Wait sorry
#groups-rings-fields
woven delta
I didn't read what you posted
And I'm actually not sure what they are trying to say
But I'll just give an exposition on the semidirect product
Okay so suppose H and K are groups, and we have a map from K to Aut(H)
smoky cypress
Ok thanks, I'm kinda lost, I have no idea what the semidirect product, whether internal or external, is trying to do
woven delta
Ie we have an action of K on H
smoky cypress
Yeah
woven delta
So we want to use the action of K on H to construct a new group
In this new group we have an embedding of H and K, H is normal, and the action of K on H by congugation is exactly the action of K on H given by our map
Does that make sense?
smoky cypress
What do you mean action of K on H by conjugation?
woven delta
Yes
smoky cypress
So we have an action on H
woven delta
Yes
If you identify the subgroups with the abstract groups K and H they are exactly the same action
Okay so let's do this
So we define the group G to be the Cartesian product
So just H x K
smoky cypress
Ok
woven delta
Now we define $(h_1, k_1) (h_2, k_2) = (h_1 \phi_{k_1}(h_2), k_1k_2)$
cloud walrusBOT
Liquid:
woven delta
Where $\phi_{k_1}$ is $\phi(k_1)$ ($\phi: K \to Aut(H)$)
cloud walrusBOT
Liquid:
woven delta
Okay so now let's do the inner stuff
smoky cypress
This is outer semi-direct product
woven delta
Yes
smoky cypress
I keep not typing the last latter in the sentence
woven delta
Okay so let's identity H and K with the obvious subgroups
So there are a couple of questions
First off why is H normal?
smoky cypress
Well you can directly verify it
woven delta
Whatever method you want to do
woven delta
Yeah
So you can take the homomorphism to the second coordinate, verify its a homomorphism, then the kernel will be exactly H
That's one way of doing it
Okay now the fact that H and K intersect trivially should also be obvious
Why is HK = G?
That part should be obvious as well
woven delta
HxK
smoky cypress
Is that equal to (h,k) tho
woven delta
Yes
woven delta
Because the automorphism corresponding to 0 is the identity
vital quail
plucky flicker
Hey guys! If i have a homomorphism between 2 polynomial rings, then if i find somehow an element of the kernel, this element will generate the whole kernel? For example f:k[x,y,z]->k[p,q] such that x -> p, y -> pq, z -> pq^2, then clearly the xz-y^2 is in the kernel of f, but is the whole kernel <xz-y^2>?
knotty mason
I think it would not necessarily be the case if your polynomial rings were over a ring rather than a field, like Z[X,Y]
plucky flicker
and if is over a field? in my case it is always over a field
an algebraically closed field
knotty mason
in that case im not really sure, i think probably not still
Every ideal is the kernel of some ring homomorphism
does your ring have ideas that are not principal?
plucky flicker
i dont think so
it's a multivariate polynomial ring over a field
i mean in this case yes, there are not principal ideals
K[x] is PID, but K[x_1, ..., x_n] is not
plucky flicker
I mean, even simpler, if we take the map from Z to Z that sends x to 2x, then 4 is in the kernel, but it doesn't generate the whole kernel.
@open torrent the kernel in this case is trivial 0, because x -> 2x is an injective group endomorphism on (Z,+)
In the example you give for example, 2(xz - y^2) is in the kernel, but the whole kernel isn't <2(xz - y^2)> since xz - y^2 isn't in it
@open torrent xz - y^2 is 2(xz - y^2)*1/2, isn't it?
This doesn't really make sense at all
@open torrent i think you are right, but i am studying some stuff here about algebraic varieties and image closure and this question just came in my mind from a notation, i guess i'm overthinking
knotty mason
so combine the fact that there are non-principal ideals, And that every ideal is the kernel of some homomorphism
glacial prism
Given two integral domains $R$ and $R'$ and a morphism $f:R\rightarrow R'$ can you easily extend $f$ to a morphism $Quot(f):Quot(R)\rightarrow Quot(R')$ of fields when $f$ has nontrivial kernel, ie is not injective? If it's injective you can just use $Quot(f):a/b\rightarrow f(a)/f(b)$, this is well defined on the classes of fractions since $f$ is morphism.
cloud walrusBOT
MrMonday:
glacial prism
so maybe make a functor from IntDom into Fld
No xD
If $Quot(f)$ is a morphism of fields, it is necessarily injective and the restriction to $R/1$ would be injective aswell. Since this restriction is basicially $f$ this doesn't make sense
cloud walrusBOT
MrMonday:
woven delta
Field of fractions is a functor from the category of integral domains where the morphisms are injective maps to field
next obsidian
Field of fractions is a functor from the category of integral domains where the morphisms are injective maps to field
@woven delta wait the maps are injective maps into other integral domains no?
next obsidian
Yeah
next obsidian
It suggests the only morphisms we can consider are injective with target a field
Ohhh
I see
woven delta
The morphisms are between Integral domains
next obsidian
Lmao
Gotcha
So it’s integral domains with morphisms injective maps
To Field, the category of fields
I was just really confused why you seemed to only be looking at injective maps into fields
When that’s really oddly specific and not at all the most general case one can consider haha
glacial prism
Yeah if you have an injective map f between integral domains and want to extend to fractions over that domain, ie send a/b to f(a)/f(b) this works because Ker f is trivial and you don't divide by zero
shy bluff
@elder valley continuing from ysetrday then I get a sum of the form $$0 = \sum_{i = 0}^{n} (\sum_{j = 0}^{i} r_{(i), (i - j)})y^i)$$ I think? ANd we basically want to show that the only time that we have that we have that $\sum{j = 0}^{i}(r_{(i), (i - j)} = 0$ is if said term is a multiple of s - t?
cloud walrusBOT
Liria ^(;,;)^:
elder valley
first sentence looks good. for the second sentence it's not clear what you mean by term
shy bluff
each coefficient of y^i
elder valley
or what term you're referring to
shy bluff
in order for that whole mess to equal to 0 we have to have that each $\sum_{j = 0}^{i} r_{(i), (i - j)} = 0$
cloud walrusBOT
Liria ^(;,;)^:
shy bluff
$\sum_{i = 0}^{n}(\sum_{j = 0}^{i} r_{(i), (j - i)} s^{i} t^{i - j})$
cloud walrusBOT
Liria ^(;,;)^:
elder valley
in order for that whole mess to equal to 0 we have to have that each \sum_{j = 0}^{i} r_{(i), (i - j)} = 0
so solve this sum for, say, r_{i,0} and substitute that into your expression for r(s,t)
elder valley
r_{j,i-j} i think
actually you have it right except with the expression you wrote your outer sum should be variable j and inner sum should be i
like in that last picture solve the 4th equation for r_{4,0}
shy bluff
$r_{i, 0} = \sum_{j = 1}^{i} (- r_{(i - 1), j})$
cloud walrusBOT
Liria ^(;,;)^:
elder valley
i think your notation is all messed up
shy bluff
Yea it's flipped around lol
but o k I see it now lol
Thank you!
