The only things I have on the centre is this and the property I just proved that it is non trivial for groups of order p^n
#groups-rings-fields
subtle granite
subtle granite
a group G is simple if the only normal subgroups it contains are {e} and G itself
upper pivot
right, so we want to show it has a nontrivial normal subgroup yes
subtle granite
yessir
upper pivot
do you see how the center could connect to this
subtle granite
I can see that the centre is nontrivial
but is it also a subgroup?
I'd prob have to prove that then
upper pivot
yeah test out if its a normal subgroup
subtle granite
oo okay
mild laurel
how's AM going btw
mild laurel
I enjoy them, you didn't really need any topological space stuff right
other than definition?
upper pivot
yeah
mild laurel
okay there's an idea of a basis that I think you have to use but
subtle granite
right so,
a). identity is always in the centre
b). to check its closed if x1, x2 are in Z(G), then for any g in G, (x1x2)g = x1(x2g) = x1(gx2) = g(x1x2) so
that should do it
right
sorry for lazy notation
been doing this since 6am
now its 9am
π
mild laurel
Oh I didn't realize that basis was in Rudin
upper pivot
mhm thats inverses and normality in one go, good job
upper pivot
or wait nvm lol x in Z(G)
upper pivot
do the normality one, it should be easy
upper pivot
yep
subtle granite
ok thank you!!
mild laurel
yes
mild laurel
looks fine
chilly ocean
solve x^2+3x+2=0 in Z2
x^2+3x+2=(x+1)(x+2)
x = -1,-2
-1 = 1 (mod 2)
-2 = 0 (mod 2)
so its 1,0?
mild laurel
yeah, you can also reduce your polynomial, to get x^2 + x = 0 instead
chilly ocean
Also Z2 only has 2 elements xyz so you can just input both of them into the equation and check if it works
ohh
chilly ocean
how do i know if a polynomial is irreducible?
chilly ocean
x^2+1
mild laurel
a degree 2 polynomial is irreducible if and only if it has no roots
chilly ocean
is that true for other polynomials too?
if it is irreducble then it has no roots is true for all deg > 1 polynomiasl
but if it has no roots then it is irreducible is not true for higher ones
bleak abyss
Degree 2 and 3
But square an irreducible quadratic and you'll get a polynomial with no roots yet which isn't irreducible
chilly ocean
ok for 2
if it is reducible then f(x) = q(x)g(x)
deg q,g>1
=1
then it will be like
(x-a)(x-b)
so it has roots?
this is contrapositive
chilly ocean
you mean deg(q*g)?
latent anvil
q*g is f, so that's what I said
latent anvil
deg f = deg (q*g) = deg q + deg g
chilly ocean
ok
so q and g will be degree 1 polynomals and all degree one polynomials have roots
for 3
f(x) = q(x)g(x)
1<=deg(q),deg(g)<deg(f)
one of them must be degree 1 to get to 3
so it will have at least one root
does this proof work?
chilly ocean
thanks π
halcyon siren
@chilly ocean try using gauss lemma
chilly ocean
Find all irreduciblepolynomials of degree 2 overZ3
how do i do this?
do i just make polynomials and check or is there a better way?
ax^2 + bx + c
a,b,c in Z3
bleak abyss
Yeah just check which ones do and don't have roots
chilly ocean
okay
it seems like there would be alot though
theres 2 choices for a 3 choices for b and 3 choices for c
bleak abyss
Did they say find all? Not necessarily monic?
And well you can immediately reduce it to two choices for c
Because if c=0 you're auto reducible
So now you're checking, 12 polynomials in the non-monic case, 6 in the monic case
Which isn't that bad
Especially since for each one you really just check whether 1 or 2 is a root
chilly ocean
oh i see
willow garden
can someone confirm my madness please? suppose H β€ G. if H is normal in G, then the normalizer of H is G itself, right?
oblique river
yes
latent anvil
iff, too
willow garden
that's what I thought
I had one of those brain things for a minute, thanks for clarifying
chilly ocean
A is finite over A_0 iff integral and finitely generated
finitey generated as a ring?
mild laurel
What's A_0?
mild laurel
Yeah, I think it should be finitely generated as a ring
chilly ocean
alright thanks
mild laurel
It might mean that A is finitely generated as a module over A_0?
chilly ocean
That wouldn't make sense though because then integral wouldn't be needed
mild laurel
Ah yeah you're right
latent anvil
Finitely generated as an A_0 algebra
Is what this means
Finite (meaning finitely generated as a module over A_0) = finite type (meaning finitely generated as an algebra over A_0) + integral
halcyon siren
How does it make sense to take the inner semidirect product of G and a map acting on G?
mild laurel
What's the definition of a Steinberg endomorphism?
halcyon siren
@mild laurel An endomorphism on a linear algebraic group G-> G such that for some power m, F^m maps (a_ij) - > (a_ij ^(p^k)), for some k. (p is the characteristic of the field over which G is defined)
Oh wait..
mild laurel
Oh
Anyways, this makes sense because you're taking the cyclic group generated by F
And this cyclic group acts on G in the obvious way
halcyon siren
Do inner semidirect products rely on some action? I thought it was more that you had to have G and the cyclic group with some compatible operation
I have a theory though...
mild laurel
This is an outer semidirect product
halcyon siren
π€ Is the inner semi direct product of H and N: HN, and H n N is the identity ?
