groups-rings-fields

[hushed flume]

but (123)-->(4123)

[delicate orchid]

so if you're able to move one into another, you have to be able to fix some terms in the cycle, which is impossible

[hushed flume]

and (412)-->4123

[delicate orchid]

I'll worry about that later

[hushed flume]

if am choosing between two elements to add to each k cycle sometimes I get the same cycle

[rocky cloak]

I feel like this is easier to do directly than by induction

[hushed flume]

yes that's what I thought too

[delicate orchid]

agreed, induction is a REALLY dumb way of doing this

but that's what was asked of me

[hushed flume]

the professor basically says that it's the rigorous way of doing it

but I understand the formula intuitively

[delicate orchid]

the professor is full of shite.

[hushed flume]

I won't sleep if I don't understand how to do it with induction

[delicate orchid]

the general conjugacy class size formula is proven directly

[hushed flume]

literally

[delicate orchid]

I think if we restrict ourselves to just looking at cycles in S_k then there is exactly one way (minus the, quite literal, edge cases) to generate each cycle in S_k+1

[hushed flume]

hm yeah i guess

but other than just adding the new element somewhere how else would I do it

I feel like I have to show that I'm going to get k+1 more cycles than I'm supposed to

[delicate orchid]

I think your way is better than my idea of adding something already not in the tuple to the end and then accounting for cyclic permutaions of everything

[hushed flume]

yes but I'm still stuck

[delicate orchid]

yeah ok, since our method double counts adding things to the beginning and the end - just don't allow yourself to add things to the end

so there are only k places to add things, not k+1

[hushed flume]

no wait you misunderstood me

it chose k in the first place

but even if you don't add to both edges you still get (k+1) more cycles than you are supposed to

[delicate orchid]

how

[hushed flume]

12 replicas of each cycle

(123)-->(4123)

(412)-->4123

[delicate orchid]

I said we're restricting our cycles to S_k

actually that's a point, are we inducting on n as well? or is n fixed

[hushed flume]

the induction for n is easy

just multiply by n+1

[rocky cloak]

I guess if you have to do it by induction:

In a cycle (x1, ..., xk) there are n-k things you could insert, into k spaces. And in hindsight any of the k+1 elements could have been inserted. So that's a change by a factor of (n-k)k/(k+1)

Then you just have to check that
(k-1)! * (n choose k) * (n-k)k/(k+1) gives you the desired formula

[delicate orchid]

I'm asking if we can assume the number of k-cycles for S_m with m < n

[hushed flume]

its for k-cycles of Sn where k<n

[delicate orchid]

so that's a no then

[hushed flume]

yeah I guess

[delicate orchid]

this is a smart way to bypass needing to check for double counting

"is the formula you get by naively putting things in correct"

[rocky cloak]

Well, it is checking for double counting

The divide by (k+1) is the double counting

[delicate orchid]

is it encased in the tomb of the labrythine statement "in hindsight any of the k+1 elements could have been inserted"

because I have no idea what that means

[rocky cloak]

(x1, .., xk+1) could have been obtained by adding any of the xis to a k-cycle

So that's an overcounted by a factor of k+1

[delicate orchid]

I mean dividing by k+1 would account for it cause you spin it around by a k+1 cycle

ah so both sources of over counting we were thinking about are actually the same source

[hushed flume]

I don't quite understand it could you explain it again?

[hushed flume]

what does this mean exactly?

[delicate orchid]

that "in hindsight" phrasing is still tripping me up

[hushed flume]

I undestand the k(n-k) part but I don't understand the justification for dividing by k+1

[delicate orchid]

we're dividing by k+1 because, if we start with a k+1-tuple (x_1, ... x_k+1) then this could have been achieved from just appending x_1 to (x_2, ..., x_k+1), or x_2 to (x_1, x_3, ... ,x_k+1)

so on all the way up

so there are k+1 ways to make each k+1-doodad

[hushed flume]

oh shit

thank god

[delicate orchid]

and then (k-1)!*(n choose k)*(n-k)k/(k+1) does indeed simplify down into what we want

[hushed flume]

thinking backwards!

that's so smart thanks so much!

[hushed flume]

wait I just realised I don't know how to induct on n either

[hushed flume]

nevermind I'm tripping

[wide brook]

Since the extension contains only one root of the irreducible polynomial x^3-3 the only automorphism is the identity map. And it’s unique, got it! I’m having doubts about whether $ /mathbb{Q}(sqrt[3]{3}) $ is a splitting field or no. It contains all the roots of the irreducible polynomial x^3-3 but not all the roots of (x^2-3)(x^3-3). Based on which of these polynomials we decide?

[rocky cloak]

It does not contain all the roots of x^3 - 3, only one of them.

[wide brook]

Ah sorry I meant this extension $ /mathbb{Q}(sqrt[3]{3},i) $

With the i

[rocky cloak]

I see, well I still don't think that contains all the roots of x^3 - 3

[wide brook]

The roots are sqrt[3]{3}, and I’ll send a pick of the other one

[wide brook]

Am i doing something wrong here?

[rocky cloak]

Where does the sixth root come from?

(3^(1/3))^2 = 3^(2/3)

If that's the confusion

[wide brook]

Oh shit! It’s a freaking typo. I was losing my mind here. Sigh* anyways…….So sqrt(3) is not in this extension(but it’s in the roots) and I’m using the irreducible x^3-3 to show that the extension is not a splitting field, got it! Thanks for the assistance, appreciate it

[hollow mica]

in lemma 1 it looks like they don't assume integral domains to be unital

But then in lemma 2 don't you need integral domains to be unital for the statement to be true?

Like to even speak of "invertible elements" you need a unity

[dire siren]

they say "integral domain with identity"

[hollow mica]

ooh I can't read thanks

[wispy light]

Let $F,G,H \in k[x,y,z]$ be homogeneous polynomials of degree $n+1$ such that $xF + yG + zH = 0$. Is it obvious that there exist homogeneous polynomials $A,B,C \in k[x,y,z]$ of degree $n$ such that the following equation holds?

$$F , dx + G , dy + H , dz = \begin{vmatrix} dx & dy & dz \ x & y & z \ A & B & C \end{vmatrix}$$

We may assume that $k$ is of characteristic $0$ or even algebraically closed if necessary.

[cloud walrus]

Eduardo León

[wispy light]

Alternatively, is it obvious that $A,B,C$ have no common factors if and only if $F,G,H$ have no common factors?

[cloud walrus]

Eduardo León

[cobalt heath]

Note that terms of F, G, H divisible by xyz can be considered separately, so we may reduce to the case where no terms have all of x, y, z.
Then, put z = 0, x = 0, y = 0 to see how the other terms work.
For instance, x F[z=0] + y G[z=0] = 0, from which one can show F[z=0] = yC, G[z=0]= xC, etc.

Think characteristic 0 would be required, otherwise you can have F = yz, G = zx, H = xy for char 3.

[languid trellis]

I've been thinking about this exercise for a little while now and I'm completely stuck. Can someone give me a small hint in the right direction please? (FYI Thm. 2.18 is "finite subgroups of the multiplicative groups of fields are cyclic", which they prove by considering x^exp(G) - 1 = 0, and pointing out that it has at most expG solutions in G, and ex.5 is "if the order of a finite field is even, then it is char 2, and p-1 | |F| - 1".

[cobalt heath]

You can compute the product from F^* cyclic, I think

[vivid tiger]

Note: if q=p this is Wilson's theorem

[languid trellis]

I was thinking something along the lines of WLOG a1 generates F*, then writing a_j in terms of a1 or something

[languid trellis]

Wilson's theorem is the next exercise 🙂

but yes I do see it

[vivid tiger]

Strange

[languid trellis]

(I see that it's wilson's theorem, not how to prove it)

[cobalt heath]

Yeah, you can do that

Can you compute the product under the settings?

[vivid tiger]

Well.

Is this just vieta?

Actually you only wanted a hint so I'll shut up

or use the magic of spoiler tags

note that i might be wrong. whether this is just vieta is left to you

||x^{q-1}-1 has x a root for all x besides 0. Then the product is -1/1=-1||

[languid trellis]

it'll be a1a2a3...a^{q-1} = a1a1^2a1^3 etc..., = a1^{1+2+3+...+q-1}. Now consider the equation a1^{1+2+3+...+q-1} + 1 = 0

[vivid tiger]

in general i see F_q and then look at the secret polynomial

[cobalt heath]

Oh, the hint said "proof of.."

Works in any dir, tbh

[vivid tiger]

and this successfully gets me like all of the properties of it.

[languid trellis]

dyi

[vivid tiger]

I like my proof better

[cobalt heath]

Yeah, maybe it is talking about f(x) = x^expG - 1

[vivid tiger]

The hintier part comes from the way Lang writes this proof: ||roots of a^m - 1||

also guys why do y'all keep using exp G

[cobalt heath]

Still, hard to spot

[vivid tiger]

what is this

is there a lie group

[languid trellis]

Yeah, in this context x^expG + 1 = 0

[vivid tiger]

gimme gimme gimme

[cobalt heath]

Huh?

