Darkrifts:
#groups-rings-fields
cloud walrusBOT
topaz solar
where i is something in I or whatever
i'm a bit tired for this so excuse some typos here and there
bleak finch
by a ~ b you mean a, b are in the same coset?
bleak finch
ok
brisk granite
um, so, this is my answer to the question above
Could this please be verified?
Let the sylow $p$-subgroup $P$ (where $|P| = p^n$) act on $\text{Syl}{p}(G)$ by conjugation. Then, we can write $\text{Syl}{p}(G)$ in terms of $n$ distinct orbits: $\text{Syl}_{p}(G) = O_1 \cup O_2 \cup O_3 \cup \dots \cup O_n$. Let a representative of the orbit $O_i$ be $P_i$. \
To begin, $|O_i| = |P: \text{N}{P}(P_i)| $. Now, consider an element $x \in N{P}(P_i)$, and note that $\langle x\rangle P_i$ must be a $p$-subgroup, implying $\big|\langle x\rangle P_i\big| \leq p^n$. Because $|P_i| = p^n$ and $P_i \leq \langle x \rangle P_i$, we have $\big|\langle x\rangle P_i\big| = p^n \implies \langle x \rangle P_i = P_i \implies x \in P_i$. Further, since $P \cap P_i \subset \text{N}{P}(P_i)$ and $\text{N}{P}(P_i) \subset P \cap P_i$, we have $P \cap P_i = \text{N}_{P}(P_i)$ and, thus, $|O_i| = |P: P \cap P_i|$\
Now, we can say $|\text{Syl}{p}(G)| = |P: P \cap P_1| + |P: P \cap P_2| + |P: P \cap P_3| + \dots + |P: P \cap P_n|$. Let $P_1 = P$. Then, we have $|\text{Syl}{p}(G)| = 1 + p^{\alpha_2} + p^{\alpha_3} + \dots + p^{\alpha_n}$ where $\alpha_i \neq 0$ (If it did, then that would imply $P_i = P$). If we assume that all $\alpha_i > 1$, then we have $n_{p}(G) \equiv 1 \mod p^2$. Thus, there must exist some $\alpha_i = 1$, and, hence, some $P_i$ such that $|P: P \cap P_i| = p$
cloud walrusBOT
∀ScoopityPoop, Scoop ≠ Poop:
spark bear
I forgot what are p-Sylow groups... 😭 😂
brisk granite
Essentially, they're just the biggest possible p-subgroups
Like, by order I mean
And, a p-group is a group that only consists of elements that are a power of p
from cauchy's theorem, it's also easy to see that these groups must be have an order p^m.
so, more simplistically, p-groups are groups such that their order is a power of p
brisk granite
ok, thankyou
brisk granite
Could I have an example of a 2-transitive group?
Other than S_n (and things that are isomorphic to it)
hot lake
do you mean a 2-transitive group action on a set ?
uncut girder
@placid pond
placid pond
@uncut girder that argument wasn't what i had in mind, if that's your question
uncut girder
How did you prove C+(x^k+1) is Noetherian
placid pond
i did it using a variation of euclidean division
pick any ideal I of this ring
then, pick polynomials in I having degree c mod (k+1) for every residue class c such that they have minimal degree
then you can do a variant of euclidean division to show that these guys must generate the ideal I
so in fact you get an explicit bound on the number of generators
namely, k+1
i think this is somehow equivalent to their argument actually
i just did it a bit more concretely
uncut girder
So every ideal is finitely generated
placid pond
yeah
in fact we have something stronger, they are "uniformly finitely generated"
that is, we have a uniform bound on the number of generators required
that's not always the case
uncut girder
How did you know l would be generated by smallest deg polys with degree = c mod k+1
placid pond
i just thought of what i would need to make euclidean division work
it's the same idea as in the hilbert basis theorem
basically
@uncut girder what i'm doing is just rehashing the proof that direct sums of noetherian modules are noetherian in this specific case
finite direct sums
uncut girder
I mean
That's actually a corollary in this book
Its before hilbert basis theorem
If M_i are Noetherian A modules, then so is
(+)_i M_i (1<=i<=n)
So lets just stay with 2 modules M and N
0->M->M(+)N-> N->0 is exact (why?)
placid pond
yeah, the more general statement is that if you have an exact sequence 0 -> A -> B -> C -> 0
and A, C are noetherian
uncut girder
Then it follows M(+)N is Noetherian
placid pond
yes
bleak finch
What's the algorithm for multiplying two permutation cycles?
bleak finch
That involves translating them into the two row form
placid pond
it's an algorithm
i don't know of an easy way to deduce the cycle structure of the product from the cycle structure of the individual factors
unless you're dealing with a trivial case
bleak finch
It's an algorithm in the same sense that converting numbers from base ten to tally marks, adding them and converting back is an "algorithm"
placid pond
like one where the cycles are disjoint or whatever
uncut girder
So I dont understand the relationship between noetherian modules and noetherian rings 
placid pond
@uncut girder a noetherian ring is a noetherian module when viewed as a module over itself
that's the definition
and finitely generated modules over noetherian rings are always noetherian
(as modules)
that's about all you need to know
uncut girder
The submodules of a ring when viewed as a module over itself are the ideals of the ring correct?
placid pond
yes
uncut girder
What about subrings
placid pond
subrings you can't say anything about, they may or may not be noetherian rings
again, take a trivial example
any integral domain embeds into its field of fractions, which is trivially noetherian
since it only has one proper ideal
but there are many domains which aren't noetherian
so about subrings you can't say much
you can think of a noetherian ring as a weaker kind of principal ideal domain
many of the results carry over, a PID is like a one dimensional noetherian ring in some sense
that's not exactly true, but whatever
just like every finitely generated module over a noetherian ring is noetherian
every finitely generated module over a PID is free
well
torsion-free
is required
most of the module theoretic properties of principal ideal domains are also present in noetherian domains, just in some weaker form
uncut girder
So we have C + (x^k+1) is Noetherian as a C module
But we want it Noetherian as a ring
Is that a stronger or weaker statement
@placid pond
placid pond
it's not noetherian as a C-module
it's noetherian as a C[x]-module
well
C[x^(k+1)] module
which is the same thing
and yes, if you prove it's noetherian as a C[x^(k+1)]-module you're done
as a C-module (vector space) it's infinite dimensional
uncut girder
Oh so you're thinking of C as a C[x^(k+1)] module?
