Why?
#groups-rings-fields
shy bluff
next obsidian
Or hmm
Oh jjst
I think theyβll remain as disjoint cycles
Actually Iβm not too sure
Lol
shy bluff
I think they'll remain as disjoint cycles too
next obsidian
Yeah but you might need to break up
s_i
Into like two cycles
If itβs even and you squares it
Then you get two different cycles
Of half the length
And maybe it still works out the same but idk
In general if xy = yx
elder valley
you should have that <t,s_i> is the internal direct product of <t> and <s_i>
next obsidian
And their orders are coprime then the order is the product
shy bluff
wait why Auvera?
elder valley
trivial intersection and they commute with each other
shy bluff
Oh ok
next obsidian
Oh true
shy bluff
Interesting
shy bluff
Then the order of <t, s_i> is just |<t>| * |<s_i>|?
next obsidian
Of t^n and s_i^mβs order
shy bluff
Ah
And then <t>'s order is just 2 and <s_i>'s order is n - i
Maybe off by 1
What happens if we have that i = 1 or 2?
elder valley
i think when i=1 you generate S_n but that might be completely wrong
shy bluff
I don't konw if that's true
Like say we have (1 2) and (1 2 3)?
(12)(123) = (132), if i'm doing this right? (which I might very well nto be)
elder valley
that's the Herstein way π€£
shy bluff
Pardon?
elder valley
assuming you evaluate your functions from right to left you get (12)(123)=(23)
in Herstein's book he does left to right
shy bluff
OH
Idk I'm jus tfollowing this
But uh yea if i = 1 then I see that <s_i> is just S_n but how does that interact with <t>?
elder valley
<s_1> is not S_n, it's only order n. but <s_1, t>=S_n
shy bluff
wait why
shy bluff
hrmmm
elder valley
for i=2, i've no idea
shy bluff
hrmmm ok
Well I emailed the prof about the nilpotent matrices, thing and he said Suppose G acts on a set X, and x is an element of X. Then we say that the orbit of x is the set { gx : g in G }, and we say that a subset S of X is an orbit if S is is the orbit of x for some point x in X.
I still don't really understand how we're supposed to show that the set of nilpotent matrices of rank i is an orbit under the conjugation action of GL_2(K)
err what that actually means that we're trying to show
Oh I found a proof for <s_1, t> = S_n
Oh hey wait https://kconrad.math.uconn.edu/blurbs/grouptheory/genset.pdf both s_1 and s_2 generate s_n
shy bluff
Yea
And then corollary 2.6 is the i = 2 case
But question
why does this work?
Err
Where did the $\sigma^{-1}$ go?
cloud walrusBOT
Liria ^(;,;)^:
elder valley
i think you just evaluate it in cases. the only numbers in {1,2,...,n} that aren't fixed are sigma(1) and sigma(2), so just plug those in and see where they go
shy bluff
wdym by aren't fixed?
elder valley
aren't fixed by the function sigma * (12) * sigma^-1
fixed meaning mapping to itself
shy bluff
oh ok
shy bluff
Oh that's pretty cool
(12)(2...n) = (1...n) and so we end up generating S_n again
That took me a while to get π
shy bluff
Question: In general, for a group, can we say that its idenitty generates the group?
sour plume
Usually when you say "[thing] generates the group", you mean "every element in the group is a product of elements in [thing]"
Hence, when you say "the identity generates the group", you're saying "every element in the group is some product of identities"
but since identity times identity equals identity, this is never true when your group has some elements which are not the identity
wind steeple
product of elements in [things] U [thing]^-1*
sour plume
Yee, you're right actually
But in any case, even with inverses, the identity still doesn't generate much π
smoky cypress
How do I show this: Let p be the smallest prime dividing the order of a finite group, then any subgroup of index p is normal
wind steeple
look at the action of G over G/H
or H over G/H I don't remember
where H is of order p
smoky cypress
Oh thank you that's a good hint
shy bluff
hrm
pure crest
How can i show that the union of all Surds is a feild?
elder valley
how are you defining surd here?
knotty mason
any element of that union would just be a single surd right?
pure crest
sorry i meant to say
union of all tower of number feilds such that $B_{i+1} = B_{i}(\sqrt{\alpha})$ where $\alpha \in B_i$
cloud walrusBOT
boat:
vestal snow
What are good books for ANT (with good exercises, in particular)?
I got a copy of Lang's book, but it doesn't have exercises
shy bluff
Hrm
vestal snow
Thanks, I'll check it out
knotty mason
im hoping to start on it soon
shy bluff
How do I show that for every commutative R, there is an isomorphism like this?
elder valley
you just prove it's an isomorphism
shy bluff
Hrm
I think that part of my problem is that I don't really understand the procedure for generating such a homomorphism I guess
So like I get taht phi should map r(s, t) to r(y, y)
Right?
elder valley
yeah that's it
shy bluff
And then I need to prove that that is a surjective homomorphism with kernel (s - t) yea?
elder valley
yup
shy bluff
I guess that the hard part is showing the bit with division
Like I can show that clearly (s - t) is in ker(phi)
But I don't know how to show that it is all of ker(phi)?
elder valley
the direct approach would be to write out r as a polynomial in s and t, determine r(y,y), and set it equal to 0
you'll get some system of equations in the coefficients of r you can relate back to r(s,t)
shy bluff
wdym by "Determined r(y, y)"?
shy bluff
hrm
So I can go and show that like <s - t> is in ker(phi) right? And then I go and show it the other way around?
elder valley
well you posed the problem as R only commutative, so no
elder valley
i'm not sure you even need commutative
shy bluff
hrm
Do you be like "For any r(s, t) we have that phi(r) = r(y, y) = a_ny^n + ... a_1y + a_0", and then show that I can...