[languid trellis]

well because we're solving ..... + 1 = 0

perhaps

[cobalt heath]

I don't think it will change

[vivid tiger]

It will. Only one of those two options works

[cobalt heath]

Welp, I dunno.

[vivid tiger]

What are you trying to do here? You have a pooynomial.

You could try to express the thing you want as a polynomial and hope it obviously has a root.

[cobalt heath]

I do not understand this part, wdym here?

[languid trellis]

we want to show a1a2...a{q-1} = -1. so, a1a2...a{q-1} + 1 = 0

[vivid tiger]

So, do you remember the construction of F_q?

[languid trellis]

who are you asking sorry

[vivid tiger]

The construction is why I do what I do every time I am asked a question about F_q

[cobalt heath]

Ah. I guess the "nifty proof" is quite difficult to think of, I dunno how to give a hint about it

Brute-force computation is easier to achieve

[languid trellis]

hm

[vivid tiger]

huh???

so F_q is defined as?

splitting field of X^{q-1}-1 over F_p where q=p^n

[cobalt heath]

Imo, it is easier to just compute the product.

[vivid tiger]

how do you do that

[cobalt heath]

You do it like this

[vivid tiger]

i don't understand

[cobalt heath]

Then you sum up from 1 to q-1

gtg for a while

[vivid tiger]

so like now do you see it

Oh, wait since F_q is cyclic you could try 1+2+...q-1

That's what you meant?

[languid trellis]

I suppose to compute the product we could pair up multi inverses and use that finite fields have char p

So any multiplicative subgroup is a multiple of p-1

Which is congruent to -1 mod p

Something like this

[vivid tiger]

you want -1 in F_q

oh but -1 in F_p \subset F_q

I definitely think this is harder and feel that you will feel that the one line proof is staring you in the face

[languid trellis]

And The one liner starts with 'consider x^m +1 = 0 in F_q'?

[vivid tiger]

Do you want me to answer?

The start of the one liner has been written by at least one of the three us

(why is ":cheeky:" on the physics server?)

[wispy light]

Thanks.

[cobalt heath]

Pair up..?

Isn't there straightforward way of just computing it?

[dire siren]

that's already pretty straightforward

but the characteristic is not relevant

after pairing, we are left with the elements of order 2, but -1 is the only one

[short spire]

Can you provide the statement of theorem 2.17?

[languid trellis]

This is what I ended up doing after thinking about it a bit more

[cobalt heath]

Oh. Did not know that is so straightforward

[languid trellis]

Characteristic is slightly relevant though I feel because for char F neq 2, F* has even order so an element of order 2 exists

Whereas if char F = 2, F* is odd, so I'm not sure we csn say anything about having elements of order 2

[languid trellis]

Polys of degree n have at most n roots in F

[cobalt heath]

The hint mentions Exercise 5

For when q is even

[languid trellis]

Exercise 5 says |F| even implies char F = 2

Oh wait I see. Even in the char 2 case after we pair up multi Inverses, we have ajak left over which must be it's own inverse and hence equal to -1 even if aj or ak aren't equal to -1

I beliebe

[cobalt heath]

Well, could be even simpler

Compare -1 and 1 in char 2

[languid trellis]

I'm happy with my solution lol

But 1=-1 in char 2

[cobalt heath]

Yeah, so it should work out either way.

[dire siren]

good point with char 2

[short spire]

Oh I get it now. I am sorry.....

[orchid iron]

Hi,

Does anyone know of a reading group based on Dummit and Foote's abstract Algebra?
(Active/Inactive anything is fine)

Please share link/invite if you do?

[pseudo sierra]

Hello can anyone help me understand what an elementary divisor and rank is and its relationship to the smith normal form? I know what a rank is in linear algebra but not in the context of this question that I assume is more abstract algebra

Given a matrix A, i calculated its Smith Normal Form

And i am being asked to determine the elementary divisors & rank of this

Based on the fundamental theorem of finitely generated modules in PID an elementary divisor and rank is defined like this

Does the “di” have anything to do with the d1, …, dl entires of the smith normal form?

[delicate orchid]

They’re exactly the entries in the smith normal form

Moreover, if we stop crying and recognise that R/(0) = R the same holds true even for those columns which are all 0s

And the number of such 0s is the rank

[languid trellis]

Lol

[vivid tiger]

Okay so here's my solution
x^{q-1}-1 has every element but 0 as a root

By vieta the product of roots is -1/1=-1

[dull ginkgo]

Sup chat

[vivid tiger]

well this is literally just because even F would have to be F_{2^n}

[dull ginkgo]

What are we trying to do

[vivid tiger]

and those are extensions of the char 2 field Z/2Z

[vivid tiger]

the product of all the things is (F_q)^× is -1

proof: vieta

[dull ginkgo]

proof: each is a solution to x^q - x

factor out 1 and x

[vivid tiger]

x^{q-1}-1

yes

we have the same mind

a melding of the minds

[dull ginkgo]

but ye, characteristic

[vivid tiger]

oh shit

(-1)^(q-1) (-1/1)= (-1)^q

so we do indeed have to use that if q is even then 1=-1

[dull ginkgo]

-1 = 1 mod 2 lol

But yeah, you can use Vieta's formulas or just, go right off the poly shit

I need to make note of what exercises I have basically done already in this chat ahead of time

because I know damn well when I get to it I'm gonna draw blanks

this happens a lot more than it should

@vivid tiger I actually think pretty much every proof of this kind of circles back to polynomials lmao

[vivid tiger]

that is indeed due to the definition of F_q

[languid trellis]

Vietas formula are coeffs interms of roots right

I only half know what it is

But that's a nice solution xela

[dull ginkgo]

I tried proving it another way and just cycled back to F_q^x being cyclic

[rotund aurora]

You can prove it by the formula $\sum_{d\mid n} \phi(d)=n$

[cloud walrus]

croqueta3385

[rotund aurora]

well it's pretty much the same

[vivid tiger]

$\prod_i(x-r_i) = X^n + (-1)^{n-i} E_i(r) X^{n-i}$ where $E_i(x) = \sum_{\sigma \in S_n} \frac{1}{i!} \prod_{j \le i} x_{\sigma(j)} =$ elementary symmetric polynomials

[cloud walrus]

Xela

[languid trellis]

Symmetric polynomials are the next section

AAAAAAĄAAAA

[delicate orchid]

https://tenor.com/view/kitty-gif-27687064

[barren sierra]

Symmetric polynomials 🔥🔥

[vivid tiger]

you don't need any fancy properties here

define the elementary symmetric polynomials via Viéta's formulas

note that they are
x1+x2+...xn
x1x2+x1x3+...x2x3+...

etc

that is, sum of single products, sum of double products, sum of triple products, etc.

note that they are fixed by S_n acting in the obvious way

[languid trellis]

i was never gone sweaty

[languid trellis]

I'll reconsider this when I get a proper introduction to sym polys, but I appreciate your time

[dull ginkgo]

@languid trellis did Jacobson show F^X is cyclic?

[languid trellis]

what do you mean by F^X

[dull ginkgo]

Multiplicative group

(Group of units)

[languid trellis]

of a field?

then yes

[dull ginkgo]

Yes

You can just sum up all the numbers from 1 to the order of the cyclic group lol

[vivid tiger]

so you don't need even that, you just need the first and last. it's where the well known fact that for a quadratic the sum of roots is -b/a and the product is c/a

it is as simple as the constant term of \Prod (X-r_i) being (-1)^n \Prod r_i

[alpine pasture]

What does the notation G>H (or G<H) mean, where G and H are groups?

[vivid tiger]

H ≤ G usually means H is a subgroup of G, and H \lhd G means H is a normal subgroup of G

do note that while $\le$ is transitive, $\lhd$ isn't

[cloud walrus]

Xela

[vivid tiger]

it instead obeys the "tower law" thing that many things in algebra obey: if $K < H < G$ and $H \lhd G, K \lhd G$, then $K \lhd H$

[cloud walrus]

Xela

[alpine pasture]

thank you

[cobalt heath]

What is Vieta's formula? I was never introduced to it.

[vivid tiger]

this

so for an n degree polynomial

[dull ginkgo]

Oh my god jacobson leaves the newton identities as an exercise

of course he does

why wouldn't he

[vivid tiger]

the product of roots is (-1)^n a_0/a_n

where we are saying a_n x^n + ... a_0

the sum of roots is - a_{n-1}/a_n

and the sum of double products is + a_{n-2}/a_n

and the sum of triple products is -a_{n-3}/a_n

so like for example in x^3+3x^2+7x+2

the sum of roots is -3

[cobalt heath]

Ahhh, idk why was not named in my lec

[vivid tiger]

the sum r_1r_2+r_2r_3+r_1r_3 = 7

and so on

glad i don't have to keep explaining because this specifically is annoying to type

[cobalt heath]

Hell is newton's identities?

