groups-rings-fields

[chilly ocean]

we have (0,1) and (1,0) in ZxZ

(a+bj)^2 = a^2 + b^2 + 2abj = a+bj

[wind steeple]

1/2x + 1/2

[chilly ocean]

fractions not allowed

[wind steeple]

oh yeah lol

[chilly ocean]

there are no nontrivial elements right?

[wind steeple]

there aren't then

[chilly ocean]

that would prove it

good

why do we care about x^2 = x lol
there will be 4 possible homomorphisms since j has to be mapped to something that squares to (1,1)

that also works lol

although would it really be easier

[wind steeple]

(1,0) works

but not (0,1)

[chilly ocean]

(1,0) doesn't square to (1,1)

[wind steeple]

(1,0) squares to (1,0)

[chilly ocean]

yes

what's ur point

[wind steeple]

you want to solve x² = x

count the numbers of solutions to show that they are not isomorphic

x² = x have 2 solutions in Z[j]

wait wtf

I bug

[chilly ocean]

0 and 1 no?

ZxZ has (0,0), (0,1), (1,0), (1,1)

already more

[wind steeple]

yes

thus they aren't isomorphic

[chilly ocean]

yes

🎊

QED

🎊 🎉

[clever flax]

@hazy torrent בוא לווייס יחרא

[hazy torrent]

אללה אקבר

[bitter mauve]

is this jesus language

[dawn kiln]

still more readable than most maths i see

[minor void]

Maths in Hebrew always stresses me out

[thorn flint]

Hi does this proof make sense?

[small aurora]

Not really @thorn flint. First you want to say something like “Observe that I \in N thus N is nonempty” rather than “Let...”. This is kind of a nitpick, but the language is confusing otherwise. As for a bigger problem, you’ll want to make sure that α is not equal to β where you prove N is closed under products. (you have b as the 1,2 position in both α and β). Also, the line where you take the determinant line is unnecessary. Giving an explicit inverse like you did is enough.
For the proof of normality you need more details. The line where you have \frac{1}{g^{-1}}Ng = N leaves me unconvinced. Since matrices are not commutative, you should write out an arbitrary g\in G and n\in N and explicitly multiply everything out in order to show this. You have the idea of what you need to do correct, you just need to make sure your wording is clear, and you need to make sure there are not missing details

[thorn flint]

@small aurora Thanks for the feedback!
This was a draft version. I wanted to just get the main ideas right and get some feedback. I'll clean it up now

So for the part where its closed under products, should I just put an explicit integer? x = (10 01) and y = (10 11) and show that it still holds in N?

Or maybe show x=(10 b1) y= (10 c1), then xy = (10 bc1). Since bc is an integer, it is in N

[chilly ocean]

Someone please do it

[small aurora]

Do the latter Mac

[thorn flint]

Okay, thanks @small aurora !

[chilly ocean]

I want to get the logic

[limber horizon]

hey guys hello, im a bit of a new comer here and wanted to know if you guys could help me with a math problem

[mild laurel]

Just ask

[leaden finch]

can someone help me with irreducible plolynomials

[golden pasture]

typeinlatex
anyway, Im u=Ker v does not imply v is onto, it is the final morphism that implies v is surjective since every element is in the kernel. Cokernel of a map f:A to B is defined as B/f(A)=B/im f, doesn't matter if it is in some exact sewuence.

v is a morphism, cokernel is a ring, v•u=0 is implied by the fact it is exact
its super messy:p

Sequence is exact iff coker u=M''

also i dont quite understand the last part

if want to be explicit, usually we go about proving it by first defining ubar and vbar, then let f is in Im vbar, show it is in Ker ubar, if f is in Ker ubar, show it is in Im vbar

and that vbar is injective

[last ether]

I have been working on it for a few days, my assignment is to give a complete explanation for the two sequence. (An exact sequence of modules induces an exact sequence of homomorphisms)

[golden pasture]

rip uh are you familiar with module homomorphisms :p

[last ether]

A module homomorphism preserves addition and scalar multiplication i believe

[golden pasture]

yes so it is similar to a group homomorphism and ring homomorphism

So when we some $A-$modules $M,N,P$, and we have the (homo)morphisms $f:M\to N$, $g:N\to P$, we call the sequence

$$M\overset f\to N\overset g\to P$$

of morphisms exact iff im $f=\ker g$. Image and kernels are defined identically to their linear algebra counterparts

When we write $0\to M$ this implicitly means the map that maps the $0$ module to $M$ by sending $0$ to $0$ in $M$. The image of this morphism should be clear

Similarly $M\to0$ is the map that sends all elements of $M$ to $0$, the kernel of this morphism should be clear.

Now going back to the exact seq above, first define $\bar u$ and $\bar v$. Note that elements Hom$(M'',N)$ are morphisms itself.

Now we simply show that $\bar v$ is injective by showing if $\bar v f=\bar v g$, then $f=g$ and show that im $\bar v=\ker\bar u$ by showing both are the same as sets($f\in$im $\bar v\implies f\in\ker\bar u$ and the converse)

howisimnotdefined

[cloud walrus]

Ariana:

[last ether]

This just seems like we are applying the inverse operations to get from the first line to the second line

[golden pasture]

that is generally illdefined

and doesn't work

rmb Hom(A,B) is a bunch of homomorphisms from A to B

so it is basically functions

[mild oxide]

am I doing primary decomposition correctly here?

[solemn rain]

whats :

well for u to just check

[mild oxide]

I can't decompose further if elements aren't relative prime, right?

[solemn rain]

yea Z_nm ~ Z_n x Z_m iff (n,m) = 1

[mild oxide]

like, I can't break Z_8 down further into Z_2 x Z_2 x Z_2

[solemn rain]

they wouldnt be isomorphic

[mild oxide]

I'm just kinda unsure if the process is really that simple

[solemn rain]

nah it is ig

[mild oxide]

okay :)

ig?

[solemn rain]

i guess

if G is an abelian group

uh nvm

[stone fulcrum]

Za × Zb = Zc
If a and b are coprime

I think I'm trying to find it to be sure

[solemn rain]

what

find what

[mild oxide]

what is c there

[solemn rain]

ab

[mild oxide]

gotcha

[stone fulcrum]

Oop yes that should be ab my bad

So note there's multiple ways to break a group down

[mild oxide]

Consider the abelian group Q/Z. Show that this is a torsion abelian group which is *not* finitely generated.
I'm sort of confused by this. If a group is torsion, doesn't that mean that its elements have finite order? doesn't that mean it is finitely generated? or am I mixing things up here..

also kind of unsure how to go about showing it is torsion in the first place.... I need to show that it's elements are all finite?

[mild laurel]

You're getting things mixed up

Z itself is not torsion, but is finitely generated

Generated by just 1

You need to show all the elements have finite order

[solemn rain]

What is the smallest value of n for which S_n has an element of order 60?

is it 30

[mild laurel]

No

[smoky cypress]

Is it ||12|| zoph

[solemn rain]

20?

