groups-rings-fields

[sour plume]

https://en.wikipedia.org/wiki/Galois_theory#A_non-solvable_quintic_example this is at least an explanation how one proves this about the standard quintic example

maybe that helps in understanding how to creat a degree-6 example

[brisk granite]

yea, I've seen it done for the quintic before

[sour plume]

aw alright

[brisk granite]

I just need a 5-cycle and a transposition, and, then I can conclude that Gal(K/Q) is S_5

it doesn't work for S_6 because when I use cauchy's theorem to note that an order 6 element exists in the galois group, it can be either the product of 3-cycle and a 2-cycle or a 6-cycle

but idk how to find a polynomial where I can conclude that it's a 6-cycle

[latent anvil]

What if you try something stupid and just guess an irreducible polynomial?

also I think there are nonsolvable transitive subgroups of S6

Exotic S5 maybe?

[wind steeple]

maybe reduction modulo p works to, if it's reduction modulo P isn't solvable then your polynomial isn't solvable

bc computing the Galois group isn't easy on Q

[brisk granite]

@latent anvil I did try that, but it was rly hard for me to nail down the Galois group

[sour plume]

https://planetcalc.com/8372/ this looks like a really useful tool for your question

[brisk granite]

Is this true?

What If we consider Z/3Z in S_5?

Z/3Z is solvable (cuz it's abelian) but obv not in a group of order 20

@sour plume uhm, how will it help me?

[sour plume]

You can try fixing some random nice degree-6 polynomial and see all the different factorizations in different finite fields without having to do it all by hand

And those factorizations potentially tell you what kind of cycles you have in the Galois group, right?

But tbh I've not engaged with Galois theory in a long time, your question just motivated me to stumble through it today

e.g. it seems that this Dedekind theorem is applicable to $x^6 - x - 1$ modulo 5 and 3; mod 5, it splits into a linear and a degree 5 factor, and mod 3 it's irreducible, hence the Galois group contains a 5-cycle and a 6-cycle; is that somehow enough to deduce something good about the Galois group?

[cloud walrus]

Lartomato:

[sour plume]

(i think you also have to make sure that your polynomial is separable mod 3 and mod 5, which needs some arguments with the formal derivative, but i think that's mostly standard)

(sorry that that doesn't answer your second question)

[mild oxide]

I feel like this should be pretty simple and easy, but I'm struggling here trying to figure out what exactly the first ring looks like in the first place

[scarlet estuary]

R[x] is, intuitively, R with an added set of elements x^n for natural n

what happens if you say "ok, but now let x^2 + 1 belong to the same equivalence class as 0, so x^2 + 1 = 0 in our quotient set"

"hence x^2 = -1"

"wait a second"

so the intuition behind this is that x = i

up to equivalence classes

this isnt a proof, but that should help you get an idea what it "looks like"

[this isomorphism imo is a really nice example of how "powerful" quotients are]

[in terms of versatility with how you can use them as "rules" to make new rings]

[shrewd halo]

i guess it should be mentioned that the isomorphism, namely, is plugging in i for x in the modded out polynomials and sending it to the resulting number in C

[mild oxide]

sorry, doing some cooking; gimme a sec

[scarlet estuary]

yeah thats what the isomorphism looks like, i just wanted to explain what the ring looks like

so they coudl use that to construct the isomorphism

[shrewd halo]

yeah u right

[mild oxide]

then it sounds like I just send x to i?

didn't even think about x^2 + 1 being in the same equivalence class as 0. I was trying to think about what belongs to the set, which would only consist of polynomials of degree <= 2 right?

that assumption seems a bit off

I think my understanding of quotient groups is a bit rusty

[scarlet estuary]

well, every element of the set is equivalent to an element of degree 0 or 1

more precisely: every equivalence class in the quotient has a representation of degree 0 or 1

[mild oxide]

oh right, strictly less than 2, not or equal to

[scarlet estuary]

since you can just eat any multiples of x^2

and spit out -1 instead

(or 1 if it's an even multiple, i.e. a multiple of x^4)

[vital quail]

Or just establishing that P(x) mapsto P(i) is a ring homo and then first iso theorem @shrewd halo

[mild oxide]

so an isomorphism f: R[x]/(x^2 + 1) -> C would be f(a_0 x^0 + ... + a_n x^n) = a_0 i^0 + ... + a_n i^n

maybe that's a bit verbose

[vital quail]

@mild oxide Well the domain is modded out so

[scarlet estuary]

its not verbose enough

you dont need to write everything symbolically, you can just

explain what it does

[vital quail]

It's more like ax + b

[scarlet estuary]

"it maps x to i"

if you'd like, you can say "pick a degree 0 or 1 representation for your equivalence class, it always exists because [reason]"

"then f(a+bx)= f(a+bi)"

[vital quail]

Maybe neatest is what I suggested

[upper pivot]

you can map R[x] to C and find its kernal to be (x^2+1), thats how i personally do these anyhow

[vital quail]

R[x] to C by x mapsto I

Yea

i

[mild oxide]

okay, yea that makes sense

[vital quail]

Verify this is a homo

Then bam, first iso theorem

The ker is (x^2+1)

[scarlet estuary]

yeah the first iso theorem is definitely the "fastest" way

in fact, iirc this exact problem was used in my undergrad to motivate the first isomorphism theorem

i mean we'd already seen it for vector spaces

[mild oxide]

first iso theorem?

[scarlet estuary]

but this was used to demo it for rings

[vital quail]

R/ker phi iso phi(R)

Coimage iso image

Guys when you shake your hand really fast it feels strangely like rubber

Like, claw your hand and wave, rotate your hand really fast back and forth

[scarlet estuary]

is this the first time youve discovered this

[vital quail]

No but I just remembered

Probably when I was 5 or so I did it for the first time

Or my dad told me about it

[mild oxide]

I'm sitting here like an idiot shaking my hand all over the place trying to figure out what you're talking about

define "claw" and "wave" and "rotate" and "feels like rubber"

[upper pivot]

yes archsys

pls explain

[vital quail]

Like literally move your hand in any way but do it really fast

You can flap it back and forth too

Like, loosen your wrist and just flap it up and down

[upper pivot]

why

[vital quail]

It feels rubbery

[mild oxide]

lol

[vital quail]

You have to extend your fingers though

Idk I guess it's the changing force experienced by your hand

It makes it feel, I mean, pretty much rubbery is a good description

[mild oxide]

like, limp? almost numb in a way?

[vital quail]

Sure

[mild oxide]

I think I know what you're talking about. Sometimes I'll shake my hands very vigorously after washing them to dry them when there's no towels anywhere.

[vital quail]

Mfw actually a dog

I just wipe on the sides of my pants lol

[mild oxide]

I'd rather shake 'em dry than get my clothes wet

[vital quail]

I guess I have a bad habit of wiping my hands on my clothes

I do it all the time

It's a habit when I eat I guess

[mild oxide]

or do both, but get them not very wet at all because I shook off most of the water

also, is this what y'all were referring to as the first iso theorem?

