groups-rings-fields

[opaque finch]

No but then that is finite. So it doesnt work

[vagrant zinc]

I have to do that demo, that exercise of yours and mine is in the dummit book.

[tough raven]

Yeah

I realised it's obvious, sorry.

[tardy hedge]

Q

[vagrant zinc]

what?

closed and bounded implies compact in R^n
in metric space must be paracompact i.e. sequentially compact eha

[spare orbit]

can I get a hint for showing ker(phi) has inverses
https://media.discordapp.net/attachments/1253301321541619744/1304367300119494706/897772568000471050.png?ex=672f2236&is=672dd0b6&hm=05ccba81bc91b3c42dfe966ce88c17e90e176f6706452423743822448d60f7fe&

[coral spindle]

This is equivalent to showing that if phi(g) = e then phi(g^-1) = e too.

Use properties of homomorphims you are familiar with

[serene dune]

quotient out and ||try to see whether the cosets form a group or not||

or u can use the ||conjugation|| to overkill it

[spare orbit]

wdym by quotient out

[coral spindle]

It's unclear to me how you think this would work

But in any case, I think you are suggesting methods that are a bit out-of-reach

[serene dune]

essentially the homomorphism bit i wanted to picture a different process for u

u can also prove via contradiction, if u are comfortable with that

[serene dune]

i mean that's a tool to check the property of commutativity of an element or a subgroup, no ?

[coral spindle]

No, but also that was not at all the question

You are suggesting some very strange things

[serene dune]

well, i thought one would like to get as many pictures as possible, but sorry

[void cosmos]

here is a cool algebra problem that i would have never been able to do

suppose you have a ring that has the property that any direct sum of injective modules over this ring is injective

prove that this ring is noetherian

[rocky cloak]

Even more fun, this is actually equivalent to being Noetherian

[void cosmos]

Yeah and its much essier to do

Even i could do it

but the other sude is

Side is

Much more involved

chmonky helped me with this around a year ago or smth

[rocky cloak]

IIRC the main trick is that a module is injective iff it has the lifting property with respect to inclusions of ideals.

Since ideals of Noetherian rings are finitely generated any map from them factors through a finite direct sum, so then you're done

Proving the trick is like easy/medium difficulty Zorn proof, but realizing that you should use that trick if you haven't seen it before seems close to impossible

[rocky cloak]

And for commutative rings you also have the dual: product of projectives is projective iff ring artinian

[void cosmos]

wow now thats cool

[void cosmos]

Yeah exactly

the details themselves arent hard

i also really really like the question of

Figuring out which rings have the property that proj <-> free

[rocky cloak]

Is there like a nice characterization of this?

[void cosmos]

idk honestly

Assuming ur asking the question if a ring has this property the must it be X

i meant like over poly rings

[rocky cloak]

Yeah, polynomial rings and local rings and PIDs work. And for artinian rings it's iff the ring is basic, so in particular commutative artinian rings

Is it preserves under localization? Hmmm...

[void cosmos]

yeah i just gave you an idea for a paper

yw

haha

[lucid shadow]

I feel dumb
How can I find the number ring of L in Exercise 13

[opal trail]

Hello ! Can someone explain to me why, in the last paragraph it's obvious that "there exists at most one homomorphism ψ of F into G which makes the following diagram
commutative"

[void cosmos]

suppose there exists more than 1

and arrive at a contradiction

if they agree on the basis set must they agree everywhere?

[opal trail]

yes it was indeed obvious sorry for my dumb question

[void cosmos]

no bro its cool

[opal trail]

yes it's just that

[void cosmos]

when a textbook says its obvious its meant to be bait for the reader to do it by themselves

so this is not a dumb question or anything this is literally how ur supposed to learn

the fact that it is enough to define a function on the basis elements and then extend is the whole point of free groups

the construction is just a nuisance to show it exists

[opal trail]

Okay thank's for the help !

[void cosmos]

yw

[spare orbit]

i’m having trouble understanding what the conjugate definition of a normal subgroup is trying to say

[serene dune]

try to sketchout whatever u have digested then try to prove various results from the sketch

[serene dune]

have u gone through conjugacy classes yet ?

[spare orbit]

I don’t think so, no

[coral spindle]

You will see later the notion of a quotient group. It will turn out that normal subgroups help with this notion by "behaving like the identity." In this sense the fact that normal subgroups are invariant under conjugation is saying that we can treat it like the identity of a different group -- since of course the identity is indeed invariant under conjugation in any group.

You will later see things called ideals of rings, which are comparable to normal subgroups of groups, and you will be able to see those as things that "act like zero."

[bitter rover]

What is the "conjugate definition?" That a subgroup $H < G$ is normal if $gHg^{-1} = H$ for all $g \in G$?

[cloud walrus]

Cufflink

[rotund aurora]

I don't think you are suppoesd to find the ring of integers of L

[lucid shadow]

Thank you
Well I get it now, disc(Z[α,i]) is an integer multiple of disc(L)
So proving that p doesn't divide the larger discriminant is sufficient

[rotund aurora]

I think the problem of finding the ring of integers of Q(fourth root m, i) is difficult

I think you can't even find the ring of integers of Q(n^(1/4)) in polynomial time

https://www.math.leidenuniv.nl/~lenstrahw//PUBLICATIONS/buchlen/jtnb.pdf

See remark 6.10, page 33

[lucid shadow]

Wow

[tough raven]

There's something extremely gratifying about (i) finding it really difficult to compute something (ii) thinking you didn't learn it properly / you only know how to prove things about it and not to compute with it, then (iii) finding out it's actually computationally difficult and it's not on you.

Is it true that any ℝ-linear map between ℂ-vector spaces V and W can be represented as the sum of a unique pair of a ℂ-linear and a ℂ-conjugate linear map? Is there a way to interpret ℂ-linear and ℂ-conjugate linear maps as "eigenvectors" (or 1-dimensional subrepresentations) of something with "eigenvalues" (or "weights") the identity and complex conjugation respectively?

[rocky cloak]

If f is R-linear, then
[f(x) - if(ix)]/2
is C linear and
[f(x) + if(ix)]/2
is antilinear.

[tough raven]

I see. IG you can define f*(x) = f(ix)/i, getting a representation of C_2, and this is the usual way to split into eigenvectors for the trivial and "sign" characters respectively.

[rocky cloak]

Yes, I was just about to type this out

[tough raven]

Thanks, this was nice!

I wonder if this generalises to other finite Galois extensions from ℂ/ℝ.

(With degree coprime to the characteristic for semisimplicity if we want the decomposition).

If L = K(a), the obvious natural extrapolation of the above is a representation of ℤ defined by f*(x) = f(ax)/a for K-linear f (so that f* = f iff f is a-linear iff f is L-linear). But this doesn't lead to a representation of the Galois group.

[rocky cloak]

I'm thinking maybe L^*/K^* is the relevant group, as opposed to the Galois group...

Except, that doesn't quite work for C/R, so nvm I guess

[tough raven]

Right, but ℂ^⨯/ℝ^⨯ = S^1, not C_2. I think we might be able to make both perspectives work.

[tough raven]

Alternatively, we get a L (⨯)_K L module structure on Hom(V, W) coming from the L-vector space structures. To be concrete, let (l1 (⨯) l2)(f) (x) = l1 f(l2 x). We can view Hom(V, W) as a L-vector space by (l . f)(x) = l f(x) (i.e. use the first factor of L), then this is a representation of the L-algebra L (⨯)_K L (algebra structure from embedding into the first factor). We have L (⨯)_K L ≡ ∏_{g in G := Gal(L/K)} L by l' (⨯) l ↦ (l' g(l))_{g in G}, so a rep of L (⨯)_K L is a direct sum of reps of L.

To get at the ∏_{g in G} L module structure, we need the inverse of the above isomorphism, which is painful (or at least I don't know a way of writing it down easily), so I'm stuck for now.

[rocky cloak]

Yeah, this makes more sense, since C(x)C = C[C2]

[toxic zephyr]

can someone tell me how egregious this notation is? i need to show that 2+13Z generates the multiplicative group mod 13 and <2>mod13={1,2,4,8,3,6,12,11,9,5,10,7} seems like the least bad way to verify 2 has order 12.

my prof was like "use a calculator" to verify it has order 12 and i'm like... uhhhhhhhh... how??? i wanted to avoid writing out every power, but it seems she doesn't think
Z13^x=<2+13Z> iso Z12^+ is enough

[tough raven]

Can you use that the order of 2 must divide 12?

[toxic zephyr]

she also said proving that verifying that a^((p-1)/q) not equivalent to 1 mod p for all primes q that divide p-1 is "too much work!" 😭

I WROTE OUT THE WHOLE PROOF

sorry didnt mean to shout at you

[rocky cloak]

To be fair, writing out 12 numbers is not that much, and you don't even need a calculator as it's easy to do in your head.

[toxic zephyr]

yeah its not that bad, i just didn't want to have to write 1+13Z,2+13Z,4+13Z etc. etc. that would take up so much space. hence my attempt to make it much more compact with <a>mod p

[rocky cloak]

Oh yeah that's totally fine. I wouldn't even bother explaining the notation

Just say "the powers of 2 modulo 13 are
2, 4, 8, 3, ... etc"

[tough raven]

I would just write "we compute that 2^0, ..., 2^11 mod 13 are respectively 1, 2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7, which includes every element of Z13^⨯"

After all, if you were supposed to do it by calculator, whoever grades it can check by calculator I suppose.

[toxic zephyr]

true, i'm sure that's fine. i guess just because i have to do this for 5+ different primes, i think the defined notation is slightly preferable imo. if it was just 13 then i think i would just write it in words.

[tough raven]

You have to list all the powers in any case, right?

You could say at the beginning "for each case, we list the powers of the claimed generator starting from the 0^th until every invertible residue has occurred" and then repeat nothing for the individual cases except the list of powers

[tough raven]

This can be done with CRT, which in this case becomes Lagrange interpolation. Let p_g(X) = f(X)/(X-a_g)f'(a_g) for each conjugate a_g = g(a). We have p_g = g(p), where p = p_1. It can be verified that p(g(a)) = 1 if g = 1 and 0 otherwise and so p_g(a_h) = 1 if g = h and 0 otherwise. If p_g(X) = ∑_{n ≥ 0} a_n X^n, let e_g := ∑_{n ≥ 0} a_n (⨯) a^n; in other words, e_g := p_g(1 (⨯) a). Then e_g is mapped to the idempotent corresponding to the g^th factor. Hence we can define a "g-type" K-linear map to be one for which e_g acts as the identity, and every map is uniquely a sum of "g-type" maps for each g in G.

[tough raven]

For any (L, L)-bimodule V (for which the K-actions are the same), e_g ⋅ v = p_g(R_a) v where R_a is the right action of a and we use the left action to make V into an L-vector space to interpret p_g(R_a) (R_a is left L-linear). But p_g - 1 is coprime to (X-h(a)) for h ≠ g, so (p_g - 1, f) = (X - g(a)) as ideals. Hence, v is of "g-type" iff R_a = L_{g(a)}, i.e., v ⋅ l = g(l) ⋅ v for all l in L. (This can also be seen by how an L-vector space structure pulls back by the g^th map from L (⨯) L to L.)

[tough raven]

These formulas are (as they must be) special cases of the more general one: for a = i and f(X) = X^2 + 1, p_1(X) = (i+X)/2i = (1 + X/i)/2 and p_*(X) = (i-X)/2i = (1 - X/i)/2, so p_{1,*}(f) = (f ± f(i ⋅)/i)/2.

But I'll stop hogging the channel now.

