groups-rings-fields

[upper pivot]

no so by (m) i mean the ideal generated by m

mZ if u will

[pastel cliff]

ahhh okie

[upper pivot]

do u understand what the sum of two ideals is

[pastel cliff]

this is where things start to get murky

bc most ring stuff i know is selftaught mostly, and not well

[upper pivot]

i see, well i guess make a guess, what do u think I+J should mean where I,J are ideals

[pastel cliff]

but i can kinda follow if you wanna just write it out

ok just to be sure im not being silly first

and ideal is like an absorbent subring right

like how a normal subgroup is to a group

i would assume that I+J has that property as well then

[upper pivot]

yeah, they are precisely the "kernels" if you will. and when we say subring, it doesnt need to have the unit inside it

[pastel cliff]

okay

so what is the significance of summing two ideals

im assuming it'd be an ideal as well

[upper pivot]

its just a way to write down a new ideal

its kind of like, the smallest ideal that contains both I and J

yeah let me just tell you what it is

$I+J={i+j: i\in I, j\in J}$

[cloud walrus]

JohnDS

[upper pivot]

so now does the original statement (m)+(n)=(1) make sense

[pastel cliff]

not entirely but i dont wanna take too much of your time

[upper pivot]

oh well, do you see what I+J means

[pastel cliff]

it seems similar to direct sum?

that could also be very wrong

[upper pivot]

kinda yeah

so do you see why (m)+(n) is the whole ring (i.e (1))

remember am+bn = 1 for some a,b

[pastel cliff]

yup definitely hadnt forgotten that, totally

(m) and (n) then like, uniquely partition the ring ig?

[upper pivot]

well partition isnt the right word

cause they have intersection yeah

[pastel cliff]

https://math.stackexchange.com/questions/552173/if-an-ideal-contains-the-multiplicative-identity-then-it-is-the-whole-ring

needed this rq

but yeah i loosely see what you mean

[upper pivot]

right so (m)+(n)=(1) bc we have, am\in (m) and bn \in (n) and am+bn=1 right

so the coprimity gives u this

[pastel cliff]

[upper pivot]

so now lets pass

to a general ring

so i will make the statement

[pastel cliff]

oh shit

he gon do it

[upper pivot]

let $I,J$ be coprime ideals in $R$, i.e $I+J=R$, then $R/(I\cap J) \cong R/I \times R/J$

[pastel cliff]

https://tenor.com/view/obama-mic-drop-largar-o-microfone-gif-7413222

[cloud walrus]

JohnDS

[upper pivot]

but yeah this is like, exactly the same proof as b4

[pastel cliff]

ah ok that makes more sense

(I) confused me

[upper pivot]

also remember (m)\cap (n) = (mn) ig

but yeah thats it, thats the CRT

generalized to all rings

[pastel cliff]

this is the bridge i was missing too i think

so the picture is coming into view

[upper pivot]

mhm

[pastel cliff]

thank you john gonna let this marinate and come back to it soon

[upper pivot]

yeah np

[pastel cliff]

im being asked to prove that the map f:D_8 -> C_2 x C_2 given by f(a) = (c, id) and f(b) = (id, c)

intuitively i can see why this is a homomorphism

but to prove it does it suffice to check f(a^4), f(b^2), and f(ba^3)

we learned early on that those are like the defining/governing relations of D_8

so i figure just checking those should be enough?

[median pawn]

Here "sum" means direct sum, right?

[chilly ocean]

In proving that a group whose elements are all powers of p must have an order which is a power of p

I argued that |G| = p^a*m, and m must equal 1, otherwise there would be a cyclic subgroup composed of the prime factorization of m, which implies that one of the elements of G is the order of that prime. Thus m = 1 and |G| = p^a

Would that be correct?

[next obsidian]

No

Direct sum would imply that they intersect trivially

This means sum, as in like N + K = {n + k| n in N, k in K}

In this particular case it does end up being a direct sum I think, because minimal submodules intersect trivially

[median pawn]

Why do minimal submodules intersect trivially?

[next obsidian]

Think about it a bit more

[cloud walrus]

Hausdorff

[median pawn]

Does that make sense?

[median pawn]

There is no harm in replacing sum by direct sum in (a) and (b) both, then? @next obsidian

[next obsidian]

Yes

I mean it doesn’t really matter

The result is the same, it’s a finite sum of elements drawing from the minimal submodules

[median pawn]

Yes, thanks!

[broken stirrup]

I'm trying to show that Q is not a projective Z-module. Suppose Q is projective and there is a submodule K such that F=~ K+O Q just like in (iii). If F is a free Z-module, does it imply that Q must be a submodule of F?

[tribal moss]

What are you trying to prove about that map?

[ember field]

If in a semi group G ax=b and ya=b for any a,b in G. Show that such seni group is a group

[mint seal]

you can take a and itself for that condition

so for a, there exists x such that ax = a

and there exists y such that ya = a

then you'll want to show the same x works for every a, and is the identity element

once you have the identity, call it e

and take any a and e, as in the condition again

there exists x such that ax = e

[ember field]

didn't see that thanks man

[mint seal]

sure

[pastel cliff]

is a homomorphism

lol my b

[tribal moss]

I see. But those two equations don't define a map from D_8 at all unless we also state that the map must be a homomorphism. So what you really need is to prove that the definition manages to the define anything at all.

If is clear that f(a) = (1,0) and f(b) = (0,1) defines a homomorphism from the free group on a and b to C2×C2. But you need to show that when you quotient out some relators in the free group in order to get D8, your homomorphism factors through the quotient.

That is, that everything in the same coset in the free group maps to the same element of C2×C2.

In particular, the relators themselves need to map to the identity in C2×C2.

Then the fundamental theorem on homomorphisms does the rest.

[tribal moss]

However ba³ is not a relator for D8. You probably have something like the relation ab=ba³, but the two sides of this are not relators individually. If you rewrite it to aba^(-3)b^(-1) = e, then the LHS will be a relator.

[coral shale]

How are the automorphisms/symmetries of the cayley graph related with the automorphisms of the group it represents?

https://i.imgur.com/RexWqu7.png

An example 🤔

[wooden ember]

i know that cayley graphs are nice with the left multiplication action but apart from that im not too sure if you always can relate symmetries of the graph to automorphisms

well i guess one easy thing you can say is that flipping all the arrows corresponds to the automorphism of taking the inverse of each generator but that's not very interesting

[coral shale]

😢

[tribal moss]

If you require symmetries of the Cayley graph to preserve the direction and labeling of the edges, then each of them is uniquely determined by where it takes the identity node. That doesn't sound promising for relating them to group automorphisms.

But if we don't require edge labels to be preserved, then the Cayley graph of the free group on two generators has continuum many symmetries fixing the identity, again far too many to usefully relate to the countably many group automorphisms.

[modern hawk]

hi, quick question, is $\mathbb{Z}[x]/(x^n + 1)$ mean the ring of integers which are divisible by $x^n + 1$?

[cloud walrus]

fisu

[coral shale]

doubt.

integer polynomials mod x^n+1 ?

[modern hawk]

oh

[tribal moss]

Yeah.

[modern hawk]

I see

thanks, now I have something new to look up

[coral shale]

thanks, rip that

[coral shale]

Was going to try writing down all the automorphisms of this and try to see if there is any kind of correspondance

But none at all huh

ig I was just after something visual to represent an automorphism

[tribal moss]

Your dihedral-group example has very few graph symmetries no matter whether we explicitly preserve labels or not. A more interesting finite example might be (C_2)^n. There are (2^n-1)···(2^2-1)(2^1-1) at least 2^(n^2-n) group automorphisms, but I believe just n! graph symmetries ignoring labels but preserving the identity node, or n!2^n graph symmetries in total.

[strong yacht]

Take n = 2 for the sake of an example. Then Z[x]/(x^2+1) is like adding an element x to Z with the property that x^2 + 1 = 0. In other words, Z[x]/(x^2+1) is nothing more than Z[i]

Since the defining property of i is that i^2 + 1 = 0

(This is not very formal but it gives an intuition into why Z[x]/(x^2+1) is isomorphic to Z[i] = {a + bi : a, b are in Z})

[modern hawk]

hmm, yea I'm still trying to learn about rings etc so that sounded a little confusing, I'm reading "abstract algebra by david dummit and richard foote"

[strong yacht]

If I were you, I'd spend a long time thinking about this

This type of thing comes up again and again, you'll be extending fields all day by quotienting out by an "irreducible polynomial"

[modern hawk]

my entire bachelors thesis will essentially be around this so yeah.. I will.. D: the ring is for ring learning with errors it's essentially for finding the shortest vector in a lattice

so now I'm trying to wrap my head around this without having taken a single abstract algebra course...