uncut girder
C[x^(k+1)] is a subring of C + (x^(k+1))
Also C + (x^(k+1)) is finitely generated as a C[x^(k+1)] module
Therefore C + (x^(k+1)) is a Noetherian ring
I finally understand it
It's just the proof from that image earlier
Ok you dont even need the direct sum thing from earlier
You literally just need this proposition:
Let A be a subring of B; suppose that A is Noetherian and that B is finitely generated as an A-module. Then B is Noetherian as a ring.
chilly ocean
For cyclic group 3, we have {I, R_{(2\pi)3}, R_{(4\pi)/3}, 0}
mild laurel
Is this a question
hollow flume
Understand what?
chilly ocean
@hollow flume how to get the rotation of a C_n object
hollow flume
I have no clue what that is
chilly ocean
Well, an N dimensional rotation for C_n
In this case, C_4
This has to do with the symmetry of the dimensional object
Would it just be
{I, R_{2pi/2}, R_{3\pi/2}}
hollow flume
Sort of makes sense too me
But I'm not that interested in math
so this probably won't matter much too me
chilly ocean
How the fuck do these rotations happen
tame bear
what are you confused about
chilly ocean
How do I calculate them?
@tame bear because for C_3, we were given {I, R_{2pi/3}, R_{3\pi/3}}
Where I = identity
The 2nd element is the 2nd possible rotaion, and the 3rd element is the 3rd rotation
But I don’t understand what I’m supposed to do — how do I get 2pi/3
tame bear
thats just the 3rd roots of unity
chilly ocean
Uh, yes, yes— what does this mean @tame bear
tame bear
this is the 5th roots of unity
theyre complex numbers with length 1, and an angle that lets you spin around to 1 after n powers
multiplying complex numbers is easy, you just multiply the lengths, and add the angles
chilly ocean
But the professor was showing actual rotations?
tame bear
same thing
different ways of writing it down
rotating by 2pi/3 is just 120 degrees
and 3 120 degrees is 360, which is no rotation
chilly ocean
So what does it mean to change C_3 to C_4 @tame bear
tame bear
c4 is just 90 degree rotations instead of 120
chilly ocean
Oh, so 360/4?
chilly ocean
Determine the set of complex vertices of the regular n-gon inscribed in the unit circle f the conplex w/ vertex at 1. Prove a Bijection between this set of complex numbers under multiplication, the rotation group C_n, and the additive group Z_n, for n is an element of the natural numbers_>2
@tame bear
Any idea?
tame bear
Determine the set of complex vertices of the regular n-gon inscribed in the unit circle f the conplex w/ vertex at 1
these are just the n roots of unity that i was talking about
cos(2pi / n) + i sin(2pi / n)
and all n powers of this
uncut girder
@placid pond are you around
The question this time is to determine if the followijg ring is noetherian: the ring of polynomials in z and w with complex coefficients, all of whose partial derivatives with respect to w vanish for z=0.
It turns out the ring is C[z] + zC[z,w]
uncut girder
(You can obtain this by writing am element p in C[z,w] as a polynomial in w with coefficients from C[z], then taking a partial derivative wrt w, and setting it equal to 0, gives that p is in C[z] +zC[z,w] )
Finally the ring is not Noetherian because the ideal zC[z,w] is not finitely generated.
latent anvil
Can someone help me with some homological algebra?
I'm working on problem 2.2.1 in weibel: P is a projective object in Ch(R-mod) if and only if it is split exact and P_n is projective for each n
I got the if direction
Weibel says to consider the surjection g : cone(P) -> P[-1], which is g(x, y) = -x. So I defined γ : P -> P[-1] by γ_n = (-1)^n d_n (the sign weirdness is to make it a chain map, since the differential of P[-1] is the negative of the one for P).
This gives me a unique map φ : cone(P) -> P[-1] such that g ° φ = γ, i.e. φ_n(x) = (-γ_n(x), ψ_n(x)) = ((-1)^(n+1) d_n(x), ψ_n(x)) for a family of homomorphisms {ψ_n : P_n -> P_n}
By messing about with the chain map square for φ, I got that ψ_n(d_(n+1)(x)) = d_(n+1)(ψ_(n+1)(x) - (-1)^n x), but I might have screwed up the signs
I'm not really sure how to proceed? I wanted to show that the identity extends to a map on cone(P) but it's not popping out
solemn hollow
why did they only take cycles as the representatives? S_3 is the set of all permutations on 1,2,3.
mild laurel
What do you mean
H is just a subgroup within G?
Do you understand what the (1 3 2) notation means?
tame bear
permuations are cycles
solemn hollow
I thought they were trying to get all the right cosets
zoph, 1 goes to 3, 3 goes to 2, 2 goes to 1
stone fulcrum
Every element is there! What's missing?
tame bear
those are the right cosets
solemn hollow
oh, isn't (1)(2 3) for example in S_3?
solemn hollow
oh I see now after trying the other elements of S3 as representives, thats cool
ty everyone
smoky cypress
On ordered set is just set where order is defined right?
Which means that for an ordered set $S$ and for all $a,b\in S$, only one of
$$a<b,a=b,b<a$$
is true.
cloud walrusBOT
Whoever:
smoky cypress
Got this definition from Principles of Mathematical Analysis by Walter Rudin
I feel like that's not right for some reason
And also I've heard that there are these partially ordered set and totally ordered set
What are the differences
And what are all kinds of ordering?
fickle brook
ok so uh
what you listed is the law of trichotomy, which is true for total orders, but is not the only thing defining them
a partial order is a relation that is reflexive (a ≤ a), antisymmetric (a ≤ b and b ≤ a => a = b) and transitive (a ≤ b ≤ c => a ≤ c)
a partially ordered set is a set equipped with a partial order
a total order is a partial order that is also total (i.e. for all a and b, a ≤ b or b ≤ a)
an example of a partial order that isn't total is the inclusion relation defined on 2^X for some set X
smoky cypress
Ok
Oh the total (for all a and b, a ≤ b or b ≤ a) is just the trichotomy law right?
fickle brook
no
smoky cypress
xD
fickle brook
i mean, i guess it's equivalent to trichotomy
smoky cypress
That's what I meant
fickle brook
but i'm not using the corresponding strict order at all
smoky cypress
Ok, why do we use ≤ instead of <
In the definition?
and can I just define < to be ≤ but not =
chilly ocean
@fickle brook for any set S, a one-to-one and onto mapping is called a permutation of S.
