groups-rings-fields

[pulsar idol]

Wait

sorry I'm having trouble understanding the meaning of that

so basically that means like

every element in that subgroup satisfies that relation?

but then using that notation is a bit misleading tho isnt it

[rapid cave]

The point is that for a,b in G this product "measures" how close a,b are to commuting. If its trivial, they commute.
When you quotient by G' which is generated by all such products, in the quotient group G/G' all such products are trivial and hence this quotient grouo is abelian.

[pulsar idol]

ok great

so just to make sure, im not supposed to read $<aba^{-1}b^{-1}>$ as the cyclic subgroup generated by $aba^{-1}b^{-1}$

[cloud walrus]

lifelong dumbass

[pulsar idol]

but rather

[rapid cave]

Not cyclic, its generated by all product aba-1b-1

Oh i read your message wrong

[pulsar idol]

the subgroup containing all elements $g$ such that $g=aba^{-1}b^{-1}$ for some $a,b \in G$

[cloud walrus]

lifelong dumbass

[rapid cave]

Not quite

[rapid cave]

This is not a subgroup

So you take the generated subgroup of that

[pulsar idol]

the generated subgroup?

[pulsar idol]

sorry whats the difference

oh wait you mean

[pulsar idol]

the subgroup generated by these elements

is that what you mean?

[rapid cave]

Yeah

Its the smallest subgroup that contains all of them

[pulsar idol]

ok

thank you

[ebon prairie]

can someone help me with q21 and 24

[rapid cave]

for 21, assume the direct sum of Ri has a unity, can you build a unity for R1?

for 24, if u_i is a unit in R_i then (u_1, u_2, ..., u_n) is a unit in the direct sum of R_i.

[tardy hedge]

What book is that from

[ebon prairie]

gallian

[ebon prairie]

i don't understand ab = pk part

[limber tapir]

Well what does it mean for some number to be zero modulo p?

[ebon prairie]

multiple of p

[limber tapir]

Exactly

[ebon prairie]

oh,

thanks

[fringe marten]

I remember asking for help in this channel like 5 or 6 months ago for help to understand something, it was for a very important presentation i needed for my exams. Turns out i got a really good grade and got the uni i wanted which allows me to continue to do maths in the future. So i want to say a very big thank you to the people who helped me then if i hadn't got that good of a grade on this project i wouldn't be where i am rn

It might seem like nothing but it really helped me

[rocky cloak]

This?

[fringe marten]

Yes

And you helped me

I did ask multiple questions tho

[rocky cloak]

Then np, and @south patrol thx too

[fringe marten]

Wait lemme

This too

[thorn jay]

that's amazing!

great job to you too

[fringe marten]

It was a presentation about a personal initiative

And turns out i needed algebra

[white oxide]

Is K the kernel of M -> M (x) B?

Tbh I'm trying to see why this helps

[rocky cloak]

K is the kernel of B(x)M -> B(x)N

(This is just the definition of exactness)

You want to use that tensoring with C if faithful

[white oxide]

Let $h: M \to N$ be a monomorphism of $A$-modules. Let $M' = \ker(M \otimes B \to N \otimes B)$. Since $C$ is flat over $A$, the following sequence is exact:
[0 \to M' \otimes C \to (M \otimes B) \otimes C \to (N \otimes B) \otimes C]

Is this basically what you said?

[cloud walrus]

okeyokay

[white oxide]

So we want to show that (M (x) B) (x) C -> (N (x) B) (x) C is injective

Now M (x) B is a B-module so that M (x) B -> (M (x) B) (x) C is injective by assumption but I don't think this really tells us anything lol

[white oxide]

This problem must be up there with the Riemann hypothesis

[long obsidian]

Is there a name for functions p on a vector space over a field k to k whose kernel is a vector subspace but are not linear?

For example let V=k[x,y,z]_2 be the vector space of two dimensional multivariate polynomials of total degree 2.

Let W=k^3. Then for each w in W define the function w:V->k by w(f)=f(w). Now w is not a linear functional on V but ker(w) is a linear subspace of V. I don’t know what I’d call such a thing, all I know is that w defines a hyperplane in V.

[south patrol]

I don't really think there is a name for this - it seems quite a specialised notion cause you can just map any subspace to 0 and extend almost arbitrarily

[thorn jay]

for such a notion of a kernel we immediately just kinda need the function to be a homomorphism of abelian groups

unless you simply require the fiber of 0 (everything that gets sent to 0) to be a vector space

in that case, the notion isn't really, useful at all I can imagine

it corresponds to a choice of subspace U and a function f : V \ U -> W \ { 0 }

[south patrol]

Lol

[long obsidian]

Oh wow false flag I am pretty sure this evaluation map is linear idk what I was thinking. Sorry for the confusion and thanks for the comments.

Idk why I thought this wasn’t a linear map

[thorn jay]

lol

its either a linear map, bilinear map, or youre doing algebraic geometry and its a regular map

all the others maps either dont appear or you try to not work with as much as possible

pro tip

[rocky cloak]

Notice that M(x)B(x)C = M(x)C

[pulsar idol]

is there a way for me to know if ive got all of them

[rapid cave]

If you give a proof that what you found is everything

[pulsar idol]

hmm

ok then

[rapid cave]

Every homomorphism should depend on its value on some fixed generator

If by Z24 you mean cyclic of order 24

[radiant knot]

Are those supposed to be Z/24Z and Z/18Z?

[pulsar idol]

i mean this

[pulsar idol]

not this

i dont mean the quotient groups

[radiant knot]

So you mean cyclic groups okay

[rapid cave]

.

[radiant knot]

My intuition says if Hom_Grp(Z_24, Z_18) ~ Z then it checks out,
In other words if you can always find maps f: Z_24 → Z_18 for each integer then it is 100%

(or equivalently if it checks out as another group in GRP)

Revisiting this is kinda nostalgic lol

[pulsar idol]

ok so it basically just depends on what i map $\phi (1) $ to (im letting $\phi$ be the homomorphism)

[cloud walrus]

lifelong dumbass

[pulsar idol]

because $\phi (k) $ = $k \phi(1)$

[cloud walrus]

lifelong dumbass

[pulsar idol]

so now

if i consider mod 18

ok so basically depending on my choice the possibilities are $\phi(1) =0, 1, 2 \cdots 17$

[cloud walrus]

lifelong dumbass

[pulsar idol]

in mod 18

[rapid cave]

✅

[pulsar idol]

so

basically

thats all the possibile choices

[pulsar idol]

ok great thanks for the hint on generators

and using linearity

abstract algebra is lowkey kinda cool

[rapid cave]

Agreed xD

[pulsar idol]

its hard but like i just learned the isomorphism theorems and my brain i feel has expanded 3 times in size

i still kind of dont get it fully

[rapid cave]

They are all insanly useful

[pulsar idol]

yea ik its all so broad

im trying to just wrap my head around it

[twilit wraith]

maybe its only because im still kinda new to algebra but the first isomorphism theorem is extremely useful

its also not too difficult to understand

basically its saying that if you have a surjective homomorphism from G to H, you can mod out the useless stuff from G to get an isomorphism to H

[pulsar idol]

the first one is fine

the third one is really unintuitive

[quiet pelican]

Which one is the third one for you?

Pretty much the only one that I’ve seen consistently numbered is the first

[pulsar idol]

uh

the correspondence theorem

[thorn jay]

You can see the set of normal subgroups of G as a partially ordered set. This means that, for a normal subgroup N ⊂ G, there are normal subgroups N ⊂ M ⊂ G, which are "above" N. The correspondence theorem states that the normal subgroups of G above N correspond to the normal subgroups of G/N

my universal algebra book has this picture

(Con A here is a generalisation of the poset of normal subgroups)

θ is such a generalised nornal subgroup

[twilit wraith]

this would mean that if N is maximal normal in G, then G/N is simple?

[cedar vault]

Is order of G p^n here?