[upper pivot]

no

first iso is

[vital quail]

I said it above

[upper pivot]

if $A\overset{f}{\longrightarrow}B$ is a homomorphism, then $A/\ker(f)\cong f(A)$

[cloud walrus]

JohnDoeSmith:

[vital quail]

R a ring, phi a ring homomorphism, R/ker phi iso phi(R)

[mild oxide]

oh okay

[vital quail]

Here phi(R) is the image of R under phi

For example the map from R[x] to C by x maps to i is a ring homo

[mild oxide]

gotcha; thanks

[chilly ocean]

I am not sure if this will be considered abstract algebra, but I would like to have someone check through this working, because it seems to be incorrect

[cloud hill]

lmao wasnt this just posted on sci

[sour plume]

1. the given translation matrix is not invertible
2. there is no reason to consider 3x3-matrices rather than 2x2-matrices, because this is a fully 2-dimensional problem
3. it's a bit ambiguous with respect to which point you should scale by 3, but i would assume you're scaling it with respect to the centre of the rectangle. hence, the order of operations should be: first translate, then scale, then rotate, then translate back

these are the issues i see off the top of my head @chilly ocean

ah, the 3x3 vs. 2x2 problem is probably because they're using homogeneous coordinates; that makes more sense then, but still the given matrix is not invertible

then, in any case, it's a bit of a problem that the person who wrote this doesn't mention that they're using homogeneous coordinates

(also this is rather linear algebra than abstract algebra)

[chilly ocean]

right, this working was given by my lecturer, and i just don't understand it, the T matrix part specifically

[sour plume]

ye, so they're using a trick; if you translate a 2-vector by some other 2-vector, that is not actually a linear operation, so you cannot straightforwardly express it as a matrix

(because clearly: (a + b) + v =/= (a + v) + (b + v))

so they're using homogeneous coordinates, so instead of considering a 2-dimensional vector (x,y), they consider a 3-dimensional vector (x,y,1)

and there you can actually express translation as a linear operation, as described here: https://en.wikipedia.org/wiki/Translation_(geometry)

only problem: the matrix in your screenshot has a zero row at the bottom, and hence is not invertible. if you followed the rule in the wikipedia article I linked, then your translation matrix should be given by the same matrix, but with last row (0,0,1)

does that help?

[chilly ocean]

Yes, that does :)

I'm not surprised the lecturer made a mistake tbf.. he always does that lol

and confuses people @.@

How did the -3 come about? if i may ask

[sour plume]

The center of the rectangle is in the point (3,3), so to move things into the origin of the coordinate system, you have to subtract (3,3)

[chilly ocean]

right, but after scaling it would be 9,9 right?

so we translate first then scale

right then it makes sense

[sour plume]

Yeah might be a good idea to pay close attention to the final translation

But I think you get the idea

[chilly ocean]

In the last working line, the other T matrix is -3,3 is that a typo?

[sour plume]

haha yeah has to be

the person who wrote this was really not on top of their game

[chilly ocean]

Alright haha, thank you very much for clarifying :)

[sullen island]

what does it mean by respecting the natural pairing between V* and V?

how i understood the notation $\langle v^\ast , v \rangle$ is simply $v^\ast$ acting on $v$

[cloud walrus]

xy:

[sullen island]

but im not sure why $\rho^\ast (g) (v^\ast)$ acting on $\rho(g) (v)$ has to equal $v^\ast$ acting on $v$

[cloud walrus]

xy:

[sullen island]

am i misinterpreting something?

<@&286206848099549185>

[old nimbus]

well an orthogonal projection is a projection thats also orthogonal

[sour plume]

@sullen island what they mean by "G should respect the natural pairing" is exactly the equality below that

if you apply the group element to both elements of the natural pairing, both transformations should somehow "undo" one another

it's like when you apply an orthogonal transformation to both vectors in an inner product, the inner product stays the same

[sullen island]

uhm

im not understanding why, because a group element in general is not a planar isometry (which preserves length)

so the equality shouldnt hold in general 🤔

[sour plume]

I just meant to use it as an analogy

If you define your group actions in the "right" way, this is what they fulfil

e.g. if you defined G to act in some way on the vector space V, usually how you define G to act on the dual space V* is by setting, for each functional $\alpha$, the functional $g \cdot \alpha$ to be $(g \cdot \alpha)(v) := \alpha(g^{-1} v)$

[cloud walrus]

Lartomato:

[sour plume]

And then, transforming alpha and v simultaneously by the same g is like doing nothing at all

[sullen island]

but why is g required to do nothing to alpha and v?

😮

[sour plume]

It isn't required to do that, but that's what happens

because with the above definition you find $\langle g \alpha, g v \rangle = \alpha(g^{-1} gv) = \alpha(v) = \langle \alpha, v \rangle$

[cloud walrus]

Lartomato:

[sullen island]

could you give some intuition/motivation behind the definition? like why does the action of g on V correspond to the action of g^{-1} on V*

its pretty strange

[sour plume]

One reason for that is, if you started with a left action on $V$, and you want to get a left action on $V^*$, you need to take inverses. If you defined $(g \alpha)(v) = \alpha(gv)$, then this would give you a right action

[cloud walrus]

Lartomato:

[sour plume]

So if you don't permanently want to switch between left and right actions, you need this inversion

But of course you don't absolutely have to do it this way, you could define different actions on the dual. It just very often turns out that this is what's useful to you

Or, that this is somehow just the most "natural" definition for the setting that you're in

[sullen island]

uhhh

how is that a right action?

its still just alpha acting on some element of v transformed via left action on V

[sour plume]

You could just try it out on a sheet of paper

[sullen island]

which is also what $\alpha (g^{-1} v)$ is

[cloud walrus]

xy:

[sullen island]

hmm okay

[sour plume]

The inversion turns around the order

See if g(h\alpha) = (gh) \alpha or (hg) \alpha

[sullen island]

@sour plume hey sorry, i've been thinking about it for the past hour and this is what i've got: $g(\alpha (v)) := \alpha (gv)$, then we have $g_1 (g_2 (\alpha (v))) = g_1 ( \alpha (g_2 (v))) = \alpha (g_1 g_2 (v)) = (g_1 g_2) (\alpha (v))$

[cloud walrus]

xy:

[sullen island]

which is still a left action? 🤔

[sour plume]

You're applying the definitions wrong

(g_1(g_2 \alpha))(v) = (g_2 \alpha)(g_1 v)

you start by acting on the functional g_2 \alpha with the element g_1

You might have gotten that wrong because your bracketing looks misleading; for example, g_2 (\alpha(v)) makes no sense, because we have not defined how g_2 acts on the real number \alpha(v), it can only act on alpha or on v

[sullen island]

ohhhhhhhh

that's right

[sour plume]

it's good to struggle with these things, they're so fundamental that it's good to have them burnt into your head, lol

[sullen island]

sorry, i just started representation theory

[sour plume]

ye no problem

But basically the philosophy is that whenever you pull an action into the argument of a function, this reverses the behaviour of the action

i.e. turns left-actions into right-actions

And the way to counter that is by applying another thing which reverses the order of things

namely the inverse

[sullen island]

okay, thank you!

[sour plume]

np!

[chilly ocean]

how is this lagrange

?

theorem

Like how does lagrange theorem show this

[chilly ocean]

if gcd(w,m) = 1

[vital quail]

@chilly ocean Take (Z/mZ)^x. It has order phi(m). The order of w, d, divides phi(m) hence phi(m) = sd for some s. Then w^sd = (w^d)^s = 1 mod m

[shy bluff]

So for thtis question , part b), I'm confused so as to what counts as a subgroup of D_8

In our notes we have that

And

Does this mean that {e} by itself is a subgroup? What about {e, s}? Or do I need {e, s, s^3}?

[sour plume]

First question you'd have to ask is: do {e} and {e,s} fulfil all the axioms from your screenshot? Have you tried checking those?

[shy bluff]

{e} fulfills the non-emptiness

ee = e = e^-1 and all of those are in {e}

I think?

