groups-rings-fields

[small bison]

can you construct a-bar from a and Q?

actually jk it depends on whether Re(a) is rational or not

[dull root]

Well, lets suppose Re(a) and Im(a) are both rational

[small bison]

in that case i think you can make a- once you have a in your field

cause like a - Re(a) = i * Im(a)

and then you can do a - 2i * Im(a) to get a- right?

[dull root]

yea, then the degree is 1.

[small bison]

hmm in general i think it depends on the rationality of a

actually in the general case idk if it's clear that the degree is 1 or 2

cause you would need |a|^2 to be in Q(a) right?

[dull root]

yea, nvm then.... I don't know if that degree is even bounded then

[tribal moss]

We can use the Lindemann-Weierstrass theorem to find an a whose conjugate is transcendental over Q(a), if only we can find an algebraic a where the degree is more than 1.

[chilly ocean]

Is there a list of presumed knowledge to learn Galois theory?

[dull root]

Hmm, so in my specific case, I want to find the extension of Q by all complex roots of x^3 - 2. It is not hard to see there are two such complex roots, namely a = (-1 + i \sqrt 3) /(2^(2/3)) and its complex conjugate. The issue I have is how to see what [Q(a,\bar{b}: Q(a))] is

I know for sure the degree is either 2 or 3, but I can't see which one is impossible here

[coral shale]

how

as said earlier look at the complex conjugate root thm

[dull root]

Well, that would say that the minimal polynomial of a and its complex conjugate are the same over Q

[inner acorn]

basic group theory, linear algebra, and some elementary number theory?

[coral shale]

ring thry

[inner acorn]

most texts that I know on intro galois theory cover ring theory

[coral shale]

there's barely any linear algebra

[dull root]

But how does that help us here. We know the minimal polynomial of \bar{a} over Q(a) only divides the minimal polynomial over Q, so it only bounds the degree of the extension above

[inner acorn]

I mean, you often consider field extensions as a vector space over the base field

sorta where the whole tower law comes in etc...

[coral shale]

Yes but that's it

[inner acorn]

also most of your examples of rings/field/etc... are matrix examples xD

not necessarily for galois theory, but again

[coral shale]

You need stuff in lag up to what a basis is

[inner acorn]

introductory texts use galois theory to introduce rings/fields

[coral shale]

You need stuff from ring theory to understand what K[x]/(f) is

[inner acorn]

My point is, if you're going to learn Galois theory from scratch, 9 times out of 10 the text will also cover rings & fields too
and authors almost always expect a knowledge of linear algebra & basic number theory

you'll see all sorts of matrix rings etc...

[coral shale]

ah nvm, I guess I misthought

Now I think about it, I don't think you can say much

I think I can construct examples for every degree

or maybe not

I believe if you take f(x) = x^n - 2

and a to be one of the non-real roots

Then [Q(a, a-) : Q(a)] should be n-1............................

Q(a) only adjoins one root of f

Q(a, a-) splits f

@dull root^

Oh wait I need n prime up there

But that still generates a lot of examples

You certainly can have examples of Q(a) that splits the minimal polynomial of a, in which case the degree will be 1

And these examples should exist for every degree of polynomial

I am thinking about the primitive element theorem, here.

[upper cape]

Does the absolute galois group being cyclic imply that any finite extension is cyclic?

It's obvious if the algebraic closure is a finite extension, but I just want to make sure that it is true in general, since I haven't done much infinite galois theory

[chilly ocean]

guys i want to show that the product of 2 cyclic groups results in a cyclic group implies the gcd(m,n) = 1

[next obsidian]

Figure out a formula for the order of an element (a,b) inside the product of two groups in terms of the order of a, and the order of b

The reason this matters is that for Z/nZ x Z/mZ to be cyclic you require a generator, aka an element of order mn

[chilly ocean]

hmmm

a formula?

[next obsidian]

Yuh

If you can’t think of what it might be, just try a couple explicit examples

Or take m,n coprime, and find the generator of Z/nZ x Z/mZ

You might notice what’s going on

[coral shale]

https://i.imgur.com/t2MzGDk.png

This comes from my algebraic number theory notes

In my Galois notes, it talks about K-homomorphisms L -> L (which are subsequently proven to be automorphisms)

I'm thinking and comparing the 2 🤔

And uhhhhhh not much progress - is there something obvious about the difference between the 2 I can say?

(Note a Number Field K is a finite field extension K : ℚ)

===
I'm thinking there is a 1 to 1 correspondence between:
The K-embeddings of L in C
The K-homomorphisms M -> M where M : L is the splitting field extension (not sure if I phrased the idea I had rightly)

But not sure

(An aside - what does Gal(C : Q) look like???)

[next obsidian]

That’s not a finite extension, I think more than likely nobody knows

But I’m not certain

[barren sierra]

need help on 7 but 3 and 5 have the relevant definitions

really the first part of 7 is what I need help with

so if x in N(R) then we have x^n = 0 for some n

so then consider 1 + xr in 1 + xR

I want to show that 1 + xr is a unit

clearly x^n is in J(R) because 1 is a unit and 1 + x^n R = 1

[next obsidian]

N(R) is the intersection of all prime ideals

Or well all you need is 6 for 7

Once you show J(R) is contained in all maximal ideals, N(R) < J(R) is automatic from@6

Because any maximal idea is prime

Oh well

You actually need to know that J(R) = intersection of all maximal ideals

Not just contained in it

Okay for the start of 7

Show that unit + nilpotent is a unit

[barren sierra]

I haven't considered the intersection part

[next obsidian]

Note that xr is nilpotent

And clearly 1 is a unit

[barren sierra]

yes

[barren sierra]

well isn't that the whole point of the problem

[next obsidian]

Well show it lol

It’s just like

An identity that you do, I can’t really tell you it or there’s no work left

Just like

a is a unit, b is nilpotent

[barren sierra]

hmmmm

oh wait

[next obsidian]

Then b^n = 0 for some large n

[barren sierra]

yea I think I know what identity to use

[next obsidian]

Yeah idk

[barren sierra]

lemme try to whiteboard it out

[next obsidian]

What’s your idea?