[vivid tiger]

relation between

uhh

one other kind of symmetric polynomial

and these ones

[vivid tiger]

so like these things are invariant under the obvious action of S_n

so they are called elementary symmetric

they are elementary in the sense that every polynomial is a polynomial in the elementary symmetric ones

[cobalt heath]

That be annoying

[vivid tiger]

so there's also the power sum symmetric polynomials

1+1+1+1...1

x1+x2+...xn

x1^2+x2^2+...xn^2

etc.

now these generate as a ring the symmetric polynomials over the ratioanls

The ring of symmetric polynomials over ℚ in n variables equals ℚ [p_1,...,p_n] where p_i is the ith power symmetric polynomial in n variables

There's another kind.

The complete homogenous symmetric polynomials

$h_k = \sum_{1 \le i_1 \le i_2 ... \le i_k \le n} X_{i_1} ... X_{i_k}$

[cloud walrus]

Xela

[cobalt heath]

Is this symmetric?

[vivid tiger]

Yes. Uhh...look at it?

$=\sum_{\sum_i^n l_i = k} X_1^{l_1} ... X_n^{l_n}$

[cobalt heath]

Ah, it's vieta + power sum

[vivid tiger]

We also have that

[vivid tiger]

...it is?

I didn't see that

maybe I need to look at it?

[cobalt heath]

Vieta is basically one with i1 < i2 < .. < ik, iirc

Oh, right. Not abt power sum.

[cloud walrus]

Xela

[vivid tiger]

yes, choose k out of n without double counting

power sum is i1=i2=...=ik

here we say ≤

this is a good observation

so then isn't this literally the sum?

[cobalt heath]

Yeah, I missed that something

[vivid tiger]

Right and vieta and power sum are both homogenous

[vivid tiger]

So did I, but you missed it closer

[cobalt heath]

I mean, methought i1 < i2 < i3 or i1 = i2 = i3 <=> i1 <= i2 <= i3 lol

Wish my brain functions

[vivid tiger]

So like Newton's identities are some relation or whatever between elementary symmetric and power sum symmetric

There's also some similar relation that =0 for elementary symmetric and complete homogenous symmetric

and like combinatorics can be done

[vivid tiger]

methinks "methought" is spreading.

[sonic coral]

does anyone know what the smallest positive integer n for which there is a transitive subgroup of S_n that has not been realized as a galois group

i believe it’s 23 with M_23 being the group in question but i figured id ask

[hardy musk]

i'm trying to think up proofs of elementary facts about free inverse semigroups, but i'm having a lot of trouble with the word problem here

for instance: how do i show that abb*a* is not equal to baa*b*?

i get the sense (from what little ive read) that getting a sense of structure in a free inverse semigroup is difficult, but even just playing with basic toy words i don't really see how something can be done

ping me as much as youd like

[untold garnet]

I would guess that it is equivalent to finding some inverse semigroup H with two elements x,y such that xyy*x* not equal to yxx*y*.

[torpid knot]

the lemma seems extremely obvious but im not sure what substitutions to specifically make to prove it

op nvm found smth online for it

[terse crystal]

an abelian group A is direct sum of n many cyclic groups Z/p^ei Z where 1<=i<=n (p a prime number e1,…,en positive integers)
How can we calculate how many subgroups of order p^r there are ?
(if it’s hard to solve the general case I am particularly interested in the case where r=e1+…+en -1, namely number of index p subgroups)

[rocky cloak]

The case r = e1 + ... + en - 1 is pretty straight forward at least.

Since any subgroup of order p^r is the kernel of a surjective map to a group of order p.

The only map A -> Z/p that isn't surjective is the 0 map, so we're looking at |Hom(A, Z/p)| - 1 surjective maps. And to maps have the same kernel if one is the other but composed with an automorphism of Z/p. So that's an overcounted by a factor of p-1.

There are p maps from Z/p^ei to Z/p, so |Hom(A, Z/p)| = p^n

In total (p^n - 1)/(p-1) subgroups

[terse crystal]

Thanks a lot
If I understand your method correctly, can we generalize that
The number of index p^r subgroups is |epimorphisms from A to Z/p^rZ|/|Aut(Z/p^rZ)|?

Oh shit… should it be
Σ|{ epimorphisms from A to B}| /|Aut(B)| where the sum is over all distinct order p^r abelian groups B…？

[delicate orchid]

I’d say this is correct, all the other sporadics have been shown to be Galois

I’d be very surprised if there’s one acting on fewer than 23 elements

[terse crystal]

And I have another question. Abelian group A is Z/p^e1Z oplus … oplus Z/p^enZ, e1>=…>=en
subgroup B of A, is isomorphic to Z/p^f1Z oplus … oplus Z/p^fmZ, f1>=…>=fm
We have partitions λ=(e1,e2,…,en)
μ=(f1,…,fm). Is it necessary that young tableaux of λ contains entirely young tableaux of μ?

[rocky cloak]

Yes, it would be that. Which quickly becomes a very tedious computation. So for bigger index you might want to switch to a different method

[terse crystal]

I see, thank you

[topaz gale]

Question about euclidean rings and their valuations. I'm having a hard time solving problems concerning them, so any insights would be much appreciated.

So let say we have an euclidean ring R where the only inversible elements in R are 1 and -1. If a is an element of minimal valuation from all the non inversibles elements of R*, then R/<a> is isomorphic to Z/2Z or Z/3Z.

So I start by constructing a set E=min{phi(a) | a in R, a non inversible}. This set is non-empty by hypothesis, and thus has a minimal element. Say /phi(s)=min(E) where s is non-inversible.

Here's where I'm not sure how to proceed. If R is euclidean, can I say that R/<a> is also euclidean, and check the valuation there? Otherwise, I say, R is euclidean, so I can divide here, and so there exists q,r in R such that a=qs+r with r=0 or phi(r)<phi(s). I can then deduce a couple of things : if r=0, then a=qs and thus s|a. If r is not 0, then r is inversible, otherwise we would have phi(r) >= phi(s) by the minimality of phi(s), a contradiction.

From here I'm not sure where to go. Should I work in the quotient, and if so, how so?

[rocky cloak]

Remember that you assumed the only invertible elements to be 1 or -1. So if r is invertible, then it must be either 1 or -1

[glass zealot]

probably stupid question, but why do free abelian groups exist? doesn't the commutativeness imply relations between the generators which makes it non-free?

[rocky cloak]

Use this to deduce what the cosets of <a> can be

[glass zealot]

i.e <a,b | ab=ba> shouldnt be a free group no?

[rocky cloak]

It's 'free (abelian group)' not 'free group which is also abelian'.

In general you can thing 'free X' as an X with no relations beyond the definition of X.

So a free abelian group is an abelian group with no added relations

[south patrol]

It is an unfortunate coincidence of terminology lol

[south patrol]

I should also jokingly add that the trivial group and Z are both free groups and abelian :)

[rocky cloak]

I think 'commutative graded ring' is even worse

[glass zealot]

ahh lol

alright thanks

[south patrol]

Actually i talked to my advisor about this since it is unfortunate in more than one way

Lol

Well like

Graded rings are not graded objects in the category of rings

That is unfortunate enough kinda lol

[topaz gale]

Okay, so I get that if r is invertible, then it must be either be 1 or -1. I'm not quite sure how to use this to deduce what the cosets of <a> can be in this context.

Should I just plug it back in the euclidean division formula?

[rocky cloak]

The cosets of <a> are of the form
<a> + r
for some r. So how many cosets could there be?

[topaz gale]

I would say only two, right? <a> + 1 or <a> - 1?

I feel like I'm missunderstanding something

So yeah, if I plug it back into the euclidean division formula, I have three choices :

(i) a = qs
(ii) a = qs + 1
(iii) a = qs - 1

So in the quotient, a can only be 0, 1, or -1.

Is that correct?

[rocky cloak]

There is also <a> + 0.

But yeah, that's right. There are at most 3 cosets, so R/<a> is a ring with at most 3 elements

[topaz gale]

Getting closer here! Thank you for sticking with me!

So I would suppose now that what left to be seen is under which conditions 1 and -1 are showned to be the same (mod 2) or distinct (mod 3)?

[rocky cloak]

Yes, so that's a question of whether 2 is in <a> or not

[topaz gale]

Perfect, thank you for the help!!

[barren sierra]

wait what's the issue with this?

[rocky cloak]

A commutative graded ring is a graded ring where
ab = (-1)^|a||b| ba
when a and b are homogenous

[barren sierra]

yea I know

[rocky cloak]

Hence, not commutative

[barren sierra]

oh

I never thought about that

ugh

I hate math

[south patrol]

well it is commutative for the correct definition of commutative

hehe

[hidden wind]

oh my god i’ve completely neglected my abstract algebra course

exam i believe coming up very soon

a lot of catching up to do

[celest furnace]

wtf 💀

[next obsidian]

no this is graded commutative

I think

I swear there is a difference

https://en.wikipedia.org/wiki/Graded-commutative_ring

[hidden wind]

commutative graded ring vs graded commutative ring?