[mild laurel]

No

Yes whoever I can't add

[smoky cypress]

Oh ok

I was kinda confused for a sec lol

[solemn rain]

any hints 😄

[mild laurel]

I mean, how are you getting your number

[solemn rain]

im just thinking of lowest common multiples

so in S_20 i can have a cycle of length 15 and 20

lcm(15,20) = 60

[solemn rain]

thats what i thought of

[smoky cypress]

Why does the permutation has to have only two cycles?

[mild oxide]

okay, this is what I've come up with. I'm not super confident in the air-tightness of my argument, however.

[solemn rain]

it oesnt but like

the more numbers i find the common multiple of the less it gets no?

or like

nvm

[oblique river]

deku you need to review the definitions

Z has infinite order but is finitely generated

[mild oxide]

oh finite order =/= finitely generated

[oblique river]

no

that's why I said to go review the definitions

[mild oxide]

man, the brain fog is real right now

am I even remotely on the right track though?

[oblique river]

well

in the sense that the words you're using are the words that would be in a correct solution

I guess so

but you just need to be more clear with what you mean

[mild oxide]

which definitions am I confused about here? finite order and finitely generated?

[oblique river]

like "Q/Z is torsion because its elements are finite" doesn't erally make sense

[mild oxide]

oh, "because it's elements have finite order" I meant

but I suppose I can't say that yet, can I

[oblique river]

well once you prove that the elements have finie order you can

[mild oxide]

To show something is torsion, I need to show that its elements have finite order.

[oblique river]

yes

[mild oxide]

I show that the order of the elements of Q/Z are at most n, which is finite, so the elements of Q/Z have finite order, therefore it's torsion.

[oblique river]

yes

but the n will depend on which element

differen elements will have different orders

[mild oxide]

so I'm just off on wording for that part?

[oblique river]

yes

[mild oxide]

what do you mean the n will depend on which element? does that matter? isn't n always finite?

[mild laurel]

@solemn rain but that formula only works if your cycles are disjoint, which they can't be if you're taking a cycle of length 15 and a cycle of length 20

[solemn rain]

mm yea

[oblique river]

it's just weird to say "the orders of the elements are at most n"

[mild oxide]

the order will always be at most n, which is always finite, so the order will always be finite

[oblique river]

that makes it sound like you're saying there's some number n

which is a bound on the orders of every element

but in reality, the n is different for each element

[mild oxide]

so I should be saying something like "the orders of the elements are at most some arbitrary n, which is finite" to be clear that the n is not fixed?

[oblique river]

no

just prove that each element has finite order

then say "each element has finite order so the group is torsion"

you don't need the n in that sentence at all

also the n is not arbitrary

it's determined by each element

[mild oxide]

but how else am I proving that each element has finite order? I'm proving they have finite order because they're each bounded by n, which is finite. they're all finitely bounded. I don't mean to imply they're all bounded by the same n.

[oblique river]

yes but you are implying that

"consider an arbitrary m/n + Z. notice that n(m/n + Z) = m + Z = Z so the order of m/n + Z is at most n which is finite. since m/n + Z was arbitrary, we conclude that lal element have finite order and thus teh group is torsion"

you prove taht each element has finite order

full stop

therefore the group is torsion

[mild oxide]

so should I just say "each element is bounded by some finite element n (not necessarily the same for each element)" or something? like just add that little proviso?

[oblique river]

the best way to phrase it that way would be "bounded by some n which depends on the alement

but you shouldnt phrase it that way

prove each element has finite order

period

then forget about the n's

then say "therefore the group is torsion"

that's the clearest way to say it imo

[mild oxide]

okay, I think I've got it.... let me think through this once more.

[oblique river]

yheah your "math" is right I just don't think you're being clear in your explanation

[mild oxide]

What about the other part? Is my showing that the order of Q/Z is infinite sufficient to say it is not finitely generated? Or was my use there also confused about the definition?

[oblique river]

look at Z

that's infinite

and finitely generated

[mild oxide]

Right. So no. I'm trying to look at my definitions to see how I go about proving a group is not finitely generated and I'm kind of lost here. I'm guessing by some proof by contradiction but I'm not sure what theorem or corollary I can use to reach one.

[oblique river]

use the nike method

just do it

there's not heorem to use

[mild oxide]

lol

[oblique river]

assume you have a finite generating set

find a contradiction

[mild oxide]

okay, just capping things off here. I think I've got it

[oblique river]

one sec i'll look

*i'll look in one sec

each order is bounded by some finite n

not each element is bounded

[mild oxide]

oof, you're right lol

[oblique river]

also

say what your generators are

like

"suppose Q/Z is finitely generated by x_1, ..., x_j

or somethign

[mild oxide]

okay. thanks a lot

[round igloo]

who here is good at spanish

[smoky cypress]

[solemn hollow]

from https://proofwiki.org/wiki/UFD_is_GCD_Domain. How did they go from 1st line to 2nd line? I tried assuming, for a contradiction, that all irreducible elements in the factorization of f divides d

nvm I think I found another way to do it

[mild oxide]

okay, I'm just absolutely lost here. looking for any kind of pointers for a direction of attacking this one.

[latent anvil]

Do you know what conjugating a cycle does?

Like, if σ = (1a) for a > 3, what is σ^(-1) (123) σ?

More generally if σ(1) = a and σ(2) = b and σ(3) = c, what is σ^(-1) (123) σ^(-1)?

[mild oxide]

@latent anvil No, I guess I do not. I understand what a conjugate is, but I guess I'm having trouble with what it means to "conjugate a cycle." I know if I'm trying to show that c_1 and c_2 are conjugate in S_n, I need to come up with some g in S_n such that c_1 = g c_2 g^(-1). But I'm confused about the information about c_1(x) = x iff x is not in X, etc. and how to use that information to find g....

[rapid tinsel]

If A, B are free modules over a finite ring, is A / B free?

[mild oxide]

uhhh, we haven't talked about modules or rings at all yet

[rapid tinsel]

If A, B are free Z-modules, certainly it's possible that A / B is torsion, but I can't find counterexamples when the ring is finite

[mild oxide]

so I do not know what that means

[rapid tinsel]

It's analogous to the case where A, B are just powers of a fixed cyclic group C_n, then A / B is also (isomorphic to) a power of C_n

[jolly crypt]

a friend gave me this question: let's suppose that R is a commutative unital ring, and that we have an isomorphism $\phi: R[x] \to R[x]$ with the property that $\phi(x) = ax + b$ (where $a$ is a unit). Then prove that $\phi$ is uniquely determined.

[cloud walrus]

4TL:

[jolly crypt]

For this problem, it would suffice to show that the map phi restricted to R is just the identity map, since R and x generate R[x]. But I'm struggling to show that R even maps to itself.

[latent anvil]

@rapid tinsel it's not just analogous, it's actually true, right? I think if the underlying abelian groups of some Z/nZ modules are iso, they're iso as Z/nZ modules

[rapid tinsel]

Doesn't the module structure present some form of obstruction to this?