[barren sierra]

1/(1 + xr) -> geometric series

[next obsidian]

Swag

[barren sierra]

finite sum

ya

ok cool

[next obsidian]

Or wel

[barren sierra]

well I gotta show it's finite

but that's easy

[next obsidian]

You have to replace 1 with an arbitrary unit

But this is the idea

You can also look at difference of powers

[barren sierra]

do I?

[next obsidian]

Oh well

For your problem, no

[barren sierra]

I was thinking about this but couldn't come up with a suitable difference

[next obsidian]

But unit + nilpotent is unit is always true

Wel

[barren sierra]

x^n r^n - x^n?

[next obsidian]

Okay so this is easy right

a^-1(a + b) = 1 + a^-1b

a^-1b is still nilpotent

So now 1 + a^-1b has an inverse, call it u

Then a^-1u(a + b) = u(1 + a^-1b) = 1

Oh for that one

So if b is nilpotent

So is -b

So look at a^n - (-b)^n = (a + b)(Sum a^n-i-1 b^i)

But notice that the LHS is just a^n

So we can multiply by a^-n

And now you have an inverse for a + b

I think this is like exactly what the geometric series thing gives you lol

[barren sierra]

yea lol

that's what I was realizing and writing out on a whiteboard as you were typing it lmfao

[median pawn]

yes

i think this isn't phrased very well

so i'll rephrase it

[cloud walrus]

Hausdorff

[next obsidian]

Yeah so it’s just the roots of p

[median pawn]

yes

[next obsidian]

Oh!

Hmm

[cloud walrus]

Hausdorff

[median pawn]

so essentially we care about the maximal ideals in R, i think

[next obsidian]

Yes

So that’s what I was missing!

The maximal ideals aren’t the simple things

It’s the quotients by them

And by the nullstellensatz the maximal ideals of that ring are (x-a,y-b) where (a,b) is a root of p

[median pawn]

hmm so idk what nullstellensatz is

[chilly ocean]

your new favorite theorem

[next obsidian]

Oh

Uh well it says that maximal ideals of k[x1,…,xn] are of the form (x1 - a1,…, xn - an) when k is algebraically closed

And then f is in that ideal iff f(a1,…,an) = 0

So the statement follows

So it’s clear from algebraic closure that C[x]’s maximal ideals are of the form (x - a)

Because it’s a PID so every maximal ideal is generated by 1 thing

And that has to be irreducible, but the only irreducible polynomial are linear

But to go to two variables…

I’m not sure if the 2-variable case has a nice proof

[median pawn]

here R is not k[x,y] tho, it is k[x,y] quotiented by an ideal

[next obsidian]

Yes but

Maximal ideals of the quotient are the same as maximal ideals containing that ideal

And once you know the maximal ideals look like (x-a,y-b) you can show that f is in that iff f(a,b) = 0

[median pawn]

ok cool let me try that. thanks!

[next obsidian]

The biggest hiccup is proving the Nullstellensatz for C[x,y]

I don’t know how to do that without just proving the nullstellensatz

[chilly ocean]

maybe there's a nice complex-analytic way

that'd be cool

[next obsidian]

2-variables

[chilly ocean]

there is an analytic version of the nullstellensatz. just slice off a few terms from the series and you're good

[chilly ocean]

I need sanity check

maps from one localization of R-modules to another are uniquely defined by where they send 1/a_i where a_i are generators of domain?

this is a hunch

idk if this is true or not

[hidden haven]

1/a_i is not a well defined thing

a_i is an element of the module

you can only put elements of the set you are localising at in the denominator

[chilly ocean]

yeah my big oopsie

say we are localizing by elements

uh

i need to be more exact with my words

wait what?

i thought when you localize in modules the denominators are elementa in multiplicatively closed set of the ring

so a_i are elements of some closed set in the ring?

because at first I was thinking in specific context when we localize by elements

what if there is a notion of generators of multiplicatively closed sets

and we just let those be a_i

i was trying to generalize R_f->R_g where f,g are elements in R

[hidden haven]

So you need to know what all the elements of the module itself go first

[hidden haven]

This is for ring homomorphisms right?

As long as you know what happens to m_i/s_j you should be good, where the m_i are the generators of M and s_i of S

[chilly ocean]

i was hoping it translates the same for module homomorphisms

the initial context was exercise 3.23ii) in atiyah macdonald

the restriction homomorphism

notation wise X_f=Spec(A)-V(f) so like if you have U= X_f,U’=X_g then define A(U)=A_f=S^-1A where S={1,f,f^2,…} then A(U)->A(U’) given by restriction homomorphism is uniquely defined by where 1/f gets sent

I have proof id like verified ill write down and send pic

also you are familiar with notation im assuming?

idk how much atiyah macdonald strays from standard notation

[median pawn]

what's that

[chilly ocean]

rate my handwriting

out of 5

[chilly ocean]

nullstellensatz but for analytic functions

the version i know from complex analysis is: if f_1, ..., f_k are analytic and have no common zeroes, then \sum f_i g_i = 1 for some analytic g_i.

the proof uses some mittag-leffler type stuff. i wonder if you could look at the proof in the case that the f_i's are polynomials and somehow get that the g_i's can be chosen polynomials. that'd be the (weak) nullstellensatz over C

but this is the algebra channel, so i shouldn't go any further

ok yeah im pretty sure im not crazy

besides for me saying multiplicative closed sets should have generators

[next obsidian]

This just follows by the nullstellensatz

Oh I see

[chilly ocean]

is there a shorter way to say nullstellensat

[next obsidian]

You’re saying maybe you can prove the Nullstellensatz that way

[chilly ocean]

like nss or something?

[next obsidian]

I’ve never seen anyone shorten it

[chilly ocean]

its goofily long even though its someones name

[next obsidian]

It isn’t someone’s name

[chilly ocean]

is it a word in german?

[next obsidian]

Yes

[chilly ocean]

wow

is it descriptive?