Hrm
elder valley
i think you need to write out r at the start, before applying phi
shy bluff
that seems awful to write out
How would you write out r? Like for regular polynomials it's a_nx^n + ... a_1x + a_0 but for here we can have differing degrees of x and y yea?
elder valley
yeah
often the definition R[s,t] = (R[s])[t] is used, so r(s,t) is a polynomial in t with coefficients that are polynomials in s
shy bluff
hrmmmmmmmmmmm can I just be like "For any r(s, t), if we take r(y, y) we get an element of R[y], and so we can write a_ny^n + ... a_1y + a_0"? the problem is that I don't know how to write r(y, y) as the product of something + (s - t)
Like it's not euclidean
So we don't have a division algorithm right?
elder valley
yes but also r(y,y) is a polynomial in y alone. the s and t are gone
because you've mapped into R[y] now
shy bluff
Wait does R[y] have a division algorithm?
elder valley
if R is a field it does
shy bluff
ok but R is just a commutative ring
elder valley
you're assuming r(s,t) is in ker(phi) right?
elder valley
So I can go and show that like <s - t> is in ker(phi) right? And then I go and show it the other way around?
i thought you were on the "other way around" part now
elder valley
right. so you're choosing r(s,t) in ker(phi)
shy bluff
Oh ok, so I'm choosing r(s, t) already in ker(phi)
And I want to show that it takes on the form of g(s, t)(s - t)
elder valley
yes
shy bluff
ok but how do you go and do that without a division algortihm 
elder valley
that's what i was outlining above π€£
shy bluff
So set r(y, y) = 0?
elder valley
yeah although i don't think it will do you any good without knowing how the coefficients of r(y,y) relate to those of r(s,t)
elder valley
if you write it out you will see
i'll get you started. say $r(s,t) = \sum \limits_{i=0}^{m} \sum \limits_{j=0}^{n_i} r_{i,j}s^it^j$ for some $r_{i,j} \in R$
cloud walrusBOT
Auvera:
shy bluff
Somewhat like so but the r coefficients are wrong
yea that
So you end up with $r(y, y) = \sum_{i = 0}^m \sum_{j = 0}^{n_i}r_{i, j} y^{i + j}$
cloud walrusBOT
Liria ^(;,;)^:
shy bluff
wait
can I go and use the fcat that like R[s, t] = (R[s])[t] to go and write like.. r(s, t) = r_1(s) + r_2(s)t + ... r_n(s)t^n, and then go and show that if f(r(s, t)) = 0, then we have that 0 = r_1(y) + r_2(y)y + ... + r_n(y)y^n? Or something like that?
elder valley
yes that's all true. where do you go from there though
elder valley
seems like you're trying very hard to avoid this double summation lol
shy bluff
Ah rip it doesnt' help
The double summation is hard for me to read , and thus I feel like using it is impairing my ability to understand what's going on
Like the hard part I think is that we don't have a diviison algorithm?
elder valley
hmm. even assuming R[y] is euclidean i don't see how that helps
like what do you divide by
shy bluff
Well then I can write it as being p(s - t) + r
And then I just need to show that r = 0
Right/
elder valley
no, s-t is not a polynomial in R[y]
shy bluff
Err
We can write r(s, t) as being
p(s - t) + q for some p, q in R[s, t]
Then clearly from the first part, p is in ker(phi)
hrm
Actually yea that doest' help much
elder valley
you could probably use grobner bases to get it pretty easily but i don't know much about them
shy bluff
Never heard of those either 
elder valley
it's the closest thing you get to a division algorithm in a multivariate polynomial ring
shy bluff
Well I found this, http://www-users.math.umn.edu/~felix077/download/sec9.pdf exercise 14 in it but I'm having a hard time understanding their arguement for why the remainder goes to 0, and this also doesn't apply to this question because R is not necessarily an integral domain
elder valley
yeah that should work
if you've proved the more general version of division algorithm, i think you dont need R to be a field, you just need the leading coefficient of the polynomial to be a unit
so you could apply that to the ring R[s][t]
shy bluff
I don't think I've necessarily proved a more general division alogrithm
Like this is what we have?
elder valley
how did you prove R[x] is euclidean domain?
shy bluff
Yea lol
I mean I have a second question that's very similar imo and if I can prove this oen I think I can prove the second one
vale flint
can we prove wolstenholme's theorem ($\dbinom{pa}{pb} \equiv \dbinom{a}{b} \pmod{p}$) using the frobenius endomorphism? i'm trying to right now by expanding the function $(1+x)^{ap} = (x^p + 1)^a$ in $\mathbb{F}_p$ and fiddling around with binomial coefficients, but i just can't get the result to stick. further, can we extend this to modulos $p^2$ and $p^3$?
smoky cypress
Using the equation $(x+1)^{ap}=(x^p + 1)^a$, the coefficient for $x^{pb}$ on the left side is $\binom{pa}{pb}$, and the coefficient for $x^{pb}$ on the right side is just the $b$th term from the right, which is $\binom{a}{b}$, and this proves the congruence mod p
cloud walrusBOT
Whoever:
vale flint
wait but doesn't FLT complicate the proof above...? i kept reducing powers modulo p-1 so that all i could get was an equivalence between sums (which said nothing about term-by-term equivalence, obviously)...
vale flint
like actually you have $\sum _{i \equiv b \pmod{p}} \dbinom{ap}{i} = \sum _{j \equiv b \pmod{p}} \dbinom{a}{j}$, right?
cloud walrusBOT
deiknymi0:
vale flint
i know i already took away the x
so like i'm taking the coefficients of x^1, x^p, x^(p^2), etc... because all of these are equivalently just x
smoky cypress
The equation is $\sum_{k=0}^{ap}\binom{ap}{k}x^{k}=\sum_{k=0}^a\binom{a}{k}x^{pk}$
cloud walrusBOT
Whoever:
vale flint
faceplam ahaha the irony i can't even spell it right. thanks for responding β€οΈ
smoky cypress
Lmao
delicate bloom
π very nice
sly bramble
Hi, I have a question pertaining to isomorphisms between general and special linear groups and symmetry groups. It is very easy to observe that an isomorphism exists between GL(n, Zp^k) or SL(n, Zp^k) and the symmetry groups S3, S4 and A4, and S5, but I was wondering if there was a generalizable isomorphism to a symmetry group Sn for all n. More specifically I am looking for an isomorphism from a matrix group with finite field entries to the symmetry group of any n-dimensional simplex.
supple marsh
Regarding GL_n (Z_p) - the determinant map is going to have a kernel thats of order p, so it would need to be p = 2 to be an S_m for some m. My sense is that except for small n the answer will be no? I think calculating the orders of these groups probably gives a proof of no, but haven't checked it. (The orders of GL_n are easy to count, you just describe ordered basis by picking a nonzero vector, then picking a vector independent to it, etc. and count the number of choices at each level. I think the order of SL_n is that divided by |F|, from first isomorphism using the determinant map.)