Err as far as I can tell

[sour plume]

Yeah! So the neutral element alone always gives you a subgroup in any group

[shy bluff]

So {e} is a subgroup of D_8

[sour plume]

Does the same hold for {e,s}?

[shy bluff]

Oh so {e, s} isn't true

Because s^-1 = s^3

[sour plume]

Yeah exactly

[shy bluff]

Oh ok

[sour plume]

That should already give you an idea for what you have to look for in a subgroup

[shy bluff]

And then I could also go and make {e, s, s^2, s^3} a subgroup because s^2 is its own inverse

Thank you!

[sour plume]

Exactly! Good luck finding all others. No problem

[shy bluff]

Oh and it should always include identity

[sour plume]

Yep, that's also something that very quickly follows from the axioms

[shy bluff]

Is a group a subgroup of itself?

[teal perch]

yes

[shy bluff]

For this question, what does "H = mZ := {mk : k in Z}" mean?

[sharp sonnet]

$H$ is equal to $m\mathbb{Z}$, which is defined as all the integer multiples of $m$

[cloud walrus]

Lochverstärker:

[shy bluff]

Ah

Thank you

[vital quail]

{f(a) | a satisfies some conditions} is set builder notation, read as "the set of all f(a) such that a satisfies some conditions"

for example {2a | a in Z} is the set of all 2a such that a is an integer, i.e., the set of even numbers

or { : }

but in my opinion { | } looks cleaner

[light spindle]

What is the canonical injection of Z/3Z into Z? Is it just [0] -> 0, [1] -> 1, [2] -> 2?

[mild laurel]

Uh, there is no canonical injection

[upper pivot]

^, check to see if that is a homomorphism

[light spindle]

I see ... f: Z/3Z -> Z as I described is not a homomorphism since 3=f([1])+f([2])=f([1]+[2])=/=f([0])=0

Are there no injections at all?

[mild laurel]

no

[dawn kiln]

ooops

mb

[halcyon siren]

What is the canonical injection of Z/3Z into Z? Is it just [0] -> 0, [1] -> 1, [2] -> 2?
@light spindle The obvious injection is what you say

it's an injection as kernel is 0

[smoky cypress]

Is it even a homomorphism?

@halcyon siren

[halcyon siren]

No, f(1)^3 =/ = f(1^3) (wrt group operation +)

But as far as I know injections don't need to be group homomorphisms

[upper pivot]

in the context of structures like this, all maps are homomorphisms

it does not have any meaning to just set-inject Z/3Z into Z

[halcyon siren]

@upper pivot well then it's not a map

[upper pivot]

ok, who talks about set maps when it comes to groups/rings/top spaces what have you

[halcyon siren]

Well if you're asking if it's possible to set inject, then yes it is

That's all I'm saying

[upper pivot]

also if its a set-map and not a homo

then ker=0 doesnt imply injective

[halcyon siren]

True

You're right

[upper pivot]

anyhow my point is, clearly in context we are talking about homomorphisms

[halcyon siren]

Sure

[upper pivot]

anyhow you can see that injections dont exist @light spindle , take a homo$ \bZ/3\bZ \overset{f}{\longrightarrow} \bZ$, and look at $f(x)$, then $0=f(3x)=3f(x)$ so $f(x)=0$ for all $x$

[cloud walrus]

JohnDoeSmith:

[upper pivot]

(extend this reasoning as to why you cannot embed finite groups/torsion groups into Z)

[mild laurel]

There's also nothing canonical about that injection, even as sets

[mild oxide]

I'm lost here. Any homomorphism of (Q, +) is of the form x -> rx ? how am I supposed to find r?

[mild laurel]

Think about where 1 gets sent to

[mild oxide]

also, wouldn't it still be an automorphism if r = 0, since 0 is in Q?

[mild laurel]

say 1 gets sent to m

where does 2 get sent to?

[mild oxide]

.... 2m?

[mild laurel]

right

where does 1/2 get sent to

[mild oxide]

m/2

[mild laurel]

Yeah

so for any rational number p/q, where does it get sent to

[mild oxide]

pm/q. so m would be our r. but that seems like defining a specific homom?

[mild laurel]

why?

Think about what we assumed to start

[mild oxide]

idk, would sending x -> r + x not be a homomorphism?

[mild laurel]

That doesn't send 0 to 0, for one

[mild oxide]

if that's just kind of obviously true that it has that form for any homomorphism, I'm struggling to see how I'm supposed to "show" that it has that form

[mild laurel]

also it doesn't satisfy the homomorphism property at all

[mild oxide]

oh, duh

[mild laurel]

I mean, look at what we assumed when we did this "proof" that every rational number p/q is sent to pm/q

[mild oxide]

only assumptions are that we're dealing with a homomorphism over (Q, +)

[mild laurel]

Right, and any homomorphism must send 1 to something

which we denoted by m

[mild oxide]

so that's just how we're defining our general homomorphism?

[mild laurel]

Any homomorphism must send 1 to something

[mild oxide]

okay, I just thought that if we define a homomorphism, we should be doing something I guess more general (like, p/q -> pm/q). that just sounds like sending 1 to something is specific

[mild laurel]

We're not defining a homomorphism

If you give me any homomorphism, 1 gets sent to some rational number

which we denote by m

[mild oxide]

okay, but we're dealing with them generally here

I meant defining an arbitrary one that we can work with

if that makes sense

[mild laurel]

not really

[mild oxide]

okay

Sorry if I'm struggling to get it here. Also, I'm really confused about the automorphism part. If r = 0, and so x -> 0, how is that not still an automorphism? 0 is an element of Q, so it's still a map from Q onto itself.

[mild laurel]

What's the definition of automorphism?

And why is it different from a homomorphism?

[mild oxide]

automorphism is a bijection f: G -> G with f(x*y) = f(x)*f(y)

homomorphism is not necessarily bijective

[mild laurel]

And if r = 0, is that a bijection?

[mild oxide]

nope

okay, that one was pretty obvious and simple lol thanks

[cosmic crag]

http://prntscr.com/sjcae8
can anyone help me with this?
what would the points be

[scarlet estuary]

this is not abstract algebra.

[brisk granite]

how would I determine the number of order two elements in (Z/nZ)*

[mild laurel]

Chinese remainder theorem probably

[kindred mist]

Can someone verify if my primary decomposition is correct ^^^?

[kindred mist]

<@&286206848099549185> ^^^

I'm mostly confident about everything except the last step

(But I haven't done this before so verification with respect to my whole solution would be appreciated)

Thanks in advance

Sorry the 3rd step in my matrix transformation is hard to read, it says "change of sign" (I just changed the sign of the first row)

(that that first plus in the last line should be one of these)

[oblique river]

looks fine to me 👍

[kindred mist]

tysm!