[grim hawk]

hi all, I am looking for resources on how to better understand characters of representations. Any suggestions?

[upper pivot]

I need help with a simple looking question for my research

Suppose $G$ is finite and $H,H'$ non-conjugate subgroups

[cloud walrus]

JohnDS

[upper pivot]

is it true then that $$\bigcup_g gHg^{-1} \neq \bigcup_{g'} gH'g^{-1}$$

[cloud walrus]

JohnDS

[upper pivot]

its easy to see if one contains the other by a counting arguement

but I havent gotten much progress otherwise oof

[topaz solar]

Are there any finite groups where the conjugation action of G on G is transitive?

[rocky cloak]

Well the identity is always fixed

[south patrol]

ye

[topaz solar]

Good point

That was a stupid question

[south patrol]

well the orbits are conjugacy classes by definition and ye 1 is distinct from other tings

so there is precisely one example

up to unique isomorphism

[topaz solar]

And the other ones have to divide so no silliness there

[golden turtle]

I was wondering if someone could look over my work for this

[lucid crag]

A silly question everyone, if you are working on these formulas, are you good at coding this up

in other words

would you know whether this is a matrix or a vector and what matrix operation is applied

I worked with a mathematician for a deep learning project

she knew the maths stuff ok

but zero when it came to actually programming them, I wanted to implement those in a code

[eager tide]

Step 1 Install SymPy

[eager tide]

Step 2
from sympy import symbols, Eq, Function, simplify
from sympy.abc import x, y

Define symbolic elements to represent basis vectors and operations

v = symbols('v0 v1 v2 v3 v4') # Extend as needed
a, b = symbols('a b') # Coefficients

Define hypothetical projections and operations

We'll define functions that apply x-1 and y-1 to vectors

def apply_x_minus_1(v):
"""Represents the operation (x-1) on element v."""
return (x - 1) * v

def apply_y_minus_1(v):
"""Represents the operation (y-1) on element v."""
return (y - 1) * v

Define module-like structures

V = set(v) # Set of basis elements representing the module V
P1 = set(v[:3]) # Example subset of V for P1
P2 = V - P1 # Remaining elements for P2

Check subset operations (if (x-1)P1 = (y-1)P1)

def check_equal_projections():
result = []
for vi in P1:
eq = Eq(apply_x_minus_1(vi), apply_y_minus_1(vi))
result.append(eq)
return result

Print results for the equality of projections

print("Checking if (x-1)P1 = (y-1)P1 for each element in P1:")
for eq in check_equal_projections():
print(eq)

[eager tide]

step 3

Simple simulation of induction step

def induction_step():
# Start with v0 in P1 and apply operations
current_element = v[0]
path = [current_element]

for _ in range(4):  # Simulate a few steps
    current_element = apply_x_minus_1(current_element)
    path.append(current_element)
    if current_element not in P1:
        print("Contradiction found! Element left P1.")
        return False

print("Path of elements in P1:", path)
return True

Run the induction simulation

if induction_step():
print("All elements stayed in P1.")
else:
print("Proof encountered a contradiction.")

[golden turtle]

Why are you pinging me

[eager tide]

Limitations
Abstract Algebra Structures: This code provides only a symbolic simulation of the proof, rather than a rigorous mathematical proof. Implementing full module theory would require creating classes for modules, submodules, and more complex operations.
Automation of Proofs: Proving indecomposability rigorously is challenging in typical programming and requires automated proof assistants like Coq or Lean, which are designed for formal proofs.
Higher-level Libraries: Python and SymPy are useful for symbolic computations but are limited for full proof verification. Consider using specialized proof assistants if a rigorous, complete formalization is required.

Summary
This code simulates the structure of a mathematical proof rather than proving it rigorously. For formal proofs, tools like Coq, Lean, or Isabelle would allow you to define modules, submodules, and formal proofs of decomposability or indecomposability.

[eager tide]

Austin — Yesterday at 22:20
I was wondering if someone could look over my work for this

[narrow wagon]

Can you please explain the part b) of the corollary

The part where alpha = alpha’ is proven

[tough raven]

Did you understand it up to alpha' = ±alpha?

[narrow wagon]

yes

[tough raven]

Right

[narrow wagon]

Also understood the part which says B has rotation with some beta about some pole

[tough raven]

Define a matrix C_t to be the matrix corresponding to rotation about the same pole as B by an angle of beta t.

Then (i) C_0 = Id_3 (ii) C_1 = B (iii) C_t in SO(3) for all t.

[narrow wagon]

C_1 means rotate same pole as B with 1 degrees?

[tough raven]

C_t means rotate about same pole as B by angle beta t

[narrow wagon]

ok, so C_1 is rotate by beta which is B

[tough raven]

Actually, I got a bit stuck too.

[narrow wagon]

It’s okay, please take your time

[tough raven]

OK, so by the same argument, if u_t = C_t u, then C_t M C_t^T is a rotation about u_t by angle alpha_t which is ±alpha.

Also note that (the entries of) C_t and hence C_t M C_t^T are continuous functions of t.

Intuitively, if alpha_t suddenly changed from alpha to -alpha (or vice versa) for some t, C_t M C_t^T wouldn't be continuous at that t. So alpha_t has to be the same for all t, which means alpha' = alpha_1 = alpha_0 = alpha.

[narrow wagon]

Last line should be C_t M C_t^T is a rotation about u_t by beta_t which is +-alpha right?

[narrow wagon]

Here we defined C_t to rotate b_t

[tough raven]

No. C_t is a rotation by beta t, but C_t M C_t^T is a rotation about u_t by ±alpha (by repeating the argument used to show that B M B^T is a rotation about u' by ±alpha).

[indigo yew]

https://cdn.discordapp.com/attachments/700604209736974406/1304653978620923954/image.png?ex=67302d33&is=672edbb3&hm=b82703ddca4f6e31d9a44c1ecc2de2f5d4ba4624e693547b8b7f128bbd46f54e&

What function does the "image of the standard basis of R^m" relate to in this case? A goes from R^n to R^m, so does part i) mean the image of the canonical map from R^m to A R^n?

[tough raven]

Part (i) means the image under the canonical map from R^m to its quotient R^m/AR^n.

[indigo yew]

thank you

[tough raven]

The part where you claim that ∑ b_i v_i lies in either V_1 or V_2 doesn't seem correct to me. The hypothesis is that V = V_1 + V_2, which means any element of V is a *sum * of an element of V_1 and an element of V_2, not necessarily directly lying in one of them.

[narrow wagon]

Can you check if there is some error here

here [B] should be base change from B to E right?

Because columns of [B] are the coordinate of bases elements of B expressed in E

@tough raven

[restive birch]

,rotate ccw

[restive birch]

stuck on 18(b), ive written out the formula but its really gross and i cant get anywhere

[narrow wagon]

Are we all doing the same book?

[restive birch]

maybe

no yours is different

[void cosmos]

if two cycles are conjugate they must have the same cycle structure

[restive birch]

cycle type? i agree

i dont see where to go from that tho

[indigo yew]

michael artin: the bane of my existence

[void cosmos]

if an element in the conjugacy class of phi then it must have the same cycle structure

of phi

[restive birch]

yes but how do you show that no two conjugacy classes are the same size

if one of them is the one of compositions

[opaque finch]

Suppose I have $L/\mathbb{Q}$ a radical extension. If I attach a primitive cube root of unity $L(\omega)$, will it also be a radical extension of $\mathbb{Q}$ ?

[cloud walrus]

mycroftholmes1703

[opaque finch]

I guess it should be true but am unable to prove it

[rocky cloak]

Do you have a definition of radical extension were this is not just true by definition?

Like omega^3 is in L...

[tough raven]

Depending on what you mean by base-change matrix, it's either the base-change matrix from E to B or vice versa.

[narrow wagon]

I see it now. If I define base change as changing between the hyper vector of one base to other

then it will be base change from E to B

Defining base change as changing between coordinate, it will give B to E

[tough raven]

If A is any linear map, A_E its matrix with respect to E, and A_B its matrix with respect to B, then (Ae_1, ..., Ae_n) = E A_E, which we can write as A E = E A_E, and (Av1, ..., Avn) = B A_B, so A B = B A_B ⇔ A E [B] = E [B] A_B ⇔ A E = E [B] A_B [B]^{-1} ⇒ A_E = [B] A_B [B]^{-1}.

Personally, I'd call it base change from B to E then.

[lucid crag]

Many thanks!

[wispy light]

Since orthogonal matrices make sense over arbitrary fields of characteristic not 2, do SVDs also make sense over arbitrary fields of characteristic not 2?

[dull ginkgo]

I mean the singular values themselves are eigenvalues of the matrix MM^T but that depends a lot on how algebraicly closed the field in question is

[wispy light]

I'm willing to accept irreducible factors of characteristic polynomials as “generalized eigenvalues”. So lack of eigenvalues in the ground field itself isn't a problem.

[dull ginkgo]

Separability might be an issue

[wispy light]

Oh, okay, now that's an actual problem.

[dull ginkgo]

Over a field of char p you might get a char poly of f(x^p^n) where f is irreducible

[velvet steeple]

Why is this? How can we know we have an isomorphism G' -> H and G'' -> G/H? G -> G is clear

[wispy light]

Certainly G' is isomorphic to H = ker(g), and then G'' is isomorphic to G/H.

[velvet steeple]

H = ker(g)?

How do we know

[tardy hedge]

Im(f) = ker(g) cause its an exact sequence

[velvet steeple]

Yes

[tardy hedge]

Im(f) = G' because its exact at G'

so f injective

[wispy light]

I'm defining H to be ker(g).

In your original image, it's more precise to replace the second line of text with “there exists a subgroup H of G such that the first exact sequence is isomorphic to the second one”.

And then, to prove that claim, we can (an in fact, have to) define H.

[velvet steeple]

I arrive at this with a different justification. Let the map from 0 -> G' be called k. Then im(k) = ker(f) = {0}, thus f is injective

[tardy hedge]

yea

[wispy light]

You don't even need to give the map 0 -> G' a name, because there's only one such map. And the only information you get from it is that f is injective, as you said.

[velvet steeple]

..so?

[tardy hedge]

so what

[velvet steeple]

Does this have anything to do with what Eduardo said?

Why is it useful that f is injective

[tardy hedge]

yes it makes G' isomorphic to ker(g)

[velvet steeple]

Ah

[wispy light]

Not only is the map G' -> G is injective, but, if you take H = ker(g), then you can construct a commuting triangle whose other two sides are the inclusion H -> G and a carefully chosen isomorphism G' -> H. This gives you the left half of the sought isomorphism of exact sequences.

[velvet steeple]

I agree with the arrow I drew in

k is an isomorphism

I don't yet notice how H -> G' is an isomorphism

[wispy light]

I'll elaborate. Let $H = \ker(g) = \operatorname{im}(f)$. By the first isomorphism theorem, $f : G' \to G$ factors as $G' \twoheadrightarrow G' / \ker(f) \xrightarrow {\tilde f} H \hookrightarrow G$, where $\tilde f([g']) = f(g)$ is an isomorphism. But $\ker(f) = 0$, hence $G' = G' / \ker(f)$, hence $\tilde f : G' \to H$ is an isomorphism.

[velvet steeple]

Ok, I think there is the cause: I'm trying to avoid using the first iso theorem because my book hasn't proved it at this point yet

[cloud walrus]

Eduardo León

[wispy light]

Why on Earth are you talking about exact sequences if you don't have the first isomorphism theorem?