Is there some implicit action behind HN? Like h acts on n in a certain way π€ in this case by mapping it to itself
mild laurel
You need N to be normal, but yes
H acts on N by conjugation
Which works out because you require N to be normal
halcyon siren
@mild laurel so how does that give me a different group and not just N again?
mild laurel
Uh, HN is bigger than N?
halcyon siren
Sorry. I'm not explaining myself clearly.
HN is bigger if it's elements are just hn. But in that case we need to define what gF means.
If hn means conjugation by h, then won't hn always be in N since N is normal.
mild laurel
No sorry
Maybe I wasn't clear
The elements of HN are the normal hn under group multiplication
The only issue is how you multiply two elements of the form hn and get something of that form
Give me a sec let me get on a computer
halcyon siren
Yeah, I'm just not seeing what the obvious multiplication here is?
Are you saying it's F(g)?
π€
halcyon siren
Please :)
mild laurel
So yeah, elements of HN are just the normal hn multiplication of G
but if you have hn, and then h'n', how do you multiply them?
So their product would be hnh'n' but this isn't of the form hn yet
We can write this as hnh'n' = h h' h'^{-1} n h' n'
and then, since N is normal, the element h'^{1} n h' is in N and so h'^{-1}n h' n' is also in N
and h h' is an element of H
so we've rewritten our element as the product of something in H and something in N
halcyon siren
That's so useful to know! Thank you very much
mild laurel
And that's why I talked about the conjugation action
because you have h' acting on n by conjugation here
And in general that's how you get semidirect products
halcyon siren
Thank you so much!
And does this really work to take the semidirect product of a group with an automorphism of that group
mild laurel
Yeah so for outer semidirect products, it works basically the same way
If you have one group acting on another then you can take the semidirect product
halcyon siren
@mild laurel and is it implied that gF = F(g)?
mild laurel
hm, its better to think of outer semidirect products as
G x <F> with a weird multiplication structure
halcyon siren
How do you know its an outer semidirect product?
halcyon siren
π I am super new to them
mild laurel
Like the thing is, G and <F> give you subgroups of the outer semidirect product since the outer semidirect product is G x <F> as a set
And you can check that G is normal in this new group, and the intersection of these two subgroups is trivial
so you can check that the outer semidirect product is isomorphic to the inner semidirect product of G and <F> here
cloud walrusBOT
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quiet cave
Did you just read excerpts of Wikipedia articles to royalty free music and stock footage
rich patrol
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coarse stag
My book says that if P is a p-subgroup of G, for any p-subgroup Q, there exists an element g such that Q<=gPg^-1. What if |P| < Q?
wind steeple
your book says that P is a p-Sylow of G
wind steeple
x)
coarse stag
Is there an example of a group with more than one p sylow subgroup?
coarse stag
Mm
solemn rain
Problem : Show that H = { x in D_2n | x^2 = 1 } is not a subgroup of D_2n (here n>= 3)
im not that good with dihedral groups so i decided to try
to say that we require commutivity in order for ssuch a subgroup to exist
i generlized ig and said in order for a subgroup with such structure ( all elements have order 2 ) to exist then we must have commutivity
then just said D_2n is not abelian for n>=3 and thats it
is that a right argument?
chilly ocean
No that's not a right argument.
solemn rain
why
chilly ocean
Because you can potentially have a nonabelian group G where {x in G | x^2 = 1} is actually a subgroup of G
there's something more specific about dihedral groups that you need to exploit, not just "nonabelianness"
solemn rain
fuck really
im not that good with dihedral groups tbh
D_2n = < r,s |r^n=s^2=1 , rs=sr^-1 >
chilly ocean
The easiest example that I can think of is a semidirect product Z semidirect Z/2 with {1 -> -1} twist (not sure if you know what semidirect product is but let me just put it up here)
in this group, {x | x^2 = 1} is {e}, which is a subgroup
solemn rain
= {1,r,r^2,.....,r^(n-1),sr,sr^2.....sr^(n-1)}
i dont but okay i believe you
can u help me with the problem then
chilly ocean
So what you need to do is come up with examples. Think about how you can check that H is not a subgroup
solemn rain
show that xy^-1 is not in H if x and y is in H
chilly ocean
go back to the first principles
solemn rain
?
chilly ocean
Right ok you can use that
solemn rain
if x and y is in H
it doesnt mean that (xy)^2 is 1
since any element in H can be written as sr^k for k=0,1...n-1
chilly ocean
So you'd be done if you picked two elements in H whose squares are 1 and yet their product doesn't have a order dividing 2.
Don't be afraid to pick some elements and start experimenting
solemn rain
r doesnt have order 2
wait
s has order 2
idk
lmao
let x and y be in H
x = sr^k
y = sr^j
(sr^k)(sr^j) = (sr^j)^-1(sr^k) ?
is that right
using the relation given in the presentation of D_2n?