But if you are looking inside, then I believe there are faithful representations given by acting on the standard basis in F^n.
sly bramble
Ok that makes sense. I have found SL_n(Z_p) seems to work up to S_6 but I cannot find any matrix group with order 7! I included GL because it might've been possible but I see that it isn't. And yeah |GL_n(Z_p)/SL_n(Z_p)| = p-1
supple marsh
oh yeah by |F| - 1, lol. thanks
You always have the permutation matrices as a representation of S_n
and I think that works in any field
I didn't actually check the orders, but I would calculate them before believing anything I wrote above lol
vital quail
i was wondering why there were so many new faces talking math in chill
until i realized this wasn't chill
latent anvil
And it's faithful since you can look at the action on basis elements in a free module
supple marsh
@vital quail one day the time of great unification will come, and all channels will merge together and achieve global consciousness
latent anvil
This servers true purpose is the human instrumentality project
latent anvil
One time I was in my schools lounge and this grad student told me the empty ring wasn't a ring
supple marsh
well its not the empty ring (theres no such thing), its the ring where 1 = 0
latent anvil
So the empty scheme wasn't affine (or maybe not a scheme at all?)
supple marsh
the zero ring corresponds to the empty scheme though
latent anvil
oh sorry I meant the zero ring
latent anvil
he's like a 5th year algebraic geometry grad student who's very smart and capable
But he insisted it wasn't a ring
sly bramble
the textbook I used required a ring to contain the identity, but it's apparently a very controversial topic?
supple marsh
ppl gotta have opinions
well
the ring where 1= 0 does have an identity
its the ring {0}
sly bramble
the empty ring though
supple marsh
oh yeah there's no empty ring (at least in usually commutative algebra)
you need an element
like how there's no empty group
he's like a 5th year algebraic geometry grad student who's very smart and capable
@latent anvil does he also think every mathematician should switch to categorical foundations?
latent anvil
I'm not sure, but his thesis is focused on the derived category of the category of qcoh sheaves on a scheme, so he's definitely A Fan of category theory
bleak abyss
I'm cool with the zero ring but Z/8Z is just straight up sus
latent anvil
That's victim blaming
supple marsh
no victim blaming would be hating on 9 for getting eaten
supple marsh
8 just sat idly by while 7 used its homophone to eat 9
latent anvil
The monster
bleak abyss
@latent anvil circular reasoning smh
latent anvil
circular reasoning is valid as long as circular reasoning is valid
supple marsh
a funny thing about circular reasoning is that it's funny
bleak abyss
Wait what if we consider a truth system where truth values are on a circle and call that circular reasoning?
Logicians get on this
latent anvil
Sounds like Jan's thing
supple marsh
there's closed time like curves
circuit model of time travelling computation
Scott Aaronson has some interesting papers on it
latent anvil
Let G be a lie group. Consider a theory whose truth values lie in G
supple marsh
PSPACE becomes Polynomial sized circuits, because (philosophically) the difference between time and space disappears
sly bramble
Sorry just to go back to my question before, I guess I could use permutation matrices to represent S_n+1. I'm trying to visualize a rotation of an n-simplex corresponding to an edge-edge rotation of a tetrahedron. If the subspaces are literally just the axes of the dimension corresponding to the number of "symbols" in the permutation, then I guess that's easier than trying to find a generalized SL_n(Z_p).
supple marsh
well if you embed the simplex in R^{n+1} in the right way
the isometries /symmetries will be linear
and coincide with the permutation matrices
not sure if that answers your question tho, just free associating mostly, but maybe makes it easier to visualize
right way = convex hull of the standard basis vectors
im not completely sure what you are asking now though
whats an edge edge rotation?
like the induced action on the edge labels?
sly bramble
an edge-edge rotation is the rotation through the midpoints of the edges of the tetrahedron, so the elements of V4 I guess
I was thinking of potentially using partial permutations too to describe the rotation but I'm not sure if that would be helpful
supple marsh
helpful for what?
sly bramble
to understand the particular rotation of an n-simplex of order 2
supple marsh
what do you want to do with the understanding?
sly bramble
I need to explain a bit more to answer that question. I am dealing with "kaleidocycles" which are a "ring" of connected tetrahedrons that can be rotated. Right now I have a solid idea of how that works in R3, and a fairly decent idea of a parallel using 5-cells in R4. I'm trying to expand it to any dimension to then create a program that will visualize the rotation.
Permutations are far easier to work with than a time-variable matrix, so I figure it would be easier to deal with projective spaces rather than all of R^n
supple marsh
I'm not sure I follow why you are going to projective spaces
sly bramble
To answer simply, the rotation of an object of dimension 4 or greater is extremely difficult to visualize traditionally
supple marsh
(just set n = 4?)
sly bramble
But on the other hand a permutation of order 2 doesn't encapsulate the whole rotation, since it is both a rotation and translation
wdym?