[velvet steeple]

Serge Lang's ordering is perhaps a bit weird haha

The first iso theorem comes on the next page(s)

[wispy light]

Okay. At least you know that $f$ is injective. Letting $H = \mathrm{im}(f) \subset G$ as above, you certainly have a set theoretic bijection $\widetilde f : G' \to H$ defined by $\widetilde f(g') = f(g')$. Moreover, you can readily check that
\begin{itemize}
\item $H$ isn't just a subset of $G$, but actually a subgroup.
\item $\widetilde f$ is a group homomorphism.
\item Any bijective group homomorphism is actually an isomorphism, hence $\widetilde f$ is an isomorphism.
\end{itemize}

[cloud walrus]

Eduardo León

[opaque finch]

I got it. But thanks anyway

[wispy light]

Now let's look at the right half of your original exact sequence. Since $\mathrm{im}(g) = \ker(G'' \to 0) = G''$, we deduce that $g$ is surjective, and we get a set-theoretic bijection $\widetilde g : G / \sim \to G''$, where $\sim$ is the equivalence relation on $G$ that identifies elements with the same image in $G''$. Moreover, you can readily check that the following are equivalent:
\begin{itemize}
\item $x_1 \sim x_2$, i.e., $g(x_1) = g(x_2)$.
\item $g(x_1 \cdot x_2^{-1}) = g(x_1) \cdot g(x_2)^{-1} = e''$, where $e'' \in G''$ is the identity element.
\item $x_1 \cdot x_2^{-1} \in \ker(g)$... but remember that $\ker(g) = \mathrm{im}(f) = H$.
\end{itemize}

To continue, we need the fact that $H$ is a \textbf{normal} subgroup of $G$. Do you know that kernels of group homomorphisms are normal or do you need to prove that too?

[cloud walrus]

Eduardo León

[velvet steeple]

No, my book proved that already

[wispy light]

Okay, good! Then your quotient set $G/\sim$ is actually the quotient group $G/H$, and you can readily check that $\widetilde g : G/H \to G''$, $\widetilde g([x]) = g(x)$ is a group isomorphism.

[cloud walrus]

Eduardo León

[velvet steeple]

That completes it, thanks a lot!

[wispy light]

You're welcome. 🙂

[dull ginkgo]

Short five lemma in a trench coat moment

[glad osprey]

This paragraph from wikipedia seems to imply that every field of characteristic 0 is contained in the complex numbers. Is this true?

[wispy light]

Definitely not.

There are arbitrarily large fields of any characteristic.

[glad osprey]

Yeah, I thought so. But what are they saying though? The roots of unity in a field of char 0 are complex?

[wispy light]

Yes. In fact, the roots of 1 in $\mathbb C$ are actually elements a much smaller subfield of $\mathbb C$ called $\bar {\mathbb Q}$.

[cloud walrus]

Eduardo León

[glad osprey]

The algebraic completion of Q?

[wispy light]

The algebraic closure of Q. Although there's a purely algebraic way to construct \bar(Q), the easiest way to think about it is as the subset of C consisting of those elements that are roots of some nonzero polynomial in Q[x]. By some facts about integral extensions (see Atiyah-Macdonald chapter 5 or Dummit & Foote chapter 15), this subset is actually a ring, and, in fact, a field.

[glad osprey]

I see. So if F has char 0 and p(x) = x^n - 1 is in F[x], then all roots of p(x) lie in bar(Q)?

[wispy light]

That statement needs to be more precise.

If F is an algebraically closed field of characteristic 0, then there's a unique subfield K of F that's isomorphic to \bar(Q). Moreover, if p(x) \in Q[x], then any roots of p(x) that are in F are, in fact, in K.

Now, since K is unique with the above property, sometimes we abuse notation and say that K itself is \bar(Q).

[glad osprey]

I see, that makes sense. So we can say that any field of char 0 either contains a subfield isomorphic to bar(Q) or has a field extension containing a subfield isomorphic to bar(Q)?

[wispy light]

If F isn't algebraically closed, then there's no guarantee that F contains a subfield isomorphic to \bar(Q). For example, the field of real numbers doesn't.

But you can take F's algebraic closure instead. And, since F and K are both subfields of F's algebraic closure, you can consider the intersection of F and K.

[glad osprey]

Yeah, I just realized it follows from what you wrote by just taking the algebraic closure. Thanks for the insights 🙏

[south rain]

Hi, I just stumbled onto a relation while calculating normal subgroups of $D_4$.

Let the $C_i$ be the conjugacy classes. Does it, in general, hold that
$$H_i = Z(G) \cup C_i$$
is a normal subgroup?

[cloud walrus]

dellinger

[south rain]

And $Z(G)$ being the center of $G$.

[cloud walrus]

dellinger

[south rain]

I mean, all elements stay in $H_i$ under conjugation with any element of $G$. But is it always a subgroup?

[cloud walrus]

dellinger

[coral spindle]

Typically it is not a subgroup, no.

[south rain]

Thanks, that's what I figured.

[coral steeple]

Is there a unique subgroup of S6 isomorphic to S3? A unique subgroup of S_n! isomorphic to Sn?

[rocky cloak]

No, not even up to conjugation.

The diagonal in S3xS3 < S6 for example

[velvet steeple]

I proved the first part. Currently thinking about finding the isomorphism $G'' \to G/\ker g$

[\begin{tikzcd}
G & {G''} & 0 \
G & {G/\ker g} & 0
\arrow["g", from=1-1, to=1-2]
\arrow["{\text{id}}"', from=1-1, to=2-1]
\arrow[from=1-2, to=1-3]
\arrow["{\text{can}}"', from=2-1, to=2-2]
\arrow[from=2-2, to=2-3]
\end{tikzcd}]

g is surjective, yes I agree

[coral steeple]

Hmm, thanks

[velvet steeple]

Why do we get a bijection?

[dull ginkgo]

Is this not just FIT lmao

[velvet steeple]

Perhaps, but haven't proved that yet, most likely

[dull ginkgo]

Haven’t proved FIT?

[velvet steeple]

What's FIT?

[dull ginkgo]

First isomorphism theorem

[velvet steeple]

No, haven't proved it yet, I know the ordering is weird

[dull ginkgo]

[velvet steeple]

We take an element of $G''$. We determine its preimage in $G$. Then we send that to $G/\ker g$.

[dull ginkgo]

That’s literally FIT

[dull ginkgo]

First isomorphism theorem

[velvet steeple]

We just need to show it's an isomorphism

Any idea on that?

[dull ginkgo]

G/ker[f] ~= im[f]

Which can be verified fairly simply by considering preimage bs

[velvet steeple]

Yeah, but I shouldn't use it if my book hasn't proved it before

[dull ginkgo]

You’re basically proving it

[velvet steeple]

Yeah

That's the plan

Ok, consider the other direction

Take something in G/ker g

Map that to the exact element in G

This map is injective but not surjective

Then apply g onto it, which is surjective

We need to prove that the composition is bijective

[velvet steeple]

Though IIRC the composition of an injective and surjective function can't be bijective?

So I messed up somewhere

[dull ginkgo]

false

pretty sure

[velvet steeple]

You're right

[wispy light]

Ignoring algebra, if you have a surjective function $f : X \to Y$, then you can impose an equivalence relation on $X$, where $x_1 \sim x_2$ iff $f(x_1) = f(x_2)$. Then obviously each equivalence class corresponds to its image in $Y$.

[cloud walrus]

Eduardo León

[velvet steeple]

Basically

Take something in G/ker g
Map that to the exact element in G
This map is injective but not surjective
Then apply g onto it, which is surjective
We need to prove that the composition is bijective

Right?

I think you did it in the other direction You did it the same direction

[wispy light]

The map $g : G \to G''$ is surjective but not injective. But if we factor it as $G \twoheadrightarrow G / \sim \xrightarrow {\tilde g} G''$, then $\tilde g$ is a bijection by construction.

[cloud walrus]

Eduardo León

[wispy light]

Because, in G/~ we have identified any elements of G that would be mapped to the same element of G''.

Then it turns out that the equivalence relation $x_1 \sim x_2$ can be expressed as “$x_1 H = x_2 H$ are the same coset”, i.e., the same element of $G/H$, with the $H$ defined previously.

[cloud walrus]

Eduardo León

[wispy light]

$$
\begin{tikzcd}
0 \rar & G' \rar{f} \dar["\widetilde f"] & G \rar{g} \dar{\mathrm{id}} & G'' \rar & 0 \
0 \rar & H \rar[hook] & G \rar[two heads] & G/H \rar \uar["\widetilde g"] & 0
\end{tikzcd}
$$

[cloud walrus]

Eduardo León

[wispy light]

@velvet steeple ^ I have constructed your sought isomorphism specifically like this.

[velvet steeple]

Ok this is great

tilde g is obviously injective, yes

How do we know it's still surjective?

[wispy light]

Given $y \in G''$, there exists $x \in G$ such that $g(x) = y$, because $g$ is surjective. Then $\widetilde g([x]) = g(x) = y$.

[cloud walrus]

Eduardo León

[velvet steeple]

How do you know $g(x) = \tilde g([x])$?

[cloud walrus]

Kepe

[wispy light]

I've defined $\widetilde g$ it that way.

[cloud walrus]

Eduardo León

[wispy light]

Of course, you need to check that $\widetilde g$ is well-defined.

[cloud walrus]

Eduardo León

[velvet steeple]

Ok I think what I really mean is

Showing that it's well-defined

Yes exactly

[wispy light]

Using my earlier diagram as a blueprint, we proceed as follows:
\begin{itemize}
\item Let $H = \ker(f)$.
\item Let $\widetilde f : G' \to H$, $\widetilde f(x) = x$.
\item Let $\widetilde g : G/H \to G''$, $\widetilde g([x]) = g(x)$.
\end{itemize}

We must check that $\widetilde g$ is well-defined. If $[x_1] = [x_2]$, then there exists $h \in H$ such that $x_2 = x_1 h$, and then $\widetilde g([x_1]) = g(x_1) = g(x_1) \cdot e'' = g(x_1) \cdot g(h) = g(x_1 \cdot h) = g(x_2) = \widetilde g([x_2])$.

[velvet steeple]

How do you know x_1 * h = x_2?

x_1 * h = x_2 * h^2

[wispy light]

Because that's the definition of quotient group?

[cloud walrus]

Eduardo León

[wispy light]

Oops, I accidentally swapped $x_1$ and $x_2$ in the equation $x_2 = x_1 h$.

[cloud walrus]

Eduardo León

[velvet steeple]

Ah, that's why

Yeah

Thank you!

[velvet steeple]

Thanks a lot!

[velvet steeple]

How can we be sure that this explicit form is really unique?

[wispy light]

There's certainly not a unique set-theoretic function $G' \to H$, or even a unique group homomorphism $G' \to H$. But there is a unique function $\widetilde f : G' \to H$ such that $f = \iota \circ \widetilde f$, where $\iota : H \hookrightarrow G$ is the inclusion, and $\widetilde f$ turns out to be a group isomorphism.

Similarly, there's not a unique set-theoretic function $G/H \to G''$, or even a unique group homomorphism $G/H \to G''$. But there's a unique function $\widetilde g : G/H \to G''$ such that $g = \widetilde g \circ \pi$, where $\pi : G \to G/H$ is the quotient map, and $\widetilde g$ turns out to be a group isomorphism.

[velvet steeple]

Thanks a lot!

[cloud walrus]

Eduardo León

[wispy light]

And from now on I'll be away for a few hours.

[restive birch]

?

[tough raven]

Well, what are the possible cycle types of a permutation of order 2?