[thorn jay]

exactly

if your group G is finite, then this process can be repeated to get a subnormal series
0 < N1 < N2 < ... < G
i.e Ni is normal in Ni+1, such that Ni+1 / Ni is simple. This is called a conposition series for G, and it turns out that each composition series has isomorphic factors

Jordan-Hölder my beloved

[twilit wraith]

i gotta do more group theory

i was thinking about taking a course on it but it required grad alg 1 and 2 so i didnt think i was ready

it just seems really cool tho

[rocky cloak]

What is Cl_H(A)?

[thorn jay]

use \;

[cedar vault]

Im not sure what this is, but I think you should look at elements in the center to get proper containment. The center will be nontrivial . This is what I would try:
||If there is an element of the center not in H, then you are done. Otherwise if H contains the center, pass to the quotient G/Z(G) where you know the corresponding subgroup of H will be properly contained in the normalizer by induction and then you can come back to G. ||

[rocky cloak]

Allright, I don't completely see how you would use this to prove the exercise.

I can give you some hints for alternate approaches if you like though

What Herzog said would basically be my hint I guess.

And also ||the statement is true for nilpotent groups in general||

H is always contained in NG(H) though

That's just immediate from the definition

[cedar vault]

It is what makes the problem nontrivial, yes

[thorn jay]

anything algebra related is really cool

[cedar vault]

Not sure what you mean by that?

[rocky cloak]

The idea is that simple groups don't have many normal subgroups

[cedar vault]

I know that, but what do you mean by kernel to subgroup is always 1

Kernel of what?

Which map?

Yes it involves an idea like this but you need to be more specific

This works yes

You don't need to compute the kernel. It will be trivial like you said. So the map G --> S_G/H is an injection which is what you want

I personally would use the two interchangeably, so it's no problem for me

[crystal vale]

Any hint?

[quiet pelican]

||x^-1 y can be written in the form hk||

[crystal vale]

Yes but I don't see any idea from this

[quiet pelican]

So $H^x K^y = xHx^{-1} yKy^{-1} = xH h^{-1} x^{-1} y k^{-1} K y^{-1} = xHKy^-1$
Can you finish from there?

[cloud walrus]

mico

[crystal vale]

Yes

Got it, thank you

[crystal vale]

I proved that \sigma X(h) ≥ 2 | H |, now how do I use it?

[velvet hull]

think about what elements in G could vanish on X

i.e. what elements are in the kernel of the action

[crystal vale]

Which are not in any conjugation of H

[crystal vale]

Kernel is H

[topaz solar]

Do we ever have cases where G \wr H (reduced wreath or full wreath) is mysteriously finitely generated even if H isn’t?

[topaz solar]

I may simply be stupid, but I meant H infinite but finitely generated

[tidal schooner]

No, G \wr H is f.g. iff G and H are both f.g. See Theorem 1 here for a reference

[topaz solar]

I see I see, checks out

[twilit wraith]

i have to prove using group actions that the stabilizer of i for i in {1,2,...,n} is a subgroup of Sn

what im thinking of doing is that having the group action be sigma * x = i for any x in {1,2,...,n}

so then the kernel of this group action is exactly the stabilizer

you know i had a question coming in but im pretty sure just saying this solved my own problem

[crystal vale]

I don't want a hint, I just want to know what I have to prove in part a?

[velvet hull]

what part of the question statement do you not understand

[crystal vale]

What does it mean by invariant under right multiplication by H, does that mean the coset Hg such that HgH = Hg

[crystal vale]

So the number of orbits of size 1 under group action restricted to H ?

[covert cliff]

how can we make C(X) naturally an abelian group?

[rocky cloak]

Yes,
(f+g)(x) = f(x) + g(x)

[covert cliff]

im actually stupid

or tired

or both

but thank you

[covert cliff]

I forgot addition exists ngl, I was multiplying

[pulsar idol]

hey somewhat irrelvant but what are your opinions on judson

im going through it but im worried im not learning anything deep (as in im worried the problems are too easy as ive heard)

[thorn jay]

that even makes it a ring

sheaf of continuous functions 🙏

[crystal vale]

Is that true when G is not finite

https://math.stackexchange.com/questions/4770051/textbfn-ghh-is-the-number-of-right-cosets-of-h-in-g-invariant-unde

I don't understand the counterexample given there

[rocky cloak]

HgH = Hg
if and only if
gHg^-1 is a subset of H

But g is in NG(H) if and only if gHg^-1 is equal to H.

If H is infinite it can happen that gHg^-1 is a proper subset, witch would mean HgH = Hg without g being in NG(H). So you don't have this natural correspondence between elements of NG(H)/H and these cosets.

However it seems plausible to me that in the cases where this natural correspondence fails, both of these sets will be infinite, so then they will probably have the same size even though there's no natural correspondence between them

[crystal vale]

Okay

[white oxide]

Does this work? Let $h: M \to N$ be a monomorphism of $A$-modules. Let $M' = \ker(M \otimes B \to N \otimes B)$. Since $C$ is flat over $A$, the following sequence is exact:
[0 \to M' \otimes C \to (M \otimes B) \otimes C \to (N \otimes B) \otimes C] Moreover, the following diagram commutes, where all vertical arrows are isomorphisms:
[
\begin{tikzcd}
&& {M \otimes C} & {N \otimes C} \
0 & {M' \otimes C} & {(M \otimes B) \otimes C} & {(N \otimes B) \otimes C}
\arrow["{h \otimes 1}", from=1-3, to=1-4]
\arrow[from=1-4, to=2-4]
\arrow[from=2-1, to=2-2]
\arrow[from=2-2, to=2-3]
\arrow[from=2-3, to=1-3]
\arrow[from=2-3, to=2-4]
\end{tikzcd}
]
Since $h \otimes 1$ is injective, we see that $(M \otimes B) \otimes C \to (N \otimes B) \otimes C$ is injective. Therefore, $M' \otimes C = 0$, and since $C$ is faithfully flat over $A$, $M' = 0$, as desired.

[cloud walrus]

okeyokay

[rocky cloak]

I think maybe you need to label your tensor products if I'm to follow this.

From you're surrounding text it doesn't seem like you're tensoring over A vs B at appropriate times

[white oxide]

Let $h: M \to N$ be a monomorphism of $A$-modules. Let $M' = \ker(M \otimes_A B \to N \otimes_A B)$. Since $C$ is flat over $A$, the following sequence is exact:
[0 \to M' \otimes_A C \to (M \otimes_A B) \otimes_A C \to (N \otimes_A B) \otimes_A C] Now,
$(M \otimes_A B) \otimes_A C \cong M \otimes_A (B \otimes_A C)$. Since $(B \otimes_A C)$ is a $B$-module (via $b'(b \otimes c) = bb' \otimes g(b) c$) we have
[M \otimes_A (B \otimes_A C) \cong M \otimes_A C] Similarly, $(N \otimes_A B) \otimes_A C \cong N \otimes_A C$. The following diagram commutes, where all vertical arrows are isomorphisms:
[
\begin{tikzcd}
&& {M \otimes_A C} & {N \otimes_A C} \
0 & {M' \otimes_A C} & {(M \otimes_A B) \otimes_A C} & {(N \otimes_A B) \otimes_A C}
\arrow["{h \otimes 1}", from=1-3, to=1-4]
\arrow[from=1-4, to=2-4]
\arrow[from=2-1, to=2-2]
\arrow[from=2-2, to=2-3]
\arrow[from=2-3, to=1-3]
\arrow[from=2-3, to=2-4]
\end{tikzcd}
]
Since $h \otimes 1$ is injective, we see that $(M \otimes B) \otimes C \to (N \otimes B) \otimes C$ is injective. Therefore, $M' \otimes C = 0$, and since $C$ is faithfully flat over $A$, $M' = 0$, as desired.