[modern hawk]

is x then just any number in Z

[strong yacht]

Nah x is just a formal symbol, it doesn't have any meaning attached to it

[modern hawk]

oh

[strong yacht]

Except, when you're quotienting by x^2+1, what you're saying is that x has this property that x^2+1 = 0

[modern hawk]

I see

[strong yacht]

So now you've got all the integers, and this new special symbol x

It's a bigger ring than what you started with since it contains all of Z and this new x

[modern hawk]

oh

but doesn't Z include all of x^2 + 1 anyway?

[coral shale]

🤔

Z[x] is a ring that contains all integer polynomials

-5, x+1, x^2+3, ...

each of these things are different.

[modern hawk]

right

[coral shale]

x is an indeterminate.

[modern hawk]

so basically here we simply specify the polynomial it should use

[strong yacht]

Z is {..., -2, -1, 0, 1, 2, ... }, but which element of Z satisfies x^2 + 1=0?

The answer is none of them do. So we can just chuck in this element and call the ring formed by this Z[x]/(x^2+1)

You can also think of it in terms of polynomials too, I'm just saying that this is an intuitive way that I see the ring Z[x]/(x^2+1)

[modern hawk]

I think I'm starting to understand

this is quite confusing..

[strong yacht]

It really is!

[modern hawk]

but slowly I'm starting to understand how all of this works

[coral shale]

if you glue on pi instead, perhaps it will be more obvious (turns out the 2 rings will be isomorphic)

Z[x] is isomorphic to Z[pi]

Z[x]/(x^2+1) is isomorphic to Z[pi]/(pi^2+1)

Z[pi] contains all the integers as usual

as well as 'polynomials in pi'

this means stuff like 1 + pi - 5pi^2

will be in it.

[strong yacht]

I think that's a potential source of confusion, since pi^2+1 = 0 doesn't look good (pi is typically referring to something else that is transcendental)

[modern hawk]

I don't know what isomorphic is...

I apparently jumped over the chapter that covered isomorphisism as I wanted to save some time...

[coral shale]

Then Z[pi]/(pi^2+1) means you take these 'polynomials in pi' mod pi^2+1.
pi^2+1 will be considered to be 0

You cant really skimp out on the concept of isomorphism.

It means 2 algebraic structures are considered to be the same

[coral shale]

Not really, when you take integers mod n, you are doing the same thing

[modern hawk]

yea, I just went back and I'm reading about it now

[coral shale]

mod 5 means multiples of 5 are now all 0

[modern hawk]

yeah

[coral shale]

mod pi^2+1 mean all multiples of pi^2+1 are 0.

but back to the original, really we're dealing with x, not pi

but my point is there is no difference in concept.

All multiples of the polynomial x^2 + 1 are now 0

this includes stuff like -10(x-1)(x^2+1)

So the polynomial x + 1 will be the same as the polynomial x + 1 + x^2 + 1 and so on (in the quotient ring you specified)

[strong yacht]

Apologies yes I think you're right

[modern hawk]

ohhh

[modern hawk]

I didn't somehow realize this was a thing

somehow thought that (x + 1)(x^2 + 1) wouldn't be a multiple of x^2 + 1, didn't cross my mind...

[coral shale]

truly, Z[x]/(x^2+1) is isomorphic to Z[i], it turns out

[modern hawk]

sure when doing it with simple numbers it makes sense but polynomials kinda passed me

[coral shale]

i and -i are exactly the solutions to to x^2+1=0

[modern hawk]

what's i there though

[coral shale]

sqrt(-1)

[modern hawk]

oh

right

[coral shale]

The x behaves exactly like i

(or -i, if you wish)

[modern hawk]

thought it was like x, a formal symbol or w/e

[coral shale]

it is.

[modern hawk]

oh...

[coral shale]

But I'm saying that the 2 rings are isomorphic

because of how x behaves

Z[x]/(x^2+1) === Z[i] === Z[-i]

[modern hawk]

wait, I'll quickly read up on isomorphisism so I can understand why better

as I thought I understood but now I'm confused again

[coral shale]

You don't have to fully understand this example at the beginning - but it will be important later on

[modern hawk]

alright, can you explain why they're isomorphic though?

[coral shale]

If the polynomial you quotient by isn't irreducible, this example breaks

x^2+1 does not factor in Z[x] so it is irreducible

ie. x^2+1 = (x+a)(x+b) for some integer a, b

[coral shale]

The isomorphism you choose is a + bx -> a + bi

Note all polynomials in Z[x]/(x^2+1) can be written in the form a + bx

If you have 1 + 2x - 3x^2, you can put it in that form by adding on an appropriate multiple of x^2+1

[strong yacht]

The isomorphism can be seen by sending x to i as well

[coral shale]

Yeah. Replace every x you see with i

[modern hawk]

oh I see

[strong yacht]

The take away intuitive message is that they're the same ring because i is defined by i^2 + 1 = 0 and Z[i] is Z with this i chucked in, and Z[x]/(x^2+1) is defined by chucking in an x into Z with x^2 + 1 = 0: notice I've just merely defined the same ring in a different way

[modern hawk]

yea, okay that makes sense

thanks for the help, I think I'm starting to wrap my head around this somehow now

[frank fiber]

is there a classification of groups of order 1024?

[sharp sonnet]

apparently yes

https://people.clas.ufl.edu/davidburrell/files/NoteGroups1024.pdf

[next obsidian]

@obsidian sleet were you at Univ of Florida?

[obsidian sleet]

i am at university of florida

i know david burrell (he made my latex seminar a pain in the ass to schedule)

hes a computational group theory guy and seeks to classify finite groups of rly fucking big order

his work is very interesting

but i know nothing about it he tried to give a talk on it once and i understood very little

@next obsidian

[next obsidian]

I C

[obsidian sleet]

naturally he turns the problem into linear algebra problem

[fast stratus]

is a cyclotomic field, extension of Q with a nth root of unity or with all the nth roots of unity?

[mint seal]

if you adjoin one nth root of unity, you get them all

so, yes

[fast stratus]

Thank you... Make senses... I gotta check, if it gives them all for myself!

[south patrol]

if you mean a primitive root

[mint seal]

oh, yes

[fast stratus]

Yeah, I feel like, I missed something in highschool... what makes $e^{2\pi i / n}$ primitive?

[cloud walrus]

Helium

[mint seal]

it means by taking powers of it you generate all the roots of unity

[fast stratus]

ah... got it

they form an cyclic group... and it generates it!!!

[mint seal]

yup

[strong yacht]

I'm not sure if you're interested, but this is true for arbitrary fields. The nth cyclotomic extension for an arbitrary field K is the splitting field of t^n - 1 (the 1 here is the multiplicative identity of K, not to be confused with the integer), it's a good exercise to show that the roots of t^n - 1 form a cyclic group.

[white nymph]

i feel like maybe im missing something obvious, but i'd like a hint on how to show that if Q_1 direct sum Q_2 is an injective R mod, then so are Q_1 and Q_2.

i have a TFAE theorem regarding injective modules, but I can't seem to figure out which part to use. I feel like this should be easy and follow-your-nose, but im missing something.

[latent anvil]

try using the defintion about extending maps along injections. If you have an injection i : M -> N and a map f : M -> Q1, can you get a map g : N -> Q1 with f = g°i?

@white nymph

[pastel cliff]

does this look okay?

[south patrol]

There are counterexamples even if you force rings to have identity tbf

[thorn delta]

No there isn’t

[thick abyss]

Proving these groups aren't isomorphic: $D_4 \cross \mathbb{Z}3, D{12}, A_4 \cross \mathbb{Z}_2, \mathbb{Z}_2 \cross D_6, S_4, \mathbb{Z}_12 \cross \mathbb{Z}_2$

Is the easiest way finding the number of elements of order 2, and noticing they're different, or is there a nicer way?

[cloud walrus]

ryаn

[thick abyss]

Since that seems to be the tactic used by most people.

[next obsidian]

That’s the first thing I’d do

Count up how many elements of each order there are

[patent girder]

Let $H<G$ be a subgroup containing $S_{q_i}$ for some primes $q_i$ and let
$q$ be an other prime dividing $n$. Is it true that $H\cdot Sq$ only
contains $S{q_i}$ and $S_q$ but no other minimal normal subgroup?

[cloud walrus]

ohNoiAmHere

[fallow plume]

is the significance of this proposition that $p(x) \in E$ such that $p(x) \not\in K[x]$ is automatically reducible?

[cloud walrus]

Thomas

[median pawn]

Trying to understand this proof

Why is every free left R-module semisimple?