[restive birch]

theyre all some number of disjoint 2 cycles

[tough raven]

Right. How many permutations are there with k disjoint 2-cycles?

[restive birch]

$\frac{n!}{k! 2^k (n - 2k)!}$

[cloud walrus]

henryduke

[restive birch]

right?

[tough raven]

Yep.

[restive birch]

i tried working with this

i got nowhere

[tough raven]

Call this number C_k. Then C_{k+1} = C_k ⨯ (n-2k)C2 / (k+1). As k increases from 0 to floor(n/2) - 1, the numerator (n-2k)C2 decreases while the denominator increases. Hence C_1 > C_2 > ... > C_k ≤ C_{k+1} < ... < C_{floor(n/2)} for some intermediate k. Thus, it is enough to check that C_{floor(n/2)} < C_1.

Oh, you're just supposed to show that no C_k is equal to C_1, they're not necessarily all smaller.

[restive birch]

whats C2?

also at some point we have to use the assumption that n is not 6

since n = 6, k = 1 and 3 return the same value

[tough raven]

The group PGL_2(ℂ) acts on the Riemann sphere by Mobius transformations. This determines a (maximal?) compact subgroup PSU_2(ℂ) of PGL_2(ℂ), consisting of those elements that are isometries (equivalently, preserve antipodality of points). Is there a similar geometric description of other compact subgroups (say those conjugate to PSU_2(ℂ)) of PGL_2(ℂ)?

Maybe an equivalent way of asking the same question: given a Mobius transformation g, can we describe when two points p, q are the images of antipodal points under g?

[tough raven]

Binomial coefficient: (n-2k)C2 = (n-2k)(n-2k-1)/2.

[tough raven]

Oh, I see.

[tough raven]

Wait, I got the orders wrong here: it's C_1 < C_2 < ... < C_k ≥ C_{k+1} > ... > C_{floor(n/2)}. In particular, if C_{floor(n/2)} > C_1, then C_k > C_1 for all k > 1.

Let k = floor(n/2); then n = 2k or 2k+1 and C_k / C_2 = (1! 2^1 (n-2)!) / (k! 2^k (0 or 1)!) = 1/k ⨯ [ (n-2)! / (k-1)! 2^{k-1} ]. The factor in [] is the number of permutations in S_{n-2} with floor((n-2)/2) transpositions (i.e., the analogue of C_k for n-2) - call it a_{n-2}. It suffices to show that a_{n-2} > k for n ≥ 7. For odd n ≥ 7 this is easy, because for any l, a_{2l+1} = (2l+1) a_{2l} so that a_{2l+1} ≥ 2l+1; hence a_{n-2} ≥ n-2 > n/2 > k. It must be true for even n ≥ 8 as well because a_n ought to grow pretty fast, but I don't have a proof available.

In fact, a_{2k} = (2k)!/2^k k!, s a_{2k+2} = (2k+1)(2k+2)/2(k+1) a_{2k} = (2k+1) a_{2k} = a_{2k+1} (there's a combinatorial proof of this as well). So a_{2k+2}, a_{2k+1} ≥ 2k+1. In particular, a_{n-2} ≥ n-3, and n-3 > floor(n/2) for n > 6.

[tough raven]

bumping my question,

[tough raven]

since I buried it myself 💀

[cloud walrus]

mycroftholmes1703

Can someome help me solve this?  
Let $L$ be the splitting field of a cubic over $\mathbb{Q}$, and $\omega$ be a primitibe cube root of unit. Prove that the extension $L(\omega)$ over $\mathbb{Q}$ is a radical extension. 


My aporoach is the following: Given a polynomial $f$ and a root $\alpha$ then $L = \mathbb{Q}(\alpha,\sqrt{D})$ where $D$ is the discriminant. I got to show $L(\omega)$ is a radical extension $\implies \mathbb{Q}(\omega,\alpha,\sqrt{D})$ is a radical extension. Now I know that $\omega^3 \in \mathbb{Q}$ and $\sqrt{D}^2$ \in \mathbb{Q}(\omega)$, it just remains to show $\alpha^k \in \mathbb{Q}(\omega,\sqrt{D})$, for some $k$. Can anyone help me out from here?
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                $
l.53 ...ega^3 \in \mathbb{Q}$ and $\sqrt{D}^2$ \in
                                                   \mathbb{Q}(\omega)$, it j...
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you left one out. Proceed, with fingers crossed.

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[opaque finch]

I just need to find such a $k$

[cloud walrus]

mycroftholmes1703

[toxic zephyr]

i think i'm driving myself insane. does saying that p prime does not divide r and p does not divide s imply that p does not divide rs require proof? in order to prove that Zp* is closed under its multiplication operation.

[opaque finch]

No it doesnt. If a prime p does not divide r or s then it can never divide rs. Since its a prime

[rocky cloak]

I mean, for many purposes this is the definition of prime.

For integers specifically it's called Euclid's lemma

[toxic zephyr]

i actually wrote out a god damned proof of it omg

okay im just gonna say "by euclid's lemma" lol

not knowing how much detail my prof wants is driving me insane

[opaque finch]

@toxic zephyr I think I understand your confusion perhaps.
6 doesnt divide 9 and 6 doesnt divide 4 but 6 divides 36. Here the anatomy of factorisation suggests that primes factors of 9 and and prime factors of 4 multiply in some sense to produce a 6.
But for primes it cannot be the case as you have to see that primes can only be produced by multiplying 1 with itself. So either one of the factors has to have the whole prime

[toxic zephyr]

yeah exactly

[toxic zephyr]

i like this definition of prime, because it's how we define prime ideals. but the one i'm used to being the standard is that it has no nontrivial divisors. pretty sure it's exactly equivalent, but the former is much more useful to group/ring theory haha

it feels too obvious to need to prove. but i was scared and wrote the whole proof anyway

[opaque finch]

Good job anyways

[rocky cloak]

In a general setting this is usually called irreducible, which is equivalent to prime iff your ring is UFD

[toxic zephyr]

ah yeah, right. that makes sense.

would be funny to teach primes in primary school using the terms "irreducible" and UFD haha

i missed algebra a lot actually. this is fun.

and thanks again jagr ❤️

[sonic coral]

0 should be a prime since it generates a prime ideal

[rocky cloak]

I mean, for people who know English, irreducible makes more sense than prime I guess

[toxic zephyr]

and mycroft ❤️

[sonic coral]

anyone wanna sign my petition

[opaque finch]

True

[toxic zephyr]

i will

[toxic zephyr]

buhp, still unanswered

[opaque finch]

As far as I understand the motivation for defining "ideals" comes from Kummer's so called "ideal numbers" when Kummer was trying to investigate the special case of Fermat's Last Theorem for regular primes (in simpler terms, determining the properties of ring of integers over cyclotomic extensions over rationals). The philosophy essentially was to translate every properties of integers to ideals as every number can be thought of as an ideal (now we know this is true for integers as the ring is a PID). Hence such a definition of prime ideals makes sense.

[opaque finch]

Actually I am not sure, but as I think more deeply, I think the Cardano's formulae directly helps here with k = 3.

[wispy light]

Yeah, it works because, for Dedekind domains, there's a “prime factorization for ideals”.

[tidal torrent]

about irreducability i dont understand eisenstein's criterion at all

actually learning field theory before ring theory but thats probably not a good idea

[delicate bloom]

like how to prove it or just how to apply it

[tidal torrent]

how to apply it

[delicate bloom]

maybe helps to just see an example

x^2-2

the constant term is divisible by p=2 exactly once

and the 0x term (hiding but there) has coefficient 0 which is divisible by 2 at least once

and the x^2 term has coefficient 1 which is not divisible by 2, so it satisfies eisenstein

so x^2-2 is irreducible by eisenstein, which means we have proved sqrt(2) is irrational

[tidal torrent]

does the converse of eisenstein criterion hold true

[delicate bloom]

I don't think so

we could probably prove it doesn't

also worth mentioning people will often times translate the polynomial in order to get the eisenstein criteria to pop out

example for you to work out, f(x) = x^2+1 for instance, translate that to f(x+1) and prove it's irreducible

[tidal torrent]

yeah figures since if f(x) is irreducible over Q doesnt always mean that there exists a prime p s.t p is not a divisor of each of the coefficients

dont know how to show that though

[delicate bloom]

well working this out as a specific example should suffice

you just need one counterexample to knock down a claim

although based on what I said about translation I could see why that might not be the best example if you want one that's irreducible after translation but never by eisenstein

I like $\Phi_p(x) = \frac{x^p-1}{x-1}$ to $$\Phi_p(x+1) = \frac{(1+x)^p-1}{x}=\sum_{k=1}^p \binom{p}{k}x^{k-1}$$
Since $\binom{p}{k}$ is divisible by $p$ for all except the highest degree term and only the constant term once, it's irred by eisenstein

[cloud walrus]

Merosity

[delicate bloom]

spook'd yerself 😛

[tidal torrent]

not too familiar with cyclotomic polynomials but is this a nontrivial or trivial factorization

[wispy light]

Yeah. The way you wrote it is clearer anyway.

[tidal torrent]

Over Q

[wispy light]

The cyclotomic polynomials are defined by total induction in such a way that $x^n - 1 = \prod_{d \mid n} \Phi_d(x)$.

[cloud walrus]

Eduardo León

[tidal torrent]

Hm ok

[wispy light]

How do you make it so that your TeX'd output has a white background?

[delicate bloom]

I forget lol

#latex-help 😅

or #bots idk

@smoky cypress showed me eons ago

[narrow wagon]

Can you explain why do we have such a definition for D_n?

[wispy light]

It's the group of symmetries of a regular $n$-gon in $\mathbb C$. Say, with vertices at $\zeta^k$, for $\zeta = e^{2\pi i/n}$ and $k = 0, \dots, n-1$. Then $x$ corresponds to multiplication times $\zeta$, and $y$ corresponds to complex conjugation.

[cloud walrus]

Eduardo León

[narrow wagon]

What is complex conjugation?

[wispy light]

Complex conjugation sends $a + ib \mapsto a - ib$, where $a, b \in \mathbb R$.

[cloud walrus]

Eduardo León

[toxic zephyr]

the fact that (p-1)+pZ is not equal to 1+pZ (if p>2) is like... so freaking obvious it doesn't even really need to be explained, let alone proven, right? 😭

[opaque finch]

It is not needed to be explained that they are not equal. I dunno how obvious it is, but maybe even if you want to prove it then consider the group $\mathbb{Z}/p\mathbb{Z}$. Your two expressions are just 2 distinct elements here

[wispy light]

How is it not obvious? Any two distinct elements of ${ 0, \dots, p-1 }$ belong to different equivalence classes modulo $p$, because their difference (in the correct order) is $> 0$ and $< p$, hence not a multiple of $p$.

[cloud walrus]

Eduardo León

mycroftholmes1703

[wispy light]

So... It's obvious, but if you're taking your first algebra course, then you still need to prove it anyway.

Just to make sure that it's indeed obvious to you.

[bitter rover]

The 😭 makes me skeptical. In a context where everyone would agree "Yes, that's obvious", nobody would really balk at having to justify it.

After all, if it's obvious, spend the 30 seconds it takes to write the proof.

If it takes more than 30 seconds then you were just fooling yourself about how clearly you understood it.

Here 30 seconds is just a placeholder time, but if someone really understood it I think it'd take less time to write the proof than it would to debate whether or not it's worth writing the proof.

So just write it.