=(r^-j)s^-1sr^k = r^(k-j) ?
thiss doesnt have order 2 cuz r has order n
n >=3
so if my algebra is right im done
@chilly ocean
chilly ocean
You are giving rs = sr^{-1}
solemn rain
yea
chilly ocean
So you need (sr^k)(sr^j) = s(r^k s)r^j = s (s ^{-k})r^j = s^2 s^{j-k} = s^{j-k}
solemn rain
oh fuck
chilly ocean
there are some steps you're missing and the final answer that you have there is not the same as the calculation right above
solemn rain
i thought in this equation
rs=sr^(-1)
this holds for r and s in D_2n
not THE r and s
chilly ocean
no
It's specific to r and s. You use those generators and relations to generate other elements and see if there are further relations using those relations.
solemn rain
okay so
lets clean this up
let x and y be in H
we want to show xy is not in H
x and y are in D_2n
so x =sr^i and y = sr^j
for i and j = 0,1,.....,n-1
xy = (sr^i)(sr^j) = s(r^is)r^j = s(sr^-i)r^j = s^2(r^(-i+j))
= r^(-i+j)
which does not have order 2
hence not closed hence not subgroup
?
is that right?
chilly ocean
No
solemn rain
why
chilly ocean
What if i = j = 0?
chilly ocean
Be more concrete yoh
solemn rain
wdym
chilly ocean
experiment with values of i,j
solemn rain
shouldnt i be general and not see special cases?
cuz assume i get that at i=1 and j=1 they are not in H
chilly ocean
You're trying to come up with a counter example to show that {x | x^2 = 1} is not a subgroup
solemn rain
maybe the subgroup doesnt have those
chilly ocean
Right, you should check which ones are in it and then pick some which don't have order 2 after multiplying
chilly ocean
Ok so it suggests what elements to try playing with
chilly ocean
You picked x = s. Now give yourself y = ?
chilly ocean
Ok cool. Can you now pick k so that (xy)^2 is not 1?
chilly ocean
Yep
solemn rain
so
chilly ocean
So, what did you get out of working on this problem?
solemn rain
ig i just like
exhaust some cool cases
in problems that i do not know how to do
maybe i get something/
tyms
tysm
chilly ocean
NP.
latent anvil
@chilly ocean isn't (0,1) also an element of order 2 in that semidirect product?
chilly ocean
Ah you are right. My bad. Take any nonabelian group of odd order :)
latent anvil
Well the example still works
Oh wait maybe not
There's an obvious presentation <a, b | aba^(-1) = b^(-1), a^2 = 1>. But if a^2 = 1, then a^(-1) = a, so this is equivalent to <a, b | aba = b^(-1), a^2 = 1>, or <a, b | abab = 1, a^2 = 1>. Changing variables c = ab, this is <a, c | a^2 = c^2 = 1>, or the free product of C2 with C2
So there's at least two nontrivial elements of order 2
Yeah ac has infinite order, so it's not in the set of elements of order <= 2, so that set isn't a group
chilly ocean
GL_n(Z/p) for any odd prime p should work (n >= 3) as an example
for mo2men's "G nonabelian and {x | x^2 = 1} is a subgroup}"
latent anvil
Yeah
chilly ocean
Thanks for spotting that error
latent anvil
I'm curious about the case where there's at least one element of order 2 though
So not just any odd order nonabelian group
GL_n(Z/p) definitely has one, take -I
In the case n = 1 you get a subgroup because this is the unique element of order 2
Why do you say it works for n >= 3?
chilly ocean
Nvm I made a mistake again.
latent anvil
Oh and n = 1 doesn't work because it's abelian
The quaternion group works
It has a unique element of order 2 and so that forms a subgroup, but it's nonabelian
Further this is an example of minimal order, since the transpositions generate S3
chilly ocean
Yes that works
chilly ocean
How 'bout upper triangular matrices in M_3(Z/3) with diagonals +/-1?
or in M_3(Z/p) for p odd anyhow
latent anvil
Is there a nice way to compute the order of such an element?
chilly ocean
I don't know, throwing up here for now
chilly ocean
"characterizing nonabelian groups where { x | x^2 = 1} is a subgroup"
{x | x^2 = 1} is a direct sum of Z/2's so if we have any group extension of the form 0 -> Z/2 (+) ... (+) Z/2 -> G -> H -> 1 with H an odd order group then should be ok (do finite case first)
maybe there's a nice way to extend Q(sqrt(2), sqrt(3), ... sqrt(p)) / Q (which has galois group (Z/2)^(number of primes from 2 to p)) to a galois extension L of Q while keeping L/Q(sqrt(2), ...., sqrt(p)) an odd order and Galois as well
with extension satisfying those above
anyway gotta get back to my own work. Let me know if you find a nice way
chilly ocean
It is. But you also want to keep L/Q(sqrt(2), ..., sqrt(p)) to be Galois as well
yes find a bigger L
latent anvil
Okay yeah that makes sense
bleak abyss
{x | x^2 = 1} should be a normal subgroup right?
chilly ocean
You can probably just take some big galois closure L where L/Q(sqrt(2) ,..., sqrt(p)) may not necessarily have an odd order, but you can try to take a subgroup of Gal(L/Q(sqrt(2) ,..., sqrt(p)) and massage around a bit
latent anvil
So you always have such an extension
chilly ocean
Oh well then that works π
latent anvil
Wait maybe not, you said H was odd order above
chilly ocean
I want that yes
bleak abyss
Oh wait hmm
Oh it's closed under conjugation but if you're not abelian it's not a subgroup
chilly ocean
Quaternion, some semidirect product of Z/p with Z/q (q dividing p-1) should work too and so on
latent anvil
I was saying by dami's thing that you always have an extension 0 -> (Z/2Z)^((+)n) -> G -> H -> 0, but H might not have odd order
chilly ocean
We're trying to characterize groups (let's start with finite groups G) where {x | x^2 = 1} is a subgroup of G
(nonabelian obviously)
latent anvil
Actually if H has odd order than this is the 2-sylow, so all elements of order 2^k have order 2
chilly ocean
After this one can naturally naturally think about when {x | x^3 = 1} is a subgroup, etc. etc.