[next obsidian]

I think so

[chilly ocean]

wikipedia says theorem of zeros

not bad ig

theorem of zeros sounds better in english

[chilly ocean]

I have a question on Group Rings if anyone is familiar with those. Let A be a commutative Ring. Let G be a finite group and let K be a field. I'm interested in the group algebra defined by K[G]={a{1} b{g{1}} + ... + a{n} b_{g_n} | a_i in A}. For the usual definition for addition and multiplication, see here : https://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/Passman.pdf

Now I cannot for the life of me solve this easy following problem. I want to say that the sums of b_g are a divisor of zero in KG, ie there exists a d in A such that ad=da=0. If G is cyclic, (1-g) would be a good candidate, but what if G is not cyclic? Let's look at easy and small exemples. If G = Z/2Z, it feels like (b_1-b_0) is a good candidate, I think? But what if G = Z/2Z x Z/2Z? The article is trying to explain it in that case but I cannot understand the algebra he's quickly skipping over. Maybe I'm just being dumb and not understanding the product definition? Any help is appreciated!

I also got another hint : If I write S=the sum of b_g, I should maybe evaluate b_g times S? But I don't see how that can work.

[tribal moss]

Let g and h be any two different elements of G, you can then set d = 1g+(-1)h.

(You don't need d to be in A; it merely has to be in K[G]).

(Hmm, actually it looks like you haven't decided firmly whether A is an alternative name for K or for K[G]).

[chilly ocean]

It's K[G], sorry for the confusion on the notation. I merely used A for the definition. Let me try your d.

I would be interested to see how you picked up that intuition for d however?

[tribal moss]

The key fact is that Sg = S for every g in G -- it's the sum of all the same elements, just in a different order.

[inland otter]

it litterally means null=zero, stellen=points and satz=theorem lol

[chilly radish]

Is there a case where the tensor product of nonzero algebras over an infinite field ends up being 0?

feels like this shouldn't happen

[hidden haven]

Dimensions multiply

[maiden ocean]

What about the infinite dimensional case

[hidden haven]

Cardinal multiplication

[maiden ocean]

do u get some wacky cardinality argument still

Ah

Yeah

[chilly radish]

right, I forgot that algebras were vector spaces lmao

I was thinking of the case of like, tensor of modules

which can vanish

but vector spaces are vewy nice

also tensor product of vector spaces having a basis implies that there is no way to write zero from nonzero terms right? In particular, if $a\otimes b=0$ then $a=0$ or $b=0$

[cloud walrus]

ShiN

[hidden haven]

And just as importantly, tensor of algebras happens to have underlying vector space as tensor of their underlying spaces

[chilly radish]

yeye ofc

actually that 'in particular' doesn't actually follow

[hidden haven]

Ye

[chilly radish]

but it's still true right

yea

ok great

[hidden haven]

You construct bilinear map for any given (a,b), so that this pair is non zero

You do this by extending a and b to bases of the 2 spaces

And then we know how to construct bilinear maps on those

[chilly radish]

bilinear map into what space

[hidden haven]

Any

Take it to be underlying field

Like if you want to show that a given tensor is non zero

[chilly radish]

oh yea I see what you're saying

[hidden haven]

It suffices to find a bilinear map out of direct product which maps that tensor's preimage under the universal map to non 0

[chilly radish]

yes yes

thanks!

I forget how well-behaved vector spaces are lmao

[hidden haven]

😼

[wooden ember]

Can we generalize p-adic integers by considering the discrete valuation $\nu_p(p^i\frac{a}{b})= i$ on the field of fractions of a general principal ideal domain with p a generator of a prime ideal?

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

oh is that what localization is all about

[delicate orchid]

when I think about generalising the p-adics my mind immediately jumps to localisation/completion at a prime ideal

maybe because I got taught about completion before p-adics

[wooden ember]

im a bit confused as to why people online say that Z_p and Z_(p) arent the same

i mean Z_p being the valuation ring of v_p, it's the rationals a/b such that b is coprime to p no?

and that's the same as Z_(p)?

[chilly radish]

wiki says Z_p is the completion of Z_(p)

[wooden ember]

maybe i have the wrong definition of p-adics

because they say it's uncountable

but i have it as a discrete valuation ring on Q

which can only be countable

oh yeah the post i read even says completion

i just didnt read properly never mind

alright so if i can understand p-adics i might be able to intuit localization

[delicate orchid]

I feel like localisation is pretty easy to intuit from the definition alone

going via the p-adics means you'll also have to get comfortable with completion as well which is far more wacky as a concept imo

[wooden ember]

i mean p-adics in the sense i was thinking

[median pawn]

okay soooo

[cloud walrus]

Hausdorff

Hausdorff

[median pawn]

so how do i take it from here?

[next obsidian]

Show that f is in (x-a,y-b) iff f(a,b) = 0

[prisma shuttle]

do u guys know how i can get links to the references in the bibliography in overleaf

like the red letters here

i tried use the normal \label and \ref commands but it didn't work

[ornate arch]

Ineed to learn how to do this:
find some examples of outer automorphisms, and try to prove that indeed they are not inner.

can somoene help me find some example problems

[next obsidian]

Take any abelian group

It has no inner auto morphisms besides the trivial one

Take any automorphism that isn’t trivial

[ornate arch]

the identity mapping is the trivial one im assuming

So like take Z_3

[prisma shuttle]

i thought trivial was everything going to 0

[next obsidian]

Not in this setting

That’s not an automorphism

[prisma shuttle]

oh right

[ornate arch]

who are u replying to

[prisma shuttle]

was replying to this

[next obsidian]

I was replying to JustKeepRunning saying the zero map

The trivial automorpjism is indeed the identity

[ornate arch]

okay

so lets say I take Z_3 group

what are the automorphisms

im a little lost tbh

[next obsidian]

You can write down every single map from Z_3 to itself just by brute forcing it

And you need bijective ones

There’s only 6 of these

[ornate arch]

Z_3 = {0,1,2} what would these maps look like

@next obsidian

[next obsidian]

Just write down every bijective map and see which ones are grpup homomorphism

Hint: 0 needs to go to 0 so there’s actually even less than 6 you need to look at

[ornate arch]

idk what the bijective maps are

[next obsidian]

My man

Just send 0 to something

Send 1 to something else

Send 2 to something else

[ornate arch]

i know

[next obsidian]

There’s only 6 ways to do this

[ornate arch]

but how are there 6

[next obsidian]

Because 3! = 6

Just write them down

3 options for the first element go to

2 for the next

1 for the last one

[ornate arch]

0->0, 1->1, 2->2
0->1, 1->2, 2->0
0->2, 1->0, 2->1
0->0, 1->2, 2->1
0->2, 1->1, 2->2
0->1, 1->2, 2->2