I'm just worried that, since Z/nZ has zero divisors, some things might be killed by accident

[latent anvil]

Suppose φ : M -> N is an iso of abelian groups. Take x in M and [k] in Z/nZ

Then [k] x = (Σ [1]) x = Σ [1] x = Σ x

Where the sum is taken k times

Then φ([k] x) = φ(Σ x) = Σ φ(x) = [k] φ(x)

Does that make sense?

Basically since the underlying additive group is cyclic with generator 1, and we now how 1 acts, the entire action is determined

[rapid tinsel]

Oh right, yeah, because the action is induced by the natural Z-action

Thanks - do you mind if I give you my full context / the whole problem?

https://math.stackexchange.com/questions/1951502/bijection-between-submodules-of-m-that-contain-l-and-the-submodules-of-m-l

@latent anvil If I replaced "submodules" with "free submodules" or "projective submodules", it's not true in general especially for Z-modules (since 4Z < 2Z are both free, and the corresponding thing 2Z/4Z = Z/2Z is torsion, hence not projective). What if I considered Z/nZ-modules instead? Are the free submodules of a Z/nZ-module M containing a fixed free submodule N in bijection with the free submodules of M / N (which is free)?

https://math.stackexchange.com/questions/1951502/bijection-between-submodules-of-m-that-contain-l-and-the-submodules-of-m-l
Hmm I guess this is the third isomorphism theorem or something

[toxic zephyr]

https://puu.sh/FA9vL/2f02e56133.png

could someone walk me through the second half of this proof

I'm not quite sure what 5.14 is saying

$X^tY-X^tA^tAY=0 \implies X^t(I-A^tA)Y=0$

[cloud walrus]

nix:

[toxic zephyr]

i think i understand that

but then what does 5.14 mean

[golden pasture]

idk it seems like a trivial tautology e_i is your ith basis element

[toxic zephyr]

Oh okay I think i might be seeing it now. 5.14 is just a statement of fact.

if the right vector has only a 1 in the jth column, then when we left multiply B, we just get the jth column of B.

Since the left row vector has only a 1 in the ith column, the dot product just gives the ith row of the jth column

so since $e_i^tBe_j=b_{ij}$ is in the form $X^tBY$ which we know equals zero, then by extension every entry of B, $b_{ij}$ must be zero. So $I-A^tA=0 \implies A^tA=I$

[cloud walrus]

nix:

[toxic zephyr]

is that the idea?

[tranquil ruin]

hello is anyone here right now that knows how to do abstract algebra

Can you help me with my exam tomorrow online LOL

[golden pasture]

please do your exam yourself-.-

[tranquil ruin]

rip

if anyone wants $20 for some exam help lmk

[mild oxide]

I'm sure people here would love to help you study for it.

[steep hull]

No

[golden pasture]

ban?

[tranquil ruin]

its in 12 hours

[golden pasture]

thanks

[mild oxide]

No one here is going to take your test for you.

[tranquil ruin]

kk

[golden pasture]

ban in 12 hrs

[steep hull]

Yeah I’ll report

[tranquil ruin]

im not forcing anyone

chill

[golden pasture]

it isnt about forcing someone

[mild oxide]

I did not imply force of any kind.

[golden pasture]

it is about academic misconduct

[mild laurel]

You're not going to get really far in math with that attitude anyways

[tranquil ruin]

lmao you noobs and your arbitrary morality on my case, probably don't even uphold your own code

get off my case

[steep hull]

We do actually

[tranquil ruin]

lmao im sure

[mild oxide]

oh boy lol

[mild laurel]

Just make assumptions about us to make yourself feel better okay

Whatever helps

[tranquil ruin]

assumptions of you having not lived according to your morality once huh

seems like a basic fact about humans to me

lmao how arrogant to even make that claim

[mild laurel]

That says a lot about you

And your worldview

Sad

[tranquil ruin]

Only sad thing here is you convincing yourself you havent ever once made a decision your regretted

[scarlet estuary]

my worldview consists mostly of 🐱

[outer meadow]

the world is a hard place, you aren't gonna get far sticking to your morals

[scarlet estuary]

wait what

that doesnt make sense in context

[tranquil ruin]

Its not about whether you should stick to your morals or not

[mild oxide]

people being flawed doesn't mean they shouldn't aspire to be better. say, for argument, that people here have cheated before. does that mean that they have to help someone else do it, otherwise they're a hypocrite? what a childish position.

[scarlet estuary]

you're... preemptively regretting cheating tomorrow?

but still planning on doing it?

[outer meadow]

we all make hard decisions, that we regret

[tranquil ruin]

Thats not the point im making, im saying that what you believe to be wrong isnt the same for me

I have a class ill never work on again

It doesn't matter to me

It doesn't affect anyone if I pass with a 50

[outer meadow]

are you a math major?

[golden pasture]

then it shouldn't matter if you fail it

[tranquil ruin]

ahaha no

I'd rather not fail

lmao

Im assuming you think thats weird too

[mild laurel]

You're not going anywhere in life with this attitude buddy

[rich patrol]

[tranquil ruin]

Its not about going anywhere in life

[mild oxide]

regardless of morality or whatever, the point is that you're just fundamentally misunderstanding the point of this server and the kind of people who occupy it if you think people will directly help you during a test.

[tranquil ruin]

you've got a long ways to figure that out

[outer meadow]

wait i'm curious what major you are @tranquil ruin

[inner acorn]

We're not just forcing our morals on you, it's one of the rules of this server which you should have read
Go elsewhere if you want to cheat

[tranquil ruin]

No im not, you don't seem to understand what even happened

I asked if someone wants to help me with my exam

if you don't want to don't reply

if its not allowed

say so

instant ban

[scarlet estuary]

it's not allowed

[tranquil ruin]

lmao get off my case

[scarlet estuary]

as mentioned in #rules

[mild oxide]

lol it is said so, in the rules

[scarlet estuary]

and #❓how-to-get-help

and probably other places

[tranquil ruin]

Nice let me read the apple terms and agreement before I make my account to

Delusional

[outer meadow]

the rules say "asking for help /during/ an exam is not ok"

[tranquil ruin]

I won't ask if its not allowed

Its that simple

[scarlet estuary]

so you're expecting people to help you on-demand, but you're not willing to put in the minimum modicum of effort to understand the rules of the space?

[outer meadow]

lol

[scarlet estuary]

the entitlement

[mild oxide]

bruh reading like a dozen rules for a discord server is not the same thing as a fucking dicken's apple novel

[tranquil ruin]

Help me on demand?

are you illiterate?

[golden pasture]

go join another discord server lol

[tranquil ruin]

Why? Unless you'll ban me I wont, and I'm not going to ask for exam help again if its not allowed

lmao

[gentle pendant]

It is not.

[tranquil ruin]

You've said that 3 times now

I understand

[gentle pendant]

This is the first thing I have said here.