sly bramble
lol
supple marsh
for 4 dim polytopes people like to use schlegel diagrams for visualization
sly bramble
yeah I've seen schlegel diagrams used but for anything beyond that, it seems like Coxeter diagrams and Cayley graphs are far more useful
which inspired the idea of using symmetry groups
latent anvil
Yeah Alex Voet
next obsidian
gotcha
next obsidian
chhhhhmonkey
wind steeple
a_v is zero almost everywhere
vestal snow
I thought it was something like $a_{(k_1,...,k_n)} X^{(k_1,\cdots,k_n)} = a_{k_1}{(k_1,...,k_n)} x^{k_1} + \cdots + a{k_n}_{(k_1,...,k_n)} x^{k_n}$
cloud walrusBOT
The formerly edible banana:
Compile Error! Click the 
reaction for details. (You may edit your message)
oblique river
it's times, not plus
also, it should be x_i^(k_i)
it's a polynomial in n variables
so those n variables should appear somewhere
vestal snow
$a_{(k_1,...,k_n)} X^{(k_1,\cdots,k_n)} = a_{k_1}{(k_1,...,k_n)} x_1^{k_1} \cdots a{k_n}_{(k_1,...,k_n)} x_n^{k_n}$
cloud walrusBOT
The formerly edible banana:
Compile Error! Click the reaction for details. (You may edit your message)
vestal snow
So this?
vestal snow
Gotcha
oblique river
a polynomial in n variables is a linear combination of terms of the form x_1^(k_1) * (x_2)^(k_2) * ... * x_n^(k_n)
oblique river
which is abbreviated as x^(k1, ..., kn)
and the coefficvient on that term is called a_(k1, ..., kn)
vestal snow
Oh wait
I actually came up with the same notation a couple of months back I think
I think I get it now
Thanks
Fellow 'nana
oblique river
haha
vestal snow
I have several questions:
- How do we chose m?
- Why does this choice of m guarantee monic in x_n?
- How exactly does induction kick in?
next obsidian
I feel like this proof isnβt a very well written one
Which book are you pulling it from?
vestal snow
Why does s_x = 0 imply s|_U_x = 0?
vestal snow
s_x = <s|_U,U>
What does this mean?
next obsidian
an element of the stalk at x
is an equivalence class
<s,U> where s in F(U)
and x in U
and <s,U> = <t,V> iff s|V\capU = t|V\cap U
so by definiton
next obsidian
I guess I'm not too sure with that construcitno of the direct limit
I just use the one I stated
haha
then s_x = <s,U> when s in F(U)
and by definition 0 in the stalk is just <0,U> for all U
So if <s,U> = <0,V> which means s_x = 0
then s|_U\cap V = 0|_U\cap V
vestal snow
Wait so shouldn't that give us that s|_U_x's image in the direct limit is 0?
Which would imply s|_U_x_x = 0?
wind steeple
yes
vestal snow
but why does the book say s|_U_x = 0 then?
wind steeple
s_x is the set of all sections over neightbourhoods of x mod equality on one neightbourhood
wind steeple
uh yeah mb
next obsidian
But banana
wind steeple
s_x is one of these classes
next obsidian
they're saying, depending on x
you have a nbd U_x
such that s|_{U_x} = 0
but since you have one such U_x for all x in X
that they cover
vestal snow
Yeah, but I don't get how s|_{U_x} = 0
next obsidian
because if s_x = 0
wind steeple
If s_x = 0 that means that 0 is in s_x, since s is in s_x by definition, we have that s and 0 are equal on a neightbourhood of x
next obsidian
then you're saying <s,U> = <0,V> for some V
then take U\cap V = U_x
then s|_{U_x} = 0
by definition of that equality
vestal snow
How do we know here that the intersection is not empty?
next obsidian
both have to be nbds of x
vestal snow
Okay and that's because we are taking the limit over the neighborhoods of x right?
wind steeple
That's fun that your book states that lemma and proves it when the course I have set this fact as obvious
smoky cypress
How do I show that if the action of G on S is primitive and effective, then the action induced by any normal subgroup H (β 1) is transitive
pure crest
Hey guys
im trying to prove this problem
show that $x^2 - \sqrt{2} $ is irreducible over $\mathbb{Q}(\sqrt(2))$
cloud walrusBOT
boat:
pure crest
what i did was assume that $\alpha \in \mathbb{Q}(\sqrt(2))$, then$ \alpha = a_1 + a_2\sqrt{2} $
cloud walrusBOT
boat:
pure crest
then i substitued alpha into x^2 - \sqrt{2}
so i get sqrt{2} = (a_1^2 + 2a_2^2) + 2sqrt{2}a_2a_1
then a_1 + 2a_2^2 = 0 so a_1 = a_2 = 0
so we have sqrt{2} = 0
a contradiction
so its not reducible
oh right
i am kinda confused about the next part
i dont see how this being irreducible fits into it
find the elements of gal(E/Q(\sqrt{2})
E is the splitting field of f(x)
so E = Q(\sqrt[4]{2})
since the roots are \pm \sqrt[4]{2}
i was just going to use the fact that f(x) has 2 roots
in E
so it must be isomorphic to Z_2
but i am not sure how to make your arguement
I know that | Gal{E/Q} | = | E: Q |
which is 2
but that doesnt really tell me anything about the elements right
^ thats only true if f(x) is seperable
but irreducible polynomials are always seperable in char 0
im not sure if E is char 0
you have the idenity
transformation
right
and you have the conjugate transformation
those are 2
ohh so thats how we use irreducible
like if a is a root -a is a root
in this case
since there are two
sorry i mean
a --> -a
-a --> a
is a map
and you have the identity map
idk how i can argue that it is a map
a, -a is a root of f(x)
so we have two elements in Gal(E/F) from before.
so the elements must be: 1. automorphism taking a - > a and -a -> -a
2.taking a - > -a and -a -> a
and these are in Gal since they fix F and take a root to a root
is that enough to say these are all the elements?
ohh i see
i just wasnt sure if the arguement was enough
thank you very much Viburnum
pure crest
hey guys
How would i find the order of a galois group
why is Q(\sqrt[4]{2}) / Q not Gaois?
solemn rain
some texts
do not define a galois group if its trivial
for the order of the galois group
pure crest
right
because i just found the galois group of that a few mins ago
which was Z_2
ohh right
hmm
but then i am not sure i understand
the minimal polynomial would be x^4 - 2
\pm \sqrt[4]{2} and \pm i \sqrt[4]{2}
ohh right
since they are imaginary
so its not a splitting feild
Thats why its not galois
ohh i see
we just defined galois feild over a feild extention
how would i find the order of Q{\sqrt[4]{2}, i}?