[next obsidian]

Now verify which ones are group homomorphisms

You even know for any group homomorphism that the identity has to go to the identity

So this rules out 4 of these immediately

[ornate arch]

homorphisms defined by: f(xy) = f(x)f(y)

so

[next obsidian]

Shouldn’t be a , unles that’s what you use for the operation

[ornate arch]

wait wym

oh

yeah

typo

0->1, 1->2, 2->0
0->2, 1->0, 2->1

@next obsidian

[next obsidian]

No, you want the maps where 0 -> 0

Because you need f(1) = f(0 + 1) = f(0) + f(1)

You can cancel the f(1) to see that f(0) = 0

Depends on what you consider interesting

[ornate arch]

oh

[next obsidian]

There is work with them, yeah

[ornate arch]

0->0, 1->1, 2->2
0->0, 1->2, 2->1

that makes sense

[next obsidian]

But it is a bit weird, it’s different than the other structures with more axioms

So now you just have to verify if those both are group homomorphisms

[ornate arch]

how come f(1+2) doesnt matter

[next obsidian]

It does

You have to test if that’s the same as f(1) + f(2)

[ornate arch]

f(1+2) = 0

[next obsidian]

That’s what I meant by verify if it’s a group homomorphism

I don’t know. I just know that there is work with them, but I don’t know any of the details

[ornate arch]

f(1) + f(2) = 0 too

in Z_3

So for both they are group homomorphisms

what does that let us imply

until now we've: Taken a group, found every bijective map, and found which are homomorphisms

these are the automorphisms?

now if we want inner automorhphisms 𝑔∈𝐺 , 𝑖(𝑥)=𝑔𝑥𝑔^−1 for all 𝑥∈𝐺.

@next obsidian

[next obsidian]

So you've found a non-trivial automorphism

the one that sends 0 -> 0, 1 -> 2, 2 -> 1

[ornate arch]

yeah

[next obsidian]

but notice that given an inner automorphism i, that i(x) = gxg^-1

but Z_3 is abelian

so gxg^-1 = gg^-1x = x

[ornate arch]

yeah exactly

[next obsidian]

so every inner automorphism is trivial

[ornate arch]

if its abelian then all automoprhisms are always inner

[next obsidian]

no

no

every inner automorphism is trivial

[next obsidian]

this is an example of a non-inner automorphism

because if it were inner, 1 would have to map to 1

[ornate arch]

so which is the actual inner automorphism, the one that sends it to a copy of itself?

even though trivial

So to sum up:

[next obsidian]

wdym the actual inner automorphism? all the inner automorphisms are just the identity

[next obsidian]

via this computation

[ornate arch]

To find the automorphisms of a group:
Find every bijective map that sends it to itself

and to find the inner automorphisms: check the homomorphisms. However, if abelian, then every inner automoprhism is trivial

does this work for all groups

[next obsidian]

I don't understand what step 2 means

and step 1 is probably bad

I only suggested that method because Z_3 is very small

[ornate arch]

my bad dude seems to be going over my head

[next obsidian]

if your group is even like, size 10

then there's 10! bijective maps

[ornate arch]

true

[next obsidian]

in general you have to be more clever to classify the automorphisms

[ornate arch]

that was gonna be my next question lol how could u do it when u cant brute force it

[next obsidian]

Depends

there's not an algorithm

so I can't tell you what to do

[ornate arch]

i see

could u possibly tell me where to find some eg problems so i can work on it

i cant find anything myself

[next obsidian]

about what?

[ornate arch]

well just to be able to do this

[next obsidian]

what is "this"

[ornate arch]

"given outer automorphisms, try to prove that indeed they are not inner."

[next obsidian]

uhhh, idk

that's really too specific a question

at this point I think you should just like

make your own problems

Take some small groups you know

[ornate arch]

lol how do i check if theyre right tho lol

[next obsidian]

That's a part of learning math

you have to be able to sniff out if you're right

that being said

you're in luck with this specifically

there's a website, https://groupprops.subwiki.org/wiki/Main_Page

This has a lot of information about a lot of finite groups

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

[spice whale]

i like that wiki

[next obsidian]

for example here's the quaternion group

it includes the automorphism group

so you can check your answers here

[spice whale]

it's very helpful
i just used it to get all the elements of A4 for this

[next obsidian]

here you see that it lists its automorphism group, inner automorphism group, and outer automorphism group

[spice whale]

is there a better way to do this than explicit computation

[next obsidian]

Maybe you could go omega big-brain

But I'm just gonna say no

[spice whale]

yeah

[next obsidian]

this is a very specific subgroup and asks you to explicitly write out the cosets

[spice whale]

ok good

[next obsidian]

this is just gonna be a computation

I don't think you can really avoid a computation

[spice whale]

i didn't write out a bunch of permutations of 1234 for nothing

[next obsidian]

You can at least tell how many cosets there has to be

so at least like

once you list out all of them

you can be sure you're done

so rather than multiplying in all 12 elements of A_4

[spice whale]

yeah

[next obsidian]

you can stop early

but besides that, I think you just bash em out

[spice whale]

what i did

oh wait yeah nvm

i was gonna say the same thing

[next obsidian]

lol

[ornate arch]

ok thanks

[spice whale]

I've gotten 5⁵/5! for this because each element can be in one of five subsets, but these subsets are only unique up to permutation
only thing is, 5⁵/5! isn't an integer

not sure where I've gone wrong tbh

[delicate orchid]

Groupprops my beloved

[delicate orchid]

This is more combinatorics but I’ll give it a think, I think your error comes from the fact that you have to have x ~ x

Either that or transitivity

[spice whale]

oh hm

[delicate orchid]

It’s definitely not a free choice though

[spice whale]

no it's not that

i did it by partitions

[delicate orchid]

Yeah, I’m saying it’s not a free choice where you can put an element in that partition

It’s just how it’s not free

Wait the number of equivalence relations is in bijection with the number of partitions

So it’s just P(5) = 7 surely

[spice whale]

there's more than 7 partitions

[delicate orchid]

And then you multiply by something cause rearrangements of elements within those partitions so we’re not done

[spice whale]

ah

[delicate orchid]

Hmmm

[spice whale]

what i did was all 5-digit base 5 numbers

5⁵

but then divide by permutation of specific digits

5!