[fallow plume]

me who thought i did module theory but hasn't even heard of semisimple

[median pawn]

Well, an R-module is semisimple if it can be written as the direct sum of simple modules

https://math.stackexchange.com/questions/2055190/why-r-is-semisimple-ring-iff-every-r-module-is-semisimple

Found the answer here

[fallow plume]

oh that's not nearly as bad as I thought it'd be

[median pawn]

Proposition 4.1 says that a module is semisimple iff every submodule is a direct summand

[cloud walrus]

Hausdorff

[next obsidian]

Identify E with a submodule of B, then write B = E (+) M, the splitting map is given by projecting the first coordinate

This is just a general statement for modules, look up what the splitting lemma is

[median pawn]

but M need not be equal to C is my problem

[next obsidian]

It literally is isomorphic

Because both are the cokernel

Also this doesn’t matter, left split = right split

Again, look up what the splitting lemma is

[cloud walrus]

Hausdorff

[next obsidian]

The way that E sits Inside B = E (+) M is via including into the left component

Once you identify E with a submodule this is just what it means to be a direct summand

[median pawn]

Ah, I see! I'm just confused because there is no reason for the map f (which is what I specified in the original s.e.s.) to equal the inclusion map?

[next obsidian]

Yes, that’s why you have to just identify E with its image via f

Using the first isomorphism theorem

In reality you’re going to be taking pi and then composing it with f^-1

0 -> E -f> E’ -> E’ (+) C = B

E’ being the image of E, then you take
E’ (+) C -pi> E’ -f^-1> E

This is the splitting map if you want to avoid identifying E with E’

[cloud walrus]

Hausdorff

[next obsidian]

Uhhh… well… I guess?

I feel like writing B = E (+) M is wrong

If that = is supposed to be an internal direct sum

Like as a set B = f(E) (+) M

Where that (+) means {a + b| a in f(E), b in M}

[cloud walrus]

Hausdorff

[next obsidian]

Yeah but those two (+) kinda mean different things like

The right side is an external, the left side is internal

This totally doesn’t matter

But if you’re confused about what’s really going on here it’s worth noting

[median pawn]

Yep, I understand the difference, but I think we can afford to be loose with notation here

[next obsidian]

Yeah, I just wanted to be clear what’s really going on

[cloud walrus]

Hausdorff

Hausdorff

[thorn delta]

this would work, yeah. Of course you could also shrink the direct sum to be only over a set of generators of M

[foggy merlin]

how hard is the proof of $\mathbb{C} \cong \overline{\mathbb{C}(x)}$ ?

[cloud walrus]

Dr. J. Stockfish

[foggy merlin]

isomorphic just as fields and not as $\mathbb{C}$-algebras

[cloud walrus]

Dr. J. Stockfish

[pastel cliff]

moldilocks

[hidden haven]

You need to know about transcendence bases and about when you can extend field homomorphisms

[foggy merlin]

it's supposed to be a corollary of the result that all uncountable fields with the same charcteristic and the same cardinality are isomorphic.. but this seems like it would require a very ugly proof
edit: *algebraically closed fields

[chilly ocean]

Suppose i have sequence of groups $A_n \xrightarrow{f_n} A_{n+1}$, if these eventually stablize, i.e, the $f_n$ are isomorphisms after some $n$

[cloud walrus]

lime_soup

[chilly ocean]

is the colimit of this system just whatever group A_N

[hidden haven]

You need them to be algebraically closed in that theorem

[hidden haven]

Yes

[pastel cliff]

really quick: what can i conjugate s_3 by to get s_1

in S_4

[hidden haven]

What is s_i

[chilly ocean]

is this obvious

[pastel cliff]

oh right my prof is a crank using crank notation

[hidden haven]

Yes, check the universal property

[pastel cliff]

s_i = (i,i+1)

[hidden haven]

(13)(24)

[pastel cliff]

can u spell it out a stupid little man

,ti

[cloud walrus]

The current time for nitezba is 06:13 AM (EDT) on Thu, 14/04/2022.

[pastel cliff]

guess how much ive slept

[hidden haven]

bruh

If you start with the permutation (i,j)

and you conjugate that by some permutation p

Then you get the permutation (p(i), p(j))

[pastel cliff]

damn

im so used to the little fucking s_i thing that i didnt see that

[hidden haven]

F

[foggy merlin]

ah yeah forgot to menion that

[hidden haven]

The proof of that isn't too bad

Same uncountable cardinality implies same transcendence dimension

So there is a bijection map of transcendence bases

You extend this to a field homomorphism of the purely transcendental parts

And then everything remaining is algebraic over this

And you need a lemma which says that homomorphisms from a subfield into an algebraically closed field can always be extended

[foggy merlin]

so you'd just look at C and colsure of C(x) as fields over the algebraic closure of Q in the example I mentioned?

[hidden haven]

Which you prove by proving it for simple extensions and then using Zorn's lemma

[hidden haven]

As fields over Q should be enough

[foggy merlin]

but this means that the isomorphism cant really be constructed in the general case right?

as this uses zorn's Lemma (or some weakened version of it idk)

[hidden haven]

What do you mean by general case

Math assumes Zorn's lemma

[foggy merlin]

I mean for the mentioned example, can that isomorphism be constructed or be given by some explicit formula?

[hidden haven]

No I don't think so

Zorn's lemma is required to prove that Aut(C) is non trivial, and it feels like Aut(C(x) closure) should be non trivial without Zorn's lemma, so they shouldn't be isomorphic without it

Maybe we already used it to prove that C(x) closure exists

[foggy merlin]

it's usually already used (or some weakened version of it) to prove that an algebraic closure of any field exists

I thought

[hidden haven]

ye

[chilly radish]

wait why do you need zorn to prove that Aut(C) is nontrivial? Isn't conjugation clearly a nontrivial automorphism?

[hidden haven]

oh yeah other than conjugation

lul

[chilly radish]

oh ok lol

[hidden haven]

Aut(C) is uncountable

[chilly radish]

yeye I get what u mean

[hot tinsel]

any hints for what such an non-frobenius automorphism looks like

Source: Morandi Fields & Galois Theory

[hot lake]

you could try to glue together automorphisms of Z/p^nZ

[pastel cliff]

moldi & wew lads remember how my prof is a literal fuckster when it comes to grading

just got my first 100 on an exam of my undergrad (and im a junior )

so excited to walk into his office hours and wipe my ass with that exam

[next obsidian]

I think I’ve only aced a single math test in my entire undergrad lol

[frank fiber]

is there an implementation in gap of the p-group generation algorithm?

[delicate orchid]

cracked beyond all recognition

[ember field]

What's the idea behind for having the condition f(xy)=f(x)f(y) for isomorphism

[sharp sonnet]

respecting the group operation

[pastel cliff]

they call me rad-aline

[shell brook]

just want to make sure im not going insane, but in this proof of eisensteins, when they do say "So there is a least integer t", they really mean t = r, right?

\

like by assumption t must be r?

oh wait okay maybe not

Gallian

[delicate orchid]

yeah looks like r is just some index for b but t is the minimal integer such that blah blah

oh no there it is again

hold on lemme keep reading

[shell brook]

yeah i dont get why he did this lmao

i thought u want the rth coeff of f(x)

[delicate orchid]

I don't think it always has to be r is the problem

[shell brook]

oh is the idea that this handles the case when you have s>r or smth

maybe that inequality is backwards

[delicate orchid]

well the degree of h is bounded by "n", is n the degree of g? I can't see it defined

[shell brook]

n is the degree of f sry

[delicate orchid]

no worries

[shell brook]

f(x) = a_nx^n + ... + a_0

standard eisenstein setup

[delicate orchid]

ohhh I think it's because p divides b_0 but not b_r there must be some element inbetween them such that it's the largest element p divides

[shell brook]

[delicate orchid]

if I'm visualising this polynomial multiplication right anyway

[shell brook]

how is that guaranteed

oh okay this is some cosnequence of poly multiplcation yes

[delicate orchid]

polynomial multiplication my one (of seven) weakness(es)

yeah like if you look at the form of a_t then it factors like b_tc_0+p(nonsense) which then leads to a direct contradiction regardless of if t equals r or not

I feel like it's just a bit of an extra general statement, I'm struggling to see it as well

I think it plays into the fact that we have to have p|a_i for i between 0 and n-1

[azure plinth]

Okay so guys I am stuck on this problem from character theory, seems like I might be missing something? I can't seem to be making any progress on this. C_x is the conjugacy class of x in G

[delicate orchid]

that's the lifted character, right?