[elfin wraith]

Eh I’m not convinced by that, maths isn’t a spectator sport and all, I think it’s completely fine to skip over details that can be verified in that short of a time

Of course there needs to be a balance though

And different people draw the line in different places, in a comalg homework recently I said a quotient ring clearly contained no elements of degree >=1 and the marker complained saying never to say something is clear unless it’s truly the most obvious thing in the world

I’d argue that R[x]/<x> having no degree 1 or higher components is the most obvious thing in the world but ya know here we are

[kindred vector]

I would honestly complain about that in a comm alg class tbh

Maybe even in a first course on rings I'd complain

Does more damage to a student long term by asking them to write that out in full imo

[spare orbit]

what does $[G:H]$ mean

[cloud walrus]

sushi

[spare orbit]

thank you

[hidden wind]

i’m so confused by ideals in quotient rings

[elfin wraith]

oh I do intend to lol, this is no where near close to a first algebra course at my uni its kinda insane, this same marker also took a mark off me claiming I should elaborate on the statement "the quadratic formula reveals x^2 + x + 1 has no rational roots" so idk whats happening there

maybe an extreme example for the point I was trying to make but you get what I mean

[tardy hedge]

Wdym?

What are you confused by?

[kindred vector]

Jesus Christ who is marking your work? Hopefully something comes from your complaint

[bitter rover]

It's annoying but let's not pretend it's beyond the pale lol

[void cosmos]

should have said the rational root theorem haha

idk the context, but if you are working on a general field then the quadratic formula may not work

take any field that has cahr 2

char 2*

the rational root theorem however is always there

[hidden wind]

honestly it’s probably primarily the notation

[tardy hedge]

Notation for what?

Sorry im just trying to understand which part youre talking about

[hidden wind]

like it’s just so inconvenient to write down explicitly the elements of an ideal in a quotient ring

[void cosmos]

how

[bitter rover]

Give a specific example of what you find confusing.

[opaque finch]

Can someome help me solve this?
Let $L$ be the splitting field of a cubic over $\mathbb{Q}$, and $\omega$ be a primitibe cube root of unit. Prove that the extension $L(\omega)$ over $\mathbb{Q}$ is a radical extension.

My aporoach is the following: Given a polynomial $f$ and a root $\alpha$ then $L = \mathbb{Q}(\alpha,\sqrt{D})$ where $D$ is the discriminant. I got to show $L(\omega)$ is a radical extension $\implies \mathbb{Q}(\omega,\alpha,\sqrt{D})$ is a radical extension. Now I know that $\omega^3 \in \mathbb{Q}$ and $\sqrt{D}^2 \in \mathbb{Q}(\omega)$, it just remains to show $\alpha^k \in \mathbb{Q}(\omega,\sqrt{D})$, for some $k$. Can anyone help me out from here?

Edited : Apologies for the previous malcompilation

[cloud walrus]

mycroftholmes1703

[coral steeple]

Is there a nonabelian group where every element cubed is 1?

My first thought was A3 but apparently this is abelian 😦

One also has the identity ab=(b^-1a^-1)^2=(b^2a^2)^2

[bitter rover]

https://en.m.wikipedia.org/wiki/Heisenberg_group

[coral steeple]

New group unlocked

Thanks

[chilly sparrow]

So, a bit of a clueless question: With the 3rd iso thm for rings. Say you've got:
$R/J \cong (R/I) / (J/I)$ and some isomorphism $\tilde{\varphi}$ so $R/I \cong R'$
Can you then say $(R/I) / (J/I) \cong R' / \tilde{\varphi}(J/I)$?

Or do you have to make a specific argument if that happens to be the case?

[cloud walrus]

Saoirse

[toxic zephyr]

no, the proof is completely trivial, and I was able to write it easily. i just keep doubting myself about what I do and do not need to just for these proofs I'm doing (been happening all week)
the situations I want to avoid are

• writing the proof for it and the Prof being annoyed that I would waste their time or they would think I'm artificially lengthening my proof
• getting marked down because I didn't justify what I thought was too trivial.

[toxic zephyr]

no this is grad algebra lol

[tough raven]

You can say that.

[bitter rover]

Oh

[toxic zephyr]

driving myself insane, I did a justification using that (p,1-(p-1))=(p,2)=1
and then also a general proof that
a+nZ=b+nZ, where 1≤a,b<n iff a=b
but then I read it again and thought "why the fuck are you proving this"

[opaque finch]

Its actually good thing sometimes.

[stuck musk]

I want to show that a maximal ideal J in a UFD is principal. I know that by Zorn’s lemma I can find a principal ideal in J that is maximal among all proper principal ideals in J, and this principal ideal should just be J, but I’m not sure how to show this. (I also know this principal ideal is prime) Suggestions?

[opaque finch]

are you sure you want to show this ?

[stuck musk]

I was afraid this would be the response haha

[opaque finch]

Cause $\mathbb{Z}[x]$ is a UFD but the ideal $(2,x)$ is maximal but not principal. As far as I know
All maximal ideals of a UFD is principal iff it is a PID

[cloud walrus]

mycroftholmes1703

[stuck musk]

Ah shoot, you’re totally right

I should probably add a condition then

Everything’s clear now. I just didn’t read the problem correctly. Thanks for the example Mycroft!

[toxic zephyr]

when talking about abelian groups/free abelian groups, is it bad to use the Z-span or Z-linearly independent terminology if we haven't talked about modules yet? or should i just use "span" and "linearly independent" and it's clear from context that we're talking about "integer scalars"?

[wispy light]

I can't imagine talking about Abelian groups without linear algebra terminology.

[opaque finch]

So actually free abeliam groups as far as I understand are just free $\mathbb{Z}$-modules. So if you have a basis then
$G = \mathbb{Z}b_1 + \mathbb{Z}b_2 + \dots + \mathbb{Z}b_n$. So I hope it can be understood that you have coefficients from integers and there is no need for a seperate specification..

[cloud walrus]

mycroftholmes1703

[toxic zephyr]

yeah abelian groups are just Z-modules, basically. i just hesitate to start using that terminology prematurely in the class

[opaque finch]

Exactly as the fact. But it doesnt matter, how one sees it as long as one is comfortable with it

[toxic zephyr]

fair enough. i do kind of like being specific by saying Z-span.

ty mycroft ❤️

[opaque finch]

I dunno why but \mathbb{Z}-span somehow sounds a bit awkward to my ears.....,.............., but it doesnt matter.

Maybe problem with my ears or something 😆

[crystal vale]

By using Zorn's lemma we can prove that every ideal I of a ring R is contained in maximal ideal.

But let ring Q with usual addition and multiplication defined as ab = 0. Then Q has no maximal ideal.

Then how in this case can I say any ideal is contained in the maximal ideal?

But I think Zorn's lemma says the existence of maximal elements so it can be ring R itself but then it is not maximal ideal

[lapis latch]

ring with identity

if you look at the proof that every proper ideal is contained in a maximal ideal to ensure that the upper bound for a given chain is a proper ideal you use the fact that it cannot contain 1. As that example shows if you dont have 1 then that might fail

[crystal vale]

Ah right, thank you ❤️

[late marsh]

Suppose $G$ is a finite group, $H$ is a subgroup of $G$, and $N$ is a normal subgroup of $G$. Assume that the order $|H|$ and the index $[G:N]$ are relatively prime. Prove that $H$ is a subgroup of $N$.

[cloud walrus]

Emma

[late marsh]

I'm a little stuck on this question, so far i was trying to use that HN is a subgroup of G and |HN| = |H||N|/|H cap N|

and to show it's a subgroup i could show intersection of H and N has to be H but i'm a bit stuck on that part

i guess since |HN| divides |G|, |H|/|H cap N| has to divide |G|/|N|, right? and if the intersection isn't just H then they wouldn't be relatively prime

i dont know if that makes sense

since then the order of the intersection would be something else divisible by both,

is that correct?

[marble crypt]

Yes. You could also look at the coset $hN$ for $h \in H$, by writing $1=a|H|+b|G/N$|.

[cloud walrus]

RickC137

[fading summit]

Cayley’s theorem states that every group of order n is isomorphic to a subgroup of Sn.
Consider G = Z2 × Z4. According to Cayley’s theorem, there is a subgroup of S8 that is isomorphic to G.
What is the smallest value of n such that Sn has a subgroup that is isomorphic to G? Justify your answer (the justification should have two parts: that value of n works; no smaller value works).

im guessing the answer is one of the subgroups of S4????? or at least, n is between 4 and 8.

wait actually doesnt it have to be a subgroup of S8? because G has 8 elements, so a group isomorphic to G must also have 8 elements, but there is none such subgroup for Sn where n < 8

[fading summit]

oh wait, i found out that Z4 × Z2 is a subgroup of S5

and obviously Z4 × Z2 is isomorphic to Z2 × Z4

so n = 5 must be the answer

[grave sedge]

Are you sure about this?

[fading summit]

dangit idk

[south patrol]

Also like

S_m is a subgroup of S_n whenever m < n

Just consider permutations which don't affect m+1,...,n

[fading summit]

the thing is that the subgroup also has to be abelian, besides have 8 elements

and i think the only such subgroup is Z8 in S8

[grave sedge]

by cayley you know there is a subgroup of S8 isomorphic to Z2xZ4 though

so that cannot be the case

[fading summit]

isnt Z8 that subgroup?

[grave sedge]

Z8 is not isomorphic to Z2xZ4

[fading summit]

oh snap Z2 x Z4 is not cyclic....

[opaque finch]

Ok let us see it like this perhaps. HN is a subgroup of G of course, and you have $[HN:N] = |HN/N| = [H:H \cap N]$ and hence divides $|H|$. We also have $[HN:N]$ divides $[G:N]$ right ? But they are relatively prime and it divides both so it should be $1$. And hence $|HN/N| = 1 = [H:H \cap N]$ and so $H = H \cap N$ and you are done.

It think this is the solution that you tried to explain in your following answer. If I am not wrong.

[cloud walrus]

mycroftholmes1703

[fading summit]

okay so the only abelian non-cyclic groups of order 8 that i can think of are:

• Z2 x Z4
• Z2 x Z2 x Z2 (not isomorphic to Z2 x Z4 so it doesnt matter)
  and afaik neither of these are in a subgroup of Sn where n < 8

[grave sedge]

again, why are you so sure about this?

(if A is a subgroup of Sn and B a subgroup of Sm, think of a way to get a subgroup of S(n+m) isomorphic to AxB)

[fading summit]

so Z2 is a subgroup of S2 and Z4 is a subgroup of S4, so i need to think of a way to get a subgroup of S6 isomorphic to Z2 x Z4 ?

[grave sedge]

yes

[fading summit]

like the subgroup generated by (1 2) and (1 2 3 4)?

[grave sedge]

nope

not abelian

[opaque finch]

this is actually a very nice wondering. But this is actually a theorem I studied. Let $\mu(G)$ be the minimal such $n$, you want. Then if $G$ is an abelian group then decompsoing $G$ by Fundamental Theorem of Abelian groups, you should have $\mu(G)$ as the sum of values of all the $Z_k$'s. It is a well established theorem. In your case your $n = 4 + 2 = 6$.

[cloud walrus]

mycroftholmes1703

[opaque finch]

Nothing minimal than that

[grave sedge]

i don't think saying it's "a well established theorem" helps here

[opaque finch]

Well, it does. Atleast if you know that it is isomorphic to a subgroup of $S_6$, then you have a huge progress to finding the exact subgroup $S_6$ without pondering for higher powers. Now if you mean you want a proof, then sorry I will cite the reference but cannot write the whole proof here.

[cloud walrus]

mycroftholmes1703

[fading summit]

oh, how about the subgroup generated by (1 2) and (3 4 5 6)

[grave sedge]

yes

that works

[grave sedge]

do you see how this works in general?