[red imp]

$F$ is a field, and $T$ is a finite subset of $F$. $$I(T):={f\in F[x],\vert,f(a)=0,\forall a\in T}$$ is an ideal of $F[x]$

[cloud walrus]

Bobbicals:

[red imp]

I am trying to show that $I(T_2)\subseteq I(T_1)\implies T_1\subseteq T_2$ but I am having trouble formulating an argument

[cloud walrus]

Bobbicals:

[red imp]

I understand the intuition, like if p(x) has roots at all points in T_2 then it has roots in all points in all subsets of T_2

[upper pivot]

what trouble are you having

that is the arguement

[red imp]

I mean I assume that the intuitive argument I had ^ can be applied here but I don't know how to apply it

[stone fulcrum]

It's practically immediate by definition haha

[upper pivot]

i mean what you said

if p(x) has roots at all points in T_2 then it has roots in all points in all subsets of T_2

would suffice for a proof lol

[red imp]

oh

[mild laurel]

i dont think you guys are reading this right

[upper pivot]

well sorry thats the other side actually

[red imp]

yeah phew

because I used that to prove the other way (it's an iff statement really)

[upper pivot]

idea is very similar to this

[red imp]

oh I think I might have it

1 sec I'll post it here to get it checked

is this valid

[upper pivot]

i mean yeah this works, although you did not need to go do all that

[red imp]

:D fuck yea

[stone fulcrum]

Displaying the factorizations themselves convey no info, lol. But yes you've got it

[upper pivot]

you could just have said that if $I(T_2)\subset I(T_1)$, then all polynomials in $I(T_2)$ are also in $I(T_1)$, so if $T_1$ annihalates every poly of $I(T_1)$ then it also annihalates all of the $I(T_2)$ poly

[cloud walrus]

JohnDoeSmith:

[red imp]

oh yeah that's nice

I'm not familiar with algebraic proofing techniques yet, so I have trouble formulating these arguments even if I understand them

[upper pivot]

oh thats fair, practice and you'll get used to em

[red imp]

orthonormal bases are what's normally used

pre sure it will always be hermitian

[chilly ocean]

I don't know how to use chinese remainder to conclude this

like the theorem seems to be a bit different

a corollary I'm assuming I need to use is: if gcd(p,q) = 1, and a,b in Z such that x = a (mod p) and x = b (mod q), then suppose x0 is a solution then set of all solution is x= x0(mod pq)

I don't see how to use it

like when I use the theorem do I say let x = w be the solution to the system?

[mild laurel]

letting x0 = w^(de) satisfies both

[chilly ocean]

I see

thanks

[chilly ocean]

still I don't see how using chinese remainder theorem helps @mild laurel

Like can't you just derive it directly, chinese remainder theorem seems to be directed to another problem

Letting x0 = w^(de) only draws the conclusion that the set of all solutions to x=w (mod p) and x = w (mod q) is x = w^(de) (mod pq)

[simple agate]

does the second part to this question have anything to do with the structure theorem? sounds like a basic group theory question to me

[chilly ocean]

<@&286206848099549185> can anyone explain how to use chinese remainder thm here?

assuming that you already know w = w^(de) (mod p) and w = w^(de) (mod q) and to conclude that w = w^(de) (mod pq)

https://cdn.discordapp.com/attachments/496784958430380033/712252286507941962/Screen_Shot_2020-05-19_at_6.35.40_PM.png

[mild laurel]

Chinese remainder theorem gives you a bijection between (mod p), (mod q) and (mod pq) right

In one direction, if you have some number x (mod pq), then this gets mapped to x (mod p), x (mod q)

So in this case w^de (mod pq) gets mapped to w^(de) (mod p), w^(de) (mod q)

Since it's a bijection, that must mean that the inverse, in the reverse direction

w^(de) (mod p), w^(de) (mod q) gets mapped to w^(de) (mod pq)

[chilly ocean]

we haven't learned the definition of the bijection but I think I know what you mean earlier

like I could replace x0 with both w and w^(de) to get the result

[thorn flint]

Hi can someone help me to understand to understand why for this group, the list of generators are [E,H,K,M,N,O]?

I understand why E is a generator but I don't see why H,K are as well

[hot lake]

why are you saying that E is a generator

[thorn flint]

Since A,B,C,D,F can be written in terms of E

[hot lake]

are you saying that E generates the whole group just by himself ?

so anyway

being a generator

shouldn't be understood as a property that an element can have

instead

it's a property that a subset of the group can have

and 2.

there are several subsets of the group that generate the group

so there is no "the" list of generators

and you don't say a list of generators because that will confuse everyone

[thorn flint]

In my understanding, E is just a subset of this group G that contains some of the elements. And I need to find other subsets that contain the rest of the elements. Then this subset is the generator?

[hot lake]

instead you say a generating list

or a generating subset

E is clearly an element of the group here

[gritty oriole]

Is the original question: “show that the set {e,h,k,m,n,o} generates this group” or something similar?

[thorn flint]

This is the question they gave us

[gritty oriole]

A list

[hot lake]

the question should be "find a subset of the group that generates the group"

or possibly

"find a minimal generating subset"

[gritty oriole]

Yes

[thorn flint]

Isnt a generating list the minimal subgroup?

[hot lake]

no

a generating subset of a group G is a subset S such that every element of G can be written as a finite combination of elements of S (and their inverse if G is infinite)

[gritty oriole]

So trivially G is a generating subset of G

[thorn flint]

Okay yup I have that definition

[hot lake]

yes

and everytime someone writes "list of generators" instead of "generating subset", god kills a kitten

[thorn flint]

Haha well I'm not sure if it was my professor or the TAs who wrote these questions but maybe it's cause I'm studying in EU

[hot lake]

so when we say that {E H K M N O} is a generating subset of the group

we DO NOT SAY that E is a generator

or H or K for that matter

[thorn flint]

Okay sorry

[hot lake]

because {E} isn't a generating subset and probably neither are {H} and {K}

[thorn flint]

So can you please explain how the answer is correct?

[hot lake]

well, can every element of the group be written as a finite combination of elements in {E H K M N O} ?

[thorn flint]

I guess but why is G or J not included?

[hot lake]

well you could include them

[thorn flint]

G^2 = D, G^3 = J
H^2 = D, H^3 = K

[hot lake]

if {E H K M N O} is a generating subset, then {E G H J K M N O} also is a generating subset

[thorn flint]

So is it safe to say that G^2 = D and H^2 = D, thus G^2 = H^2

That's why we don't need to include them

[hot lake]

I suppose but I'm not sure where you're going with that

wait what

no ???

can you find a way to write G as a finite combination of elements in {E H K M N O} ? or even a finite combination of elements in {E H} ?

[brisk granite]

Is it true that any 2 transitive subgroup of S_n with a transposition is the full group?

[hot lake]

yes

[brisk granite]

Is this something easy to prove?

[hot lake]

yes

[brisk granite]

Could I have a hint lol?

I dont know where to start

[hot lake]

write down what 2-transitive means

[brisk granite]

A subgroup H is 2 transitive if, for all a,b,c,d there exists f in H such that f(a) = c and f(b) = d?

[hot lake]

almost

but yeah it means something like that

and you have a transposition

so what can you make out of that

(you have to require a<>b and c<>d in the definition)

[brisk granite]

Ah, right lol

[hot lake]

wait actually it was correct

wait no

you can't find f such that f(a) = c and f(b) =d if a=b and c<>d

anyway that's petty matters

[brisk granite]

yea, I still don't really see what to do

[thorn flint]

can you find a way to write G as a finite combination of elements in {E H K M N O} ? or even a finite combination of elements in {E H} ?
@hot lake
I see it now!

Thanks

a lot

[hot lake]

nice

[brisk granite]

oh, wait

I thought of somehting

cool

I think I got it

basically, conjugation was the key

if we have a transposition (x,y) in H and some f such that f(x) = a and f(y) = b, then f^-1(x,y)f = (a,b)

since we have all transpositions, we have the whole group

[hot lake]

yes

[outer monolith]

so a polynomial can be defined over any field right>

[sharp sonnet]

sure

[brisk granite]

so, why does this imply the galois group is double transitive

[hot lake]

if it wasn't it couldn't factor as 1 and (n-1) under any specialization

that second factor would always be reducible

[brisk granite]

uh, sorry, that doesn't seem clear to me

are you using a theorem?