[azure plinth]

Hmmm let me think about that. I can see the resemblance, however this was a problem given to us a month ago, when lifted characters wasn't taught to us yet. My point being, sure, I can give some thought to that, and that might lead to a nice and quick solution, but I believe it can be solved without any concepts from the theory of lifted characters, which is what I am wondering about

[delicate orchid]

yeah I'll need to think about this one

just to confirm this permutation character is the one given by representing G through permutation matrices corresponding to how G acts on X, right?

[azure plinth]

I believe that's correct

[delicate orchid]

ok so the character of x would correspond to the number of cosets fixed by left multiplication by x
and the fixed cosets are exactly the ones such that ga_i \in a_iH <=> a_ig \in a_iH - might be something there?

[azure plinth]

Indeed, that's it. And the RHS is essentially (|G|/|Cx|)/(|H|/|Cx \cap H|), which is neat but I am not quite sure what to make of it

Okay this might lead somewhere

[delicate orchid]

just trying to get at the connection between x commuting with elements of H and why that would make the number of fixed cosets smaller/larger

[azure plinth]

This is the quotient of centralisers of x in G and x in H essentially. And ah, go on, by all means. That makes sense and I was thinking about just that right now. That a and g will have to "commute", abusing terminology, in that sense. Which should require g to be in H in addition perhaps?

Essentially that should mean gaH =ahH for some h in H I guess

[delicate orchid]

that is the implication I was trying to see but I'm not sure if it's true

[delicate orchid]

I think this is correct

[azure plinth]

I am trying to think along the quotient of centralisers line for now, hmm...

[delicate orchid]

I also recalled this which may be of use, X^g is the number of fixed points i.e. the character and and |X/G| is the number of orbits

I think it's burnside's lemma?

[azure plinth]

I believe it is

[delicate orchid]

it's giving me some inner product of characters vibes from the 1/|G| and the summing all over G

[azure plinth]

Oh yes, that it does. Again, I know about inner products of characters now, but it wasn't taught at that point, so there might be a solution without that. But that being said, I wouldn't mind a solution involving inner products either, wouldn't make a difference really

[delicate orchid]

no worries I think I've found something different

as we discussed before a_iH being fixed implies that ga_i = a_ih, but if g is in H then g is in C_g \cap H = C_g

ah nevermind that's trivial, unfortunately

it does mean that g is in C_{a_i} \cap H though which is a bit more interesting

[azure plinth]

Aye, precisely

[delicate orchid]

ah have you found all of this already

[azure plinth]

Oh I have been thinking so yeah, some of what we have been talking about :P Anyway, I am thinking of the left cosets of C_H(x) wrt C_G(x), where C_G is the centraliser of x wrt the group G, and similarly we can define C_H

Each coset would have elements ah, where a and h are both in the centraliser of x, so we essentially have a xah=axh=ahx situation

[delicate orchid]

where a is the coset representative?

[azure plinth]

Which would make sense, x would fix ahH=aH then

Oh yes

Or more precisely a is in the centraliser of x wrt G

h in that of x wrt H

[delicate orchid]

I seeeee

So are those elements exactly the ones that fix aH?

Or rather, are they exactly those aH that get fixed

[azure plinth]

That is the only part of the question that remains to be solved I suppose. We know that these are fixed by x, now we have to show that these are the only ones that get fixed

[delicate orchid]

Yeah cause then if you do that you’ve basically done it

The number of those cosets is just |C_G|/|C_H| right?

[azure plinth]

Precisely, it should be quite doable hereon I suppose. And oh yes, in that case it would be that. And that equals the RHS

So xaH= aH = ahH for some h in H

So xa = ah for some h I guess?

[delicate orchid]

I believe so yeah

[delicate orchid]

You’ll have to say that this holds for all h in H though

So then you can pick a specific h’ such that xa = ah’, the existence of which is guaranteed by xaH = aH

[azure plinth]

Oh yes, ignore the last ahH thing maybe? And oh that's a better way of going about it, I like that!

[azure plinth]

I meant this. This should do and thus finish this solution. Dang, thinking out loud and seeking help does wonders. I think we are done with this problem

Thanks a bunch for your help :) I really appreciate it and can sleep better without this gnawing on my mind so to speak :P

[delicate orchid]

No worries, I’m just glad my area of expertise was finally useful

[azure plinth]

XD And I appreciate it mate

[strong yacht]

Does anybody have examples or a reference to some examples of determining Galois groups of a polynomial f with coefs in Q by reduction modulo p? I know that the Galois group of f_p (f reduced mod p) is cyclic and embeds in the Galois group of f, and I think you can say a bit more (the lengths of irreducible factors of f_p correspond to a particular permutation with cycle type relating to those lengths in Galois group of f viewed as a subgroup of S_n).

[merry island]

a quick and potentially dumb question about Nakayama's Lemma - i've seen two different versions, statement 1 and statement 2 on this page: https://en.wikipedia.org/wiki/Nakayama's_lemma

if i have a proof of statement 2 without using statement 1, is it possible to prove statement 1 or is it actually a stronger result?

[desert dome]

Hi, I'm reading through this proof. Could anyone help explain a bit why at least one of P->P' and P' -> P is surjective? Thank you!

[next obsidian]

I don’t buy this

I think this is wrong, namely if M = 0 then you can let P = P’, and consider the only map P -> M and P’ -> M

You can factor this using the 0-map,
P -0> P’ -0> M and P’-0> P-0> M

[chilly ocean]

What does subdomain mean

[next obsidian]

@desert dome

[latent anvil]

No, bc the map P -> M isn't zero

Also the lifting thing isn't a universal property bc it's not unique

[next obsidian]

eh?

[latent anvil]

we lift φ' along φ to get a map ψ : P -> P' with φ' ° ψ = φ

if ψ = 0 then we'd have φ = 0

yeah?

[next obsidian]

I mean both maps P -> M and P' -> M are 0

you can lift these via the 0-maps

[latent anvil]

No??

They're surjective

[next obsidian]

M = 0

In my example

I'm just showing that this setup isn't sufficient to conclude that one of P -> P' or P' -> P is surjective

[latent anvil]

Oh I didn't see that lol

[desert dome]

Yeah I see...

[latent anvil]

yeah okay I think your example shows that this won't work

[desert dome]

[next obsidian]

I don't know if it's possible to conclude that some factorization exists with a surjective mpa

but just invoking the definition of projective modules isn't good enough because of a lack of uniqueness on lifts

[desert dome]

I see. Thanks!! I'll think about this more

[mint gulch]

Hi

Someone knows how to interpret tge sum that is in the proof?

Phi is a module endomorphis but a_(i, j) is an element on the ideal and then all is multiplied by x_j 🤔

[next obsidian]

So there’s a couple things

This is actually applying this map to x_j

so the map inside of the parentheses is going to be delta_{ij}phi - a_ij, which means it sends x_j to delta_ijphi(x_j) - a_ijx_j

we can view any scalar as an endomorphism by just multiplying by it

so in total this expression sends x_j to

$\sum_{j=1}^n \delta_{ij}\phi(x_j) - a_{ij}x_j$

[cloud walrus]

CHMOLUMBIA

[mint gulch]

I see

Thank you 👍

[next obsidian]

👍

[mint gulch]

I think he should add at least () then

[next obsidian]

¯_(ツ)_/¯

It is common practice at a certain point to just drop parentheses

Like when talking about function composition to just write gf

And something like gfx = g•f(x)

[mint gulch]

Yes, but this time there was a scalar then

Seema confusing

[next obsidian]

It kinda is bad, but once you know that it’s a convention you can usually figure out what is going on by context

Yeah, but one way to think of scalars is as endomorpjism

One way to define an A-module is via a ring map A -> End(M) where the latter is endomorphisms as an abelian group, with multiplication given by composition

The way you do this is, if you have a module structure as you are familiar with, you define f:A -> End(M) by saying f(a) is the endomorphism
a-

Which is to say x -> ax

And conversely given such an f, you can define ax to be f(a)(x)

So really you can think of scalars as endomorphisms in a very very natural way which makes this notation make more sense

[mint gulch]

Ok. Thank you. I'll continue reading then

[chilly ocean]

Why is I ideal

I know now

[toxic zephyr]

in the ring $\bZ^{n\times n}$, would the ideal $(2\bZ)^{n\times n}$ (the set of all matrices with even entries) be generated by $2I$?

[cloud walrus]

nix

[toxic zephyr]

ive only heard my professor talk about principal ideals being generated in commutative rings. 2I does commute with every matrix though... so would this be a principal ideal?

and could i denote it as $(2I)$ or $\langle2I\rangle$?