[mild oxide]

nice, productive conversation

[tranquil ruin]

the group

I enjoy this conversation

It's funny seeing you incapable you lot are

how*

[scarlet estuary]

lol

"help me cheat on this test"

"btw yall are incapable"

[chilly ocean]

bruh reading like a dozen rules for a discord server is not the same thing as a fucking dicken's apple novel

i think i read this wrong

[rich patrol]

I won't ask if its not allowed

[scarlet estuary]

what is a "dicken's apple novel"

[scarlet estuary]

actually curious

[tranquil ruin]

cheating on a test for a class that doesn't mean much to me
Incapable of having a conversation

[mild oxide]

could have phrased better

[scarlet estuary]

i know what a dicken's novel is obviously

[mild oxide]

apple terms + dicken's novel

[scarlet estuary]

i see

[mild oxide]

verbose and long

[golden pasture]

ahh

[gentle pendant]

good one

[tranquil ruin]

lul

[mild oxide]

thanks I worked real hard on it

||actually I barely worked on it at all||

[chilly ocean]

[scarlet estuary]

wow im shocked

[golden pasture]

good job im proud of you

[scarlet estuary]

i thought this was your magnum opus

the culmination of a life of work

[tranquil ruin]

Asked for help on exam
No someone ban him

Ok if it's not allowed I won't ask again
proceeds to talk about getting far in life like thats even a thing

meme reactions as opposed to a response

[scarlet estuary]

yeah getting far in life is actually a capitalist myth

rise up against the bourgeoisie oppressors brother

[tranquil ruin]

Its not about capitalism LMAO

[scarlet estuary]

the only ones with wealth are those born into it

[golden pasture]

half true lel

[tranquil ruin]

It's about what you are striving for, and what you think the purpose of life is

[steep hull]

This guy is trying to get us off topic now

[mild oxide]

was my actual reaction though

[chilly ocean]

[tranquil ruin]

Off topic? I started the conversation

[mild oxide]

this is my favorite subgroup

[scarlet estuary]

fan the flames of revolution comrade

you have nothing to lose but your chains

[gentle pendant]

There's not much of a topic to discuss, the rules and consequences have been made clear.

[mild oxide]

workers of the world: unite!

[tranquil ruin]

kk

[golden pasture]

or could just just come back in 12 hrs and give some abstract nonsense answer

[scarlet estuary]

what is the most communist ring

[golden pasture]

F_1

[scarlet estuary]

i guess that works since

• everything is equal

[mild oxide]

red one

[scarlet estuary]

• true F_1 doesnt exist

[leaden finch]

question

does nonconstant mean nonzero?

[outer meadow]

huh?

[scarlet estuary]

it means... nonconstant

ie has a term of degree greater than 0

the zero polynomial is a constant polynomial

but so is, say, the polynomial 4 in Z/5Z[x]

[outer meadow]

you are not reading the problem @scarlet estuary

[golden pasture]

x+1 is nonconstant poly

[leaden finch]

but isnt this the same as that one?

[scarlet estuary]

@outer meadow what do you mean

also what do you mean "isnt this the same"

those are different theorems

[leaden finch]

oh okay i see

[outer meadow]

i guess i haven't read it either, but these are sort of trivilaities, i figured there is something tricky in the problem

[scarlet estuary]

...

[outer meadow]

i guess i expected the question to be something tricky

eg theorem 4.12 is true for nonconstant, or zero polynomials, except that (3) is messed up

[golden pasture]

lol yea (3) is screwed

[steep hull]

Just say nonzero tho

[golden pasture]

x^2-1=(x+1)(x-1), neither of which is nonzero constant, may it meant to mean nonzero nonconstant?

[stone fulcrum]

x² - 1 isn't irreducible

[golden pasture]

oh whoops

[steep hull]

Yea was v confused

[golden pasture]

didnt read first line

lel

ok (3) is ok

[mild oxide]

mmmm. I'm having trouble with this one. I'm sure I have to put A in Smith normal form, which would make the det stuff easier, but I can't tell how that connects to Z^n /L being finite.

Also, how would I represent A arbitrarily here? just use an arbitrary smith normal form type matrix?

[leaden finch]

can someone explain too me how they got these numbers

[scarlet estuary]

rational root theorem

[outer meadow]

but there are errors, eg there is no p/q=8 in it

[leaden finch]

im confused how they got 4 and 2?

[scarlet estuary]

4 divides 8

and 2 divides 8

[shrewd wasp]

How do I check if an operation is well defined?

[mild laurel]

An operation + is well defined if given a = b, c = d, then a + c = b + d

[shrewd wasp]

My exercises are to show that if J is a subring of a ring, then addition and multiplication are well defined on the set of cosets of J

[mild laurel]

Then do that

[shrewd wasp]

How

[mild laurel]

Basically the problem here is that addition and multiplication are defined using the coset representative

In other words, (a + J) + (b + J) = (a + b) + J

if this is the notation you use

But (a + J) = (c + J) for some other c's

[shrewd wasp]

yeah it is but what happens to the second J on the right?

left*

[mild laurel]

uh, this is just notation to represent a coset?

There's no addition of J's happening

[shrewd wasp]

Oh Im dumb af ya know

I dont think I understand what addition means in terms of cosets

[mild laurel]

This is how it's defined

(a + J) + (b + J) = (a + b) + J

a + J is some coset, so is b + J

[shrewd wasp]

Does this mean if I take any a in one coset, add it to any b in another coset, I will always end up in the coset for (a+b)?

[mild laurel]

Well, that's essentially what you're trying to prove

that addition is well-defined

[shrewd wasp]

but to make it clear to myself, a isn't fixed but (a+J) is a specific coset and a could be any element in that coset? is that right?

[mild laurel]

Yeah that's the right idea

[shrewd wasp]

Okay I think I understand the addition part then

[mild laurel]

a + J represents a whole coset, and a + J = c + J whenever a and c are in the coset

[shrewd wasp]

So how would I write a proof that addition of cosets is well defined?

I tried and put (a+J) + (b+J) = (c+J) + (d+J) = (a+b) + J = (c+d) + J

but it doesnt seem like much of a proof of anything

I feel like I just assumed it tbh

[mild laurel]

remember, a +J is a whole set, how do you show that two sets are equal?

[shrewd wasp]

subsets of each other

so I need to show that a + J and c+J are subsets of each other if a and c are in the same coset?

[mild laurel]

you're assuming that a + J = c + J (that a and c in are the same coset) and that b + J = d + J

you're trying to prove that (a + b) + J = (c + d) + J

[shrewd wasp]

Ive tried to prove it but I don't know how

This is what I did for trying to show that (a+b) + J is a subset of (c+d) +J:

Let $n\in(a+b)+J$, then $n\in a + (b+J) = c + (b+J) = (c+b)+J$. As R is a ring, addition is commutative and c+b = b+c, so $n \in (b+c)+J = b + (c+J) = d + (c+J) = (d+c)+J = (c+d)+J$ So $(a+b) + J \subset (c+d)+J$

[cloud walrus]

Shipreck:

[shrewd wasp]

Im not sure about how okay it is to assume a + (b+J) = c+(b+J)

[mild laurel]

how do you know that a + (b +J) = c + (b + J)?

[shrewd wasp]

a and c are in the same coset

[mild laurel]

but that just means that a + J = c + J

A couple hints maybe to make this easier

a + J = c + J means that a - c \in J, so that a - c = j for some j \in J

This goes both ways as well, so you can transform what you want to prove

(a + b) + J = (c + d) + J

into a + b - (c + d) \in J

[shrewd wasp]

a+J = c+J means a -c \in J?