normally i would try to find a irreducible polynomial
but i only worked with one element
when i have two, can i just set alpha to sqrt[4]{2} + i
and find it
right so Q(sqrt[4]{2},i) \subset Q(sqrt[4]{2}) \subset Q
but i need to show that Q(sqrt[4]{2},i) is a splitting feild of f(x) and f(x) is seperable first right
but i would have to calculate an f(x) using sqrt[4]{2} and i
i see
thank you viburnum
pure crest
can i use eisenstien's criterion on polynomials with just two terms say x^2 + 2
i can take all a_i = 0
so any prime would divide it
and take any other prime
but then it would hold for x + 2 as well for p = 3
but its clearly reducible
woven delta
Yes
(to your first question)
Idk what you're saying after that
What do you mean you can take all a_i = 0
pure crest
a_i is all the other terms of 2 + a_1 x + x^2 + a_3 x^3 and so on
it would work in this case
woven delta
Lol you're doing it wrong
reread the statement of eisensteins criteria
Oh wait nevermind
next obsidian
wait what
woven delta
I don't understand why you have an a_3 there?
next obsidian
eisensteins requires seomthing about the leading coefficient
and then all the others
woven delta
Yeah
oblique river
and the constant coefficient
woven delta
Here's the statement
next obsidian
yeah
pure crest
woven delta
Okay so what's your question boat?
pure crest
right
woven delta
Why is x+2 clearly reducible?
oblique river
for example, you can use eisenstein's criterion to prove that x^5 - 2 is irreducible
oblique river
that's correct
eisenstein's criterion says "if this property holds then f(x) is irreducible"
pure crest
right
oblique river
it does not say "if f(x) is irreducible then this property holds"
pure crest
ohh right
good point
that was a dumb question
sorry guys
im just trying to cram everything
woven delta
Supposedly you can do some substitutions to apply eisenstein to even more cases
woven delta
Lol
next obsidian
f(x) is irreducible iff f(x + a) is
woven delta
Well we are working over Q
next obsidian
Also not claiming it fails in other places, but I proved it for integral domains for a hw haha
There's something about like x^{p - 1} + x^{p - 2} + ... + x + 1
pure crest
woah thats a good point
woven delta
A natural question would be "what's a polynomial you can't show is irreducible using eisenstein with substitution"
next obsidian
then apply eisenstein
oblique river
do you have any examples of how it fails in a non-domain?
next obsidian
I don't
oblique river
isn't x --> x + a always an automorphism of k[x]?
next obsidian
but I only proved it for integral domains
next obsidian
sorry so
k[x] is a bad example here tight?
Like I proved it in that case
we'd want a like R[x] for R arbitrary
oblique river
and since it's an isomorphism, it takes units to units and factorizations to factorizations
next obsidian
Is x --> x + a an automorphism of R[x] for arbtirary rings?
oblique river
it clearly has an inverse
bleak abyss
x-a
oblique river
namely, x --> x - a
bleak abyss
Oh lol
next obsidian
hmm
bleak abyss
Snipe? Bullets collide in midair? idk
next obsidian
Then maybe it's for arbitrary
I proved it really hands on
and think I used integral domainness
oblique river
sure, I'm just suggesting that there seems to be a very easy proof which works in general
pure crest
how do you guys remember all these proof techniques
next obsidian
just cuz
I had a problem
and was like "wtf lol"
and someone was like "haha do x -> x + 1"
and it just stuck in my brain
bleak abyss
You tend to remember the ones that come up a lot
next obsidian
or if u come up with it yourself
on a problem that gave you a lot of trouble it's memorable
next obsidian
Is that from freshman year of UG?
oblique river
yeah
next obsidian
damn
bleak abyss
This in particular is how you prove that the cyclotomic polynomials Phi_p are irreducible
next obsidian
I'm finishing up year 2 of doing math
crazy I still got a loooong ass way
before I'm even looking at post docs haha
oblique river
don't wanna scare you or anything but there are people who have been doing math for 5 decades
bleak abyss
I'm still surprised you're so young lol, in my mind you're at the very end of undergrad/a grad student
next obsidian
me?
bleak abyss
Yeah
next obsidian
lol
pure crest
are you a mathematician?
next obsidian
so like, age wise I'm at that end of UGish age
bleak abyss
Oh I know it consciously just like, when I talk to you it feels like I'm talking to a 4th year or a grad student
oblique river
what's a "mathematician"
pure crest
you probably did your grad school and got a phd by now, since you been doing it for 10 years
next obsidian
I'll take that as a compliment
pure crest
what do you do after?
pure crest
congratulations!
oblique river
thanks :)
next obsidian
Wait Buncho do you have a Post doc lined up?
oblique river
not a postdoc, but I do have an academic job
next obsidian
oh?
pure crest
i just define mathematician as those crazy smart guys who get paid to write mathematical papers
oblique river
I decided to transition into teaching-focused position instead of a research-focused position
oblique river
so I have a permanent position
bleak abyss
What about the ones who aren't crazy smart and get paid to write papers?
pure crest
damn nice
oblique river
"tenure track but for teaching"
solemn rain
wait
bleak abyss
waiting
next obsidian
At a 4 year uni?
bleak abyss
Math professors
pure crest
well they gotta be pretty smart to be mathematicians
oblique river
yeah, washington university in st louis
next obsidian
Oh nice!
pure crest
i mean math phd's
solemn rain
yea math phds
next obsidian
congrats buncho
oblique river
I'll still have some time to do research but I decided I wanted to focus on teaching and do research on the side instead of vice versa
thanks :)
pure crest
what did you do your thesis on?
oblique river
and yes mathematicians get paid lol
bleak abyss
boat you say that but I'm a grad student and it took me forever to see why max(yr,y^{-1}r^{-2}) \ge y^{1/3}
oblique river
I work in number theory
next obsidian
also
oblique river
but idk how much taht means to you haha
bleak abyss
π
next obsidian
swtich to discussion maybe?
solemn rain
meh i can just go rant about like how ppl around me think studying math is the stupidest thing to do
bleak abyss
Yeah prob
solemn rain
as i wont get paid and end up as hs teacher ( who is underpaid here )
wtrf
?
but this isnt the channel
so let me ask an algebra question
pure crest
i think they are right @solemn rain
oblique river
lol
solemn rain
a facebook problem i was very interested in but did no effort whatsoever ins olving
cuz i am bad
pure crest
is it one of those with fruits?
solemn rain
no
Let \phi_n be the nth cyclotomic polynomial. Suppose that \phi_n splits into powers of d irreducible factors mod p for some prime p. Show that all these factors have the same degree and multiplicity.