[next obsidian]

Yes

[barren sierra]

I forgot about this as well 😭😭😭

Math is awful

[celest furnace]

Real

[frank cosmos]

is this proof that if Aut(G) is cyclic then G is abelian correct?

Suppose Aut(G) is cyclic. By the previous exercise, the set of inner automorphisms is a subgroup and hence cyclic as well, say generated by $c_x$ where $c_x g=x g x^{-1}.$ Let $y \in G.$ Then $c_y=(c_x)^n$ for some $n \in \mathbb{Z}.$ Thus for all $g \in G, ygy^{-1}=x^n g x^{-n} \implies g=y^{-1}x^n g x^{-n}y=c_{y^{-1}x^n}g$ thus $y=x^n.$ So $G$ is cyclic and thus abelian.

[cloud walrus]

Hello1

[next obsidian]

Why does that final thing imply y = x^n?

It isn’t true that c_e = c_y^-1x^n => e = y^-1x^n

A priori

Also don’t think it’s true at all

[frank cosmos]

you're right it just shows that y^-1x^n is in the center

do you think i can fix this into something correct

[barren sierra]

Inn(G) is cyclic

You got that first step correct

Can you think of some subgroups / quotients of G that are isomorphic to Inn(G)?

Off the top of my head idk a way to repair the proof you have

But maybe there's a way

[dull ginkgo]

||Are you suggesting to use Inn(G) ~= G/Z(G)?||

[next obsidian]

Yes

What if y = x^n-1

What do you conclude

Then you get more stuff in the center

[frank cosmos]

wdym

take this conjugate?

[next obsidian]

And you can then conclude all inverses in the center

So GG

What if y = x^n-1 lol

Then what do you get to conclude

[frank cosmos]

dosent n depend on y

[next obsidian]

you’re so right

Damn okay I think this won’t work

You can try y = e

[frank cosmos]

smart

[next obsidian]

But then you’re kinda hozed

[frank cosmos]

ye

[next obsidian]

Because of this dependency of n on y

[frank cosmos]

n could be 0 lol

taking y=e

[next obsidian]

True

Yeah this probably doesn’t lead to anything then

[frank cosmos]

i thought i cooked

[next obsidian]

Also you could never conclude G was cyclic

Take G non cyclic abelian and then InnG is cyclic

(Trivial)

[barren sierra]

Think about what I said here @frank cosmos

[frank cosmos]

ok this should be G/Z

probably

so that is cyclic

[barren sierra]

And what if G/Z is cyclic 👀

[cloud walrus]

Hello1

[frank cosmos]

so what i had before pretty much

rip

wait is this troll

[barren sierra]

No

[frank cosmos]

so then $y=x^n z$ and we're done commutativity e

zz

[cloud walrus]

Hello1

[frank cosmos]

z

does this not work

[barren sierra]

Uhhh it's a little hard to follow what you've written

Let's be very direct

You know G/Z is cyclic

I give you x and y in G

Show me xy = yx explicitly

I think what you have is ok but it's hard to follow (as evidenced by your uncertainty as well)

[frank cosmos]

im 100% sure it works

thx

[barren sierra]

lol ok

np

[lucid wigeon]

I'm trying to understand Quotient Rings built from Polynomial Rings but I'm struggling with some examples I see. How is <x+a> an Ideal of K[x] for $K = \mathbb{Z}/7\mathbb{Z}$? How is x an Ideal of $\mathbb{R}[x]$? Neither seems to absorb multiplication by all elements of their parent Ring. Are these invalid examples or what am I missing?

[cloud walrus]

StackCanary

[celest furnace]

I think you are a little confused by notation. (x+a) is, by definition, the smallest ideal containing x+a. <x+a>, which I think you mean as the abelian subgroup generated by x+a, is not an ideal, or even a ring

[lucid wigeon]

Apologies for being careless with notation. The example I'm referencing is specifically $K[x]/\left<x^2 + a\right>$ from https://math.stackexchange.com/questions/2054968/when-quotient-ring-is-a-field. I forgot the square

[cloud walrus]

StackCanary

[celest furnace]

So yeah a major abuse of notation would be also witting <x^2+a> as the smallest ideal containing x^2+a

So one is a subgroup and the other is an ideal, try to use the () notation instead to make it more clear

But in general for elements a_1,…,a_n, (a_1,…a_n) is the smallest ideal containing the ai

Meaning the intersection of all ideals containing the ai

He writes “since we are quotienting by a polynomial of degree 2, every equivalence class in the quotient ring can be represented by a linear polynomial of the form ax + b”

I think this is where you were getting confused?

Well maybe I should’ve asked this before but what do you think <x> is?

[lucid wigeon]

Well now I'm not sure how to interpret the <> notation. There's a lot of notations going on. I'm confused about what is and isn't an Ideal and what defines an Ideal. I thought they had to absorb multiplication by all elements of their parent Ring. Quotient/Factor Rings can only be constructed from Ideals, correct?

[celest furnace]

Yes exactly. S, a sub ring of R, is an ideal if r*x is in S for all r in R

And yes quotient and factor rings can only be constructed from ideals

[lucid wigeon]

So R[x]/x is a simpler example. x doesn't seem to be an Ideal of R[x] but perhaps it's notation for x as the generator via modulus? x itself isn't the Ideal though, right?

[celest furnace]

Yes exactly

People are just abusing notation.. in this case x would actually be the ideal (x) = {p(x) * x for p(x) in R[x]}

It would be more correct to write R[x]/(x)

[lucid wigeon]

Ah ok now that makes sense! So would R[x]/xR[x] also be an ok notation? I'm new to the () notation.

[celest furnace]

Yes it would

That’s why you see Z/nZ sometimes

[lucid wigeon]

Wonderful, thank you for clarifying. In the end, notation turns out to be the problem. A simpler solution than I thought. Much appreciated.

[celest furnace]

Yep notation can be a pain lol

When people get lazy you get stuff like this https://media.discordapp.net/attachments/1150849425288208394/1232139785817292881/image.png?ex=662a5958&is=662907d8&hm=e86fd6c72d43114944f00788687bb6924e66adb4a2539a1daa975b8514f437ce&

[exotic verge]

is this proof proofy enough

https://i.imgur.com/ki4qIcB.png

[tame fable]

is hard to read

maybe add some more whitespace?

[exotic verge]

uhh i meant more like the content

[chilly radish]

You didn't prove the order is d

You only proved it divides d

[exotic verge]

do you mean d divides the order?

nvm what i said doesn't make sense

but im not sure what you mean

oh I have to show for any m less than d f^m does not equal the id?

[summer path]

can you not just write it out in cycle notation?

[molten rivet]

I think I've figured out how to solve this problem, but I don't see how F not having characteristic 2 is relevant. My argument is just induction on the degree of f:

If f has degree 1, clearly the splitting field of g has degree at most 2.
If f has degree n, we let alpha be a root of f in its splitting field, so that f=(x-alpha)h for some h. Then, the polynomial h(x^2) has a splitting field of degree at most 2^(n-1)(n-1)! by the induction hypothesis, call it K. Since g=(x^2-alpha)h(x^2), g splits over the extension K(sqrt(alpha)). Since sqrt(alpha) has degree at most 2n, [K:F] is at most 2^nn!.

I never used the fact that F doesn't have characteristic 2 here. Is there something in my logic that breaks down for fields with characteristic 2?

[eager willow]

in characteristic 2 g(x) = f(x^2) = (f(x))^2 so the splitting field is the same as that of f(x), which has degree at most n!, but clearly this is less than 2^n * n!

[molten rivet]

Ok, thank you

[exotic verge]

is this the correct way to go about the proof?

https://i.imgur.com/tHtpbQ8.png

i cant see a way through without a bunch of casework

[lusty marlin]

Seems fine

[tough raven]

Personally I do this but it's always because I got tired of writing the parentheses.

[tough raven]

That's not true unless the coefficients of f lie in F_2.

[tough raven]

I'd rather phrase it as [K : F(α)] <= 2^{n-1}(n-1)!, [F(sqrt(α)) : α] <= 2 so [K F(sqrt(α)) : F) <= 2^n n! so [K F(sqrt(α)) : F] <= 2^n n! since [F(α) : F] <= n.

[tough raven]

As far as I can see, this remains true in characteristic 2.

If the coefficients of f are squares (e.g. if F is perfect), we can write f(X^2) = g(X)^2 where the coefficients of g are the square roots of those of f.
Then as @ datorangeguy pointed out, you get a stronger bound n! on the degree of the splitting field of g. But the original bound should still hold.

[tough raven]

Yes, exactly.

[tough raven]

Also this

[tough raven]

You can simplify the casework by the following:

if g commutes with any x, y it commutes with xy and x^{-1}. (Exercise.)
So if something commutes with the generators x, y, it commutes with the entire group D_n.
Thus you can take just the equations
x^i y^j commutes with x
x^i y^j commutes with
instead of all k, m.
In fact, you can further simplify this to
y^j commutes with x
x^i commutes with y
(why?).

Beyond this I think it's better to actually do the casework.