[fading summit]

sure. the guessing-and-checking kinda sucks though lol.

thanks for the help though

[grave sedge]

by the way, to show n=5 does not work, try looking at which elements in S5 can commute with a given element of order 4

(if you find less than 8 elements which commute with it then you cannot have an abelian subgroup containing it)

[whole jetty]

How can I show that they are isomorphic? This probably is more of a linear algebra question.

[cloud solar]

Let A be a nxn matrix with integer entries. Prove that det(A^4+I)≠13

[cloud solar]

I want to prove that A^4+In is invertible in M_n(Z/pZ) if A is in M_n(Z/pZ)

[coral spindle]

Well that would in general be false, for example when p = 1 mod 8 you can choose A to be a diagonal matrix having primitive 8th roots of unity as entries, whence A^4 = -I.

[vivid birch]

Let G be a finite group. Can we guarantee that there exists a total order on G that makes G into an ordered group?

[coral spindle]

What's the definition of an ordered group to you?

Because the one I know implies that no finite groups are ordered

[vivid birch]

if a <= b then ac <= bc and ca <= cb

[coral spindle]

And is the order partial or total?

Oh, you said total. My bad.

[vivid birch]

total

[coral spindle]

Suppose $g \in G$ such that $1 < g$. Then $g < g^2$ also, and so $1 < g^2$, etc. Since $G$ is finite, $g$ has finite order and so we conclude $1 < 1$. Contradiction. This also happens if $1 > g$.

[cloud walrus]

$\mathbf{Boytjie}$

[coral spindle]

In general any such group must be torsion-free, so in particular no finite groups are orderable.

You don't seem to have much to say about this

In what context were you seeing a finite ordered group?

[vivid birch]

I was looking into groups definable in IR^n, i.e. a subset X of IR^n and a map f: X x X -> X such that (X,f) is a group (the "definable" just gives us some information about the map f and the set X , but that's not important rn). Then I noticed that when X is finite, we can pick any subset Y of IR with the same number of elements as X, and we can define a map g: Y x Y -> Y such that (Y,g) is a group isomorphic to (X,f). As Y is arbitrary, I was trying to permute the elements of Y in IR so that the resulting group would be ordered, but yeah, as you mentioned, this is impossible.

[coral spindle]

Sounds like cool stuff

[vivid birch]

if we were to consider a partial order, would this be possible @coral spindle ?

[coral spindle]

Well, the trivial partial order works! 😁

I.e., equality

I don't know if you can imbue a finite group with an interesting partial order

It's not something I've thought about before

Hm

Well OK here's something

The only thing comparable to 1 must be 1

Now suppose that a < b

then ab < b^2, ... ab^n < b^n+1 etc

So there is some m such that ab^m < 1

So a^-1 = b^m for some m.

(In particular, m is the order of b, minus 1)

So in fact a = b.

Well yeah I guess there was some logic there I didn't inspect carefully, but it seems to me that for a finite group there is only the trivial partial order

[vivid birch]

ok yeah, I get that

thank you for your help

[coral spindle]

No worries man

[elfin wraith]

Wondering if anyone can point me in the right direction for this, kinda going in circles

I’ve got ring R, PID, comm with 1, and non zero non units a,b,c,d. We’ve got that aR+bR=cR+dR \neq R and I’m trying to show that abcdR isn’t semi prime/radical

Basically what I’ve got so far is that there’s another ideal <e> equal to the sum of the other two and such that gcd(a,b)=gcd(c,d)=e, so Im guessing im using this for some sort of contradiction but I’m unsure how

We get that ne^4 \in abcdR I think for some n (since e divides each of a b c d), which if we’re talking about a semi prime ideal seem like a nice thing to have, but I’m not quite sure how it’s helpful yet

[rocky cloak]

Just write
a = a' e
b = b' e
c = c' e
d = d' e
n = a'b'c'd'

Then abcd = ne^4, so ne is not in abcdR, but (ne)^4 is

[elfin wraith]

Yeah that’s nice thanks jagr, I basically have that written down but in considerably worse notation which made it much harder to see haha

[tough raven]

Then 1 < a^{-1}b, no?

[coral spindle]

Oh yeah that's a simpler way lmao

[stark gull]

hi

How can I proceed to solve this exercise?

[wraith cargo]

What's the order of the Galois group for starters

[opaque finch]

Step 1: The extension is Galois, since $\mathbb{Q}$ has characteristic $0$ and your extension is just the splitting field of $p(x) = x^4 - 2$.

Step 2: $|\text{Gal}(\mathbb{Q}(i,\sqrt[4]{2})/\mathbb{Q}) = |\mathbb{Q}(i,\sqrt[4]{2}):\mathbb{Q}|$. Compute the extension degree we get 8.

Step 3: The problem can be ended directly from here if one refers to the subgroup structure of $S_4$ as the unique group of order $8$ embedded in $S_4$ is just $D_8$.

Let us say we don't go into that. Let us compute then

Step 4: Let $\sigma \in \text{Gal}(\mathbb{Q}(i,\sqrt[4]{2})/\mathbb{Q})$, then it can be uniquely determined by it's action on the roots. Observe that $\sigma(\sqrt[4]{2})$ is a root of $p(x)$ and $\sigma(i)$ is a root of $x^2 + 1$ right ? Hence you must have a total of $8$ automorphisms and voila this is exactly same as the order of Galois group !

Step 5: Then you realise that the only kind of automorphisms for your Galois group is done just by permuting the roots of of $p(x)$ and $x^2 + 1$, seperately, and by that I mean that you have to send roots of $p(x)$ to itself and $x^2 + 1$ to itself as well. Let $r(\sqrt[4]{2}) = i\sqrt[4]{2}, r(i) = i$ and $s(i) = -i, s(\sqrt[4]{2}) = \sqrt[4]{2}$. Last step I leave it to you to check that all other automorphisms that I mentioned above is generated by $r$ and $s$, and moreover in the sense that you definitely have the following relations $r^4 = e, s^2 = e, rs = sr^{-1}$, and finallly this is $D_8$.

[cloud walrus]

mycroftholmes1703

[stark gull]

Sorry for my inconvenience, I have some questions:
a) In step 4, why do you conclude that you must have 8 automorphisms? (it is because of the result that says |Gal(L/F)|=[L:F]?)
b) What is D_8? (I swear I have no idea. I probably haven't touched on that topic yet)

[opaque finch]

You have all possible permutations of the roots of $x^4 - 2$ and the roots of $x^2 + 1$. First one gives you $4$ permutations and second gives you $2$ and hence $4 \times 2 = 8$.

$D_8$ is the dihedral group of order $8$.

[cloud walrus]

mycroftholmes1703

[stark gull]

really thanks, tomorrow I'll tell you how my solution goes.

[tough raven]

I just want to mention (for the benefit of those reading that don't know this) that the argument you used, namely that if there are 4 possible images of one generator and 2 of another then all 8 combinations must be realised by some automorphism, is not valid in general (ℚ(sqrt(i), 2^{1/4})/ℚ is an instructive counterexample). It works here because it is true that there is at most one automorphism realising each combination (because i, 2^{1/4} generate the extension) and we have already found out that there are 8 automorphisms.

[old hollow]

Let R be a ring with an ideal A. Suppose there are finitely many ideals between A and R, that is, finitely many ideals B such that A ⊂ B ⊂ R. Does this mean A has finite index in R?

[mighty kiln]

No

[old hollow]

hm ok

im trying to understand a particular justification in this book

[mighty kiln]

Eg take the ideal (x) in the ring of rational functions over C such that x does not divide the denominator

[old hollow]

oh i see

[mighty kiln]

Oh wait you could also just do (0) ⊂ C

[old hollow]

you could pick a maximal ideal which has infinite index

that too

this is what im trying to understand

maybe im just not seeing something lol

[rocky cloak]

R is a number ring, so is an (finite) integral extension of the integers.

So when you mod out a prime you get an (finite) integral extension of Z/p for some p.

Hence a finite ring

[old hollow]

why is the second sentence true

sorry I haven't done much commutative algebra

is it like, A ⊂ B integral extension, then if q ⊂ B prime, then B/q is an integral extension of A/p for some prime ideal p ⊂ A

[rocky cloak]

Yes, intersecting q with A gives you a prime ideal of A.

[old hollow]

ah

[rocky cloak]

The only thing you have to be a little careful with is that that prime isn't (0).

[old hollow]

is there a 'concrete' way to see it, in this situation

[old hollow]

wait why would that be an issue

like, maybe (0) isn't prime in A? you mean

[rocky cloak]

Because Z/(0) isn't finite

Anyway, for any nonzero element in R it's minimal polynomial will have non-zero constant term. So any non-zero ideal contains a non-zero integer.

[old hollow]

oh this makes sense

so when you quotient by a non-zero ideal, you are killing all multiples of that non-zero integer, effectively making a finite ring

[restive birch]

would someone mind looking over a couple of proofs:
-show that a group of order 24 is not simple.

-show that a group of order 90 is not simple.

for the first one

the only option for n3(G) is 4

let P be a sylow 3-subgroup of G

then since n3(G) = 4, |G : N(P)| = 4

therefore there is a homomorphism pi: G -> S4, which is clearly nontrivial and must have trivial kernel since G simple

therefore G iso S4 which has A4 as a normal subgroup

for the second one:

we know that n3 must be 10

and n5 must be 6, and n2 is 15 or 45

say all sylow 3-subgroups of G have trivial intersection

then there are 8 * 10 = 80 elements contained in sylow 3-subgroups of G, and this is a problem since there are also 4 * 6 elements of order 5 in G

therefore there are P, Q distinct sylow 3-subgroups of G with intersection of size 3

since they are both abelian they are (proper) subgroups of N(P cap Q), which therefore

has order divisible by 9, > 9 so its either 18 or 45

if its 45 it has index 2, contradiction

if N(P cap Q) has order 18 and therefore index 5, then there is a homomorphism pi: G -> S5 by conjugation of P cap Q

it must have trivial kernel so then G is iso to a subgroup of S5, a contradiction since 90 does not divide 120

those work?

[tough raven]

If it's 18 then P, Q are normal in N(P \cap Q) as they have index 2, but also conjugate.

[restive birch]

oh thats also true

[tough raven]

Sorry, just noticed that and had to say it.

[restive birch]

you get there either way

i mean actually just by counting you have an issue, since like

[tough raven]

AFAIK you haven't used the n2 thing anywhere (and I don't see where it came from either).

[restive birch]

right yea its not necessary

i got it from

[tough raven]

Seems legit.

[restive birch]

n2 is 3 or 5 or 9 or 15 or 45 right

it cant be 3 or 5, since then by taking the representation of conjugation you get that G <= S3 or S5

i cant remember how i eliminated 9...

but it doesnt matter

[restive birch]

if you look at sylow 3-subgroups

oh wait this might not work

hmm

i was gonna say they each have to like 'add' 6 unique elements but thats not true

or at least i cant assume it

they have pairwise intersection <= 3

in general i found a cool result

for checking if groups of order n are simple

if you write out all the np(G)s

[tough raven]

You could take cases on whether they're cyclic or not, I suppose.

[restive birch]

and any of them has n not dividing n_p(G)!

[restive birch]

true

[restive birch]

then G not simple

so for instance, cant have a simple group of order 36

since n3 is forced to be 4

and 36 does not divide 4! = 24

the larger problem was:
let G be a nonabelian simple group of order <100. show that |G| = 60

[tough raven]

oof

[restive birch]

was fun actually

[tough raven]

That has a lot of cases along the way.

