is the defn of a multivariate polynomial ring given

oh wait i can just use the case for n = 1

lol

[oblique river]

If you know how to do it for 1 variable you can do it for any number

Inductively

[cursive temple]

like R[x_1, \ldots, x_{r-1}] is a noncommutative ring

so yeah it generalizes naturally

[oblique river]

Yeah exactly

[cursive temple]

this is like the problem with tech support

you ask for help and then it fixes itself

thank you anyways

[oblique river]

This is why you always keep a rubber duck on your desk

Ask the duck first

And then if that doesnt help, then you ask other people

:^)

[cursive temple]

would really be a worthwhile investment

[oblique river]

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

[urban acorn]

lmao I used to mess with computers and I did that with a stuffed elephant toy

[somber marsh]

what is a $(1/2,0)\oplus (0,1/2)$ representation?

[cloud walrus]

Muffin Man

[oblique river]

what is the context

[prisma ibex]

are these the weights for a direct sum of spin representations?

[tropic spade]

Is this false: "Let A be a commutative ring with unity and let J be an ideal of A. Then A/J is a field iff for any a in A, if a is not in J then for some b in A, ab-1 is in J"?

I'm thinking it fails for A=J since A/J will be the trivial ring hence it will not have a distinct unity from zero making it not a field.

And the second half of the biconditional seems like it will be vacuously true since every a will be in J.

My book says to prove it is true tho .

[next obsidian]

I think it probably wants you to assume J is a proper ideal

Is ab-1

Supposed to be ab minus 1

[tropic spade]

Yes

[next obsidian]

Or a(b^-1)?

[tropic spade]

It is subtraction not multiplicative inverse.

[next obsidian]

Do you already know about a quotient being a field iff the ideal is maximal?

[tropic spade]

Yeah that was proved a few exercises earlier.

[next obsidian]

Actually

Scratch that

ab-1 in J

Implies that mod J, ab = 1

Aka b is an inverse to a

[tropic spade]

Yah

[next obsidian]

Oh, is your question simply about the case J = A?

[tropic spade]

I used that for one direction but reasoning towards A/J being a field at the last step i couldn't rule out that A/J wasn't trivial.

[next obsidian]

Right

Just assume J is proper

[tropic spade]

Which is where I came upon considering the case A=J

That makes sense.

[next obsidian]

There’s a lot of times some statements just fail for the really dumb cases

Like something is 0 or it’s the whole thing

I think ppl just forget to write that condition down

[tropic spade]

Lol, yeah that's probably what happened.

[tropic spade]

Idk anything about constructible numbers but is S1 a proper subset of S2?

If so could the identity function give you that |S1|<=|S2|?

Then could it be possible to rule out that |S1|=|S2|?

(Perhaps S1 is countable but S2 is not or something?)

Just a random guess.

[fickle tapir]

i dont think so, since S1 is the set of points on S where both coordinates are constrictuble, and S2 is the set of points on S where at least one of them are not constructible

[tropic spade]

So if (x,y) is in S1 then both x and y are constructible hence at least one of x or y is constructible therefore (x,y) is in S2. <- Why doesn't that work to show S1 is a subset of S2?

[next obsidian]

S2 is where one of them isn’t constructible

If both are constructible then you aren’t in S2

S1 and S2 form a partition of all possible points I think

[tropic spade]

I missed the not in "not constructible" in the defn of S2 lol.

[next obsidian]

@fickle tapir this is a ridiculous way to do it

But S = S_1 disjoint Union S_2

So for S to have cardinal out c one of S_1 or S_2 must have cardinality c

And the cardinality of either set is bounded by c

Actually this won’t even work lol

I was going to suggest showing S_2 has cardinality c

But then S_1 could still have the same cardinality

And then you’d only get a weak inequality

So you could instead prove S_1 has cardinality < c

And that forces S_2 to have cardinality c

then you get what you want

How would you prove that S_1 has cardinality < c? Fuck if I know, it probably has to do with the definition of constructible number

[tropic spade]

Are the constructible numbers countable?