Think I just in general need to go recap my group theory notes because im obvs very very rusty haha

thanks for the help though pal

[mild laurel]

its if and only if

[leaden finch]

can someone help me with this one

[smoky cypress]

[cerulean siren]

I'm super close to getting this problem and yet I don't have it

[cloud walrus]

jynelson:

jynelson:

jynelson:

jynelson:

[cerulean siren]

but I can't get the algebra right!

[cloud walrus]

jynelson:

[smoky cypress]

You can write \sqrt{2}

[cerulean siren]

ah well that's nicer

the closest I've gotten is $2 = (a^2 + 2ab\alpha + 2b^2)^2 / b^4$ which is just to ugly to think about

[cloud walrus]

jynelson:

[cerulean siren]

and it's going to be degree 4 anyway so I'm not sure if it even helps me

[oblique river]

what is $\sqrt{2 + \sqrt{2}} \cdot \sqrt{2 - \sqrt{2}}$

[cloud walrus]

Buncho Bananas:

[cerulean siren]

well that's the thing I don't know if $\sqrt{2 - \sqrt{2}}$ is in $Q(\alpha)$

[cloud walrus]

jynelson:

[cerulean siren]

if I did I'd be done already

[oblique river]

lol

that is the fact that I'm helping you prove

[cerulean siren]

hmm multiplying them gives me sqrt{2}

[oblique river]

great

is sqrt(2 + sqrt(2)) in Q(alpha)?

(yes by definition)

[cerulean siren]

yes, it's alpha squared

[oblique river]

is sqrt(2) in Q(alpha)?

[cerulean siren]

ugh

yes it is

[oblique river]

yes ti's alpha^2 - 2

so rearranging the equation that you just wrote down.....

[cerulean siren]

so then I can divide sqrt(2) by alpha and I get sqrt(2 - sqrt(2))

thank you!

[oblique river]

yeah lol

[cerulean siren]

and this works because Q(alpha) is a field

[oblique river]

yes

[cerulean siren]

how did you know $\sqrt{2 + \sqrt{2}}\sqrt{2 - \sqrt{2}}$ was going to be useful?

[cloud walrus]

jynelson:

[oblique river]

if something works once, it's a trick, and if it works more than once, it's a technique

this is a technique

which is to say that I've seen it used before several times to prove these kinds of statements

[topaz solar]

Unsurprisingly, $ab=c$ means $c/a=b$ in fields, nonzero things

[cloud walrus]

Darkrifts:

[oblique river]

let K be any field. A common fact that can be very useful is the following: K(sqrt(a)) = K(sqrt(b)) if and only if ab is a square in K

(it's also true if you use "a/b is a square in K")

so in this case let K = Q(sqrt(2))

and a = 2 + sqrt(2), b = 2-sqrt(2)

we want to show that sqrt(b) is in Q(sqrt(a))

[cerulean siren]

ab is a square in K
what do you mean by this?

[oblique river]

which is equivalent to saying that Q(sqrt(2), sqrt(a)) = Q(sqrt(2), sqrt(b))

a*b is an element of K

and I want it to be a square

like

c^2

for some c

[cerulean siren]

for some c in K?

ok

[oblique river]

yes

a square in K

it's always a square somewhere

(like Q(sqrt(ab)) haha)

[cerulean siren]

yeah that's why I was confused lol 😆

[oblique river]

so I knew to solve this problem basically we needed to check that (2+sqrt(2))*(2-sqrt(2)) was a square in Q(sqrt(2))

which we could rephrase as

the sqrt of the left hand side was an element of Q(sqrt(2))

[cerulean siren]

do you mean $Q(\sqrt{2 + \sqrt{2}})$?

[cloud walrus]

jynelson:

[oblique river]

no

I mean, that was enough

[cerulean siren]

I don't think \sqrt{2 + \sqrt{2}} is in Q(\sqrt{2})

I mean, I could be wrong

[oblique river]

it's not

I never said it was

but (2+sqrt(2))*(2-sqrt(2)) is

and that's what was important

[cerulean siren]

I'm a little confused where the Q(\sqrt{2}) came from but other than that I think I follow

[oblique river]

because

this question is equivalent to asking if the extension K(sqrt(a)) = K(sqrt(b)) where K = Q(sqrt(2))

and a = 2+sqrt(2)

b = 2-sqrt(2)

and I know from experience that working iwth quadratic extensions is easier

[cerulean siren]

but K here is Q

not Q(sqrt(2))

[oblique river]

what do you mean

I defined K to be Q(sqrt(2))

I don't remember seeing K previously defined

but if so, my bad, call it L then

[cerulean siren]

it wasn't previously defined

hold on let me read again

[topaz solar]

He meant extending Q rather than Q(sqrt 2)

[oblique river]

the problem is about proving that a certain degree-4 extension of Q is normal

I observed that that was equivalent to proving a statement about degree-2 extensions of a different field

namely, K = Q(sqrt(2))

which is easier to do because quadratic extensions are easy

[cerulean siren]

ok I think that's enough math for tonight 😆

[leaden finch]

anyone wanna tutor me for modern algebra? I am willing to pay you. Pm pleaaseee

[stone fulcrum]

@leaden finch
I don't need pay haha but feel free to ping if you ever want to discuss something you don't know

[solemn rain]

Let G be a simple group, and let f : G → H be a nontrivial group homomorphism. Show that if N is a normal subgroup of H which does not contain f(G), then f(G)∩N = {e}.
i am p sure im wrong but any help?
tahts what i tried
suppose for the sake of contradiction f(G) intersects N is not trivial
and N does not contain f(G)
let x be an arbitarily element in N
there doesnt exist an element g in G such that f(g) = x --> not in f(G) ( if x is arbitarily in N then x is not in f(G) then itnersects in 1 )
then f(G) intersects N is the identity
idk wtf
fuck this shit any help?
idk how would u use G is simple
i cant imagine a waay
all i thought was like
mayube showing f(G) itnerssects N is normal in H
hence H is nonsimple
hence contradiction as G and H are homomorphic
but how the fuck would i do that idk
if N does not contain f(G)
then if x is in f(G) then x is not in N ( x is arbitrary
f(g) = x for some g in G
f(g) is in H tho hence in N tho?
idk im losiong my mind
any help

i did not even use normality of N lmfaaao

[upper pivot]

jesus pls latex this no one is gonna want to read through that lol

[solemn rain]

Let G be a simple group, and let f : G → H be a nontrivial group homomorphism. Show that if N is a normal subgroup of H which does not contain f(G), then f(G)∩N = {e}.

suppose for the sake of contradiction f(G) does not intersect N in identity and N does not contain f(G)

let x be an element in f(G)

since x is arbitrary x cant be in N

x = f(g) for some g in G tho

oh okaay

okay can u help me with this

[upper pivot]

"since x is arbitrary x cant be in N"

?

[solemn rain]

yea thats totally wrong

cuz N doesnt contain

f(G) i thought

[upper pivot]

doesnt contain

but can still intersect

[solemn rain]

yea

my bad

now any hints with this?