[cloud walrus]

nix

[foggy merlin]

anyone know a nonabelian group of order 7^9 by chance? or know that there exist none?💀

wait nevermind, I think just using the construction of a nonabeliangroup of order p^3 should work when I product it with some abelian group of order 7^6

[oblique leaf]

I have to find what $$\mathbb{F}{q^s}\bigotimes{\mathbb{F}p}\mathbb{F}{q^r}$$ is and then show why

[cloud walrus]

alyosha

[oblique leaf]

I know that for a polynomial of degree m, say r, we have that $$\mathbb{F}{p^n}[X]/\langle r\rangle\cong\bigoplus_i\mathbb{F}{p^n}[X]/\langle r_i\rangle$$ with $$r_i$$ the factors of $$r$$

[cloud walrus]

alyosha

[oblique leaf]

Does anyone know how to proceed?

[lethal dune]

is the supposed to be over F_q?

[oblique leaf]

yes

[lethal dune]

say V and W are VS over F with dim m and n, then V \otimes W is a vector space over F with dim mn

you can expect the tensor prod to be a vector space to be of dim sr

so try finding a bilinear map from that to F_q^(rs)

might help IDK

say v \in Fq^s be coordinates, same for w\in Fq^r, then try the function vw^T

[oblique leaf]

hmm but what does that achieve?

because the our tensor product of finite fields isn't in fact isomorphic to F_q^{rs}

it is isomorphic to the direct sum of gcd(m,n) copies of F_{p^l} where l=lcm(m,n)

@lethal dune

[lethal dune]

then idk

[barren sierra]

Why are seperable extensions called seperable

I don't get the name

I feel square free might be a better name

But maybe there's an intuition for these extensions that lead to the name seperable

That I don't quite get

[south patrol]

according to keith conrad (and ig intuition) it's because the roots all sort of 'separate out'

[uncut girder]

If you're a normal subgroup, then you are the only element of your conjugacy class of subgroups, right?

[delicate orchid]

yur

can't conjugate cause then gHg^-1 = gg^-1H = woahhhhh

[devout crow]

is it more common to write "I is an ideal of R" or "I is an ideal in R"?

[tribal moss]

"of"

[patent girder]

Let $H<G$ be a subgroup containing $S_{q_i}$ for some primes $q_i$ and let
$q$ be an other prime dividing $n$. Is it true that $H\cdot Sq$ only
contains $S{q_i}$ and $S_q$ but no other minimal normal subgroup?

I'm fairly sure that the answer is yes, since $S_{q_i}$'s are contained in H. The group specifically that im working is G(m,p,n)

[cloud walrus]

ohNoiAmHere

[patent girder]

Since the group is defined for primes dividing m, there should be some link there right?

I just can't prove it

[strong yacht]

In my notes, a poly is separable iff its irreducible factors are separable (and for irred polys it means having distinct factors in splitting field), e.g. x^2 (over Q say) is separable but not square free

So separable does not necessarily coincide with square free (which is already a coined term that means something else) depending on your defs

[uncut girder]

@delicate orchid @latent anvil here's the heawood map obtained as a quotient of the upper half plane by an index 42 torsion free normal subgroup of the modular group

[delicate orchid]

this is pretty mental in the mind imo

[shell brook]

$\mathrm{Gal}(\bQ(\sqrt[3]{2}, \sqrt{-3})/\bQ) \cong \bZ_2$?

[cloud walrus]

jesse

[shell brook]

Gal(Q(cbrt(2))/Q) = {e}, surely.

[next obsidian]

Definitely not

I mean what does Gal mean to you?

[shell brook]

Automorphisms

[next obsidian]

Is that only for Galois extensions or is it for automorphisms that fix the base field?

It must mean the latter

[shell brook]

In my course it is for Gal extensions only but I am doing an exam from another version of the course

does this change the answer

[next obsidian]

Well yeah

Q(cbrt(2)) isn't Galois over Q for example

but assuming that the one you originally asked about is Galois

then it can't be Z_2, because the size of the Galois group has to be equal to the degree of the field extension

and that isn't a degree 2 field extension

[shell brook]

which is 6

ah

[next obsidian]

I think it's degree 6

[shell brook]

yeah

[next obsidian]

But idk if that extension is Galois

[shell brook]

okay wait my book says Q(cbrt2)/Q is trivial Gal group so it must be the second definition you wrote

[next obsidian]

And I am shit at determining if extensions are Galois so

tfw

[shell brook]

hm well the exam just says 'Compute the galois group'

[next obsidian]

anyway, I don't really know the answer

[shell brook]

[next obsidian]

[shell brook]

I thought I could just determine the uatomorphisms independently

and then be like. okay its the product

[next obsidian]

I think this is true given that like you know a bunch of stuff are Galois

But I am shit at field theory / Galois theory so

idk

[shell brook]

devastating

hm

[desert dome]

Hi everyone, I asked about this question yesterday. I discussed it with my friend and our thought is like: from s.e.s, we have M = P/ker(phi) and M = P'/ker(phi'). Then by universal property, we kinda have either P is a submodule for P' or P' is a submodule for P. WLOG, suppose P is a submodule of P', we naturally have a surjection. (The surjection mentioned in the last line is not the same homomorphism as from universal property) I'm still pretty confused, could anyone give me some suggestion? Thanks!

[next obsidian]

I don't see how you can conclude that P is a submodule of P' or that P' is a submodule of P

I just think this proof is fundamentally flawed, and that you should ditch it and try to prove it yourself

[desert dome]

I thought this could come from the s.e.s... but yeah I don't quite get it. Do you have any suggestions about how to approach this problem?

[next obsidian]

It either
1: Follows from some very general nonsense (not recommended unless you have seen said nonsense)

2: you can define a map adhoc, to define a map from ker phi' (+) P -> ker phi (+) P' it is equivalent to give four maps

P -> ker phi, P -> P', ker phi' -> ker phi, ker phi' -> P'

using the fact that P/ker phi = P'/ker phi' you can figure out a way to just sort of define a map that works

I guess here you do use that you get a map P -> P' and P' -> P

you just don't know that either of them are injective

[desert dome]

Thanks!! I'll try to figure it out

[next obsidian]

Also one more thing I guess

Because P' is projective, surjections onto it split

so you really want to find an SES like this

0 -> ker phi -> P (+) ker phi' -> P' -> 0

you can use all the info you have to make that SES, and then because it's split you get your desired isomorphism

[desert dome]

Thank you!! I'll try

[next obsidian]

GL! FWIW this is how I did this problem in the past, and the maps I made are kinda icky, but if you just keep trying to make the maps, you should basically be forced to stumble on the right thing

there's kinda only one way to do this I think

[desert dome]

Sounds good!

[shell brook]

Any galois gregs in chat want to help :v

very tricky

[desert dome]

Hey! Sorry to bother again. By universal property, I have nu: P -> P' and nu': P' -> P. I can show that nu(ker phi) is subset of ker phi'. I am currently thinking define the morphism ker phi to P oplus ker phi' as a -> (a, nu(a)). However, I had a hard time thinking of a reasonable morphism from P oplus ker phi' to P'. Do you mind giving me another hint?

[next obsidian]

To define that map it suffices to give a map P -> P’ and one ker phi’ -> P’

(The map from the direct sun will just add the two things together)

There’s kinda two obvious maps here, but I believe you’ll need a sign change in one side in order to have this actually be an exact sequence

Also one sort of thing that is subtle, I wouldn’t really call it a universal property for projective modules

Universal properties should involve unique maps

But the maps from a projective module are not unique

This is a really important point because it can sometimes complicate things somewhat

[desert dome]

Thanks! I'll pay attention to that. I'm thinking about adding them together too, but my map P to P' is nu and ker phi' to P' is embedding. However I have a really hard time proving this is surjective

[next obsidian]

Yeah, so you want to subtract those two maps

In that case you’ll see that this is a chain complex because a goes to (a,nu(a)) which goes to nu(a) - nu(a)

To show that it’s surjective is when you use nu’ I think

And probably to show exactness in the middle as well

[desert dome]

I see. Thanks!!