[cloud walrus]

okeyokay

[rocky cloak]

Then I don't see why
M(x)B(x)C -> N.. would be injective, nor would it help since C is not assumed faithfully flat over A

So you can't guarantee M'(x)C = 0 implying M' being 0

[white oxide]

Oh wait oops I got g o f and g mixed up lol

i give up

[rocky cloak]

A hint could be that

$M \otimes_A B \otimes_B C = M \otimes_A C$

[cloud walrus]

Not jagr2808

[white oxide]

Oh I thought that was the hint you gave me before

[rocky cloak]

I mean, I guess it was. But you never used it

So yeah, I guess your proof is essentially what I had in mind except
$M \otimes_A B \otimes_A C = M \otimes_A C$ is not correct

[cloud walrus]

Not jagr2808

[white oxide]

You meant the way that I showed that is not correct?

[rocky cloak]

What I'm trying to say is that it would be correct if you just switched which ring you're tensoring over at appropriate places

[toxic sapphire]

I think I've shown that i. implies ii and that ii implies iii but I'm struggling to show that iii implies i

working the algebra out component-wise I think I can show that the squares of the diagonal entries of A^TA sum to n, but I'm not sure how to show that they're all necessarily 1

(that is by taking v to be the vector with 1/sqrt(n) as each entry)

[thorn jay]

try taking v to be the unit vectors (0, ..., 1, ..., 0)

[toxic sapphire]

that is a fantastic idea wow. ok I'll try again

[thorn jay]

pro tip: matrices like unit vectors

[toxic sapphire]

yeah idk what I was thinking. I'll report back shortly 👍

[thorn jay]

good luck soldier o7

[rocky cloak]

A trick that can come in handy is that v and w are orthogonal iff
|v + w|^2 = |v|^2 + |w|^2

[toxic sapphire]

ok thanks

[toxic sapphire]

pretty sure I got it! some messy algebra in this problem but I just needed to think of the obvious choice of vectors 😅
thanks!

hmm actually not quite. this argument shows that each diagonal entry of A^TA is 1, but not that the remaining entries are 0. in fact, I'm not sure any choice of vector will show that non-diagonal entries of A^TA are 0, since the only vector with norm 0 is the 0 vector. but when v is the 0 vector all terms in ||Av|| vanish anyway

are you sure? I'm talking about the entries in A^TA not A itself

[rocky cloak]

Sorry, I misread what you wrote

[toxic sapphire]

actually I think I solved it, though

[rocky cloak]

But yeah the other entries being 0 means the columns of A are orthogonal

So you want to show that A preserves orthogonality of vectors

[toxic sapphire]

I'm not sure exactly what you're getting at, but I think noting the diagonal entries as being 1 and choosing v = (1, ..., 1) proves that non-diagonals are 0

oh wait nvm, I actually think you get it from v being a vector 0 everywhere except 2 entries

[white oxide]

Why is this line true? My original thought was that we have as x 1/s = a, but q is an ideal of A, not of S^{-1}A so this doesn't make sense

[thorn jay]

there's nothing being done in this line with localisation yet

by definition of primary, a*s in q implies a in q or s in q or s,a in sqrt{q} = p. As s is not in p, and therefore also not in q, a must be in q

primary ideals essentially say that in the quotient every zero-divisor is nilpotent, and as s isn't nilpotent in A/q (it isnt in p, the set of elements which would become nilpotent in A/q), it must be the case that a becomes 0 in A/q, i.e. is in q

[proud vigil]

hydrogen bomb vs coughing baby but you could use the fact that A^T A is diagonalizable from spectral theorem and show from 3 that the only possible eigenvalue of A^T A is 1

[toxic sapphire]

I'll think on that but we have not covered the spectral theorem, thanks for the idea though

[proud vigil]

valid lol

oh i do see a really straightforward way of doing this though

proving that i and ii are equivalent isn't difficult, i -> ii by some manipulation and ii -> i by choosing appropriate unit vectors

proving ii -> iii is also easy, that's just a special case

to prove iii -> ii, we know that for arbitrary v, w,

|A(v + w)| = |v + w|
which expands to
|Av|^2 + |Aw|^2 + 2<Av, Aw> = |v|^2 + |w|^2 + 2<v, w>

and iii simplifies this down to

<Av, Aw> = <v, w>

which is exactly ii

i guess in other words iii -> ii -> i -> iii might be a tiny bit more straightforward than i -> ii -> iii -> i

since i -> ii and i -> iii are kind of the same principle anyway

[toxic sapphire]

yes I actually thought about this briefly (not in detail) after proving i -> ii -> iii but I figured I'd be better off getting through it with the current route since I'm pretty short on time today

but I will probably return to this problem in the future

[thorn jay]

ITS BACK

[toxic sapphire]

the problem?? 😨

[thorn jay]

no it's just that mathcord was down for like half an hour for me

[toxic sapphire]

oh lol. not sure if mine was, I was in lecture

[velvet hull]

ok its not just me

ive been like periodically semi-banned and unbanned from this server for the past hour

[thorn jay]

I was so scared I got shadowbanned

lol

[velvet hull]

its not just this one, it seems to be an issue with "official" discord servers

because the official obsidian server which im in experiences the same shutdown at the same time

[thorn jay]

hm

maybe an issue on discord's side then

[velvet hull]

some shits going down at discord HQ i assume

[thorn jay]

shit's boutta be on the news fr

[glad osprey]

the mods have had enough of you talking about quandles and universal algebra and stuff

[thorn jay]

it appears so

😔

[toxic sapphire]

I'm not really sure how to approach this problem

it's been a while since I did linear algebra as well, so I'm shaky on important results and intuition

what does a matrix of rotation of R^3 around l by theta look like?

[sonic coral]

what’s your definition of SO_3(R)

[toxic sapphire]

the subgroup of GL_3(R) consisting of n x n invertible matrices A with A^TA = Id and det(A) = 1

[toxic sapphire]

but we did prove these conditions were equivalent to that definition of orthogonality earlier in the problem set

[sonic coral]

since you know it’s elements are orthogonal (or unitary) operators, it’s eigenvalues must be roots of unity

[toxic sapphire]

a unitary matrix is one whose transpose is equal to its inverse, right? are orthogonal matrices necessarily unitary?

[sonic coral]

unitary is conjugate transpose = inverse. orthogonal is transpose = inverse, so orthogonal implies unitary

[toxic sapphire]

ah ok. I thought we could have A^TA = Id but AA^T != Id, since matrices don't generally commute with their transpose

[tardy hedge]

Lol

THIS MAN AND HIS QUANDLES!!!

[untold belfry]

Matrices generally commute with their inverse though.

[languid trellis]

Quick question, \Omega denotes the set of algebraic integers, (roots of monic polynomials with integer coefficients). I don't see how this proof fails in the more general case of alpha being an algebraic number (roots of polynomials in Q[x]). Am I correct in thinking the result holds in the case where alpha is an algebraic number?

[rocky cloak]

Yes, Q(alpha) = Q[alpha] iff alpha is algebraic

[languid trellis]

Awesome thanks : )

[crystal vale]

I showed that the group of order 30 is not simple, but I want to show that if G has order pqr, p < q < r, then r-Sylow subgroup is normal.

In general i showed that the group of order pqr is not simple

[velvet hull]

what can you say about the quotient group?

(suppose that we are quotienting by some p or q sylow subgroup for now)

[crystal vale]

Can we do that?

[velvet hull]

well how did you show that a group of order pqr was not simple

[crystal vale]

By counting argument

[velvet hull]

yes, you count the number of elements of each order right?

a corollary of the argument you used should be that G has to have a normal sylow subgroup

[crystal vale]

Yes

[crystal vale]

Yes

[velvet hull]

so WLOG let it be a sylow p or q subgroup

what can you say about the quotient

[crystal vale]

It has order pr

[velvet hull]

pr or qr, yes

what do we know about groups of order pr?