Isn’t this just the same as a bijection?
thorn delta
yea
chilly ocean
Then why the fancy wording ;c
thorn delta
one to one is injective
onto is surjective
one to one and onto is bijective,
but I dont think there is a one word equivalent for bijective :c
chilly ocean
Yes, but there must be a difference between a bijection and permutation? @thorn delta
Or is it the same and depends on your style of writing???
solar vessel
I'm pretty sure a permutation on S is a bijection from S to S
mild laurel
You can have bijections between different sets
a permutation is specifically a bijection from a set to itself
chilly ocean
Oh, I see! Thanks Zopherus.
uncut girder
Can you have modules A and B so that A\oplus B is free while B is free but A is not?
oblique river
yes, such modules A are called "stably free". https://kconrad.math.uconn.edu/blurbs/linmultialg/stablyfree.pdf
uncut girder
Thanks
brisk granite
given a group of order p^n where n > 2, how do I show that it isn't simple?
fickle brook
show that it fails to satisfy the definition of a simple group
brisk granite
well, yeah
I can't figure out how to create a non trivial normal subgroup
Is there another definition?
fickle brook
well yes you want to create a nontrivial normal subgroup
try to think about how you may do that
and what order it might have
thorn delta
wouldn't any group of finite order have a cyclic subgroup? 🤔
fickle brook
any group has a cyclic subgroup
thorn delta
well, a non trivial one. Cuz wouldn't that be an example of a non trivial normal subgroup (since its abelian)?
fickle brook
order p^n doesn't imply abelian
you can have a sdp like $\bbZ_p^2 \rtimes \bbZ_p$ with the action given by $$\bbZ_p \ni x \mapsto \bigg( (a,b) \mapsto (a + bx, b) \bigg)$$
cloud walrusBOT
Ann:
fickle brook
p sure that won't be abelian
and yet the order of this group is clearly p^3
anyway that's a tangent
thorn delta
im a bit of a noob to group theory, so bear with me for a second. Lets say you have a non cyclic finite group G of order m>1. Let g != e be in G. wouldn't <g> be a non trivial cyclic subgroup of G? If so, then it would also be a normal subgroup, and scoopity's thing follows.
faint elm
I can't see why it is also a normal subgroup. Can you explain please?
fickle brook
yeah no that doesn't hold up at all @thorn delta
brisk granite
It doesn't have to be normal. As a counter example, consider A_5 and <(123)>
brisk granite
oh wait, the center is always normal but it's never trivial for p-groups
so, yeah, that's it right?
placid pond
@brisk granite sure
brisk granite
k
Also, I think I managed to show that any group who's order is a product of three distinct primes isn't normal simple
Is this fact true?
brisk granite
Is it true that all groups of order p^n * q^m are not simple?
oblique river
I believe that is true
well maybe nto, I wanted to apply zassenhaus but maybe you need to do a little more work
magic owl
I vaguely remember seeing that statement or a similar statement though
brisk granite
I'm not sure how to start tho
To prove that statement
It would appear that my sylow subgroup tricks don't really work here
btw, can I have the number of sylow p subgroups be q?
oblique river
I mean, what do the sylow theorems tell you
oblique river
I mean I could look them up, I just don't remember the exact congruences
do the congruences say that the number of p-sylows could be q?
if so then yes its possible, if not then no it's not possible
brisk granite
Well, the sylow theorems simply guarantee that n_p(G) = 1 mod p
oblique river
great, so if q is not 1 mod p then n_p(G) can't be q
but if q = 1 mod p then it could be q
brisk granite
Oh
Ok
I'm not sure I understand because all the theorem states is that the group must be solvable(I'm not sure what this means )
stone fulcrum
A group G is solvable if you can create a chain of groups:
e → A → B → ... → G
Where every group is a normal subgroup of the last, and the quotient of any two connected groups is abelian
Note A5 is not abelian and has no normal subgroups, so it is not solvable
placid pond
nontrivial
thorn delta
what does it mean for two group elements to be conjugates of each other?
thorn delta
i already checked, but Ill double check real quick lol
mild laurel
What book is it
magic owl
There exists g in G such that x=gyg^-1
tame bear
lots of elementary group books introduce important stuff like conjugation in exercises, for some awful reason
$x = g * y * g^{-1}$
cloud walrusBOT
Sigma:
magic owl
I said that 😭
tame bear
not with $\LaTeX$
cloud walrusBOT
Sigma:
tame bear
conjugation is really neat though, it has tons of nice properties
magic owl
There’s a bijection between covering spaces of a top space X and the conjugacy classes of its subgroups
thorn delta
oh i think i got it now, but I am now bamboozled by how this is reflexive. x is conjugate to itself?
magic owl
Of the subgroups of its \pi1
Yes
Can you think of a specific element of g
Such that gxg^-1
Is still x
thorn delta
the identity i guess. I thought g was a fixed element though?
magic owl
No
tame bear
could also write it as $xg = gx$
cloud walrusBOT
Sigma:
tame bear
g can be anything
magic owl
As g varies across all of G
tame bear
x and y are conjugate if there is ANY g that works
where g is just another element of the group
~~ conjugacy classes of x are just images of x under an Inner Automorphism ~~
thorn delta
alright i will ponder this. thanks
thorn delta
https://media.discordapp.net/attachments/496784958430380033/607725077424898068/unknown.png?width=1440&height=140
so on part b of this problem, it just wants you to see that these are the group elements that only commute with the identity, correct?
thorn delta
i noticed that a is also conjugate to itself via a (in addition to the identity)
beat me to it lol.
chilly ocean
Why do that line through thing
What’s the point of it
Why are you cheating on us Ethan
thorn delta
go bak to ur hobbit hole ilmd 
chilly ocean
stone fulcrum
@thorn delta
Elements that only commute with the identity? I don't feel like that's correct
These are elements that always map back to themselves via conjugation
thorn delta
hmm okay but when does that happen, i.e. when does a = gag^-1? Presumably whenever g = e or g = a.