[next obsidian]

Also

At some point I think you need to take a in P’

Map it to some m in M

Then you know there’s a b in P mapping to m

I think that is needed as well

[desert dome]

Thank you! I'll note that

[next obsidian]

Also @desert dome I came up with an example to show the statement about the projective modules embedding into each other is false

Note that any two modules surject onto 0, so if you could do what that proof tried to do, you’d get that for any two projective modules P,P’ that one of them embeds into the other

In a semisimple ring (like for example CG if you’ve done representation theory) any module is projective

So you can take two nonisomorphic nonzero simple modules and they won’t be able to embed into each other

So that proof is unsalvageable because there’s no way to guarantee that there’s an injection from P to P’ or vice versa

Shamrock gave me a better example that’s more hands on

Consider R = k (+) k, and P = k (+) 0 and P’ = 0 (+) k, both considered a module via the obvious map from R

P and P’ are projective since P (+) P’ = R

But you can see that there’s no way to embed either of P or P’ into the other because of R-linearity

[desert dome]

I see. Tysm!! I really appreciate it

[strong yacht]

One characterisation of Galois extensions is that they are precisely the extensions which "have all the automorphisms they need" i.e. this is a degree six extension, so you want six automorphisms. Proving that it is Z_2 is actually a proof that it isn't a Galois extension (I think)

[strong yacht]

This might work: the min poly of $\sqrt[3]{2}$ and $\sqrt{-3}$ are: $x^3 - 2$ and$ x^2 + 3$. The roots of the first min poly are $\sqrt[3]{2}, \sqrt[3]{2}(-0.5+0.5\sqrt{-3}), \sqrt[3]{2}(-0.5-0.5\sqrt{-3})$ and the roots of the second are $\sqrt{-3}, -\sqrt{-3}$. So actually an automorphism is completely determined by where you map $\sqrt{-3}$, and since its min poly is $x^2+3$ you have two choices for that, hence the automorphism group (I don't like the term Galois group for non-Galois extensions) is $\mathbb{Z}_2$.

[cloud walrus]

Greenman

[strong yacht]

This doesn't feel right actually

[strong yacht]

Is this statement actually true?

I don't think so, I mean what I wrote is complete nonsense, isn't this the splitting field of x^3-2 so it's Galois (splitting fields are Galois) so it has at least as many auts as its degree (which is 6)?

[next obsidian]

I just don’t think this is the splitting field of it

The three roots are cbrt(3), wcbrt(3) and w^2cbrt(3) where w is a primitive third root of unity

This has argument 2pi/3

But everything inside of Q(cbrt(3),sqrt(-2)) has an argument a multiple of pi/2

In fact I think this is a proof this isn’t Galois, because this extension isn’t normal

[strong yacht]

Wait hold on just a small point, did you switch your threes and twos around

[next obsidian]

I just don’t know how to compute the automorphism group because I suck ass at field theory

Idk did I?

[strong yacht]

It's cube root 2, and square root -3 right?

[next obsidian]

Whatever

It doesn’t really matter

[strong yacht]

I think it may make a difference

[next obsidian]

You don’t have any multiple of a third root of unity which is all that matters I’m pretty sure

[strong yacht]

Wolfram Alpha says these are the roots of x^3 - 2 = 0

What you said is correct, but you can write those roots in terms of sqrt(-3)

[next obsidian]

Wtf

Literally how does that work

Oh I’m so fucking dumb

Yeah

If you just add something to i its argument is no longer pi/2 lmao

I’m so bad at field theory

Yeah idk maybe this is the splitting field lol

[strong yacht]

Same man, I'm highly motivated for Galois theory rn cos of exams and a project

[next obsidian]

And the Galois group can’t be abelian

Because Q(cbrt(2)) isn’t Galois

And if my memory serves me the only groups of order 6 are S_3 and Z_6

So this pins it down as S_3

[strong yacht]

Another way I think of seeing it: the Galois group are the permutations of the roots of x^3-2 so embeds in S_3 so must be S_3 (can't be Z_6, that acts on six things)

[uncut girder]

I have a question.

Say G is a finite group. Consider two surjective homomorphisms
h_1, h_2 : PSL(2,Z) -> G
Suppose that ker(h_1), ker(h_2) are torsion free. Is it true that ker(h_1) = ker(h_2)?

[coarse storm]

Any examples of such maps?

[harsh anvil]

If I want to use any 2 generators to produce a hamiltonian path through a symmetric group, does anyone have any ideas where I might begin?

[delicate orchid]

I know of a way to generate any symmetric group from just two elements but I have no idea what a Hamiltonian path is

[harsh anvil]

I think that would be good enough! got a hint?

[delicate orchid]

you can generate any symmetric group S_n with a transposition and an n cycle, specifically ||(1,2) and (1, 2, 3, ..., n)||

[barren sierra]

Ok so

I've determined the roots are $\frac{1}{\sqrt[4]{2}}(\pm1\pm i)$

[cloud walrus]

Spamakin🎷

[barren sierra]

Obviously then the splitting field is $\mathbb{Q}\left( \frac{1}{\sqrt[4]{2}}(\pm1\pm i) \right)$

[cloud walrus]

Spamakin🎷

[barren sierra]

if I can show this is equal to $\mathbb{Q}(\sqrt[4]{2}, i)$ the degree becomes immediate

[cloud walrus]

Spamakin🎷

[barren sierra]

but showing this equality is proving to be a pain.

One direction is immediate but I can't get the other direction

[chilly ocean]

Which direction have you proved so far?

[barren sierra]

very obvious

[wooden ember]

dont you mean the other direction

[chilly ocean]

well, now you just need to prove that the generator of the right hand side belongs to the left field

(I'm too lazy to type them out)

I think that the one that you proved is the least obvious of the two

[wooden ember]

yeah which is why i think you got the inclusion the wrong way around

cause the reverse inclusion is very obvious

[barren sierra]

oh wait

yea I typed that backwards

[wooden ember]

yeah then for the other inclusion i can see how the algebra would be tedious

ill pull up some paper and see if i can figure anything out

[barren sierra]

$\mathbb{Q}\left( \frac{1}{\sqrt[4]{2}}(\pm1\pm i) \right) \subseteq \mathbb{Q}(\sqrt[4]{2}, i) $
this is what I have showed

and yea maybe there's some algebra I am missing but I can't get it to work

[wooden ember]

just to be sure your +- means the extensions are (1+i)/2^1/4 and (-1-i)/2^1/4 or also (1-i)/2^1/4 and (-1+i)/2^1/4

[barren sierra]

yea

[wooden ember]

dont mind my lack of parentheses

[barren sierra]

4 roots

[wooden ember]

aight thanks

[chilly ocean]

well if you multiply any of those roots by i enough times you can get all of them

[wooden ember]

alright i got 2^(1/4)

take (1+i)/(2^1/4) * (1-i)/(2^1/4) to get sqrt(2)

[barren sierra]

so fast?

man I am bad at algebra

[wooden ember]

add these same roots to get 2^(3/4)

then divide the two

and i follows easily

take 1+i/(2^(1/4)) + (-1+i)/(2^1/4) and divide by 2^(3/4)

i think this works

[barren sierra]

yea once you get 2^(1/4) or i

the other would be easy to get

it checks out

damn

[wooden ember]

in general with these i think it's good to randomly take products and sums and see what intermediary values you can produce

and only divide by the end

oh if you think about them geometrically i comes straightforwardly as (-1+i)/(2^(1/4)) divided by (1+i)/(2^(1/4))

[barren sierra]

I must be missing something

how is this not just

i is in L, i is not in K, QED

[shell brook]

Ohh this helps a lot

thank you

[shell brook]

Okay nvm I dont get it

I think an automorphism of this group has to fix cbrt2

like let s be an aut, and since s fixes Q,
2 = s(2) = s((cbrt2)^3) = s(cbrt2)^3
and so s(cbrt2) = cbrt2

So you can only determine the aut by where it sends sqrt(-3), no?

[chilly radish]

This shows the degree is at least 2. Now you have to show that the degree is at most 2

[barren sierra]

Ah

I see

[chilly radish]

Like it's not 100% immediate but it should be pretty easy

[barren sierra]

Yea

[chilly radish]

Just gotta see that putting i in you're not missing any complex algebraic numbers

That you couldn't otherwise write in Q^alg\cap R(i)

[barren sierra]

Simple contradiction

[shell brook]

guh maybe this is wrong, theres a similar question in my notes that is talking about the gal group of the sf of x^3 - 2/Q

[strong yacht]

What you've written here is mostly correct

But your conclusion does not follow

[shell brook]

I do not see why

If cbrt2 is fixed we can only permute the other two roots, no?

[strong yacht]

s(cuberoot2)^3 = 2 does NOT imply s(cuberoot2) = cuberoot2

[shell brook]

Oh

I see

its the only real root but yeah

[strong yacht]

Yeah but what it does imply is that x^3-2 is fixed by an aut

[shell brook]

I see where I was led astray

Okay this makes sense then

thank u :d

[strong yacht]

No probs!