[shy bluff]

How do I go and show

Given that I just have

And

Do I first show that identity is in S?

[woven delta]

Do it by induction lmao

[shy bluff]

Like we already have that the operation is commutative for all elements of S

Is that not enough to show that it's abelian?

Pardon?

what do you mean do it by induction?

[woven delta]

Okay so first add in inverses and identity to your subset

Then you want to show that any 2 finite products of elements in your subset commute

And you do this by induction on the number of elements you are multiplying together

@shy bluff here's an outline for the proof. First you want to show that if you take the set S and consider the set S' = S U {x^{-1}| x \in E} U {e} you need to show S' has the same property that all the elements in it commute

Next suppose that n is the smallest number st there is x= a_1 a_2 ... a_n and y = b_1 b_2 ... b_m (where n >= m) where x and y do not commute. You need to show that x and y commute

Then you need to show that <S> is exactly the finite products of elements in S' by showing that that forms a group and that any subgroup that contains S must contain all finite products of elements of S'

And that's it

This is really the sort of statement that is entirely trivial but requires a bit of bookkeeping to do

[shy bluff]

hrm

Ok

[gritty oriole]

one thing to remember, S is not necessarily <S>

[shy bluff]

S is just a group of elements while <S> includes all the elements and all of tehir products rigght?

Like if S = {e, s}, and |s| = 3 or somethingg then <S> = {e, s, s^2} right?

[sharp sonnet]

S is not a group (in general)

[shy bluff]

Err sorry

S is just a subset

[sharp sonnet]

ye

<S> now includes all finite products of elements of S

[shy bluff]

Yea

[sharp sonnet]

and their inverses

i.e. it is the smallest subgroup that includes all of S

[shy bluff]

Oh so that's what they meant

Question about this

[sharp sonnet]

Like if S = {e, s}, and |s| = 3 or somethingg then <S> = {e, s, s^2} right?
@shy bluff this is correct, but it's kind of a special case

[shy bluff]

wait why is it a specila case?

[sharp sonnet]

because you never have to include e in a generating set

because s^0 = e for all s

[shy bluff]

Oh

[sharp sonnet]

so this is really just a subgroup generated by a single element

i.e. a cyclic subgroup

[shy bluff]

Ah

[sharp sonnet]

which is a special case

[shy bluff]

Whereas say for example D_8 can't be generated by a single element right?

[sharp sonnet]

and its again a special case because s has finite order

so you dont have to include negative powers

e.g. s^2 = s^(-1)

[shy bluff]

Yea

[sharp sonnet]

D_8 is the dihedral group?

[shy bluff]

Yea

[sharp sonnet]

yes, it can't be generated by a single element

[shy bluff]

You need at least 2 elements

[sharp sonnet]

because then it would be cyclic, and cyclic groups are abelian

D_8 is not

[shy bluff]

Oh ok

So uh about

Ok so b = gcd(a, n), then both n and a are multilpse of b, so there is k such that a = kb, and there is j such that n = jb

Why does a = kb imply that [a] is in <[b]>?

Is it because if a = kb then [a] can be generated by [b]?

Or rather that b^k = a?

[sharp sonnet]

can you tell me what [b] is?

[tribal pasture]

I think the equivalence class of b

[steady axle]

yes if a=kb then $[a]=[b]+\cdots+[b]-$ k times

[cloud walrus]

bertwit:

[snow flint]

is abstract algebra by lee good?

[patent phoenix]

YO

SOM1 HELP

[chilly ocean]

Def: A plane curve $F\in K[x,y]$ has a cusp at 0 if it has exactly one tangent and 0 and the intersection multiplicity $\mu_0(F)$ with that tangent is 3 and the multiplicity $m_0(F)$ of 0 (that is the the degree of the smallest homogeneous part that doesn't vanish) is 2.
I want to show: If $F$ has a cusp at 0, then there can only be one irreducible component that passes through $0$ (I guess by some casework using the additivity of the intersection multiplicity?) and i want to find a lower bound for the intersection multiplicity in 0 of two curves that have cusps in 0.

[cloud walrus]

seeker of knowledge:

[glossy crater]

is 7 irreducible in gaussian integers?

[fading wagon]

@glossy crater Yes, otherwise, if I can find 2 complex numbers that multiply to 7, their norms must divide 49, hence norm of 7 which is impossible because 7 is not a sum of 2 squares

[fierce perch]

If R/I is simple, does it follow that I is a maximal ideal?

[steady axle]

matrices over division ring are simple right?

[fierce perch]

So if R/I is a division ring it follows that I is maximal?

[stone fulcrum]

R/I is simple ⇔ I is maximal, yes. @fierce perch
If R is commutative, then that's Field ⇔ Maximal

[fierce perch]

I just argue that if J is an ideal that contains I then it is either 0 or R/I in R/I, which means it is is either I or R in R.

[glossy crater]

given an integral domain R and a unital subring of it S, if they have the same field of fraction, does that imply that R=S?

[fierce perch]

If im not mistaken you should be able to get an injection of Q(R) into Q(S) by the universal property which should give you an injection of R into S. Since S is a subring it ought to be iso?

[glossy crater]

sorry can u elaborate a bit

[stone fulcrum]

@glossy crater
No, since Q and Z have the same field of fractions

[glossy crater]

q and z u mean?

yeah i found that out too

thx for confirming though

[stone fulcrum]

Sorry I was a bit late haha

[halcyon siren]

I'm reading my notes on representation theory and after computing the character of the representation of the permutation representation of S4 acting in the edges of a tetrahedron and showing it is isomorphism to the trivial representation plus one of the 3D reps of S4 and the 2D rep it asks as an exercise to compute what these reps are as subspaces of the tetrahedron

Is a 3D rep of a tetrahedron then going to be one generated three different edges

The 2D one by two

And the trivial by one?

I'm pretty confuddled

[hot lake]

so you have a vector space of dimension 6 and you need to describe 3 of its subspaces ?

" what these reps are as subspaces of the tetrahedron" really not sure this makes any sense

or maybe find some obejcts that S4 acts on that are like those representations

[halcyon siren]

" what these reps are as subspaces of the tetrahedron" really not sure this makes any sense
@hot lake mmm? They are subrepresentations so they must act on a space that is invariant under their action

The question is what are these spaces as subspaces of the edges of the tetrahedron I think

This isn't homework or anything dw, but I'm looking at exercise 4.19

[hot lake]

yeah, find the subspaces of V

so you want to find subspaces of V that are stable by the action of S4

[halcyon siren]

So will these look like three edges for the 3D subrep

🤔

[hot lake]

no ????

[halcyon siren]

And the space is actually the Three D complex space generated by these edges

[hot lake]

being a 3 dimensional subspace of V has absolutely nothing to do with 3 edges

[halcyon siren]

Mm? Well isn't V just the vector space generated by formal sums of the edges

[hot lake]

yes

you are sounding as if the only lines in R² are the horizontal line and the vertical line

being 3 dimensional means you are generated by 3 vectors, and each vector is a linear combination of the 6 edges

[halcyon siren]

Noooo but as far as I understand this is like labelling the edges of the tetrahedron 1-6 and then just thinking of V as the C space that is generated by complex sums of them

Aaaah I see your point tho

Yeah

I don't necessarily need to take subsets of the generators

Like <1+2> works, right as a 1D subspace

[hot lake]

yeah that would be a line

I don't think it would be stable under the action of S4 no matter how you assign numbers to the edges

[halcyon siren]

Hmmm

[halcyon siren]

Aah wait, should I not be expecting these subspaces to be subrepresentations of the permutation representation?