stone fulcrum
Not only those cases. For example, imagine a commutes with any g
mild laurel
There are more cases
stone fulcrum
Also, remember that g should be allowed to be any element in the group
So one example is a = e, as e always maps back to itself
thorn delta
so this is a complete reversal of before, would this mean that a commutes with any g in general? I mean the only way a's equivalence class could be {a} is if b = gag^-1 => b = a for any g.
yea, i suppose thats what it means to "map back to [itself] via conjugation" means.
stone fulcrum
Indeed, let's say that you're working with addition on the integers. What is 5 conjugated by 3?
thorn delta
5
stone fulcrum
On any abelian group, gag⁻¹ = a
So in an abelian group, every element is alone in its conjugacy class
thorn delta
right, that makes sense.
stone fulcrum
More generally, any element that commutes with all of the other elements are alone in their conjugacy class
tame bear
the Center of a group
lapis turret
hello guys, i've been doing a bit of research but i would like a second opinion, what's a good resource for learning abstract algebra at a beginner's level, aka pls don't recommend me the napkin cuz i tried and failed miserably due to lack of maturity
so i would like to know what are the best resources to build math maturity in abs alg
mild laurel
The napkin is definitely not meant for learning a subject in depth
If you haven't yet
You should probably read a book on proofs
I enjoyed How to Prove It by Velleman
I think usually people recommend Herstein's algebra book as the easiest
bleak abyss
I used Herstein when I was starting and it was fairly easy but there's an issue
So I philosophically agree with using x(f) instead of f(x) notation
It makes everything work nicer
But it's not standard and as a result it's kinda not worth learning
e.g. I was multiplying elements in S_n backwards and it took a while for me to break that habit, which I eventually kinda had to
I hear Artin's not too hard it seems. He also does some linear algebra if you don't know it/only had a computational treatment, and is generally good at making algebra seem interesting
I mean you might be like me and just like algebra right off the bat but if not you'll wanna see it situated within math as a whole more to get the point
thorn delta
Harvard university has a series of free video lectures that follows Artin as well
elder valley
is my argument good here? showing: for k an algebraically closed field, A := k[x,y]/<xy-1> is not isomorphic to k[t].
suppose they were isomorphic. k[t] has maximal ideals such as <t>, and so A must also. maximal ideals in A correspond to maximal ideals in k[x,y] which contain <xy-1>. maximal ideals in k[x,y] correspond to points in affine 2 space, so say <x-a, y-b> is an ideal of k[x,y] containing <xy-1>. then there are polynomials f,g such that xy-1 = f(x,y)(x-a) + g(x,y)(y-b). evaluating this equation at x=a and y=b separately give that f is polynomial in y and g is polynomial in x, so xy-1 = f(y)(x-a) + g(x)(y-b) = xf(y) + yg(x) - af(y) - bg(x). left hand side has no non-mixed variable terms, so f and g must be constants. but then right side has no xy term, so contradiction
uncut girder
Interesting
elder valley
is it? xD
it doesn't really seem correct. essentially proving that A has no maximal ideals right? but assuming zorns lemma isn't every ideal contained in a maximal ideal?
uncut girder
ay-1=g(a,y)(y-b)
How does that imply g is a polynomial in x
ay=g(a,y)y
lmao
1=g(a,y)b
a=1/b
Yeah I dont see how g is a polynomial in x
elder valley
yeah that was my mistake then
i agree lol
i should be able to get that if I := <x-a,y-b> contains <xy-1>, then I must be principal since it corresponds to some ideal in k[t] through the isomorphism
elder valley
that's the contradiction
teal lodge
hey guys do the terms ring and integral domain refer to the exact same thing? a set that is closed under addition and multiplication?
and if so is integral domain still used or not?
teal lodge
no? to which
fringe nexus
its not the same thing
so integral domains are always rings
but not the other way round
teal lodge
hmmm i see
fringe nexus
integral domains have to satisfy the condition where
we have a*b = 0, then either a or b has to be 0
teal lodge
ohh i see now
yea i just compared my definitions a bit more closely
so an integral domain has bigger restrictions then
chilly ocean
yep
a very likable restriction
like everyone loves to say ab=0 => (a=0 or b=0)
however that's not true in the ring of 2x2 matrices over IR for example
magic owl
t o r s i o n f r e e
thorn delta
I'm not exactly sure how to "give an example" here. Is it not enough to show that complex invertible matrices do not commute with real invertible matrices?
thorn delta
err, im not sure I see what normal subgroups have to do with this?
mild laurel
what does it mean if the right and left cosets of GL_2(R) were always equal?
thorn delta
That for any AB there exists some B'A such that AB = B'A. (cosets with respect to a matrix A).
mild laurel
uh what
tame bear
less words
mild laurel
what are A and B
thorn delta
$A \in GL_2 (\bb C)$ and $B \in GL_2 (\bbR)$. Left coset is $AGL_2 (\bbR)$ and right coset is $GL_2 (\bbR)A$.
cloud walrusBOT
kxrider:
Compile Error! Click the 
reaction for details. (You may edit your message)
thorn delta
is it all wrong 
bleak abyss
You wrote \bb C instead of \bbC
thorn delta
yea not worth editing
mild laurel
but this is exactly the definition for a subgroup to be normal
solar vessel
\bC
mild laurel
So what I was saying, is that what you were saying, isn't enough to show that this isn't a normal subgroup
bleak abyss
There's a useful characterization of normality which is fairly close to an idea in matrices
tame bear
do you mean similarity
bleak abyss
Yeah
Okay so originally I wanted to Socratic dialogue through this problem but I'm gonna take a shower so I'll just tell you what to do
So you wanna show GL_n(R) isn't normal in GL_n(C), that means that there's a real invertible matrix that's similar to a complex matrix
You can do it either directly by trying matrices and seeing what happens, or you can give a more conceptual description
Think Jordan form/triangularization
Anyway see you
thorn delta
okay that makes sense. thanks
fading wagon
What answer?
brisk granite
gimme a min
\textbf{Claim:} Given a subgroup $H$ of $G$ such that $|G: H| = p$ where $p$ is the smallest prime that divides $G$, then $H$ is normal.