Realyl I should be looking for $r_{0, i}$ I think ?
cloud walrusBOT
Liria ^(;,;)^:
shy bluff
Because r_{i, 0} gives me that t - s is the kernel, not the othe rawy around lol
elder valley
it should work no matter what you solve for
its divisible by t-s if and only if its divisible by s-t
shy bluff
Yea same thing
chilly ocean
So you can show that for primes p,q, groups of order pq are not simple. You can show this by showing that one of sylow p or q subgroup is unique and thus normal. You can show that groups of order 56 are not simple. So my question is, is there a general statement P(n) such that all groups of order n are not simple if and only if P(n)
next obsidian
So you can show that for primes p,q, groups of order pq are not simple. You can show this by showing that one of sylow p or q subgroup is unique and thus normal. You can show that groups of order 56 are not simple. So my question is, is there a general statement P(n) such that all groups of order n are not simple if and only if P(n)
@chilly ocean if you want a nice P(n), I’m almost certain the answer is no. However, we have a classification of finite simple groups (now who actually knows the classification? I’d reckon only a handful of people) but theoretically we know every single finite simple group so P(n) could just be “according the classification of finite simple groups, no simple group of order n exists”. Now this is not really what you had in mind I think, but I don’t imagine that there’s anything very easy you can check, as the number of groups of each order tends to get really big.
shy bluff
@elder valley that was very messy but I finally got it 
elder valley
yeeeah you should have just used sum notation xD
the proof is like 4 lines then
glad you got it though
shy bluff
Idk I could'nt get the sum notation to work in my ehad
I had to manually write it out and even then I could'nt go and get the sum notation to work
Like Inever figured out how to write this as a sum
I tried and the indices got too messy
LOL
elder valley
you had r(s,t) written as a sum from the beginning. you should be able to keep it that way throughout the whole proof
so you don't have to convert
shy bluff
hrmm
ok
Well anyways
how does this work?
I don't really understand how they get that f + I = p(x) + yq(x)?
elder valley
because in the ring R[x,y]/<x-y^2> you have y^2=x, so you trade all the powers of y for x's except for the y^1 term
elder valley
yeah pretty much
in the beginning they separate f into even and odd powers of y
in the middle rather
shy bluff
hrm
o k
I have a different similar question that I'm trying to solve and I think that I can use this method
WAit question
Why is it that between these two lines
they go from f_0(x) + q(x)y + (f_2(x)y^2 +... f_{2k}(x)y^{2k}) but in the bottom lien, the second term is in terms of x... x^k?
Is it because we've set x = y^2?
elder valley
yeah. if that last polynomial is g, they are saying f+I=g+I
shy bluff
So is showing isomorphism in this general way really just figuring out how to write it in a specific way that lets you factor out your quotient easily?
elder valley
i think it's mostly just because the ideal is principal
shy bluff
wdym
Err if you have a ring tha tyou're taking the quotient of with a principal ideal
Then it's basically just "Figure out how to write it so that you can factor out the quotient"?
elder valley
yeah, because if the ideal is principal and you want to show something is in the ideal then that amounts to showing divisibility by the generating element
shy bluff
hrmm
ok
Frustrating, the questions that we've been given are all in multiple variables 
elder valley
that makes it more interesting!
elder valley
polynomial rings are never fields since the variables aren't invertible
elder valley
only with 1 variable
shy bluff
hrmmm why
elder valley
i think the best you get is "R is a UFD -> R[x] is a UFD"
which you can apply iteratively to extend to finitely many variables
shy bluff
Then so is R[x][y]
yea
Can I DM you with my solution for this one later and ask you to check it over 😅
shy bluff
ya no rush
charred pewter
Is this asking what the order of Z/12Z is or how many units (as in invertible elements) there are
chilly ocean
Is the Lemma below valid? (I just thought of it for another proof)
The additive subgroup generated by any nonzero element in a finite ring of order p, p prime, has itself order p.
Proof:
Let A be a finite ring with |A| = p, p a prime number
Any nonzero element a in A under the additive operator generates the subgroup <a>. By Lagrange's theorem |<a>| k = p.
But |<a>| > 1 because a is not the zero element.
Hence it must be that |<a>| = p
(I can't post in the abstract algebra channel so I'm posting here)
charred pewter
This is the abstract algebra channel
elder valley
Is this asking what the order of Z/12Z is or how many units (as in invertible elements) there are
@charred pewter the number of invertible elements in the ring
charred pewter
@elder valley The weird thing is this question is from a chapter that's like 5 chapters before ring theory lol
in my abstract algebra class, I mean
That's why I was really confused
elder valley
@chilly ocean yes that looks good
chilly ocean
ya true
charred pewter
Idk what I should do about this question
I'm tempted to answer it as an invertible elements question but then again this was like.. weeks before we started talking about ring theory
elder valley
you must have talked about invertible elements at least, for that question to make sense
chilly ocean
could they mean additive inverses?
charred pewter
Then again, this isn't the first time he's posted a question that has nothing to do with what we've discussed so far
Rip
shy bluff
If I have a group action of G on an element x, then it has to be for every element g of G on x rigHT/
sly storm
Hello. I asked the following question in some places without receiving the answer. I guess the problem is equivalent to solving a quintic equation, but I cannot prove such a claim. Can that claim be proved by using som abstract algebra methods or preferably some simple math?
Suppose we know $A=a^2+b^2+c^2+d^2+e^2$, $B=a^2b^2+a^2c^2+a^2d^2+a^2e^2+b^2c^2+b^2d^2+b^2e^2+c^2d^2+c^2e^2+d^2e^2$, $C=a^2b^2c^2+a^2b^2d^2+a^2b^2e^2+a^2c^2d^2+a^2c^2e^2+b^2c^2d^2+b^2c^2e^2+c^2d^2e^2$, $D=a^2b^2c^2d^2+a^2b^2c^2e^2+a^2c^2d^2e^2+a^2b^2d^2e^2+b^2c^2d^2e^2$, and $E=a^2b^2c^2d^2e^2$. How can we find $a+b+c+d+e$?
cloud walrusBOT
MathPhysics:
nova plank
a, b, c, d and e >= 0?
sly storm
Yes
nova plank
Well, I can't find a + b + c + d + e, but I can find abcde, does that help? /s
scarlet estuary
that seems miserable
glacial prism
If we are in characteristic 2 then
scarlet estuary
and i dont expect there to be a good abstract algebraic method
scarlet estuary
maybe something combinatorial in flair but even that seems unlikely
glacial prism
you could plot each equation seperately as a variety and then look at their intersection
or just all equation as a big variety
like a^2+....+e^2-A=0
scarlet estuary
what does that get you
glacial prism
Geometrical intuition :D
scarlet estuary
for... what?
im not sure how to geometrically interpret a + b + c + d + e given that
or indeed, how to interpret any given variable
glacial prism
true
well you could add one more varibale to the system and take the whole variety including a+b+...+e-X=0. Then each nonempty slice with X constant is a solution to the system with a+...+e=X
although I dont know how to do that lol
scarlet estuary
and this is supposed to help intuition how?
kinda feels like its doing the opposite at this point
there might be a more elegant approach but sums of squares in R^5 fucking suck to work with
glacial prism
well you could add one more varibale to the system and take the whole variety including a+b+...+e-X=0. Then each nonempty slice with X constant is a solution to the system with a+...+e=X
This wasnt for intuition, this was supposed to be a way to solve it :D
scarlet estuary
ah i see
blergh
i guess thatd work if you jam it out
but it seems like a right pain
and your solutions would not be nice whatsoever
glacial prism
seems more elegant to me than jamming out 50 combinatorial identities xzd
scarlet estuary
just figuring out how to write down the solutions you extract from that method could probably flesh out a few textbook chapters on its own
rational points on elliptic curves style
sly storm
<@&286206848099549185>
scarlet estuary
i still think i want to try a combinatorial method, but i have no clue whether it actually goes anywhere
and its too early for me to think about this
ill sleep on it though
weird problem
sly storm
Should I repeat my question?
sly storm
Thanks.