Daminark weren't you at Chicago? Chicago people are good at this stuff π
bleak abyss
Yeah trying to ignite my old skill
chilly ocean
You can take some extension L as mentioned above and try to take some odd index subgroup in Gal(L/Q(sqrt(2), ..., sqrt(p))
bleak abyss
Okay so yeah if that's a subgroup then it's normal, so we are just dealing with certain group extensions. And what's your thing about H in particular?
chilly ocean
but gotta keep everything normal and such
(err not as mentioned above, but some big one L' and cut it down)
bleak abyss
You're trying to extend just be odd order H?
chilly ocean
Sure
Was just looking for examples like this in the "wild"
where {x | x^2 = 1} is somewhat large despite it being a subgroup in a nonabelian group.
bleak abyss
Hmm, at best you can do a third right?
Like take (Z/2)^n and non-trivial semidirect product with Z/3
chilly ocean
yea you can do stuff like that too. Trying also to come up with non-split examples
What's a good "Ext" replacement tool for nonabelian group extensions?
chilly ocean
sorry, the extensions went the wrong way. Take any odd order nonabelian galois extension K over Q, then take some linearly disjoint quadratic extensions K(a), K(b), ... over K and Gal(K(a,b,...)/Q) should be a big class of examples (non-aplit too).
latent anvil
My hw has me proving that if D is a division ring and V = D (+) β¦ (+) D the unique simple module over M_n(D), then End_(M_n(D))(V) β D
But this confuses me
Oh wait I'm unconfused
Okay wait now I'm confused again
It seems like it's iso to D^op
bleak abyss
Alright it's Weibel time
fading wagon
Try changing the ^2 to something simpler
erm, wait, this is #groups-rings-fields
not #precalculus , go there instead
latent anvil
Oh hell yeah weibel time
I hope it's still weibel time
It's always weibel time in my heart
latent anvil
It's usually an integer
bleak abyss
Oh I ranted elsewhere actually
chilly ocean
Well yeah
chilly ocean
But what
bleak abyss
Hmm I'll work through theory here and do exercises there
latent anvil
n by n matrices
bleak abyss
Spread the love of Weibel
chilly ocean
Ahh
latent anvil
I'm pretty sure my homework problem is misstated
And that there should be an op somewhere
@bleak abyss there's a dead homological algebra server that you could rant in
Where are you in weibel?
bleak abyss
I'm starting from 0, my class is somewhere in chapter 2 right now but I slept through too many classes lol
latent anvil
Oh that's a mood
I'm gonna try and keep doing Hartshorne problems but I'm two sections behind
But apparently most of the class is too
latent anvil
Is M_n(R^op) iso to M_n(R)^op iso to M_n(R)?
I think that might be what I'm missing
I'm pretty sure M_n(R) is iso to its opposite ring by transpose
chilly ocean
Does End_(M_n(D))(V) mean all endomorphism over D? And the equal, does that mean a isomorphism?
latent anvil
No
chilly ocean
Cause you wrote V=D
chilly ocean
Oh nvm missed the next line
latent anvil
the β¦ is n times
This is the unique simple module over the matrix ring, assuming I did the previous problem correctly
chilly ocean
It's fine as it is. Why do you think it should be D^{op}?
x -> (multiplication by x).
chilly ocean
Daminark: Doesn't that have a funny preface something to the effect that "Thanks to my wife, family, etc. etc." and then "... without whom I would have finished this book much earlier?"
chilly ocean
Let's see, End_{M_n(D)}(D^n) should be just scalars becaause they're just matrices A in M_n(D) that commutes with everthing in M_n(D) right
scalar multiplications anyhow
chilly ocean
if you take an element f in End_{M_n(D)}(D^n) then that's a map f: D^n -> D^n such that f(Ax) = Af(x) for all A in M_n(D), or fA = Af for all A in M_n(D).