\begin{proof}
Let $G$ act on the cosets of $H$ by left multiplication. Then, we can define the homomorphism $f: G \rightarrow S_p$ where $\ker(f)$ is clearly in $H$ (because all elements of the kernal fix $H$). Now, we know that $p$ divides $|G: \ker(f)|$ and that $|G: \ker(f)|$ divides $p!$. Let $|G: \ker(f)| = pm$, and, hence, we know that $m$ divides $(p-1)!$. If $m$ divides anything other than $1$, then we contradict the minimality of $p$. Thus, $|G: \ker(f)| = p \implies \ker(f) = H$
\end{proof}
cloud walrusBOT
∀ScoopityPoop, Scoop ≠ Poop:
fading wagon
I think your proof is too long
brisk granite
yea, it's pretty long
fading wagon
try breaking it into chunks of <2000 characters
brisk granite
I have a lot of cases
I can send it on discord but it texit just doesn't wanna print it
Lemme just take a pic
Also, is there a better way to do this?
fading wagon
Well, probably not, unless you use classification of simple finite groups, which is probably not the main idea here
brisk granite
are the cases correct tho?
fading wagon
Well, you went through all the cases, but my group theory knowledge is not enough to verify
brisk granite
ok, thankyou tho
brisk granite
I dunno, it was in my textbook
But, uh, there is a lot of repetition in there.
repetition of the same argument
placid pond
@brisk granite well, there are like 3 arguments
in introductory group theory
so that's inevitable
brisk granite
The sylow subgroup stuff, the normalizer stuff, and the kernal stuff, right?
placid pond
i didn't actually count when picking the number
lol
it's just that it's very constrained
most of what you prove you prove using sylow
brisk granite
yep
Does group theory have any applications beyond number theory, combinatorics, and topology?
Non math fields I mean
placid pond
i mean, is physics a non math field
placid pond
in my experience you see a lot of lie theory in physics
finite group theory is not as popular
brisk granite
ok
placid pond
but lie theory is important because of a lot of reasons, one is that it's important for solving certain kinds of PDEs which come up
two is it's important for analyzing symmetries of physical systems
you get a connection between observables and symmetries given by noether's theorem, so it's important to understand the representation theory of the gauge group of your physical theory if you want to better understand it
it's important for quantization of classical fields, because when you do that you have a collection of observables which are tied to one another through certain commutation relations, and you want to construct a representation of the lie algebra generated by those guys from scratch
finite group theory i haven't seen as much
but that's probably because of my own ignorance
brisk granite
tbh, this kinda gave rise to more questions, but I don't wanna really go down that line of questioning
probs cuz I know basically nothing about physics
placid pond
ahah
chilly ocean
quick question
do the parenthesis signify composite functions or is it simply multiplication
magic owl
without context hard to be sure but I'd be shocked if that notation wasn't composition
chilly ocean
hmm its the lie bracket of vector fields X and Y used on f(x^u) where x^u are coordinates on a manifold
actually yeah it should be composition since you can show its linearity with X,Y= a[X,Y]f + b[X,Y]g
thanks
fallen bluff
This exercise is bullshit right? Take a = a matrix (0 1 ; 0 0) and b = 0, then a^2=0, a^3=0 but a is not 0
fallen bluff
What retardo puts wrong exercise in their book
wind steeple
♿
hollow peak
@fallen bluff it works if they are invertibles
tame bear
check the errata to see if theyre more specific
latent anvil
Can I ask for feedback on some problem sets I'm writing for an undergrad algebra course/study group next year?
golden pasture
yea
latent anvil
Here's the schedule and problem sets for the first quarter/first two weeks of the second
It's for two ten week courses, first on groups and second on rings/fields/Galois theory
The old spring problem sets are under revision, but can be found by modifying the url, like https://math.berkeley.edu/~tb65536/algebra_materials/spring2.pdf
(also that site is not mine, but the person hosting it helped organize this year's version of the course with me)
Students aren't expected to have any prior experience with algebra
Practice problems are supposed to be easy and not super interesting, presentation problems are supposed to be moderate to hard and interesting, and tricky problems are supposed to be extremely hard
lapis turret
@bleak abyss @mild laurel thnx for your insight will look into those boosk
and sry for the bit tardy reply
fossil vapor
Do all theorems in the theory of real numbers that aren't true in the theory of real closed fields start with an existential quantifier?
Related https://math.stackexchange.com/questions/151184/model-of-theory-of-real-closed-field/151206
I.e. in the theory of the reals - that is a real closed field but lacks some more axioms - are all theorems we can prove there "just" existence statements for certain numbers with certain properties? Or are we robbed of more?
oblique river
weak flower
A proof in the textbook for in a Euclidean ring R which is not a field, and whose degree function is a norm, x ε R is a unit if deg(x)=1 looks a bit strange to me.
The textbook divides x by x^2 to give x=q*x^2+r, where deg(r)<deg(x^2)=1 or r=0, and since deg(r) cannot be 0, r=0 and x=qx^2 and x cancels to give 1=qx and x is a unit.
It looks to me that I can divide 1 by x and fill in everything else the same way.
So 1=qx+r and r=0 or deg(r)<deg(x)=1, since deg(r) cannot be 0, r=0 and 1=qx. Is there an error in this proof?
hot lake
your proof looks fine to me
bleak finch
The definition of a submodule is a subset of N such that for every x, y in N and r,s in R rx + sy is in N.
can this be relaxed to rx + y is in N?
bleak abyss
So, in general please don't bump stuff
Now, the point is that you're trying to find a way to think about V, which is a priori just an F-module, into an F[x]-module
bleak finch
Whoops I meant Problem 6
bleak abyss
You already know how F acts on V so the question is how x acts on V. And the point is it acts via T
Product of two sets should just be the set of products
bleak abyss
So that's already relevant for the first problem
But like
A+B
= {a+b}
Modules are additive groups
So it works even there
bleak finch
So I'm proving a property for groups and then saying that it also works for modules.
corollary
bleak abyss
Yup, any time you prove something works for all groups the corresponding additive statement is true for modules
(Or for all abelian groups even)
So the usual definition of modules basically just writes out all the axioms, and to be fair it's not hard to see (or it's explicitly stated) that those of addition make it an additive group
But there's a nice way to think about modules
Here's an exercise for you: find suitable operations that make the set of endomorphisms of an abelian group into a ring
And when you do that, show that (and I'm being intentionally vague but once you play around I will tell you if you can't figure it out) the usual definition of a module has the same data as the following: Let R be a ring. An R-module is an abelian group M and a ring homomorphism R->End(M)
magic owl
all abelian groups are Z-modules anyway right
bleak abyss
Yeah
magic owl
the muddying of the waters of Z-module vs Ab was very confusing
for someone who had not studied either properly
(in the context of homology)
bleak finch
Daminark does the ring homomorphism have to be onto?
bleak abyss
Nope
magic owl
where you go back and forth between thinking of your chain complexes as modules but you homology groups as abelian groups
bleak abyss
Yo you should totally read Jacobson's Basic Algebra
I'm working through that rather slowly (I'm like, probably a bit over halfway done with the rings chapter) since distractions but it's fucking good
I'm basically ready to declare it the correct algebra book
magic owl
Oh shit, I'm interested in that
bleak finch
is this a recommendation for me 😄
magic owl
what does HAlg use?
bleak abyss
It uses a sleeping pill called "Dummit and Foote"
I mean if you've already started reading yours and it's fine then I'm not gonna make you start over
Jacobson's development is very...