[next obsidian]

If you know constructible numbers have cardinality < c, then S_1 is a subset of the product of constructible numbers by itself, and the product of two sets with cardinality < c has cardinality < c

So that’s one way to do it

[ivory dust]

Whats the splitting E of f(x) over Z3
where f(x) = x^4 - x^2 - 2 e Z3[x]?

(x^2+1)(x^2-2) is its factorization (i think unless it can go further)

so would it be Z3(alpha, beta)?

where alpha beta are roots of those factors?

i guess i, and sqrt2?

[chilly ocean]

hm

i think Z3[i] has a square root of 2

i^2 = -1 = 2

so i would think the splitting field is just Z3[i] (or really, including sqrt(2) should be equivalent as well i think, Z3[i] = Z3[sqrt(2)] = Z3[i, sqrt(2)])

[dusty river]

i = √2 here because -1 = 2

They're not the i and √2 of ℂ

[chilly ocean]

ah ok right

[dusty river]

You should be able to check if they factor further or not

You've done these kinds of problems before

[ivory dust]

Oh

I dont think they do factor

its (x^2+1)^2

x^2+1 evaluated at x = 0,1,2 is 1,2,2 respectively =/= 0 so its not reducible

so the splitting field is Z3(alpha) w alpha is the root of that ^

so i or sqrt 2 i guess?

and [Z3(alpha) : Z3] = 2

[dusty river]

Yeah

[dusty river]

As long as you don't mean i or sqrt2 of C

[ivory dust]

questions asking me about Gal(E/Z3) so thatd have order 2

Oh true

[dusty river]

Yeah

[ivory dust]

ill just use alpha as arbitrary element thats a root

and Gal(E/Z3) iso to Z2 = {id, phi}

phi mapping alpha to -alpha?

[dusty river]

Yep

[ivory dust]

yay tysm

[dusty river]

assuming extension is galois

It's easy to check

[ivory dust]

wait

whats the requirement for htat

[dusty river]

Normal + separable

[ivory dust]

[Z3(alpha) : Z3] = 2 = ord(Gal(E/Z3)) was a theorem if E is splitting field f(x) of Z3 and f is seperable i think

[dusty river]

Yeah, splitting field of a set of separable polynomials is always Galois

So you have to prove separability of the stuff you're adjoining

[oblique river]

Finite extensions of finite fields are always separable

[ivory dust]

Uhh

oh seperable means all the polynomials w their coefficients are seperable (?)

i was confused cuz seperable describes polynomials

[dusty river]

Separable means that minimal polynomials of all elements in the extension are separable

This happens iff minimal polynomials of the stuff you're adjoining are separable

So it's enough to check the minimal polynomial of alpha here

Or you can use what buncho mentions if you've seen it

All finite fields are perfect

[sullen island]

given that i have proven that if k is an algebraically closed field then k must be infinite, how do i show that for $n \geq 2$, any nonconstant $f \in k[X_1, \hdots , X_n]$ must have infinitely many zeroes ?

[cloud walrus]

xy

[sullen island]

im thinking induction but i cant even prove the base case n=2

[gritty sparrow]

Well here is a hint: Let the variables be x and y. If we substitute in a particular value from k in for y ,say a, we get a polynomial in just x. If this polynomial is not constant, then this will have a solution, b. Then (a,b) is a solution. But we can do this for any a in k. (You have to be a bit careful about choosing the a so that the polynomial is non constant in x, there will be some casework you’ll have to do)

[sullen island]

ok lemme try

[sullen island]

how can the resulting polynomial in one variable be constant? f is an element of $k[X_1, \hdots , X_n]$ and there are no additional relations (equations) imposed on it

[cloud walrus]

xy

[sullen island]

so you always get a nonconstant polynomial

[gritty sparrow]

Say f(x,y)=y+1.

If we substitute 0 for y, we just get 1

xy+1 if you want an example that has an x in it

[sullen island]

ahh i see, ok lemme continue working on it

[sullen island]

okay i cant figure it out :S

say $f(x, y) = a_1 x^{p_1} + a_2 y^{p_2}$ and $f(x, a) = a_1 x^{p_1} + a_2 a^{p_2} = \phi$ for some constant $\phi$, then we have a solution where $x$ is the $p_1$-th root of $$\frac{\phi - a_2 a^{p_2}}{a_1}$ ?

