groups-rings-fields

[tardy hedge]

where are those questions coming from. a textbook?

[karmic moat]

Once you work out the solution, you can generalize this to saying that if b is nilpotent (i.e., b^n = 0 for some n), then a + b is a unit

[vast verge]

Oh wait, I know what you mean I think

[vast verge]

Yeah, I'm just seeing how it is asking questions on this server compared to just checking the solutions when I'm stuck

[thorn jay]

we are of course way better

[knotty badger]

no promises from me

[tardy hedge]

what textbook is it

[vast verge]

Gallians contemporary abstract algebra

[tardy hedge]

ah yea cool

[vast verge]

We broke up during chapter 5, but we ended up getting back together again

"welcome home cheater" kind of business

[thorn jay]

we make bad jokes tbough

thats gotta give some brownie points

[south patrol]

Real

[crystal vale]

How can I prove the centre of GL(n,R) is a set of scalar matrices in GL(n,R).

I know the centre of M(n,R), as the ring structure is a set of scalar matrices in M(n,R)

[candid patrol]

You already know the center of Mn(R) ?

[crystal vale]

Yes

[lavish ferry]

What do you mean by scalar matrices?

[vast stump]

aI

[thorn jay]

λI where I denotes the identity

[south patrol]

Ig you mean "the" set of scalar matrices

[desert verge]

if AB = BA for all B maybe choose B to have 1 in one entry and 0 everywhere else

[thorn jay]

not in GL_n

[desert verge]

oh maybe that works for M_n(R) then my bad

[lavish ferry]

Consider some arbitrary element in the centre, say A. Try fixing some row/column of a matrix to contain all ones and use that A will commute with such a matrix

[thorn jay]

maybe consider the permutation matrices?

[lavish ferry]

(All other entries 0)

[thorn jay]

that is again not an element of GL_n

[lavish ferry]

Whoops, yes you’re right

[thorn jay]

:P

[desert verge]

yet again, this is not in GL_n.

[thorn jay]

that statement is also not in GL_n I am afraid

[rocky cloak]

I think this will basically work, just add 1s along the diagonal to get something in GLn. Should work out about the same

Essentially: do exactly the same you would to prove stuff about the center of Mn, but replace B with I+B

[lavish ferry]

I think I have an idea. Consider the matrices formed from the identity matrix but with a single row scaled by some non-zero $\lambda$, denote such a matrix as $B$. $B$ is indeed in the general linear group as $\det B = \lambda \neq 0$. We then get, assuming $A$ is in the centre of the general linear group, that $AB = BA$. By some calculation - assuming the $i$th row of $B$ is scaled by $\lambda$, we eventually get by the arbitrariness of $\lambda$ that anywhere not on the diagonal A must have 0 as an entry

[cloud walrus]

Ellie

[lavish ferry]

This gives us eventually that $A$ is indeed a scalar matrix

[cloud walrus]

Ellie

[south patrol]

Probably also lol. Consider the operator f(B) = AB - BA If this vanishes on GL_n then it vanishes on M_n by density

[lavish ferry]

I like this idea

Whichever way is least computational lol

[south patrol]

Only works over R at firsr but you can use the Zariski topology to extend lol

[tidal schooner]

Ooh

[south patrol]

Psrsonally this is my favourite / cleanest

Cause you can test being in the centre with matrices with one nonzero entry, and hjen adding identity makes it invertible

[rocky cloak]

Guess you need to be a little careful if your field has characteristic 2, but otherwise this just works

[south patrol]

Sorry, why is this the case?

||And who said we had a field 😏 ||

[rocky cloak]

I + [0, 0; 0, 1] might not be invertible

[south patrol]

Ah lol

I assumed all of the other guys would be nilpotent for some reason even though this is clearly false lol

[lavish ferry]

I wonder if my dumb way works for arbitrary fields,,,

[south patrol]

If you are over F2 then you are only considering the identity matrix

unless I am misreading

[rocky cloak]

You would need lambda to be different from 1.

Which might be hard of 0 and 1 are the only elements of your field

[lavish ferry]

Ahh, yes I can see that it follows apart quite quickly there. Thank you

[tulip otter]

question 29, but i included 28 for the definition of T_x

i am going to write my attempt to solve 29, and if someone can check my work i will be grateful

a) Let $x\in K$ and $A\in S$, then $T_x=\imath$ where $\imath$ is the identity map. So the $T_x(A)=A\implies xA=A\implies x\in A\implies x\in H$ (since $A\subset H$). Hence $K\subset H$\ \b) $K=\ker T$ so there exists an injective homomorphism $\varphi: G/K\to Perm(S)$ and so $G/K\cong\Im\varphi$. Now $(\Im\varphi:1)\mid (Perm(S):1)=n!\implies (G:K)\mid n!$

[cloud walrus]

pirateking0723

[tulip otter]

as for part c i was thinking about something wrong so i will give it a thought again

[rocky cloak]

Why do you say A is a subset of H?

Or xA = A implying x in A for that matter

[tulip otter]

ah wait it need not be a subset of H

A is just a coset of H and H might not be normal

[rocky cloak]

If you want a hint ||it might be easier to just consider when A=H||

Being normal doesn't really play into it

[tulip otter]

yea nvm that was stupid of me

[tulip otter]

I will wait a bit longer before seeing this hint

i want to try correcting myself

[tulip otter]

on a second thought that seems reasonable to me tbh, if x is not in A but xA=A, then for any y in A there exists z in A such that xy=z so that x=zy^(-1) in A which is a contradiction no?

[rocky cloak]

Why would zy^(-1) be in A, I guess is the faulty reasoning here

A is just some coset

[tulip otter]

ah yes for some reason i was working under the assumption that A is more than that lol

hmmm alright then i will think about some other approach

[tulip otter]

alright so from xA=A multiply both sides (from right) by H to get xH=H. Now since H is a group, this argument (with A replaced by H) shows that x in H

is that correct now ?

[rocky cloak]

Not really no. Multiplying by H on the right doesn't change the coset.

It might be better to write down the general form of a coset and how it's related to H

[tulip otter]

so something like A=yH for some y in G, so xA=xyH and xA=A implies xyH=yH implies y^{-1}xyH=H implies y^{-1}xy=z for some z in H implies x=yzy^{-1} in H?

ohhh i see

[rocky cloak]

Now, yzy^-1 might not be in H for all y, that's related to H being normal.

But remember you're the one choosing y, so you can pick it whatever you like, even something very simple

[tulip otter]

ah yes, the goal is to prove that any x in K is in H, so i take x in K and then consider any convenient coset of H like the coset H=eH (where e is the identity of H), then x=z for some z in H

is that it ?

so all of my attempts were wrong till the very end

but no problem, eventually things will be better (hopefully)

[tulip otter]

what about b?

[tulip otter]

now that i see your hint, it wouldve led to the answer directly

[rocky cloak]

b looks good

For c a hint could be that ||b may be useful||

[tulip otter]

alright, maybe i got something for c

so we are given that $(G:H)=2$. Now $(G:K)=(G:H)(H:K)=2(H:K)$. But $K$ is a subgroup of $H$, so $(G:K)\leq (G:H)=2$. by this and the previous equation, it follows that $(G:K)=2$ so that $(H:K)=1$. Hence $H=K$ and $H$ is normal (since $K$ is normal)

[cloud walrus]

pirateking0723

[tulip otter]

maybe now it is more readable

[tulip otter]

now that i am seeing your hint and rereading what i wrote, it doesnt seem that i used b much. I wonder how you wanted me to use it

[south patrol]

Looks good

[rocky cloak]

I guess I maybe misread your proof.

K being a subgroup of H would mean
(G:K) >= (G:H)
not the other way around

[south patrol]

Oh wait yeah lol

[tulip otter]

ohhhh

[south patrol]

I trusted jagr and will just blame him

Joking.

[tulip otter]

hmmm then i will think about a way to correct this

[rocky cloak]

But the idea that if (G:K)<=2 then you're done is correct

If only there was an earlier exercises giving a bound on (G:K)...

[tulip otter]

how to give a hint in a roundabout way:

[tulip otter]

hmmm i am not reaching anywhere

[tulip otter]

so 2=<(G:K)=< 2n (n is given to be the order of H), but i dont think this is the bound that you are looking for

and in fact (G:K) is even

but thats everything ive got

[tidal schooner]

Not sure where the 2n bound is coming from (remember that n = (G:H) = 2 in this problem). But you actually have everything you need already: ||you showed earlier that (G:K) ≥ (G:H) = 2 since K is contained in H and H has index 2 (by assumption), and part (b) gives you (G:K) ≤ (G:H)! = 2! = 2||

[tulip otter]

ah wait a sec, i was working under the assumption that the order of H is n, ie (H:1)=n

thats why i got the 2n bound

ah i am too stupid

alright now that you notified me about this let me tell you what i came up with before checking your hint so that you can correct me

[tidal schooner]

No it's all good, I'm not sure why the problem introduced another variable n instead of just using (G:H)

[tulip otter]

nvm, maybe i was hallucinating

I will just check the hint

[tulip otter]

ah wait what, it is just that?

and i was going in circles all this time

alright now i see

tysm everyone, have a great day/night

[unreal sedge]

hello beautiful people

Guess whos back

I had gotten a headache and dreamt about this exercise

wait let me double check my work before presenting it

[cloud walrus]

whyhello

[balmy vector]

would (the first part of) this exercise in dummit and foote (section 4.3) still hold for infinite groups G?

since the approach outlined in the exercise involves division and thus would (seemingly) only work for finite groups, and i cant think of another method that could work for infinite groups at the moment

[lavish ferry]

Do you mean everything up till the first full stop?

[tidal schooner]

I think it does hold. Let $H$ act on $\mathcal{K}$ by conjugation, and let $x,y\in \mathcal{K}$. We have to show that the orbits $H\cdot x$ and $H\cdot y$ (i.e., the conjugacy classes of $x$ and $y$ in $H$) have the same size. By the orbit-stabilizer theorem, we have an isomorphism of $H$-sets $H\cdot x\cong H/C_H(x)$, and similarly for $H\cdot y$. Thus, it suffices to show that $C_H(x)\cong C_H(y)$. Since $x,y\in\mathcal{K}$, there exists $g\in G$ such that $y=gxg^{-1}$. Since $H$ is normal in $G$, the map $h\mapsto ghg^{-1}$ defines an isomorphism from $C_H(x)$ to $C_H(y)$.