You can't easily jump into the middle I feel
For example, his chapter on groups does a lot of stuff more generally for monoids
So when he does stuff for rings he basically double cites stuff
magic owl
I already own a copy of DF
but thats insane
DF is the dryest way possible
to learn algebra
bleak abyss
You don't quite have the same notion of a "normal submonoid", he just defines a quotient on a monoid to be an equivalence relation such that multiplication descends to equivalence classes
And then proves that if you're a group, then the equivalence class of the identity is a normal subgroup and the equivalence relation is precisely the cosets
And also that if you're a normal subgroup then the usual quotient group construction is a "congruence" (sorry that was the name of the equivalence relation, not a quotient)
Anyway so he does that in its generality
When it's time for rings
He's like okay a ring is an abelian group with another operation for which it's a monoid
magic owl
this is a neat approach
maybe i'll use it alongside Halg
although I'm not honestly that worried for the class
bleak abyss
And he's like okay a congruence for a ring is a congruence for both operations. By the group theory side it's a subgroup, and that it's a multiplicative congruence shows that it's an ideal
And that quotients of ideals are congruences
So there's a lot of handling things in one shot which makes it hard to jump in the middle
But it minimizes repeated work
Which I think is kinda good especially when said work is very axiomatic
Oh so
magic owl
is there a reason you're going through this
bleak abyss
Review for quals
magic owl
Right, makes sense
bleak abyss
I don't have to do quals now but if I can it's nice
bleak finch
you can just publish original quality research and you will be regarded as a professional mathematician without a ph d
bleak abyss
Algebra, analysis (2 options), topology (2 options), applied math, computational math, and logic (3 options)
magic owl
or E) read weibel
bleak abyss
@bleak finch within reason that strat won't succeed
Unless you're divine
In which case you're succeeding for sure
bleak finch
well yeah because divine is dead
magic owl
tbh its hard enough to publish research with a phd
bleak abyss
But what if you went to your algebra prof and tried to say that you sorta know a fair bit of algebra already, could you be more independent?
bleak finch
although you're right if you publish OR that's good straight out of uni that's a miracle
bleak abyss
I know Maxime was okay with one guy who knew rep theory of finite groups kinda doing some side stuff on Lie groups
magic owl
what if i just give nothing but category theory defs of algebra stuff
a group action? just a functor
bleak abyss
So if you just go to your algebra prof like yo fam, could I meet with you on the side to do more advanced algebra stuff, especially stuff that's relevant to your interests
magic owl
Actually
bleak abyss
For instance the group theory quarter you could start to teach yourself group cohomology
magic owl
youll get a kick out of this maybe
bleak abyss
Or something like that
bleak abyss
Daniil is p good, kinda sad I never took anything with him
magic owl
So I think I could def get him to do side stuff with me
He works in a subject that is ironically close to and very far form my interests
bleak abyss
So yeah maybe that could even replace the normal stuff so you don't have to do like, double psets for a class on top of 5 classes
magic owl
he does mixed tate motives
bleak abyss
Like yeah instead of a pset on the axioms of a group could I meet with you and do some group cohomology or algebraic groups or something and I just talk to you weekly about what I'm learning
magic owl
I'd be a little worried that the finer points of like
finite group theory and sylow stuff
are stuff I need to actually spend time on
tbh I wouldn't mind doing double psets + reading group + 4 classes
bleak abyss
I guess it depends on which classes you're doing
bleak abyss
That doesn't say much by itself
bleak abyss
Like idk for instance there was one quarter where I took CS 151 + Classics + Multiscale Bio + 207
In fact I almost did 161 CS instead of 151
magic owl
I hear its the least non-trivial
magic owl
ugh I've heard
but I have a pretty extreme buff/nerf to my productivity based on interest
like when I'm bored I'm fucking useless
bleak abyss
I don't wanna be the one to encourage taking joke classes for the core in general, like I took HBC and Classics and it was great
But I will say that I blew off Ancient Empires so fucking hard
I'm not proud of that
But I did
And I got a B+ each quarter
magic owl
I got an A is every quarter or sosc/hum and have never done a single reading
you're not getting judged by me is my point
bleak abyss
Lol fair
I mean I'm not worried about getting judged so much as, on principle of the matter I feel guilty saying this
But if you don't care anyway I'll say that I missed so many fucking CIV classes
magic owl
I can easily participate as well as other students without doing any readings
so like why bother
honestly some of those discussion sections are so lame
and the takes are so bad
bleak abyss
Spent some classes in second quarter ancient empires writing up the one page things that were due that same day
magic owl
ok wait wait
magic owl
this isn't abstract algebra
magic owl
what was the author tho
woven delta
👀
magic owl
Jacobson?