[crystal vale]

Cyclic, because p can't divide r-1

[velvet hull]

is "p cant divide r-1" an assumption of the question you were given?

because it can

[crystal vale]

p < q < r

[velvet hull]

doesn't mean p can't divide r-1

p=2, r=13

[crystal vale]

Oh

[velvet hull]

there is still something you can say about it

[crystal vale]

the r Sylow subgroup in quotient is normal

[velvet hull]

yes, now try and figure out the question from here :)

[crystal vale]

I can say subgroup of order qr is normal in G

Yes

Yes

but p,q,r are all prime

[velvet hull]

very good, think more about that

that is very close

this question is very hard to do without using the Sylow theorems, mq

you need all 3

[crystal vale]

I am thinking that subgroup must be abelian

[velvet hull]

not necessarily

[crystal vale]

Group of order qr, q < r, q Sylow subgroup is already normal

[velvet hull]

yes, but groups of order qr do not need to be abelian

unless q does not divide r-1, but that is still not necessarily the case

[crystal vale]

q is normal, r is normal

Isn't make sense?

[velvet hull]

dihedreal groups are like that

they are not abelian

[crystal vale]

How?

[velvet hull]

take any prime diherdral group

[crystal vale]

But then subgroup of order 2 is not normal there

But why do you think it is not true

Group of order 2 is not normal

[velvet hull]

wait, what do you mean by this

[crystal vale]

Say G has order qr, where q < r and q has normal subgroup.

Now take H be normal subgroup of order q and K be normal subgroup of order r.

Now H commutes with K, so G = HK is abelian group

[velvet hull]

why are you sure H and K exist

[crystal vale]

By Sylow

I assumed G has q-Sylow subgroup which is normal

[velvet hull]

can I see your sylow proof?

[crystal vale]

G has order qr, right?

Now i assumed G has q-Sylow subgroup which is normal, say H

Now by Sylow we have existence of r-sylow subgroup

Say K

K is normal, right?

[velvet hull]

okay, that is fine, but why did you assume G has a normal sylow q subgroup

[crystal vale]

Here, we have q Sylow subgroup which is normal so that will be normal in subgroup too

[velvet hull]

sorry, I do not understand what you have done here

can you pretend I did not hear anything just now and repeat all the steps that leads to this

[crystal vale]

Forgot it

Now I have subgroup which have order qr and it is normal in G

Any further hint?

[velvet hull]

yes

now look at that subgroup

what can you say about that subgroup

[crystal vale]

I am thinking that subgroup will be abelian

[velvet hull]

it's the same issue with the quotient group

if you do not tell me that q does not divide r-1, then it can be nonabelian

[crystal vale]

Okay

Simply say, H is subgroup of order q, K is subgroup of order r

We know H is normal in G

So HK makes subgroup of order qr

Which has index p

Which is smallest divisor of | G |

So HK normal in G

Now H is normal in G, H is normal in HK

[velvet hull]

okay, that's an alternate proof to show why G has a normal subgroup of order pr

so what next

[crystal vale]

So HK is abelian, right?

Oh K is normal in HK

HK cyclic

So every subgroup must be normal in G

So K is normal in G

Every subgroup of HK

[velvet hull]

okay, I see what you do there

I think this works, not the proof i had in mind but I think this works

[crystal vale]

Thanks @velvet hull

[velvet hull]

here was what I was thinking - any group of order pr has a unique sylow subgroup of order r.
but normal sylow subgroups are characteristic, so r char in pr, pr normal in G implies r normal in G

[crystal vale]

I see

Normal Sylow subgroups are characteristic is the main thing, yeah great

Can we prove that every group of order p^m q^n, is not simple, m,n≥1? Without using any advanced tool? I don't want a solution

[velvet hull]

you're thinking of burnside's theorem, which has a very nontrivial proof

but it is true

in fact groups of order p^mq^n are solvable

[crystal vale]

professor has not introduced about solvable yet

[quiet pelican]

Yup

[crystal vale]

Yes

[quiet pelican]

Every element generates some subgroup, then Lagrange implies subgroups are either trivial or the whole group

[rapid cave]

The first proof found used representation theory if I remember correctly

[velvet hull]

yee

[crystal vale]

Oh I don't know the representation proof

[crystal vale]

Any other hint?

[velvet hull]

it's an advanced proof that's beyond me

[quiet pelican]

And also iirc there are known proofs without, but they were found much later?
Either that, or they don’t exist
Burnside’s theorem and “Frobenius kernel is normal subgroup” both originally needed Rep theory, and I remember one still ‘needs’ it, and the other only had another proof found a long time after it was originally proven

[rapid cave]

lcm(p, n) = pn if p doesn't divide n

otherwise lcm(p, n) = n

[karmic moat]

A lot of algebra is motivated by number theory lol

[velvet hull]

naive number theory is 90% abstract algebra

you are essentially studying number theory right now

wilson's theorem and fermat's little theorem are easy group theory proofs

[south patrol]

And then non-naive is 50% depending on whether you do analytic stuff or not

[karmic moat]

Chinese remainder theorem

Potato i have a joke

Ready

[south patrol]

Wilson's theorem has a funny proof

Yeah I am ready

[karmic moat]

What did the mathematician name their fabrics shop

Fiber products

[south patrol]

What like "group theoretic" way do u have in mind

[south patrol]

Yeah so I discovered this joke the hard way

[crystal vale]

[velvet hull]

lmao which one?

[south patrol]

When googling

[velvet hull]

bc they're both p straightforward

[south patrol]

Wilson

[karmic moat]

Aw shit

I came up with it yesterday

Fawk

[south patrol]

It is funny!

[velvet hull]

product of all the elements of a finite abelian cyclic group is the identity iff the group is not isomorphic to Z/2Z

[karmic moat]

Theft is no joke

Preinage of a point basically

Preimage

[south patrol]

What about Z/4 (additively)? The sum is 1 + 2 + 3 + 0 = 2

[velvet hull]

oh, wait it should be cyclic

[south patrol]

Yeah sure

[velvet hull]

if it's abelian it needs to be modified to talk about the set of 2-torsion elements

[karmic moat]

Fiber product is different I think

[velvet hull]

but you dont need that for wilson's anyway

[karmic moat]

I think of it as base change

The intuition for it comes from taking fibers tho

[south patrol]

The proof I know is quite funny. Consider the polynomials (x-1)...(x - (p-1)) and x^(p-1) - 1 over Fp. The difference is of degree < p-1 and has p-1 roots, so it is the zero polynomial. Now take constant coefficients

[velvet hull]

ahh, that's spicy

[south patrol]

It's cute

[velvet hull]

with that proof you can modify to work for F_p^n

[south patrol]

O nice

[velvet hull]

which is also cool

[south patrol]

Yeah if you have no elements of order 2 you just pair up with inverses but then idk how it works with them

Like for a converse

[karmic moat]

It’s called product because it’s like the cartesian product

Anyway your statement doesnt sound like it should be true

[south patrol]

To like avoid doubt like fibre product of maps A -f-> C <-g- B is the set of points (a,b) in A x B with f(a) = g(b)

[velvet hull]

the set of all order 2 elements in an abelian group is a subgroup

[south patrol]

Yeah

[velvet hull]

so it's just n copies of Z/2Z

and you look at that

[south patrol]

True that is nice

Yeah

[karmic moat]

who says your map satisfies f(ab) = f(a)f(b)

[south patrol]

Lol been thinking about smth for a while and found a paper that does what I need

[karmic moat]

W

[south patrol]

Only thing was I was like happy to nearly have it and then this guy has beasted me but that's very ok

[karmic moat]

that's how i feel about my fiber product joke

[south patrol]

It was gonna be a remark that I like only sketched the proof of cause I never use it

[karmic moat]

conversely, my friend had a paper he wanted to read

[south patrol]

Great way of doing it

[karmic moat]

but it was in french and it was old so chatgpt couldnt really translate it

[south patrol]

Cause like I don't feel bad cause I never use it idk

[karmic moat]

so i said "okay i'll translate it" so i spent like 2-3 days putting it through chatgpt and other stuff and fixing errors and stuff

[south patrol]

Dawg just read the French

[karmic moat]

he's an international student 😭 that guy dont know french

so i sent the translation to him

and he was like "ahhh this paper actually doesn't have what i need"

heartbreaking

[south patrol]

No I mean like a lot of non-french-speakers read French math including me

[south patrol]

LOL rip

😭

One time I had to translate a bit of a German paper for a friend but I am glad I didn't have to like translate the whole thing

[karmic moat]

anyway in case anyone ever needs it attached is my translation (Denef-Sargos.pdf) and the original (asdf.pdf)

[south patrol]

Asdf

I like trains.