[hot lake]

no they are subspaces of the edge representation

they are subspaces of V

not of other random representations out there

[halcyon siren]

I'm sort of confused because this seems to contradict what my notes say

https://math.stackexchange.com/questions/2025544/find-irreducible-representations-of-a-reducible-representation-of-s-4

[hot lake]

????

what contradicts what ?

that question isn't even about the same representation as yours

[halcyon siren]

He is looking at the induced representation on the edges

Isn't he

[hot lake]

but he has ordered edges

and your edges are unordered

so have you found a line inside V that is stable by the action of S4 yet ?

[halcyon siren]

But I think it is fine for him to do that

[hot lake]

it's fine for him to do what ?

[halcyon siren]

Orientating the edges

[hot lake]

he doesn't have a choice

and you don't have a choice either

and since you want the 1-dimensional representation to be isomorphic to the trivial representation, you even need to find a vector that is fixed by the action of S4

[halcyon siren]

so have you found a line inside V that is stable by the action of S4 yet ?
@hot lake nope 😔

I guess maybe an axis? 🤔

[hot lake]

so, finding a vector like that is a basic exercise in linear algebra

no

[halcyon siren]

Idk

[hot lake]

you have a bunch of linear maps from V to V

find a vector that is fixed by all of them

[halcyon siren]

Does the sum of all of them not work

[hot lake]

yes the sum of all the edges is invariant by everything

[halcyon siren]

Huzzah

Now I want a space that is generated by two vectors that is invariant under everything

🤔🤔🤔🤔🤔

[hot lake]

not invariant, stable

[halcyon siren]

Yeye

[hot lake]

well

those words are always a bit vague

and yeah now that one is harder

[halcyon siren]

Nah I get ya, fair play

🤔

[hot lake]

have you proved mashke's theorem

in those notes

[halcyon siren]

Ye

[hot lake]

there is a tool in there that you might be interested in

in the proof

assuming it's proved the same as what I remember from the proof I had

[halcyon siren]

The funky projection

[hot lake]

yes

[halcyon siren]

So are you thinking I use it with the subrep I already have

[hot lake]

uh no you want to project V into the subreps that you haven't found yet

you compute the projection, then you look at its image and you win

because the image is what you want

[halcyon siren]

Aaah we did it by projecting into the rep we have and using the kernel

[hot lake]

that would give you the sum of the 2 irreducible representations

you would still need to decompose that

[halcyon siren]

Yeah

But how do I project into exactly one irreducible subrep

Do I not need to repeat this?

[hot lake]

=_=

[halcyon siren]

Do it into the direct sum and then again

[hot lake]

the two you want have different character tables

just look carefully at what those projections are, what they are made from and what they do

[halcyon siren]

@hot lake what my notes say is if you know W, you take a basis for W, extend to a basis of V to get a projection, do the averaging trick to get a G homomorphism and then the kernel is a subrep

I don't see what I'm doing with characters atm

I really don't see how I'm supposed to get a projection into a stable space based off the character table

[hot lake]

I need to see the details of those projections again I guess

I might be misremembering them then

you could still try to guess some stable subspaces

for example

the trivial rep + the 3-dim rep should be isomorphic to the standard representation of S4 on {1;2;3;4}

if you could embed that into V

aka find 4 vectors "1" "2" "3" "4" inside V such that S4 permutes them

then you could find - after removing the trivial 1-dim rep from it like you say - the 3-dim rep inside V

[eternal furnace]

Do the integers act on the reals by multiplication ?

[oblique river]

yes

the integers act on any abelian group by multiplication

[eternal furnace]

thought so lecturer must've made a mistake

thanks

[latent anvil]

Well the integers also act on the reals in lots of other ways

[woven delta]

they act as a ring in exactly one way

[vital quail]

I think you act on the reals

Transitively

[mild laurel]

uh

if sr^j = sr^{-j}

then that follows

[brisk granite]

are there any techniques to prove irreducibility over field extensions of Q?

specifically, I'd like to prove (x^6 -10x + 5)/(x-a) is irreducible over Q(a) where a is a root of x^6 - 10x + 5

[mild laurel]

I'm not sure you want to prove irreducibility directly, as its not easy

I think it'd be easier to look at the degrees of the field extensions. I.e., if that polynomial is irreducible, then the degree of the splitting field must be divisible by 5, and this is a sufficient condition as well

[brisk granite]

I see

specifically, I'd like to prove (x^6 -10x + 5)/(x-a) is irreducible over Q(a) where a is a root of x^6 - 10x + 5
btw, does this result seem true to you?

nvm, I'm pretty sure it's true

[mild laurel]

yeah I have no clue

but maybe this isn't easier, because you'd want to show that the galois group contains an element of order 5, but reducing your polynomial mod 5 doesn't help

[brisk granite]

uh, what would reducing my polynomial mod 5 do?

[mild laurel]

Ah you might not have learned it, but there's a theorem that essentially says that

Given some integer monic polynomial f, you can reduce it mod p to get a polynomial f' in F_p[x]

[brisk granite]

ik that irreducibility in F_p means irreducibility in Q (that doesn't apply for extensions of Q tho)

[mild laurel]

Then, if f has no multiple roots the galois group of f' over F_p embeds into the galois group of f over Q

[brisk granite]

oh

[mild laurel]

Yeah, so you can try other prime mods, but I'm not sure they'd be helpful

[brisk granite]

actually, I think this helps. I don't rly care about this specific polynomial per say.

I should probs be able to find this theorem in dummit and foote, right?

[mild laurel]

Uh, I know it's in Lang, not sure about dummit and foote, but yeah its probably in there

https://math.stackexchange.com/questions/2223818/computing-a-galois-group-by-reducing-mod-p

the first answer here gives a couple examples on how this works, and the question cites the theorem from lang

[woeful flint]

Im trying to show that if $I$ is an ideal of $R$ and $\overline{J}\subset R/I$ is an ideal then there exists $J$ an ideal of $R$ such that $J/I = \overline{J}$. My idea is to use the natural homomorphism $\sigma:R\to R/I$ and let $J = \sigma^{-1}(\overline{J})$. It seems to work but I'm not entirely sure. Any thoughts?

[cloud walrus]

Kraft Macaroni:

[upper pivot]

why are you not sure?

[woeful flint]

it comes out nicely but it feels finicky

Plus i just doubt myself when it comes to working with quotients more directly

[upper pivot]

well write out the proof, if it works it works right

[woeful flint]

yeah i guess

would you mind giving it a read for me?

[upper pivot]

sure

[woeful flint]

Sorry I cocked that up

[upper pivot]

,rotate

[upper pivot]

right that should work

ill add two remarks

if we have a homo $A\overset{f}{\longrightarrow}B$ and an ideal $\mfk{b}$ of $B$, then $f^{-1}(\mfk{b})$ is always an ideal(we dont need surjectivity), and is called the contraction of $\mfk{b}$ in $A$, often denoted $\mfk{b}^c$

[cloud walrus]

JohnDoeSmith:

[woeful flint]

Ah thanks so much man

[upper pivot]

and second is that once you found J, you could just say by defination $\sigma(J)=J/I=\bar{J}$

[cloud walrus]

JohnDoeSmith:

[upper pivot]

and skip most of the working you did in the last part

(altho working through it is good if you are uncomfortable with the ideas)

[woeful flint]

ah thats a very good point

thanks again

[upper pivot]

yeah np

[shy bluff]

For a cyclic group, does # of generators = # of subgroups? Is there a relationship between the two?