[delicate orchid]

So for the class where everything is equivalent to everything else we’d have one choice, for the class with a 4+1 partitions we have 5 choices, for the class with 3+1+1 we have

Ah I think it’s to do with 5 choose something

[spice whale]

ohh

[delicate orchid]

(The numbers in the sum are the sizes of the set btw)

It’s just partitions like 2+2+1 that are weird

[spice whale]

1 + 5C4 + 2•5C3 + 3•5C2 + 1

and then i think it's symmetrical

not positive

[delicate orchid]

Hmm

[spice whale]

ohhh

wait

it's not 3 as a coefficient on the 4th entry

it's 2

because 2+2+1 is equivalent to 2+1+2

[delicate orchid]

Which is equivalent to 1+2+2

[spice whale]

yeah

but i didn't include that

[delicate orchid]

So we divide the choose function by the number of subsets factorial

[spice whale]

no

that's wrong

because then we end up with 5/5!

[delicate orchid]

We could just go through each partition of 5

There’s only 7 ;)

[spice whale]

yeah

that seems like a better plan

[delicate orchid]

For 5 = 5 there’s exactly one way to do it
5 = 4+1 there’s 5 ways (determined by the choice of the lone element)
5 = 3+1+1 10 ways? 5c2 is 10 probably
5 = 3+2 is 5 I think, we get 10 ways but we want {x1, x2} = {x2, x1} so we divide by 2?

And so on

[spice whale]

oh yeah 5C3 = 5C2

the one difficulty is 2+2+1

i believe this is 5C2 • 3C2

yeah

[delicate orchid]

Yeah 2+1+1+1 is just uh

Wait

[spice whale]

5C2

[delicate orchid]

Yeah

[spice whale]

1+5+10+10+10+30+1

[delicate orchid]

5!/(2*3!) = 5C2 I believe it

And 1+1+1+1+1 is obviously 1

[spice whale]

yep

67

weird number but whatever

[delicate orchid]

It’s 52

[spice whale]

uh

did you look it up

[delicate orchid]

Noooooooo

[spice whale]

i calced it as 67

[delicate orchid]

I looked up the answer but I literally only know the number

I like to know where I’m heading in this dark turbulent world

[spice whale]

here's my working

[delicate orchid]

We’ve over shot by 15

Did you take into account the 2+2+1 = 2+1+2 in that partition?

[spice whale]

yes

ohh

[ornate arch]

ok so im doign Z_4 rn

Automorphisms are:

trivial one

[delicate orchid]

5!/(2!*2!*1!*2)

[ornate arch]

and 0->0, 1->3, 2->2, 3->1, right? @next obsidian

[spice whale]

we have to divide the thing for 2+2+1 by 2! because the 2+2 can be in any order

[delicate orchid]

Yeah

[delicate orchid]

When in doubt just start with 5 factorial and start dividing

[spice whale]

yeah

this was interesting

[delicate orchid]

Now

Time to find the formula for n

[spice whale]

uh the partition function for n is already hopelessly complicated

this seems very difficult

[delicate orchid]

I'm thinkin right

I'm even thinkin left

[spice whale]

hm

[delicate orchid]

idk lol it might be an recurrance relation like the partition function is

it seems really hard and I have actual work to do unfortuantely

those two do not mix

[spice whale]

the function would involve the partition function and a million other things i imagine

[upbeat swift]

So, we just started Quotient Groups and I am still confused as to what it means (kinda). I am currently working on this problem, would anyone kindly guide me?
G is a group and H is a normal subgroup

[next obsidian]

An element of G/H looks like xH where x is in G, and xH = yH when xy^-1 is in H

so to know if xH is the identity you can plug in e for y, and you see that the condition is just that xH is the identity when x is in H

So now, to determine that xH has finite order, we want to know there’s an n so that (xH)^n = identity

But (xH)^n = x^nH

This is most of the work, do you see how the rest of it will probably go?

[upbeat swift]

I recognize the first part as a theorem

[next obsidian]

I would say it’s more a definition

But it doesn’t matter how you want to describe the result

Does the part about when xH is the identity make sense?

[upbeat swift]

Well I guess my book addressed it as a theorem and proved it. Anyways, it does make sense, that part

[next obsidian]

Okay

Then the next step?

About when xH has finite order?

[upbeat swift]

Yes, by the given part of the problem, that if we use the definition, we want to prove that x^n has finite order

[next obsidian]

Right but really I want to aim for a stronger statement than this problem

This is about “every element of G”

Let’s focus just on a single x

I guess really I want to show that xH being of finite order is equivalent to having some n so that x^n is in H

The actual problem there follows from this

[upbeat swift]

Okay, so it's like saying if xH has finite order, then so does x^nH?

[next obsidian]

Or well

It says that xH having finite order means x^nH = eH

That’s the definition of xH having finite order right?

But now, definitionally (or by a theorem in your textbook), what does x^nH = eH mean?

[upbeat swift]

Yes

That x has finite order, so it is in H. Then by the theorem, xH=H

[next obsidian]

No, it won’t actually say that x has finite order

When is x^nH the identity?

This was the second point I initially made in my response

[upbeat swift]

xH

[next obsidian]

I mean just apply your theorem about when xH = yH

Except here x is really x^n and y = e

[next obsidian]

I want to circle back to these points I made

[upbeat swift]

Isn't G/H a set that consists of all cosets in H?

[next obsidian]

Yes

But I don’t see how that’s relevant to what’s at hand

Just because xH has finite order doesn’t mean x has finite order

[upbeat swift]

I was just confirming that because yeah 😅

[next obsidian]

Ah okay

[next obsidian]

Let’s just handle this

[upbeat swift]

Okay

[next obsidian]

So you have a theorem / definition about what it means for two things to be equal as cosets

xH = yH if and only if…

[upbeat swift]

ab^-1 is in H

[next obsidian]

Right

So let’s apply that to x^nH = eH (remember, this is equivalent to xH having finite order)

[upbeat swift]

So then x^ne^-1 is just x^n?

[next obsidian]

Yup

[upbeat swift]

x^nH=H

[tribal moss]

Shouldn't it be x^-1y in H, with the inverse on the left side, by the way? (I accept this is immaterial once we assume H is normal, but since we're talking about basic definitions....)