^ facebook from ' actual good math problems ' page
bleak abyss
Finite field extension of Q
bleak abyss
Hmm
pure crest
sorry, the question isnt the same, it just looked very familiar
solemn rain
any hints for Let \phi_n be the nth cyclotomic polynomial. Suppose that \phi_n splits into powers of d irreducible factors mod p for some prime p. Show that all these factors have the same degree and multiplicity.
bleak abyss
Still thinking
solemn rain
damn i wish i could think
tbh
like actual mathematical thinking
there is a remark made on the post
that i do not understand maybe you do
'Actually this is also true over the p-adic integers by Hensel's lemma if the multiplicity is one in the mod p split. This happens when p==2 does not divide n. I don't wanna make claims about p=2.'
bleak abyss
My heart tells me that it's got something to do with the business about like, x^{p^n} - x being the product of irreducible polynomials of degree dividing n? Because you factor out x and now you're kinda getting something a bit cyclotomic
But my head tells me that my heart is stupid
solemn rain
wait really
let me check it
where is x^[p^n]-x
field
nvm ih ave no idea
i cant remember what ycclotimc polynomial even means
i think a polynomial having nth root of unity as a root
next obsidian
bro if you wanted a solution or to understand the problem
first look up what a cyclotomic polynomial is kek
woven delta
Lol
solemn rain
and once is aw the remark i just got put off
i hate it when i attempt a problem i dont know its out of my reach
or maybe unsolvable
jusut aw aste of time
next obsidian
nah
woven delta
What problem are you talking about?
next obsidian
progress in math isn't cuz you solved x amount of problems
solemn rain
any hints for Let \phi_n be the nth cyclotomic polynomial. Suppose that \phi_n splits into powers of d irreducible factors mod p for some prime p. Show that all these factors have the same degree and multiplicity.
next obsidian
you thinking about it teaches you stuff
solemn rain
yea yea ik
but im just not in the mood for learning math nowadays at all
but yea ur right
i just caught the problem on fb :d
woven delta
Doing random problems is silly
If you're not willing to put in the time to think about them
solemn rain
yea just caught my eye ig
knotty mason
it can be deduced from Galois theory
first a theorem in group theory
Assume $G$ acts transitive on a set X. Let $H \lhd G$, let ${O_i}_r$ be the orbits of $H$ on $X$.
i) $G$ permutes the sets ${O_i}_r$ transitively
ii) Each $|O_i|$ is equal
iii) If $x \in O_i$ then $|O_i| = |H : H \cap G_x|$
iv) The number of orbits is $|G : H G_x|$
cloud walrusBOT
skymoo:
knotty mason
Now if you bring that through the Galois correspondence you will get the result
Let $f(X) \in F[X]$ be irreducible of degree $n$ over $F$. Let $L$ be the splitting field of $f(X)$ over $F$. Let $K$ be any Galois extension of $F$ contained in $L$ then $f(X)$ splits into a product of $m$ irreducibles of degree $d$ over $K$ and $m = [F(\alpha) \cap K :F]$, $d = [K(\alpha) : K]$
cloud walrusBOT
skymoo:
knotty mason
Well I suppose this is not quite relevant to what you asked about mod p, this is more about extensions of Q.
from Cox - Galois Theory pg 404
The cyclotomic polynomials have a cyclic Galois group
So I suppose every cycle type you get from decomposition mod p would be cycles
Any simpler ways to deduce it?
(x^13-1)/(x-1) splits mod 53 
https://math.stackexchange.com/questions/376668/is-there-a-pattern-of-the-factorization-of-a-polynomial-modulo-p-as-p-varies interesting link with some relevant stuff
knotty mason
OK I see how to solve this problem more directly
If you look at how Phi_p(X) factors mod q you can notice some patterns
it's determined entirely by the congruence class of q mod p
next you can look at field extensions of Z/qZ
in particular, when Phi_p(X) occurs as a factor of X^(q^d) - X
teal lake
I need some help with this https://cdn.discordapp.com/attachments/269573202018041856/742387259680227458/515168765189816321.png
cloud walrusBOT
craft:
undone bay
this is the general idea
teal lake
@undone bay how do you interchange the indices of B and change it to + or - accordingly. I mean how do you know it is symmetric or anti-symmetric?
If we can show that $(B^\mu_\alpha B^\gamma_\beta + B^\mu_\beta B^\gamma_\alpha)$ is symmetric, we can use the fact that product of a symmetric tensor and an anti-symmetric is zero. But how do we prove that it is symmetric?
cloud walrusBOT
ElonMoist:
teal lake
(finally typed that right)
undone bay
@teal lake I'm not assuming anything about the Bs, the only indices I change for the B tensors are the indices being summed over, which are arbitrary
undone bay
no problem!
neat ginkgo
how does the parity of a permutation behave with respect to permutation composition?
even * even = even
odd * even = odd
odd * odd = even
?
dusty oracle
What's permutation composition
neat ginkgo
the operation in Sn
dusty oracle
Well I guess that group is isomorphic to $S_2$
cloud walrusBOT
oblique river
yes @neat ginkgo that's one of the reasons why it's named "even" and "odd"
dusty oracle
Where odd is the function where you swap the elements and even is the identity function
oblique river
because the operation acts like how even and odd numbers behave under addition
neat ginkgo
nice, thanks
that's all i needed to confirm, idk why its so hard to find this on the internet
maybe i just dont know how to ask questions
oblique river
I just googled "product of even and odd permutations"
and found several sites giving that answer haha
also it's on wikipedia https://en.wikipedia.org/wiki/Parity_of_a_permutation
that's usually a good place to look first
neat ginkgo
im just trying to deduce the parity of Ο^(-1)Ο^(-1)ΟΟ for n>=2 Sn
apparently its always even
where pi and omega are any permutation
oblique river
yes
neat ginkgo
Ο^(-1)Ο^(-1)ΟΟ = (ΟΟ)^(-1)ΟΟ
oblique river
yes
oblique river
yes and you can prove it with the fact you just asked about haha
neat ginkgo
(ΟΟ)^(-1) = (ΟΟ)^(n-1), right?