[upper pivot]

there is a relation, but thats not it

[shy bluff]

Oh

[upper pivot]

well not number of generators i mean

think about the orders

[shy bluff]

Orders of?

[upper pivot]

group and subgroup ofcourse

[shy bluff]

Ok so the order of (Z/nZ, +) is n

The order of the subgroup is just the number of elements within itself

[upper pivot]

well can you say more about the orders of subgroup?

[shy bluff]

It's at most n

[upper pivot]

i guess you havent learnt lagranges yet? if so thats fine

[shy bluff]

We haven't, no

[upper pivot]

ok so lets be more direct

what will a subgroup of this look like

[shy bluff]

What do you mean?

[upper pivot]

exactly what i said, think about what the subgroups will look like

[shy bluff]

Well say for example we have (Z/3z, +), then the whole group has {[0], [1], [2]}, and we can make (Z/2Z, +) I think? {[0], [1]} is non-empty, closed under productsand closed under inverses? And we can make (Z, 1Z, +) too but that one is the trivial subggroup?

[upper pivot]

its not closed under addition nor does [1] have inverse there

note Z/2Z has a different + operator

[shy bluff]

Oh wait

[upper pivot]

so it doesnt make sense to say this

[shy bluff]

Oh

Ok so (Z/3Z, +) only has the trivial subgroup?

[upper pivot]

yeah

but thats cause 3 is prime

look at like. Z/12Z

[shy bluff]

Ok Z/12Z, we have {[0]... [11]} as the actual group

Trivial subgroup is always one of the subgroups, but otherwise I don't think that there's anything that satisfies closed under products?

[upper pivot]

there is

mac channel under use

[shy bluff]

Because say we want [10], then we'd need [10] to be in the subgroup right?

[thorn flint]

I’m sorry I’ll wait

[upper pivot]

yeah

ok ill umm guide you in the right direction

first prove this

subgroups of cyclic groups are always cyclic

[shy bluff]

I think it has something to do with divisors

[upper pivot]

yeah

[shy bluff]

Divisors form subgroups?

[upper pivot]

not quite

think about what i said with the cyclic subgroup

[shy bluff]

So the Is <2> a subgroup of Z/12Z? It's closed under addition and inverses and contains identity

[upper pivot]

yep

[shy bluff]

<5> is not a subgroup of Z/12Z

[upper pivot]

it is

[shy bluff]

Oh

[upper pivot]

<5> generates the whole group though

[shy bluff]

Oh

If it doesnt' divide it then it ends up generating the whole group?

[upper pivot]

yeah

infact, when are <a>=<b>

[shy bluff]

<a> = <b> when both a and b generate the same set right?

[upper pivot]

yeah

(subgroup but since operation is same sure set)

[shy bluff]

I'm not sure

[upper pivot]

consider ||gcd||

[shy bluff]

Wait can you give me an example of <a> = <b>?

[upper pivot]

<2>=<10> in Z/12Z

actually let me give a better one

<5>=<10> in Z/25Z

yeah this is better, now try proving this

[shy bluff]

gcd(10, 2) = 2, gcd(10, 5) = 5; gcd(2, 12) = 2, gcd(10, 12) = 2; gcd(5, 25) = 5, gcd(10, 25) = 5

gcd(a, b) = gcd(a, n) = gcd(b, n)?

[upper pivot]

forget the first example btw

just the last equality is needed

try proving it

ill be going to sleep now, ill just tell you this then, all subgroups of cyclic groups are also cyclic, and as you showed is that if they have the same gcd then they induce the same subgroup. so the possible values of gcd uniquely determine the subgroup, how many values of gcd are there?

[shy bluff]

what do you mean by "how many values of gcd are there"?

[upper pivot]

given n, how many different values can gcd(a,n) take for arbritary a

(no need for explicit answer, just a description will suffice)

[shy bluff]

Isn't there only one value that it can take on?

That's the definitionof gcd isnt' it? The largest number that divides both?

[upper pivot]

note i said arbritary a

i.e what are

like for 6

gcd can take on 1,2,3,6

depending on the value of a

[shy bluff]

that's the number of divisors of n

I feel that it has something to do with coprimeness

But not sure

[upper pivot]

ok so umm lemme show you one case

we know Z/nZ = <1> right

generated by 1 that is

take any a that is coprime to n

we know there is some integer k such that ak=1

because inverses

can you carry this on

(i am actually going to sleep this time, sorry if my explanation was messy i was p sleepy the entire time)

[shy bluff]

It's ok

Sorry for being obtuse, I think that I just need some more sleep

[upper pivot]

nah you are fine, you werent obtuse i realize my explanation was a bit messy

[chilly ocean]

let $H$ be a normal subgroup of $G$, and $f,g,f',g'\in G$ so that $g g'^{-1}\in H$ and $f f'^{-1}\in H$. I'm having trouble showing that $gff'^{-1}g'^{-1}\in H$

[cloud walrus]

exele:

[chilly ocean]

can anyone give me a hint?

[solemn rain]

g and g'^-1 are in G

ff'-1 is in H

gg'^-1 is in H

use normality

[chilly ocean]

but g and g' might not be the same

[upper pivot]

heres a hint

how do u get

$gff'^{-1}g^{-1} $ times something is that

[cloud walrus]

JohnDoeSmith:

[chilly ocean]

sorry what?

[upper pivot]

umm multiply something to this to get the thing in your problem

[latent anvil]

okay, so you can use normality to get that g f f'^(-1) g^(-1) is in H, right?

[chilly ocean]

oh right yes sorry misread

yes

[latent anvil]

So JohnDS is asking how you get from that to g f f'^(-1) g'^(-1)?

[chilly ocean]

right multiply by g'g^(-1)

right?

[latent anvil]

I think the other way around? But that's the idea

[upper pivot]

yeah its the other way round

but you should get the idea now

[chilly ocean]

oh right k ty

[thorn flint]

Hi I have this question

And I did this

But I’m confused why 12 bar goes to 65 and why 5 bar goes to 31 instead of the other way around? Is it because we do the multiplicative inverse so both swap?

[hot lake]

????

[thorn flint]

At the bottom of the 2nd picture

[hot lake]

31 * 21 - 10 * 65 = 1, yes

[thorn flint]

The equivalence class part

[hot lake]

ah well the equivalence class part is pulled out of thin air

[thorn flint]

Where I put in brackets next to it (12 bar goes to...)

[hot lake]

not sure what any of that means

[thorn flint]

These were the answers given

By the professor

[hot lake]

that looks more like numerology than mathematics

[thorn flint]

??

I don't know but it's in my Abstract Algebra class

[hot lake]

I mean it's an application of some kind of magic rule that we are not told about

so I can't really comment on the "so the class is ..." part

[thorn flint]

Oh

okay

[hot lake]

if you are going to obtain the result by computing 12 * a + 5 * b for some magic numbers a and b, what should a and b mod 31 and 65 be ?

[fading sparrow]

anyone have a hint on this?