[karmic moat]

fair enough

forgot about asdfmovie

wow

[south patrol]

But nah I mean depends like if u don't have time etc

Just for AG a lot of important stuff is in French

[karmic moat]

unfortunately that is the case indeed

but french mathematicians have such an elegant way of writing

like serre is a great writer

[south patrol]

Real

Goated...

[karmic moat]

there's some undergrad math club meeting tonight and i went to the one last time bc my phd friend presented some stuff

if i go to this one it'll be kinda a waste of time because i have stuff to study

[south patrol]

I liked reading some of BOURBAKI ALGÈBRE

[karmic moat]

but also free pizza 🤤

[karmic moat]

at some point in time i would really like to read EGA or SGA

i know there's some incomplete english translations online

[south patrol]

Go for free pizza

[karmic moat]

but i wanna read the french one

[karmic moat]

ye

[south patrol]

Lol

Tomorrow I am going to a conference

[karmic moat]

W

what's it about

[south patrol]

Algebraic Geometry

Lol

Heh

[karmic moat]

oh yeah i guess that's obvious

lol

[south patrol]

But quite broad

It isn't a silly question at all

[karmic moat]

there's the AMS fall sectional in st louis missouri

I'll probably go

[south patrol]

Could be more specific or another area etc

Pog...

[karmic moat]

true

i went to some last year on like moduli stuff

understood none of it

[south patrol]

Sectional when the retractional appears

[karmic moat]

but it was fun anyway

[south patrol]

Valid

This conf has stuff that is not my area at all

[karmic moat]

conferences always got awesome snacks

like the morning snacks with the coffee and the afternoon snacks and stuff

[south patrol]

Hopefully

[karmic moat]

i went to some utah conference last year and it had the best brownies ive ever had

[south patrol]

How many has bro been to

[karmic moat]

traveled for 2, one was hosted at my school, and one is this small conference between my school and nearby schools

and then some undergrad symposium to present a poster

those 2 i traveled for were more like "i want to go kiss some ass and try to get a phd offer"

and it gave me an excuse to visit some friends who were nearby

[south patrol]

Nicee

[south patrol]

LOL

[south patrol]

Posters scare me ngl so gg

[karmic moat]

actually not even "more like" it was purely to do that

like i had just "finished" chapter 1 of hartshorne (ie read it and pretended like i knew it and then realized i didnt know shit)

and i wanted to go kiss some ass

[karmic moat]

yea someone asked me a question and i was like "ummm idk actually"

"i just wrote what my advisor told me to write"

lesson learned, think about everything you put on a poster

[south patrol]

Valid

Just claim big conjectures on ur poster

[karmic moat]

i also finished the poster like 12 hours before my flight

[south patrol]

And say proof is secret

[karmic moat]

true

my friend said that mohan kumar's papers are really short because he has some huge statement that he handwaves in a few sentences

and he said that each of those handwaved statements can be made into a whole paper

[crystal vale]

Let | G | = p^3q, we have to show G has normal sylow subgroup.

it is clear that if p > q, then p-sylow subgroup is normal.

Now, say p < q, so if i write n_p as number of sylow p subgroup, similiarly n_q, then by assumption there is no sylow normal subgroup, n_p should be q.

Now n_q has two choices p^2 or p^3.

Case 1 : Say n_q is p^2, that means q divides p^2 -1 , but since p < q implies q divides p +1, it is possible when p = 2, q = 3.

so now explore the group G which has order 24. Since G has subgroup of order 8 which has index 3, by group action we get non-trivial normal subgroup contained in H.

i have to think how can i handle this case. any hint?

case 2: then n_q is p^3 so p^3(q-1) + (p^3-1) + 1 = p^3q, but since n_p = q, so total elements in G will be more than p^3q

[crystal vale]

i think it is not necessary that if G has order 24, then G has normal sylow subgroup

say S4

[south patrol]

lol

[crystal vale]

potato, can you verify one argument?

[crystal vale]

this one

[karmic moat]

root vegetable

[thorn jay]

capitalisation necessary

[velvet hull]

wait, why can't n_q be p

[crystal vale]

then q must have to be divide p -1

[velvet hull]

what's wrong with that

[rocky cloak]

If p < q that would be hard

[rocky cloak]

I mean, S4 does not have any normal sylow subgroups

[crystal vale]

Yes

I tried this problem last 30 minutes, and what I get, the tutor made a typo 🙂

i am in post graduation

maybe i don't understand what do you mean? In our school they assign phd students or post doc students as our tutor

[velvet hull]

wow, nice school

[crystal vale]

isn't common in other university?

[velvet hull]

wait, do you mean advisor

[crystal vale]

not advisor, advisor for phd students, right?

[thorn jay]

in our uni weve got TA's which are just 2nd/3rd years

i guess they arent direct tutors but during study groups you can ask them stuff

[crystal vale]

in phd 2nd year/ 3rd year?

[crystal vale]

exactly

[thorn jay]

no

bachelor

[crystal vale]

so you are in bachelor?

[thorn jay]

probably in higher years youll get master students

[thorn jay]

first year, yes

[crystal vale]

are you kidding?

[thorn jay]

no? why

[crystal vale]

is it true that every group of order p^m q^n, m>=1, n>=1 is not simple?

[crystal vale]

you know lots of advance stuffs

[thorn jay]

and my knowledge and skills are severely lacking in other places lol

so it balances out

[crystal vale]

may i ask, in which school you are in?

[thorn jay]

Radboud university

in Nijmegen

[crystal vale]

okay

[thorn jay]

english people wouldnt be able to easily lol

it has a soft g

N-I-m wagen

the combination "me"

your mouth naturally makes a "w" sound as it transitions from the m to e

at least, when you pronounce it the dutch way

close enough

good job

not nice to live in

Nijmegen is much better

[glad osprey]

Is this pronunciation from wikipedia not correct? https://upload.wikimedia.org/wikipedia/commons/transcoded/5/52/Nl-Nijmegen.ogg/Nl-Nijmegen.ogg.mp3 Or are there different dialects?

[thorn jay]

yeah thats correct

i tried to get the ij sound in english

[glad osprey]

that sounds like an m-sound more than w to me

but it also sounds like an alien, non-human language, so what do I know

[thorn jay]

ah yeah it does there

how i pronounce it is with a very clear w sound though

maybe cuz im from Brabant?

[tardy hedge]

everyone here is so european

[coral spindle]

in de schipol. straat op "gejorking het"

[south patrol]

Lmao

[cedar vault]

It is true by burnsides theorem

[rocky cloak]

It is indeed
https://en.m.wikipedia.org/wiki/Burnside's_theorem

Burnside's lemma is also funnily enough not due to burnside in any way

[delicate orchid]

my favourite finite group is
"

G"

[thorn jay]

burnside's lemma, also know as "the lemma that is not burnside's"

[rocky cloak]

Looks good to me

Idk what's in df, but there are a few different ways to prove this yeah

[covert cliff]

d&f does tend to be a yap fest lmao

[delicate orchid]

pick the right map and it's a one-liner

[languid trellis]

bro when i discovered this i literally transcended. first iso is the only thing that requires any work

[crystal vale]

any hint to show group of order p^nq^m, n>= 1, m>= 1, is not simple, burnside lemma?

[languid trellis]

and the second and third iso are just first iso with the right map

shits wild

[covert cliff]

first iso is the only reason I learned group theory

[languid trellis]

if by 4th iso you are referring to the correspondence theorem then you are simply wrong. that is a goated thm

dawg that shit is so useful. get it in ur veins

[delicate orchid]

correspondence theorem tends to hold even more generally than first iso so imo it's even more goated

[languid trellis]

trust me its worth

I just rolled my eyes

eat ur vegetables

silly boy

did your parents never tell you that to grow strong and healthy you need to eat ur veggies

[delicate orchid]

the 4th is just as easy to prove as the first I don't get it

I'm gonna stick with subgroups for this one but like. By quotienting out by some H you're collapsing everything under H to a point without touching anything above H

so there's an obvious correspondence there

like think about the posets given by inclusion of subgroups, then what I'm saying is literally what happens in that poset when you quotient

[thorn jay]

this tbh

like its not even anything fancy its just a lattice correspondence

not even the amount, but youve got maps between the subgroups of G containing N and the subgroups of G/N:
H ↦H/N
K ↦π^-1(K)
where π : G → G/N is the canonical projection, and the theorem says that these maps are inverses and preserver arbitrary intersections and joins

[glad osprey]

It's even a lattice isomorphism, right?

[thorn jay]

or, equivalently, they are isotone (bijective and X < Y implies f(X) < f(Y) )

[thorn jay]

thats what I meant haha

well the bijection respects inclusions

thats the real important part

and K < K' => π^-1(K) < π^-1(K')

?? π for projection

its really common

[thorn jay]

read

[limber tapir]

The map that sends g to its coset in G/H

Canonical just means it's the most "obvious map"

[peak condor]

Hey sorry to interrupt, but i was curious if anyone knows any good tutors for group theory?

[thorn jay]

its canonical because its obvious and uniquely defined by the properties youd expect a quotient to have

[coral spindle]

You could have many more maps G -> G/N that aren't nice. For example the map sending everything to the identity always exists.

[thorn jay]

i.e. its the morphism satisfying the universal property of the quotient

[delicate orchid]

just couldn't help urself could u

[thorn jay]

abstract nonsense is a formalisation of what we handwaved and called "obvious"

uh

familiarity with general algebra like groups, rings, vector spaces n stuff

and id recommend a good grasp on at least the basics of order theory

although there is a chapter on the necessary lattice theory in the book i used

so youll probably learn enough from there

[velvet hull]

hint - ||recall that the product of a normal subgroup and a subgroup is a subgroup||

[chilly radish]

@hushed granite

so

You can get C from R by adding 1 number

which number would that be

[hushed granite]

i

probably

[chilly radish]

right

a square root of -1

If we want to be more agnostic about it, we can say that we are taking a root of x^2+1

The way to do this formally is to take a ring quotient of R[x] by the ideal generated by (x^2+1)

don't worry if you don't know what that is

basically it means we are introducing a variable and forcing the relation x^2=-1 on it

[hushed granite]

oki

[chilly radish]

Ok so now we can ask what kind of "symmetries" does C have as a field extension of R

in other words, we're looking for isomorphisms C->C which fix R in place

so this would be determined by where we send i

right?

[hushed granite]

to some degree yeah

[chilly radish]

this is an isomorphism as fields so it preserves addition, multiplication etc.

[hushed granite]

mhm

[chilly radish]

but for every x\in R, f(x)=x

so for a general element we have

f(a+bi) = f(a) + f(bi) = a+f(b)f(i)=a+bf(i)

so to understand this automorphism we only need to understand f(i)

[hushed granite]

mhm

[chilly radish]

can you tell me where f(i) is sent?

there's not many options

[hushed granite]

f(i)=i

?

[chilly radish]

that's one option

what's the other

[hushed granite]

f(i)=-i

[chilly radish]

right

because we still have that f(i)^2 = f(i^2)=f(-1)=-1

[hushed granite]

yep

[chilly radish]

so the group of automorphisms, which is just the set of isomorphisms C->C fixing R pointwise is just two elements

so the group with 2 elements

call it {Id,f} where f o f = Id

this is the Galois Group of C/R

the / here isn't a quotient

[hushed granite]

what was f?

[chilly radish]

it's read as "over"

f is f(i)=-i

[hushed granite]

ah ok

yeah i remember seeing this with deltoid

[chilly radish]

ok so now, the reason why polynomials in R factor as quadratics and roots

and then i'll say something about galois correspondence and then i'll go back to grading

[chilly radish]

If you have a polynomial p(x)=a_n x^n + ... + a_1x + a_0 with coefficients in R

suppose that z \in C is a root of this polynomial

so p(z)=0

but I can apply f to this number

what is f actually

you know it by a different name

[hushed granite]

meaning you can apply f(0)? or wdym

or f(z)

[chilly radish]

f(p(z))

yea

which is f(0)=0

what is f as a function

what does it do to a general complex number

[hushed granite]

.

with f(i) = -i

so it is the conjugation?

[chilly radish]

yea

exactly

so we get that f(p(z))=p(f(z))

because p has real coefficients

i.e. if z is a root of p, then so is its conjugate

[hushed granite]

yeah

[chilly radish]

and those give a quadratic polynomial in R

ok so now in general

you wanna study galois extensions

which are when you take a polynomial with distinct roots, then you take the smallest extension containing all of its roots

so for example, Q(sqrt(2)) is Galois

because it contains all of the roots of x^2-2

over Q

but Q(cbrt(2)) is not Galois

where cbrt is the cube root

because it does not contain all of the roots of x^3-2

there is one complex root

you still with me?

[hushed granite]

yeah

so essentially you try to make C "smaller" in a way to fit for your problem?

[hushed granite]

by wanting to achieve this

[chilly radish]

It's less of that and more of, when the polynomial has real coefficients, I can "descend" my splitting from C to R

oh

I misunderstood

[chilly radish]

It's just that you want just all the roots you need

and nothing more

[hushed granite]

yeah

but... why? C does the job no?

[chilly radish]

well

[hushed granite]

it's just.. well.. big

[chilly radish]

because then something magical happens

[hushed granite]

o.O

[chilly radish]

So, if you havea galois extension K/F, i.e. you add all the roots of a separable (all distinct roots) polynomial f(x) with coefficients in F

then you can form the Galois group, which is the group of automorphisms of K/F. That is, isomorphisms K->K fixing F pointwise

like we did before

[hushed granite]

ok

so you can characterize possible mappings?

[chilly radish]

yea

and it all comes down to permuting the roots

[rapid cave]

Galois theory ❤️

[chilly radish]

like we sent i to either itself or -i

we can move the roots around

and the ways we can do that in determine the structure of the galois group

[hushed granite]

So essentially we can wiggle around the roots, but the space doesn't budge too much?

[rapid cave]

recently there a few more

devs of expedition 33 ( I think they are french???)

[chilly radish]

Idk wjat you mean by space exactly

there's constraints on which roots can go where

when you get to higher degrees

[hushed granite]

Well the galois extensions gives me possible functions f:K->K, right?

[chilly radish]

but the thing is, the structure of this group tells you the structure of the field and vice versa

[chilly radish]

yea that preserve F

[hushed granite]

where i can wiggle around where my roots are being "placed"

[rapid cave]

and are homomorphisms

[chilly radish]

Subgroups of the Galois group are in one-to-one correspondence with subfields of K containing F

that's the magic

You can understand the lattice of subfields by studying the group, and udnerstand the group by studying the lattice of subfields

[hushed granite]

So essentially there exists a bijection?

[chilly radish]

yes

and it has nice properties

[hushed granite]

hmm

now i see where we are going

[chilly radish]

like the degree of the field extension being equal to the index of the group

[rapid cave]

(and its even true in the infinite galois extension case using Krull Topology!)

[chilly radish]

right

[hushed granite]

mhm

mhm

very nice

[chilly radish]

and this can be used to even do explicit things

like there's constructability which is a nice example

[rapid cave]

and solvability of quintics

[chilly radish]

right, i'm getting to that

[tardy hedge]

LET HIM COOK

[chilly radish]

do you know what a constructible number is?