[cloud walrus]

harmacist

[tidal schooner]

So basically the idea is to use the statement of the orbit-stabilizer theorem in terms of $G$-sets, which is more general than the one I think is given in DF (iirc they give the one in terms of sizes of finite sets)

[cloud walrus]

harmacist

[tidal schooner]

Oh yeah and there's the other part of the exercise, the one about index. Let $\mathcal{K}/H$ be the set of orbits of $\mathcal{K}$ under the action of $H$ (i.e., the partition of $\mathcal{K}$ into conjugacy classes of $H$), and let $G$ act on $\mathcal{K}/H$. Let $H\cdot x\in \mathcal{K}/H$. Then by OST, we have an isomorphism of $G$-sets $H\cdot x\cong G/\mathrm{Stab}_G(H\cdot x)$. Show that this stabilizer is $HC_G(x)$, and we're done. No order division necessary

[cloud walrus]

harmacist

[tidal schooner]

@balmy vector

[balmy vector]

ohh that makes sense

thank you!

[minor gazelle]

Guys is this proof correct? Sorry it’s not latex since I am traveling 😭😭

I initially thought it was false because it only gives the condition for right identity and inverse however it can be manipulated to show that it is a group after a couple of attempts

[crystal vale]

Yes, it is correct

[crystal vale]

But just mention that you assumed that for fixed g there exists h such that gh = e

[languid trellis]

Does anyone know of any place where I can read more about the mobius function and mobius inversion? In particular how its introduction is motivated. I find myself using it a lot, but I am a little unsatisfied because ireland and rosen's presentation is a little 'out of the blueish'.

[river heart]

I think you can try as an exercise to derive the inverse of (1,1,1,...) in the convolution algebra.

Like from scratch.

[languid trellis]

What is the definition of the convolution algebra?

[cloud walrus]

FakeMMAP

[river heart]

^.

You can show that this operator makes the set into a commutative ring, with the obvious addition.

As an exercise, I'm asking you to compute the inverse of any element, and derive a formula. This will be the Möbius inversion formula.

As an alternative, you can take the Dirichlet Series approach.

[languid trellis]

Okay, I'll have a think. Thanks

[vast verge]

Still stuck on this

Any idea how to start it?

[crystal vale]

Write what is m•a and n•a and then use distribute property

[thorn jay]

@vast verge

[vast verge]

I ended up doing it by double induction, hopefully that's correct?

[cedar vault]

that works, yes

although you can get away in two(albeit comparatively dense ) lines with what .enpeace_music suggested

[regal mango]

Can anyone tell me a good book to start ring theory

[karmic moat]

atiyah macdonald

Introduction to Commutative Algebra

[regal mango]

I have a good grasp on group theory and wanted to start ring theory

[regal mango]

How is it like for theory and practice?

[karmic moat]

awesome for both

concrete examples, good exercises

[regal mango]

I was thinking for gallian or dummit and foote

[karmic moat]

idk anything about those two

but atiyah macdonald is awesome

[crystal vale]

[regal mango]

I will look for it

[karmic moat]

what monkagiga it's a great book

[regal mango]

Thanks for the suggestion

[karmic moat]

when i have children i will read it to them as a bedtime story

np

[stark mist]

I liked the chapter on it in Apostol Analytical Number Theory. Although to see applications/motivation, I needed the 2 chapters after it, which deal with proving things towards (but not quite) the prime number theorem.

[karmic moat]

actually no i wont. because it's such a good book my children will be restless and yearning for more knowledge

[tardy hedge]

dummit and foote is a more friendly introduction to rings

chapters7-9

[tardy hedge]

W

[regal mango]

Can you recommend a lecture series too ?

[languid trellis]

Thanks, I'll take a look

[cedar vault]

you can look at the mobius function on a poset as the inverse of the incidence function in the incidence algebra of that poset. the classical mobius function then just becomes the mobius function on the poset of divisors of a number

if i remember correctly camerons book "combinatorics, topics and techniques" introduces it this way

[vast verge]

Does this look ok?

I don't know how I'd get 1 in the ring

[unborn terrace]

This is not the correct answer

[vast verge]

Hmm okay

In that case I don't know what the smallest subring S would be

I just went with my (incorrect) intuition

[rocky cloak]

Do you not define rings to be unital?

Cause if you do then 1 should be there just by definition

[vast verge]

Right now, rings aren't necessarily uniral

[unborn terrace]

Then the subring doesn't need to contain 1, no?

[cedar vault]

For example, the set of Z linear combinations of natural powers of 2/3 should work as a subring that doesnt have 1 but contains 2/3

[vast verge]

Oh so what you're saying is if a ring contains a unitary element, then every subring of that ring must contain the unitary as well?

[rocky cloak]

Well, if having a unit is part of the definition of ring it should also be part of the definition of subring

[tardy hedge]

Why doesnt this have 1

Oh natural powers

Z[2/3] has 1 tho yeah

[cedar vault]

Is you question: if a ring has an identity 1, will any subring also have the same identity 1? If so thats not true

Thats how your question reads to me atleast

[vast verge]

It's not included in the definition of a ring that I'm using I think

It requires an additive identity, not a multiplicative identity

[rocky cloak]

Then you would not expect subrings to have 1

[tall igloo]

well first, what are the subrngs of Z?

in particular, whatever the smallest subrng of Q containing 2/3 is, will be a ring extension of some subrng of Z

hint: having 2/3 requires you to generate one of the subrngs of Z automatically

[vast verge]

The subrings of Z are nZ where n is natural I think

[tall igloo]

correct

[vast verge]

I haven't done ring extensions yet though

This is the introduction to rings exercise

[tall igloo]

that's alright, ignore that bit then. also isnt really helpful, that was just a bad motivation to get you to consider nZ

if a rng contains 2/3 (really, any rational number) i claim it must contain a subrng of Z -- which one is it?

[vast verge]

Well it must contain 2Z, like I said above

[tall igloo]

oh sorry, i did not see that part. but yes so you just need to force 2Z to contain 2/3

[vast verge]

I think, I was previously trying to find the largest subring, rather than the smallest

Thank you

I'll try finishing my solution now

Wait, but 2/3 is not included in 2Z

[tall igloo]

correct - you start with even numbers and want to add more elements, starting with 2/3, until you get a rng

as in, you have 2Z; then you have 2Z, 2Z + 2/3, 2Z + 4/3 - but thats not a ring since its not closed under multiplication; so you then have 2Z, 2Z + 2/3, 2Z + 4/3, 2Z + 4/9, 2Z + 8/9, and 2Z + 16/9; etc etc

[vast verge]

My first and original guess for the smallest subring containing 2/3 is
$$ R = { \frac{2a}{3^b} | a,b \in \mathbb{Z} } $$

[cloud walrus]

Tropical Greens

[tall igloo]

it is literally this but with 2Z

how would 2/9 be in the rng?

[vast verge]

It wouldn't my bad

[tall igloo]

no worries!

you are on the right track thinking about powers though

make sure you also are able to hit all even integers

[vast verge]

How about $$R = \left{ \frac{2^a}{3^b} | a \in \mathbb{Z}, b \in \mathbb{N} \right}$$

That's not going to hit all even integers...

[cloud walrus]

Tropical Greens

How about $$R = \left{ \frac{2^a}{3^b} | a \in \mathbb{Z}, b \in \mathbb{N} \right}$$
```Compilation error:```! Missing delimiter (. inserted).
<to be read again> 
                   {
l.49 How about $$R = \left{
                            \frac{2^a}{3^b} | a \in \mathbb{Z}, b \in \mathb...
I was expecting to see something like `(' or `\{' or
`\}' here. If you typed, e.g., `{' instead of `\{', you
should probably delete the `{' by typing `1' now, so that
braces don't get unbalanced. Otherwise just proceed.
Acceptable delimiters are characters whose \delcode is
nonnegative, or you can use `\delimiter <delimiter code>'.```

[tall igloo]

correct it won't, it also still hits 2/9 for example which it should not

(you can formulate the answer without explicitly making reference to even numbers, by the way. they will just naturally appear if you choose some correct parameters such as a, b here)

[rocky cloak]

2/9 does equal
2/3 - 4/9 though

[vast verge]

Does that mean 2/9 is in the subring?

[tall igloo]

oh shoot you're right, ignore me!

[vast verge]

Okay, I will try and prove it then

[tall igloo]

i can do basic math 👍

[vast verge]

Thank you for your help

[tall igloo]

wait your proposed rng is still not including all even numbers though (for example, how do you get 6?)

[vast verge]

Oh you can get six

$9 \cdot \frac{2}{3} = 6$

[cloud walrus]

Tropical Greens

[tall igloo]

your definition here will never allow for 6

even if you allow b to be negative, you won't get, 10 for example

[vast verge]

That's the wrong definition

It's the one above that that I'm using

$S = { \frac{2a}{3^b} | a \in \mathbb{Z}, b \in \mathbb{Z}_0^+ }$

[tall igloo]

which one? thats the last definition i see

[cloud walrus]

Tropical Greens

[tall igloo]

okay yes now i agree!

[vast verge]

So you can get 6 by doing a = 3, b = 0

Right now I'm just figuring out how I'd generate 2/3^b using only ring operations on 2/3

[tall igloo]

try to use this as a base step for induction

well, sorry i guess 2/3 is base step. use this as intuition for what the induction should be

in particular you cannot multiply by 1/3 in your rng but you can for 2/3^b

[vast verge]

This is where I am currently

[tall igloo]

there is another way to write down 2/27 that will make your life easier

[vast verge]

Hmmm

Does it include 8/27?

[tall igloo]

nope

convince yourself you can multiply the first line by 1/3

[vast verge]

Hmm yeah, you got a point

Thanks

[tulip otter]

I have a question, if a group is of order 4, then any group of order 4 is isomorphic to it right? if thats the case then why would a group of order 4 be either isomorphic to the group in a or isomorphic to that in b

i mean both of these groups should have order 4 and thus be isomorphic no?

[tidal schooner]

No, up to isomorphism, there are two groups of order 4

To see this, try finding the orders of the elements in these two groups. This will show that they aren't isomorphic

[tulip otter]

so if 2 groups have the same order then there exists a bijection between them but not necessarily an isomorphism?

[tidal schooner]

Yes

[tulip otter]

ohhh i see this would make sense

[tidal schooner]

An isomorphism has to be a bijective homomorphism. A bijection by itself isn't enough

[tulip otter]

yes, but i thought that same order would imply isomorphism between both groups

i compared that to finite vector spaces with same dimensions, thats why i arrived to this conclusion

i thought that it may be analogous

[tulip otter]

any hint on how to begin with this?

hmmm wait, maybe i came up with something

so let G' be a group of order 4, if G' is cyclic then G' is obviously isomorphic to G in part b and nothing needs to be proven (the isomorphism being the map that sends the generator a' of G' to a).

Now suppose that G' is not cyclic, then it can be generated by at most 3 elements. Assume that its least generating set consists of 3 non-zero elements a',b' and c' (non-zero here means that non of them is equal to the identity element) , then G'={e',a',b,c'} where e' is the identity element of G' (no other elements can be in G' since |G'|=4).

But a'c',b'c' and a'b' are elements of G' so that one of a',b' or c' can be generated by the other elements of G' which is a contradiction. Thus G' can be generated by 2 elements

let these 2 elements be a',b'. Then G'={e',a',b',a'b'}, now a' neq b' neq e' so that a'^2=e'. A similar reasoning shows that b'^2=e', finally b'a'=a'b' for otherwise a'=e' or b'=e' could be reached

Hence G' is isomorphic to the group in a

i am not sure if this is formal enough or too wordy or even correct, so can anyone check it and tell me how to correct it?

[mental flint]

i was looking for some thoughts on my publishment
https://divisionbyzeroaxiom.github.io/dbz-algebra-site

please check my axiom consistency.

[tidal schooner]

I think it's a valid proof. Nice work! I will say that the non-cyclic case can be done a lot more quickly with no need to consider generating subsets. Namely, if G isn't cyclic, then by Lagrange's theorem, it has three distinct elements of order 2, say a, b, c. Similarly to how you did above, we can show that ab = c = ba, which completes the proof

[velvet hull]

looked over it very briefly, the construction is consistent because this the same construction as the dual numbers but over an arbitrary ring

[mental flint]

so you do not see any inconsistency? it’s a system that will allow calculation around divbyzero

[velvet hull]

It’s just R[x]/<x^2>, where R is any commutative ring

And you’re not diving by zero, not really

Just introducing a nilpotent

[mental flint]

ye ur right, the underlying object is the dual numbers
A = R[ε]/(ε^2).

its not making 0 invertible. instead i just define a total operator

x/0 := x·ε    (ε central, ε² = 0)

it’s semnatics for expressions that contain
“/ 0”

while keeping ring laws (assoc/comm/dist and x·0=0).

• linearity: (x+y)/0 = x/0 + y/0, 0/0 = 0
• homogeneity: (λx)/0 = λ(x/0)
• product bookkeeping (for the “remnant”):
Res(xy) = Re(x)Res(y) + Re(y)Res(x) if x=a+bε, y=c+dε

i set f(x0)ᴰᴮᶻ = A + R·ε where R = p(x0)/q'(x0) (residue) and A is the finite part.

the “division by zero” is the explicit totalization x/0:=x·ε
and the rule for evaluating singular formulas without breaking the algebra

[velvet hull]

Sure, calling it multiplication by 0 is a bit misleading but there’s nothing inherently problematic with the dual numbers

[mental flint]

this is what it’s calculating

[mental flint]

its not multiplying by 0

the rule is:

x / 0  :=  x · ε

with ε central and ε^2 = 0. so it’s a totalized “/0” operator inside A,

not a reciprocal of 0. (x·0 = 0 still holds; 0/0 = 0 by linearity.)

on the graph im plotting

• a rational f(x) at a simple pole x0 (vertical asymptote).

• the DBZ value is f(x0) = A + R·ε,

  where R = p(x0)/q’(x0) is the residue if f = p/q and q(x0)=0, q’(x0)≠0,

  and A is the finite part after removing R/(x−x0).

so in the label “div by 0: 100_5”:

• x0 = 100,
• the ε–coefficient (the “remnant”) is R = 5,
• the printed form A_R is “A_5”; if A=0 it’s “0_5”.

example checks:

• f(x) = 5/(x−100) → DBZ at x0: 0 + 5ε (prints 0_5)
• f(x) = g(x) + 5/(x−100) → DBZ at x0: g(100) + 5ε (prints g(100)_5)

thats the reason i say “division by zero”

it gives a consistent value to formulas with a /0 and carry the residue through algebra properly without breaking math

(linear in the numerator and with Res(xy)=Re(x)Res(y)+Re(y)Res(x))

[velvet hull]

Division by zero, that’s a typo

And also, what LLM are you using?

[mental flint]

??? i just typed this for 3 mins

nd bro is not even reading it

gg

[velvet hull]

I mean I think it’s obvious that I understand what you’re saying

But sure, I’ll take it back

[mental flint]

well, then you aren’t properly seeing the utility of it

[velvet hull]

I don’t see any reason why your algorithm can’t be implemented using normal real numbers

If anything introducing epsilon like that sounds like more work

But sure, it’s something

[mental flint]

if u start using plain reals then you still need TWO channels at a pole:

A = finite part,
R = residue.

if you hand carry (A,R) through code, you must re-derive and maintain:
(A,R) + (C,D) = (A+C, R+D)
(A,R) * (C,D) = (AC, AD + BC)

thats exactly the dual-number ring

R[ε]/(ε²): A+Rε

so “ε” isn’t extra work..it’s the minimal algebra that guarantees the propagation laws by construction and keeps every op total (no NaNs )

[velvet hull]

It’s the same as multiplying real polynomials and truncating the terms past x^2

[mental flint]

1. compatablity:

• with reals you must branch at every /0 and push flags around

• with A+Rε, algebra composes automatically (assoc/comm/dist hold)

2. no magic limits/epsilons:

ε is formal (ε²=0). no step sizes, no tuning.

3. product rule:
   Res(xy) = Re(x)Res(y) + Re(y)Res(x) is built in.

with reals you’d have to remember and reimplement it everytime…

4. NaNless pipelines:

a/0 becomes 0_a (A=0, R=a), not NaN/Inf

so the rest of the system keeps working and can decide based on R.

5. higher-order poles immediately via jets R[ε]/(ε^{N+1}).

[velvet hull]

I just telling you that I dont see a use for this, you don’t have to justify it to me if you think its good then its good

I’m not that interested in this, doesn’t mean you can’t be

[mental flint]

not even.. multiplying real polynomials and truncating past x² looks similar, but in standard ℝ[x]/(x²) you don’t attach an explicit semantic to ε as “1/0”

ε isn’t just a formal variable
it’s a structured placeholder for division by zero that preserves algebraic laws and keeps every op total (no NaNs), while carrying the finite part + residue automatically..

with normal polynomials you’d still have to manually guard /0 and handle branches

with A+Rε, assoc/comm/dist all compose without casework, and the “division by zero” step is just another total algebraic operation

no special limits or undefined zones.

[mental flint]

well i’m just refuting ur claims cuz they arent true

like ur saying its the same as real but that’s just not true

[velvet hull]

I think you misunderstand me when I say the same

But sure

[unborn terrace]

Why is epsilon 1/0? Shouldn't 1/0 be very large, and epsilon is meant to represent a small quantity?

[tall igloo]

is there any sense in which a solvable group is a group admitting a filtration with associated graded in Ab? it looks like this if you squint your eyes but i cannot seem to force the category the filtration exists in to be abelian (i am guessing the answer to this question is "no")

[south patrol]

I guess there is the issue that "associated graded" does not quite make sense unless you are already subnormal (in which case you do get the definition)

Like i think the usual definition of solvable is as close to what you want as makes sense

[unreal sedge]

am i tweaking

because this proof from my book seems to be missing a couple of things

or im just not comprehending the english

[tall igloo]

what do you believe it is missing? it looks fine to me

[unreal sedge]

I'm not sure what it means that the associative law of * " follows

[tall igloo]

l*(x*z) = (l*x)*z = 1*z = z

you can put parantheses wherever you want

[unreal sedge]

wait so

$(l \star x) \star y = (l \star x) \star z$ let $e = l \star x$. so then we have $e \star y = e \star z = y = z$

[cloud walrus]

whyhello

[unreal sedge]

ohhh

wait nevermind i'm just tweaking out

Yeah i understand now

[tulip otter]

Ohh I see, I suspected that there would be a shorter way because I felt mine was too long for the asked question, tysm for reading my long answer and giving me a shorter way. Have a great day/night

[tidal schooner]

Np, you too

[lavish ferry]

I find it weird that they omit writing the identity for the monoid? But other than that it seems okay

(Although it’s very very brief)

[tulip otter]

For parts a and b, let $G=\langle x\rangle$.
\a) Let $y,z\in G$, then $y=x^k$ and $z=x^m$ for some $m,k\in{1,2,...,n}$. Now $f_a(yz)=(yz)^a=(x^kx^m)^a=(x^{k+m})^a=x^{a(k+m)}=x^{ka}x^{ma}=y^az^a$ so that $f_a$ is a homomorphism of $G$ into itself.\\b) First suppose that $f_a$ is an automorphism, then in particular it is injective. Ie, $f_a(y)=e\implies y=e\implies x^k=e\ \text{for some}\ k\in{1,\dots,n}\implies n\mid k\implies n=k\implies (a,n)=1$.\ \conversely, suppose that $(a,n)=1$ and consider the equation $f_a(y)=e$, then $f_a(y)=e\implies x^{ka}=e\implies n\mid ka\implies n\mid k$ (since $(a,n)=1)\implies k=n$ so that $f_a$ is injective.\Finally, it is clear that $f_a$ is surjective after looking at $a\mod n$

[cloud walrus]

yassine

[tulip otter]

is this proof correct?

[velvet hull]

a looks right.
For b, your presentation could be better, I understand what you mean with all the arrows but it’s confusing to read.
e.g. “f_a(y) = e -> y = e -> x^k =e” should really be “(f_a(y) = e -> y = e) -> x^k =e”, if you want to write it out formally
Also, an injective group homomorphism from a finite group to itself is necessarily an isomorphism (why?), so you dont need to explain why it’s subjective.

[glad osprey]

The first part of b) can't be right, because you haven't used the definition of f_a. You've just used the fact that an injective homomorphism has trivial kernel (which constitutes the first 4 arrows), then you conclude (a, n) = 1 without justification

[velvet hull]

I also don’t exactly follow how you’re able to conclude that (a,n) = 1 from x^k = e:
Lagranges theorem tells you that k divides n, not that n divides k

[tulip otter]

ohhh now i understand what you mean by "your presentation could be better (after i read the edit), as your second point i will think about it a bit and then respond

[tulip otter]

well k|n for sure, but since x^k=e and n is the order of the cyclic group G, then |x|=n, so x^k=e would also mean n|k hence n=k

[velvet hull]

But x is not fixed here

[tulip otter]

(assuming x is the generator of G)

[velvet hull]

Oh, wait

If x is the generator then that’s also not right

[tulip otter]

I started with let G=<x> at the very beginning, mb i shouldve mentioned it in part a and b separately

[velvet hull]

Because you’re only showing that the generator is not sent to e

How are you sure nothing else is sent to e?

[tulip otter]

because f_a is injective

[glad osprey]

are you talking about part b? you can't assume f_a is injective, that's what you're trying to prove

[velvet hull]

Sorry, can you explain what you mean by the “x^k = e” line then

[velvet hull]

They’re taking about the forward direction lol

[glad osprey]

oh, okay

[tulip otter]

part b wants f_a is an automorphism iff (a,n)=1

so is part b now correct (aside from the bad presentation)

[velvet hull]

Hold on I’m still trying to understand your argument, sorry

[velvet hull]

.

[tulip otter]

nw i am not in a rush

[glad osprey]

the forward direction cannot be correct, as I said here. You don't even refer to a at all until the conclusion (a, n) = 1

[tulip otter]

ok so f_a(y)=e implies y=e since f_a is injective, now y=x^k for some k in {1,2,..,n}

so x^k=e

[velvet hull]

So why does n divide k then

When k is between 1 and n by assumption

[tulip otter]

because x^k=e

[glad osprey]

the fact that f_a(y) = e where y = x^k implies n = k follows just because injective homomorphisms have trivial kernel. It's just a reformulation of f(y) = e implies y = e

[tulip otter]

and |x|=n

[velvet hull]

But you haven’t used the function

[tulip otter]

sure

but i am not seeing the problem

[glad osprey]

the problem is that this is true for any injective homomorphism between any two groups, not just this particular homomorphism between cyclic groups. So since you haven't used the assumptions in the problem, the proof cannot possibly be correct

[velvet hull]

Your wording is confusing me, let me restate: what you want to prove is that if (x^k)^a = e implies that x^k = e, then (a,n) = 1

[tulip otter]

I mean i used the assumption that f_a is an automorphism of G, and then used that G is cyclic

[velvet hull]

In other words, your goal is to show that if n divides ak implies n divides k, then (a,n) = 1

[tulip otter]

so which assumption in the problem am i missing

[tulip otter]

yes

[glad osprey]

well, you used it, but you didn't have to, because it's true more generally

[velvet hull]

So, pretend I’m stupid (not that hard) and walk me through it again.
So if f_a is injective, then f_a(y) = e implies that y =e. What next?

[tulip otter]

ah no you are not
ok so now y=e implies x^k=e for some k in {1,2,..,n}. Now |x|=n so n|k

but 1<k=<n so k=n

[velvet hull]

Sure, so what next

[tulip otter]

well i was thinking that this would imply [a,n]=an where [a,n] is the lcm of a and n

but maybe i am wrong let me recheck

[velvet hull]

But you haven’t mentioned a in your proof up until this point

So I don’t see how that would imply that

[tulip otter]

it would imply it from x^{ka}=1 (assuming that this implication is true to begin with)

but i agree that even if this is the case it would my fault for not writing the proof in a better way

[velvet hull]

“If ‘n | ka => n | k’ for all k then (a,n) = 1” is the core statement that’s missing that I’m looking for

Which, by the way if you want to be really complete you should prove why that’s true

[tulip otter]

I think i see what you are trying to say, you want me to explicitly use this particular homorphism and not just use properties that apply to any homo/automorphisms in general?

[tulip otter]

hmm i see

brb i will go for a moment

[glad osprey]

yep 👍

[glad osprey]

btw, looking at the lcm is a good idea, that's how I solved it. There are probably other ways too, not sure

[tulip otter]

i am back

[south patrol]

Lcm.

[velvet hull]

Since you understand what I was originally trying to say, I’ll just give you my approach for what I would do for this question (proof by contrapositive):

Let d = (a,n).
Suppose that d > 1.
Then n divides an/d.
Hence f_a(x^(n/d)) = e, but x^(n/d) cannot be e. So f_a is not injective.

[tulip otter]

is that related to what i just deleted or is it regarding the initial question in general

[south patrol]

Idk what you deleted

I just said lcm

[tulip otter]

ohh i see

[south patrol]

My take on these questions is that if you swap to additive notation they are much easier

Like this is a statement about modular arithmetic

[tulip otter]

ohhhh, pretty neat

[tulip otter]

ok so what i came up with is the following:
f_a(x^k)=e implies x^k=e so that k=n (since k in {1,2,..,n})
Assume that there is k<n such that x^{ka}=e
Now x^{ka}=e implies k=n by the first statement which is a contradiction. So [a,n]=an and (a,n)=1

is this good enough?

[coral spindle]

I have literally never heard of it lmao

That's so funny

I like to think I have heard of most of the standard textbooks by this point lol

Not saying it's bad! Just never heard of it

[glad osprey]

yeah, that works, but maybe you can explain in more detail why the contradiction implies [a, n] = an?

This is the standard textbook for a couple of courses at my uni. I think it's fairly decent, I like that it has a lot of worked-through examples, but the exposition is a bit weak some places

[tulip otter]

as in, the contradiction implies [a,n]>=an along with a|an and n|an?

or should i add also the detail of why [a,n]>=an in the proof too?

[glad osprey]

yeah, from the contradiction you conclude that there doesn't exist k < n such that x^(ka) = e. Why does this imply that [a, n] = an?

I'll answer in the other thread

[tulip otter]

k>=n so k|a|>=|a|n so [a,n]>=an, now a|[a,n] and n| [a,n] so that an| [a,n] but an is a common multiple of a and n so [a,n]=an

using |a| allows to avoid separating into 2 cases where a>0 and a<0

is that sufficient now ? or is it still lacking some details?

[tulip otter]

also i would like to know your proof if you have the time

[glad osprey]

hmm, maybe I'm just really slow, but I can't see how k = n implies [a, n] = an. I think the problem is that the assumption that k < n such that x^(ka) = e is too weak, so when you conclude that k = n, you can't proceed. Here's how I did it:

Let ka = lcm(n, a) where 1 <= k <= n. Then x^(ka) = 1 since ka is a multiple of n.
By injectivity of f_a we get that x^k = 1, so k = n, thus lcm(n, a) = na.

It's basically the same as yours, but I assume that k is a factor of the LCM, so when I get k = n I can conclude that lcm(n, a) = na

[tulip otter]

so what i've shown is that if k<n then it isnt possible for ka to be a multiple of n since if it was then it would imply that x^k=e but that cant be possible

thus the [a,n] must be >=an

[tulip otter]

from my pov i am seeing that this contradiction does lead to [a,n]>=an, but maybe i am wrong

and if this is correct, then it would lead to [a,n]=an since [a,n]>=an and an is a common multiple of a and n

so to rewrite my claim in full:
first f_a(x^k)=e implies x^k=e and so k=n by the injectivity of f_a (keep in mind that k in {1,2,..,n})
Next assume that there exists k<n such that x^{ka}=e, or in other words, assume that [a,n]=ka where k<n. From the injectivity of f_a it was concluded that k must be equal to n, which gives a contradiction. Hence [a,n]>=na. which would mean that [a,n]=an

now that i reread and rewrote my supposed proof, i can see that i did something weird which is saying "assume that x^{ka}=e where k<n" right after proving that k=n in this case in the previous line. Maybe this is where confusion arose

[glad osprey]

yeah, I think that works 👍 I don't think you need contradiction btw, you can just assume that [a, n] = ka where k <= n, then you get k = n directly

[tulip otter]

yea, your proof is basically the ideal version of what i was trying to do, nice work

sorry for any confusion, and tysm both you and HChan. have a great day/night both of you

[glad osprey]

thanks, you did nice work too!

[austere pecan]

Let R be a commutative ring with unity and let a, b [ R. Show that
ka, bl, the smallest ideal of R containing a and b, is I 5 {ra 1 sb |
r, s [ R}. That is, show that I contains a and b and that any ideal
that contains a and b also contains I.

How to solve this

Let R be a commutative ring with unity, and let a, b ∈ R.

Show that:

⟨a, b⟩  (the smallest ideal of R containing both a and b)  
= { r*a + s*b | r, s ∈ R }

In other words:

1. Show that the set I = { ra + sb | r, s ∈ R } contains a and b.
2. Show that any ideal of R which contains both a and b must also contain I.

My problem is in 2nd part

[tidal schooner]

It's a direct check. An arbitrary element of I has the form ra + sb with r, s ∈ R. If J is an ideal of R, then J is an abelian group under addition, and it's closed under multiplication by elements of R. Can you take it from here to show that if a, b ∈ J, then I ⊂ J?

[austere pecan]

Understood👍

[coral steeple]

Am I missing something?

[velvet hull]

M1 oplus 0 is not equal to M1, but isomorphic

although I guess looking at the question statement it seems like the question itself is ignoring that detail

but yeah, it's just a straightforward application of 1st iso for modules

[coral steeple]

Oh did I write that somewhere

Oops yeah typo and I also repeat myself there

[rocky cloak]

If (+) means internal direct sum then they are equal

[velvet hull]

true, I didn't think about that

[south patrol]

And writing = for iso is common too

As is identifying M (+) 0 with M in a direct sum (i would rather write M anyway)

[rocky cloak]

at least for cannonical iso

[south patrol]

Yeah

[narrow temple]

Given the Ideal J=(Y^2-X^4,X^2-2X^3-X^2Y+2XY+Y^2-Y) inside K[X,Y] , I want to show that the element A=(Y-X^2)(X-1) does not lie in J.
One can simplify J to (Y-X^2)(Y+X^2,Y-2X-1) now if A does not lie in the intersection of both ideals, it does not lie in J.
Clearly A lies in the first factor, so it remains to show that it doesnt lie in the second factor denoted by B.

What are neat ways to see this or how else can one approach the problem nicely? I think im a bit clunky with this, so feel free to share neat ways that are good to try when it comes to questions like this. This is what i came up with:

1. I think elements in B vanish on (-1,-1) so if A doesnt vanish on (-1,-1) its not in B. A doesnt vanish on (-1,-1) so its proven. Here it was easy to find a common zero though.

2. I think we can pass to the quotient K[X,Y]/(Y-X^2,Y-2X-1)=K[X]/(-X^2-2X-1)=K[X]/(-1*(X+1)^2)=K[X]/(X+1)^2
   And If the image of A is zero, then its contained in the ideal. The image of A is -2X^3+2X^2 which using the relations is equal to -10X-6. Now this element is not contained in (-X^2-2X-1) for degree reasons and I could also check that -1 is not a root of the image of A. So again, A is not contained in B.

Adding the ideal in question, so no second guessing needed if I transcribed it properly.

[rocky cloak]

I think your factorization of J is incorrect, (-1, -1) is not a common root of J. The common roots seem to be y=x^2 and (1, -1).

Anyway, fun suggestion:

Try to localized with respect to the element y-x^2. Then A turns into (x-1) while J turns into
(y + x^2, x^2(x-1)^2)

this I'm pretty sure shouldn't contain (x-1), even though y-x^2 is a unit.

[brazen ravine]

In general can we always find the explicit inverse function (that is an explicit construction for the inverse) of any group or we need more assumption, or being more general can we always find the inverse function of any sets reparameterised?

[narrow temple]

Oh youre right, I fucked up the factorizing so its (Y-X^2)(Y+X^2,2X+Y-1) and 2X instead of -2X. I hope thats correct now.

We localize at y-x^2 then the question boils to whether x-1 is contained in (Y+X^2,2X+Y-1). Then Y+X^2-2X-Y+1=(X-1)^2 and i think we can then say that (Y+X^2,2X+Y-1)=(Y+X^2,(X-1)^2), right? Where did u get the extra factor of x^2?

[rocky cloak]

I didn't factor out y-x^2 just simplified the ideal, but you're right, there is a factor of y-x^2 in the second term that can be factored out

[narrow temple]

Okay nice, thanks. "Morally" localization is nice here, because it lets us ignore the y-x^2 part and work with x-1 which is easier to handle or how would you phrase it?

[teal hedge]

Hi guys! I am a newbie and I have a hard time understanding this definition from Kenneth Ireland's book.
Definition: If a1, a2... an belonging to integers, we define (a1, a2,... an) to be the set of all integers of the form a1x1+a2x2 +.. anxn where each xi belong to integers,
Let this set A = (a1, a2,.. an)
Sum and difference of two elements in A are again in A. A is an ideal in ring Z.
Can someone help me understand this and sorry if this is too trivial, I am a beginner

[thorn jay]

what dont you understand

[teal hedge]

For a given set B = (x1, x2,. xn) do they mean that for some a1 to an all the possible weighted sums belong to set A? Is each element in set A is a scalar integer?

[karmic moat]

the notation $(a_1, \dots, a_n)$ for $a_i \in \mathbb{Z}$ is just the set of all $\mathbb Z$-linear combinations of the $a_1$ through $a_n$

[cloud walrus]

anamono

[karmic moat]

this is called the ideal generated by a_1, ..., a_n

so for example if i looked at the ideal (2) in Z, then elements of (2) are of the form x2, where x is in Z (i.e., all even integers)

[karmic moat]

(since we are taking the ideal in Z)

[teal hedge]

Got it! Thanks A lot! Ideal(2) = all linear combinations of 2,so all the even numbers.

[karmic moat]

yeah

in general given a ring $A$, we call a subset $I$ an ``ideal'' if it's a subgroup under addition and absorbs under multiplication, i.e., for any $a \in A$ and any $i \in I$, we have $ai \in I$

[cloud walrus]

anamono

[teal hedge]

Got it!

[south patrol]

Absorbs... woah

[karmic moat]

yeh it's like a sponge

[south patrol]

(Left ideal)

[karmic moat]

yeah yeah left ideal blah blah noncommutative aaaaaahhhh

[south patrol]

😼

[south patrol]

I just dont like how this seems to suggest commutatjve rings r uninteresting gr

But i love the meme too

[thorn jay]

commutative rings are awesome i just like poking fun at people who assume properties because i am an envious universal algebraist

[south patrol]

Ofc

Why would you assume ur rings are associative

[karmic moat]

don't assume anything

assuming only makes an ass out of u and me

[austere pecan]

Have you ever doen any mathematics projects?

I want to do some math project but dont have any idea what to do

[karmic moat]

i've done a small project with some phd students. they were my advisor's phd students. culminated in a small poster presentation

usually professors have a lot of small questions floating around their mind

just go to one you find interesting and ask

[austere pecan]

No i mean i want to do projects so should i go to teacher and ask them what?

I never did something like that ..

[karmic moat]

jsut be like "hi i find your field of research interesting, do you have any small projects i could work on"

my experience with special topics courses are that professors will usually mention some open questions in passing

[austere pecan]

I am undergrad

[karmic moat]

yeah me too

[austere pecan]

Then should i ask that to my teacher?

[karmic moat]

yeah

not much to it, not worth overthinking

just ask them directly

[austere pecan]

Ok then

[karmic moat]

usually helps if you know that topic somewhat well and/or have taken a course with them, but the latter isn't too important

[tall igloo]

"do not assume your ring has elements"

[thorn jay]

nono see the entire point of a ring is that its the set of endomorphisms of an affine algebra/ternary group

so you generalize that part to make the "underlying group" a general algebraic structure

of course after that you assume the ring to be commutative anyways because i fancy things actually working when defining localization

[south patrol]

Real

I don't understand when people ask whether a ring is commutative. It is structure not a condition

[thorn jay]

well it has to otherwise it doesnt have 0 and 1

[tall igloo]

you can state all the ring axioms in terms of morphisms, without ever saying anything about elements

[thorn jay]

(it is a genuine very big issue in universal algebra that some algebras have a constant operation and therefore cannot be empty and some algebras do not)

[thorn jay]

ring viewed as a category?

then really what youre assuming is the existence of morphisms

[tall igloo]

thats a good example, then a ring is a unital monoid object in Ab

[thorn jay]

(taking the monoidal bifunctor to be (x) )

[tall igloo]

if you have higher category brainrot you can replace Ab with Sp and then such things still deserve to be called rings too!

despite having no sensible way to define "element"

[thorn jay]

Sp?

[thorn jay]

RELATIVE ELEMENTS RAHHH

[south patrol]

"Spectra"

[thorn jay]

fear

[fossil shuttle]

Lie algebras are rings

[south patrol]

Real.

[rocky cloak]

Let's stick with complex lie algebras please

[thorn jay]

the only Lie algebras i accept are that over F_2

no.

only old people call "nonunital rings" (mental illness) rings instead of rngs

hmm i think im just gonna resort to growling at people who claim that rings do not need a unit

no, they are an ideal

❌

whats the matter nonunital ring enjoyer? cant guarantee a maximal ideal?

how do you prove it for nonunital rings?

the Zorn's lemma argument only works if your top ideal (so R) is compact, i.e. finitely generated as an ideal

this is guaranteed in unital rings but not as a rng im pretty sure

i mean, take any abelian group, put the trivial multiplication on it, and then the ideals would be just the subgroups, and im sure that for abelian groups a maximal subgroup isnt guaranteed

exactly :P

my voice sounds angelic i know

ew mainstream progressive house

[rocky cloak]

Looking at you Q

[thorn jay]

i was contemplating on saying that but i actually wasnt sure

figured Q would show its rough ass head

because studying actual music is stressful af, and im just interested in fucking w it as a hobby

[south patrol]

Curious

[tall igloo]

is this book any good btw? came across it while trying to find a book to study for my prelims lol

yeah from what i can tell it is mostly examples and exercises, lol

damn

my undergrad used Algebra by artin but i feel like if i should study to test out of a grad algebra class i should not use my undergrad book lolol

[south patrol]

If you have trouble falling asleep it's good

[thorn jay]

may be just a skill issue

[thorn jay]

i mean its a prelims thingy right

it should test you on undergrad stuff

[tall igloo]

yeah true

[thorn jay]

i think dummit and foote is good for looking up stuff though

[elfin wraith]

Dummit and Foote is good because it’s very gentle and comprehensive

Dummit and Foote sucks because it’s very gentle and comprehensive

Also they don’t require rings to have unity

[cursive spindle]

Yes

[tall igloo]

messed up

[cursive spindle]

One reason why D&F is bad

[tall igloo]

as everyone knows "rng" is not a backformation, it was already the word for a wedding rng belonged to someone in a now divorced (= non-unital) couple

[south patrol]

this is my main criticism of the book lol

because whenever I help someone on this server I have to guess whether they are using D+F or not

Lol i just saw a funny restatement of the Nullstellensatz

If $k$ is an algebraically closed field and $R$ a finite type $k$-algebra, then $k \to R$ admits a left inverse

[cloud walrus]

Prismatic Potato

[tall igloo]

whenever someone asks a question about rings and says do not assume they are unital, I always reply saying rng because there is no such thing as a ring I cannot assume to be unital

[south patrol]

lol

[tall igloo]

wdym by finite type

[south patrol]

finitely generated as a k-algebra

[tall igloo]

ah makes sense

[south patrol]

but i am an ag person so i say this

Lemme think how this relates to the other forms of the nullstellensatz

well it will be mostly the "weak" nullstellensatz

Clearly it implies that any finitely generated k-algebra which is a field is k (as there aren't non-trivial retracts in fields), which is itself called a form of the nullstellensatz

[south patrol]

And uh if J = (f_1,...,f_n) then a retraction of k -> k[x_1,...,x_m]/(f_1,...,f_n) is equivalently a choice of common root of the f_i

[south patrol]

Yes, guess who I heard this from

But now I realise this is actually standard like this is more or less how i saw nullstellensatz proved in Atiyah-Macdonald

[cursive spindle]

Nullstenellsatzian is such a word

[south patrol]

So is nullstellensatzian

[tall igloo]

Shortest German word

[south patrol]

kinda annoying though how it is capitalised

it's an adjective smh

[tall igloo]

I am guessing this is because Nullstellensatz is a noun

So it's capitalized in German and then everyone else was like. Guess we're capitalizing this now

[rocky cloak]

A bit weird that it's not Hilbert's zero locus theorem or whatever

[elder wave]

There is an argument to be made because it’s a Wortneubildung (creation of a new word)

From a fixed name

So it could reasonably be capitalized

[tall igloo]

no clue lol, is it the person I'm thinking of in your department

[rocky cloak]

English was like, nah too much work translating this

[elder wave]

This always confused me

Like I’m German so sure call it that but English speakers do not have an easy time pronouncing it in my experience

And it has an easy translation

[elfin wraith]

There’s a couple of theorems in ring theory people just didn’t bother to translate and I’m not quite sure why

I don’t think it happens literally anywhere else in maths, or in any other language

[rocky cloak]

I mean they don't voice the st, but otherwise I don't see why they couldn't pronounce it

[rocky cloak]

You got some more examples?

[elfin wraith]

I’m going to guess that’s a big fat, it depends. Essen teaches their masters in German, Bonn does not

[elder wave]

There’s no point where this necessarily happens

[tall igloo]

dear lord

[thorn jay]

bro has one insult

[elfin wraith]

The Hauptidealsatz is the first example that comes to mind but I’m sure there’s others

[karmic moat]

i think that's commonly called Krull's principle ideal theorem

[rocky cloak]

I've just seen that as principal ideal theorem

[thorn jay]

its not a true generalisation unfortunately, because the hard part is really the fact that the intersection of all primes is the nilradical

[karmic moat]

or principal or whichever one

[tall igloo]

correct me if I'm wrong but aren't most German (math) PhDs taught in English?

[thorn jay]

i.e. ISP of the class fields/integral domains is the class if reduced rings

[elder wave]

masters ig

idk

[elfin wraith]

I’ve seen it as both, certainly not as common as Nullstellensatz no but I have seen it

[elder wave]

here we have some english and some german courses

[tall igloo]

this caused me quite some confusion a few weeks ago when I was on the train. Unfortunately I was already hungry and this did not help

[elder wave]

at some point the uni does not care anymore so the profs decide

Essen kannste vergessen

depends as well

some prefer english some forcibly translate an entire english source into german

[elfin wraith]

I often wonder what the standard maths textbooks are in languages which aren’t English, like I imagine for most languages the “standard” UG curriculum has a book in that language

[karmic moat]

Translations maybe?

But idk

[elfin wraith]

I would’ve thought, at least for languages in the developed world, that the core courses would’ve had their own specially written books

But I guess just translating is probably easier, there’s enough maths specific words you likely don’t have to do too much

[tall igloo]

this is making me realize I am only math fluent in one language despite being ~trilingual

[elfin wraith]

Yeah my UG did that, I only had to read like 2 textbooks

They were of deeply mixed quality though and having secondary sources was useful

[icy totem]

Is there a channel for galois theory or can I ask here?

It's the closest to galois theory that I found

[karmic moat]

Probably #advanced-algebra

[icy totem]

Thanks

[chilly ocean]

Does anyone have a recommendation for good book(s) or resources to start learning ring theory? I learned group theory mainly from Jacobson's BA1 + Dummit and Foote, would those be good to continue with? I found Jacobson a bit terse, and I'm worried that this will only get worse as there's more and more concepts to track

[karmic moat]

atiyah macdonald

[elfin wraith]

A&M is good but possibly similarly terse no?

[tardy hedge]

anamono be like: atiyah macdonald

[wraith cargo]

is this really good for basic ring theory

[wraith cargo]

what are we talking about here

did you finish the ring theory section of dummit and foote?

[karmic moat]

Hmm true AM is a bit terse

But I would say it’s really good for ring theory

[fervent solstice]

i found it better as a reference book than as an introduction one

it doesn't do much for motivation, does it

[karmic moat]

I’d say the examples are good and the exercises are very very good

I found the motivation to be fine

But as always textbooks are personal preference

[wraith cargo]

this is true

[karmic moat]

But yeah it is terse at times

[elfin wraith]

I think if theyre finding Jacobson too terse and dont know any ring theory A&M is probably not the call, but yeah lol

[wraith cargo]

I personally never liked Atiyah and Macdonald because I felt like the later chapters aren't that good
The first few chapters are very good but it falls off
Personally I find their dimension theory chapter total dogshit

that's why I always promoted Matsumura's ring theory that I found much better explains a lot of the more advanced concepts that A&M is too short to cover

[karmic moat]

I felt like the dimension theory chapter is standard no?

[karmic moat]

Fair tbh idk anything about Jacobson’s book

Eisenbuds got some book called “commutative algebra through ag” or something like that right?

Is that one good

[elfin wraith]

Im pretty sure thats a second text on comalg

[karmic moat]

Ahh okay

Yeah that makes sense

[wraith cargo]

not at all it basically covers nothing
It barely mentions regular local rings let alone something like CM rings or other important classes of local rings that have stuff to do with dimension
Tho my main gripe is that they don't talk about kahler differentials that are very fundamental in algebraic geometry

plus it assumes no homological algebra so you can't do stuff like local criteria for flatness that's also important in AG

[karmic moat]

For the CM rings and stuff fair but also maybe that’s out of scope for an “intro to CA book?”

[wraith cargo]

well ig then you'd have to argue what an intro to CA should be

which is a fair point to make

[karmic moat]

That’s true

[elfin wraith]

Actually it looks like you could use it as a first course, but its probably not ideal

[karmic moat]

I’ve heard good things about Gathmann’s CA notes

[wraith cargo]

those are very good

also Milne's CA notes

[karmic moat]

Nevermind also assumed basic ring theory lol

[elfin wraith]

Im honestly just glad I had course notes for all my algebra at uni because I dont love any of the big textbooks out there

[wraith cargo]

for commutative algebra or just regular algebra?

[elfin wraith]

Bit of both but mainly just regular algebra, the CA offerings are better

[karmic moat]

yeah milne's notes look nice

it's just a bit odd that he doesn't define what a ring is but defines what an ideal is

lol I guess matsumura doesn't define neither a ring nor an ideal

maybe i'm the odd one

[coral spindle]

Iirc matsumura starts proving things about modules before defining them

but I suppose he does eventually define them

[karmic moat]

weird

yeah this is the first paragraph of matsumura commutative ring theory

yeah lmao

this seems like a really weird choice to me

[coral spindle]

I remember reading this and thinking that he just started writing without a plan lol

[karmic moat]

according to preface originally one of matsumura's friends was supposed to write the book but the friend passed away

so he took over

[fervent solstice]

maybe he thought it fine to assume the base definitions are known since it's a commutative ring theory and not a general ring theory book (i have no idea)

[karmic moat]

i thought appendix A would be like "a ring is... an ideal is... a homomorphism of rings is..." and not "tensor product of modules and limits and colimits"

anyhoo the later chapters look nice

[elfin wraith]

That’s a crazy appendix lol, I should read mastsamura I hear great things

[elder wave]

i have that book

it's nice

[elfin wraith]

So much maths and so little time in the issue

And a waning desire to suffer for the next 10 years in the hopes of maybe making 45k

[fervent solstice]

in the US?

45k sounds like a lot actually but i guess yall have more stuff to pay for compared to western europe

[chilly ocean]

I'll take a look at this, thank you!

[chilly ocean]

I'm just starting out on ring theory- I've worked through the first bits of the chapter in Jacobson (up to 2.5, which is to say covering basic definitions, matrix rings, ideals and quotient rings)

[barren sierra]

I am sadly stuck on fiddly details of this

I only want to use the definition of nilpotency given here

not the one with commutators

lemme type up where I'm at + establish some notation

[thorn jay]

derive the equivalence

[elfin wraith]

I’m in the UK

[barren sierra]

Say $G$ is a $p$-group, $\abs{G} = p^a$ and $a \geq 3$ and $G$ is non-abelian (so that there's actually stuff to do).
So then $Z(G)$ is non-trivial and so $G / Z(G)$ is a $p$-group of order $\leq p^{a - 1}$.
So then $\overline{G} = G / Z(G)$ is nilpotent by induction and so we have
\[
  1 = Z_0(\overline{G}) \trianglelefteq Z_1(\overline{G}) \trianglelefteq \cdots \trianglelefteq Z_{\overline{c}}(\overline{G}) = \overline{G}
\]
for some $\overline{c} \geq 1$.
Then from here I want to prove by induction that $Z_i(G)$ is such that $Z_i(G) / Z(G) = Z_{i - 1}(\overline{G})$ and I'm struggling with this for the case of $i \geq 2$.
So fix $i \geq 2$.
So I know that $Z_i(G)$ is such that
\[
  Z_i(G) / Z_{i - 1}(G) = Z(G / Z_{i - 1}(G)) = Z(\overline{G} / Z_{i - 1}(\overline{G}))
\]
but then from here I am stuck

[cloud walrus]

Spamakin🎷

[barren sierra]

last equality comes from one of the isomorphism theorem, gun to my head I could not tell you which number

[elfin wraith]

That looks fiddly and annoying

[barren sierra]

if this comes on the qual I will be quite annoyed hence why I want to fully figure it out now

[barren sierra]

Because then if I can show this then $Z_{\overline{c} + 1}(G) = G$ as desired

[cloud walrus]

Spamakin🎷

[thorn jay]

there is a much nicer definition from which it follows that, if you have a group extension
0 → G → E → H → 0
then if G and H are nilpotent of degree n and m respectivelt then E is of degree ≤ n+m

idk you can try to do the same using this definition

[barren sierra]

cool, on the qualifying exam this is the definition I would be given and so I'd have to work with this

I think proving the equivalence AND proving the statement for the equivalent definition would be much harder in a short period of time

hence why I want to work with the given definition

[thorn jay]

yes thats why i added this

[barren sierra]

yes thats why i said this

[thorn jay]

i meant to say that you could maybe prove the group extension bit directly from the given definition, but maybe not

[barren sierra]

Oh I see

[barren sierra]

Hm I mean it seems like the same question essentially just with G = Z(E) and H = E/Z(E) and I'd get stuck much in the same way

[thorn jay]

yes i see lol

I think youre on the right track by considering G/Z(G). We have the projection π: G → G/Z(G), and if
1 < Z1(G̅) < ... < Zn(G̅) = G̅ = G/Z(G)
is the series for G̅, then
1 < Z(G) < π^-1(Z1(G̅)) < ... < π^-1(Zn(G̅)) = G
is the series you should investigate, rather than proving that the series for G divided by Z(G) is the series for G̅, maybe try to prove that this series "pulled back" along π is the series for G?

[barren sierra]

are those not just the same thing?

[thorn jay]

youre trying to show that they are the same thing

[barren sierra]

no no I meant like the series you wrote that I should be investigating is the series that I wrote that I should be investigating no?

like just unpacking what pi^-1(Z_i(Gbar)) is

[thorn jay]

no what you want to show is that Z_i(G) / Z(G) is Z_i-1(G̅), rather than that Z_i(G) is π^-1(Z_i-1(G̅))

maybe someone more awake than me should be doing this tbh

lmao

[barren sierra]

Hmm yea got nowhere with that idea either 😵‍💫

[crystal vale]

My idea was, if G has order p^a, and if G/Z_i(G) is not {e}, then Z_{i+1}(G) has strictly greater order than order of Z_i(G).

So at some point Z_j(G) has to be equal to G.

[barren sierra]

That's the idea yes but actually working out the details is where I'm stuck

[terse umbra]

This answer in the book (Pinter's a book of abstract algebra) is wrong right? For associativity

[velvet hull]

it is true, where's the wrong part?

[terse umbra]

What is true

How he came to the conclusion that it is not associative, or that it is associative

[velvet hull]

oh, I see

[terse umbra]

When having answers in the book is less helpful than not having them😭

[tribal moss]

The hand-written answer (yours?) looks more convincing. The printed solution skips a lot of algebra, and you make a good case that there must be something wrong in the skipped parts.
To show this more concretely, try plugging in x=y=z=1. Doing 1*(1*1) concretely gives (1/3)/(7/3) = 1/7, but the final equals sign in the book solution says it should be 3/7 instead...

[terse umbra]

Yeah, thanks then I'm not crazy :^) yes the other solution is mine

[untold torrent]

I wanted to know elements of S4

So okay first I make all the partition

1+1+1+1

1+1+2

2+2

1+3

4

Total 5 partition

Now i focus on each partition elements which will be

First 1111 is identity element

How do I make rest permutation?

E.g.--112 type permutation

[rocky cloak]

A permutation is just a way to rearrange the elements of a set, in this case the numbers 1,2,3,4.

So to make a permutation of type 1+1+2 you just pick two numbers that are mapped to themselves and two that are swapped

Depending on exactly what you want it might be easier to just describe all permutations first then sort them by cycle type later

Like
1234
can be the identity and
1243
sends 1 to 1, 2 to 2, 4 to 3 and 3 to 4
Etc

[untold torrent]

Yes I want to understand that permutation formula actually I have read something similar thing in group of permutations in High School@rocky cloak

[brazen ravine]

What are conjugations really, how can we think of them? x -> axa^-1

When rationalising fractions, we say take the conjugate of the denominator by changing the sign (I don't know how to visualise this, in R^2 maybe)

In C we say conjugate by flipping the sign as well similar to fraction case. This is reflection across the x-axis.

I take that a change of basis is also somekind of conjugation? So what are conjugations?

[glad osprey]

I don't think the group theoretic conjugate (ie. x |-> gxg^-1) and the complex conjugate (ie. x + yi |-> x - yi) are directly related, except in name

[tribal moss]

They're both automorphisms, but I'm not sure it can be taken much farther than that.

[thorn jay]

these are all different notions of conjugation. Really what conjugation means is changing it in some way buy keeping properties the same.

in the case of groups you can see axa^-1 as how x looks "from the perspective" of a. This van for example be seen in group actions, where, if x stablizes an element r, then axa^-1 stabilizes a*r. Two conjugate elements in a group share many properties, which is because conjugation is a (rather nice) automorphism

[karmic moat]

I think you can write a complex conjugate as a conjugation of matrices by sending the standard basis e1, e2 of R2 to e1, -e2

Doesnt the Galois group of C/R have something to do w complex conjugate too?

My galois theory is not very good so dont quote me on that but I’m pretty sure it’s given by an automorphism on Gal(C/R)

[tribal moss]

Gal(C/R) is cyclic of order 2, and the nontrivial automorphism is indeed complex conjugation.

[knotty badger]

Yes I think conjugation can just be thought of as a change of perspective

[tribal moss]

This connects to speaking about "conjugates" when rationalizing denominators: for a nondegenerate quadratic number field Q[sqrt(d)], the single nontrivial automorphism takes a+b·sqrt(d) to a-b·sqrt(d).

[knotty badger]

I’ve talked about it here before:

#linear-algebra message

[thorn jay]

conjugation so important they made a whole algebraic structure out of it

[knotty badger]

Wait which one

[thorn jay]

granted they have the most stupid name

(quandles)

[tribal moss]

C*-algebras?

[thorn jay]

(although actually quandles came from knots and links and the link with group conjugation came after)

[thorn jay]

thats complex conjugation

[barren sierra]

If people have ideas for this plz help (I'll probably also ask on math SE)

[barren sierra]

My work so far

[tribal moss]

So if I read you right, you have all the fancy bits in place and just need a solid argument for "and the upper central series of G/Z(G) corresponds exactly to the upper central series of G itself, other than G_0"?

That is, if we set H=G/Z(G) then H_k = G_(k+1)/Z(G).
That feels like it should just be induction.

[rocky cloak]

In Galois theory two elements are called conjugate if there is a Galois automorphism taking one two the other. The identity and complex conjugation are the two only Galois automorphisms of C/R

[rocky cloak]

I'm not sure I understand where you're stuck.

The nilpotency class of G/Z(G) is one less than that if G, so the result follows by induction.

[vast verge]

Is it possible to have an integral domain of characteristic 2?

[tall igloo]

maybe im missing something obvious but doesnt Z/2 work?

[tribal moss]

It does.

[karmic moat]

just take any field of characteristic 2

[copper ledge]

Let G be a finite group and K,H subgroups such that their direct product KxH is also a subgroup of G (not necessarily normal), is there some natural set bijection between G/(KxH) and G/(HxK) ?

[vast verge]

Can someone explain the highlighted part, I don't understand

[tribal moss]

If K×H is not a normal subgroup, then does G/(K×H) even mean anything?

[copper ledge]

left cosets

[karmic moat]

if a subgroup isn't normal then the quotient isn't always a group

[copper ledge]

That's why I asked for some natural set bijection

Not a group isomorphism

[karmic moat]

probably comes through the bijection of sets K x H to H x K

sending (k, h) to (h, k)

[tribal moss]

"Contains" is sloppy wording -- they mean that a² divides (n-1)!

[karmic moat]

because direct product is commutative anyway

[copper ledge]

Probably, I couldn't get a map through that. It is weird that the map factors "downwards"

[tribal moss]

Since the two sets have the same size, I think much of it comes down to what you'd accept as "natural".

[vast verge]

Oh it's because a(a-1)*a(a-2) divides (n-1)! right?

[karmic moat]

given g in G, wouldnt the coset g(H x K) be identified with g(K x H) purely because direct product is commutative?

[copper ledge]

I guess as natural as it can be. Probably as natural as the isomorphism HxK = KxH if possible

[karmic moat]

if that's what you want

yeah

[copper ledge]

was probably a dumb question xd I am trying to find some nice reformulation for shuffle permutations which are representatives of quotients of symmetric groups

[tribal moss]

Not necessarily -- if g1(H×K) = g2(H×K) it doesn't follow that g1(K×H) = g2(K×H), so your map isn't necessarily well defined.

[karmic moat]

why not?

[tribal moss]

For the time being mostly because I don't see a reason it would. Let me see if I can figure out a concrete counterexample.

[thorn jay]

Z over 2 what??? apples?? bananas??

[tribal moss]

Z/(2)

[thorn jay]

we dont really tend to identify subgroups with eachother when they are isomorphic

[tall igloo]

hey my notation is better than Z_2 lol

[karmic moat]

we're working over only sets

is my impression

[thorn jay]

still

[fading acorn]

Z_2 somehow feels like localization at 2

[thorn jay]

Z_(2)

[karmic moat]

i guess i'm not seeing your point

like are you saying i shouldnt use the word "identify?"

[thorn jay]

im probably missing context

in which case ignore me

[tall igloo]

i will always use Z/n for ring of integers mod n and no one will ever convince me otherwise (well, \mathbb{Z}, lol)

[karmic moat]

hmm i'll think about it. for me it's not clear why it shouldn't be lol but yeah i'll think about it

[proud vigil]

g1(H x K) and g1(K x H) are isomorphic right

[tribal moss]

Oh, I see your point now. If we say the direct product H×K is a subgroup of G, then it contains (ek)(he)=kh for every h, k, so it is the same subgroup as K×H -- and obviously has the same cosets.

[copper ledge]

Oh my bad I think I messed up the question

[tribal moss]

(And if it's not an internal direct product but just a subgroup that happens to be isomorphic to the external direct product, then everything depends on how we choose to embed H×K and K×H into G).

[copper ledge]

I meant HxK as containing pairs (h,k) but then H and K can't be subgroups of G

[karmic moat]

yeah

i think external direct product gets a bit weird tho

because of what you said

[copper ledge]

I think I abstracted the question a bit too much (which is still an improvement) xD Sorry about that

[karmic moat]

all g

[vast verge]

What are these sets?

[karmic moat]

where is this from? the only notation i've seen for K* is the group of units of K

idk how you endow it with a field structure tho

[tribal moss]

Do you have more context? Immediately I think "K* is a subfield of F*" is probably a typo for "K* is a subgroup of F*".

If so, your screenshot would be (part of) a viable answer to, "prove that the field with 32 elements has exactly one proper subfield".

[vale oyster]

Yo gang, i stumbled upon this question and have no clue how to solve it. i've shown that L/K_i separable but that doesn't help me much, anyone can explain?

nobody in the help helped me

and someone told me to ask here so

[rocky cloak]

For the first equality you may want to consider
prod[sigma(a)] (x - sigma(a))

(product over conjugates of a)

[velvet hull]

Ks(Ki) = Ki(Ks) since both are by definition the subfield of L generated by Ks and Ki

and as a hint, if l is in L, and sigma is an element of the automorphism, then the product of sigma(l) where sigma ranges over the automorphism group is always in Ki

the proof I figured out after staring at your question for a while involves the following lemma (which is quite hard to prove actually if you don't know the right trick):

Suppose that f is an irreducible not not separable polynomial with coefficients in a field with characteristic p.
Then there exists a unique g(x) and integer k, with coefficients in the same field, such that f = g(x^p^k), such that g(x) is an irreducible separable polynomial.

[vale oyster]

Thanks @velvet hull @rocky cloak that made sense, have a nice day

[tidal schooner]

Given a ring A, we can consider the subset A* whose elements are units (i.e., anything having a multiplicative inverse in A). You can verify that A* is a multiplicative group.

A commutative ring F is a field if and only if every nonzero element of F is a unit, meaning that F* = F \ {0} as a set. In this example, F is a field of order 32, so F* is a multiplicative group of order 31

[south patrol]

Kinda amusing how it works like this (to me)

Well like 32 = 2^5, and the result would be true for any field of order p^l where p,l are primes

[tribal moss]

But this particular argument isn't available for p=2, l=11.

[south patrol]

But here miraculously you get a prime after reducing 1

That's what I mean lol

[tribal moss]

Even though the conclusion is still true.

[south patrol]

Funny thing where it is rigged so that you can use a more ad hoc proof

[tribal moss]

I'd even find it more satisfying to say something like, suppose there were a field L strictly between F2 and K (of order 2^k). Then K would be a vector space over L ...

[south patrol]

This is how I would argue, yee

[proud vigil]

does anyone recommend a supplementary text to aluffi, I'm rereading it/working towards the ring and field theory i missed to review some of the things I forgot but I kind of want a non-category theory approach to this too

[kind temple]

d & f is good for that

[proud vigil]

ooh ok

[elfin wraith]

Remember that D&F doesn’t require rings to be unital though

[proud vigil]

:0

[elfin wraith]

Which is not entirely nonstandard but certainly not the norm either

[elfin wraith]

They typically do just say let R be a unital ring or whatever but just bear that in mind if you’re coming from another text

[proud vigil]

ooh ok thats good then

[barren sierra]

All the stuff wrapped up in "so the result follows by induction" is where I am stuck. You need to show that the series witnessing the nilpotency of G/Z(G) lifts to be the tail of the series witnessing the nilpotency of G.

[rocky cloak]

That's just immediate from the definition though.

The next step in the series is just Z(G/Z(G)) and you apply Z until you reach the trivial group.

If you apply it once you now have to apply it one fewer times

[barren sierra]

is it immediate by the definition?

the next step is not just the center of the previous step that's the thing

it's a lift of it

but then you need to lift it again from G/Z(G) to G (by lifting I mean by subgroup correspondence stuff for quotient groups)

and then you actually need to show that the result of that second lift is what we want

[rocky cloak]

A subgroup is everything iff the quotient is trivial, so you don't really need to think about the lifting much.

Like the definition is just you take
Z(G/Z(G)) several times then lift those to subgroups of G

[barren sierra]

Hmmm I guess

[rocky cloak]

Like if it helps you:

G -> G/Zi -> G/Zi / Z(G/Zi)
Is the same as
G -> G/Z[i+1]
by the second isomorphism theorem

Or perhaps by the definition of Z[i+1] if you will

[rancid timber]

I think I’m misunderstanding how they’re proving cauchy’s thm

If we take an example group like G as the cyclic group of order 12

P will be 2

Wait nvm

I got it

[south patrol]

Interesting as usually Cauchy's theorem is done for not necessarily abelian stuff

Like it is cool you can go the whole way just from a weak form of Cauchy

[rancid timber]

I actually haven’t read through the whole proof yet

Idk how they’re going to extend this to nonabelian groups

[elfin wraith]

I really need to get a better grip on the proofs of Sylow, I think I always get about 3 lines in before I get bored start skimming and decide to trust they work

[kind temple]

i have been telling myself this for the past 3 years

[rocky cloak]

You can get it fairly short with a simple fact about group actions of p-groups:
If a p-group acts on a set, then the number of fixed points is congruent to the size of the set mod p.

Theorem 1
Now for a p-subgroup P < G, let P act on G/P by left multiplication. Then the fixed points are N(P)/P, so
[N(P):P] = [G:P] (mod p).

Induction+Cauchy's theorem applied to N(P)/P then gives existence of sylow p-subgroups.

Theorem 3
If P is a sylow p-subgroup, then [N(P):P] = [G:P] mod p means that [G:N(P)] is 1 mod p, so the number of conjugates of P is 1 modulo p.

Theorem 2
Let P and P' be two different sylow subgroups. Then P acts on G/P' by left multiplication. Since the number of fixed points is not 0 mod p, it is nonzero. So there is a g with
PgP' = gP'
In other words
g^- P g = P'

[tulip otter]

let $G=\langle a\rangle$ and fix $z\in Z$. Define a map $f:G\to Z$ by $f(xy)=f(x)f(y)\ \forall x,y\in G$ and $f(a)=z$. Then this map is well defined and is the $\bf{unique}$ homomorphism that satisfy $f(a)=z$. \Indeed, given $x,y\in G$ such that $x=y,\exists k\in{1,2,\dots,n}$ such that $x=y=a^k$ so $f(x)=f(a^k)=f(y)$ and $f$ is well defined (is this really sufficient to prove that $f$ is well defined?). Now $f$ is a homomorphism by construction.\\Finally, let $g:G\to Z$ be a map defined just as $f$, ie $\forall x,y\in G, g(xy)=g(x)g(y)$ and $g(a)=z$, then $g=f$. Indeed, given $x\in G,\exists k\in{1,2,\dots,n}$ such that $x=a^k$ so that $g(x)=g(a^k)=g(a)^k=z^k=f(a)^k=f(a^k)=f(x)$ which completes the proof of the hint

[cloud walrus]

yassine

[untold torrent]

Troposphere is nice name👌

[gray parcel]

Might be a relatively elementary question for this channel, but uh. There's this theorem that says you can't have finite-dimensional normed division algebras over the reals of dimensions other than 1, 2, 4, and 8. But what about the infinite dimensional case?

[tidal schooner]

There exist infinite-dimensional normed division algebras over R that don't have units, but there aren't any unital ones. See https://doi.org/10.1090/S0002-9939-1960-0120264-6

[kind temple]

that is cool

[gray parcel]

Someone in a different server brought up the field of formal Laurent series.

Wouldn't that work?

It's still unital, no?

Or is that...not normed?

[untold torrent]

I'm trying to check how many elements of (abc)(d) type will be there in S4

But this formula is not working why?

4!/(1^1×2^0×3^1)(1!)(3!)

4/3?

[tidal schooner]

It's a field, so it's certainly unital and a division algebra. You can even equip it with a discrete valuation. The issue is that it can't be equipped with a normed algebra structure, afaik

[gray parcel]

Ahh

[sonic coral]

it’s also like this in Gallian until chapter 24

[humble dirge]

someone help in this one.

[elfin wraith]

Oh that’s incredibly slick, much nicer than the “standard” proofs I’ve seen

[vast verge]

The definition of zero divisor only works for commutative rings. Is there a reason for this?

It also seems like (on the surface) that every non-commutative ring has zero divisor. (My only examples are matrices over the integers)

[knotty badger]

How about quaternions?

[vast verge]

I've never studied quaternions

But you make a good point

[rocky cloak]

You can define zero divisors just fine, but there is a distinction between left, right and twosided zero divisors. You can have that yx = 0, but there not being any z with xz = 0.

And noncommutative rings need not have zero divisors. For example the quaternions, the Weyl algebra, the noncommutative polynomial ring k<x, y>

[vast verge]

Is k<x,y> over the complex numbers so that it is not commutative?

[rocky cloak]

You can put in whatever commutative ring you want for k. Could be the complex numbers.

The reason it's non-commutative is just because xy is not equal to yx

[vast verge]

Hmm okay okay

The way you wrote the brackets made me think you were doing an inner product

[rocky cloak]

No it's just the notation that distinguishes it from the commutative polynomial ring k[x, y]

[velvet hull]

the fixed under 1/x implies that f,g have to be of the same degree such that the coefficients are reflectionally symmetric, i.e. a_k = a_n-k. (this is necessary and sufficient)
the fixed under 1-x implies that f(x) = f(1-x) and g(x) = g(1-x) separately

that's all have for now

[tall igloo]

is this like the free k-algebra in 2 variables? havent seen this notation before lol

[south patrol]

This is fairly standard notatiom i think but maybe not in our part of the world

[rocky cloak]

Where is y'all's part of the world and what notation do you use?

[tall igloo]

yeah most americans dont know what a free k-algebra is, its really embarassing

[south patrol]

Actually idk, by that I meant in homotopy theory, but a lot of the rings that appear there are commutative honestly lol

I would use your notation tho jagr

I think i have used this to mean slightly different things tho too, sad

[rocky cloak]

Yeah, only so much notation to go around

[humble dirge]

Thanks.

[velvet hull]

it's not a complete solution lol, far from one

just what I could figure out

[tall igloo]

are there even non-commutative ring spectra lol

[south patrol]

But this notatiom appears even in like, say, arithmetic geometry because of Dieudomné modules

[south patrol]

Lol i mean E1 stuff appears but I mean many ordinary rings r commutative right

[tall igloo]

like H F_p^{n, n} or something isn't a thing bc the higher EM spaces dont exist

this is a bad response but E1 stuff is still E1-commutative lol

[humble dirge]

Yeah.. but thanks for your response that's what I should do.. same progress I have made. But that didn't work

[elfin wraith]

I thought potato was in my part of the world and k<x,y> is the notation I’m used to

[south patrol]

By my part of the world i meant mathematically aha

But i also do AG stuff anyway so eh

[elfin wraith]

Ahh that tracks haha

[barren sierra]

I've seen that notation as well a lot in algebraic complexity theory

Cause a relatively popular technique for proving circuit lower bounds is proving them over non-commutative polynomial rings and then lifting those lower bounds to the commutative setting with some (hopefully not too much) weakening of the lower bound

Cause a big obstruction in algebraic complexity theory is that we don't have good control on when things cancel out to zero

And non-commutativity gives some control (in that less things cancel)

[vast verge]

11 is meant to be a solution

How did I miss it?

[tribal moss]

Solve it separately in Z4 and Z3, and CRT the solutions back to Z12.

[rocky cloak]

(x-2)(x-3) = 0 does not imply that
x-2 or x-3 is 0, because Z/12 has zero divisors.

(11 - 2)(11 - 3) = 9*8 = 0 (mod 12)

[glad osprey]

He accounted for that already, on the left side he enumerates (x-2) = {2, 3, 4, 6} which are the zero divisors in Z_12

[tribal moss]

It seems like you only looked for cases where (x-2)(x-3) is exactly 12 rather than ones where the product is a multiple of 12.

[glad osprey]

The problem with his approach is that for (x-2)(x-3) doesn't have to be 12, it can be any multiple of 12. So for example, (x-2) = 6 doesn't imply (x-3) = 2; you could have (x-3) = 4 or 6 or 8

[rocky cloak]

Well 8 and 9 are also zero divisors

[glad osprey]

oh yeah, good point

[rocky cloak]

(and 10)

[glad osprey]

so basically, you haven't enumerated through all the zero divisors properly

[vast verge]

What's the CRT?

[tribal moss]

The Chinese Remainder Theorem.

[rocky cloak]

Chinese remainder theorem
(e.g. Z/12 ~= Z/4 x Z/3)

[vast verge]

I thought Chinese division theorem (or euclidean division) was a=bq+r

[tribal moss]

All of 2,3,4,6,8,9,10 are zero divisors in Z/12, and almost all of them has more than one possible partner divisor.

[rocky cloak]

I haven't heard anyone call euclidean division for Chinese division before, though I'm sure Chinese people knew about it before Euclid

[south patrol]

Ye lol

[vast verge]

Oh right, basically any number that isn't relatively prime to 12

[tribal moss]

Indeed.

[vast verge]

Huh, i thought that at least most people at least knew about the term "euclidean division algorithm", even gallians textbook calls it that

[tribal moss]

What Jagr said was that he hadn't heard anyone call it "Chinese".

[vast verge]

I thought it was euclidean division algorithm was sometimes called chinese remainder theorem

Or maybe chinese remainder theorem is something else...

Also, does being a zero-divisor imply not being a unit?

[tribal moss]

The Chinese remainder theorem, elementary form:

Suppose n and m are coprime positive integers. Then for every a and b, the modular equation system
x == a (mod n)
x == b (mod m)
has exactly one solution for x with 0 <= x < nm

[south patrol]

it is

Lol

[tribal moss]

Suppose some element a were both a zero divisor and a unit -- that is, there exist nonzero b and c such that ab=0 and ac=1. Then abc = b(ac) = b but also abc = (ab)c = 0·c = 0, which is absurd.

[vast verge]

Okay thanks

[south patrol]

x is a (right) zero divisor iff (left) multiplication by x is non-injective, and it's a unit iff multiplication by x is a bijection

[next obsidian]

You know it’s fleema when you see glenbo, skewskew, and roltorb

[vast verge]

Gallian just uses this star notation without defining what it is :(

[rocky cloak]

Well it is the group of units

[tribal moss]

It's the multiplicative group of each of the fields.

[rocky cloak]

But yeah, bad Gallian for not defining stuff

[tribal moss]

At least it says "subgroup" here instead of writiting it as "subfield" in item 41, as was quoted yesterday.

[vast verge]

Ohh okay

It's like how Z6* excludes 0,2,3,4 right?

Or the U-groups in group theory

[glad osprey]

Gallian actually uses F^* to mean the non-zero elements of F, and U(R) for the group of units of a ring. Kinda weird convention

(for a field they are the same of course)

[vast verge]

I'll be real with you, I just skip the notation pages

Because >90% of it is stuff I've already seen

[tribal moss]

But other than that mention (before even the title page), the notation is only ever used offhand in examples or exercises -- sometimes with an explanation attached, at other times not, with no real rhyme or reason I can see.

[vast verge]

Normally it would show up in some example

[tribal moss]

(Why on earth does he not write Z and Q in bold, when he does for R and C?)

[vast verge]

Oh yeah I agree

I right blackboard Z and Qs myself

[south patrol]

Tbh i have seen this. I think R^x is most commom for units tho

[south patrol]

Yeah lol writing bare Q and Z looks very ugly

[glad osprey]

ah yeah, R^x is what I've seen the most

[tall igloo]

me when im in a horrible notation competition and my opponent is this