woven delta
This isn't Algebra
bleak abyss
But point is there's a good chance that including class time civ took 30 hours in the whole quarter maybe so if you need a light class, Ancient Empires is an option
Yeah Jacobson
There are two volumes
Volume 1, chapters 1-4 (also 0 but lol) are most relevant
0 = preliminary stuff, 1 = groups and monoids, 2 = rings, 3 = modules over a PID, 4 = Galois theory "of equations"
magic owl
have you done any aluffi
stone fulcrum
I'm reading it, he's wordy. It would be nice if it were split into paragraphs and not a big wall introducing concepts randomly
bleak finch
Daminark what do you think of algebra 0 by aluffi
magic owl
but don't like clearly deliminate
cool stuff to learn alongside learning category theory
when you already understand all the algebra
Chapter 0 is not in my opinion the right introduction to algebra
bleak finch
so I have kind of a question on an application to group theory to real life
bleak abyss
I hear Aluffi is good for introducing category theory in stuff
But his exercises are bad
bleak abyss
But I haven't personally verified this, this is just hearsay
Actually lemme send the contents of Jacobson so you can see how fucking hype it is
bleak abyss
Probably the only intro algebra book whose table of contents I'd call sexy
bleak finch
what is the name of riehl's book
magic owl
Sorry this is for category theory not algebra
but Category Theory in Context
great great book
bleak finch
which algebra books have good exercises
magic owl
probably the definitive text on category theory
DF has a ton
Pinter has good exercises
maybe Dami can speak to Jacobson
idk I feel like in some sense
and maybe this is because of its age
DF doesn't "see the light"
it makes me feel like I'm being taught algebra by an analyst who knows the material perfectly well
but doesn't love it
Pinter, on the other hand
might love it too much
because he is so fucking wordy
but honestly its a great gentle intro to algebra imo
bleak abyss
bleak finch
then you might shouldn't have recommended DF if it isn't the greatest
magic owl
does he mean like Aut(S) for a set S?
bleak abyss
magic owl
Oh ok
bleak finch
yeah I did; so I guess the exposition is unideal but the exercises are good?
bleak abyss
magic owl
and Aut(S) is the group version
magic owl
Not that interested in 9-11
But everything else looks cool
Oh shit
Chapter 6 is lit
bleak abyss
Sounds like u suck
bleak abyss
There's stuff on libgen but none of the copies are nice
magic owl
I might ask daniil to help me with weibel
bleak abyss
So I decided on these because I felt that having physical copies of the books would make me less likely to just download the pdf and never actually get around to working
If I spend money on it I'm more guilty about not using it, you know?
But if I was going with the online one
bleak abyss
There was another book I was considering
magic owl
I have a nice collection going
topaz solar
@bleak abyss that jacobson books looks lit
bleak abyss
Jacobson or the one I linked?
topaz solar
I assume the one you linked, since the 2
bleak abyss
Jacobson also has 1 and 2
topaz solar
true
smoky cypress
Oh my goodness
chilly ocean
@bleak abyss book name?
smoky cypress
Basic algebra?
I guess it is actually basic
Nvm
The second volume looks hardcore
But first volume doesn’t look at bad
libgen time
lyric nacelle
I have a stupid question. In basic group theory, how do I refer to $g^2$, $g^3$ and so on. Saying the second or third power of $g$ doesn't sound right. I think I am blanking on something obvious.
cloud walrusBOT
StealthTouch:
shrewd halo
I just say g cross g however many times
That’s just me though, there’s probably some standard
woven delta
g to the n
fading wagon
g squared, g cubed, g to the n, I just treat it like multiplication
woven delta
Is what I say verbally
lyric nacelle
Thanks!
If I want to refer to several of them at once, should I say something like "the first few powers of $g$"?
cloud walrusBOT
StealthTouch:
fading wagon
That sounds okay
bleak abyss
g shift six n
lyric nacelle
@bleak abyss do you think I am more likely to gain friends or lose friends if I use this?
bleak abyss
If you do it once or twice you may gain friends
If you keep your friends afterwards then you should be concerned
smoky cypress
Lol @magic owl I’m not really formally learning topology I’m just struggling with rudin chapter 2
And holy fuck the exercises are brain fucking me
And now it’s late and I’m not functioning correctly
brisk granite
I think it's true that the normalizer of G_i is G_i
brisk granite
nvm, I got it
H intersection G_i is normal in G_i but G_i is simple and H intersection G_i isn't just the identity
rain burrow
I'm not sure if this is the right place to ask the question.
From this part of the video, does it mean that a vector can represent area of many surfaces because there are many surfaces that the vector is perpendicular to.
https://youtu.be/f5liqUk0ZTw?t=94
smoky cypress
Ah tensors
chilly ocean
tensors are dank
brisk granite
@smoky cypress do you actually have a phd in topology?
smoky cypress
What do you think @brisk granite
brisk granite
idk
smoky cypress
Well you sounded like you don’t believe it
rain burrow
well, no one?
fading wagon
context?
twilit pawn
I need a sanity check on something from Artin's algebra book
the matrix isn't the correct one for the permutation he gives right?
bleak abyss
You're insane
bleak abyss
That matrix is fine
placid pond
^
twilit pawn
ok so I'm not understanding something basic then lol
bleak abyss
Yeah I imagined you were thinking about x_3 ending up in the x_1 coordinate as 1->3
twilit pawn
I was doing a dumb thing where I was like "the first entry became x_3, so x_1 -> x_3"
bleak abyss
Since side by side it looks like that
bleak abyss
np
twilit pawn
also @bleak abyss I think you recommended artin to me. very good call
bleak abyss
Also lol when people say sanity check and are correct I love to say "Right but you're still insane"
twilit pawn
I wish I had used this in the first place lol
bleak abyss
Yeah Artin's pretty good at giving you a reason to care about algebra
twilit pawn
no sane person would do math
bleak abyss
Compared to Dummit and Foote being all like "Here's a list of dem tru fax"
twilit pawn
I got backtracked to this because of a problem to show that A_n is generated by 3 cycles
and the first part of that problem was related to permutation matrices, and I wanted to use a similar idea
but then I realized I was misunderstanding them in a dumb way lol
bleak abyss
Oh, I don't think about that in terms of permutation matrices really
You can just write a double transposition as a product of 3-cycles
twilit pawn
oh ok that works as a solution lol
the first part was showing that S_n is generated by transpositions, and I think that's kind of nice to think about with permutation matrices
you can just describe an algorithm for that if you want
bleak abyss
Hmm, how do you think about it in terms of permutation matrices?
twilit pawn
so Sn is isomorphic to the group of permutation matrices. transpositions correspond to row swaps
you can take any permutation matrix and reduce it to the identity with row swaps. the algorithm would be something like "get the first row correct, then you have an n-1 x n-1 permutation matrix so you can induct"
so then if you reverse that, you'll have your original permutation matrix using only row swaps
which is the same as writing the permutation using only transpositions
bleak abyss
Ah
Clever
I think I just reduced it to the cycle case
Since product of disjoint cycles
twilit pawn
artin presents a lot of matrix things with row operations. I like this presentation a lot, it feels like a very clever way to do lots of these things.