Also here is some good information for you, these types of extensions are called Kummer extensions and are splitting fields of x^n-a

[chilly ocean]

coomer extensions

[strong yacht]

Literally realised they sound like coomer extensions yesterday

[next obsidian]

Tbh we don’t need that here

Goodbye gif

[chilly ocean]

does every exact sequence of abelian groups split

or

is there a mild condition on the groups that makes this true

[devout crow]

$0 \to \bZ_2 \to \bZ_4 \to \bZ_2 \to 0$ is a nonexample, right?

[cloud walrus]

xdres

[next obsidian]

Not really, you can assume that either the first is injective or the last is projective

But in general, there’s absolutely no reason to expect this, and splitting is measured by an Ext group in some sense

One mild condition I guess is also that the first and third are finitely generated, and the middle is isomorphic to the direct sum of the outer two

Just having the middle be the direct sum does not suffice unless you assume finite generation

[chilly ocean]

wow

i did not expect that

[next obsidian]

It is a surprising result IMO

But the proof is not too bad

I can’t find a reference for it tho oof

https://math.stackexchange.com/questions/135444/a-nonsplit-short-exact-sequence-of-abelian-groups-with-b-cong-a-oplus-c

Check the first answer

[long obsidian]

Are subgroups of free groups necessarily free?

[chilly ocean]

https://en.wikipedia.org/wiki/Nielsen–Schreier_theorem

yes

it's usually proven using some basic algebraic topology

[long obsidian]

Thank you so much!

[languid walrus]

well, F[x] is not a field

[latent anvil]

maybe $E = \Q(\sqrt{1 + \sqrt{2}}, i)$

[cloud walrus]

tr.deg.Shamroc/k

[latent anvil]

so let's think

the galois group is order 8

we have a non-normal subgroup of order $2$ and a normal subgroup of order $4$

[cloud walrus]

tr.deg.Shamroc/k

[latent anvil]

the auts fixing sqrt(1 + sqrt(2)) and the auts fixing i

bc of the non-normal subgroup of order 2

that rules out Q8 or any abelian group so it's Q8

maybe $\eta = \sqrt{1 +\sqrt{2}} + i$?

[cloud walrus]

tr.deg.Shamroc/k

[latent anvil]

$\eta(\sqrt{1 + \sqrt{2}} - i) - 2 = \sqrt{2}$

[cloud walrus]

tr.deg.Shamroc/k

[latent anvil]

so $\Q(\eta) \subseteq F$ and $\Q(\eta, i) = \Q(\sqrt{1 + \sqrt{2}}, i) = F$

[cloud walrus]

tr.deg.Shamroc/k

[latent anvil]

but $\eta(\eta - 2i) = \sqrt{2}$ so $\Q(\eta, i) = \Q(\eta, \sqrt{2})$

[cloud walrus]

tr.deg.Shamroc/k

[latent anvil]

observe that $\sqrt{1 + \sqrt{2}}\sqrt{1 - \sqrt{2}} = \sqrt{1 - 2} = i$

[cloud walrus]

tr.deg.Shamroc/k

[latent anvil]

F = E lol

$[E : \Q] = 2[\Q(\sqrt{1 + \sqrt{2}} + i, \sqrt{2}) : \Q(\sqrt{2})]$

[cloud walrus]

tr.deg.Shamroc/k

[latent anvil]

$E = \Q(\eta, \sqrt{1 + \sqrt{2}})$

[cloud walrus]

tr.deg.Shamroc/k

[latent anvil]

$\eta^2 = 1 + \sqrt{2} + 2 i \sqrt{1 +\sqrt{2}} - 1 = \sqrt{2} + 2 i \sqrt{1 +\sqrt{2}}$

[cloud walrus]

tr.deg.Shamroc/k

[latent anvil]

$\eta^2 - \eta = \sqrt{2} - i + (2i - 1) \sqrt{1 +\sqrt{2}}$

[cloud walrus]

tr.deg.Shamroc/k

[latent anvil]

we know $[\Q(\eta, i) : \Q(i)] = [\Q(\eta, i) : \Q(\sqrt{2})] = 4$

[cloud walrus]

tr.deg.Shamroc/k

[latent anvil]

$[\Q(\eta) : \Q(\eta) \cap \Q(i)] = [\Q(\eta) : \Q(\eta) \cap \Q(\sqrt{2})] = 4$

[cloud walrus]

tr.deg.Shamroc/k

[latent anvil]

if $[\Q(\eta) : \Q] = 4$ then $\Q(\eta) \cap \Q(i) = \Q(\eta) \cap \Q(\sqrt{2}) = \Q$

[cloud walrus]

tr.deg.Shamroc/k

[round jay]

Prove that if g and h commute and
gcd(|g|, |h|) = 1, then |gh| = |g| |h|.

Any tips guys?

g and h are elements of a group G

[latent anvil]

What does it mean for gh to have a certain order (like |gh| = k means what?)

[round jay]

This means k is the smallest positive integer so that (gh)^k=e

[latent anvil]

right

[round jay]

Where e is identity

[latent anvil]

so what would you need to show in the case k = |g| |h|

[round jay]

I understand the problem statement, but do not know how to prove it

[lethal cipher]

Question: given $G_{i-1}\subset G_i$, why does $[G_{i-1}:G_i]$ being prime imply the quotient group is cyclic?

[cloud walrus]

dackid

[chilly radish]

Groups of prime order are automatically cyclic

Because of Lagranges theorem

[lethal cipher]

If I may, how does Lagranges theorem imply this?

[chilly radish]

Let a be a nontrivial element of a group of prime order. The order of a is either p or 1 by Lagrange's theorem, but a is not the identity, so the order of a is p

I.e. it generates the group

[barren sierra]

Given $H \leq G$ then you must have $|G| = [G : H] |H|$

[cloud walrus]

Spamakin🎷

[chilly radish]

So the group is cyclic

[barren sierra]

set $[G : H]$ prime and it's pretty clear

[cloud walrus]

Spamakin🎷

[lethal cipher]

Oh, of course. Thank you

[chilly radish]

This doesn't really answer the question in the case that G and H are infinite spamakin

[barren sierra]

oh

right that's a thing

[chilly radish]

e.g. Z and pZ for p a prime

[barren sierra]

yea

well I can't count to infinity so it doesn't exist

[lethal cipher]

Luckily for us, it's not a bother too much for me rn. I only care about finite groups

[chilly radish]

Anyways the prime groups being cyclic thing is really important to keep in your back pocket

[round jay]

One idea I had was to prove that |h||g| divides |gh|

[barren sierra]

I'm at the point where I want to show that the degree is at most 2 (showed that it's at least 2). So for contradiction suppose that the degree is greater than 2 meaning there is some element in $\mathbb{Q}^{\text{alg}}$ that is not in $\left( \mathbb{Q}^{\text{alg}} \cap \mathbb{R}\right)(i)$. This element $\alpha$ has a minimal polynomial in $\mathbb{Q}[x]$ with $\alpha$ as a root and then from there not sure where to go.

[cloud walrus]

Spamakin🎷

[barren sierra]

Can I just say that $\alpha = a + b i$ for some $a, b \in \mathbb{Q}$. Then we have $m(a + bi) + a + b i = a + b i = \alpha$ but $m(a + bi)$ is entirely made of rational numbers and $i$ and so must be in $\left( \mathbb{Q}^{\text{alg}} \cap \mathbb{R}\right)(i)$?

[cloud walrus]

Spamakin🎷

[barren sierra]

that feels off

but I don't know why

what?

[chilly ocean]

How can phi_alpha[F[x]] be a field and not a field

those are two different cases

[barren sierra]

oh I thought you were answering my question, I was so lost lol.

[chilly ocean]

it's a field if and only if alpha is algebraic over F

[chilly ocean]

i'm also lost trying to answer your question...

[barren sierra]

I feel

this should be easy

and yet I'm missing something

[languid walrus]

i would suggest doing this directly: suppose alpha = a + bi in Q^{alg}. what do you want to show about a and b?

[barren sierra]

they must be rational

so I know there is a nonzero polynomial m such that m(alpha) = 0

[languid walrus]

not quite, K = Q^{alg} cap R

[chilly radish]

You just have to prove every algebraic number is of the form a+bi where a,b are real algebraic

[barren sierra]

oh

[chilly radish]

This gives you exactly what you want

You can do this by the contrapositive if you want, or by degree arguments

[barren sierra]

wdym degree arguments

[chilly radish]

You know that an element is algebraic iff the extension it generates is finite

You can use that fact

I don't wanna say anymore cuz that would spoil the question

[round jay]

Can someone help me pls

[chilly radish]

Final hint, spoilered ||if a complex number is a root of a polynomial with real coefficients, so is its conjugate, you can use this to show that the real and imaginary parts are elements of an algebraic extension generated by finitely many algebraic elements and hence, algebraic themselves||

[chilly ocean]

what have you tried so far?