[chilly radish]

That would do yea

[crystal vale]

Since we have G which is generated by x and G is finite so if |G| = n , where n is composite number then there must exist a subgroup of order of a, where 1<a<n, where a is divisor of n, which contradicts that there is non-trivial Subgroup

Is it correct?

If I want to show that H is a subgroup of G such that if Ha≠Hb then aH ≠ bH. Then H is a normal subgroup of G.
Ha≠Hb implies aH ≠ bH, means if ba^(-1) not in H then b^(-1)a not in H.
I want to show that gHg^(-1) is a subset of H for all g in G.

If g is in H then it is done.
If g does not belong to H and let gHg^(-1) not in H, then take b=g and a^(-1)=hg^(-1) then b^(-1)a not in H which implies that g^(-1)gh^(-1) not in H. It contradicts that H is a subgroup.

Is it correct?

[pastel pier]

is a binary operation always closed ?

here it says H is closed if its a binary operation

so G Is assumed to be closed by birth ?

if a binary operation is defined on a Set, then its closed under that binary operation ?

but then its not a binary operation aswell, look at the definiton

i could include 💀 , i am not a mathematician

[crystal vale]

They define binary operation image in G so yes G is closed under operation

[stark helm]

Can anyone tells me how can I know if there is a root on unit circle?

[dire siren]

@stark helm the coefficients are symmetrical, so to solve p(z)=0 you can divide by z^2 and make the substitution t=z+1/z
you end up with a quadratic in t

oh, funnily the polynomial is just (z^2-z+1)^2

[random pasture]

I feel like i keep missing some equivalence

[jagged gate]

wdym ? what have you got right now ?

[topaz solar]

What have you tried

[delicate orchid]

3 implies 1 and 2 trivially, now use the fact that if z is invertible we have some z' such that zz' = 1 - take the norm of both sides of that expression to show 1 => 2. 2 => 3 is also quite straightforward, for a hint: when is x^2+y^2 = 1? And then we're done as we have 1 => 2 => 3 => 1

[topaz solar]

Time to !nosols wew

[delicate orchid]

by my standards that's practically opaque

[stark helm]

Can someone tell me how to determine if a and b are constructible?

[delicate orchid]

see if the field extenstion has degree equal to a power of 2

if it doesn't, then you know that they aren't constructable

if it does then you'll need to show/disprove that it can be constructed out of quadratic extenstions

which, given the cube roots, is probably unlikely

[stark helm]

for a, I can clearly see that 7+5*sqrt(2) is in tower(sqrt(2)), but the cube root implies that its adjoining root of next tower is of degree 3, so it is not surd implies not constructible, is that true?

[topaz solar]

Surd is a weird word

But uhh you went up Q -> Q(\sqrt 2) -> Q(\sqrt 2, a) = Q(a) as extensions right?

[stark helm]

how will you argue this one?

[topaz solar]

???

What do you mean

[Q(sqrt 2)/Q] is 2, and [Q(a)/Q(sqrt 2)] sure looks like 3 to me

[delicate orchid]

,w minimal polynomial (7+5\sqrt(2))^(1/3)

Ermmmm… what the deuce?

Oh

[cloud walrus]

Results provided by WolframAlpha

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

Requested by wewladstbh

[topaz solar]

Ermmmm what the ^1/3

Awful

[delicate orchid]

Ermmm… what the deuce?

[topaz solar]

By evil sorcery

This is 1+sqrt(2)

I hate cube roots

Most constructible number alive

[delicate orchid]

It actually is as well. Wtaf

[topaz solar]

Yeah, stacking roots is awful always

b is definitely not though since ya know

Please don’t be :3:3:3

[delicate orchid]

I don’t trust anything anymore

[topaz solar]

Well if it were, cube root of 2 would be

And uh

[delicate orchid]

,w minimal polynomial sqrt(1+4^(1/3)/2)

[cloud walrus]

Results provided by WolframAlpha

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

Requested by wewladstbh

[delicate orchid]

We’re all good…

[topaz solar]

The trick was [Q(a)/Q(root 2)] was secretly 1….

[sonic coral]

where does this result come from?

[delicate orchid]

Wikipedia

[topaz solar]

Definition of constructible things is coming from quadratics or whatever right?

Stacking 2’s

[delicate orchid]

Well the proper definition is that it’s the stuff you can make with straight edge and compass

And it takes some work to show that you can’t get cubic shit

[topaz solar]

Who cares about geometry fr

[delicate orchid]

Agreed

[topaz solar]

But yeah showing that the constructions are the same as the good definition is nontrivial

However: I’m assuming they have at least something to work off of

Before asking that question

[sonic coral]

wait so if i wanted to know if roots of a polynomial were constructible, could i check if the galois group was a power of 2?

[topaz solar]

Mfw not every extension is Galois

[delicate orchid]

Who caare

But you just need to check the degree

If it’s not a power of 2 there’s no chain of quadratic extenstions because that’s how primes work

[sonic coral]

i see okay

[stark helm]

So a and b are all not constructible right?

[topaz solar]

a is

[cobalt heath]

,w min poly sqrt^3(7 + 5 sqrt(2))

[cloud walrus]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by abstia

[cobalt heath]

Oh wow

[topaz solar]

,wolf (1+sqrt 2)^3 - 7 - 5 sqrt 2

Foiled again by decimal approximation

[cloud walrus]

Results provided by WolframAlpha

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

Requested by sharp2229

[cobalt heath]

Ah, I wrote that wrongly

So literally Q(a) = Q(sqrt(2))? Lol

[topaz solar]

Yep

[stark helm]

you mean a is constructible? Can you tell me how you will argue that a is constructible?

and also why b is not constructible?

[topaz solar]

I will not write the proof for you

[topaz solar]

Look at what was written above

[cobalt heath]

I do wonder how one could "prove" 7 + 5 sqrt(2) is cube of something

[topaz solar]

Finding the cube

[cobalt heath]

Physically construct a cube of side length 1 + sqrt(2)

Measure volume

???

Not profit because it cannot be exact 🫠

[stark helm]

I think a is constructible while b is not, because we can know cube root of 4 is not constructible, so the stuffs in square root not constructible implies b not constructible, while a can be argued by radical extension( since all surds are constructible)

[stark helm]

Does anyone know how to solve this question?

[chilly ocean]

bust out your compass and straightedge

[sonic coral]

does this work?

[topaz solar]

Well you can’t construct cube roots, this one just happens to work as square roots

[stark helm]

so a is not constructible, because it is a cube root of a constructible root right?

[topaz solar]

It is constructible

Because it’s equal to 1 + sqrt 2

[dull ginkgo]

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

[stark helm]

ok I think I should suppose that stuff equals a+b*sqrt(r) first

[stark helm]

do you mean the question I gave?

[stark helm]

I can see that it does not have rational root

but is that adequate to prove here? I am somewhat doubting on it because it seems to let x^2=x here

[winter shore]

What were (are?) the motivating examples for introducing associative algebras? I've see them a couple of times pop up so I'm wondering if they appear naturally somewhere

[dull ginkgo]

Maybe try to eliminate the square term via a shift

wait show it has no constructable root

Hmm

[stark helm]

i think I can see it now, because x not rational, not constructible root of this polynomial of degree 3

[dull ginkgo]

Yeah

[stark helm]

then sqrt(r) not constructible implies r not constructible as well

[dull ginkgo]

I think?

[random pasture]

how exactly can i prove that rs can't be a cube if r and s aren't both cubes

[stark helm]

does anyone know how to determine if it is constructible? I think cos(7.5) constructible, so square root(1-sin(7.5)^2) is constructible, then we know sin(7.5) construtible. but how can we know if sin(7) is not constructible?

[topaz solar]

what have you tried

[winter shore]

bump

[dim widget]

Are you going to go through and ask every homework question in this forum?

[steel wolf]

Is there standard notation for a representation. Like how the cross product algebra is a representation of so(3)

[random pasture]

I tried representing r and s as products of irreducible elements

and then assuming neither is a cube

in order to examin what happens when you multiply them together

oh i see

Is it that by the definition of irreducibility, no irreducible element can be a cube?

[dim widget]

Yes

[random pasture]

So based on that, and the idea that products are unique up to associates, I can confirm that rs can't be a cube if one of the two isn't a cube>

[stark helm]

this is actually our practice problems

[dim widget]

sounds like you should practice them

[stark helm]

then do you think my argument for sin(7.5) constructible correct? And how can I know if sin(7) is constructible or not?

[dim widget]

You didn’t give an argument, you just said cos(7.5) is constructible

[stark helm]

I know that cos(60) is constructible, and given any angle, its bisector is constructible

60->30->15->7.5

[hidden wind]

A5 = PSL(2, 5) = PSL(2, 4)

wait what?

[delicate orchid]

surprise!

[hidden wind]

composite numbers are weird sometimes

[delicate orchid]

the PSL(2,4) one isn't so bad to see - it has a natural action on {0,1,x,1+x,\infty} which gives an isomorphism to A_5
PSL(2,5) is the bizarre one

[dim widget]

yes this is a good argument

[hidden wind]

it would be nice if i could make sense of klein's work on the icosahedron but eek i am so lost

[stark helm]

then I can know that sin(7.5) construtible by wrting cos(7.5)=sqrt(1-sin(7.5)^2)

[stark helm]

then how can I show that sin(7) not construtible?

[hidden wind]

how tf do you go from the rotations of the icosahedron to here's a general solution to the quintic 😭

[dim widget]

Do you know galois’ theorem on constructible numbers?

[stark helm]

I know galois's, but don't know how it applies to construtible

[stark helm]

do you mean tower of field?

[dim widget]

I guess? It should be that the real and imaginary parts of a number a are constructible if and only if Q(a) is an iterated quadratic extension

Or more simply if Q(a) is not of degree 2^n then a is not constructible

[stark helm]

Then how can it apply to angle?

[dim widget]

you should think about that

hint: consider the number a = cos(7) + i sin(7)

[hidden wind]

ok seing these referred to as "exceptional" and "exotic" isomorphisms i think i will just drop trying to make sense of this until some hypothetical future

[stark helm]

i only know number a can be written as e^i(7 degree)

[dim widget]

what is the galois group of that extension?

[stark helm]

Q(a) over Q

[dim widget]

yes

[alpine iron]

guys can anyone help me with these questions

[stark helm]

okay, then a^2-1=0

[dim widget]

you have to at least indicate which ones you have tried and what you have trouble with

[dim widget]

no

[alpine iron]

okay so q2

[stark helm]

a^2 not equals 0? I am considering the norm of a should be sin(7)^2+cos(7)^2

[dim widget]

the norm and the square are different

But you're right that the norm is 1

[stark helm]

oh it should be a*conjugate(a)=1

[dim widget]

yes but this doesn't help you find the galois group

so what is true is that a^{365} - 1 = 0

but that is not the minimal polynomial of a since it is not irreducible...

maybe you know something about cyclotomic fields?

[stark helm]

I think I can write it as e^i(7) while e^i180=-1, , so I need to multiply a 180 times such that 180 divides 7k

it can have a^180+1=0

[dim widget]

surely, but this is still not irreducible

did you learn about cyclotomic polynomials in class?

[dim widget]

it's insulting to people who care about math to try to get them to do your homework while you show no effort

[stark helm]

I didn't

[dim widget]

hmmm

okay there are easier ways

[stark helm]

and this class is actually not abstract algebra

[dim widget]

what is this class

[stark helm]

its name is abstract mathematics

[dim widget]

ok

well can you tell me an algebraic number which you have shown in class is not constructible, and how you have shown this?

[stark helm]

this one is definitely algebraic, but it is not constructible because cube root of 4 is not( by using RRT), then we can know stuffs in square root not constructible, then b is not constructible

[dim widget]

what is RRT?

[stark helm]

rational root thm

[dim widget]

how does the rational root theorem tell you that it is not constructible

[stark helm]

x=cube root 4, x^3=4, x^3-4=0, suppose x=m/n and by rrt I know that m divides 1, n divides 4, so rational root can be positive or negative(1,2,4), but plugging them will not let polynomial be 0

[dim widget]

this doesn't explain why the number is constructible or not

this just tells me that x^3 - 4 is irreducible

which is good

but you're REALLY using something else, no?

[stark helm]

i am actually consider that for a cubic polynomial, if we have constructible numbers root, then we have rational root, but it doesn't have rational root, so roots are not constructible

[dim widget]

okay, why is this true?

it is not only for cubic polynomials!

[stark helm]

it is a cubic poly with rational coefficient, and I consider a tower of field Q subset of T1 subset... and I proved before that cubic polynomial with rational coeff has sum of roots equals rational numbers, therefore I consider two cases: if we have root of form a+bsqrt(r), then we have another root a-b*sqrt(r), the sum of rational roots tells me that there must exist one rational root at least. The second case is trivial case, all three roots in Q

[stark helm]

since all surds are constructible

[dim widget]

i don't know what that means and it sounds racist

[dim widget]

what if the roots are all irrational?

[stark helm]

surds are constructible should make sense, because surds should refer to elements in a tower of field, which is starting from Q and keep adjoining square root such that Q subset of T1... And r in Q constructible, square root of construtible are still construtible

[tardy hedge]

Groups rings or fields which is yo favourite

Algebra structures tier lskt

List

[stark helm]

three roots sum is rational, so how can three roots be all irrational?

[dim widget]

you'd be surprised! it can easily happen

[dim widget]

e.g. X^3 - 3, or X^3 + 2X^2 + 2 have this property

for three random irrational numbers you can consider \pi, -3\pi, 2\pi.

[stark helm]

ok, I know that, I can have two -sqrt(2) and one 2sqrt(2)

but then I need to know if constructible numbers are surds

I mean if these two terms are the same

[dim widget]

well they are

you should prove this! think about what x-coordinates you can get by intersecting circles with circles and lines

[delicate orchid]

consider the fact that (2a)^2-2a = 0

[celest furnace]

How to show x is not in k(x^3+x)

Feel like it should be easy …

[delicate orchid]

any algebraic combination of x^3+x is going to be degree at least 3, x has degree 1

[celest furnace]

What about like 1/(x^3+x)?

[delicate orchid]

I ain't even got shit to say on that one...

[rocky cloak]

Might be a bit of a cheat, but k(x^3 + x) is the field of fractions k[x^3 + x], which is a UFD. Hence by Gauss lemma t^3 + t - x^3 - x is irreducible over k(x^3 + x) iff it is irreducible in k[x^3 + x]. It clearly doesn't have any roots in k[x^3 + x], but x is a root.

Hence k(x) / k(x^3 + x) is a degree 3 extension.

[rocky cloak]

This is basically this argument, just first moving to k[x^3 + x]

[delicate orchid]

yeah showing k(x) as an extension is degree 3 is smart

other way

[rocky cloak]

And I guess the same reasoning shows that k(f) cannot equal k(x) unless f has degree 1, 0 or -1.

[delicate orchid]

oh yeah, just extend the valuation lol

[stark helm]

I think all intersection of points on surd plane will always generate equation with surd coeff, and therefore we can know any geometric construction on surd point should be surd point as well, so we know every constructible points in this plane has surd coordinates, so every constructible numbers are surd

[celest furnace]

Fantastic argument! Just amazing

[dire siren]

assume x = f(x^3+x)/g(x^3+x), for some polynomials f,g in k[t], say m=deg(f) and n=deg(g)
then f(x^3+x) = x*g(x^3+x)

the degree of LHS is 3m

the degree of RHS is 3n+1

[celest furnace]

Darn i totally thoguht i tried that

I just had to combine my fractions !!!!

[hollow mica]

In PIDs people say that “to divide is to contain”, i.e that b | a iff (a) is contained in (b)

I don’t see where this breaks down for general commutative rings with unit

if (a) is contained inside (b) then a is an element of (b) so b | a

and if b | a then a = bc for some c and thus (a) = (bc) and for any r we have (bc)r = b(cr) which is an element of (b) and thus (a) is a subset of (b)

nowhere did we use the PID property

[rocky cloak]

That is correct. For principal ideals, containment is the same as one being a multiple of the other.

So the only thing special about PID is that all ideals are principal.

[dire siren]

The saying "to divide is to contain" is not about that.
In a Dedekind domain, the ideals factor uniquely into prime ideals, and so we can talk about divisibility of ideals. And it turns out that an ideal I divides an ideal J if and only if I contains J.

[stark helm]

i have an idea here, n is construtible iff n is a multiple of 3, because if n mod 3 is not 0, it should be 1 or 2, and 1 and 2 both should be constructible at the same time, so it can't. hence degree 7 is not able to be constructible

[vocal fern]

Let R be a UFD such that every invertible element is a cube. Let r, s ∈ R, and suppose
that every element of R which divides both r and s is invertible. Show that if rs is a cube,
then r and s are cubes.

Proof:
Suppose rs is a cube, but r = p0p1 · · · pk and s = q0q1 · · · ql are not cubes for irreducible elements pi, qj . Then none of pi, qj are cubes, implying that they are not invertible. However, gcd(r, s) = pa1 pa2 · · · pam qb1 qb2 · · · qbn for some a1, . . . , am ∈ {1, . . . , k} and b1, . . . , bn ∈ {1, . . . , l}. But since each of these irreducibles divide both r and s, they are invertible; a contradiction.

is this an adequate? i feel like its missing something

[sonic coral]

Let F = F_{5^6} be the finite field with size 5^6. Suppose that α ∈ F^{×} and the multiplicative
order of α is 31.

How can i determine [F(α) : F]?

i think it would be at most 6 but not much other than that

[dire siren]

isn't it just 1?

[dull ginkgo]

I am going to go back to proving if every element has an n >1 such that x^n = x that the ring is commutative eventually

It’s just hard as fuck and I get stuck every time

[coral spindle]

Remember that the group of units of a finite field is cyclic. So what is the order of the generator of the group of units of F_{5^6}? Think from there.

[dire siren]

but doesn't a∈F imply F(a)=F?