oblique river
the product of sigma and sigma^(-1) is the identity, which is even
if sigma and sigma^(-1) had opposite parities then their product would be odd
but the product isn't odd, it's even
this all follows from the fact you were asking about
neat ginkgo
wait my brain
is there not an X>0 such that sigma^X = sigma^-1
omg im so stupid
not every inverse of a permutation Ο is a power of Ο
right?
im probably making this way harder than it is
oblique river
yes but why does that matter
neat ginkgo
the reason i was confused was because i thought the inverse of a permutation is always a power of that permutation, so an odd Ο would change parity every time it composes with itself
oblique river
yes but just look at the proof I gave you
already
if pi and pi^(-1) had opposite parities then their product would be an odd permutation. but we know their product is the identity which is an even permutation. therefore pi and pi^(-1) have the same parity
neat ginkgo
from your proof i can conclude that (ΟΟ)^(-1)ΟΟ = identity
but not (ΟΟ)^(-1)ΟΟ
right?
oblique river
but we separately know that sigma*pi and pi*sigma have the same parity.
neat ginkgo
you're right
oblique river
for exactly the same reason. no matter what the parity of sigma and pi are, we know that odd * odd = even, odd * even = odd, even * odd = odd, and even * even = even
neat ginkgo
thank you!
oblique river
π
smoky cypress
How do you show that there is no simple group of order 56?
Or 148
I know I probably need to use Sylow theorem to find a normal subgroup
elder valley
probably give a classification of all groups of that order using sylow theorems and semidirect products, and show that every possible group is non simple
shy bluff
I still haven't figured out how to do that quotient 
Have you had any further thoughts on how to do the quotient thingy from a couple day sago @elder valley
elder valley
what part are you stuck with
shy bluff
Well I managed ot show that <s - t> is a subset of ker(phi)
But not the othetr way around remember
For uh phi(r(s, t)) = r(y, y)
where R is just a commutative ring but not euclidean or anything
smoky cypress
@elder valley Oh I think I got it, classifying all groups of that order is a bit too hard for me, but I can show that it only has 1 sylow subgroup, which shows that that sylow subgroup is normal
woven delta
So the factorization of 56 is 8*7, so we look at the number of 2 sylow and 7 sylow subgroups. We know that n_2 is odd and divides the order of the group so it is 1 or 7. We know that n_7 is 1 mod 7 and divides the order of the group and so is 1, 2, 4, or 8
Obviously it must be 1 or 8
Our goal is to show that n_2 or n_7 is 1
smoky cypress
Yeah
woven delta
This tells us that the 2 sylow or 7 sylow subgroups are normal
elder valley
@shy bluff lets go to another channel so there's not overlap
woven delta
Okay so if there are 8 7 sylow subgroups then there are 48 elements of order 7.
It's easy to see why this isn't compatible with there being 7 2 sylow subgroups
@smoky cypress
woven delta
Yeah sorry I'm going pretty fast and not explaining what I'm writing
Which part are you not sure why is true?
smoky cypress
It's easy to see why this isn't compatible with there being 7 2 sylow subgroups
This
woven delta
Okay so each 2 sylow subgroup has 7 elements of order a power of 2
smoky cypress
Oh got it
woven delta
So there can be at most 1
smoky cypress
For some reason I thought the sylow 2 subgroups have order 2
When they have order 8
woven delta
Yeah
Another nice trick is there is a map from G to S_{n_p}
That's a common one you use for these kinds of problems
But just straightforward counting arguments work very often
(that trick doesn't work here unfortunately lol)
smoky cypress
Are you referring to the proof that if p is the smallest prime dividing |G| then any subgroup of index p is normal?
woven delta
I'm just saying that G acts on the p sylow subgroups
woven delta
Yes
woven delta
So maybe it will work on 148
148 is 37*4
So we look at the 2 sylow and 37 sylow subgroups
n_37 is 1, 2, or 4 so it must be 1
So this problem is just trivial actually
pure crest
hey guys, how do i prove that if $x | f(x)$ then $\sigma (x) | \sigma (f(x)$
cloud walrusBOT
boat:
pure crest
where sigma is an isomorphism Gal(E/F)
pure crest
what do you mean?
next obsidian
If x divides it you can write f(x) as
xg(x)
Push that through the isomorphism
Then you get sigma(f) = sigma(x)sigma(g)
Unless I am misunderstanding lol
pure crest
hmm that makes sense
pure crest
right
next obsidian
Also I didnβt mean to be a dick when I said definitional
I just mean if you unwind the definition of βdividesβ itβs basically just that
So I meant to say it as like
Youβre probably just overthinking it
pure crest
yeah dont worry i think think you were being mean haha π
i was over thinking it haha
thank you very much
next obsidian
NP
charred pewter
Do I have to prove all of these to show R is a ring?
smoky cypress
Yes
scarlet estuary
(though there are some theorems that can speed this process up; for example, if you can prove that multiplication is commutative, then it suffices to show only one of D1 or D2)
charred pewter
But that would only work if R is a commutative ring so I guess that only applies in specific cases
scarlet estuary
yeah, only in specific cases
smoky cypress
Also if you want to show H is a subgroup of G you can just show that H is non-vacuous and a*b^(-1) is always in H for any a,b in H. So if you want to show something is a subring, instead of showing A1,A2,A3,A4, you can just show that the subset is non vacuous and a-b is in the subset for any a,b in the subset
A group can have more subgroups than it has elements
That's pretty interesting
For $n>2$, each element of the additive group of $(\bZ/2\bZ)^n$ has order $2$, so each element with the identity is a subgroup. This gives $2^n-1$ subgroups, and of course there are more proper subgroups which gives more than $2^n$ proper subgroups
cloud walrusBOT
Whoever:
pure crest
if $\sigma$ doesnt fix any elements in E. let $\alpha \in E$, and that i have a set ${\sigma(\alpha) : \sigma \in Gal(E/F) } = {a_1 , a_2, ... a_n}$ Show that for any $\sigma \in Gal(E/F), \sigma $ applies to this set gives the same set.