So f commutes with everything in M_n(D).
in particular f is in M_n(D) (should note early on as well, by taking A to be scalar multiples of I).
latent anvil
Okay, so I claim that if f(x) = ax for an element a not in the center, then f won't be linear
chilly ocean
so it should be scalar multiples of elements in the center of D
latent anvil
You think it's just the center?
chilly ocean
so it should be center of D involved here.
latent anvil
Here's the thing
There is an injective ring map f : D^op -> End(V)
given by f(x) = v |-> vx
f(x)(Av) = (Av)x = A(vx) = Af(x)(v) by associativity of matrix multiplication, f(x)(v+w) = (v+w)x = vx+wx = f(x)(v) + f(x)(w) by distributivity in each component, so f(x) is linear. Then for any x, y in D and v in V, f(x+y)(v) = v(x+y) = vx + vy = (f(x) + f(y))(v) and f(yx)(v) = v(yx) = (vy)x = (f(x) Β° f(y))(v)
So End(V) at least contains a subring isomorphic to D^op
chilly ocean
Right that makes sense
latent anvil
Oh I figured it out. My professor defined End(V) as acting on V on the right
Which is stupid and dumb
chilly ocean
that was my next question. But hah
latent anvil
And I wasn't there the first day of this section
Ughhh
Wait no, I just misread it
It says End_R(R) β R^op
By the argument I was trying to make
And End_R(R^n) β M_n(R^op)
Which makes sense
Maybe
Idk I'm confused
chilly ocean
so the map is D -> End_{M_n(D)}(D^n), x -> (multiplication by x on the left)
x -> (y -> xy)
Given thaat M_n(D) acts on the right of V = D^n
latent anvil
Yeah, that makes sense to me
I think I just need to go to OH and clear up which direction stuff is happening on
chilly ocean
and what is the action on the right?
latent anvil
I think M_n acts on the left but End(V) acts on the right
In lecture he writes v^Ξ±
So sending r to right multiplication by r actually gives a map D -> End(V), because we swap direction twice
latent anvil
Posting here even though I'm working out of a book on manifolds
Call an R-algebra F geometric if the only f in F such that x(f) = 0 for all x in Hom_R(F,R) is f = 0
Let V be a finite dimension R-vector space and G a linear operator in V. Let F be the algebra generated by G. The problem is to figure out when F is geometric
If G is nilpotent then F isn't geometry
If m(x) is the minimal polynomial of G, then F β R[x]/(m(x))
We know that m factors as a bunch of linear terms times some irreducible quadratics
Group together the powers
Powers of distinct irreducibles are coprime and intersect trivially, so you can apply the Chinese remainder theorem
So you have a bunch of products of rings of the form R[x]/(p(x)^r)
If any r is greater than 1, you get that the thing isn't geometric
Since any homomorphism into R needs to kill that off
Oh and this happening says the minimal polynomial has a repeated root, so the characteristic polynomial has a repeated root, and thus your get an eigenvalue with multiplicity > 1
But it doesn't go the other way because if you take G = I you get F = R, which is geometric
Maybe it's only nilpotent matrices?
Oh consider V = R^2 and G(x, y) = (-y, x). This has characteristic polynomial Ξ»^2 + 1 (and since this is irreducible its the minimal polynomial too)
Then there's no homs F -> R
And more generally if you have an irreducible quadratic in your factorization, this will be a problem
So I guess the characterization is that the minimal polynomial of G is a product of distinct linear polynomials
latent anvil
@vital night what do you know about the ring K[x]?
vital night
I just know the general axioms
latent anvil
What's the context of this problem?
There's a theorem which is usually covered in an introduction to ring theory which makes it much easier
vital night
No
latent anvil
Do you know what Euclid's algorithm is in the context of integers?
For computing the gcd
vital night
No, I don't know what that is
latent anvil
Hmmm
So my first instinct is to divide f by g, or g by f
deg f <= deg g or deg g <= deg f, so divide one of bigger-or-equal degree by the one of smaller-or-equal degree
And by divide I mean divide with remainder
vital night
okay so use #2 for that
latent anvil
Right
This gives you something to work with at least
Assume g has bigger degree
Then you get g(x) = q(x) f(x) + r(x)
Can we say anything about that?
vital night
deg r < deg q
latent anvil
Oh yeah, sorry
vital night
we know r is non-zero
latent anvil
Yeah, that's a good line of thought
Even better, we know g and r are congruent mod f
Right?
Oh wait have you seen congruence? Probably not if you haven't seen ideals
Sorry
vital night
yeah sorry I don't know what that is π¦
latent anvil
Np. I'm just used to thinking about this problem with some more machinery
So my thought is that maybe we need to argue this inductively
And by dividing by remainder, we're able to get something of smaller degree than g
But we need to make sure we have a similar situation as at the start of the problem
vital night
okay so divide g by r
latent anvil
Hmm, I'm not sure that'll help
vital night
oh sorry I thought that was what you meant by "dividing by remainder"
latent anvil
Sorry I meant dividing with remainder
What we did above
We were able to get something of strictly smaller degree than g
Right?
latent anvil
Well we might have deg f = deg g
But I was thinking more about r
It's sort of the bit of f left over when you take out q
vital night
ah right okay
latent anvil
So hopefully some of the information we started with is preserved
vital night
what follows from this?
latent anvil
I'm not really sure how to help without giving it away
But I think division with remainder+induction is the way to go
latent anvil
Well the idea is that when you divide by remainder, the degree goes down
vital night
Right
latent anvil
Induct on deg f + deg g
vital night
what is the new degree though? deg qf + r still equals deg g
latent anvil
Right, that's true
So you have to figure out what to apply the induction hypothesis to
You started out with two coprime polynomials
How can you get two new coprime polynomials with the sum of degrees reduced?