[hushed granite]

no

[chilly radish]

so it's basically a number that can be constructed as a length via straightedge and compass constructions

so there's 2 constructions you can do

a line between 2 points

and a circle determined by 2 points

this is either a linear polynomial, or a quadratic one

so every step either adds a square root, or doesn't add anything

if we do a lot of these in a row, we get a tower of field extensions $F\subseteq F(\sqrt{a_1})\subseteq \sqrt F(\sqrt{a_1},\sqrt{a_2})\subseteq \cdots \subseteq F(\sqrt{a_1},\ldots,\sqrt{a_n})$

[cloud walrus]

ShiN

[chilly radish]

and any number in one of these types of field extensions is constructible

[hushed granite]

hm

[chilly radish]

because (and this is a non-trivial fact) constructible numbers form a field

so the sum and product and quotient of constructible numbers is constructible

so, the degree of this extension (dimension as a vector space over F) is 2^n

assuming I threw out all the square roots that were already in the field

You can actually prove using this that if cos(2pi/p) is constructible, where p is a prime, then p is a fermat prime

i.e. of the form 2^k+1

the proof is a bit technical but it uses the the structure of the galois group to prove something about the degree

also think of cos(2pi/p) as the length of the side of an equilateral p-gon

The other cool thing is the abel-ruffini theorem, which states that you can't solve a degree 5 polynomial

[rapid cave]

using radicals

[chilly radish]

yea using radicals and regular operations

so the way the proof goes is you prove a polynomial is solvable by radicals iff its splitting field (adding all the roots) is a radical extension, i.e. you can make it using a tower of taking n-th roots

now you use a fact about the structure of such extensions, and the kinds of subfields it can have

together with a fact about the galois group of the splitting field of a specific degree 5 polynomial

the fact is that there's a polynomial with galois group S_5 which is not solvable

ok back to grading, hope you enjoyed this

sorry for being vague, I just didn't wanna dump a tonne of definitions on you

[rapid cave]

grading what

[chilly radish]

analsis

[rapid cave]

probably a pain

right?

[chilly radish]

yea

very much so

[thorn jay]

have you no dignity...

By the hypothesis, there must exist a subgroup H < G of order p^n-1. By Sylow and induction, H must contain groups of all orders p^k for k < n-1, so in fact by the hypothesis must contain all proper subgroups of G. Therefore, considering any a in G that is not in H, it must generate G, as else it would generate a subgroup of order p^k for some k < n, and therefore be in H.

bro has one joke

its not an element?

G \ H means the difference of sets G and H

not G/H

zero reading comprehension smh

right

whatever you say, bud

[tardy hedge]

Lol

[tough raven]

Unital joke ring

[dim wagon]

im confused about this exercise

Isn't it true that every R-module is an abelian group and every R-module homomorphim is a group homomorphism

and thus every R-module homomorphism is an abelian group homomorphism?

[quiet pelican]

“Show that Z module homomorphisms are the same thing as abelian group homomorphisms” is probably a better way of phrasing it
Ie it also wants the reverse direction

[dim wagon]

ahhh, i see

thanks for the clarification

[dim wagon]

could i check if I am correct here?

that the forward direction would actually hold for any R

[quiet pelican]

Yes

[dim wagon]

okay thank you

[languid trellis]

I'm struggling with 5. Showing Gauss' Lemma wasn't much issue (question 4). So far, I've seen that we can pick g(x) in Z[x] of minimal degree such that g(a) = 0. Then we also know that g(x) is primitive (as it's monic by definition), and irreducible in Z[x]. But, by definition of f, we know that f(x) | g(x). So g(x) has a nontrivial factorisation in Q[x], say g(x) = f(x)h(x). I don't really know how to proceed from here. I know I need to use that g is primitive and Gauss' lemma but the way forward isn't so clear. Can anyone help?

[languid trellis]

Okay wait. we have our factorisation g(x)=f(x)h(x). We can clear denominators then divide by the gcd of the new coeffs to get that g(x) = a/bf'(x)h'(x), where f' and h' are in Z[x] and are primitive. Then bg(x) = af'(x)h'(x).

I want to conclude that a/b is an integer and then we get a nontrivial factorisation but I'm unsure how. I also haven't used gauss' lemma yet

[rocky cloak]

So f and h are monic polynomials so there is an integer c such that
cf is a primitive polynomial, and similarly for h

Then g times an integer being primitive means that integer is ±1

[languid trellis]

We can clear denominators of f, then divide by the gcd of the new coeffs to get something primitive, and I understand that this has to be an integer because if it is a rational then our leading coefficient won't be an integer. Is that the reasoning?

[rocky cloak]

Yup, if you multiply a monic polynomial by a rational number q, then the leading coefficient is q

[languid trellis]

a, b are integers such that af and bh are primitive. so then we have abg = afbh, with af, bh primitive. So then (ab)g is primitive, but the gcd of the coeffs of (ab)g is ab, so ab = +-1, which shows we have a nontrivial factorisation of g in Z[x], contradicting our hypothesis.

Thank you!

[marsh fulcrum]

Let R be an integral domain with infinitely many elements, of which only finitely many are irreducible.
Suppose every non-unit has an irreducible factor. Show that R has infinitely many units.

[hexed cargo]

This feels like Euclid's proof that there are infinitely many primes

Like, let $p_1, \ldots, p_n$ be all the irreducible elements of $R$. Then $(p_1 \ldots p_n)^k + 1$ for $k = 1, 2, 3, \ldots$ are all distinct elements without an irreducible factor, so they must be units.

[cloud walrus]

Adayah

[rapid cave]

I don't think this works.

becuase if $N_k = (p_1 p_2 p_3 ... p_n)^k + 1 $ then all we know is that $N_k \notin (p_i)$ for all $1 \leq i \leq n$, this generally doesn't mean that $N_k$ is unit or equivalently it is not in any maximal ideal. I think your proof will work when R is a PID

[cloud walrus]

ExpertSqueeSQUEE

[rapid cave]

.

[quiet pelican]

Doesn’t it work because we’re assuming we’re working in a ring where “not divisible by any irreducible” => “is a unit”?

[rapid cave]

Oh I missed that line

hmmmm let me think

ok yeah I think it works

[tardy hedge]

what are the minimal prime ideals of an ideal? Ideal contained in p for p prime and no other prime contained in between?

is this related to primary decomposition in some way? I forget

So for I = cap qi a primary decomposition, rad(qi) is a prime ideal containing I

[tardy hedge]

Ohhh

Wait im thinking about it again this is making sense

if I is a subset of cap ai then I is a subset of ai for each i right

just for sets

[thorn jay]

alternatively, looking at Spec R as a topological space, they correspond to the maximal irreducible components iirc

[tardy hedge]

that shi all so cool. I like math

its hard though

[thorn jay]

definitely lol

[tardy hedge]

I wonder if ill find the time to learn it once i start working

I will definitely miss it tbh

but at the same time learning it while being in academia was more stressful than fun

for me

It also doesn't allow me to grow other aspects of my life as effectively

I'm almost 25 yk, I don't want to be stressed about money, want to settle roots down somewhere ....

everyone is different though i guess

[marsh fulcrum]

Guys please do this one I M badl y stuck

[marsh fulcrum]

Lol I have an exam in 6 hours
Slept 3 hours barely
I know what you saying

[tardy hedge]

are you in undergrad

[marsh fulcrum]

Msc @tardy hedge

[tardy hedge]

cool same im almost done

[marsh fulcrum]

Bro please do it if you can this might come in exam

[tardy hedge]

im taking differential geometry this term and just finishing my thesis

[marsh fulcrum]

Closer to freedom eh

[tardy hedge]

yeah lol

[thorn jay]

damn, wouldnt imagine you taking that

[marsh fulcrum]

Finally
You can rejoice

[thorn jay]

thought you were way more algebrapilled

[tardy hedge]

Me? Yeah man, it wouldnt be my first choice. However my school isnt very good

[marsh fulcrum]

Guys.
Please do this one
Like
This may come in the exam

[thorn jay]

right i see

[tardy hedge]

they have category theory ppl and sometimes they offer that sort of thing

[marsh fulcrum]

Prof made this question himself in the sheet of practice problems for midsem

[tardy hedge]

wait in fact let me check if they have cat theory rn

i mightve just missed it

[thorn jay]

i looked at the master options for radboud and there are like three options for math

maybe 4

[knotty badger]

Cat theory :D

[tardy hedge]

ya like why would i take diff geo idc abt that

i remember there being not many other options tho

[glad osprey]

do what?