[upper pivot]

right ok

so if G is simple

what can you say about f

[solemn rain]

( please be gentle with me i am very stupid )

whats f

[upper pivot]

the homomorphism

[solemn rain]

has kernel 1

[upper pivot]

yeah

[solemn rain]

injective

[upper pivot]

so f is injective.

[solemn rain]

okay

[upper pivot]

now f(G) cap N is a subgroup of f(G) right

[solemn rain]

yes

[upper pivot]

is it just a subgroup tho? maybe its stronger than that

[solemn rain]

idk

i wouldnt have thought of this

fuckl

m,y flie

N alone is normal in f(G) right?

orn o?

N is normal in H

f(G) is in H right?

[upper pivot]

i mean no? N doesnt need to be contained in f(G)

so that doesnt make sense

[solemn rain]

yea

[upper pivot]

however we can take the component of N in f(G)

aka N cap f(G)

[solemn rain]

whats cap

intersect?

[upper pivot]

ye

[solemn rain]

yea

yea yea

its normal

[upper pivot]

yeah prove that N cap f(G) is normal in f(G)

[solemn rain]

isnt f(g) in H

for any g in G?

[upper pivot]

yes

[solemn rain]

and N is normal in H

so f(g)N=Nf(g)?

elements in f(G) are of the form f(g) for g in G

right?

am i right? im sorry if i sutpid

[upper pivot]

right

[solemn rain]

cool

[upper pivot]

this isnt proof tho

[solemn rain]

yea yea

[upper pivot]

u need to show N cap f(G) is normal in f(G)

[solemn rain]

let x be in N cap f(G)

x is in f(G) ---> f(g)xf(g)^-1 is in f(G)

x is in N ---> f(g)xf(g)^-1 is in N

[upper pivot]

is in N* for the last line

but yes

[solemn rain]

yea yea

fuck me

f(g)xf(g)^-1 is in f(G) cap N

for all g in G

[upper pivot]

so if f(G) cap N is a normal subgroup of f(G), what can you say about it

keeping in mind G is simple

[solemn rain]

its normal in H

[upper pivot]

forget H

[solemn rain]

whyuy

f(G) is a subgroup of H

[upper pivot]

what do the normal subgroups of f(G) look like

[solemn rain]

ea nvm

[upper pivot]

if G is simple

[solemn rain]

idk

umm

f(G) is a subgroup of H

so they should be tirvial

as G is simple

oh fuck

okayy

okay now i would have never thought of this shiit

i am no problem solver whatsoever

[upper pivot]

yeah either trivial or the whole thing(latter is impossible as imposed by question)

[solemn rain]

okay

tysm

[upper pivot]

np

also dw again just takes practice

[solemn rain]

bro i am so bad

i should just stop serious

fuck this

[mossy dagger]

Could someone please explain the last two sentences of this proof to me, please?

Could someone please explain the last two sentences of this proof to me, please?
@mossy dagger In particular I'm having difficulty understanding how the author is bounding the order of M/aM.

[oblique river]

because M is isomorphic to Z^d

and so M/aM = (Z^d)/a(Z^d) = (Z/aZ)^d which has a^d elements

[maiden ocean]

hmm

so i have an exercise in this book to show that

If $I_1, I_2$ are relatively prime ideals and $I = I_1 \cap I_2$ then $R/I \cong R/I_1 \oplus R/I_2$

[cloud walrus]

Sloth:

[maiden ocean]

could i send each $x \in R/I$ to $(\bar{a_1}, \bar{a_2})$ where $\bar{a_i}$ is the conjugacy class of x in $I_i$?

[cloud walrus]

Sloth:

[maiden ocean]

and then show that the kernel of this is $I_1 \cap I_2$?

[cloud walrus]

Sloth:

[oblique river]

that doesn't look like point-set topology

[maiden ocean]

:c

[oblique river]

:P

yeah you're on the right track -- do you mean to say "equivalence class" instead of "conjugacy class" though?

[maiden ocean]

oh

yeah wtf lol

[oblique river]

haha

[maiden ocean]

that would be

interesting

[oblique river]

also just a little trick to make things easier, it's "hard" to define maps out of R/I because your instinct is to define them on elements x of R but then you need to argue that the map doesn't depend on the choice of element in the coset x + I

so the easier thing to do is to define the map out of R

R --> R/I_1 + R/I_2

[maiden ocean]

yea thats wut i was doing

[oblique river]

oh okay

you just said "send each x in R/I"

[maiden ocean]

oh oops

lol

sry thats my bad

[oblique river]

I see the rest of your argument makes more sense now

yeah the kernel is I and therefore by 1st iso theorem

[maiden ocean]

yep

[oblique river]

now you just need that the map is surjective

so we have one more thing to prove... and one more hypothesis that we haven't used yet...

[maiden ocean]

i already proved chinese remainder theorem

:p

[oblique river]

oh

[maiden ocean]

hehe

[oblique river]

then what is this exercise?

[maiden ocean]

the one right after it

lol

[oblique river]

oh

wait how is this not just the CRT

[maiden ocean]

it was "exercise 10 prove the chinese remainder theorem" "exercise 11 show this"

[oblique river]

hmm

okay well I guess that does that then haha

[maiden ocean]

ok wew

thanks

[oblique river]

👍

[onyx mirage]

anyone on?

[golden pasture]

[obsidian zealot]

How many elements of order 2 does dihedral group D6 have

[sharp sonnet]

just check

[stone fulcrum]

@obsidian zealot
Is that the dihedral group on the hexagon?

Imagining the hexagon makes this a bit easier. They are r³, s, sr³

[chilly ocean]

i think there are some more

[sharp sonnet]

there are

[obsidian zealot]

is Z12 isomorphic to Z2 x Z6

[solemn rain]

no

@obsidian zealot

[obsidian zealot]

How do i prove it

[solemn rain]

prove general case

Z_nm is isomoprhic to Z_n x Z_m iff gcd(n,m) = 1

[obsidian zealot]

oh ok i see

[solemn rain]

hint : think about the order of the supposed generator of Z_nm

[obsidian zealot]

gcd is 2

Z, as an additive group, is isomorphic to all of its nontrivial subgroups.

[solemn rain]

what

[mild laurel]

That's true, what's your point?

[obsidian zealot]

I wasn't sure if it was isomorphic or not

[solemn rain]

so ur done with the first problem?

i dont know if knowing this would help in proving Z_n x Z_m is iso to Z_nm

[obsidian zealot]

yes, I just had to verify

[solemn rain]

cool

anything else

[obsidian zealot]

hmm not atm , it was just those 2 questions

[solemn rain]

col

[raw moth]

Anyone have any cool groups problems I might reasonably be able to solve in my head?