[next obsidian]

Right here

[next obsidian]

Idrc, the thing is normal

technically that’s true but I think it’s better to ignore this

[next obsidian]

We want to apply this where a = x^n and b = e

[upbeat swift]

if you wanna have a look at what it says in my textbook

[next obsidian]

Oh okay so we can just apply (b)

Apply (b)

To conclude something about x^n

[upbeat swift]

This?

[next obsidian]

Right

So we know this

And now applying (b)

In theorem 5

We get to conclude what?

[upbeat swift]

x^n has finite order

[next obsidian]

No

What does (b) say?

[upbeat swift]

x^n is in H

[next obsidian]

right, so that’s the key thing we want

We have the following chain of equivalencies

xH has finite order <==> there exists n such that (xH)^n = H <==> there exists n such that x^n is in H

So we now have established a connection between when a power of x is in H, and when xH has finite order

[upbeat swift]

Quick question, how do you know (xH)^n is the same as x^nH?

[next obsidian]

That’s by definition

[upbeat swift]

(might be a dumb question but I wanna make sure)

[next obsidian]

That’s how the operation works inside of quotient groups

[upbeat swift]

Oh okay

[next obsidian]

So now you can do the original problem

It said that if for every x, there exists n such that x^n is in H, then every element of G/H has finite order

So take an arbitrary element of G/H, it looks like xH

By assumption, there’s an n so that x^n is in H

We know this is equivalent to xH having finite order

So every element of G/H has finite order

[coral shale]

Intuition:
All mathematical quotients that I know of can be thought as 'gluing'.

You 'glue' things together with an equivalence relation. The glued things are now considered to be 1 point/class.

When you calculate 12/4, you can consider drawing a grid of dots, then circle the columns to 'glue' them together. Then you have 3 groups of points.

The same idea applies to quotients in abstract algebra. You glue stuff together and want to preserve some structure in the new thing (a quotient group is a group). To preserve Group structure, you need to quotient by a Normal Subgroup (it acts like the identity).

[upbeat swift]

Ngl Shuri, you typing made me nervous xD was thinking "what's she typing"

But thank you very much for that

[upbeat swift]

So that's for the first part

?

[next obsidian]

Right

But the other direction follows similarly

The key is we know that xH has finite order is equivalent to there being an n such that x^n is in H

For the other direction you start with x in G, consider xH

By assumption xH has finite order, and so…

[upbeat swift]

Oh okay, let me write it down and try

Okay I got it! Thank you

[next obsidian]

[white jackal]

for this problem, i've proved it if gcd = 1 then it is a cyclic group. how would i go about the other way?

[next obsidian]

Find the formula of the order of (a,b) in terms of the order of a, and order of b

[tribal moss]

It might be simplest to show the contrapositive: if the gcd > 1, then the group cannot be cyclic.

[next obsidian]

Once you do this, if you have an element of order mn you will be able to conclude the gcd(m,n) = 1

Idk, I think doing it via contrapositive or directly will use basically the exact same elementary number theory facts

[white jackal]

do i use the formula for the |a| and |b| with the gcd

i dont remember the formula exactly

[chilly radish]

Derive it

Try to think about the least number k so that (a,b)^k=(e,e). What does it have to satisfy

[white jackal]

is it the |a^k| = n/gcd(n,k)

[chilly radish]

No

[white jackal]

if k = 0 then it will give the identity right

[chilly radish]

Well, that might help you later, but it's not what you need for chmonkey's hint

Yes but by definition the order is some number k>0

[white jackal]

this is how i started: let a,b be elements of group G. Then a has order m and b has order n. How do i show the ord(ab) = mn?

[barren sierra]

I have a question about a note in my homework for this week

ok so this says this is question 7.5.1 from Dummit and Foote

here's the question

so lets look at Theorem 15

How does this theorem relate at all to what was asked on my HW

[chilly radish]

They probably just made a mistake with the problem number

This is just localisation

Alternatively, they are using a different edition

[barren sierra]

this is the right edition

very strange

idk what localization is but I'll probably figure that out doing this question lmfao

[proud bear]

looks like it's a typo. should say 7.6.1 i think

[barren sierra]

ahhhh

[coral shale]

Any tips for eisenstein substitution?

$$x^3+x^2+2$$

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[coral shale]

I need to do this one, but I never got the hang of it

A hint for a systematic approach in general would be appreciated

(Or am I meant to just do this one by long division - I can see the factor must be (x+-1) or (x+-2) if it exists)

or wait factor theorem

for -2, -1, 1, 2

But anyways - is this doable by eisenstein substitution (idk how to tell)? 🤔

[south patrol]

In (x+a)^3 + (x+a) ^2 +2 note the x^2 coeff is 3a + 1 and x coeff is 3a^2 + 2a, ig those two are coprime (the second is the product of numbers coprime to 3a+1) so eisenstein sub won't work (if this is what you mean) (at least for integer a)

But it's degree <= 3 so factor theorem is usually a good bet

[coral shale]

Ok thanks

Rusty on this stuff =...=

So in general if I want to try eisenstein sub

[south patrol]

ye nws

[coral shale]

I try x -> x + a

And see if it might be possible for some a?

[south patrol]

Isn't that what you mean by Eisenstein sub?

[coral shale]

yes it is

[south patrol]

So ye

[coral shale]

I couldn't remember completely

[south patrol]

Dw dw

[coral shale]

I thought maybe you had to try x -> ax + b

[south patrol]

I mean in this case it's cute too as if there were an integer root we'd have x^2(x+1) = -2 lol

[coral shale]

aha yes

[coral shale]

But yes thanks for reminding me, I forgot you could generally sub x -> x + a (I kept just thinking specific a)

[south patrol]

Np

[kindred jay]

any suggestions on how to proceed?