latent anvil
But g/f won't be a polynomial
Since g and f are coprime, f definitely doesn't divide g
fickle brook
what are you trying to do again
fickle brook
can't you just run the euclidean algo
vital night
I haven't learned what that is yet
fickle brook
alright well
ok
strong induction on deg(f)+deg(g) then
base case would be when that sum's 0 ig
ie when they're both constant
vital night
Right
fickle brook
now take two arbitrary polynomials f and g and suppose we've already proved this result for all pairs of polynomials whose degrees sum to less than deg(f)+deg(g)
wlog assume deg(f) β₯ deg(g)
write f = gq + r with deg(r) < deg(g)
obtain polynomials a', b' such that a'g + b'r = 1
write r = f - gq
a'g + b'(f - gq) = (a' - q)g + b'f
= 1
done
latent anvil
By the induction hypothesis
You need to argue that r and g are coprime
Anybody know an easy example of a division ring which isn't iso to its opposite? My professor says that they're isomorphic as a consequence of the Artin Wedderburn theorem, but math stack exchange says that this is false
(sorry to interrupt)
fickle brook
and suppose we've already proved this result for all pairs of polynomials whose degrees sum to less than deg(f)+deg(g)
vital night
oh I see
fickle brook
the strong inductive hypothesis
vital night
and lastly, a' - q
I can simply set a new polynomial equal to that
to preserve the structure
1= a(x) f(x) + b(x) g(x)
is that right?
vital night
okay great, thanks for the help!
latent anvil
@chilly ocean there was an error on my homework
The prof thought that Artin Wedderburn implied any semisimple ring is isomorphic to its opposite
bitter mauve
@mild laurel why is ann(omega_i) in p
mild laurel
M is an A module here?
bitter mauve
yes
mild laurel
If there's something in the annihilator that's not in p
Then when you take the localization
Okay it's easier to just say that if
\omega_i = 0/1, then there exists some s \in A - p such that s \omega_i = 0
just by definition of localization of module
aka s is an annihilator of \omega_i
bitter mauve
so what he means is that if omega_i/1 not 0 in M_p then its annihilator is not in A-p
mild laurel
its annihilator not being in A-p is the same as the annihilator being in p lmao
bitter mauve
correct
:v
thanks
i didnt think of 0. i thought what if ann(omega_i) is empty, then its not in p either
which was obv bad thinking
mild laurel
empty sets are subsets of all sets lmao
bitter mauve
im dumb ok, leave me be
jade crypt
Agreed
covert perch
Hey guys. I m a 1st year undergrad math student. I have done two courses of algebra( group theory and linear algebra), two of analysis, and two of probability theory. I was thinking of doing a reading project this summer. Can you suggest some topic to me?
Thanks
Group theory fascinated me the most
clear obsidian
I keep hearing the word "action" from people online in the context of (modern) algebra but I never see this word in any books I've been reading (at least not in an undergrad context that explains what it means), does "action" just refer to a ring (or field) multiplication that gives an abelian group module (or vector space) structure (respectively)?
(I've seen what group action means, isn't group action just a generalization of these "actions"?)
bleak abyss
So if I think of a ring action on an abelian group as basically the same thing as a module
clear obsidian
Ok well that helps, but do you know where I can find a more rigorous elaboration on the matter? Thanks
worthy kindle
There's a little paragraph titled "Actions" in Bourbaki's Algebra chapter I, but it's probably not what you're looking for
clear obsidian
hmm, I'll give it a look anyway, ty
ya this is probably different context but glad I checked it out at least
uncut girder
In a field of characteristic 2, can you say anything about powers of elements?
Like is it true that x^2 = 1 for all x
worthy kindle
Actually, you may want to keep reading it because it looks like they do the most general stuff in "Actions" and then do particular cases in other paragraphs
clear obsidian
ah, ok will do, thank you!
(I meant to apologize earlier about my confustion with X^Y notation but keep forgetting to)
uncut girder
Say you have a matrix and you form its rational canonical form
Do you get a different answer based on the characteristic of the field you're doing this over?
latent anvil
@uncut girder in the finite case you get a good result about powers
For sufficiently large N, x^(2^N) = x for all x
The minimal such N determines the field up to isomorphism
And anything in the algebraic closure of F2 satisfies x^(2^n) for some n
I'm not sure what you're asking about with the canonical form question
woven delta
@latent anvil finite field or finite characteristic?
Hopefully you mean finite field
woven delta
@uncut girder I don't think your question is so well defined
latent anvil
Doesn't rational canonical form depend on the factorization of your characteristic polynomial?
If you take A= [[0 -1] [1 0]] in M_2(R) and embed this in M_2(C) those should be different
I think
Oh sorry you said characteristic
Not just the field
Yeah idk
uncut girder
For that example, they shouldn't be different
Over R or over C
Because the entries are in R
latent anvil
Oh yeah lol
uncut girder
And even over Q
latent anvil
It's a canonical form
uncut girder
But my question is
My question is really about this:
[ c 0 -1]
[ 0 c 1]
[ -1 1 c]
D&F chapter 12.2 exercise 9 asked you to find the rational canonical form of that matrix but doesnt tell you over what field. I didnt assume anything about the field and got the same answer (being careful consider characteristic 2 seperately)
uncut girder
How do you find rational canonical form of
I cant do the elementary row/column operations to get the smith normal form ;_;
latent anvil
This sucks
uncut girder
I was stuck up until I realized I could use 1 in the top right to get rid of constants anywhere in the matrix
latent anvil
god i should practice matrix nonsense for the gre
wind steeple
Hey, for A a principal ring and E an A-module, p in A an irreductible element and k a natural number, is there any canonical way to inject E (x) A/p^kA into A/p^(k+1)A ? By canonical I mean a categoric construction.
winter yoke
Show that a magma M satisfying a((ac)(bc))=b for all a,b,cΒ£M is an abelian group.