hot lake
you find it by finding the roots of a degree 32 polynomial
maybe a degree 16 polynomial then taking a square root
eager willow
for affine algebraic sets $V, W$, I am trying to show that $$k[V]\otimes_k k[W] \simeq k[V \times W].$$
I know by the universal property that mapping $f\otimes g$ to a function defined by $$(f \otimes g)(v, w) = f(v)g(w)$$ is a well defined homomorphism. For injectivity, I know that if $f \neq 0 \in k[V]$ then there is some $v' \in V$ with $f(v') \neq 0$ and so given that $f(v)g(w) = 0 \forall (v,w) \in V \times W$, we would have $f(v')g(w) = 0 \forall w \in W$ and hence $g = 0 \in k[W]$ so $f \otimes g = 0$. But I don't know if this mapping is surjective
golden pasture
@sly storm (i think you asked the qn right?) isnt a^2,b^2...e^2 just the roots of a quintic?
so what you do is
pray that A,B,C,D,E are nice numbers
then you get integer sols
oops that was literally the qn on how to convert to a quintic
||vieta||
sly storm
@golden pasture Do you have any idea for this question?
golden pasture
yea
consider the polynomial (x-a^2)(x-b^2)...(x-e^2)
edit: b^2 instead of b^1
what are it's coefficients?
scarlet estuary
i tried that but it doesnt give a nice path
from what i could see at least
maybe im just dumb though
delicate bloom
it's just some random question they came up with
so there's no reason to expect anything nice, it wasn't designed to be
scarlet estuary
it feels like there should be a nice combinatorial approach though
like the coefficients are obviously 1, -A, B, -C, D, -E
but what does that tell you about a + b + c + d + e?
maybe im missing something obvious
eager willow
@dapper nebula I am still a little confused. I know the basic statement of the universal property but I haven't used it much in a proof before.
Returning to the potential counterexample, would I want to map $x_1 + x_3$ to the tensor somehow given by $(x_1 \otimes 1) + ( 1 \otimes x_3)$? Then maybe it'd be hard to find the simple tensor notation for that element, but it's a group so one exists by the univsersal property?
cloud walrusBOT
datorangeguy:
eager willow
I think if thats the right idea for that example then I know what to do for the general case :)
Oh nevermind, it just doesn't have to be a simple tensor and that's why it's surjective. I think I got it
For some reason I had it in my head that every tensor was simple
latent anvil
That's like the most common error with tensor products lol
Everybody does it
Even after you get used to working with tensor products you'll have times where you think every tensor is simple for a couple minutes and then realize
eager willow
Yeah i was using the universal prop just to see that it was a well defined homomorphism but the mapping was put only in terms of simple tensors so I got lost
I was up pretty late last nighr scratching my head lol I guess I just had to come back to it
latent anvil
Wait is your field k algebraically closed?
I have a counterexample in the case where it isn't
I guess it probably is since you're working with algebraic sets, nvm
okay yeah nvm my concern was that the tensor product of two reduced algebras could be nonreduced, but this can't happen over a perfect field (in particular it can't happen over an algebraically closed field)
disregard my rambling lol
eager willow
These were the coordinate rings for affine algebraic sets so idk if that's a reduced or not but no assumptions are made on the field
sly storm
I missed some terms in my question, so I revise it:
Suppose we know $A=a^2+b^2+c^2+d^2+e^2$, $B=a^2b^2+a^2c^2+a^2d^2+a^2e^2+b^2c^2+b^2d^2+b^2e^2+c^2d^2+c^2e^2+d^2e^2$, $C=a^2b^2c^2+a^2b^2d^2+a^2b^2e^2+a^2c^2d^2+a^2c^2e^2+a^2d^2e^2+b^2c^2d^2+b^2c^2e^2+b^2d^2e^2+c^2d^2e^2$, $D=a^2b^2c^2d^2+a^2b^2c^2e^2+a^2c^2d^2e^2+a^2b^2d^2e^2+b^2c^2d^2e^2$, and $E=a^2b^2c^2d^2e^2$. How can we find $a+b+c+d+e$?
cloud walrusBOT
MathPhysics:
delicate bloom
you could divide D/E to get the sum of inverse squares version of A
similarly C/E, B/E, and A/E will get other similar looking ones
does that do any good, no clue, except at least now you can state you have even more similar looking equations to think of as your starting point
you could then try adding them, like A + D/E or A-D/E and write it in terms of hyperbolic cosine and sine, probably won't be helpful though
don't know, don't care tbh
golden pasture
uh
cant you literally
just solve quintic
square root the roots
and done?
if we dont want to find roots, im pretty sure no amount of manipulation of A,B,C,D,E leads to a+b+c+d+e
knotty mason
Consider the symmetrization of the ring Q[a^2,b^2,c^2,d^2,e^2], Q[a^2,b^2,c^2,d^2,e^2]^S_5 = Q[A,B,C,D,E]
and Q[a,b,c,d,e]^S_5
a+b+c+d+e is in Q[a,b,c,d,e]^S_5
I believe that Q[a,b,c,d,e]^S_5 can be given a basis in terms of Q[A,B,C,D,E]^S_5 by only extracting square roots
what do you think?
sly storm
@knotty mason Thanks for your response. Can you please elaborate what you wrote; I do not know anything about symmetrization of a ring and the rings Q[A,B,C,D,E]^S_5. There is no simpler way for this question?
knotty mason
Let G act on {X_i} then R[X_1,..,X_n]^G is the ring of symmetric polynomials R[X_1 + ... + X_n, X_1 X_2 + ..,, X_1 X_2 .. X_n]
sly storm
G is the symmetric action?
knotty mason
G is a group that acts by permuting variables
sly storm
Q is any polynomial ring, right?
knotty mason
Q is the rationals
sly storm
Now, you want to prove what?
I believe that Q[a,b,c,d,e]^S_5 can be given a basis in terms of Q[A,B,C,D,E]^S_5 by only extracting square roots
@knotty mason
Why is that true?
knotty mason
i don't have a proof for that
hot lake
I would be very very skeptical of that
knotty mason
does it sound like a good approach for the problem though?
hot lake
galois theory is probably a good approach
sly storm
How?
hot lake
galois theory is made for answering those questions
sly storm
Galois theory is made for answering unsolvability of polynomials.
hot lake
what if a+b+c+d+e was the root of a polynomial with coefficients in Q(A,B,C,D,E) ?
what if galois theory told you how hard that polynomial is to solve ?
knotty mason
Do we know a polynomial with a+b+c+d+e as a root?
sly storm
Hard question
golden pasture
isnt the qn like
Let $\alpha_i$ be roots of some quintic, can we find $\sum\sqrt{\alpha_i}$ in terms of coefficients
cloud walrusBOT
ariana:
knotty mason
@golden pasture you can solve it that way, by finding a,b,c,d,e then computing a+b+c+d+e
but I was hoping we could exploit the fact that a+b+c+d+e is symmetric to express in a simpler way than roots of a quintic
golden pasture
yea but seems like the asker wants a explicit form in terms of the symmetric polys
knotty mason
I tried to apply symmetric polys in sage a bit but I couldn't really get it to work out
golden pasture
my suspicion is no but not too sure how to show no yet
sage sucks with like a lot of abstract algebra stuff for some reason
once it isnt GF(p)
golden pasture
is ok
im trying to find minpoly of the sum rn
for some simple polynomial
but taking some time 🤔
it's gg be like a degree 50 lol
gg
knotty mason
lol
golden pasture
x = polygen(AA)
p = prod(x-i for i in range(1,6))+1
t = sum(sqrt(i[0]) for i in p.roots())
t.minpoly()
if anyone wishes to burn their cpus
but i really dont see any reason why it should be possible
like i feel it's kinda asking can we express some polynomial f(t) in terms of the coeff
which seems supeeer unlikely
hm anyone knows of a way to check if some number can be expressed as a tower of radicals
without like min poly > galois group
knotty mason
I was looking for a way to estimate a minpoly numerically
golden pasture
but then the galois group is pretty sensitive to the exact coeff of the minpoly sometimes right? esp if you encounter the case that the group is solvable poke something a little becomes unsolvabe
knotty mason
? polgalois(x^6 - 6x^4 + 11x^2 - 5)
%1 = [48, -1, 1, "2S_4(6) = [2^3]S(3) = 2 wr S(3)"]
thought this was kinda cool, your program generated this poly for range(1,4)
never seen a wreath product galois gorup before
golden pasture
owo
i wonder if we can use more basic polynomial theory
so we have the poly $\prod\left(x-\alpha_i\right)$. is it possible to construct some poly that gives what we want that is a multiple $\prod\left(x-\sqrt{\alpha_i}\right)$ that doesnt also have $-\sqrt{\alpha_i}$ as one of it's roots
feels wrong if it's possible
cloud walrusBOT
ariana:
golden pasture
lets just go with it
$\prod\left(x^2-\alpha_i\right)$ is a candidate and the coefficients are easy to find, we can split it to $\left(\prod\left(x-\sqrt{\alpha_i}\right)\right)\left(\prod\left(x-\sqrt{\alpha_i}\right)\right)$.