[restive birch]

not too many

[tough raven]

I think.

[restive birch]

i heavily used that

[tough raven]

I see.

I've never actually done it.

[restive birch]

groups of order p, pq, p^n, pqr, p^2 q are never simple

where all of those are primes

so there were like 10 cases left

[restive birch]

this cut most of them down

then just casework

[tough raven]

Hmm PSL2(field) is supposed to be simple, shouldn't that give you smaller ones

[restive birch]

PSL2?

[tough raven]

SL2 / {±Id_2}

[restive birch]

hmm

well for F2 right

what is the order of SL2(F2)

[tough raven]

Probably abelian

[restive birch]

true

[tough raven]

But I feel F3 should get out of that.

[restive birch]

okay whats the size of SL2(F3)

[tough raven]

|GLn(Fq)| = (q^n-1)...(q^n-q^{n-1})

[restive birch]

right

[tough raven]

For SLn(Fq) divide by the number of possible determinants, which is q-1

[restive birch]

so here GL2(F3) = (3^2 - 1)(3^2 - 3) = 48

[tough raven]

So change the last factor to q^{n-1}

And then divide by 2 again

[restive birch]

right so order 12

but order 12 groups arent simple

i mean we can just classify them right

[tough raven]

(q^2-1)q/2 (q^2 - 1)q/2 if q is odd, (q^2 - 1)q if q even.

[tough raven]

I assume you showed this in your |G| < 100 case bash.

[restive birch]

its not hard its basically like

say n3 = 4 right

since it would have to

then there is an injective morphism G -> S4 so G is iso to a subgroup of order 12 in S4

G has (3 - 1) * 4 = 8 elmnts order 3

[tough raven]

This is easily not prime unless p = 2.

[restive birch]

S4 has 8 elements of order 3

all of them are in A4

so pi(G) cap A4 >= 8 so pi(G) = A4

so G is iso to A4 in this case

which is not simple

(it has the normal sylow 2-subgroup {(1 2)(3 4), (1 3)(2 4), (1 4)(2 3), 1}, obviously normal because it is a conjugacy class of S4 and therefore normal in S4)

[restive birch]

huh lmao

looked this up

the two cases we looked at are the only exceptions

that is, PSL_n(Fq) is simple for n >= 2 except in the cases where n = 2 and q = 2 or 3

https://kconrad.math.uconn.edu/blurbs/grouptheory/PSLnsimple.pdf

[tough raven]

Apparently this is only true for algebraically closed fields (maybe) and finite fields except for q = 3.

[restive birch]

and q = 2

we looked up the same thing lol

and it is true for q = 2, 3 if n >= 3

[tough raven]

That's what the Internet tells me as well, but |PSL2(F2)| = 3 so it should be simple, right?

[restive birch]

oh lol

[tough raven]

Unless it's 6?

[restive birch]

its what like

2^2 - 1 times 2^2 - 2

[restive birch]

it is

you dont divide by 2 at the end

because -1 = 1

[tough raven]

Oh, it's 6. {±Id_2} degenerates to {Id_2} in characteristic 2.

[restive birch]

yep

tricky

[restive birch]

so this isnt necessarily true

[tough raven]

edited lol

[restive birch]

nice

anyways im going to sleep but sylow subgroups are so cool

very powerful tool

[restive birch]

i presume that its S3 but i have no evidence and im not going to calculate it

no way its Z6 tho

[rocky cloak]

PSL(2, 2) is isomorphic to S3

It acts on the projective line, so must be a subgroup of S3.

Similarly PSL(2, p) is a subgroup of S[p+1]

[somber sleet]

Does someone know if there is a way to show that to matrices $A,B\in M_{n,n}(K)$ are equivalent if and only if they are equivalent on their field extension $\bar{K}$?

[cloud walrus]

damn_guuurl

[mighty kiln]

Two matrices are equivalent iff they have the same rank

[somber sleet]

I thought that one has to show that the matrix P in $A = P^{-1}BP$ had to be the same both over $K$ and over $\bar{K}$

[cloud walrus]

damn_guuurl

[mighty kiln]

Oop I have a different defn for equivalent

[long swan]

i thought the defn was like

matrices are equivalent iff they're the same up to elementary row operations

i could be wrong

[vast stump]

<@&268886789983436800> same random message in multiple channels

[rocky cloak]

I guess most people call this matrices being "similar".

Anyway, by the structure theorem of PIDs a matrix is determined up to similarity by its invariant factors.

So you may think of A as the action of x on the K[x]-module

K[x]/(f1) x K[x]/(f2) ...

where fi are the invariant factors. Considering A over a field extension Kbar just amounts to tensoring with Kbar, which preserves the invariant factors.

[south patrol]

nice approach

[long swan]

Yeah damn

[glad osprey]

Are quotient modules dual to submodules, and if so, in what sense? This theorem is the dual of a similar one concerning noetherian modules, and I notice in (ii) "submodule" has been replaced by "quotient module"

[glad osprey]

I think I figured it out: the universal property of quotient modules is just the universal property of submodules with the arrows reversed. Like in topology, the quotient is characterized by maps out of it, while the submodule is characterized by maps into it

[tough raven]

What is finitely cogenerated BTW?

[tough raven]

Hmm, that is nice.

[somber sleet]

oh that is really beautiful

that's a really nice approach

do you maybe know a way that can be explained to a person that only did linear algebra

[rocky cloak]

Here's a nice little argument that works assuming K is infinite
https://math.stackexchange.com/a/57257/306319

[rocky cloak]

If you just mean explain the same proof in different language, then it just amounts to proving rational canonical form.

Which you can do with cyclic vectors, invariant subspaces, and essentially the same proof as the structure theorem for PIDs

[wispy light]

Given $n \in \mathbb N$ and an algebraically closed field $k$, how do I find the finite group embeddings $G \hookrightarrow GL_n(k)$?

[cloud walrus]

Eduardo León

[wispy light]

Actually, I'm only interested in $\overline {\mathbb F_p}$, $\overline {\mathbb Q}$ and $\mathbb C$... probably.

[cloud walrus]

Eduardo León

[rocky cloak]

Up to conjugation a group embedding is just a faithful representation.

There are lots of tools to determine what all the representations, and then you can filter out the faithful ones.

Things may get a little complicated if the characteristic of your field divides the order of the group though

[wispy light]

Okay, then I'll just focus on bar(Q) and C for the time being. The problem I'm facing is that I have basically zero knowledge of representation theory, but I'm trying to study embeddings of linear algebraic groups.

And of course nothing makes sense.

[rocky cloak]

Are you reading something that doesn't make sense, or what is the thing that isn't making sense?

[wispy light]

I mean, I'm reading something that makes sense if you know representation theory. The problem is that I don't.

[rocky cloak]

I see

[rocky cloak]

Well, I guess you'll just have to learn some representation theory, it's fun 😊

[glad osprey]

If I is an ideal of R, then I is a submodule of R regarded as an R-module, right? Is the converse true, are the submodules of R as an R-mod exactly the ideals of R?

[tardy hedge]

yeah

I think thats one reason why modules are a good thing to study, because it includes that kind of ideal-ring structure as a special case

[glad osprey]

I see, thanks

[rocky cloak]

In general (left) submodules are left ideals.

[regal zodiac]

Is it True a k[x,y] module has a natural structure of k vector space ?

[next obsidian]

Yeah this is called restriction of scalars

[glad osprey]

Thanks I've seen two definitions of a ring R being noetherian, either that it has the ACC, or that R as an R-module is noetherian, but if the ideals and submodules coincide then it makes sense that these definitions are equivalent

Man, I keep forgetting that ideals are not subrings. I was about to ask whether an ideal of a noetherian ring is noetherian 😅

actually it's kind of weird how ideals seem so much more important than subrings

[next obsidian]

Ideals of Noetherian rings are Noetherian modules

[glad osprey]

yep, I figured that, since ideals are submodules, and the submodule of a Noetherian module is Noetherian

[next obsidian]

Also this theorem is true

For (i) assume B and A are commutative

Finite over A means finitely generated as a module

[glad osprey]

What book is that?

[next obsidian]

Matsumura Commutative Ring Theory

[restive birch]

need a hint:
let G be a finite simple group where each proper subgroup is abelian, and let M and N be two distinct maximal subgroups of G, show that they have trivial intersection

tried taking M's representation by conjugation in S_|G:M|

[rocky cloak]

Hint: ||their intersect is normal||

[restive birch]

hmm how would i go about showing that

[dim widget]

well what happens when you conjugate by an element of M?

[rocky cloak]

Next hint: ||the normalizer of a subgroup is a subgroup||

[restive birch]

right i know that the normalizer of a non normal maximal subgroup is the subgroup itself

[restive birch]

hmm, conjugating an element of M cap N by an element of M will yield back an element of M which might be an element of N still but might not?

[opaque finch]

Let $H = <x,y>$ be a nilpotent group generated by two elements $x$ and $y$. Does this imply that $xy = yx$ ?

[cloud walrus]

mycroftholmes1703

[restive birch]

D8?

[rocky cloak]

Any subgroup of an abelian group is normal

[opaque finch]

Ahh my bad, sorry for not mentioning this. the order of $x$ and $y$ are co-prime

[cloud walrus]

mycroftholmes1703

[restive birch]

ohh of course, so the normalizer of M cap N contains at least M and N so it contains <M, N> and so is G

thanks

forgot the abelian condition tbh

[rocky cloak]

My Google fu tells me every finite nilpotent group is the product of its sylow subgroups, so yes.

[opaque finch]

thank you

[restive birch]

okay immediately got stuck again. show that An has no proper subgroup of index < n for n geq 5. what i have so far is that for some i if Gi is the stabilizer of i then |Gi : Gi cap H| geq n - 1. am i on the right track?

also, sanity check, if H is generated by a union of conjugacy classes, H is normal, right?

[wispy light]

Any inner automorphism of G will merely permute H's generators...

[rocky cloak]

The action of G on G/H for a subgroup of index k induces a homomorphism G -> Sk

[restive birch]

right, so we get a homomorphism An -> Sk for k < n, must have trivial kernel, and this breaks simply from order considerations

that work?

since if it didnt have trivial kernel An wouldnt be simple

is it true that for n >= 5, An is the only normal subgroup of Sn? i think i proved it but im not sure

[rocky cloak]

It is true yeah. Precluding the trivial group and Sn itself

[restive birch]

right

okay i need another hint lol bear with me

,rotate ccw

[restive birch]

problem 6

i dont get how the hint is useful am i missing smth obvious

[dim widget]

So if there is a homomorphism to another group with nontrivial kernel, and you take the simple group containing one element not in the kernel and one element in the kernel, what happens to that group under the homomorphism?

[restive birch]

i mean, it gets sent to a subgroup of the codomain that's generated by the image of the non-kernel element?

[rocky cloak]

If H < G is a subgroup and N < G is a normal subgroup, then N\capH is a normal subgroup of H.

[restive birch]

oh huh, so let N be a supposed normal nontrivial subgroup, and take any nontrivial element of it, then that element moves, say, n many elements, and so N cap An <= A has to be nontrivial normal in An, and so is An (you can just increase n to 5 or more by adding stabilized elements of the element you chose) and so every copy of An in A is contained in N, which implies N = A?