[thorn jay]

one of them is category theory + homological algebra, might take that next year

[tardy hedge]

thats nice

[thorn jay]

if im allowed though idk how that would work with study credits n all

[tardy hedge]

its funny how damn long it takes me to learn / digest some new math these days but when i do its still super cool

[marsh fulcrum]

Problem i posted

[glad osprey]

someone already gave you a solution

[tardy hedge]

Lol

[marsh fulcrum]

Where

[marsh fulcrum]

Bro as a 16-18 year old I learnt math so fast
But these things
These days
I feel tired
Exhausted
By the subject I once held so dear and sacred

[tardy hedge]

Oh tell me about it man

It's exhausting yeah

[marsh fulcrum]

I wanted to be a commutative algebraist /Algebraic geometer

I dont know what I want now

[tardy hedge]

me too my favourite is algebra

[tardy hedge]

what i want is money and stability

[marsh fulcrum]

@tardy hedge btw bro
Can you help me in one more problem
For the exam

[thorn jay]

tough job market i can imagine for that

[tardy hedge]

just post questions here and ppl answer

[tardy hedge]

yea theyre pretty sexy fields arent they

[thorn jay]

everyone wants to be category theorist algebraic geometer

[tardy hedge]

if i did combinatorial algebra it would be easier to get opportunities

[marsh fulcrum]

Same. I want a job I can work 6-7hours, Pays good, come home to study math myself

[tardy hedge]

but i dont like combintorial stuff its just not interesting

[tardy hedge]

yup so do it man thats what im doing atm

[marsh fulcrum]

Such a shame I fell in love with the 2 branches of algebra with most competition

[tardy hedge]

its ok man be confident in yourself if ur dedicated u can do it

but i mean, for me its not worth the effort anymore

[thorn jay]

in universal algebra youve got no competition but unfortunately no grants either 💔

[marsh fulcrum]

@tardy hedge give me 1 minute I am posting it

[tardy hedge]

Lol

[thorn jay]

luckily dont have to worry abt it

yet

[tardy hedge]

first year undergrad now right?

[thorn jay]

but once i finish my master's the world's gonna be burning anyways so its alrr

[thorn jay]

ye

[tardy hedge]

oh God 😭

fr what a state we live in these days

[thorn jay]

im a positivist i swear

[marsh fulcrum]

I used something good to attack this, wait

[thorn jay]

(iv) smells like krull's theorem

[karmic moat]

lots of competition but at the same time they're very general and broad fields

[thorn jay]

one of them

what am i doing up bro ive gotta sleep im already sick

[marsh fulcrum]

@karmic moat @thorn jay
I proved this

This basically kills the problem

I just need help from you guys in the very last bit of executing this problem on my problem

[thorn jay]

im gonna go to sleep tho lol

gl w this

[marsh fulcrum]

NOOOO JUST THE END OF MY PROBLEM

[thorn jay]

shouldve gone 2 hours ago

[marsh fulcrum]

It has been reduced to group theory problem now

@tardy hedge @you there

[tardy hedge]

Sorry man im prob not gonna look at these problems rn but if u post them people are willing to help eventually

Lots of ppl here they are just inactive atm

I could look at these later for fun though

[marsh fulcrum]

I have reduced the problem
ALL I NEED TO SHOW IS
that any element of the ring k[x,y,z] is congruent modulo a finite sum of elements from the basis { $X^a$, $X^a Y$, $X^aZ$} modulo the ideal P as a is nonnegative integer

[cloud walrus]

Leno

[marsh fulcrum]

@tardy hedge just hell me this one

Help me show this much and my question is done

So any element of k is obviously congruent to X^0 aka 1

[glad osprey]

Are you using this to solve 9 (i)?

[thorn jay]

I think a good way to start is to stop pinging people begging them for help lol

[marsh fulcrum]

Yes

[glad osprey]

That seems really cumbersome, I'm sure it's easier to just prove (i) directly

plus \phi in (9) isn't surjective, so I don't think 1.35 applies

[marsh fulcrum]

I
ALL I NEED TO SHOW IS
that any monomial $X^a Y^b Z^c$ of the ring $k[X,Y,Z]$ is congruent modulo a finite sum of elements from the basis ${$ $X^n$, $X^n Y$, $X^n Z$ | $n \in \mathbb{Z}$ , $n \geq 0$ $}$ modulo the ideal P

[marsh fulcrum]

I just consider the range of phi and its same thing

[marsh fulcrum]

But like
I am towards the end of it
Just need to solve this word problem aka a problem of transforming a word in X Y Z to what I want

[cloud walrus]

Leno

[marsh fulcrum]

ALL I NEED TO SHOW IS
that any monomial $X^a Y^b Z^c$ of the ring $k[X,Y,Z]$ is congruent modulo a finite sum of elements from the basis ${$ $X^n$, $X^n Y$, $X^n Z$ | $n \in \mathbb{Z}$ , $n \geq 0$ $}$ modulo the ideal $P = (X^3 - YZ, X^2 Y - Z^2 , XZ -Y^2)$

[cloud walrus]

Leno

[marsh fulcrum]

This is a word problem. My approach is to just keep swapping in $X^a Y^b Z^c$ using relations inherited from $P$

[cloud walrus]

Leno

[marsh fulcrum]

Any idea anyone

[heady snow]

whats the problem which one?

[marsh fulcrum]

I did it

[heady snow]

alright

[marsh fulcrum]

Prove that the ideal $P = (X^3 - YZ, X^2 Y - Z^2 , XZ -Y^2)$ in $k[X,Y,Z]$ cannot be generated by two elements

[cloud walrus]

Leno

[marsh fulcrum]

This is the last problem
Once this is is solved I am free from this torment

[heady snow]

well obviously the minimum element needed is 3

[marsh fulcrum]

Using monomial ideal right

Of the leading monomial using an ordering

[heady snow]

yes

[south patrol]

Should also be possible to use dimension theory and the principal ideal theorem

[velvet hull]

Does it? Because the PIT only gives an upper bound on the height of a prime ideal, and here we want a lower bound on the number of generators

I remember this exact question from ch1.1 in hartshrone, tried working at it for a bit and then giving up and looking up the solution online and all of them were pretty convoluted

[unborn terrace]

The upper bound on the height is the number of generators, so if we can show the height of the ideal is greater than or equal to 3, we are done

[velvet hull]

Well yeah, my point still stands

PIT is an upper bound, we want a lower bound

It’s not applicable

[south patrol]

Ig i meant like if it had two generators then the height would be too small

[manic iron]

If you want, you can find two generators for an ideal with the radical of that being your curve https://math.stackexchange.com/questions/5073496/is-the-affine-curve-t3-t4-t5-t-in-k-subset-mathbba3-a-set-th

[crystal vale]

So I know the problem 1A.10 gives us that p divides |G : N_G(P) |.

But how do I show N_G/P is non-trivial group?

[delicate orchid]

Let S be the Sylow subgroup of G that contains P. N_G(P) contains N_S(P), [G:P] divisible by p implies P is not equal to S, hence P < S is a proper subgroup. Normalisers grow in p-groups hence N_S(P) is strictly larger than P, and thus N_G(P) is strictly larger than P

therefore the quotient is non-trivial

fuck I just read the rest of the problem

[velvet hull]

if we can assume that such a sylow subgroup containing P exists, doesn't that

yeah lol

[delicate orchid]

that's dumb

[velvet hull]

i actually don't think it matters if N_G(P)/P is trivial or not

[crystal vale]

We have only the first Sylow theorem, so we can't say the existence of S, here

[crystal vale]

So any hint? Because if N_G(P)/P is non trivial then there exists subgroup of order p in N_G(P)/P, then I think we can make Q

[velvet hull]

wait, why does there exist a subgroup of order p in N_G(P)/P

[delicate orchid]

Cauchy's theorem

[velvet hull]

I don't see why p divides [N_G(P):P] though; we have that p divides [G:N_G(P)]