[sonic coral]

the order of the generator would be 31 right

oh i think i see

since alpha is in the multiplicative group of the field, it’s in the field, so [F(a) : F] is one

[cobalt heath]

(xy-yx)^2 = xyxy - x y^2 x - y x^2 y + yxyx

..welp

[rocky cloak]

Surely the problem is supposed to be to determine
[F5(a) : F5] right?

[exotic verge]

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

is this the same as saying

a^2 + b^2 is congruent to 3c^2 in mod 4?

well they're equivalence classes so I mean all a,b,c such that

or am I misreading?

[celest furnace]

Precisely right sir

[exotic verge]

thx

[tidal slate]

I feel like this should be easy but I'm still struggling. Why is the following true: Let $M$ be a right $M_n(R)$-module and denote $R^n_{\text{col}$ by $n$ column vectors in $R$, viewed as a left module over the matrix ring. If $E_{11}$ denotes the matrix whose $(1, 1)$ entry is $1$ and zero else, why is $$M \otimes_{M_n(R)} R^n_{\text{col}} \cong ME_{11}?$$

[cloud walrus]

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

[rocky cloak]

Notice that R^n_col = Mn(R)E11

[tidal slate]

oh oops, thanks

[thorn monolith]

What are the necessary and sufficient conditions for the reversibility of an element in the ring R[x]/x^n and how to prove it?

[rocky cloak]

Notice, that anything that is a multiple of x will be nilpotent, hence not invertible.

Things that are not a multiple of x will be of the form r + xf, which if we divide by r can think of as simply 1 + xf. Now think about geometric series.

[hidden wind]

i just realized my abstract algebra exam is in a week

been a while since last i were in a situation like this

[tough raven]

Me with multiple exams in the next three days 💀.

[thorn monolith]

i'm trying to find inverse element for $1+x+x^2 \in F_2[[x]]$. i think it is $\sum\limits_{n = 0}^{\infty}x^{3n}+x^{3n+1}$ (checked that for small $n$) but idk how to prove that correctly...

[cloud walrus]

qwert

[rocky cloak]

You can just multiple them together and see what you get.

You know how to multiply by 1, x, and x^2 separately I'm sure. Then you just add them together and collect like terms

[thorn monolith]

yeah thx

[hidden wind]

situation being i have completely neglected this course and i’ve little idea what i’m even supposed to know

[sonic coral]

i don’t think so?

it’s question 6

[rocky cloak]

Typos are possible. But I guess it could just be a trick question

[sonic coral]

we will see when he releases solutions

for question 5, is it enough to just use the fundamental theorem to get two normal subgroups, so that their intersection is normal, and then use the fundamental theorem again

[exotic verge]

D: i didnt know ur in an actual class

whats ur syllabus

[hidden wind]

good question, i will have to figure that out just about right now

ok it is definitely more than i can cover in one week haha

the standard first course in group theory is no problem but there is also quite a bit of, ring, field and galois theory

[plush pulsar]

is it necessary for rings to have two-sided distributivity, i.e. for elements $a, b, c \in R$, where $R$ is a ring with binary operations $+$ and $\times$, should $R$ satisfy both of the below two properties,
$$(a + b) \times c = a \times c + b \times c$$
$$(a \times b) + c = (a + c) \times (b + c)$$
or just one of them is enough for $R$ to be called a ring?

[cloud walrus]

tommy

[plush pulsar]

yeah

i haven't seen it in any of the examples i've encountered so far

is it not required

i know

this doesn't make sense

so, does that mean we only consider one-sided distributivity?

yeah sorry!

oh actually no

this is a question i have

it is not some homework

and i did not know how to frame it properly

but, it basically boils down to what i've asked above

yes

so, in essence, the binary operations + and * are behaving as they would in algebra

thank you!!

[thorn monolith]

what about geometric series? i didnt get it

[plush pulsar]

yes

that's what i had in my mind

[sonic coral]

if the problem was as you thought? how would you approach it?

[rocky cloak]

Have you ever seen a formula for 1/(1 - x) ?

[thorn monolith]

yeah but its for infinite series

[chilly ocean]

convergence is something made up by the analysts to prevent you from having fun

[dull ginkgo]

what if x is nilpotent :troll:

[frank cosmos]

isnt B just contained in the kernel and not actually the kernel itself?

obviously we can get this map because A is abelian so B is normal

but shouldnt the "equivalent" homomorphism be from A/ker

[last spoke]

If G is finite abelian with |G| > 2, then |Aut(G)| is even.
I found this pretty easy but I think the way I proved it was 'excessive' (i.e. invoking something unnecessarily powerful). Here's what I said:

Finite abelian means it's a direct product of Zq's with q a prime power. It suffices to check just for a (Zq)^k 'factor' of G (since the automorphisms of this factor that fix the rest of G form a subgroup of Aut(G)), where we know that Aut((Zq)^k) = GLn(Fq), which has even order except in the excluded case n = 1, q = 2. By Lagrange then the entire group Aut(G) must have even order.

The way I think I 'cheated' was invoking the fact that |GLn(Fq)| is even, which hasn't been mentioned yet in the text, altho the GL's have in general. Is there some other straightforward way of showing that |Aut(G)| is even for finite abelian groups without invoking the structure theorem like I did?

[frank cosmos]

do you mean |Aut(G)| is even

i dont think you need to invoke the fund theorem

[last spoke]

ya

fixed

[frank cosmos]

it suffices to show that everything except except one nontrivial element in Aut(G) is not its own inverse, and the other thing is its own inverse

er hmm thats not good here

oh wait just show Aut(G) has subgroup order 2 lol

[last spoke]

the one generated by x > x^-1?

lol the way I did it was so much harder

[frank cosmos]

yep thats it

its automorphism because abelian and not identiy since |G|>2

ez

[frank cosmos]

bumping this

[frank cosmos]

also, my book defines m to be an exponent of a group G if $g^m=e \forall g$ and then for an abelian group $A$ with exponent $m,$ the group $\text{Hom}(A,\mathbb{Z}_m)$ to be its dual. but dosent this depend on the choice of m?

[cloud walrus]

Hello1

[delicate orchid]

the dual Hom(A, Z) obviously has kernel Z_m because it's exponent m, and it depends on the choice of A not m

m is unique given a fixed group

[frank cosmos]

the dual is Hom(A,Z/mZ)

m is not the smallest exponent necessiarly, but anything such that g^m=e for all g

[delicate orchid]

sorry yeah, the kernel is mZ

[delicate orchid]

well that's dumb

"lets take something well defined and make it ill defined"

what the scallop

[frank cosmos]

ok i think this does make sense, pls tell me if its wrong. Suppose $A$ has smallest exponent $m.$ Suppose $A$ also has exponent $m_1.$ Then $m \vert m_1$ and for any $\phi \in \text{Hom}(A,\mathbb{Z}/\mathbb{Z}m_1),$ for all $a \in A, \phi(a)$ has order dividing $m$ thus $\phi(A)$ is contained in the unique subgroup of order $m$ in $\mathbb{Z}_{m_1}.$

[cloud walrus]

Hello1

[frank cosmos]

so the dual group is well defined?

[delicate orchid]

https://tenor.com/view/true-feel-that-thats-true-gif-7410050725709839510

[frank cosmos]

https://tenor.com/view/leonardo-dicaprio-clapping-clap-applause-amazing-gif-16078907558888063471

[indigo vale]

any recommended resources for me to learn/relearn abstract algebra? books preferred

[frank cosmos]

lol how

[cloud walrus]

Hello1

Hello1

[frank cosmos]

suppose G acts on S with 1 orbit, and k elements in G stabilize each s \in S. Then why is k*|S|=|G|

[south patrol]

This is a special case of the orbit-stabiliser theorem

But one way to prove it directly is as follows: fix any s in S and then consider the function G -> S sending g |-> g.s. You can check that preimage of each point is of size k and that this function is surjective

[frank cosmos]

how does this follows from a-s

[south patrol]

This is immediate from the statement of orbit stabiliser

[frank cosmos]

you are right lol

i solved 99% of the exercise, everything except this step, and when reading the sol it glossed over this part as trivial

glad too know it is

[south patrol]

Dw

[frank cosmos]

nice problem

average number of points in S stabilized is number of orbits

[sonic coral]

subgroups which are conjugate correspond to isomorphic sub fields right

[warped fable]

Not just any isomorphism -- that's boring. Conjugate subgroups correspond to sub-extensions related by an automorphism of the ambient extension

[sonic coral]

okay, so if i wanted to determine the number of isomorphism classes of sub fields by looking at a subgroup lattice, the normal subgroups would correspond to their own class of subfield, and if subgroups A and B were conjugate, then their corresponding fields would be in the same class ?

[tough raven]

Yes. If you continue to deal with it, you may want to think of it as the "m-dual".