[upper pivot]

yeah

show B^G is ||integral over B||

( i did this exact exercise a few hrs ago in atiyah macdonald lol)

[fading sparrow]

heh

nice

[brisk granite]

would anyone be willing to verify this?

lemma 17 btw

also, ik g could be factored slightly differently, but using a similar argument, we can show that the obtained system of equations doesn't have solutions

[fading sparrow]

@brisk granite perhaps post this in the #advanced-number-theory channel

[brisk granite]

ok

[fading sparrow]

@upper pivot I mean isn't it obvious that B^G is integral over B?

you mean integral over A right?

[upper pivot]

or sorry i meant B integral over B^G

[fading sparrow]

ah

[halcyon siren]

I'm sort of confused about this. Does G being {1} in C mean I'm only looking at scalars for X, if so isn't the only continuous map into 1 the constant map, so in this case this works?

[sour plume]

You are looking at scalars, but your argument for why the Lie algebra should be that doesn't check out

[halcyon siren]

Isn't the only continuous map into 1 the constant map, so the actual Lie algebra should be 0

[sour plume]

Yup, that's correct

But that space from your screenshot does not only contain zero

[halcyon siren]

What other scalar has exp(x) = 1

[sour plume]

2 pi i

and so on

[halcyon siren]

oooh

complex

yesss

my brain

[sour plume]

Haha, happens

[halcyon siren]

Thanks, I knew I was being stupid

❤️

[sour plume]

No problem ❤️

[sterile eagle]

When I'm calculating the degree of an extension using
[M:K]=[M:L] * [L:K]
Is [M:L] the dimension of the vector space of M in L?

🤔

ofc it is... I'm dumb

[brisk granite]

why does my book say the characteristic doesn't divide n?

I don't understand the reason for this constraint

[mild laurel]

Weird things happen if the characteristic divides n, i.e., that's how you get inseperable extensions

[brisk granite]

I see

[mild laurel]

Like this is basically the standard example of a non separable extension

Take F_p(t) for some indeterminant t

Then consider let a be the p-th root of t

Then the minimal polynomial of a is just x^p - t = (x - a)^p

and so the irreducible polynomial x^p - t isn't separable so your extension isn't galois

[brisk granite]

oh, I see

ok, I get it now

[brisk granite]

in Q(13), do you think they mean D is not the sum of squares in Q

like, D can be represented as a^2 + b^2 for rational a and b

or do they mean integer a and b?

[mild laurel]

I'd assume they mean integer a,b

Yeah, especially with the comment about the prime

[brisk granite]

oof, that makes it harder

[mild laurel]

Ah

Actually even easier, there's no ambiguity

A positive integer is the sum of two rational squares if and only if it is the sum of two integer squares

[brisk granite]

oh, that's true?

alright, I'll try proving that first then

[mild laurel]

You're going to have to use https://en.wikipedia.org/wiki/Sum_of_two_squares_theorem

[sharp peak]

Is there a reason why we never talk about the field formed by adjoining all reciprocal polynomials to a polynomial Ring?

[oblique river]

What? People talk about that all the time

it's the field of rational functions (assuming the ring was an integral domain)

[sharp peak]

You're totally right. thanks.

[brisk granite]

so, this a problem I'm working on

and some of a solution

So, I can solve the problem when n is odd, but when p_1 = 2, then I'm not rly sure how to find d

[oblique river]

notice that the only way to get more d's is to add more prime factors

just increasing the power of a prime doesn't change anything

(unless you're increasing it from 0 to 1 for odd primes or 1 to 2 for p = 2)

well actually I guess 2^3 actually gives you one more than 2^2

can you find the quadratic subfields of Q(zeta_8)?

[soft elm]

can someone give me an example of two groups having more than 1 one isomorphisms between them

[mild laurel]

Z/3Z and Z/3Z

[soft elm]

pleb two different groups

actually nvm ill try myself

ill come back if i cant

[oblique river]

but if they were different groups

then they wouldn't have any isomorphisms between them

[scarlet estuary]

S_3 and D_3

[or D_6 if you use that convention, the symmetries of a triangle]

i mean this applies to

literally every pair of isomorphic groups

but the S_3 and D_3 one, just relabel the corners

intuitively you can map S_3 into D_3 by just labelling the vertices and rotating/reflecting the triangle appropriately

change the ordering of the "starting state" and each element of S_3 is mapped to a different combination of rotations/reflections, hence a different isomorphism

[soft elm]

this is kinda an automorphism

though

[scarlet estuary]

uh

no shit

[soft elm]

arent they essentially the same

[scarlet estuary]

yes, groups are isomorphic iff they are essentially the same.

that's the whole point of an isomorphism

the point is that different elements of the domain are mapped to different eleents of the codomain

like under the first drawing, {2, 3, 1} would be mapped to two clockwise rotations

[soft elm]

yeah im just being dumb

[scarlet estuary]

whereas under the second, {2, 3, 1} would be mapped to a reflection and two clockwise rotations

[soft elm]

thanks i get it

[scarlet estuary]

ie distinct elements

r^2 and r^2s

[brisk granite]

Thanks Buncho Bananas

that rly helps

does this look good now?

[hot lake]

p* ?

[brisk granite]

g(1,p)^2

it's a quadratic gauss sum

so p* is either p or -p when p is 1 mod 4 or 3 mod 4

[oblique river]

@brisk granite yep the answer is correct!

[brisk granite]

cool, thanks

[shy bluff]

Just wanted a quick clarification, the normalizer is a uh normal subgroup of G right?

[stone fulcrum]

I see the proof there that it's a subgroup. It is also normal by its definition

[supple rampart]

If it contains a single element then yes

I mean if S has one element

[stone fulcrum]

As the definition of a normal subgroup is "closed under conjugation", and the normalizer is the largest such group.

[supple rampart]

gH =Hg is refered as conjugation?

[stone fulcrum]

ghg' is a conjugation of h by g

[supple rampart]

So that element

[stone fulcrum]

We say a subset is closed under conjugation if, when it's conjugated by anything, is still in that subset. We write this (kinda with cheating) as gSg' = S

gS = Sg works too

[supple rampart]

Okay...that exact term
We have not been introduced
We generally say a subgroup is normal
If its left cosets and corresponding right cosets are equal

So when S is just a subset..till now it means nothing special to us.

[stone fulcrum]

What's the question?

Who's "us"? Lol

[supple rampart]

Us mean all students who are studying with me in the same college

[sharp sonnet]

how is the normalizer normal by definition

@stone fulcrum

hint: it is not

[supple rampart]

He said.and i quote".its clear that its a subgroup and that property which he is calling conjugation
Is the culprit behind making it a normal subgroup"
I am pretty sure that if S has a single element its normal but not sure when S has more...need to verify

[sharp sonnet]

what

[stone fulcrum]

Yall wat?

[sharp sonnet]

how is the normalizer normal?

just check any 2 elements subgroup of S_3

the normalizer of that is itself

which is not normal in S_3

[supple rampart]

Wdym

[stone fulcrum]

I think you just said "check any subgroup that isn't the normalizer, it won't be normal therefore the normalizer isn't normal"

[upper pivot]

no he gave a counterexample

[sharp sonnet]

the normalizer of {e, (1 2)} in S_3 is not normal in S_3

you should re-read the definition

[stone fulcrum]

I'm reading it

[supple rampart]

It contains more than 1 so i am fine

[stone fulcrum]

I think you're putting in a definition not equal to that one

[sharp sonnet]

you are wrong

[supple rampart]

Did it go wrong then kaynex for more than 1 elements in S

[upper pivot]

1 element doesnt help

take SL2(R) and any singleton

{A}, then its normalizer is > {1,-1} but less than the whole group

so its not normal