Or at least can think about (ie not overly computational)

[mild laurel]

if G/Z(G) is cyclic, where Z(G) is the center of G, then show that G is abelian

Show that all groups of order p² are abelian

[raw moth]

Hmm

I have done the second one before

But that's a nice way of doing it

[outer meadow]

not an answer to the question, but i recall many asshole computational problems in D&F (prove that there are no simple groups of order 168, etc.), man, really a pain, but they were a "good" kind of pain

[mild laurel]

Let $G$ be a finite group, and $H$ be a subgroup with $[G:H]=p$ where p is the smallest prime divisor of the order of $G$, show $H$ is normal

[cloud walrus]

Zopherus:

[mild laurel]

I'd have more, depending on what you know about groups

[raw moth]

I've done like 10 weeks worth of courses on groups I think

Sylow and second/third isomorphism theorems were proven and used a bit but I don't remember that very well bc not examined on it until next year

[solemn rain]

put more please

@raw moth how didi u do the second one b4 the first?

[raw moth]

Some mess involving ||conjugacy classes|| and ||sizes of subgroups||

Idr exactly

[mild laurel]

You use class number ideas

[outer meadow]

@mild laurel i have a group theory question, it is an exercise from dummit and foote actually, and i could not solve it: recall \phi(G) as the frattini subgroup of G, ie intersection of all maximal subgroups of G, and \phi(G)=G if G has no maximal subgroups. prove that if N is normal in G, then \phi(N) is a subgroup of \phi(G)

i could prove it for fininte, abelian, maybe other crap

but genreal case was too hard for me

oh, actually i just did a google search, this was posted to mathoverflow about a year ago

(the time that i had attempted this problem was quite a while ago)

[raw moth]

For G/Z(G) you just ||write G as r^kZ(G) and use direct product theorem?||

[outer meadow]

if you are curious https://mathoverflow.net/questions/326867/frattini-subgroup-is-normal-monotone/326873#326873

[raw moth]

This reminds me actually of one my supervisor gave me last term I haven't solved

Does there exist G with G/Z(G) isomorphic to Q8

I haven't actually tried ||writing the quotient group explicitly|| tbh so maybe that works

But seems like something to do with paper

Will think about the [G:H] = p, thanks

[solemn rain]

@raw moth for the G/Z(G) cyclic ---> G abelian ||cosets of Z(G) partition G , G/Z(G) = <aZ(G)> , any element in G is of the form a^it^j for for t in Z(G) and by justg looking we get xy=yx||

whats direct product theorem

for the group G such that G/Z(G) is isomorphic to Q_8

[raw moth]

Direct product theorem is likr

[solemn rain]

||no||

[raw moth]

If H,K are subgroups of G whose intersection is e, hk = kh and for all g there exist h,k with hk =g then G isomorphic to HxK

Idk if it's standard named result, just one we named in my course last term

[solemn rain]

yea thats like

the 'identifcation theorem' or whatever

[raw moth]

Ah okay

[solemn rain]

what is K here tho

explain how u wouild use this

to prove G/Z is cyclic ---> abelian

i did the standard way in the spoilers above

if u want to see

[raw moth]

Hmm I think maybe I implicitly assumed ||finite cyclic||

[solemn rain]

[raw moth]

I'll say the method I was thinking for that and then think about a fix

[solemn rain]

cool

[raw moth]

Oh wait yeah it's kinda silly to do that way oops

As soon as you write ||G = r^kZ|| it's obvious

[cerulean siren]

can someone check my reasoning on this? I'm trying to find the smallest splitting field of $f(x) = x^3 + 1$. My thought was that $x^3 + 1 = 0$ is the same as $x^3 = -1$, so by squaring both sides we get $x^6 = 1$, and then it must be $\mathbb{Q}(\zeta_6)$. Is that right or are there solutions to $x^6 - 1 = 0$ that aren't solutions to $x^3 + 1 = 0$?

[raw moth]

x=1 for example

[cloud walrus]

jynelson:

[cerulean siren]

oh hmm

so it must be a proper subset of Q(zeta_6) then?

[raw moth]

From the epsilon I know of splitting fields, yes

Provided zeta_6 is the 6th roots of unity

[cerulean siren]

yes it is

[raw moth]

Surely you just find the roots explicitly?

[outer meadow]

x^3+1 = (x+1)(x^2-x-1), or something, but -1 is already in Q

[upper pivot]

(x+1)(x^2-x+1)

[outer meadow]

factoring is hard

[cerulean siren]

how would I find the roots of x^2 -x + 1?

oh wait quadratic formula

lol

[outer meadow]

but ah, i guess the splitting field is precisely Q(\zeta_6)

[raw moth]

Oh right that makes sense

You're going to generate the other roots anyway

[cerulean siren]

hmm

is there ever a case where the 'square both sides' trick wouldn't work?

[raw moth]

Ye

f(x) = x^3-1

[cerulean siren]

oh no I used it for a bunch of other problems 😆

hmm that's because 1^2 is one though

[raw moth]

Ye I'm just saying it doesn't necessarily work in general

[outer meadow]

oh? but x^3-1 and x^3+1 should produce the same spitting field.. right?

nvm

one is complex nvm, they are just different

[cerulean siren]

aren't they both complex?

Q(zeta_3) and Q(zeta_6)

[outer meadow]

wow, i am a dumbass

[cerulean siren]

it's ok, i just forgot the quadratic formula existed lol

[raw moth]

I think z^n = a being squared should work for a negative but not for a positive

But

[cerulean siren]

hmm any positive would make that break?

interesting

[raw moth]

I have next to no experience with this stuff

[cerulean siren]

well that's reassuring 😆

[raw moth]

Okay

I have some good news

I can guarantee that what I said is not 100% correct

But if |a|^(1/n) is rational I think it should be

[cerulean siren]

n here is the power of x?

[raw moth]

Ye

x^n = a

If a is negative, the solutions to this are |a|^(1/n) * 2nth root of unity that is not nth root of unity

I think

Ok yes

If |a|^(1/n) is rational then when we extend Q by these we can just ignore this factor

And so we just have some of the 2nth roots of unity

Meaning we generate all of the 2nth roots of unity

So it'll be Q(zeta_2n)

[cerulean siren]

wait so this would work for $x^4 +2$ as well? like $x^4 = -2 \implies x^8 = 4$?

[raw moth]

If |a|^(1/n) is irrational maybe sometimes it ends up working too but I"m too tired to think properly

[cloud walrus]

jynelson:

[raw moth]

Should do?

[cerulean siren]

I don't think I need to worry about irrational for this assignment lol

[raw moth]

You should think about how to solve x^n = c for various c btw

So that you know what's going on with this kinda thing

[cerulean siren]

if I knew how to solve those I wouldn't be doing weird stuff to be able to use cyclomantic polynomials lol

umm ok let me think about it

[raw moth]

It's simpler than cyclotomic (?) polynomials

[cerulean siren]

so you know that the n-th root of c is a solution

and you also know that you can multiply by the nth root of 1 and it will still be a solution

[raw moth]

This is for c positive

Well

Hmm

[cerulean siren]

shouldn't it work for both?