[crystal vale]

to show if G is finite cyclic group, then the subgroup {g: g^n=1} where n divides the |G|. then this subgroup is cyclic of order n.
we know subgroup of cyclic group is cyclic but how i show that it has order n.

since G is cyclic group so there is an element of order n, say g.
we have result that n= summation of all {co-prime number to d, where d is divisor of n}. so subgroup order is bounded by n, and g belongs to subgroup so at least it has n elements, is it correct?
maybe this one is easier but i don't know easier way

[rocky cloak]

I'm unsure what you're asking. Are you trying to show that if G is cyclic and n divides |G|, then G has a subgroup of order n?

[rocky cloak]

I see, I'm not sure I totally follow your argument for why it has at most n elements. But you could argue that it has exponent n, or just explicitly reason about what the elements are in terms of the generator of G.

[crystal vale]

Let H be the given subgroup. Then any element which is in H, its order divides n, say d. The number of elements of order d is the Euler function (d)( G is a cyclic group), so the order of H is n(taking summation of all d divides n).

[rocky cloak]

I see, yeah that works

[crystal vale]

Any other way to show that in a finite cyclic group, there are exactly m elements which satisfy x^m =1, where m divides| G |?

[orchid iron]

Hello

There is this Heisenberg group that comes up in Group theory,i learnt it has its origins in Quantum mechanics(explaining the Uncentrianity principle in different forms).So could someone outline how does it arise in physics or in case there is some place in mathematics when it does arise naturally,hint or give reference towards that as well?

[dim widget]

It does turn out to be the same thing

which is an important part of Galois theory

any abstract isomorphism of fields extends to an automorphism of any galois over field containing both of them

[dim widget]

yes

[warped fable]

Good point, yes.

I forgot about that.

[glad osprey]

Can't you just argue that if an element has order > n, then it by definition can't be in H?

[last spoke]

@orchid iron the 'finite field' version of the heisenberg group comes up naturally in the classification of p-groups of small order (in particular it's one of two possibilities for non-abelian groups of order p^3)

[crystal vale]

Yes

[cinder fox]

Let $x\in G$, then there $\exists m\in\mathbb{Z}$ s.t. $g^m = x$.

$\Rightarrow \exists 0\leq k<n$ s.t. $m\equiv k[n] \Rightarrow \exists p\in\mathbb{Z}$ s.t. $m=k+pn\Rightarrow x=g^m=g^{k + pn}=g^{k}.g^{pn}=g^{k}$

So we conclude that the only elements in $<g>$ are $g^0, g^1, g^2, ..., g^{n-1}$.
I think it's obvious to see that $0,1,2,...,n-1$ is exactly $n$ elements.

[cloud walrus]

VirtualCode

[cinder fox]

I haven't touched group theory in months but I remember this is how we proved it in class

Well really the definition we used for the order of an element $g$ in a group $G$ wasn't the number of elements in the generated group, but literally $|<g>|=n \Leftrightarrow (g^n=1_G$ and $\forall 0\leq k<n, g^k\neq 1_G)$

[cloud walrus]

VirtualCode

[cinder fox]

So we kind of cheated

[crystal vale]

I don't understand how I can conclude that there are exact m solution for x^m =1

[crystal vale]

By taking x=1 ?

[cinder fox]

Oh sorry, I misread 🤦‍♂️

I thought it was about the order of the generated subgroup $|<g>|$

[cloud walrus]

VirtualCode

[cinder fox]

I guess this can be proved by showing that those two sets are identical somehow though

[frank cosmos]

can i get a small hint on showing that groups of order p^2*q have a normal sylow subgroup, and are solvable

er. all sylow subgroups are normal

i think it suffices to show that there is only one of each

but i dont know how

[celest furnace]

Try showing that there is only one Sylow q subgroup

[frank cosmos]

how?

[celest furnace]

Have you tried using the Sylow theorems

[frank cosmos]

yes

[celest furnace]

Where did you get stuck?

[frank cosmos]

i dont know what the intersection of sylow subgroups look like

[celest furnace]

Don’t do that

[frank cosmos]

so like suppose there are (q+1) sylow q subgroups

[celest furnace]

Just try to show there is only 1 with the theorems 2 and 3

Here I’ll give you a hint. You know that nq = 1 mod q so nq = 1+kq. Then you know that nq divides p^2 and I think you can use this to show that q divides p^2-1

[long obsidian]

Are we able to call Z[x] the single variable polynomial ring over the integers a sub algebra of R[x,y] the two variable polynomial over the reals?

It's just confusing cause if I multiply (1/2) by (x+1) we aren't in Z[x]

[frank cosmos]

what is n

[celest furnace]

What are they álgebras over? Z?

[celest furnace]

Sorry nq is a variable = the number of Sylow subgroups

[frank cosmos]

i see. how do you know it divides p^2?

[long obsidian]

Idk if You're responding to me but Z[ x] is an algebra over Z while R[x,y] is an algebra over R.

I guess R[x,y] is also a Z algebra so in that since Z[x] is a subalgebra

[celest furnace]

It would be correct to say that Z[x] is a Z-subalgebra of R[x,y]

It would not be correct to say it is an R-subalgebra because it’s not even an R algebra

[frank cosmos]

?

[orchid iron]

Thank you very much,i see

[celest furnace]

So, you are going want to break it up into two cases: one where q > p and the other where p < q. Notice in one of these cases if nq divides p^2 then there aren’t a lot of options for nq

[frank cosmos]

q<p is easy, since i have a theorem saying if H has index smallest prime dividing |G| it is normal

i dont get why nq divides p^2

[celest furnace]

Well now either nq = p or p^2 right

[celest furnace]

This is just a Sylow theorem

[frank cosmos]

oh not in my book

i guess it is obvious from the proof though

hmm ok but now

this dosent seem to be a contradiction

its possible for p=kq+1

for some k

ah so this is where the q>p comes in

then we get q=p+1

contradiction!

very nice

[last spoke]

q = p+1 is possible

[frank cosmos]

order 12 is easy to do on its own

w/ counting argument

[last spoke]

ya

[rapid tinsel]

Let x and y be non-zero elements in an integral domain such that x^2 = zy^2 for some z. Is it true that z is also a square? Surely this is false for a horrible non-UFD but I can't seem to come up with a counterexample

[delicate orchid]

factor x^2 and y^2 into a product of primes, each exponent must be even, hence each exponent of each prime in the factorisation of z must also be even, so z is a square

I think this works

[rapid tinsel]

But in general u don't have prime factorisation?

[delicate orchid]

oh right THAT'S what you were asking about

[chilly radish]

They said non ufd

Can't you read

[rapid tinsel]

XD ahahha

[delicate orchid]

alright mate back off fucking hell

[chilly radish]

It's just banter mate calm down

[delicate orchid]

I thought they wanted a proof for UFDs and were asking a second question

[chilly radish]

Anyways

[rapid tinsel]

Chilllll 😄

[chilly radish]

It's just jokes

[void cosmos]

think about any Z[sqrt(p)] for some good p

you can probably find good counterexamples there

-p

[rapid tinsel]

I can't find a counter-example for p = -3

[void cosmos]

maybe try p=-5

[delicate orchid]

we gotta go craazzzyyy

[rapid tinsel]

I was playing around with these haha

[void cosmos]

that's not a UFD either

[chilly radish]

Try -5

That's a non ufd

-3 might be ufd

[delicate orchid]

,w -3 mod 4

[cloud walrus]

Results provided by WolframAlpha

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

Requested by wewladstbh

[rapid tinsel]

-3 isn't, that's why we have 3rd roots of unity

Wait, am I going crazy

[delicate orchid]

-3 is a ufd

is it that silly list with like -167 in it or something

[rapid tinsel]

LOL whhoooppps

I guess I was thinking of Z[sqrt(3)]

[void cosmos]

Z[sqrt(-n)) when n >=3 is always not a UFD

and n square free

[chilly radish]

It's not a ufd

[rapid tinsel]

^ nvm

[chilly radish]

Since it's not rhe ring of integers of Q(sqrt(-3))

[rapid tinsel]

Yah dis XD

[chilly radish]

UFDs are integrally closed

[void cosmos]

im not your average american

[delicate orchid]

what's the funny list with -167 (might be -163 I forget) in it then. Is that Q(sqrt(-n))

[rapid tinsel]

So yah, any obvious counterexamples?

[rapid tinsel]

It's class number 1 fields

It has nothing to do with integrally closed

[tribal moss]

Can we just wing it? Let x and y be two independent transcendentals and see which subring of C x, y, and x²/y² generates?

[chilly radish]

UFDs are integrally closed always

[rapid tinsel]

Yah

The weird list is the list of Heegner fields

[delicate orchid]

thanks boss

[rapid tinsel]

Surely there must be something obvious I'm missing 😭

[tribal moss]

More concretely, take x=ln2, y=ln3, and the ring of all integer combinations of {x^ny^m | n>=0, m>=0 or n>=2,m in Z }.

[rapid tinsel]

Lmao

[void cosmos]

that's pretty good

[rapid tinsel]

Is there like a more natural example that's not constructed just to break this

[void cosmos]

ig you won't like it unless you come up with it

[rapid tinsel]

hahahhaa

[void cosmos]

so the advice is just to try Z[somehtings]