[upper pivot]

hint is to in the multiplication of all the elements, consider inverses

[kindred jay]

to in the multiplication of all the elements?

ohh i see

thank you, will try that out

[prisma shuttle]

do ya gyus know how i can "italicize" automatically italized text in overleaf

like proposition, its auto italicized

and idk how to "unitalicize" it

[white jackal]

for this problem I've defined a map phi(g,g') = (gN, g'N'). but how do i show its surjective

[next obsidian]

I mean, everything looks like that inside (G/N) x (G-tilde/N-tilde)

Because everything in G/N looks like gN

And everything in G-tilde/N-tilde looks like g’N-tilde

Okay, Wew, you’re right. That deserved a

“How to show it’s surjective”
“Uhh, well everything is in the image”

[delicate orchid]

yeah tbf

it is kinda a proof by "look at it"

could just go for the throat and show that there's a well defined inverse straight up

[coral shale]

Algebraic Number Theory

https://i.imgur.com/9EYAe2g.png

I'm lost af

If anyone is familiar with this algorithm

I have examples in my notes and I can't follow them

[white jackal]

hahahha tyty i

[uncut girder]

Thats kinda sick

[lapis trail]

Integral domain>ufd>Ed>pid

Is that order Right? Is there a noethian domain too?

[chilly ocean]

one of them is off

[lapis trail]

Noetherian*

It's got to be Ed and pid

Right?

[chilly ocean]

yes

[lapis trail]

Because we can't even think about division

If we can't even factor uniquely

[ornate arch]

how can i find the automorphisms of Z6

i know it has to do with the generators

[chilly ocean]

any PID is a noetherian domain, but you can have a non-UFD noetherian domain, so i dunno where you'd stick those

[ornate arch]

which are 1 and 5

but im ot sure where to go from there

[lapis trail]

Those are the guys that are relatively prime to 6

I think that means they're units

[ornate arch]

whqt does it have to do with finding the automorphisms though

any auto of integers

need to send gen to gen

generator*

so therefore there can only be 2 automorphisms

but where is everything else sent

[cyan raft]

everything else can be written in terms of generators?

[ornate arch]

well the automoprhsims are

the trivial one

and then phi: z_6->z_6 s.t phi(n) = 5n

question is

how can i show they are not inner automoorphisms?

@cyan raft

we know that since its abelian all of the inner's are trivial

<@&286206848099549185>

[cyan raft]

uh, that's not trivial

so it's not an inner auto?

[ornate arch]

the inner automorphisms are trivial

all im saying

oh ur saying that since phi(n) is not trivial

it has to be outer

[ornate arch]

because of this

am i making sense im tired asf

@teal dew

[cyan raft]

yea

[ornate arch]

ok ig integer groups are p easy to find autos of

[median pawn]

I have shown that (it's actually obvious) if f is in (x-a, y-b) then f(a,b) = 0. What about the other direction?

i.e. how do I show that if f(a,b) = 0, then (x-a,y-b) is a maximal ideal of C[x,y] containing <f(x,y)>?

[next obsidian]

Express (x-a,y-b) as the kernel of a certain map to k

[median pawn]

The obvious choice is $\phi: k[x,y] \to k$ where $\phi(x) = a, \phi(y) = b$

[cloud walrus]

Hausdorff

[next obsidian]

From there it's obvious that if f does not vanish at (a,b) then f isnt contai ed in that ideal

[median pawn]

ok no it's not obvious

[next obsidian]

Oh, I guess your question was about showing that if it vanishes then its in the ideal

But from the characterization of that ideal as the kernel of that map

f is in the ideal iff f(a,b) = 0

Your map is literally "plug in (a,b)"

all you've done is replace every instance of x with a, every instance of y with b

and then fix everything else

that's just sending f(x,y) to f(a,b)

[median pawn]

ahhh i see how this solves the problem, but, i have a question still

[cloud walrus]

Hausdorff

[median pawn]

but what about the other inclusion

[next obsidian]

Well

take an f in the kernel

We want to show f is in the left ideal yeah?

Call that leftmost ideal I

Look at f’s image in k[x,y]/I

When we look mod I, x = a, and y = b yes?

Technically these should be like

x-bar

a-bar

But whatever

This lets you just conclude that f(x,y) + I = f(a,b) + I = 0 + I

But this says f(x,y) is in I

[cloud walrus]

Hausdorff

[median pawn]

Now what do you mean by this?

[next obsidian]

q(f)

You just look at f + I

Over $\overline{f}$ as it’s commonly written

[cloud walrus]

CHMOLUMBIA

[next obsidian]

The point is just that mod I we can replace x with a, and y with b

[median pawn]

ok sure

[next obsidian]

And then by assumption you can see that the thing is 0 mod I

[median pawn]

alright let me figure this out, the rest is cool

ok i think i got it, please let me know if I am right

[cloud walrus]

Hausdorff

Hausdorff

[median pawn]

right?

[cloud walrus]

Hausdorff

[median pawn]

and the sum of these two is clearly in the ideal

@next obsidian i'm just using the remainder theorem from one-variable

i hope the application is correct

[next obsidian]

Uhhh

I think of it more like this

x + I = a + I

from here, it follows that x^n + I = a^n + I

and so cx^n + I = ca^n

like...

once you know that x + I = a + I

you can just substitute a + I wherever x + I was

but you could do it that way if you want yeah

it's a clever way to reduce to the single variable case

I hadn't though of doing it this way before

[median pawn]

ahh thanks for your explanation!! it makes sense

[median pawn]

which way? the one where i split up f(x,y) - f(a,b)?

[chilly ocean]

quick question

[next obsidian]

Right

[chilly ocean]

what is difference between directed system and directed limits?

is directed limit the object itself

[next obsidian]

Yes

The system is the collection of stuff

[chilly ocean]

the system is just the tuple defining the morphism of the relation and the objects for each index?

[next obsidian]

It's like a collection of sets vs their union

[chilly ocean]

atiyah macdonald is first time i worked withem

i really like this book

the exercises build so nicely off eachother

[next obsidian]

I hate A-M lol

[chilly ocean]

what do you like better?

eisenbud or smth?

bruh

i sleep when i read thet

i tried reading it once and i took a 5 hour map

i started at 11pm tho

todd what do you study

@chilly ocean

,ti todd

[cloud walrus]

This user hasn't set their timezone! Ask them to set it using ,ti --set.