[cloud walrus]

xy

say $f(x, y) = a_1 x^{p_1} + a_2 y^{p_2}$ and $f(x, a) = a_1 x^{p_1} + a_2 a^{p_2} = \phi$ for some constant $\phi$, then we have a solution where $x$ is the $p_1$-th root of $$\frac{\phi - a_2 a^{p_2}}{a_1}$ ?
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<to be read again> 
                    
l.55 ...root of $$\frac{\phi - a_2 a^{p_2}}{a_1}$ 
                                                  ?
The `$' that I just saw supposedly matches a previous `$$'.
So I shall assume that you typed `$$' both times.

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[gritty sparrow]

What is the constant phi doing?

A solution is a solution for f(x,a)=0

[kind temple]

not being a \varphi, the only acceptable phi

[gritty sparrow]

In any case, the answer is yes there is a solution to f(x,a) (assuming a1 is nonzero)

This is because k is algebraically closed, every non constant polynomial in one variable has a solution

[sullen island]

and we can do this for every element of $k$, which yields the result?

[cloud walrus]

xy

[gritty sparrow]

for this particular polynomial yes

[sullen island]

and then we proceed by induction to prove the result for the general statement, is that right?

[gritty sparrow]

Well yes, but we haven’t finished for n=2 yet. For that particular polynomial we were able to find inf many things to substitute for y that gives us something non-constant, but we have to do this in general.

Do you see what I mean?

[sullen island]

but the polynomial is of the form $f(x, y) = a_1 x^{p_1} + a_2 y^{p_2}$ which is a general form

[cloud walrus]

xy

[sullen island]

oh oh

[gritty sparrow]

How would you deal with xy+1? This isn’t of that form

[sullen island]

yeah ok i see it

hmm

yeah im stuck 🤔

[gritty sparrow]

Ok, well here is a general form of polynomials in two variables: they will look like $p_{n}(y)x^{n}+p_{n-1}(y)x^{n-1}+…+p_0(y)$ where each $p_i$ is a polynomial in y

You can see this by “collecting” all of the x^i terms and you’ll see that you get a polynomial in y

Try using this

[cloud walrus]

saketh

[sullen island]

im not sure what you mean by collecting terms

[gritty sparrow]

Well for example yx^2+y^2x^2+xy+1

Is equal to (y+y^2)x^2+(y)x+1

[sullen island]

ah ok

i thought you meant further collecting terms in the resulting sum with p_i(y) lol

[gritty sparrow]

Oh no, I meant collecting the terms is how you get things into that form. I hope that makes sense

[sullen island]

yup

so we get $f(x, a) = p_n (a) x^n + \hdots + p_0 (a) = \epsilon$ for some constant $\epsilon$, which is a polynomial in one variable that has a solution since $k$ is algebraically closed

[cloud walrus]

xy

[gritty sparrow]

There is no need for the constant epsilon

It should just be =0

[sullen island]

i thought we had to split into cases where the resulting polynomial is either constant or non constant?

[gritty sparrow]

Well here is the catch, if p_i(a) is always zero (except for maybe p_0(a) ) then this is a constant

[sullen island]

oh right

[gritty sparrow]

Ok, should I tell you the cases?

[sullen island]

yes please 🙂

[gritty sparrow]

So the first case is if f is just a polynomial in y with no x terms (equivalent to n=0 in the form I gave). The second case is for all other polynomials (that is, all polynomials where n>0)

Notice how in the first case, no matter what I substitute in for y, we always get a constant. But that is no issue because this is just a polynomial in y which will have a soln s. And then we can choose x to be anything in k, say a, and (a,s) is a solution.

So there will be infinitely many solutions. Now we just have to do the second case

[sullen island]

and then the second case is clear because $k$ is algebraically closed and we will always have a solution, right?