Let $\prod\left(x-\sqrt{\alpha_i}\right)=\sum p_{5-i}x^i$, then $\prod\left(x+\sqrt{\alpha_i}\right)=\sum(-1)^{5-i}p_{5-i}x^i$. We know the coefficients of $\left(\sum p_{5-i}x^i\right)\left(\sum(-1)^{5-i}p_{5-i}x^i\right)$ in terms of $p_i$ and $A,B,C,D,E$, so maybe we can somehow find what is $p_1$ in a simpler way? seems super unlikely ngl
cloud walrusBOT
ariana:
golden pasture
theres one annoying thing we can do
spam resultants
and get p_1 in a polynomial
is very very ugly
knotty mason
if it works it works!
golden pasture
this is probably as good as it gets tbh
golden pasture
am trying to play with some infinite matrix groups, any program that is usually used for this other than bashing sage realllllly hard
sage: G.quotient(H)
---------------------------------------------------------------------------
NotImplementedError Traceback (most recent call last)
<ipython-input-20-ebaeef77a3b4> in <module>
----> 1 G.quotient(H)
/usr/lib/python3.8/site-packages/sage/groups/group.pyx in sage.groups.group.Group.quotient (build/cythonized/sage/groups/group.c:3226)()
236 NotImplementedError
237 """
--> 238 raise NotImplementedError
239
even constructing PSL(2,k) is a pain ;-;
probably not a good idea to do this while im half awake
scarlet estuary
yeah at this point
i dont think therell be a good approach
like, whatsoever
the best thing to do might be to jam it into mathematica
and see if mathematica can figure anything out
but im doubtful
golden pasture
mathematica feels super unnatural for groups but ig can try
charred pewter
So in a previous proof with this exact same ring, I proved R was a commutative ring since we have the additive abelian group and we have commutativity under addition. What would I need to do to improve upon that proof to prove it's an integral domain but not a field?
scarlet estuary
to prove its an integral domain, you need to show that x*y is not equal to 0 for nonzero x, y
to prove its not a field, you need to show that multiplicative inverses don't necessarily exist
(so find a nonzero x such that x^-1 doesnt exist; i.e. there is no y such that x*y = 1)
(and prove it)
smoky cypress
When talking about polynomial rings do we usually assume the coefficient ring is commutative?
next obsidian
I think that it's really hard to define polynomial rings over a noncommutative ring or something
I remember @latent anvil told me something like that
pure crest
if $\alpha_1 , \alpha_2.... \alpha_n \in E$, where E is a field extention of F, be roots of f(x) which is a irreducible polynomial over F. let $g(x) = (x-\alpha_1)(x- \alpha_2) ... ( x - \alpha_n) \in E[x]$. Then $f(x) | g(x)$
cloud walrusBOT
boat:
pure crest
how can i prove this
can i say that $x - \alpha_1 | f(x) $
i was thinking of arueing that since f(x) is irreducible, then if gcd(f,g) = 1 or f(x) since f(x) is irreducible
but i am not sure if i can still do that when i am working with elements in E
where both are not irreducible
so if x - alpha_1 | g(x) and f(x), then f(x)| g(x)
but i am still stuck in the previous step
oblique river
@pure crest if a is a root of f(x) then (x-a) | f(x)
you can prove that using the division algorithm
dividing f(x) by (x-a) you get f(x) = (x-a)*q(x) + r(x) where deg r(x) < deg(x-a), i.e., r(x) is a constant
if you plug in x = a to both sides you get
f(a) = 0 * q(a) + r
but f(a) = 0 by assumption
therefore r = 0
f(x) is irreducible in F[x] and (x-a) isn't in F[x]
because a isn't an element of F
pure crest
right
oblique river
also I just reread what you wrote
pure crest
so i am confused as to if i can make this argument: gcd(f,g) = 1 or f(x)
oblique river
is g(x) in F[x]?
I'm confused about what you're trying to prove
if you're trying to prove that f(x) divides g(x)
then whether or not x-a divides f(x) doesn't seem to be relevant
pure crest
well i was trying to work with the idea that f(x) is irreducible, gcd(f,g) = 1 or f
im kinda confused as to which field i am working with
since i am given that f(x) is irreducible over F
oblique river
f(x) is only irreducible in F[x]
not in E[x]
but if g(x) isn't an element of F[x]
then you can't really use that logic
pure crest
right
oblique river
are you doing galois theory? or do you have to do this just using field stuff?
also, the way you worded the problem is al ittle confusing to me. are alpha_1 through alpha_n all of the roots of f(x)?
pure crest
this was just part small part of my entire question i had to do
ill post the entire thing
maybe i am confused
one second
so i am doing b
oblique river
what does the conclusion of part a tell you
pure crest
any element in the galois group fixes g(x)
oblique river
great
well, fixes all the coefficients of g(x)
what do we know about things which are fixed by any element in the galois group
pure crest
well they tend to be in F
pure crest
ahh i see
i wasnt sure we could make that statement, that if an element is fixed by all elements of the galois group its in F
oblique river
I hate saying things are obvious but... it literally says that in the statement of the problem...
pure crest
oof
you are right
i am dumb
sorry
i didnt realize it haha, i used the fact earlier to prove that any element must map E to another element to prove part a, and completely forgot what it literally says
oblique river
it happens lol
pure crest
this makes the question much easier now that f(x) and g(x) are in F[x]
pure crest
if i want to show that $Q(\alpha_1, \alpha_2, ... \alpha_n) = Q(\beta, \beta_2, ... \beta_k) $, why is it enough to show that $alpha_ i$ can be written in terms of $\beta $and vice versa?
cloud walrusBOT
boat:
elder valley
@pure crest because if Q(alpha_1,...,alpha_n) contains beta_1,...beta_k, then it contains the smallest field containing {beta_1,...,beta_k}, which is by definition Q(beta_1,...,beta_k). that gives you subset in one direction. doing the vice versa gives inclusion the other way
pure crest
ohh right, it contains the smallest field
that makes sense. thank you very much Auvera!
charred pewter
If I'm proving something is not a field, then can I show that multiplicative inverses don't necessarily exist by trying to prove it does have multiplicative inverses and finding a contradiction? Or what would be the best way to prove that?
stone fulcrum
If you are ever proving that something does not have property A, you just need to produce one counter example
woven delta
Let property P be exists x Q(x)
pure crest
is there a theorem like Eisenstein criterion when working with simple extentions of Q?
how can i show that x^3-5 is irreducible over Q(sqrt{2})
woven delta
@pure crest if it was reducible one of it's roots (call it alpha) are in Q(sqrt(2)). But the degree of Q(alpha)/Q is 3
(this only works because x^3-5 is degree 3)
pure crest
like x^2 + 1 irreducible over Q(sqrt{2})(sqrt{3})
woven delta
Hmm I'm not sure this argument works for that so cleanly
But you know that that's irreducible because i is not real
A nice thing to do in this case is to look at the roots in an algebraically closed fields containing your field (usually C) and argue from there
Something else you can do is look at subfields of the extension field, which are in correspondence with subgroups of the galois group in this case
pure crest
how would i argue using the first method?
woven delta
So basically if x^2+1 is reducible in Q(sqrt(2))(sqrt(3)) this tells you both of it's roots are in that field (why)
This tells you both of the roots of x^2+1 are in the real numbers (why)
Can you answer those questions?
pure crest
Well its of degree two, so if one root is in it, then it must split
in Q(sqrt(2))(sqrt(3))
woven delta
Well more precisely if it breaks up into smaller pieces those pieces must be linear factors
Since it's degree 2
pure crest
right
woven delta
And then Q(sqrt(2))(sqrt(3)) is a subfield of R
pure crest
and it must be real because x in Q(sqrt(2))(sqrt(3)) is of the form a+ b sqrt(3), where a, b in \sqrt(2), so they are real
right
woven delta
Yeah
But like you should think of Q(sqrt(2), sqrt(3)) as the smallest subfield of F containing sqrt(2) and sqrt(3)
And clearly R is a bigger field which contains sqrt(2) and sqrt(3)
pure crest
yup you are right
woven delta
Btw figuring out if things are still irreducible in an extension is a hard problem in general
But methods like these let you figure it out in certain small cases
pure crest
i took eisenstiens for granted this whole time. I really appreciate it now
woven delta
Yeah it's great
pure crest
thank you for explaining it so clearly
woven delta
👍
sly storm
Should I repeat my question?
true zenith
Oh if it was skipped over, sure
knotty mason
Ask it on MathOverflow
true zenith
You could do that as well if you want
sly storm
I asked the following question in some places without receiving the answer. I guess the problem is equivalent to solving a quintic equation, but I cannot prove such a claim. Can that claim be proved by using som abstract algebra methods or preferably some simple math?
Suppose we know:
\begin{align*}
A&=a^2+b^2+c^2+d^2+e^2 \
B&=a^2b^2+a^2c^2+a^2d^2+a^2e^2+b^2c^2+b^2d^2+b^2e^2+c^2d^2+c^2e^2+d^2e^2 \
C&=a^2b^2c^2+a^2b^2d^2+a^2b^2e^2+a^2c^2d^2+a^2c^2e^2+a^2d^2e^2+b^2c^2d^2+b^2c^2e^2+b^2d^2e^2+c^2d^2e^2 \
D&=a^2b^2c^2d^2+a^2b^2c^2e^2+a^2c^2d^2e^2+a^2b^2d^2e^2+b^2c^2d^2e^2 \
E&=a^2b^2c^2d^2e^2 \
\end{align*}
How can we find $a+b+c+d+e$?
cloud walrusBOT
MathPhysics:
charred glacier
If we consider elementary symmetric polynomials $p_i$ and $q_i$ in $a, ...$ and $a^2, ...$, respectively, then the question is to express $p_1$ in terms of $q_i$. By squaring $p_1$ repeatedly it is possible to form an even polynomial equation (of degree 32, in this case, or non-even of degree 16) which has $p_1$ as a root and coefficients as functions of $q_i$.
cloud walrusBOT
gk:
chilly canyon
Develop the square of the sum using elementary symmetric functions
charred glacier
not sure if there is a nice looking explicit solution though
https://math.stackexchange.com/q/909122
https://mathoverflow.net/q/123058
chilly canyon
I'm feeling dumb, I'm having trouble with some 1st year shit x) what's the reciprocal of the map Z/N→Z/a×Z/b given by the Chinese Remainder ?
elder valley
reciprocal meaning inverse?
chilly canyon
Yes
elder valley
the main point of the theorem is the surjectivity of that map. meaning for any (x,y) in Z/a x Z/b you can find n in Z/N such that n=x mod a and n=y mod b. this is what you use the define the inverse
if i remember right, the proof is constructive and tells you how to find n somewhat
chilly canyon
No you prove it is injective !
elder valley
you prove both
chilly canyon
No you have equal cardinalities you need not prove it is epic
elder valley
oh, yeah that works in the case of Z/N. in the general ring version you have to do both though
makes sense that you're confused about the inverse then
chilly canyon
I need an explicit inverse for my computations :\ would you happen to have a reference you know ? 😃
chilly canyon
Thanks ! I was afraid I had to use Bezout, sadly it is the case...
elder valley
it's not so bad
if your numbers are small enough you can just guess the coefficients with trial and error
sly storm
@charred glacier Thanks for your response. Then the resultant polynomial is necessarily solvable by radicals?
chilly canyon
Thing is I don't know a nor b !
elder valley
then just define your map as (x,y) -> x m_a a + y m_b b, where m_a a + m_b b = 1
pure crest
does $\mathbb{Q}(a,b) = \mathbb{Q}(a)(b)$?
cloud walrusBOT
boat:
pure crest
is this always true?
delicate bloom
what have you tried
solemn rain
yes
olive mirage
I don't know that I would even define those two things differently.
pure crest
i dont have an intuitive idea what Q(a,b) means, but i understand that Q(a)(b) if x \in Q(a)(b) such that x = a_0 + a_1 b + a_2 +b^2 + ... + a_n b^n / b_0 + b_1 b + ... b_n b^n
we only ever worked with simple extensions so i am kinda confused what the notation even means
so the two are interchangeable? @olive mirage
olive mirage
To me, both are the minimal extension of Q containing a and b
(I'm assuming a and b are elements of extensions of Q and not free variables)
(if they're free variables, it's still true, but I acknowledge the two things are not definitionally exactly the same)
pure crest
ohh i see
i am having a hard time relating the two ideas
i see Q(a,b) as the intersection of all subfeilds of E, containing both a and b and Q
but i am not sure what the book means by E in this case
its hard to see how the idea of intersection of all subfeilds of E containing a,b and Q relates to Q(a)(b) if x \in Q(a)(b) such that x = a_0 + a_1 b + a_2 +b^2 + ... + a_n b^n / b_0 + b_1 b + ... b_n b^n
knotty mason
yes, start from the definition
nova plank
A very good place to start
When you read you begin with a b c
When you prove you begin with d e f
knotty mason
eager willow
For 0->L -> M -> N -> 0 a short exact sequence of R-modules, if there is a prime ideal P in R that is an associated to N but not M, can there also be some prime ideal that is associated to M but not L?
I know if the sequence splits then each prime associated to N is also obviously associated to M, and in that case you can get some that are associated to M but not L
and 0 -> nZ -> Z -> Z / nZ -> 0 is an example of the opposite, there are some primes associated to the quotient but Z and nZ only have the 0 ideal associated to them
this question is kind of a question of the phrasing of a textbook problem. It says to show that 'both containments can be proper' in $A(L)\subseteq A(M) \subseteq A(L) \cup A(N)$ and I don't know if that means give one example of each or an example in which both are proper at the same time
next obsidian
perhaps you can find a torsion-free L?
Can you have a quotient of torsion-free modules have torsion?
My idea at least is to find some M without torsion, so that Ass(M) is empty (or just the 0 ideal if it's over an integral domain), then consider just some submodules L, and look at 0->L->M->M/L->0
I feel like the quotient by L introduces torsion into M/L since now you're looking at stuff in R such that there's some m not in L, but rm in L
This introduces torsion, and thus you'll have non-zero ann_R(m + L), and maybe one of them just happens to be prime
@eager willow
eager willow
but Ass(L) is contained in Ass(M) so if Ass(M) is empty or 0 then Ass(L) = Ass(M) which is what I want to avoid
next obsidian
Oh shoot
right
uhh
I guess you can weaken it so
M has some torsion, L is a torsion-free submodule
alternatively actually
I think under direct sums
eager willow
I am trying for strict containment in $Ass(L) \subset Ass(M) \subset Ass(L) \cup Ass(M/L)$
cloud walrusBOT
datorangeguy:
upbeat juniper
i see Q(a,b) as the intersection of all subfeilds of E, containing both a and b and Q
@pure crest it's probably clearer if you think of it as the smallest field containing Q, a and b
next obsidian
So I think potentially
if you can show each containment separately
you can take direct sums to show the composite one
Maaaaybe
So basically like if you have
0 -> L -> M -> N -> 0
and say Ass(M) is properly contained in Ass(L) U Ass(N)
then have
0 -> L' -> M' -> N' -> 0
where Ass(L') is properly contained in Ass(M')
then I think you might be able to look at
0 -> L (+) L' -> M (+) M' -> N (+) N' -> 0
the only thing I'm not certain about
is if (M (+) M')/(L (+) L') = N (+) N', but I think that's true????
next obsidian
nice
eager willow
I'll try that out for the problem
next obsidian
And I'm certain that Ass(M (+) N) = Ass(M) U Ass(N)
Because if not then I need to revaluate some stuff LOL
but anyway, you'd need to be careful about what primes you use
since if there's some weird overlap I think you might not have proper containment of both in the final composite sequence
but if you make sure the associated primes are chosen correctly I think you can get it
eager willow
yeah that's true on the equality because again generally Ass(L) is contained in Ass(M) and so Ass(N) is contained in Ass(M + N) (an exact sequence exists in the other direction)
and so the union is contained in Ass(M + N) but the other direction comes for free
next obsidian
Oh right, that's a good way to see it. I looked at it way less high level, but failed to notice you can just reverse the arrows
Nice!
eager willow
ok yeah the direct sum totally works I came up with $$0 \to 5Z \oplus Z/2Z \to Z \oplus Z/6Z \to Z/15 Z$$
cloud walrusBOT
datorangeguy:
eager willow
(5) is in Ass(N) but not Ass(M) and (3) is in Ass(M) but not Ass(L)
steady axle
I get that Q(alpha, beta ) / Q should be of degree 6
but why should degree of Q(alpha + beta ) /Q be 6?
it couldn't be 3 or 2 ?
woven delta
Hmm
If alpha+beta had a minimal polynomial of degree 2 then alpha would be contained in a field extension of degree 4. If alpha+beta had a minimal polynomial of degree 3 then beta would be contained in a field extension of degree 9
@steady axle
woven delta
So Q(alpha + beta, beta) = Q(alpha + beta, alpha) = Q(alpha, beta)
That's what I'm using here
Hmm actually idk maybe my argument can be simplified
Definitely it can't have a minimal polynomial of degree 2, for the reason I said above
steep hull
Honestly my approach would be to show that [Q(alpha,beta):Q]=6 (it’s divisible by 2 and 3), and then consider the six complex embeddings fixing Q (by looking at whether the roots of the cubic/quadratic are real/complex, the images of alpha+beta can be shown to be distinct).
sly storm
By the way, can any one give me an example of a polynomial of the form (x^2 -a^2)(x^2-b^2) ... which is non-solvable by radicals?
oblique river
a product of things of the form (x^2 - a^2) is going to be solvable by radicals
because the roots are all rational
sly storm
a, b , ... can be any real numbers
oblique river
"solvable by radicals" is always in reference to a base field, which is usually teh rational numbers
which means the polynomials need rational coefficients
if a = pi then the roots of x^2 - a^2 aren't algebraic
so it doesn't really make sense to ask about being solvable by radicals
I guess what you could do is find some a which isn't expressible using radicals, and let b, c, d, etc be the galois conjugates of a
then that polynomial would work
sly storm
@oblique river It is not possible the roots are irrational but the coefficients of the polynomial become rational?
a product of things of the form (x^2 - a^2) is going to be solvable by radicals
@oblique river
because the roots are all rational
@oblique river
@oblique river I think the above are not right
Do you agree with me?
For example (x+ sqrt{2})(x-sqrt{2}) is of such a form, but the roots are irrational.
Am I right?
scarlet estuary
usually when we talk about polynomials in this context we're assuming integral (or rational) coefficients
if you let coefficients take any real number then
even degree-1 polynomials can be unsolvable in radicals
x - (some real number that doesnt have a radical representaiton)
voila
sly storm
My example above is a polynomial with rational coefficients, isn't it?
@scarlet estuary ^
scarlet estuary
i was referring to this line:
a, b , ... can be any real numbers
sly storm
By the way, can any one give me an example of a polynomial of the form (x^2 -a^2)(x^2-b^2) ... which is non-solvable by radicals?
@sly storm
This is my main question. ^
<@&286206848099549185>
Preferably of degree 6 or 8
Someone said it is impossible because the roots must be rational, which I cannot accept. Do you agree with me?
The base field is rational numbers.
Ok, but such polynomials can have rational coefficients while having irrational roots. For example, consider (x^2-2)=0
a product of things of the form (x^2 - a^2) is going to be solvable by radicals
@oblique river
because the roots are all rational
@oblique river
Are the above correct?
As I said a, b, ... can be real numbers
Yes, a^2 was confusing. Why did I write such a polynomial in that way?
scarlet estuary
By the way, can any one give me an example of a polynomial of the form (x^2 -a^2)(x^2-b^2) ... which is non-solvable by radicals?
this is effectively already provided by my example:
x - (some real number that doesnt have a radical representaiton)
just square each term
sly storm
The base field must be rational numbers.
scarlet estuary
so you want a^2 to be rational, even if a is not? (so a is required to take the form sqrt(q) for some rational q)
is that what you mean
if so then thats certainly solvable in radicals
since (x^2 - a^2) = (x-a)(x+a) = (x-sqrt(q))(x+sqrt(q)) giving us sqrt(q) as a solution
maybe im misunderstanding your question
sly storm
Please let me clarify.
scarlet estuary
if thats what they mean, that implies (x^2-a)(x^2-b) = x^4 - (a + b)x^2 + ab is rational, ie a+b and ab is rational; as far as i know, irrational number arithmetic operations of this form are not well-understodd
(i wrote a instead of a^2 and similar for b since it doesnt matter and makes the notation cleaner)
sly storm
I want a polynomial (coefficients are rational) of the form (x^2-a)(x^2-b) ... of degree 6 or 8 which is non-solvable by radicals.
Is it now clear?
scarlet estuary
this is eerily similar to techniques used to prove that one of pi*e and pi+e is irrational
(specifically invoking that they are both transcendental)
that polynomial is not of degree 6 or 8
but sure
sly storm
Yes, I want a polynomial of degree 6 or 8.
hot lake
polynomials of degree 3 and 4 are solvable by radicals
their composition by x² are also solvable by radicals
scarlet estuary
yeah this is essentially the crux
i was trying to think of a good way to say that
but thats a great way to phrase it
let p(x) = rx^3 + sx^2 + tx + u and q(x) = x^2, then p is solvable in radicals since its of degree 3 (say it has radical solution "m"), and (p o q)(x) = rx^6 + sx^4 + tx^2 + u is also solvable by radicals (it would have radical solution sqrt(m))
note here that every degree 6 polynomial youre constructing via (x^2 - a)(x^2 - b)(x^2 - c) can be written as (p o q) for some polynomial p [just let p = (x-a)(x-b)(x-c)]
a similar argument applies to degree 8
but note that it does not apply to degree 10 since p, being degree 5, may not be solvable in radicals
@sly storm
does that answer your question?
tl;dr such degree 6/8 polynomials cannot exist.
sly storm
So, is it true that if the polynomial $(x^2-a^2)(x^2-b^2)...$, where the coefficients are rational and $a$, $b$, ... are real numbers, is solvable by radicals then the rational polynomial $(x-a)(x-b)...$ must be solvable by radicals?
cloud walrusBOT
MathPhysics:
sly storm
@hot lake and @scarlet estuary ^
hot lake
that looks like an entirely different question
but yes, if P*Q is solvable by radicals then P is solvable by radicals
sly storm
P and Q be any polynomials?
hot lake
yeah
sly storm
Proof?
scarlet estuary
a is a root of (p o q) implies that q(a) is a root of p
clearly if a is a radical expression then q(a) is too
sly storm
- is multiplication or composition?
scarlet estuary
since q is just a polynomial
composition
oh wait fuck
you were asking about
-a and then -a^2
ok well same story but now * means multiplication lmao
in which case viburnum's statement is correct
if a is a root of p, it means p(a) = 0
so clearly (p*q)(a) = 0
since (p*q)(a) = p(a)q(a) = 0*q(a) = 0
sly storm
So, the converse of my question is also true, right?
hot lake
the converse of which question ?
sly storm
So, is it true that if the polynomial $(x^2-a^2)(x^2-b^2)...$, where the coefficients are rational and $a$, $b$, ... are real numbers, is solvable by radicals then the rational polynomial $(x-a)(x-b)...$ must be solvable by radicals?
@sly storm
cloud walrusBOT
MathPhysics:
hot lake
what is the meaning of your "so"
sly storm
By similar argument
hot lake
I can't say it's much similar
sly storm
However, the converse of it is true?
sly storm
The roots of the second polynomial are just a subset of the roots of the first polynomial
@open torrent
I meant this argument
scarlet estuary
yes that statement establishes the converse
hot lake
well no the roots of (x²-a²)(x²-b²)... aren't a subset of the roots of (x-a)(x-b)...
scarlet estuary
"the second polynomial" here means p
the first being p*q
again, this proves the converse of what you stated
the proof of what you actually asked is just that p(x)*p(-x) is certainly solvable by radicals
since p(x) is and p(-x) is
oh wait
hold on
ugh im way too tired for this tedious symbol pushing
that was backwards
the proof uses this idea though, just assume that p(x)*p(-x) is solvable by radicals, which implies that theres some radical r such that either p(r) = 0 or p(-r) = 0
voila
since p(x)*p(-x) = 0 implies p(x) = 0 or p(-x) = 0
[which is a property of real numbers]
hopefully that actually proves what you asked
and im not getting tripped up by 3 am brain
sly storm
Ok. Thank you all. Can you please help me with my original question:
I asked the following question in some places without receiving the answer. I guess the problem is equivalent to solving a quintic equation, but I cannot prove such a claim. Can that claim be proved by using som abstract algebra methods or preferably some simple math?
Suppose we know:
\begin{align*}
A&=a^2+b^2+c^2+d^2+e^2 \
B&=a^2b^2+a^2c^2+a^2d^2+a^2e^2+b^2c^2+b^2d^2+b^2e^2+c^2d^2+c^2e^2+d^2e^2 \
C&=a^2b^2c^2+a^2b^2d^2+a^2b^2e^2+a^2c^2d^2+a^2c^2e^2+a^2d^2e^2+b^2c^2d^2+b^2c^2e^2+b^2d^2e^2+c^2d^2e^2 \
D&=a^2b^2c^2d^2+a^2b^2c^2e^2+a^2c^2d^2e^2+a^2b^2d^2e^2+b^2c^2d^2e^2 \
E&=a^2b^2c^2d^2e^2 \
\end{align*}
How can we find $a+b+c+d+e$?
cloud walrusBOT
MathPhysics:
scarlet estuary
you have asked this many times here, and many times people have attempted it and not made much progress
solemn rain
wtf is this
scarlet estuary
is this from somewhere or is it just a contrivance you made up
hot lake
if you can find a+b+c+d+e algebraically you can also find a+b+c+d-e so you can find e, so you can also find a,b,c,d,e, so you can solve a quintic
so it is as hard as solving the quintic
sly storm
@hot lake Why?
scarlet estuary
thats assuming you can find a+b+c+d+e
hot lake
because you can't distinguish a+b+c+d+e and a+b+c+d-e algebraically if you only know A B C D E
hot lake
you mean galois theory
scarlet estuary
wdym
finding a+b+c+d+e?
that can be done via galois stuff or you can look at varieties
and intersection sets
sly storm
@hot lake If I can find a+b+c+d+e, then how can I find a+b+c+d-e?
hot lake
by applying the same procedure and making different choices
scarlet estuary
an assumption they had the last time they stated this is
a, b, c, d, e > 0
fwiw
sly storm
@hot lake How?
hot lake
that has no algebraic meaning
scarlet estuary
of course not
hot lake
it's why i said algebraically
hot lake
by learning galois theory
sly storm
@hot lake Can you show me the way?
hot lake
have you learned galois theory
sly storm
a little
hot lake
let L = Q(a,b,c,d,e) ; K = Q(A,B,C,D,E)
L is a Galois extension of K whose Galois group you should understand
the orbit of a+b+c+d+e under that galois group contains a+b+c+d-e and many others
so finding a+b+c+d+e is exactly as hard as finding a+b+c+d-e and those many others
scarlet estuary
this is not a very fun approach
but its probably the best one without computer assistance
well actually
remove the word "probably"
this is certainly the best approach without computer assistance
still sucks though
if you have a computer capable of doing classical ag you can plot these curves in a,b,c,d,e-space and do the appropriate charts and shit
to eventually get a+b+c+d+e in A,B,C,D,E-space
hot lake
a+b+c+d+e is a root of a degree 32 polynomial whose coefficients are in K
scarlet estuary
and as we all know, roots of degree 32 polynomials have very nice expressions
and are incredibly enjoyable to work with algebraically
hot lake
the galois group of that polynomial is the same as the original galois group
because the smallest extension of K generated by the orbit of a+b+c+d+e is all of L
so getting a+b+c+d+e is as hard as solving for a
solemn rain
can we use rep theory on galois groups
hot lake
studying representations of galois groups led to proving stuff like fermat's last theorem so it can definitely be useful
charred glacier
somebody had an attempt here https://mathoverflow.net/a/124273 but idk
steady axle
if A/I has size n then is it always true that A/I^2 has size n^2
A is ring and I is its ideal
is it true when I is principal?
wind parrot
no, take any finite ring A of size n and I = (0)
wind parrot
then you can still take any idempotent nonzero ideal, so take for instance $A = \bZ/n \times \bZ/n$ and $I = ((0,1))$
cloud walrusBOT
NielsK:
steady axle
right
so it doesn't hold in general
but i think it does when we are in nice setting
like say pid
eager willow
For a sequence of ideals in an integral domain $R:$ $$I_1 \subset I_2 \subset \dots,$$ if there is some $I_m$ such that for every $k > m$ and each $a_k \in I_k$, there is some $r \in R$ with $ra_k \in I_m$, is that necessarily enough for $I_m$ being maximal in the sequence as in the A.C.C?
cloud walrusBOT
datorangeguy:
eager willow
or enough to know that the sequence terminates eventually?
burnt pond
I don’t think this is true. For example in the ring of all algebraic integers take the ascending chain of ideals generated by the 2n-th roots of 2. This sequence doesn’t terminate but satisfies your condition. I’m not too sure tbh
eager willow
hmm. What if R is a subring of a Noetherian ring S with S integral over R? That probably changes things right?
that's how I ended up with that condition
eager willow
trying to show that R is finitely generated for each subgroup G in S_n
got a little overambitious looking at the A.C.C for R
eager willow
Another strategy that I've considered is to try Eakin-Nagata theorem, so I have to only prove S is finite over R but I don't think that's very easy either
burnt pond
Hmm I don’t think I can help with you with this particular proof but If S is the polynomial ring over a field in n unknown, then for any finite group the invariant ring R=S^G is finitely generated, so the statement you want to proof is true. This is due to a theorem about invariant rings by Noether. I think I read about it in D. Eisenbud's „Commutative Algebra with a View Toward Algebraic Geometry“ iirc. Might be worth to check it out
sly storm
@charred glacier Thanks for the link.
@hot lake Is the answer https://mathoverflow.net/a/124273 correct?
hot lake
yes
olive mirage
Oh lovely!
knotty mason
characteristic
jovial nexus
characteristic
@knotty mason ring homomorphisms don't respect the characteristic or anything
Look at Z/4Z -> Z/2Z
upbeat juniper
is there a term to describe a map f such that f^k = id for some k?
smth like nilpotent
next obsidian
Yes
wait no...
hmm
So for f^k = id f needs to be a self map, an endomorphism, and that forms a ring naturally, with composition being multiplication (actually you need a notion of addition so it needs to be like an additive category or something. In general you can say that it forms a monoid under composition, so I guess this says it has finite order?)
so if you find a term for an element x in R, such that x^k = 1 for some k, then that's what you want (for any ring R)
upbeat juniper
the original context used linear transformations so I was just looking for a general term, without having to deal with individual elements
next obsidian
Yes, but I’m saying it seems like this is a question whose answer is far more general
Either way, to say it’s nilpotent would imply that f^k = 0 I think
I wouldn’t be surprised if this just doesn’t really have a term for it
upbeat juniper
I thought idempotent or unipotent sounds natural enough but sadly they mean different things
olive mirage
be careful when talking about ring homomorphisms because there are multiple definitions
and so whether injective ring homomorphisms preserve the characteristic or not depends on what definition you choose
smoky cypress
Here D is a domain and D* is the set of nonzero element. It says that p is irreducible is equivalent to no existence of ideal I that’s between D and (p). Is this where the assumption that D is a principal ideal domain is used?
burnt pond
yes you need that D is pid, otherwise it's not true in general. For example in Z[x] the element 2 is irreducible but the ideal (2) is contained in (2,x). The equivalence is true for any domain if you replace I with principal ideals tho. So p is irreducible iff (p) is maximal amongst all principal proper ideals of D.
smoky cypress
Alright thanks
smoky cypress
Define Jordan homomorphism between the rings $R$ and $R'$ to be a map $\eta$ with the following properties: $\eta$ is a homomorphism between the additive groups of $R$ and $R'$; $\eta(1)=1'$; and $\eta(aba)=\eta(a)\eta(b)\eta(a)$ for all $a,b\in R$. Show that if $\eta$ is a Jordan homomorphism then $$\eta(abc+cba)=\eta(a)\eta(b)\eta(c)+\eta(c)\eta(b)\eta(a).$$
cloud walrusBOT
Whoever:
smoky cypress
Oh
That works
Wait does it
Hmmmmmm
(n(a)+n(c))n(b)(n(a)+n(c))=n(aba)+n(abc)+n(cba)+n(cbc)
I’m just using n instead of η
Oh yeah that works
smoky cypress
I don’t get how Lemma 1 shows that α is a unit in D
And a factorial domain is a unique factorization domain
next obsidian
Because the decomposition is unique up to unit multipliers in D
So like f(x) is one way to factor it as a gamma f(x)
And so is alpha g(x)
knotty mason
How to prove a group of order 90 has a subgroup of order 10?
next obsidian
Use cauchy’s lemma / theorem
It has an element of order 2 and element of order 5, so it has subgroups of order 2 and 5
Err
I guess you don’t know they’re normal
I was gonna say to look at the product of those subgroups but
Actually I feel like this still works because of coprime order but
Oh wait
Maybe try showing the Sylow 5 is normal
Or at least one of the Sylow 2 or 5 is normal
Maybe you need both?
I honestly forget oof
You only need one of them to be normal @knotty mason
golden pasture
||your group has a subgroup of order 45||
||groups of order 45 are abelian||
it's a bit hard to show that the sylow 5 group is normal without going through ^
next obsidian
Can you not do it by just bashing numbers
Also why on gods green earth would anyone just know the fact you put there?
It doesn’t fit some nice pq criteria with p-1 not dividing q
Is there some nice way to get that?
Also even showing there’s ||a subgroup of order 45|| seems somewhat difficult.
burnt pond
How do you use the fact that there is a normal subgroup of order 45 to show that there exists a normal subgroup of order 5? I don’t get it
golden pasture
re. ord 45, try to prove the more general fact, a group of order 2n, n odd, has a normal subgroup of order n
can be a bit more general with 2^m
next obsidian
I guess you don't even need to show ||groups of order 45 are abelian||
Sylow instantly tells you that Sylow 5 in an order 45 is unique
so characteristic, then implies normal in the group of order 90
smoky cypress
So if $S$ is a semigroup and $u$ is not in $S$, then you can simply let $M=S\cup\brc{u}$ and define $ua=a=au$ for all $a\in S$, and this makes $M$ into a monoid. So does that mean that whenever you're proving something about semigroups you can simply assume that it's a monoid? I feel like that's cheating
cloud walrusBOT
Whoever:
woven delta
@smoky cypress prove that if a semigroup doesn't have an identity then property P
smoky cypress
then what?
chilly ocean
P=it is not a semi group 
smoky cypress
Ugh
woven delta
My point is that there are certain statements that apply only to semigroups without identity
For example "S does not have an identity"
smoky cypress
I'm very confused hmmmmmmm
woven delta
I'm semi meming
Anyway in practice I don't think the identity you artificially added is so useful
smoky cypress
Oh ok
woven delta
To rephrase my point earlier, I was being a bit pedantic with what you were saying. If the statement you said was true as written, then any statement true for semigroups without an identity is true iff it is true in the corresponding monoid that you defined. But obviously the statement "S does not have an identity implies P" is always true in the monoid, but is just equivalent to P in S
Basically my point is making logical statements without working within some logical system is hard
knotty mason
? gcd(24,eulerphi(45))
%3 = 24
How would I show groups of order 45 are abelian?
oh it follows from Sylow theory
n_3 and n_5 both = 1
solemn rain
is this taken
knotty mason
Thanks a lot for the guidance I see the proof now
woven delta
@knotty mason well you also need to show groups of order 9 are abelian for that to work
smoky cypress
Oh i see liquid
Finite abelian groups written additively can’t be a Q module right?
Nontrivial finite abelian group
smoky cypress
Well, write (0), (1) for the elements of Z/2Z, then 1(1)=(1). 1/2*(1) is an element in Z/2Z that is (1) when added to itself, but such element doesn’t exist in Z/2Z
solemn rain
yea
smoky cypress
So that means Z/2Z is not a Q module
And you can repeat this argument for any nontrivial finite abelian group
What
smoky cypress
What does it mean here when it says “The role played by the infinite cyclic group (Z,+,0) is now taken by R as R-module”?
bleak abyss
@smoky cypress the point is that if you think of abelian groups as Z modules, then Z is what you're scaling things by
So you determine whether an element generates a module by taking "Z-multiples"
But if you ask whether something is a cyclic R-module, you're taking "R-multiples" of a supposed generator
smoky cypress
Oh I see
next obsidian
Isn’t this true?
Like I feel like each N_j should be described as a direct sum of the M_i
Since by assumption each M_i are like pairwise disjoint or whatever so that you can even form the direct sum of all M_i
smoky cypress
Oh I forgot part of the definition of $M=\bigoplus M_i$ assumes that $M_i$ are independent
cloud walrusBOT
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