so then do you just manually write a cycle as a product of transpositions?
bleak abyss
(a_1 ... a_n) = (a_1 a_n)...(a_1 a_3)(a_1 a_2)
twilit pawn
yeah makes sense, that's pretty simple too lol
that's how shotten did it I guess
I think I like my way a lot because it feels like you're actually doing it by hand instead of sort of just writing a formula and saying "this works"
bleak abyss
Yeah that's pretty fair
Oh speaking of Shotton
One day I was in his office and we were bored so just doing things
And in particular trying to find various ways of defining the sign of a permutation
Just to be funny
You might like this version
twilit pawn
sure lol
bleak abyss
So if you have a complex matrix, it has all its eigenvalues, and the determinant is their product
So the determinant in C of a permutation matrix is the sign of the associated permutation
Then use this to define determinants for other fields
twilit pawn
I don't think I'm following how you use this to define the determinant in other fields
placid pond
you have to be cautious here, because defining the determinant without having a definition of the sign of a permutation isn't trivial
bleak abyss
So in C you just define the determinant to be the product of eigenvalues
Which just works
placid pond
mhm
bleak abyss
Normally that's a theorem but if we have nothing better to do we call that the definition
So we use determinant in C to define sign using permutation matrices
placid pond
you have to prove that the determinant of a permutation matrix is either 1 or -1
bleak abyss
Now that we have sign we can define the determinant over a general field by the usual formula
twilit pawn
so if you define det in C as product of eigenvalues, you probably still want to check some things
placid pond
you have to prove that this definition as "product of eigenvalues" satisfies the usual properties of the determinant, that is
twilit pawn
like how row swaps change the sign. that would give you +-1 for permutations
bleak abyss
Well I think the main property that's important to show is that det is +-1 for permutation matrices and that it's multiplicative
Could you then easily rederive the original formula?
Like in principle you could define it by proxy for other fields but I do buy that the tricky part is showing that it reagrees in C
twilit pawn
so I think we can agree that with some effort this might be possible, but there are tricky things to be done lol
placid pond
there's nothing that stops you from making the same definition over an arbitrary field, by the way
you can always pass to the algebraic closure
put it in jordan normal form
blabla
we just don't know it has the properties we want the determinant to have
bleak abyss
True but I guess then there's the additional task of proving that if a matrix is defined over a subfield, the determinant also lives there
I guess Axler probably has that since I think that's how he does the determinant
But I'm a bit lazy to download and read it since this is mostly a waste of time anyway
placid pond
yes
probably you ought to find a more productive way to use your time
lol
like proving that sum_{p prime} sin(ln(p))/p converges
bleak abyss
inb4 that's some big open problem
bleak abyss
Oh lol
placid pond
the problem doesn't say it converges, but it does
i recently learned that it's still open if there are infinitely many integer x such that mu(x) mu(x+1) mu(x+2) = 1
annoying
bleak abyss
There's a part of analytic number theory which I think I'd like but probably not that stuff I must say. One guy here does like, cohomology and eigenbusiness of arithmetic manifolds
placid pond
lol
bleak abyss
Which sounds kinda fun honestly. He's on the young side and I have a bunch of people I'm more interested in so I prob won't work with him but I'd definitely like to hear him do chalkboard rants
golden pasture
i feel mildly rarted rn but im like stuck at jacobson 1.2 exercise 11
In a semigroup $G$, if $ax=b$ and $ya=b$ are solvable for all $a,b\in G$, then $G$ is a group
So rn im trying to proof a identity exists, but if we just call the solution to $ax=a$ $1_r$ i dont quite see how to proof $b1_r=b$ for all $b\in G$
cloud walrusBOT
Ariana:
hot lake
if only you could multiply a1=a by something so that it becomes b1=b
golden pasture
wait right
tame bear
could just plug in b like you did with a
golden pasture
no clue how i missed that thanks!
i shouldnt be doing this when overly sleep deprived idk why im still alive
hot lake
that would give the existence of a 1b that might be different from 1a
tame bear
but we cant really multiply by b a^1
since a^1 a = 1a
so we get b 1a 1a = b 1a
b 1a 1b = b
1a 1b = 1b
so still we have to show that 1a 1b is also 1a
golden pasture
$ya=b$ is solvable for $y$
cloud walrusBOT
Ariana:
uncut girder
I have a question
"Since the zero ideal has a primary decomposition, A has only a finite number of minimal prime ideals"
Why is this implication valid?
mild laurel
which page of atiyah macdonald is this on
uncut girder
90
mild laurel
Hm, the prime ideals belonging to 0, which is all of them, are all minimal prime ideals
uncut girder
How come all prime ideals belong to 0
uncut girder
@mild laurel
tame bear
https://yutsumura.com/prove-a-group-is-abelian-if-ab3a3b3-and-no-elements-of-order-3/
ugly ass question but i proved it 
placid pond
@tame bear that's a pretty famous problem
tame bear
oh, i hadnt seen it before
smoky cypress
Damn that proof is pretty good
Pretty smart
But the related question at the end is 
Much more trivial than the question
wow this website has problems in linear algebra and group, ring, field, galois, module theory
damn
that's pretty neat
placid pond
the problem is there aren't that many questions in group theory which are nice enough to be posed as exercises
so you see the same problems over and over again everywhere
i saw this problem 4 times or something
in different places
tame bear
i didnt like their proof so i made a different one
smoky cypress
oh ok
placid pond
nice
smoky cypress
oh
placid pond
here's another classic problem: suppose that G is a finite p-group, prove that if G has a unique proper maximal subgroup then G is cyclic
placid pond
yes
smoky cypress
what's maximal subgroup
placid pond
it's a subgroup of G not contained in any other subgroup
it's maximal with respect to inclusion
well, proper subgroup in this case
tame bear
p-group
NT and combi
the worst part of group theory
clear obsidian
So in the context of ring and field theory, when we say, 'x is an action on the set A' or 'AxA ---> A is an action where a in A acts on b in A', we just mean "multiplication" as any operation (on A) distinct from addition (but usually with convention depending on the set but not definitionally restricted to that convention)?
woven delta
I just think about an action as a map into Sym(A), or Aut(A) or whatever