[rocky cloak]

Yup

Or using the hint / what TT said more directly. Consider an element in N and one not in N and derive a contradiction

[restive birch]

one final proof if one of you doesnt mind looking it over, i think its right but my proof seems disconnected from the content of this chapter.its problem 5 here:

say H is a normal subgroup of G. if H cap Gi = Gi for all Gi, then all Gi <= H, and so G <= H, so H not proper
if H cap Gi = 1 for all Gi, then if g in G is nontrivial, then it is in some Gi, and thus not in H, so H trivial
now each Gi cap H is normal in Gi, so it is either Gi or 1. collapse this sequence so that we have only the proper inclusions, 1 < G1 < G2 < G3 ... (sorry for notation). we have two cases. First, say that when you intersect this sequence with H, you get a sequence like 1, G1, G2 ... Gi, 1, 1, 1, 1--this clearly isnt possible since Gi < G_{i + 1}. otherwise let Gi be the first of these with H cap Gi = Gi, i >= 2, and G_(i - 1) cap H = 1 which must happen since the only other possibilities were covered above. but then since G_(i-1) is nontrivial, there are elements of Gi not in H, a contradiction

[golden turtle]

I don't know what it means by eigenspaces E of g_i, since g_i are generators of C_(n_i), and I don't know why such a statement is true. It says "we have seen" but it isn't in the lecture notes previously and I don't see it in the book

Could anyone explain why this is?

[wispy light]

Your representation $\rho : A \to GL(V)$ lets you see each $g \in A$ as a linear map $\rho(g) : V \to V$. And $\rho(g)$ certainly has eigenspaces. By abuse of terminology, one calls them the eigenspaces of $g$.

[cloud walrus]

Eduardo León

[golden turtle]

sure. each g in A has an associated map p(g), I see why we can say each of the p(g) has eigenspaces yes

how do we know that V is the direct sum of eigenspaces of p(g)'s?

[wispy light]

I don't think it's claiming that $V$ is a direct sum of the eigenspaces of all the $g \in A$'s. It's claiming that, \textbf{if you fix one} $g \in A$ (that happens to generate a direct summand of $A$), then $V$ is a direct sum of the eigenspaces of this specific $g$.

[cloud walrus]

Eduardo León

[wispy light]

And here it has to be using the hypothesis that the ground field has $n$ distinct $n$-th roots of $1$. Because otherwise you could take the representation $\rho : C_2 \to GL_2(\mathbb F_2)$ that sends the generator $g \in C_2$ to the matrix
$$\rho(g) = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix},$$
which doesn't have enough eigenspaces for $\mathbb F_2^2$ to be a direct sum of them.

[cloud walrus]

Eduardo León

[wispy light]

Aha. We want $\rho(g)$ to be diagonalizable, or at least semisimple.

[cloud walrus]

Eduardo León

[wispy light]

No, we need strict diagonalizability. If we just have semisimplicity, then we could have something like $\rho : C_4 \to GL_{2n}(\mathbb R)$, sending the generator $g \in C_4$ to the matrix
$$\rho(g) = \begin{bmatrix} 0 & I_n \ -I_n & 0 \end{bmatrix},$$
which is a direct sum of irreducible invariant subspaces, but these invariant subspaces are not eigenspaces.

[cloud walrus]

Eduardo León

[cloud walrus]

yard

[errant wedge]

WLOG S is spanning (its immediate otherwise), you know there cannot be more than n linearly independent vectors, so there exist at least 2 vectors dependent

idt you need finite field for this?

unless im doing this very wrong

this holds in general for an F-vs of dim > 1 and m > 0 I think

[errant wedge]

cuz ull have dependent vectors, so u can just keep two and then cut any vectors till ur down to n

Let V be an n dim F-vs, with n>1, and consider a set of m vectors with m>n. Then clearly this can't be linearly independent right? So there exist v_1, v_2 linearly dependent. Then just choose any remaining vectors till you've built your set of n

oh hrm wait

am i being dumb

i might be

yes im being dumb huge mistake whoops

let me redo ur problem sorry one sec 😭

[errant wedge]

this isnt tru

Ooh I think I have it

one second, making sure

[errant wedge]

I tried a size argument (counted bases) but that doesnt work hrmm

Wait im overcounting

Nah still too big

I forgot to divide by n! for unordered but didnt work either

[hidden cairn]

how to construct an explicit isomorphic isomorphism from $C_{18} \times C_{15} \to C_9 \times C_{10} \times C_3$?

[cloud walrus]

pink_panther

[hidden cairn]

My attempt:
Let x, y generate C_18 and C_15. Map x to (x^2, x^9, 1) and map y to (1, y^3, y^5) but I am struggling to show it is an isomoprhism

[placid socket]

if 3b looks fun to anyone please take a stab at it

[hidden cairn]

i've found that if we define $\phi$ this way, we have $\phi(x)^{10} \cdot \phi(y)^{15} = (x^2, 1, 1)$ and $\phi(x)^{18} \cdot \phi(y)^{10} = (1, 1, y^5)$ and $\phi(x)^{81} \cdot \phi(y)^{6} = (1, x^9 y^3, 1)$ and since these elements are generators of $C_9, C_3, C_{10}$ respectively, I can conclude $\phi$ is an isomorphism. Is that right? And is there an easier way to do this?

[cloud walrus]

pink_panther

[errant wedge]

sorry im stumped. messaged a friend, ill lyk if they get it lol

[hidden cairn]

ie a way without finding the generators of $C_9 \times C_10 \times C_3$ in terms of $\phi$?

[cloud walrus]

pink_panther

[opaque finch]

Can anyone help me find an example of a group $G$ and it's subgroups $X,Y,Z$ such that $[[X,Y],Z] \neq <[[x,y],z] | x \in X, y \in Y, z \in Z>$ ?

[cloud walrus]

mycroftholmes1703

[finite turtle]

this is a proof about simple group of order 60, G being isomorphic to A_5 from dummit and foote. how to get S_5 = G A_5? and how that |P intersect Q| = 2 , why not something greater than 2?

[stuck yacht]

So I'm learning about cyclic redundancy checks, and I think they are related to galois fields...which I actually know nothing about..but is this the right channel to ask abt it? I've actually opened a help channel but I'm not sure if I'll get much help there

[sonic coral]

P and Q distinct are sylow 2 subgroups of order 4. Their intersection is a subgroup of both, and so can only have order 1 or 2.

[toxic zephyr]

is this not just R as a "left" R module where R acts on R by right multiplication?

[void cosmos]

yeah

[opaque finch]

equivalenty yes

[toxic zephyr]

ok cool

i also wanted to ask: the prof just defined the ideal generated by X, but didn't specify if R is commutative or if <X> is a left/right/two sided ideal. just the minimal ideal of R generated by X.

if i have a noncomm ring R and X is just some subset (doesn't contain 1, to keep it interesting), then is <X> ambiguous? or what kind of ideal are we assuming it is? is it possible that for some subsets X1,X2,X3, <X1> is a left (not right) ideal and <X2> is a right (not left) ideal and <X3> is a two sided ideal (all nontrivial)? or do we have to be specific about which kind of ideal we are generating?

[rocky cloak]

It is indeed ambiguous, but usually it's clear from context what you're talking about.

You should be consistent within the same context though. If you sometimes write <X> for the left ideal generated by X and sometimes for the two-sided without distinction it will be confusing

Though in situations where people talk about both left/right and two-sided ideals, the word "ideal" alone usually means two-sided

[toxic zephyr]

i see. so just to be safe, i should probably just be very explicit (at least at the start).

[toxic zephyr]

interesting. yeah im mostly familiar with comm rings, where everything is two sided and everything is nice, so this makes sense lol

[hidden wind]

am i tripping or is the final iso in the example at the section end here simply wrong

wait no nvm we’re in Q not Z

woops

[hidden wind]

fsr i’m finding it real difficult to think of integers as units

[tough raven]

It would seem so: ℚ[X]/(2X^2 - 2) = ℚ[X]/(X^2 - 1) ≡ ℚ[X]/(X-1) ⨯ ℚ[X]/(X+1) ≡ ℚ ⨯ ℚ.

[wispy light]

The author probably meant to say Q[x] / (X^2 - 2)..

Because it would be very impolite to say Q[x] / (f(x)), where f(x) is a polynomial in Z[x] of content > 1.

(Where content = gcd of the coefficients, as in the proof of Gauss' lemma.)

[hidden wind]

[upbeat dirge]

Hey sorry I had a quick Q on sylow p subgroups

in their definition (lets denote such a subgroup P) does it mean |P|= power of a prime which is that of the prime factor decomposition of |G| ?

This isn't quite clear to me, if P is a sylow p subgroup w/ |P|=5^2, does it mean then that a subgroup group |P'|=5 is also a sylow p subgroup ?

[slim kayak]

the author really liked their ctrl+v

[rocky cloak]

A sylow subgroup is a maximal p-subgroup.

It will have order the largest power of p that divides the order of G.

So you cannot have one of order 5^2 and another of order 5 in the same group.

[slim kayak]

A sylow p-subgroup is a subgroup of the order p^n, where p^n is the biggest power of p such that it still divides the order of your group.

[upbeat dirge]

thank you this is much, much clearer !

[slim kayak]

If you just mean subgroups of order some power of p, thats a p-subgroup

[upbeat dirge]

yes but the wikipedia definition was a bit misleading lol

[slim kayak]

The second part is still correct but kinda useless unless you know that there is always a sylow subgroup and that their conjugates contain any other p-subgroup

[upbeat dirge]

but to be clear, the number of sylow p subgroups isnt just the number of prime factors of |G| lol

as in there could be several subgroups with cardinality p^largest n

[slim kayak]

Well, you can have non-normal sylow subgroups

In which case you get another of the same if you conjugate it right. Thats another theorem somehwere in the void

[rocky cloak]

Yes for example S3 has three 2-sylow subgroups

[upbeat dirge]

Great to know

I had one last question on the conjugacy class of sylow p subgroups

[slim kayak]

then yes

[upbeat dirge]

the second sylow theorem tells us that two sylow p subgroups are conjugate

but in that equivalence class are there more elements than just the sylow p subgroups ?

as in P a sylow p subgroups w/ gPg^-1=K not sylow

ok this question is really dumb

[slim kayak]

That would be a good exercise

[upbeat dirge]

yes

I will figure it out on my own

thanks guys

[slim kayak]

As a sorta hint: whats the defining feature of a sylow p subgroup?

And then translate into what that translates into for conjugations in general

[upbeat dirge]

great

I will look into that

[upbeat dirge]

I was wondering how you view conjugation of subgroups

do you have some intuition you always have in the back of your mind

[slim kayak]

Yeah, but not really in terms of anything else though

[upbeat dirge]

can you give me some pieces of info as to what you think about ?

[upbeat dirge]

like what you mean in this message

[slim kayak]

Nothing, intuition as in - the sense of balance when riding a bike ig

[slim kayak]

Oh, as in thinking of the conjugation as a function for a fixed element

[upbeat dirge]

I see cool

[slim kayak]

C_a takes g to aga^-1, that sort of stuff

[upbeat dirge]

thanks

my lecturer had this remark to make in the notes

isnt it redundant given the following:

if there exists a single sylow p subgroup => G is a prime power => the sylow p subgroup in question is G itself => obviously normal

[slim kayak]

Oh

A sylow p-subgroup, so for a fixed p

[upbeat dirge]

oh I get it

[slim kayak]

Actually, ignore what he says and try to think of the useful statement by looking at Sylow's third theorem

(and maybe the fact that conjugates of subgroups are subgroups)

[upbeat dirge]

alright thanks

[restive birch]

also, its an important thm to keep in mind that a sylow p-subgrp is normal iff it is characteristic

so, in particular, normal subgrps of normal sylow p-subgrps are normal

tends to come up a lot

[sudden dust]

is my approach reasonable?

[hot wadi]

Yes