[raw moth]

I guess it still works

[cerulean siren]

so you'll have n roots, which are {root_n(c), root_n(c)*zeta_n, root_n(c)*zeta_n^2, ... }

[raw moth]

Yeah

Okay

The general gist is that for c positive, squaring introduces 2nth roots

Whereas for c negative you already had 2nth roots

[cerulean siren]

so for my $c = -4$ example you get ${ \sqrt[4]{-4}, i\sqrt[4]{-4}, -\sqrt[4]{-4}, -i\sqrt[4]{-4} }$

[cloud walrus]

jynelson:

[cerulean siren]

(for n = 4)

and I need to find [Q(all those elements) : Q]

which I'm not sure how to do without knowing the cyclotomic polynomial

[raw moth]

Q(zeta_8)

Or rather

[cerulean siren]

how do you know it's Q(sqrt(2)zeta_8)?

[raw moth]

Q(sqrt(2)zeta_8)

[cerulean siren]

there's not even a sqrt(2) in that expression 😅

[raw moth]

4th root of 4 is sqrt(2)

[cerulean siren]

in the reals, ok

[raw moth]

I basically just chose an arbitrary 4th root of -4

[cerulean siren]

how do you know that generates all the other roots?

[raw moth]

Uh

[cerulean siren]

I multiplied by zeta_4 above, not by sqrt(2)

[raw moth]

I kinda just assumed it when I wrote it - the sqrt(2) has potential to mess things up

Ye

This is why I was saying about |a|^(1/n) being rational

[cerulean siren]

yeah my professor gave a similar example in class and said you needed to adjoin multiple numbers

I think it was i*sqrt(2) or something like that

but anyway my original question was how to find the degree of that over Q

[raw moth]

I think since sqrt(2) is algebraic of order 2 you should need to adjoin at most 2 elements

Like you can generate sqrt(2)zeta_8^3 just fine

Just cube sqrt(2) zeta_8 and divide by 2

Similarly sqrt(2) zeta_8^5, sqrt(2) zeta_8^7

[cerulean siren]

hmm ok that makes sense

and then you can do the same starting with zeta_8^2

huh

[raw moth]

Wait wait wait

No, this generated all 4 of our roots

I don't know what degree means

[cerulean siren]

like $[Q(a, b, c, ...) : Q]$

[cloud walrus]

jynelson:

[raw moth]

Presumably not dimension as vector space?

[cerulean siren]

yeah exactly

[raw moth]

Oh

Then surely 2

[cerulean siren]

well just because you have to adjoin two elements isn't the same as having a dimension of 2

[raw moth]

You only have to adjoin 1

[cerulean siren]

like $[Q(\sqrt{2}, \sqrt{3}) : Q] = 4$

[raw moth]

Oh wait

[cloud walrus]

jynelson:

[cerulean siren]

(sqrt(6) is the other one you need)

[raw moth]

Lemme think

Hmm

This is too complicated for my brain right now

Linear independence over Q for these feels like something you need theory for

[cerulean siren]

yeah exactly, I have no idea where to start lol

[raw moth]

It's at most 4

I can tell you that much :P

[cerulean siren]

how do you know at most 4?

the upper bound we had in class was n!

[raw moth]

Can generate the space by sums of rational multiples of 1, sqrt(2) zeta_8, zeta_4 and sqrt(2) zeta_8^3 I think

Was n here the number of adjoined elements?

[cerulean siren]

n is the degree of the minimal polynomial

[raw moth]

Ah

[cerulean siren]

so if you adjoin all the roots , yes

[raw moth]

Okay

[cerulean siren]

but note that 'number of elements adjoined' is pretty flexible, for instance $Q(\sqrt{2}, \sqrt{3}) = Q(\sqrt{2} + \sqrt{3})$

[cloud walrus]

jynelson:

[cerulean siren]

or even simpler $Q = Q(1) = Q(1, 2, 3, ...)$

[cloud walrus]

jynelson:

[raw moth]

Ye

Ehhh

Why was the bound n!???

[cerulean siren]

no idea, I didn't understand the proof lol

[raw moth]

You can only create 2^n different combinations

[cerulean siren]

oh sorry, we had that it divided n!

[raw moth]

Wait maybe I'm being dumb

Okay yeah I was

n! Is likely not a useless bound

Hmm

But anyway yes these four elements certainly span the space

Ah

I think it's 2

Sqrt(2) zeta_8 = 1+i

So we can generate the elements I and 1-i even from integer multiples, let alone rational

🤦‍♂️

Now these two are definitely linearly independent bc they aren't scalar multiples

[chilly ocean]

Sorry to but in, I have a short question\

What is the splitting field, $K,$ of $p(x) = x^7+11$ over $\mathbb{R}?$

I feel like this isn't difficult, but I can't seem to come up with a sound argument. Any help is appreciated.

[cloud walrus]

wstayman:

[chilly ocean]

I'm able to find a splitting field for $p$ over $\mathbb{Q},$ but I get stumped finding it over $\mathbb{R}...$ wouldn't $\mathbb{R}(z),$ where $z \in \mathbb{C}\setminus\mathbb{R}$ be isomorphic to $\mathbb{C}?$

[cloud walrus]

wstayman:

[raw moth]

Hmm

If you adjoin a nonreal root then you should generate C yeah

[chilly ocean]

Yeah that's what I thought too. That's where my confusion comes from. Since the roots of $x^7 + 11 = 0$ are all imaginary besides the one real one (and since 7 is prime) wouldn't adjoining any of the nonreal ones to $\mathbb{R}$ just give us an isomorphism to $\mathbb{C}$

[raw moth]

Would've thought so

[cloud walrus]

wstayman:

[raw moth]

From a linalg perspective I'd expect so

If we take C to be R[i] then we have some linear map between them with non-zero determinant

[chilly ocean]

Determinant in what respect?

[raw moth]

Actually hmm maybe this doesn't necessarily give an iso

[chilly ocean]

R(i) is definitely isomorphic to C

[raw moth]

Ye

Determinant in the sense of considering R[I] and R[z] As vector spaces over R

Where z is the root we adjoin

And the linear map converts like coefficients in one to the other

I worry a little about multiplication in terms of this being an isomorphism tho

I could well just be saying dumb shit, I don't know much about this and it's 5am

[chilly ocean]

You just define the map $\mathbb{R}[x]/\langle x^2+1\rangle \to \mathbb{C}$ by $a+bx \mapsto a+bi$

That gives you a bijective homomorphism between R and C

[cloud walrus]

wstayman:

[raw moth]

If we considered the map between R[i] And R[z] As a change of basis on R^2 then of course everything to do with addition and scalar multiplication is preserved

And this will be bijective for determinant reasons

[chilly ocean]

I don't have much knowledge in the crossover between linear alg and abstract.. whenever I'm trying to find an isomorphism I just abide by the first theorem of isomorphisms and hope I can work one out

[raw moth]

Ah

Unfortunately my knowledge of abstract algebra beyond groups is essentially nothing :^)

[oblique river]

I think you're overthinking this a bit

the splitting field of x^7 + 11 is C

(over R)

[chilly ocean]

Just over R right?

Yeah

[oblique river]

yes

[chilly ocean]

It would be different over Q no?

[oblique river]

yes

[chilly ocean]

Okay perfect. And its C (over R) because we can just adjoin one of the imaginary roots (any of them since 7 prime) and get C?