[chilly ocean]

rip

oh wow so freaking cool

you know a little about symplectic manifolds?

what is coolest thing to you in differential geometry

oh yeah i wish i could learn more about it

but i feel like i need to relearn my basic algebraic topology b4 i move on

oh cool

seems useful

idk what metric geometry is

besides makybe looking at geomtry of different metrics

but idk what this entails

ty for overview

[wooden ember]

Grobner bases are pretty wack

[coral shale]

https://i.imgur.com/F5NlSQj.png

[coral shale]

This is the answer to the 2nd part, I think

Not sure how to proceed to the last using this (should I multiply out 🤔)

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[coral shale]

Just unsure of what the 'correct' process is meant to be

[delicate orchid]

that definitely cannot be the minimal polynomial over Q it's got a sqrt(2) in it

my guess would've been (x^2-2)^3-3 personally

but now I'm seeing how that fails hmm

[coral shale]

they cancel

theres no sqrt 2

[coral shale]

This is the answer to the 2nd part

no?

oh wait you changed

[delicate orchid]

true

[coral shale]

I multiplied, grouped the sqrt 2's then killed it by multiplying by conjugate (difference of 2 squares)

Not sure 'why' this is meant to get me the minimal poly

$x^6-6x^4 +6x^3+ 12x^2+36x+1$

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[coral shale]

When expanded its this

but 🤔

well at least wolfram tells me the roots are not integers

so minimal indeed

but im missing the why for why this process is good

[delicate orchid]

thank you wolfram very cool

[fast minnow]

Maybe you can argue that the degree of the minimal polynomial has to be 6 because of the degree of the field extension, clearly it is in Q and alpha is a root. So it has to be the minimal polynomial

[delicate orchid]

yeah uhh the real roots of this bad boy do not seem to be sqrt(2)+3^1/3

[coral shale]

wut

[delicate orchid]

lemme double check I didn't mistype it

[coral shale]

,w (2^0.5 + 3^(1/3))^6-6(2^0.5 + 3^(1/3))^4 +6(2^0.5 + 3^(1/3))^3+ 12(2^0.5 + 3^(1/3))^2+36(2^0.5 + 3^(1/3))+1

[cloud walrus]

Results provided by WolframAlpha

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

Requested by Shuri2060#9525

[coral shale]

yeah u didnt.

[delicate orchid]

hm

[coral shale]

[delicate orchid]

I do agree that it has to be a degree 6 though

[coral shale]

Well alpha has to be a root of this

All I did was multiply this out

and multiply by another polynomial

probably arithmetic error

[delicate orchid]

should work yeah

unusual

[coral shale]

yeah i made error multiplying out

[coral shale]

on this

[delicate orchid]

rip

what's the new kid on the block then

[coral shale]

2 - 3 = -1. Not -3

,w expand (x-sqrt2)^3 - 3

[cloud walrus]

Results provided by WolframAlpha

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

Requested by Shuri2060#9525

[delicate orchid]

trolled...?

[coral shale]

$x^3 + 6x - 3 - (3x^2 + 2)\sqrt2$

wait what

what is going on lul

[hidden haven]

[coral shale]

i shall hunt for my error then ask lmao

[delicate orchid]

I'll see if alpha is a root of this bad boy

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[coral shale]

[delicate orchid]

,w sqrt(2)+3^(1/3)

[cloud walrus]

Results provided by WolframAlpha

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

Requested by Wew Lads Tbh#0864

[coral shale]

,w evaluate x^3 + 6x -3 (3x^2+2)sqrt2 at (sqrt2 + 3^(1/3))

[delicate orchid]

ok alpha is a root of this

thank you wolfram very cool

[coral shale]

Ok, and then i multiplied by its uh 'conjugate'

[delicate orchid]

yeah that should still be 0 at alpha

[coral shale]

$\left((x^3 + 6x - 3) - (3x^2 + 2)\sqrt2\right)\left((x^3 + 6x - 3) + (3x^2 + 2)\sqrt2\right)$

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[delicate orchid]

I think the method is fine it's just calc errors

[coral shale]

,w expand ((x^3 + 6x - 3) - (3x^2 + 2)\sqrt2)((x^3 + 6x - 3) + (3x^2 + 2)\sqrt2)

[cloud walrus]

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[coral shale]

what i wrote earlier

ah yh fcking sign

🤦

Ok I guess expanding is dumb - doesn't help

[delicate orchid]

bingo

[coral shale]

But anyways, back to algebra

so degree 6

hmmm

[delicate orchid]

what you mean

[delicate orchid]

bro you literally just found it

[coral shale]

Im just convincing myself the minimal polynomial has to be degree 6

tower law

yes yes

ok ok ok

😂

[delicate orchid]

the uh degree of the memer is the product of the degrees of the submemers

[coral shale]

For the 1st part, I square and cube alpha

then eliminate (got 3 equations, 3 unknowns). 2^0.5, 3^0.3333 are x and y, then x, y, y^2 are unknowns

Is that da wai?

[delicate orchid]

just noticed there's no (ii)

[coral shale]

or quicker method?

the sheet is filled with errors lel (I stole it from someone)

[delicate orchid]

crimes

[coral shale]

I haven't really done practice Qs, mainly just understanding theory

[delicate orchid]

I have done neither

moldi what was that galois book you were telling me about

I know you're lurking

[pastel cliff]

practice? just understand it lol

[coral shale]

🤔

id love to just understand it, but exams exist

D:

[hidden haven]

Milne's notes

[delicate orchid]

ty

[pastel cliff]

me, who has an algebra exam in 5 days

[coral shale]

moldi can i borrow your brains

[delicate orchid]

perfect

[coral shale]

https://i.imgur.com/GKtWSEo.png

[hidden haven]

Just one more exam bro

[coral shale]

how u do this

dont tell me by some theorem

[hidden haven]

Do you know primitive element theorem

[coral shale]

i do yes

[hidden haven]

[coral shale]

but can we

not

lul

[hidden haven]

Oh nice

Oh F

[delicate orchid]

I mean cube root of three isn't a FINITE linear combination of powers of sqrt(2) so u just smash them together that's my theorem

[coral shale]

Are you suggesting to do the same method in the proof of primitive element?

[pastel cliff]

trolling

[coral shale]

I only vaguely seen it (not understood it completely yet)

[plush wasp]

[hidden haven]

The proof gives you a criterion for when a + cb generates F(a,b)

[coral shale]

yeah i didnt understand that.

[hidden haven]

It's a condition on c that is satisfied here

F