I've managed to prove that associativity implies the result. But associativity seems tricky to prove.
scarlet estuary
i'd assume that was meant to say $a \oplus n = 0_R = 1$
cloud walrusBOT
Namington:
scarlet estuary
and that it's just a typo
scarlet estuary
hm?
leaden finch
im confused why he wrote a + n = OR = 1
winter yoke
I've managed to prove a(ab)=b. Associativity seems closer and closer, but now I'm stuck again. Any ideas?
scarlet estuary
@leaden finch because that's what it means for an element to have an inverse
an element $a$ has inverse $n$ if $a \oplus n = 0_R$, where $0_R$ is the additive identity in your structure
cloud walrusBOT
Namington:
scarlet estuary
in this case, we know $0_R = 1$ from part R4
cloud walrusBOT
Namington:
scarlet estuary
again, this is literally just definitions
leaden finch
oh okay, so then he was not supoose to do -1 then ?
merry pollen
The quadratic integer ring ${\displaystyle \mathbb {Z} [{\sqrt {-5}}]}\mathbb Z[\sqrt{-5}]$ of all complex numbers of the form ${\displaystyle a+b{\sqrt {-5}}}a+b\sqrt{-5}$, where a and b are integers, is not a UFD because 6 factors as both 2Γ3 and as ${\displaystyle \left(1+{\sqrt {-5}}\right)\left(1-{\sqrt {-5}}\right)}\left(1+\sqrt{-5}\right)\left(1-\sqrt{-5}\right)$. These truly are different factorizations, because the only units in this ring are 1 and β1; thus, none of 2, 3, ${\displaystyle 1+{\sqrt {-5}}}1+\sqrt{-5}, and {\displaystyle 1-{\sqrt {-5}}}1-\sqrt{-5}$ are associate.
Could someone explain this
why does the fact that 6 factors into both 2 and 3 and the other one mean that the ring is not UFD?
cloud walrusBOT
Meet The Noob:
merry pollen
why isn't it a UFD?
It factors into those factors and they're not associate, but why does this imply that it is not UFD?
winter yoke
By definition of a UFD (Unique Factorization Domain).
scarlet estuary
intuitively, something is a UFD if everything (except 0 and units) has a "unique 'prime' composition"
prime in the sense of irreducible elements
you just found that 6 can be written as a product of irreducible elements in 2 different ways
and these ways are "meaningfully different"
(i.e. not just a simple reordering or whatever)
hence you've found a counterexample
and so this is not a UFD
is that still not clear?
golden pasture
am i being a brainlet or is $\left(\bZ^{\infty}\right)^3\cong\left(\bZ^{\infty}\right)^2$
cloud walrusBOT
Ariana:
golden pasture
$\bZ^\infty$ is a ring right
cloud walrusBOT
Ariana:
winter yoke
I managed to prove (xz)(yz)=xy, (xx)(yx)=xy, and (x(yx))x=xy. Associativity always seems just out of reach.
latent anvil
@golden pasture what's your definition of Z^infty? A countable product of Z with itself?
Because then yeah
latent anvil
That isomorphism in AM is an isomorphism of A-modules
Yours is an isomorphism of rings
I got confused by this too
cloud walrusBOT
IAmJon:
upper inlet
Yes?
(With the implicit assumption that we must show that such an x exists. If we show that x is rational, then x clearly exists)
upper inlet
I just assumed they wanted help with R12
scarlet estuary
yeah sure
just pointing out for them that they highlighted both, but only gave the info to answer one
upper inlet
Oh right
upper inlet
You determine the additive identity through a different aciom
Look at what I had above
Do you have that?
Should be 1_R
leaden finch
wym
scarlet estuary
upper inlet
Aren't we talking about R12?
upper inlet
From R11?
You would find 1_R by the axiom defining the multiplicative identity
That is, the e\in R such that e\odot x =x for any x\in R
Did you do this?
bitter mauve
what is Tor, guys
this author didnt say anything before. is theorem 4.5 the Tor thing?
(see second last line)
i see that commuting with tensor is what 4.4 is about, but commuting with Tor.... dunno. i read wiki and i dont understand shit
is it better to just ignore this for now?
latent anvil
Tor is love
Also idk what you talked about before but it's worth noting that it's a little different with Tor because Tor is covariant in both arguments
So you use projective resolutions for each argument
woven delta
I think you shouldn't worry about this too much @bitter mauve
It seems like that remark was just an aside
For the future when you know about Tor
What book is this?
bitter mauve
matsumura
woven delta
Lol that's the second version
bleak abyss
I think those aren't editions of a book actually, they happen to be similar but they were originally meant to be separate books
bitter mauve
yea, this is a first edition
but matsumura says that it grew out of the other book
in intro
again, apologies for a retarded pic, lens is scratched
clear obsidian
Ok so in a PID [k_1, ...] = [gcd(k_1, ...)] in any PID not just Z thats neat
feelsgoodman to finally be getting somewhere
chilly ocean
is x^2+1 reducible over R?
doesnβt every real polynomial factor into a product of linear and irreducible quadratic terms so that means it is?
mild laurel
its irreducible yes
mild laurel
You don't need even there, every real polynomial with degree greater than 2 is reducible
mild laurel
yes
chilly ocean
how do you know if x^4+1 is reduible over any field ?
mild laurel
depends on the field
mild laurel
what do you mean by Zp
chilly ocean
Z mod p where p is prime
mild laurel
Okay sure, then there's really nothing better to do than write
x^4 + 1 = (x^2 + ax + 1)(x^2 + bx + 1) and see if there exists a solution for a,b
but you also have to check x^4 + 1 = (x^2 + ax -1)(x^2 + bx - 1)
chilly ocean
how do you know to write it like that?
upper pivot
everything is an unit so (x^2+ax+c)(x^2+bx+c^(-1))
mild laurel
Just compare the first coefficient and the last coefficient
chilly ocean
oh you used difference of squares?
mild laurel
uh, no
upper pivot
ok suppose x^4+1 were reducable
what ways could it factor
interms of degrees of the factors
chilly ocean
then x^4+1 = f(x)g(x) where deg f,g is greater than or equal to 1
upper pivot
well tell me the possibilities
chilly ocean
2,2 or 3,1 or 1,3
chilly ocean
it has a root
upper pivot
mhm, when does x^4+1=0 have a root in F_p
chilly ocean
nowhere
chilly ocean
say it doesnβt
upper pivot
then try out the 2,2 case
this means you can write x^4+1 as product of two quadratics
chilly ocean
(ax^2 + bx + c)(dx^2 + ex + f)
upper pivot
yeah, notice since this is over a field and our poly is monic you can just make a=d=1
chilly ocean
then expand and system of equations?
upper pivot
yeah
chilly ocean
but some coefficients will be non linear
upper pivot
they will
chilly ocean
it will be like b*e x
b* e = 0 so b or e is 0
ok let me try write it out
can i also factor it by finishing the complex roots?
finding
chilly ocean
ok
mild laurel
I'm not sure exactly what you're asking
complex doesn't exactly make sense when you're working over F_p
chilly ocean
oh
it works for factoring over R though
the system of equations is ad=1 ae=0, af=0
there is a lot
but it doesnβt work
mild laurel
Yeah, but it might factor over F_p
Like you can check that x^4 + 1 = (x^2 + 1)(x^2 + 1) over F_2
mild laurel
no you get that ae + af + bd = 0
chilly ocean
oh
mild laurel
There are lots of solutions over R
abstract sandal
I have a question regarding the easiest manner in determining if two groups are isomorphic
specifically the cartesian products
Z6 x Z10
and Z2 x Z2 x Z15
all element orders seem in place so theres no clear mismatch
idk where to go from there
chilly ocean
write some sort of homomorphism between the two groups and try to show that the kernel is 0
upper pivot
do you know the general theorem $Z_{rs}\cong Z_r\times Z_s$ for $(r,s)=1$
cloud walrusBOT
JohnDoeSmith:
abstract sandal
yes
upper pivot
this will immediately show your two groups are isomorphic
abstract sandal
by (r,s) you mean gcd(r,s) right?
upper pivot
yeah
abstract sandal
if so i dont see how that makes them isomorphic
upper pivot
well first split up Z6 and Z10
upper pivot
np
this is generally how you figure it out for cyclic groups
general groups you often need to explicitly construct the isomorphism
abstract sandal
i see
latent anvil
Iirc finite abelian groups are determined by the number of elements of each order
But you need the classification to prove this
winter yoke
Let f: Zmn -> Zm x Zn be given by x -> (x mod m, x mod n). The kernel is trivial, so f is injective. Both groups have the same size, so f is a bijection, hence the groups are isomorphic.
frigid spade
Is it true that the number of inversions is related to the number of even cycles in a permutation?
@chilly ocean my bad, that actually cleared things up. Thank you.
halcyon siren
When we talk about coroot of a lie algebra associated to r do we really mean 2B(r, dot) /B(r, r)
In my book it is written as 2r/(r,r), but that would just be in the same space as r...
covert perch
How to prove that if A and B are matrices of same size such that $B^{T}A=0$
Then rank(A+B) = rank(A) + rank(B)
cloud walrusBOT
h4harshul:
gentle pendant
im(A) and im(B) are two orthogonal subspaces which implies that their sum (im(A+B)) is direct.
pure crest
determine the number of group homomorphisms from $Z_{245}$ to $Z_{175}$ whose image is size 5
i am not sure how to do this
mild laurel
Remember that a homomorphism from a cyclic group is defined by where the generator goes
pure crest
right
cloud walrusBOT
boat:
mild laurel
So where could you send 1 in Z_245
pure crest
i am not sure i understand. Cant 1 go anywhere from Z_245 to Z_175?
pure crest
1 is in the image