[cloud walrus]

xy

[gritty sparrow]

No, there is work to be done. Here is the soln in the second case: let $p_{n}(y)$ be the polynomial in the coeff of $x^{n}$This will have only finitely many roots, since k is infinite then there are infinitely many non roots. If a is a non-root, then $p_{n}(a)x^{n}+p_{n-1}(a)x^{n-1}+…p_{0}(a)$ is a nonconstant polynomial (the coefficient of $x^{n}$ is nonzero). So it has a root b. So (a,b) is a solution. Because we can do this for all infinitely many nonroots we have infinitely many solutions to the original polynomial.

[cloud walrus]

saketh

[sullen island]

oh wow

okay i see, thank you! you've helped me for about 1 and a half hours now lol

thank you so much for your patience

[gritty sparrow]

No problem

[chilly ocean]

How does it follow from this that if there is a finite trans basis then all trans bases are finite and of the same cardinality?

Like if I assume there is a finite basis A and some other basis B I don't really know how to show it's finite. If I show that then by symmetry it's immediate that they have the same cardinality

[rustic crown]

say A has n elements. B can't have more than n elements, as, if it did, you can just take out n + 1 elements and they would have to be algebraically dependent over F. But then this would contradict, that B is algebraically independent.

use this other way to get equality

[chilly ocean]

why would that contradict that B is alg independent? B is alg dependent on A if those are bases though?

wait I think im missing something obvious

I dont get the n+1 thing as well yet wait

[oblique river]

If A and B are both transcendence bases for some field K over some field F, then both A and B are algebraically independent, but they are also each algebraically dependent on the other one

[chilly ocean]

yes

[oblique river]

Now apply the proposition to each of the two pairs (A,B) and (B,A)

[chilly ocean]

but the theorem assumes both are finite

but what if A finite B infinite

not sure how to find finite B' \subset B that can be also a basis

[rustic crown]

oh oops, i didn't mean that "on A"

[oblique river]

The elements of A can only depend on finitely many elements of B

So choose those

And you still have a basis

Because with those finitely many elements of B, you can get all the elements of A (cuz thats how we chose that subset of B) and once you have A you have the whole field K

Err, sorry, if you picked redundant things you might not have a basis yet

But this is how you get B to be finite, and then just remove redundant elements if necessary

[chilly ocean]

sorrt, wdym by ' once you have A you have the whole field K'?

like becasue K is alg over F(A) right

[oblique river]

A is a basis for K

[chilly ocean]

ok

[oblique river]

Oh wait sorry, yeah up to an algebraic extension

Like, F(A) is contained in F(B) and also F(B’) where B’ is teh finite subset we chose

So K is at most an algebraic extension of F(B’)

[chilly ocean]

so the point is each a \in A is given by some b's \in B and so take those b's and name the set B' and this way we get that |A|<= |B'| but we don't know if B' is a basis since it can have some alg dependent stuff. But K is definitely algebraic over F(B') because it is alg over F(A)

just trying to make sense in my head

so if K is alg over F(B') then for sure there is a basis B'' \subset B'

[oblique river]

Im not making any claims about the relative size of A and B’ yet, but that’s what the proposition will do for us

But yes, this trick is how you turn the infinite set B into an actual finite basis B’’ which right now we dont know the size of

But then the prop tells us that B’’ and A have the aame size

[chilly ocean]

Yeah, but finiteness of A just implies there is a finite B'' that's a basis

yep I think I got it now

thanks

[oblique river]

Np and gl further

[summer falcon]

number of similar matrix of $\begin{bmatrix}
1 & 1 \
0&2
\end{bmatrix}$ in $GL_2(\mathbb{Z}_5)$

[cloud walrus]

Frostbite

[summer falcon]

this is a perticular question but is there any general method for any matrix in $GL_n(\mathbb{Z}_p)$ where p is a prime

[cloud walrus]

Frostbite

[lethal dune]

any method for what? finding number of similar matrices??

@summer falcon

[summer falcon]

kinda

[gritty sparrow]

I think this should be a partial answer, but I’m not 100% sure about the working. Let F_q be the field with q elements. If M is a nxn matrix that is diagonalisible with all eigenvalues distinct in F_q (your original matrix should satisfy this) then we can try to find the stabilisers of M under the conjugation action from gl(n,q). So if BMB^-1=M we see that BM=MB. Now multiplying M on the right will multiply the collumns of B with the corresponding element on the diagonal of M. And the left multiplication is the same but for rows. So if BM=MB then B must be 0 outside of the diagonal, but can take any values on the diagonal. Since B has to be invertible we can choose any element except for 0 on the diagonal. So there are (q-1)^n stabilizers for M. Now to calculate the orbit of M all we have to do is take |Gl(n,q)|/(q-1)^n. And Gl(n,q) has some known cardinality as well.

[gritty sparrow]

Btw something similar should work even if the elements on the diagonal repeat, but you would get kxk “boxes” for an element which repeats k times where we can choose elements of B arbitrarily. So given n=k1+..+kr as the multiplicities ki of each element of the diagonal we’ll get some formula for the number of stabilisers. For arbitrary matrices though I can’t really find anything nice.

[gritty sparrow]

Actually in the repeating diagonal entries case, finding choices st B is still in Gl(n,q) seems to be a pain, so scratch that. No nice formulae to be found I guess

[gritty sparrow]

Ok I think I have a formula for the repeating case too: |Gl(n,q)|/|Gl(k1,q)|Gl(k2,q)|…|Gl(kr,q)|

[lethal dune]

@summer falcon

[cursive temple]

how should i think about F[u]

[ivory dust]

take an element g in F[x+(f(x))] i think
then you get g(x) = b0 + b1(x+(f(x))) + ...

but (x+(f(x)))^k = x^k + (f(x)) for all k

i think

something like that

and F[x]/(f(x)) has basis {1,x,x^2...x^n-1}

so g(x) = b0+b1x+b2x^2 + ... bn-1 x^n-1 + (f(x))

if g is in F[u]

something along lines of that

Anyone know how to show x^3 - sqrt2 is irreducible in Q(sqrt2)[x]

[small bison]

s, t both satisfy 2nd degree equations over F

So as long as F is not characteristic 2 you can use the quadratic equation and that basically gives you (a)

[ivory dust]

Would splitting field of

f(x) = x^3 - sqrt2 over Q(sqrt2)

be E = Q(sqrt2, w, 6th root of 2) where w^3 = 1

cuz roots of f(x) are

2^(1/6),
-w2^(1/6),
w^2*2^(1/6)

degree of E over Q(sqrt2) is 3

idk what a basis would be

though

as a vector space of Q(sqrt2) over E

I understand Gal(E/Q(sqrt2)) tho, its just isomorphic to Z3, and permutes the 3 roots in 3 ways (identity, and 2 cyclic ways)

[fickle tapir]

i dont necessarily understand this part

in general, im having trouble seeing how to relate both the given equations and P=(S,t) to show s,t in F or s,t in F(sqrt(r))

im a bit overwhelmed by the amount of information

[small bison]

Basically you want to think about the minimal polynomials of s and t

s and t satisfy both equations and if you eliminate say s, then you’ll see that t satisfies a polynomial equation of 2nd degree

Ie there’s a 2nd degree polynomial p(x) such that p(t) = 0

So that gives you a clue about the minimal polynomials of s and t

[fickle tapir]

is this because of the ellipse formula?

[small bison]

Yep and because of the linear equation too

[fickle tapir]

the 2nd degree part

i think i can eliminate s from cs + dt = 0

[cloud walrus]

add

[fickle tapir]

and the ellipse equation is $\frac{s^2}{a^2} + \frac{t^2}{b^2} = 1$

[cloud walrus]

add

[fickle tapir]

wait whoops i messed up

[ivory dust]

If anyone has time and could bless me w help from my post above whenever^ id be very grateful

[dusty river]

Very useful lemma that you should try to prove yourself: if f is irreducible with degree m over a field F, and [E:F] is coprime to m, then f is irreducible over E

So irreducibility goes up along extensions that are co prime to degree

You can also change this slightly to get a statement about splitting, something along the lines of if f doesn't split over F, and m doesn't divide [E:F], then f can't split over E (not sure if this is correct, but there's at least a similar statement that is)

Lol I misread the problem

x³ - √2 over Q(√2)

Here you can just see the degree of the roots of this over Q

Then make a tower of extensions and you'll get a constraint on the degree of each irreducible factor

[ivory dust]

I figured out irreducibility

cuz its cubic

if it doesnt have a root we done

so i let a+bsqrt2 be a root and its impossible

and

i was just wondering what the

E would look like

Q(sqrt2, alpha1, alpha2, alpha3)?

w alpha1,2,3 are the roots of x^3-sqrt2?

[ivory dust]

wait this is cool ill save this and find the proof if its a big thm

[dusty river]

Not a big theorem, it's a typical hw problem

Another way to prove irreducibility is to see that all the roots of x³-√2 have degree 6 over Q, so they can't be contained in a degree 2 extension of Q

[ivory dust]

degree 3*?

[E:Q(sqrt2)] = 3 right

[dusty river]

Over Q

[ivory dust]

O

if the Galois group is order 3

then its isomorphic to Z3 (cyclic grp)

and Gal(E/Q(sqrt2)) = {id, f, g}

f^2 = g

f*g = id

right

and they act on the roots

[dusty river]

Yes

[ivory dust]

alpha1, alpha2, alpha3?

only 3 permutations i guess

[dusty river]

Must cycle the roots

[ivory dust]

O

true

[dusty river]

There are 6 permutations

[ivory dust]

Its not S3

Z3

[dusty river]

Only that the transpositions don't extend to automorphisms

[ivory dust]

so yea cycle roots

[dusty river]

Yeah

[ivory dust]

period tysm

Is order of Galois group

of E = Q(i, 2^{1/4})/Q = 8

Q(i, 2^{1/4}) is splitting field of f(x) = (x^2+1)(x^4-2) over Q?

and since f(x) is separable over Q [Q(i, 2^{1/4}) : Q] = 2*4 = ord(Gal grp E/Q)

[dusty river]

Do the tower thing to prove/disprove it

[ivory dust]

ill start with x^4-2

over Q

which contains 2^{1/4} root with irred polyn of degree 4

i hope its irred...

o wait yea can use RRT

err nvm

not that easy to show its irred

but anyway its degree 4

then since i is not in Q(2^{1/4})

and i is root of x^2+1 irreducible in Q(i, 2^{1/4})[x] ur done

Would Q(2^{1/4}) := {a+b2^{1/4}+csqrt2+d2^{3/4}, a,b,c,d e Q}

certainly does not contain i

O fuck i meant

E = Q(i, 2^{1/4}) over Q(sqrt2)

that would be degree 4 (?)

[trim grove]

Here i am considering function $\phi_{B}(t)=B\phi(t)B^{-1}$\ so for continuity i showed $\lim_{h \to 0 }\phi_{B}(c+h)$=$ B\phi(c)B^{-1}$ same as $B\phi(c)B^{-1}$ , is it right to say function is continious?

[cloud walrus]

TheStudent

[upper pivot]

Yeah this is continuous, follows from the fact that multiplication and inversion (hence conjugation) is continuous.

[trim grove]

the way i proved is right? or i'm missing something? ( don't know this real analysis method of proving conitinous work for matrices)?

[upper pivot]

i mean if you proved the limit that would be enough

[trim grove]

ok

[urban acorn]

I fucking love conjugation; conjugation is based.

it's the most natural way for a group to act on itself

i.e. in a group with various geometric transformations, a rotation acting on a translation by conjugation will rotate the vector by which points are translated

and translation acting on rotation will translate the point about which we are rotating

it is truly an action of a group on itself in that elements act by group automorphisms

and it provides useful structure, which in some since captures the way the group fails to be abelian

[dusty river]

Again make tower

[urban acorn]

which is already intrinsically important because abelian groups are the nice simple case

and better yet - kernels of homomorphisms are precisely those subgroups fixed by conjugation

how cool is that?

[urban acorn]

conjugation by g is like changing the "frame of reference" from which your element acts by g

and the best way to get an intuition for that is look at special cases where groups naturally have geometric actions

and it's also precisely what you need to multiply by in order to get the same result, on the other side of g

hence the relationship with commutativity

[ivory dust]

I think i got the order of Galois grp is 4

but idk if its Z2 x Z2

or Z4

I made the tower

and kinda think i get how it works

Q < Q(s2) < Q(2^{1/4})

of which none contain i so then i can ajoin i after

f(x)=(x^2-s2)(x^2+1) in Q(s2)

[terse crystal]

Can’t find your original problem... what Galois group are you calculating now?

[terse crystal]

If you are asking about G=Gal(E/Q(sqrt (2)) where E is the splitting field of f(x)=(x^2+1)(x^4-2) over Q, then G is isomorphic to Z/2Z * Z/2Z

[wooden ember]

If K is non cyclic is it always true that isomorphisms with distinct images in Aut(H) give rise to nonisomorphic semidirect products?

Actually nvm

[terse crystal]

You mean different φ : K—> Aut (H) with different images define non isomorphic semiproduct of K and H?

[wooden ember]

Yeah but it ain’t true

[terse crystal]

False I think

Yeah not true

I think choose a proper G and define φ_j: G —> Aut (G*G) where φ_1(g) is the inner automorphism defined by (g,1) and φ_2(g)......(1,g) can give a counterexample.

[wooden ember]

Am I mistaken in saying that G is just the holomorph of H? It seems too simple...

looks good on paper but something doesnt feel right

[terse crystal]

You are correct I think

[terse crystal]

Wait, I think it should be modified to be N_H(Hol(H) not Hol(H)

[wooden ember]

How so?

[terse crystal]

Since any φ from Aut(H),(1,φ) is contained in the normalizer N_H(Hol(H))

[wooden ember]

But N_H(Hol(H)) is Hol(H)

[terse crystal]

In order to make H be a normal subgroup of G. I don’t think that H is necessarily normal in Hol(H)

[wooden ember]

Since H is normal in Hol(H)

It is by definition

Hol(H) = H semidirect Aut(H)

By definition of semidirect product H is normal in this product

[terse crystal]

Oh oh my bad yes Hol(H) is correct

[wooden ember]

Coolio

Thnx

[novel parrot]

why is Rm isomorphic to R/P

i tried homomorphism R -> Rm

but multiplication doesnt work

[upper pivot]

So they mean isomorphism as modules

sending r in R to rm in Rm would be a R-module homomorphism

[novel parrot]

okeeyyy

[upper pivot]

right so think about how you would map

R/P to Rm

as modules

[novel parrot]

ive got it

[upper pivot]

and think about how this is bijective

oh nice

[novel parrot]

but just seeing that it surjects

and kernel is P

[upper pivot]

yeah first iso works too, nice

[novel parrot]

hey i have another question

will there always exist an annihaltor in a module?

{ann(m)} will this always be non empty

[hot lake]

how could that be empty

[novel parrot]

ok thanks

[ivory dust]

yeaa that was it

would it be E is the splitting field of f(x) = (x^2+1)(x^2-sqrt2) over Q(sqrt2) instead? or thats equivalent

[summer falcon]

Thanks a lot

[urban acorn]

idk if this is the better channel for this question or #point-set-topology (because it's like introductory algebraic geometry)

but just making sure right now, because I think I've got it, but I'm tired but who can tell

is it correct that - in $A^2(\mathbb{C})$ - $V(y^4-x^2, y^4-x^2y^2+xy^2-x^3) = {(1,1)} \cup {(1, -1)} \cup V(x + y^2)$?

[cloud walrus]

All groups are abelian

[urban acorn]

and that this is the decomposition into a union of irreducible algebraic sets

well, yeah, actually, assuming this identity does hold, it obviously is the decomposition into a union of irreducible algebraic sets

but the sort of computational steps I took to get to this identity were a bit elaborate, and I'm tired right now

so just if anyone knows whether this holds, tell me

[humble surge]

What is the largest order of an element in the group of permutations of 5 objects?

[gritty sparrow]

Looks good to me, hopefully I didn’t goof on the calculation

[gritty sparrow]

Scratch that, it is the largest lcm of things that can add to 5

So 6

It will be 2+3 cycle decomposition type permutation

[wooden ember]

I did everything im just stuck at the last part in the case where H_0 has order p^3

id love a hint in the right direction