[round jay]

I know that |gh| divides |g||h|

So one idea was to prove the divides relation in the reverse direction

Another idea was to try to prove that lcm(|gh|, |g|) = |gh|

But my issue is that i have no intuition foe this order stuff

[brittle quiver]

Do you not need to check that G_i is normal to form the quotient group

[round jay]

It's very number theoretic and I dunno any number theory

I'm doing 1.14

[chilly ocean]

well, what's something you haven't used yet?

[round jay]

Uhh

That |g|, |h| are relatively prime

But idk what I'm doing tbh I feel lost

[chilly ocean]

you're almost there

[round jay]

So I'm on the right track?

[chilly ocean]

yes, you just need a few more steps. what can you say about $g^N$?

[cloud walrus]

Gio

[round jay]

I tried to calculate its order

If the statement is true, it must be the case that g to the N is e

So i tried to show that its order is 1

I know the least common multiple between |g| and |h| is their product but idk what to do with that

I feel the remainder of the problem is number theoretical and idk number theory :(

[chilly ocean]

yes, and to do so we need to use the fact that |g| and |h| are coprime. what can we say about $(gh)^{| g || h |}$?

[round jay]

I know that's e

[cloud walrus]

Gio

[round jay]

It's e

[chilly ocean]

alright, so $g^{{| g || h |} = h^{- | g || h |}$ correct?

[cloud walrus]

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

[round jay]

I guess

Yeah

[chilly radish]

You're on the right track with this

I feel like you might be overcomplicating this gio. What can you say about (g^N)^|g|?

[round jay]

It's e

[chilly radish]

So consequently

What does this say about the order of g^N

[round jay]

It divides |g|

[chilly radish]

Correct!

Do you see how to procceed?

[round jay]

Hmm..

[chilly radish]

Use the fact that g^N=(h^-1)^N

Try to chew on this a bit. Hint: what can you say about the order of an inverse element in general

[round jay]

It's the same as the order of the element

[chilly radish]

Right

Do you see the idea now?

[round jay]

I'm in a car right now I have to see when I get home

[oblique leaf]

Does anyone know what the Galois group of x^4-A over Q(A) is?

I don't even really know what this question is supposed to mean. What is A in this question? A transcendental?

<@&286206848099549185>

[chilly radish]

That's more of a question to whoever assigned you that question

But i'd wager that yes

A is transcendental

[oblique leaf]

can you give me a hint about how to proceed? I know how to calculate Galois groups of polynomials over Q: finding the roots, the splitting field, calculating the degree etc.

But I have no idea how to find the Galois group over something like Q(A)

[tribal moss]

If A is transcendental, then just pretend in your mind that it's pi.

Q(A) is isomorphic to Q(pi) anyway.

[oblique leaf]

Considering x^4-pi, it seems that it would be irreducible over Q(pi), but im not sure since i know nothing of transcendental extensions ...

[tribal moss]

And so you can reason e.g. that when you adjoin one fourth root of pi, you also get its negative, but you don't get the other two roots because they are pure imaginary if you map the first root you adjoin to sqrt(sqrt(pi)).

[chilly radish]

You mean pure imaginary ye tropo?

[tribal moss]

Yes. Damn keyboard.

[oblique leaf]

So do you mean that the splitting field would be $$\mathbb{Q}(\pi,\sqrt[4]{\pi},i)$$?

[cloud walrus]

alyosha

[tribal moss]

Yes.

[oblique leaf]

So then $$x^4-\pi$$ is the irreducible polynomial for both $$\sqrt[4]{\pi}$$ and whatever root involving $$i$$

[cloud walrus]

alyosha

[tribal moss]

±(pi^1/4)i, of course.

[oblique leaf]

then the yeah

so the galois group is of order 4?

[tribal moss]

No, you have more freedom than that.

[oblique leaf]

wait actually i think it is 8?

[tribal moss]

Yes: To specify a Q(pi)-automorphism you can first chose which of the four roots to send pi^1/4 to. Once you have decided that, you get to decide which of the two remaining roots to send (pi^1/4)i to.

But this gives you an explicit description of all the elements of the Galois group, so you can write down how each of them acts on the roots.

[oblique leaf]

the way we do it for algebraic extensions is that $$x^4-\pi$$ is the irreducible polynomial for $$\sqrt[4]{\pi}$$ so $$[\mathbb{Q}(\sqrt[4]{\pi}:\mathbb{Q}]=4$$ and then $$[\mathbb{Q}(\sqrt[4]{\pi},i):\mathbb{Q}(\sqrt[4]{\pi})]=2$$

[cloud walrus]

alyosha

[oblique leaf]

so the total degree is 8

but I'm not sure this applies to transcendental extensions as well?

[tribal moss]

The transcendence is all internal to the base field, so that all still works here.

[oblique leaf]

I see what you are saying, but I'm not sure how to justify this since aren't you assuming that the automorphisms keep Q(pi) fixed? I know Q is fixed, but I'm not sure how to prove the same for Q(pi)

[tribal moss]

By definition the Galois group consists of exactly the automorphisms that keep Q(pi) fixed.

[oblique leaf]

oh yeah true

ok so this galois group is something like D_8?

[tribal moss]

I haven't actually done the calculations myself, but it sounds plausible.

Since an automorphism needs to preserve the "these two roots are each other's negatives" relation.

But it can swap each pair of opposite roots internally, and also interchange the two pairs. That sounds like the symmetries of a square alright.

[oblique leaf]

yeah i agree

wait, so actually i'm thinking that the question "calculate the galois group of x^4-A over Q(A)" doesn't really depend on what A is at all?

If A is transcendental, you just clarified that all the things for algebraic A still apply

[tribal moss]

As long as A doesn't already have a square or fourth root.

(I.e. as long as x^4-A is indeed irreducible).

Wait, not quite. It doesn't work that way for x^4+1 over Q.

[oblique leaf]

what do you mean it doesn't work that way for x^4+1 over Q?

[tribal moss]

Adjoining just one fourth root of -1 to Q gives you all of them in one go, so the degree of that extension is only 4.

[oblique leaf]

oh right it is V_4

I don't think there's a general answer to this question if we allow A to be rational, irrational, or transcendental

but lets' say if we just allow A to be irrational or transcendental, it is either the process we just discussed, or A=2nd root or 4th root correct

[tribal moss]

I think so.

Hmm, of course we'd better not take something like A = 2^(4/3) either, because then x^4-A is still not irreducible over Q(A).

[oblique leaf]

hmm ok, so it seems more complicated

[oblique leaf]

what do you mean here exactly?

"transcendence is all internal to the base field"

[tribal moss]

In context I think I just meant that the automorphisms we're interested in fix the base field by definition, so the assumed transcendence of A wouldn't give us more relevant automorphisms.

[oblique leaf]

okok that makes sense

thank you so much! I will clarify with my professor what exactly he wants from us and what A is

[chilly ocean]

Does anyone here have any tips or resources for how I should approach studying for my abstract algebra final in about 2 weeks? I’d greatly appreciate anything

[tawdry crystal]

What is the name of this diagram and what does it mean?

[chilly ocean]

are you asking what a commutative diagram is or are you asking about what's going on in this specific one?

[shell brook]

hey so Im just wondering how to prove c normally. I did it by saying that since $\sigma(\alpha) = \alpha_j, \alpha \neq \alpha_j$, and by obvious argument, $\alpha_j$ is a root of $p(x)$ too. So we get that $R = {\sigma(\alpha) : \sigma \in G}$ is a set of roots for $p(x)$ in $E$. I'm not clear on a nice way to show that $R$ is all of the roots of $p(x)$. Essentially if you suppose $\beta \not\in R$, then you want to show $R' = {\sigma(\beta) : \sigma \in G}$ equals $R$? I think this just boils down to $p(x)$ having finitely many roots but the only argument is really handwavey + bad.

[cloud walrus]

jesse

[shell brook]

the only argument *i could come up with

also (a) does not make sense to me actually lol

is it just asking s(g(x)) = g(x)?

I mean thats what I showed

Oh wait lmao

Suppose $\beta\not\in R$. Then since p(x) has finitely many roots and no automorphism fixes elements of E, there must be some $\sigma$ such that $\sigma(\beta) = \alpha$. So $\sigma^{-1}(\alpha) = \beta$. Thus $\beta \in R$.

[cloud walrus]

jesse

[shell brook]

okay well what I asked about (a) still stands

oh well I guess is \sigma guaranteed to exist? idk

probably something really basic guarantees it

ig this depends on the size of the group? i'm overthinking this.