cloud walrusBOT
boat:
stoic rose
The set is just the the set of conjugates of $a_1$
cloud walrusBOT
radiateur-man:
delicate bloom
show your attempt when you ask questions @pure crest
pure crest
oops sorry
pure crest
( i am not sure if this statement above i made is correct)
cloud walrusBOT
boat:
pure crest
im not sure ihow i can write this is a more rigorous way
is it rigorous enough
im just really bad at writing things into proof
next obsidian
Is there a good resource anyone knows of which would help me tighten up my field theory? Iβve covered the subject twice but feel shaky on it, specifically stuff about finite fields always makes me super nervous
elder valley
@pure crest i think your notation is your down fall. rather than label the elements as a_i, label them as sigma_i(alpha)
pure crest
But I want to show it's the same set
elder valley
call that set A. if you show sigma(A) is contained in A, then youll have sigma(A)=A since sigma is injective and A is finite
kindred mist
Ok silly/basic question, I understand everything up to "G/K is isomorphic to a subgroup of S_p." but why p? Card(G/K) = pk so Cayleys theorem provides its iso to a subgroup of S_pk, why would it be S_p? The thing in parenthesis about pi_H provides that G/H is would be iso to a subgroup of S_p if it H were normal in G. Its just that one part, I think I understand the rest of it, please clarify this - thanks ahead of time
woven delta
G acts on the left cosets of H by left multiplication @kindred mist
There are p left cosets
So therefore the action gives us a map from G to S_p
kindred mist
Ah ok now I see how the first iso theorem implies it
ty
I knew that was a silly question
smoky cypress
For two groups $H,N$, can we find another group $G$ such that $N$ is a normal subgroup of $G$ and $G/N\cong H$?
cloud walrusBOT
Whoever:
smoky cypress
Just a wild guess but is this semidirect product?
smoky cypress
Ok
undone bay
If we take "is a normal subgroup of" to mean "is isomorphic to a normal subgroup of" (which was probably already implied) then I can't see why G = H x N wouldn't work (x being direct product of groups). Semidirect products maybe work too?
next obsidian
I was thinking that too but
Actually yeah
I was being dumb about how N appears there
undone bay
but like yea i think i see what you mean, you technically would be doing G/{e}xN instead of G/N but that's surely fine right
smoky cypress
I mean we can just identify N with 1 x N, so N will be a subgroup of G x N
delicate bloom
have you heard of short exact sequences before
smoky cypress
Yeah
delicate bloom
oh ok
smoky cypress
Wait what about it
delicate bloom
well it's sort of what you're talking about that's all, figured that's where your question came from thinking about
woven delta
Yeah you're asking about the groups G that fit into a short exact sequence 0-> N -> G -> H -> 0
NxH is an obvious one, semidirect products also, but these cases happen when there are sections
There are lots of groups that fit into such short exact sequences which aren't semidirect products
elder valley
Exactness doesn't imply normality of N right? You have to assume that in the sequence
woven delta
No it does
smoky cypress
Oh I was thinking of something else
woven delta
All short exact sequences are isomorphic to a sequence like 0 -> ker \phi -> G -> H -> 0
smoky cypress
I was kind of thinking of classification of finite simple groups, so if we have all the simple groups, can we get all the other finite groups by building them using simple groups
That's what I was thinking about
woven delta
How would we build them?
delicate bloom
out of legos preferably
elder valley
Oh duh N is the kernel
smoky cypress
That was my question
delicate bloom
I don't build groups, I carve them out of a permutation group
like a statue out of block of granite π
smoky cypress
Welp
That was just my random thoughts
Because I don't want to do the exercises for this section
They all look too scary
delicate bloom
I agree, I'd rather do analysis
smoky cypress
Same 
delicate bloom
algebra talk is a bunch of gobeldy gook to me
next obsidian
bruh moment
smoky cypress
This is how Jacobson defines semidirect product. My question is that, is there a simpler definition, or a more intuitive definition. Then also what the semidirect product trying to do?
elder valley
It's just a more general way of making a new group from two existing ones
I learned it from dummit and foote and thought it was very clear. It's nearly the same definition but they present it a lot more explicitly than the pic you showed
woven delta
@smoky cypress there is also the internal one
If you think about the internal one it's pretty clear
Also do a bunch of examples
Something that's also nice is to think about why the semidirect product is never abelian for a nontrivial action
next obsidian
I think if you see internal ones
it makes more sense
I remember there's some like semidirect produt in a matrix group
of the symmetric matrices (acts like S_n)
and the scalar matrices (acted like (F_p)^n)
and it's like the multiplication worked normally
but the action of the symmetric matrices sort of twisted it a bit
Idk, this is just the one example that made my eyes go π
woven delta
@smoky cypress here's a nice exercise
Prove all the facts about how the internal and external semidirect products are related
smoky cypress
Wait is the semidirect product above the internal one?
smoky cypress
Oh ok
woven delta
The internal one is like "suppose H and K are subgroups of G with 1) H normal 2)H \cap K = 0 3) HK = G
Then the internal semidirect product of H and K is G
So the nice fact you can prove is that every element of G can be written uniquely as hk for some h in H, k in K
And then you're like "what is $(h_1 k_1 )(h_2 k_2)$ written in this form"
cloud walrusBOT
Liquid:
woven delta
So H is normal so $h_1 k_1 h_2 k_2 = h_1 (k_1 h_2 k_1^{-1}) k_1 k_2$
cloud walrusBOT
Liquid:
woven delta
And $k_1 h_2 k_1^{-1}$ is in H
cloud walrusBOT
Liquid: