And that's normal
#groups-rings-fields
bleak abyss
By the same logic
|S|=13
@mild laurel
Err wait actually it's not clear the action of N is primitive hmm
Maybe the subgroup of order 13 is necessarily characteristic?
Yeah it's unique by Sylow theory so it's characteristic
So we're actually done
That's a nice problem lmao
fringe nexus
How would I find an ideal I such that Z[X]/I has exactly 3 prime ideals?
I was thinking about Z_n
but i think they only have even amount of prime ideals
bleak abyss
Hmm, a product of 3 fields has 3 prime ideals, right?
fringe nexus
like (0) x F_1 X F_2?
bleak abyss
Yeah
fringe nexus
i guess that would work yea
but I can't mod by Z[x] to get 3 fields can i?
oh wait I can
Z[X]/ ((x-3)(x-5)(x-7), x)
is this isomorphic to F_3 x F_5 x F_7?
Z[x]/((x-3)(x-5)(x-7)) $\cong Z x Z x Z $ with the evaluation map at 3,5,7 right?
cloud walrusBOT
aaaaaaaaaaaaaaaa:
chilly ocean
bleak abyss
Uh, I'd need to think harder than I'm currently willing to. Product of three maximal ideals should do it by CRT
ivory dust
Really dumb question from someone who's never taken abstract algebra: is 0*a = a*0 = 0 a general property of rings?
Where 0 is the additive identity
snow bolt
yes
ivory dust
Proof?
ivory dust
Right
wind steeple
N is an N-vector space
chilly ocean
If a simple group G has a proper subgroup H s.t [G : H] = n, then G -> A_n
upper pivot
oh that sounds useful, thanks!
covert vector
oh I'm late cuz I didn't check this channel lol
@magic owl
G is isomorphic to a subgroup of S_n
if G isn't iso to a subgroup of A_n
then φ(G) \cap A_n has index 2 in A_n
which implies A_n is not simple
but it is
(for n≥5)
for n≤4, groups of order ≤4! aren't simple
so it's fine
can't do the other problem cuz no idea what a primitive action is lol
well I mean I looked it up but no intuition for it yet
fathom whale
Anyone inv
On*
Sweet
So right now I’m stuf on this subject
10 question but idk how to do them
Why channel?
What**
snow bolt
magic owl
@covert vector i was respoinding to rudy's comment
having a subgroup of index n only gives a homomorphism to Sn
covert vector
@magic owl yeah but rudy also said G is simple
so that homomorphism has no kernel
hmm wait
am i getting confused
no this is fine, tho i messed up what i said above. [G : G \cap A_n] = 1 or 2
dawn kiln
Idk man, |x| seems p abstract to me
covert vector
2 implies the intersection is normal
dawn kiln
😳
covert vector
don't shitpost here @dawn kiln
dawn kiln
Mb
covert vector
ik lol
chilly ocean
im
confusing myself
LOL
hold on
The kernel is either all of G or the trivial subgroup.
covert vector
the kernel was not the hard part lol
fringe nexus
can someone verify if i did my problem on final right
chilly ocean
post it
fringe nexus
It was to find the galois group of (x^3-2)(x^2-2x-1)
chilly ocean
and woog will check it
fringe nexus
I think its of order 12
i found the root of the quadratic to be 1+- sqrt(2)
and so are the automorphisms of the form $\sigma(\zeta) = \zeta^a, \tau(2^{1/3}) -> \zeta^b 2^{1/3} , \sigma_2(1+sqrt(2)) = 1-sqrt(2) or 1+sqrt(2))$ a is 1 or 2, b is 0 or 1 or 2
cloud walrusBOT
aaaaaaaaaaaaaaaa:
fringe nexus
I basically get a A if i got this right and if im not yikes
mild laurel
@bleak abyss Ah, thanks. I kept thinking the answer had to be 39 because you could just take a group G of order 39 and look at the right regular action of it
But this isn't primitive
chilly ocean
@fringe nexus go to #advanced-number-theory
fringe nexus
ok
mild laurel
It was on my algebra final lmao
Figured out it was 13 or 39 and kept trying to exclude 13 but couldn't
bleak abyss
Yeah actually for a second I was like uh oh with my answer
Since the restriction to N isn't necessarily primitive to redo the argument
But the order 13 subgroup is actually normal in G again
chilly ocean
What is C*/T?
ivory dust
another really basic question. If you considered the set of all ordered pairs (or more generally ordered n-tuples) of real numbers with addition defined as (a,b) + (c,d) = (a+c, b+d) and multiplication defined as (a,b)*(c,d) = (ac,bd) what properties would you be missing that something like the complex numbers or the quaternions have
like it would be a field, would it not?
mild laurel
Yes
ivory dust
are the complex numbers the only field over R^2?
mild laurel
What do you mean by over R²
ivory dust
I don't really know
lemme thing about how to make that statement more formal
something like a field where the elements are ordered pairs of real numbers
or possibly a 2 dimensional algebra over the reals that is also a field
one of the two
prisma thunder
Been stuck on an algebra problem
Suppose the polynomial x^4+ax^2+b in Q[x] is irreducible over Q with roots s,-s, t, -t. Let K denote the splitting of this polynomial.
Show that Gal(K/Q) is isomorphic to the the Klein 4-group if and only if st is rational.
I showed one direction where I first assumed st is rational.
The other direction is stumping me.
late sedge
guys
i cant think of any examples of non rings other than Natural Numbers
Natural Numbers dont have an addiditive inverse so they cant be an abelian group hence not a Ring
mild laurel
Positive rationals?
late sedge
i guess but you're partitioning Q
late sedge
woven delta
Do you mean algebraic structures?
mild laurel
What does partitioning Q even mean
woven delta
Also yeah, ideals are rings without unity
prisma thunder
That's fact.
late sedge
conditioning*
prisma thunder
If an ideal contains unity, as a consequence, the ideal is R, which isn't interesting.
late sedge
if |R| = 1, then 1R = 0R right?
woven delta
@ivory dust finite extensions are Algebraic, so a degree 2 field extension of R is going to be R[X]/(f) where f is an irreducible degree 2 polynomial. The irreducible (wlog monic) degree 2 polynomials over R are x^2+r for r>0. You can easily demonstrate an isomorphism between C and such a field.
There are lots of other arguments for showing this fact.
late sedge
to show an ideal is maximal is the easiest way really to show that R/I the quotient ring is a field?
that honestly seems difficult
woven delta
Honestly in a class your doing baby problems usually
It shouldn't be too difficult
Well not baby problems, but solvable problems
late sedge
yea i get what you're saying
prisma thunder
Depending on the institution and the type of culture your professor was exposed to, mostly "solvable" problems.
late sedge
i was just wondering bc I was struggling with a HW problem last week identifying whether an Ideal was neither, prime, or maximal and what method you guys would go about
mild laurel
There are plenty of other ways to show an ideal is maximal
It really depends on the situation
late sedge
since all Maximal Ideals are Prime, was wondering if you would check that property first
bc being maximal implies prime
true
but all maximal ideals are prime
its not a thing that works both ways
mild laurel
There are other ways
late sedge
what if I just straight up check if R/I is a field, if it is then its maximal hence also prime
is what im trying to propose
mild laurel
Like showing that if you add any element that is not in your ideal, then what's generated by your ideal and that element is your whole ring
Implies that your ideal is maximal
And an easier way to check this is that for any element r of your ring
There is some x in your ideal I such that x + sr = 1 for some element s of your ring
prisma thunder
Are prime ideals maximal ideals in division rings? I would think so.
woven delta
👀
No
It's a ring with inverses
Not necessarily commutative though
So for example the quaternions
👀
What do you mean
Why would every ring have multiplicative inverses?
covert vector
the ring of integers is commutative and unital
and is not a division ring
division ring implies you can divide
in particular that 1/x is defined for nonzero x
but this isn't the case in the integers
woven delta
Yes
covert vector
there are no nontrivial ideals in a division ring
any ideal containing a unit has to be the whole ring
the only ideal that doesn't contain a unit is the zero ideal
yes, that's a consequence of no ideals existing
woven delta
The 0 ideal is prime
covert vector
*other than that
woven delta
This is surprisingly important
late sedge
i thought a division ring is a ring with no zero divisors
covert vector
no
woven delta
No you're thinking a integral domain
Which is a commutative ring with no zero divisors
late sedge
yep you right
i have them next to each other on my white board and i confused them whoops
Integral Domains have no zero divisors, while division rings have every nonzero element as invertible
woven delta
Nope
You're wrong
Not you mortex
There's just a different definition
For noncommutative
But also who cares
👀
We only care about Spec R for commutative rings
So 
But anyway this discussion isn't interesting so I'm bowing out
covert vector
this is correct. for every a,b in the ring such that ab is in the ideal p, either a is in p or b is in p.
that's what it means for p to be a prime ideal
if p=(0) in a noncommutative division ring
and ab = 0
if both a and b were nonzero
since they are both units
then (a^-1)ab = b will be in p as well, since it's an ideal
b≠0
it's a unit, so p = the entire ring
which contradicts p = (0)
therefore at least one of a,b is 0
that is, (0) is a prime ideal
yes that's fine as well
of course ideals are subsets of the whole ring
so a product of elements in the ideal is in the ideal
covert vector
look at the definition of an ideal
if I is an ideal
and y is in I
or say, left ideal
then for any x in the ring R
xy is in I
that's a property that all (left) ideals have
yes
so if ab is in I
let r = a^-1
then r(ab) = b
alternatively
the product of two units is a unit
so if a,b are both nonzero in a division ring
then ab is nonzero
b^-1 a^-1 is the inverse
just multiply them
and see for yourself
yes
you should really try and prove some of this yourself
it's not hard
and if u get walked through everything
you won't learn anything
well it's better than nothing
keep at it
I'm going to bed
cloud walrusBOT
Kogasa:
Compile Error! Click the 
reaction for details. (You may edit your message)
bleak abyss
Yeah I think so
placid pond
i actually don't know another idea for showing a pid is not an euclidean domain
and the idea you use for this ring only works for this ring and maybe one or two others like it
it doesn't generalize at all
chilly ocean
Hey, I needed help understanding this proof
Here is the theorem mentioned in the proof
In particular, I wanted to know how if H contains one 3-cycle, it'll contain every 3-cycle
upbeat burrow
@chilly ocean you have that H = gHg^-1 for all g in A_4
So if you have a 3 cycle s in H, then gsg^-1 is in H
chilly ocean
oh okay, so the H used in both side of equation can represent different elements?
upbeat burrow
Yeah, it just means it's in the group
upbeat burrow
gHg^-1 = H means you can conjugate any element of H by an element of G and you'll remain in H
chilly ocean
So this includes any element in H, it can be same or different?
I guess they'd used h instead of H for keeping the same element
upbeat burrow
Yeah
chilly ocean
Lol my bad for confusing the notation, thanks again!
ripe basalt
@chilly ocean notice that hH = H since it's a subgroup and so it's closed for free anyways
chilly ocean
Oh, that wasn't mentioned in the book, thnx
urban acorn
I once read a book about elementary algebra and at some point they asked us to prove that x + x = 2x but they didn't mention that 1+1=2! how outrageous
oak perch
x*x = x^2 too ... in multiplicative groups.
urban acorn
that's the same thing in multiplicative notation, but i said this in an elementary algebra context, and so the equivalence wouldn't be recognized
tribal pasture
What are the conditions under which a set of linear equations is solvable in a ring
I am assuming that the system AX = b should have that det(A) modulo the ring should be non-zero
prime gale
It isn't much, but if det(A) is a unit, then it is solvable. But the condition isn't necessary.
tribal pasture
I used this argument
I just am not sure if rank nullility applies in the case of rings
<@&286206848099549185>
prime gale
Sorry--I don't know what rank of matrix not over field is.
prime gale
Maybe what you want is there is a unique solution iff b is in column space of A and kernel of A is zero. I think that is what you are trying to say, but I don't understand what you mean by dimension and rank.
smoky briar
you can define rank using vanishing of minors
if the determinant is a unit you always have an inverse using the adjoint matrix
the one with cofactors
prime gale
Well, using vanishing minors definition of rank that Dormherr suggests, it isn't true that rank(A)=m is sufficient for solution to be unique. For example, take n=m=1, b=0, A=zero divisor. But I may be misunderstanding Dormherr's definition of rank.
I just replaced "field" with "ring" in Wikipedia's non vanishing minors definition of rank.
smoky briar
yeah I'm not saying that "by extension of linear algebra, all theorems works" is anywhere near true
for uniqueness you want your minors to be units
or something like that
although as you identify, being an integral domain also makes things easier
prime gale
I see. You define rank as largest minor with unit determinant. Thank I agree--rank=m guarantees uniqueness.
smoky briar
but yeah the argument as written is very suspect
tribal pasture
Yeah but the argument doesnt rely on it coming from linear algebra or not. Merely a source of motivation to situate the reader.
smoky briar
🤔
tribal pasture
Just read it from "For A..."
smoky briar
the point is I don't think most of those hold
tribal pasture
Oh okay why so?
tribal pasture
rank being the dimension of the range
and dimension being the number of elements in the basis
All directly being taken from Artin
prime gale
What is a basis?
tribal pasture
Independent set of vectors that span the space?
prime gale
R is ring. not a field. What is a vector?
tribal pasture
You mean vector relies on the notion of a vectorspace which is not defined on a ring?
prime gale
Problem is that isn't well defined. You can have different size basises.
tribal pasture
So how would I approach this question then?
prime gale
Well, just do it as Dormherr says. Define rank as biggest minor with unit determinant.
tribal pasture
Or do it maybe functionally just using kernal and range?
prime gale
That works too.
tribal pasture
Btw the fact that A is a linear transformation, does that not induce a vectorspace somehow and thus definable notions of ranks and dimensions in terms of basis?
@prime gale
stark sigil
Let $R=\bZ/6\bZ$ and take $A$ to be some random matrix with zero divisors in it, like, dunno, $\begin{pmatrix}2&0\0&1\end{pmatrix}$ and play around with it
cloud walrusBOT
Icy001:
tribal pasture
No I do have a counterexample at hand. Just not sure what is the proper formalisms
tribal pasture
my prof didnt cover these stuff tho ):
So I doubt he wants us using these terms.
So I guess I should stick to range and kernel
chilly ocean
So you have to show that there is no Z/(6)-vectorspace?
stark sigil
Another thing to keep in mind, injective from $R^n$ to $R^n$ doesn't imply surjective
unlike with fields
cloud walrusBOT
Icy001:
chilly ocean
Or that there is no Z/(6) basis?
tribal pasture
But linear injective do imply right?
stark sigil
no
My example actually doubles as an example of that
er, wait
No need a different example
well take the same matrix but $R=\bZ$
cloud walrusBOT
Icy001:
stark sigil
Injective but not surjective
tribal pasture
Do we atleast have a definable notion of a determinant in a ring
Because I used the fact that determinant is 0 to argue that the matrix doesnt admit an inverse and thus the kernel is non-trivial.
stark sigil
Yeah, of course, because determinant involves only addition, subtraction, and multiplication
But determinant 0 implying that there is a nontrivial element in the kernel is kinda subtle
tribal pasture
Oh so the problem with rank and dimensions is only because of their definability over a vectorspace which I am not guaranteed
stark sigil
Well the problem is that the image of a homomorphism of free modules isn't necessarily free
That's the root of the entire issue
and for non-free modules you don't have a basis
Hm the kernel isn't free either if the ring isn't a PID
the Z/6Z with [[2,0],[0,1]] example shows that
Kernel is (3 * Z/6Z) x {0} which is a non-free (Z/6Z)-submodule of (Z/6Z)^2
tribal pasture
Ah I havent covered modules yet so I am gonna take your word for it. Thanks!
late sedge
guys im tryna build a relationship between unit groups, fields, and being associated
so fields are commutative rings where there exists a multiplicative inverse for each element
does that mean that for any field lets call it F = Unit Group in its entirity?
mild laurel
no
Think carefully again
and think about your favorite field
All elements have inverses?
late sedge
Q\{0}
dawn kiln
Not a field @late sedge
mild laurel
What are fields
dawn kiln
Where's your additive identity in Q\0
late sedge
I guess it's a valid point, I just struggle to understand how Q in it's entirity can be a field when 0 is there with no multiplicative inverse.
I know the definition of a field is for all non-zero elements of the division ring
dawn kiln
That's the definition of a field
It's helpful to think of algebraic structures in terms of extensions or narrowings of others like you're currently doing, but get the fundamental definitions nailed first
A field is a set with two operations, the additive one satisfying invertibility on all elements, commutativity, associativity, and requires that one element serves as the identity
The multiplicative one is invertible on all but the additive identity - thats the definition
late sedge
ok great
yea my style is i like to make connections
I feel like it makes proof writing significantly easier
dawn kiln
Have you looked at category theory before @late sedge
late sedge
no but ive heard its a useful theoretical concepts for programmers I believe?
woven delta
Lmao
upbeat burrow
/me knows nothing about category theory
brisk granite
what does it mean for an ideal to be generated by a and b? Is it the smallest ideal containing a and b?
stone fulcrum
urban acorn
@tribal pasture @stark sigil actually, linear injective does imply surjective here. limiting the domain to the integers isn't valid because even if you have that the operator maps integers to integers (in which case you can think of it as being on the corresponding module over the integers), the preimage of the integers may contain non-integer values
for example, $2I$ is clearly a bijection, despite the corresponding operator over the integers being merely injective but not surjective
cloud walrusBOT
Intel:
urban acorn
moral of the story: linear algebra is way nicer than module theory
stark sigil
@urban acorn Are you using "linear" as a synonym of "over a field"? I just ignored it because module homomorphisms are linear too
urban acorn
R^n is over a field
stark sigil
Ah-ha if you look at the LaTeX, R stood for a ring and not $\bR$
urban acorn
and the other time in which I said linear is in "linear algebra"
cloud walrusBOT
Icy001:
urban acorn
ohhhh
I see.
imaging denoting wildly different things by the same letters
this post was made by economics gang
stark sigil
$R+\mbf R+\mathrm R+\bR+\mathcal R+\mathfrak R+\mathscr R+\mathfrak r+\mbf r+r=0$
urban acorn
tfw
stark sigil
I could use a few more
urban acorn
$\mathcal{H}$ still the coolest though
stark sigil
$\mathfrak H\mathfrak N\mathfrak Y$ which one is which?
urban acorn
imagine thinking about Banach spaces when you could be thinking about Hilbert spaces
cloud walrusBOT
Icy001:
smoky briar
you are still a UFD
think about which primes can be inverted
that gives you the units
prime gale
@late sedge Things that are prime in Z, but invertible in your extension.
late sedge
this whole thing doesnt make sense, i believe there can be an infinite amount of units bc there are an infinite amount of solutions to (a+b/6)^2 = 1
but in the context of my problem (5c) i have to show 8 and 9 are associates
i guess 9/8 or 8/9 can be units in Z[1/6]
prime gale
Nothing wrong with having infinite number of units.
prime gale
Didn't you just do it?
late sedge
yea
it was just strange to me
you dont think of Z[1/6] as having infinite units
at first glance
smoky briar
you do if you think about which primes have inverses
or, really, you always do in these cases
R[1/f] has f as a unit as long as f isn't a zero divisor
late sedge
thanks
gusty ibex
Does anyone have a good reason why “faithful” is a good name for a faithful group action?
golden pasture
sounds like its from faithful functor=invective
jade sky
@gusty ibex Faithful group actions are ones where each permutation is defined by a unique element of the group. In other words, the permutations represent the group elements faithfully
[The homomorphism φ: G → Sym(X) naturally defined by the action of group G on set X is injective iff the action is faithful]
gusty ibex
Right (and thank you) I guess im just not sure why "the permutations represent the group elements faithfully" -- i.e. what is so faithful about that
jade sky
Generally we say that a representation (of one class of objects by another) is faithful if each element in the first class is represented by a unique object in the second class
It's "faithful" in the sense that given a representative, there is no doubt about which element it represents
gusty ibex
i see, i see.. thanks again!
jade sky
For example if you take the group cyclic ℤ/6ℤ = {0, …, 6} acting on the set X = {0, 1, 2}, with each element n ∈ ℤ/6ℤ moving each point x ∈ X to x + n (modulo 3), then both 1 and 4 correspond to the permutation (012) [cyclic right-shift by one step].
So the permutations {(), (123), (132)} of X do not faithfully represent the elements of the group ℤ/6ℤ in this action
For more interesting examples, you should look at representation theory, where we want to faithfully represent group elements by linear transformations or matrices; or graphical regular representations of groups, where we try to identify a given group with the automorphism group of a graph (and yes "regular" here is the same as the one in "regular action" — a faithful, transitive action)
urban acorn
using this motivation for the word faithful here, it'd make more sense in a way if the definition would be that $\phi : G \to H$ is faithful when $\phi(G) \cong G$.
because the image of the homomorphism is faithful to the original structure of the group
that's a slighly weaker condition when it comes to some infinite groups
like the group $G = \bZ_2 \cross \bZ_2 \cross ...$ when the kernel of the homomorphism is the subgroup where all copies of $\bZ_2$ are on the identity except one.
not that im suggesting actually using this definition
chilly ocean
I need to know whether the following logical inference is valid
context is group theory and cosets
{(h^(-1)g^(-1) : h in H} /\ h^(-1) = h' for some h in H => {h'g^(-1) : h' in H}
urban acorn
i cant read this. latex is a thing, you know.
i don't even think it's a valid expression
you opened regular parentheses then closed the fancy set paranetheses before closing them
also you refer to things like sets as propositions
your conclusion is literally the coset $Hg^{-1}$. that's not a proposition, that's a thing.
cloud walrusBOT
Intel:
urban acorn
also what is that /\ supposed to denote? is it logical conjunction? because the left side is - again - not a proposition
this bothers me so much
i want a clarification
with latex
jade sky
@urban acorn This is subjective, but let me explain why I don't think that would be a better definition.
Do you agree that it is similar to defining an embedding of the group G into the group H as a homomorphism φ: G → H such that φ(G) ≅ G? Which is fine for finite groups, but as in your example, you could "embed" certain infinite groups using a non-injective mapping as well.
But the idea of an embedding is to identify the domain group with a subgroup of the codomain — and this definition does part of that job. Yes, it tells us that φ(G) is isomorphic to G, but φ itself (or rather its restriction to φ(G) in the codomain) is not that isomorphism! So while it's true that G can be embedded into H if φ(G) ≅ G, it doesn't make much sense to call φ an embedding of G into H. The embedding still needs to be found.
Similarly, if we redefine the action φ of G on X to be faithful whenever φ(G) ≅ G, then the information that an action is faithful (according to this new definition) tells us that it's possible to represent the G faithfully as permutations of X, but φ itself may not be that representation, since it's possible that φ(g₁) = φ(g₂) in Sym(X) without g₁ = g₂ in G.
late sedge
question about this explanation
can only polynomials of degree 2 been in the ideal generated by (x^2+1)
mild laurel
No
late sedge
so why arent (x-i) or (x+i) in there
topaz solar
because there's nothing you can multiply (x^2+1) by to get (x+i)
late sedge
so how is Z[x]/2x-1 isomorphic to Z[1/2]
obv R/I doesnt seem to be a field so its not maximal, get that part but finding an isomorphism to R/I always seems to give me trouble
cloud walrusBOT
Kogasa:
late sedge
why 1/2
cloud walrusBOT
Kogasa:
jade sky
You can also define it ad-hoc, similar to ℚ[√2] and ℝ[i]
late sedge
so Z[x]/2x-1 are all integer polynomials divided by 2x-1?
bleak abyss
So the thing is, when you define it in the ad hoc manner
You're literally defining it as the quotient of the polynomial ring
late sedge
I understand that it's a quotient ring and the ideals partition it, but i was just wondering what could be an element of such a ring
jade sky
@bleak abyss I agree we can think about it that way (and that's what makes it useful), but you can do it without knowing about polynomial rings, simply using formal definitions
cloud walrusBOT
Kogasa:
bleak abyss
I mean I guess it's not "unsound" as Kogasa said it so much as contentless
cloud walrusBOT
Kogasa:
bleak abyss
Well your statement that this is unsound is literally false
Even if we didn't have Q
bleak abyss
You could define Z[1/2] basically as writing out in gruesome detail with localization means
bleak abyss
Now it is true that it's contentless to call that anything different from Z[x]/(2x-1)
But like
Unsound would suggest false or nonsense which isn't right
Well I'm this case we're localizing Z where? At 2Z
cloud walrusBOT
Kogasa:
bleak abyss
Lol this is a math server you gotta be careful with your words
Unsound means false not stupid
And I think he was asking about soundness in the strict sense of validity
Yeah they're kinda synonyms
Eh then it's subring of Q. But yeah regarding the localization comment since that's more interesting than semantics
So in general you can localize at an element
So in this case I was considering it at 2
It happens to be the case that since Z is an integral domain it's in Q so the two coincide
But more generally if I have a ring R and I localize at f (so S = {1, f, f^2,...}
Then this iso holds
R_f = R[x]/(1-xf)
So that's pretty nifty
late sedge
is torison an easy concept to study in like 45 minutes
bleak abyss
Definition is not even 45 seconds but it depends on what you wanna do with it
late sedge
idk i dont have much sample problems on it so im debating whether i should even look at it
does it have anything to do with finite abelian groups
bleak abyss
Torsion is just A[n] = {x\in A|nx=1}
late sedge
oh shit
bleak abyss
If A is an abelian group in general
In the fg case it's nicer tho true
And lol if you end up doing AG then it's prob worth considering other types of localizations
Commutative is fair though
I think
But if you have a variety that's not irreducible for instance. And prob some scheme stuff too
late sedge
guys does Q[sqrt(2), sqrt(3)] mean that it incorporates elements of the union of both ring extensions or the interesection?
notation question
cloud walrusBOT
Kogasa:
bleak abyss
Careful there's sqrt(6)
late sedge
i was about to say
bleak abyss
So the idea I think is like
Oh wait
You're doing ring but field
Eh should be fine actually for my idea
So like
You kinda just take some powers I think
And play around
The idea is that the minimal polynomial has degree 4
So you expect that by then you'll find something to work with. I don't have a more systematic way to do it though
chilly ocean
btw it's also true that $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \ldots, \sqrt{p_k}) = \mathbb{Q}(\sqrt{2} + \sqrt{3} + \cdots + \sqrt{p_k})$ for $2, 3, ..., p_k$ primes ($p_k$ is $k$-th prime).
bleak abyss
Rip
bleak abyss
So (sqrt(2)+sqrt(3))^2 = 5 + 2sqrt(6)
C
Then (sqrt(2)+sqrt(3))^3 = 5sqrt(2) + 5sqrt(3) + 2sqrt(12) + 2sqrt(18) = 11sqrt(2) + 9sqrt(3)
I'll let alpha = sqrt(2) + sqrt(3)
So then alpha^3 - 9alpha = 2sqrt(2)
So that's part of the way there
cloud walrusBOT
Kogasa:
bleak abyss
Well careful it's not obvious that you have sqrt(6) either
You have 2sqrt(6)
In the ring case
Right
Okay lol
cloud walrusBOT
Kogasa:
bleak abyss
Well also if you take alpha^3
That's just 11sqrt(2) + 9sqrt(3)
So that's just boom
bleak abyss
Yeah
I already passed my algebra qual lol
Gotta still do AT though
To be fair I have like forever at this point to do that but I'll try to get it done in one go this time
Yeah I probably will lol. Gonna read Bredon
We have a weird system here
So everyone is assumed to have seen point-set in undergrad
The topology qual has two options
Basically we have a semester one class which does pi_1, covering spaces, and homology
"Intro to Topology I"
And then "Intro to Topology II" does cohomology and some homotopy theory
There's also an "Intro to Smooth Manifolds" class that runs in the spring
So now the topology qual has two options, either you take the algebraic option in which case you're tested on the content of Intro to Top I-II, or the differential option in which case you're tested on the content of Intro to Top I + Smooth Manifolds
Well the weird part wasn't assuming point-set in undergrad
The weird part was having the two options
I think it's more common to have a single standard topology sequence and you have to learn everything
And like we don't do this for all quals
Algebra qual just has one
Same with applied and computational math
Topology has the algebraic option and smooth option
Analysis has a similar thing. There's Real I, Real II, and Complex
And you have to do either Real I-II or Real I + Complex for the qual
bleak abyss
1 is
cloud walrusBOT
fields:
sharp sonnet
i mean, you have access to a larger ring, R in this case
so just use multiplication from there
bleak abyss
So there are three reactions that come up right after your message
$\mathbb{Z}$
They each do different things. If I click the trash can it deletes the tex
If I click the question mark it shows the tex
And if I click the last one, it deletes my message, so there's no duplication
cloud walrusBOT
fields:
tribal pasture
@bleak abyss Mind telling me which university were you referring to above?
Because apparently my uni also has the same criterion for Top I and Top II
mild laurel
He goes to UW Madison
bleak abyss
Yup
chilly ocean
Context is cosets. Is gH = {x in G : x = gh for some h in H}?
hot lake
yes
chilly ocean
thnx
How do I find GCD of 10 and $4+2i \sqrt6$ in $\mathbb Z\left[i \sqrt 6\right]$
cloud walrusBOT
Godel:
chilly ocean
is the easiest way to just check norms and cases?
And btw, in order to use euclidean algorithm I need to know if its a euclidean domain right?
hot lake
gcd kinda only makes sense if your ring is principal, I think ?
I don't think it's euclidean, at least not for the usual norm
and I don't think that's the case ?
dawn kiln
euclidean domains are contained within the set of principal rings
sharp sonnet
there are principial rings, that are not euclidean
also, you only need a UFD to talk about gcd
this is unrelated to the original question
dawn kiln
wait, wdym by you "only" need a UFD
UFDs are not euclidean in general right? @sharp sonnet
chilly ocean
yes, they arent
topaz solar
UFDs have unique factorization
so you can uniquely write (up to a unit, ofc) them as a product of primes
so just
uh
gcd it like you'd expect
I mean, you can probably talk about GCD outside of a UFD in some cases, though what those cases are isn't something I know
dawn kiln
oh he's right actually
GCD domains are a superset of UFDs, which are a superset of PIDs, which again are a subset of euclidean domains
so apparently, GCD domains are int domains where any two elements have a greatest divisor
which is then equivalent to being able to construct a PI from some element that is equivalent to the ideal generated by the original two elements
urban acorn
unique factorisation domains have unique factorisation
the floor here is made of floor
woven delta
You're thinking the wrong direction
dawn kiln
yeah i got the inclusions the wrong way around, mb
tbh id never actually heard of a GCD domain before, only euclid
topaz solar
I mean, the idea of a GCD domain isn't that crazy
i can't personally think of a nice example tbh but it's not a stretch to imagine such a thing could show up
dawn kiln
bruh
topaz solar
clearly wikipedia doesn't either
but yeah, UFD is sufficient for obvious factorization reasons
dawn kiln
you don't care about torsion-free cancellative GCD-semigroups?
topaz solar
Specifically, it's talking about how $R[X; S]$ is GCD if R and S are both GCD (in the respective fasion) and S is sufficiently nice
cloud walrusBOT
Darkrifts:
dawn kiln
i haven't seen the notation R[X;S] in the context where S is an algebraic structure itself
topaz solar
$R[X]$ is just $R[\mathbb N]$ as a monoid ring
cloud walrusBOT
Darkrifts:
dawn kiln
i haven't seen that before
all of my algebra is self taught, so i probably have a fair few gaps
topaz solar
uh, you take your functions $f: \mathbb N \to R$ with finite support
cloud walrusBOT
Darkrifts:
topaz solar
and uh this is the same as having $\sum_n r \cdot X^n$
cloud walrusBOT
Darkrifts:
topaz solar
with the obvious $X^a X^b = X^{a+b}$
cloud walrusBOT
Darkrifts:
topaz solar
and your multiplication being convolution
actually it shoudl be r_n, but uh
only finitely many are nonzero
dawn kiln
is this X^n in the same sense as $\mathbb{R}^n$
topaz solar
boom polynomials
cloud walrusBOT
maximwebb:
Compile Error! Click the reaction for details. (You may edit your message)
topaz solar
it's X^n as in polynomial
dawn kiln
oh so just a variable
topaz solar
but if you're working with an arbitrary monoid, they're really just indices with compatible addition
i doubt you need to know much about it atm
but it's nifty once in a while
dawn kiln
i mean i probably never need to know it, i do CS lol
topaz solar
oh lmfao
dawn kiln
is this at all related to how you would form an analogue of a vector space?
dawn kiln
"vector spaces" over rings?
topaz solar
Free modules are those which are (isomorphic to) n copies of your base ring (just for addition & scalars)
and vector spaces are just free modules
But not all modules are free
and even if a set spans a free modules it needn't admit a minimal basis
like the set {2, 3} over the integers
dawn kiln
sorry to go through this slowly, but in terms of "n copies of your base ring", is this a cartesian product?
topaz solar
uh yeah
specifically with regards to the abelian group structure
because module, not an algebra
and also not the product of rings
Free modules are the nicest ones, more or less
dawn kiln
because they're simply vector spaces?
dawn kiln
i thought you said vector spaces are just free modules
topaz solar
They are, but not the other way around
dawn kiln
oh right
topaz solar
Blocks are toys
topaz solar
WEll, {2, 3} spans Z
but you can't reduce it to a one element basis
and clearly there is one, {1}
dawn kiln
oh right, i thought you meant that one doesn't exist at all
urban acorn
imagine asdf coprime integers span
death death death death death n = n3 - n2
ap + bq = 1
iff gcd(p,q) = 1
therefore
axiom of choice
dawn kiln
i like your calculation of euclids alg there
urban acorn
imagine doing number theory for abstract algebra
webb weeb
I am to weebs as a particular german man from WWII is to jews.
dawn kiln
i do error correcting codes for abstract algebra 😏
dawn kiln
urban acorn
ffs my browser isn't loadin'
dawn kiln
i do comp sci, and there are a couple of applications of algebra in that
urban acorn
so for example you could make a code such that the entropy in the output isn't fully utilized such that random errors due to imperfect networking are likely to corrupt the code so the error will get detected and handled appropriately?
like if u set every second bit to zero and only used the other bits
except less stupid
(not referring to specifically BCH codes which seem to specifically utilize galois theory, just the general idea of error correcting code)
dawn kiln
idk tbh, i haven't actually studied these yet, i just know that you can do ECC with galois
urban acorn
that sounds like algebra has applications in ECC, not that ECC have applications in algebra.
chilly ocean
Is ideal $\left(x,x^3+11\right)$ principal/maximal in $\mathbb Z\left[x\right]$?
cloud walrusBOT
Godel:
chilly ocean
would you find $(x, x^3 + 11) = (x, 11)$ helpful as a hint?
cloud walrusBOT
hochs:
chilly ocean
Why is $\pi \left(f\right) = \left(f\left(1\right),f\left(-1\right)\right), \pi: R[x] \to R xR$ an epimorphism?
cloud walrusBOT
Godel:
chilly ocean
given $(a,b) in \mathbb{R}^2$, take $f(x) = a(x-1) + b(x+1)$, then $\pi(f) = (a,b)$.
cloud walrusBOT
hochs:
chilly ocean
ok thats pretty much wath I did, but wasnt sure - if Im able to generate (1,0), (0,1) then that should also be enough?
yes
thx
im not sure if your formula is correct, just want to make sure: pi(f) should rather be (2b,-2a) right?
which doesnt really matter but just making sure
right, it should be $\frac{1}{2}f$
cloud walrusBOT
hochs:
tranquil creek
I'm trying to understand how this cyclic group is produced.
warped bay
which one
tranquil creek
I tried writing out powers of 6 and 8 and doing mod 12 but I got nothing, I cant come up with {0,2,4,6,8,10}
The first
warped bay
remember that Z_12 is a group under addition
tranquil creek
Right I tried that too
warped bay
0 is always in the subgroup
tranquil creek
Ya
tranquil creek
Multiples of 6 and 8
warped bay
multiplication?
tranquil creek
Right?
warped bay
is the operation mult'n mod 12 or add'n mod 12?
cloud walrusBOT
fields:
warped bay
addition modulo 12
cloud walrusBOT
fields:
cloud walrusBOT
fields:
warped bay
you are
cloud walrusBOT
fields:
warped bay
and we know that $0\in \langle 6,8\rangle$
warped bay
@tranquil creek get it?
tranquil creek
0 and 2?
warped bay
yes
tranquil creek
Oh I add 6 to the 8 aswell
I thought it was restricted to thier respective multiples
tranquil creek
Ya, its addition
I was think to get the multiples of 6 and 8 and then Express them as mod 12
Make sense now
tranquil creek
Now I'm having trouble with there single generators. Shouldn't 3 be fine as a generator? I seem to miss the point of the criterion of a generator
Gimme a sec, I'll keep trying
Nvm, I see it
Hard to do mod operations in head, better off writing it down to see it
tranquil creek
Oh, good point
leaden schooner
Are dihedral groups not the same as symmetry groups? I'm reading pointers abstract algebra chapter 7, and he's talking about permutations of groups and the example he uses is the square for dihedral groups
But D_4 looks just like permutations of the set of 4 elements
Am I missing something?
mild laurel
Yeah take a square
let the vertices be 1,2,3,4 starting from the top left and going clockwise
If vertex 1 goes to vertex 2 and vertex 2 goes to vertex 3
Then the movement of the last two vertices are fixed
vertex 3 must go to vertex 4 and vertex 4 must go to vertex 1
warped bay
D_n has 2n elements whereas the permutation group S_n has n! elements
mild laurel
So you cannot get the permutation that sends 1 to 2, 2 to 3, and sends 3 to 1 and fixes 4 for example
warped bay
so D_4 has 8 elements but S_4 has 4! = 24 elements
mild laurel
@leaden schooner
chilly ocean
Is ideal (x,11) in Q(x) Just the entire thing?
If 1 is in it then it should, right?
sharp sonnet
yes
urban acorn
@jade sky I agree, it's just an alternative definition that comes to mind when thinking about the possible origin of the word faithful. As I said, I do not suggest actually using it. I like your articulation of why the conventional definition is better beyond just for the sake of appealing to the existing convention.
urban acorn
So you cannot get the permutation that sends 1 to 2, 2 to 3, and sends 3 to 1 and fixes 4 for example \newline
imagine describing the mapping explicitly instead of just writing $(1\ 2\ 3) \in S_4$
cloud walrusBOT
Intel:
mild laurel
Not sure if you're trolling or not but
Someone who just is learning about these things might not have encountered the notation before
Better to write it out rather than potentially confuse them
urban acorn
I'm not trolling, but I'm not serious. I'm just meme-ing.
fading gulch
Given (a,b,c)-highest weight Verma module of Lie algebra gl(3). How do I classify its all submodules? a,b,c are three given numbers
potent lynx
Hi
basic problem
can some1 guide me through how to compute the set of automorphisms of a group
this particular 1
Compute Aut(Z_9)
idk how to narrow things down ig
ping if help
fickle brook
@potent lynx Z_9 is a cyclic group so any automorphism of it will be determined entirely by where it sends a generator
as in
well
yeah pick a generator g
potent lynx
okay
fickle brook
Z_9 will then be {e, g, g^2, ..., g^8}
potent lynx
okay
fickle brook
so the question is what can g be sent to
so that the resulting map is an automorphism
potent lynx
whats trivial
the identity
automorph
?
okay so
what are the generators for Z_9
1,2,4,5,7?
okay
something is missing with my understanding tho
why is this true
why is phi(generator) = another generator?
fickle brook
when phi is an auto?
potent lynx
yes
fickle brook
you need the image of phi to be the entire group
the image of phi is generated by phi(g)
fickle brook
well. yes an auto has to be surjective obv
potent lynx
okay so
phi(1) can equal 1
or 2
or 4
or 5 or 7
so we have 5 automorphisms
how can i find them
ik 1 is identity
fickle brook
and they end up just being multiplication by 2, 4, etc mod 9
potent lynx
okay so
so if phi(1) = 1 we know phi
if phi(1) = 2
we can know phi
right?
we can know where every element is from phi
right?
fickle brook
yes
urban acorn
no, the way to see it is you pick some generator $g$, then let $\phi(g) = g^i$, then $\phi(g^j) = (g^i)^j = g^{ij}$.
cloud walrusBOT
Intel:
urban acorn
which shows that the automorphism is multiplication of the power in each element when viewed as a power of the generator you picked
leaden schooner
Would someone mind telling me if this proof seems valid?
I think it makes sense, but I always seem to have false confidence in combinatorial things
snow bolt
sure, but it seems like overkill to me
you dont really need induction imo
just say we have n choices for phi(1), n-1 choices for phi(2), n-2 for phi(3)... which gives us n! by the multiplicative principle
chilly ocean
I need help with interpreting problem 20 in this book
How am I going to find the double-cosets of H in A_4 when they didn't say what K is?
dawn kiln
its probably a mistake
chilly ocean
i guess it's not a problem I have to solve
it's not like double-cosets are going to be in my algebra qualifying exam
topaz solar
I have no idea
mild laurel
The fact that double cosets are a thing wants me to kill myself
stark sigil
I used to hate double cosets then found out automorphic representation theory has them everywhere
bleak abyss
Icy001 perfect you can also help fuel my indecision
Hartshorne for $45 or Fulton-Harris for $30?
stark sigil
That’s a pretty tough decision
Get whichever one doesn’t have a downloadable pdf version :^)
bleak abyss
They both do tho
stark sigil
Ree
Ok next question is are you going to be using it to learn or as a reference for the odd thing here and there
bleak abyss
Both subjects are ones I wanna learn
mild laurel
Yeah, that's also what I realized, now I hate double cosets even more
bleak abyss
For what it's worth I will be taking AG next semester
But I won't be doing rep theory next semester
So Hartshorne is more immediately relevant
stark sigil
Why not both!
bleak abyss
I guess I could do the double deal actually
Have one ship here and another to my parents
stark sigil
😋
bleak abyss
Actually should I get Fulton-Harris or Serre for rep theory? Same price
mild laurel
I have a copy of Serre that I've been reading
plain sequoia
For algebraic geometry there are many better books now isn’t it?
Hartshorne has become old, and somewhat outdated? Also its still as harsh as people say it is I guess
bleak abyss
Yeah I have no idea about AG books tbh
plain sequoia
The Rising Sea by Vakil is great I’ve heard
bleak abyss
Like I'm aware of a bunch of varieties
plain sequoia
And legally free
bleak abyss
Isn't Vakil still unfinished tho? And kinda too long?
(Also I have pdfs of these books, it's just that Springer is doing $30 off so I'm going impulse shopping)
plain sequoia
I am not 
(so I don’t know)
I think Hartshorne does it the old way via varieties
Vakil does it via schemes
The more modern way I think?
bleak abyss
Hartshorne chapter 1 is varieties
But just as motivation really
Chapter 2 is schemes, 3 is cohomology, 4 is curves, 5 is surfaces
plain sequoia
But 30$ sale on springer?? I need to check!!
bleak abyss
But yeah Fulton Harris is what I'm leaning toward atm
plain sequoia
Yeah my tips are probably trash, but I’ve heard from someone very good at math that Liu’s book is great
Better than Hartshorne
I think its called algebraic curves or something
Idk if its springer tho
bleak abyss
It's not, I know of it
I guess the 3 books that I hear people cite for schemes are like
Well 4* given that Vakil is incomplete supposedly
Hartshorne, Vakil, Liu, and Gortz/Wedhorn
plain sequoia
I need to find out what to buy as well
Eisenbud for 10 bucks 
Should I get Rotman?
And in that case which of his books? I heard he is somewhat of a godly good author
bleak abyss
So I'm mostly aware of the existence of his algebra book, I've read through some of his group theory book, and less of his algebraic topology book
Group theory was nicely written but like
The stuff it covers that you need to know is covered in general algebra books anyway, so this is more useful if you're getting into the specialized stuff
Rotman's prob the best easy book on AT that I'm aware of though. He's super careful unlike Hatcher
plain sequoia
I was thinking of either his homalg or algtop book
Gonna take a look tho
Thanks!
smoky briar
I've never seen anyone call hartshorne outdated
cloud walrusBOT
Riyango:
frank haven
chilly ocean
ok, got it.
Thanks
topaz solar
O for origin
chilly ocean
I don't understand this proof of the sum-of-divisors.
mild laurel
are you going to post the proof?
solar vessel
no
jade sky
Ohio, Koga-chan!
chilly ocean
Ohio, state university
cloud walrusBOT
fields: