If you consider the set of permutations w/ composition, you can have a group, and then if you look at ({-1,1},×), you have another group
#groups-rings-fields
worthy kindle
cerulean rune
oh, sign of a permutation, thanks, i got confused assuming signature meant e.g. of a symmetric matrix
sick acorn
I have a bijection f:X->Y, and we also have the two groups of permutations S_X of X and S_Y of Y. I'm supposed to use f to construct an isomorphism F:S_X -> S_Y.
If I write F(x) = f o x, that's still just F:S_X -> (X->Y), right?
because the input x is a bijection X->X, while the output f o x is a bijection from X to Y
hmm.
oh wait
f o x o f^-1
(Y->X) -> (X->X) -> (X->Y)
errant drum
If G is an infinite abelian group with H being a normal subgroup such that all elements of H have finite order. Then the quotient G/H any element that is not the identity has infinite order
I'm sort of stuck on how to begin
My guess was a proof by contradiction
Suppose there was some g in G but not in H. Such that g^n = 1
Then (gH)^n = g^n H
which is just H
cerulean rune
By "G is an infinite abelian group", do you mean "all nonidentity elements of G have infinite order"?
errant drum
No as in G is an abelian group of infinite size
Does this mean it's isomorphic to (Z,+)
And am I supposed to use that fact?
cerulean rune
So I am somehow confused. This is false -- just let G be an infinite abelian group such that every element has finite order, for instance, and you can just let H = {e}
Oh, is H supposed to be the subgroup of finite-order elements of G
errant drum
Yeah
cerulean rune
All right, I think I understand the statement now. The beginning of your proof doesn't make sense to me. Since all finite-order elements of G are already in H, supposing the existence of a g in G with g^n = 1 is already a contradiction
If you want to proceed by contradiction (which is not necessary but probably fine), you should start by assuming the negation of the thing you want to prove -- i.e., assume G/H has an element of finite order (a g \in G with (gH)^n = 1)
errant drum
I should specify it be a non-identity element right?
cerulean rune
ye
cerulean rune
Well, what does it mean for g^n to be in H? H was constructed to be the subgroup of all elements of finite order
errant drum
g^n = h
For some h in H
And we know that h^m = 1
Because h is in H
Which means g^nm = 1
Right?
cerulean rune
yep, i agree so far!
errant drum
So then g^nm is in H which makes gH an identity element
cerulean rune
why does g^{nm} being in H force gH to be equal to H?
errant drum
Because g^nm = 1
Meaning g has finite order
Meaning g is in H
Which means gH is just combining the elements of H with something else in H
Which results in the coset H
cerulean rune
yep, that all sounds good :D
cerulean rune
:p just distracted. glad it was helpful :D
covert vector
can anyone explain what the hell the wreath product is
I'm trying to go thru the definition on Wikipedia
and it's flying over my head
this thing is too woke for me
full blaze
chilly ocean
It looks like they're taking a group and making tuples out of it
and then you can multiply group elements componentwise
or permute them
snow dew
The WRATH PRODUCT is what i read and it would make a good movie
Or smthing
Which wreath product
There are 3 different ones on my wikipedia
They say it's related to graph automorphisms, which seem like a good place to start and get an instance of the thing to manipulate while learning the general version
north tide
Can you think of a surjective homomorphism from Z[x] to Z/2Z with kernel being the ideal (2, x)?
little fox
Sorry if I seem a bit slow, but is a homomorphism when you have some f that takes one ring and maps it onto another?
north tide
I believe <2, x> refers to the ideal (2, x)
snow dew
Yeah but in R[X] that would be (x), right?
north tide
Yes a (ring) homomorphism F is a function from a ring to another such that F(1) = 1, F(a+b)=F(a)+F(b) and F(ab)=F(a) F(b)
I think the polynomial ring of a field is always a PID, but the polynomial ring of a PID may not be
In R[x], (2, x) is (1), the whole ring
north tide
In Z[x], it contains all even constant polynomials and all polynomials of degree > 0
snow dew
Even coefficients
little fox
Ok sorry back!
So from what I could tell <2, x> is an ideal, isn't it like the sum of all the multiples of 2 and multiples of x because in a different problem I saw <x, y> and it was explained as xf(x,y) + yg(x,y) and <2> was said to be equivalent to 2Z
north tide
Yes that's right
little fox
gotchya thanks, so is this Z[x]/<2, x> looking at the modular? because I don't really know how to prove that there's an isomorphic relationship between the two
north tide
It is called a quotient ring. To prove this, the most common way is to apply 1st isomorphism theorem. Have you learned that theorem yet?
little fox
My prof has gone over it, but I've been struggling with understanding it fully
But I'll go over it now, thanks for all the clarification
sick acorn
i just want to double-check this because i'm not sure if it's a trick question or not lol. Let's say we have f:Q->Q with f(r) = r/2. That's both injective and surjective, right? As is the case with g:Q* -> Q* with g(r) = r/2?
I know the former is a homomorphism, while the latter isn't one
sick acorn
Q is the rationals; Q* is just the rationals without 0
sick acorn
What does it mean for the group S_n to be generated by the set of all transpositions (a, b), where a < b? He briefly mentioned it in the notes, but it wasn't clear at all. It's part of this question:
The relevant part of the notes is:
I know we can, for instance, write a k-cycle (1, 2, 3, 4, 5) as (1, 5)(1, 4)(1, 3)(1, 2). And that we can write any sigma in S_n as a product of transpositions (2-cycles).
north tide
@sick acorn I believe you are talking about group homomorphism. The first one on (Q, +) the additive group is a homomorphism because 0/2 = 0, i.e. identity element being mapped to identity element. But Q* as the mutiplicative group has identity 1, but 1/2 is not 1. And yes both maps are bijective as a function.
sick acorn
Phew, that's what I thought. Thanks!
north tide
And S_n being generated by transposition is simply what you described, every element in S_n can be written as product of transpositions.
sick acorn
Ah, that makes sense—thanks again. I’ll think about the problem more
north tide
You're welcome
upper inlet
why is a group a groupoid with only one object
cerulean rune
The idea is to think of the morphisms as the elements of the group; composition of morphisms is the group operation
the category axiom that each object has an identity morphism (in your case, the unique object has an identity morphism) gives you the group axiom that there exists an identity element; the category axiom that composition be associative is group multiplication associativity; and then the inverses come from the groupoid condition (every morphism has an inverse)
upper inlet
but isnt a group a set of objects?
and a group is a groupoid with only one object
does t matter that its a category with only one object?
cerulean rune
you are using the word "object" in two different ways in those sentences. a group is a set, together with a multiplication on it; the elements of that set you might colloquially refer to as "objects", but that does not necessarily mean that they correspond to the category-theoretic notion of "object" (indeed, this is not a good picture)
upper inlet
ah
cerulean rune
if the unique object [edit: the unique object of your one-object category] is called A, then the underlying set of the group is most naturally identified with Hom(A,A) (the set of morphisms from A to itself)
cerulean rune
the groupoid axiom guarantees that they're the same thing, yeah
cerulean rune
i will slightly complain that "bijective" isn't a great word to use in this context (I am not thinking of A as a set, and in particular I am not thinking of the elements of Hom(A, A) as set-theoretic functions), but otherwise yes
all the morphisms in Hom(A,A) have inverses (i.e. they are isomorphisms)
upper inlet
yeah,i meant isomorphic, sorry
and so you coukd view the morphisms as the binary relations on the group?
cerulean rune
no, the morphisms literally are the elements of the group
there is only one binary relation on the group, which in this context is composition of morphisms (f, g) |---> f . g
upper inlet
that seems a bit weird to say morphisms are literally elemets
cerulean rune
i mean, it is presumably not weird to say that Hom(A, A) is a set, nor to point out that it is actually a group if every element of it is invertible
perhaps the issue is that you are trying to think of A as "the group" in some sense? this is kind of a general lesson of category theory -- which i might say vaguely and unprecisely as "the hom-sets carry the interesting information"
cerulean rune
did this come up abstractly or are you familiar with the fundamental groupoid?
upper inlet
im not sure what you mean
ive barely studied groupooids at all really if thats what youre asking
cerulean rune
a lot of times the first time someone sees the word "groupoid" is in the context of what's called the "fundamental groupoid of a topological space". if you're familiar/comfortable with this notion, it gives a better picture, as the objects of the groupoid are points of your topological space
but yeah if you just ran into this outside of this context, then it's not worth the picture atm :P
upper inlet
huh
the structure of what you said makes sense, i just have trouble seeing why it is like that
perhaps when i delve into topology it will become clearer?
cerulean rune
yeah, after you get comfortable with the notion of a fundamental group, it's worth revisiting the concept of a groupoid briefly
it's probably the canonical example
upper inlet
would you reccomend learning topology before abstract algebra?
right now im using chapter )
0*
cerulean rune
no, definitely not. it's okay to learn point-set topology before abstract algebra (and it's okay to learn abstract algebra before point-set topology), but if you are going to do fundamental group stuff (the beginnings of "algebraic topology"), then you should at least be a bit comfortable with the notion of a group/subgroup
(i don't know what chapter 0 means tho)
cerulean rune
oh i see that is quite a book name
upper inlet
yeah lol
cerulean rune
oh lol wow categories right away huh
upper inlet
yeah
i think its good to develop the concept of morphisms not just in the context of Set though
cerulean rune
yeah, i have mixed feelings on the approach
on the one hand, categorical language can be really nice for unifying concepts and isolating the "important stuff"
upper inlet
yeah
cerulean rune
on the other hand, it's unclear to me how "reasonable" or "natural" it feels if you don't have a few examples of the kinds of things it's modeling under your belt first
upper inlet
but sometimes it doesnt make a lot of sense for he new student i guess
so i have two textbooks, an abstract algebra book that goes pretty far into it and a topology book by munkres
which one do you think i shoud pursue first?
upper inlet
ive gotten farther in the abstract algebra text so far
cerulean rune
i think the parts are explicitly labeled?
upper inlet
im only in the set theory part of the topology to be honest lol
cerulean rune
yeah, that's what i mean by "point-set"
upper inlet
oh ok
upper inlet
give me a second, im going to go retrieve the books if i can
You’re right, part 2 is algebraic topology
cerulean rune
the early point-set in munkres -- up through like chapter 3 -- is good practice for, like, learning how to think precisely/mathematically, and for dealing with sets and functions. the idea of "compactness" is broadly important in mathematics, so up through ch. 3 in munkres is like, core stuff. the other chapters (4-8) are less important to other math disciplines, i would say.
i don't know this aluffi book, so i can't say whether it suits your needs well. in general learning about groups is a great place to start with upper-level/proof-based math, especially if you're more "algebraically" minded. i worry looking at the contents of aluffi that it will be slightly on the abstract side or light on examples, which i frown on a bit, but some people like it
(i learned topology from munkres so i have it around)
upper inlet
I see
cerulean rune
but i think the book to start on is the one you find more fun, if you can make such a judgement
upper inlet
Personally the set theory got a but boring aftera While when I started munkres
i didn’t take many breaks, you know what I mean?
cerulean rune
ye you don't want to be bored, that's the killer
upper inlet
so I’m inclined to choose the aluffi book, but it’s also very abstract
its already introduced universal properties
cerulean rune
do you have, like, any example of a universal property?
my experience is like, as long as i have two examples abstraction is usually fine, and if i have one it's manageable
if i have zero, i'm hosed
upper inlet
Yeah
i try to rely less on examples, but it’s definetly easier when there are some
i think I remember that {0} is universal with respect to the property of mapping sets to other sets
that’s something the book said
like many other things so far, the definition isn’t unsupported, it just seems arbitrary
cerulean rune
right, that is the problem with abstract stuff, at least for me. with nothing to tie it do the only thing you can do is memorize it
but memorizing stuff is hard
upper inlet
Agreed
on the other hand, typically as I progress it will all start tofir together
to fit*
So I’m in a bit of a dilemma
cerulean rune
well, as long as you are enjoying the aluffi it sounds good to keep going in it to me
and you can ask people stuff here i am sure
upper inlet
Yeah, I am enjoying it lol
cerulean rune
for me my recommendation is to have a few examples of categories in mind and make sure you translate basically every definition into them
cerulean rune
gl :D
sick acorn
I was wondering if anyone could provide some guidance for this first sentence:
i've done the other parts
So I'm starting with... Let H be a subgroup of Sn that contains all products of two 2-cycles (i, j)(k, l), not necessarily disjoint.
so i need to show that that set can generate An hmm....
errant drum
If you want (12345) = (12)(13)(14)(15)
I guess you have to use that some how
Oh nvm
sick acorn
Hmm... (12345) is in An, and it is generated by the set of all products of two 2-cycles...
north tide
What is your definition of A_n? If it is defined to be the set of all even permutation (i.e. It is a product of even number of transpositions), then clearly you can group them into pairs of transpositions, and each pair belongs to the set H you described.
sick acorn
It's the set of all even permutations. I ended up writing:
Let P be the set of all products of two transpositions (i, j)(k, l), not necessarily disjoint.
Let H be a subgroup of Sn containing the set of all products of two transpositions (i, j)(k, l), not necessarily disjoint. (So H contains P.)
Let σ in An. We know that any element of An is the product of an even number of transpositions, not necessarily disjoint. Then σ can be written as a product of 2x transpositions, where x in Z. We know that each element in P is a product of 2 transpositions, not necessarily disjoint. σ is a product of 2x transpositions, so it can be written as a product of x elements of P.
So any element in An can be written as a product of elements of P. (This means that any element in An can be written as a product of products of two transpositions, not necessarily disjoint.) So An is generated by the set of all products of two transpositions (i, j)(k, l), not necessarily disjoint.
north tide
Seems good to me. I think you don't need to mention "not necessarily disjoint" everytime though, since it should be well-understood. Also I'm not sure what's the point of H.
charred lance
could someone explain what the Gy=gGxg^-1 means
like its probably a stupid question but what does gGx g^-1 mean because Gx is a set right?
cerulean rune
yeah, in this context $$g G_x g^{-1} = \lbrace g h g^{-1} : h \in G_x \rbrace$$
mellow vaporBOT
cerulean rune
(this notation of writing like, (element)*(set) to mean the set of all things that look like (element)*(thing in set) is pretty standard for groups)
charred lance
thanks
restive kiln
Hey peeps. If I know that $$\langle x^ a \rangle - \langle y^b \rangle $$ is a prime ideal, do I also know that $$\langle x^a - y^b \rangle $$ is a prime ideal?
mellow vaporBOT
cerulean rune
i am not sure i understand the notation, but $$k[x,y]/(x-y-xy)$$ seems like the obvious choice for a counterexample: the quotient by $$(x) - (y) = (x, y)$$ is just $$k$$, but the quotient by $$(x - y)$$ should be $$k[x]/(x^2)$$, right?
mellow vaporBOT
restive kiln
The notation I believe is supposed to indicate the difference between the ideal generated by $$x^m$$ and the ideal generated by $$y^n$$
mellow vaporBOT
restive kiln
I'm confused because the hint for this problem I'm working on tells you to find a map from the ring $$R[x,y]$$ to $$R[z]$$ whose kernel is $$\langle x^m \rangle - \langle y^n \rangle$$ in order to prove that $$\langle x^m - y^m \rangle$$ (the ideal generated by the polynomial x^m-y^n) is prime.
mellow vaporBOT
Rendering failed. Check your code. You can edit your existing message if needed.
cerulean rune
oh gosh like the set-theoretic difference? that's not an ideal in general, right?
restive kiln
No, I'm pretty sure it's just a difference of polynomials
cerulean rune
getting confused, $$\langle x^m \rangle - \langle y^n \rangle$$, whatever it means, is supposed to represent a prime ideal, right?
mellow vaporBOT
restive kiln
Like $$4x^2my^7-4x^2y^n+7$$ should be an example of an element of $$\langle x^m -y^n\rangle $$
mellow vaporBOT
restive kiln
Yeah, it is a prime ideal
The differece, I believe, is supposed to be just taking an element generated by $$x^m$$ and subtracting something generated by $$y^n$$ to form a polynomial.
mellow vaporBOT
cerulean rune
so that's no different than $$(x^m, y^n)$$, right? like, the word "subtracting" is funny since you have additive inverses in R
restive kiln
Ring of interest is R[x,y], R an integral domain
mellow vaporBOT
restive kiln
Yeah, sorry, I guess it isn't different.
cerulean rune
I may be confused because I do not see how $$4x^{2m}y^7-4x^2y^n+7$$ is an element of either $$(x^m, y^n)$$ or $$(x^m - y^n)$$, though that is a side confusion
mellow vaporBOT
mellow vaporBOT
mellow vaporBOT
cerulean rune
yeah that's definitel yin ther
restive kiln
I'm confused because the hint for this problem I'm working on tells you to find a map from the ring $$R[x,y]$$ to $$R[z]$$ whose kernel is $$\langle x^m \rangle - \langle y^n \rangle$$ in order to prove that $$\langle x^m - y^m \rangle$$ (the ideal generated by the polynomial $$x^m-y^n$$) is prime.
mellow vaporBOT
cerulean rune
ok so the question is like, suppose you find a map $$R[x, y] \to R[z]$$ whose kernel is $$(x^m, y^n)$$. How do you conclude from this that $$(x^m - y^n)$$ is prime? (I don't see the answer at the moment, but I think I am more clear on the question.)
mellow vaporBOT
restive kiln
Yes
cerulean rune
i definitely had the wrong idea/context at first
restive kiln
Well that's reassuring
All I know at this point is that $$(x^m - y^n)\subset (x^m, y^n)$$
mellow vaporBOT
restive kiln
And I'm wondering if the former not prime somehow violates primality of the latter.
cerulean rune
i don't even see the map that kills x^m for what it's worth
i mean other than the dumb one
i am somewhat feeling like this is a misinterpretation of the hint? like if x^m is in the kernel of a map to k[z], then so is x, and likewise for y, so your map is the stupid one that just zeros out x and y
which isn't gonna tell you much about (x^m - y^n)
restive kiln
the map I used is $$\phi:; R[x,y] \rightarrow R[z^n,z^m]$$
mellow vaporBOT
cerulean rune
right, that seems like exactly what you want to do
(i must not understand the notation because that really does not have (x^m, y^n) as its kernel)
restive kiln
hmmm
my argument is too long to summarize
But I'm very persuaded that it is the kernel.
cerulean rune
we must disagree about what the notation means. like x^m - 0y^n is not in the kernel of this map, nor is x^m - 2y^n
i suspect the kernel of that map is (x^m - y^n), which is what you want, although i am currently unclear on how to make it 100% obvious
restive kiln
I was initially aiming for (x^m - y^n) being the kernel, but it didn't appear to fall out of the work
Again, the hint for the problem was to find a map with kernel (x^m)-(y^n)
cerulean rune
i am getting a little lost in the middle (and didn't realize gcd(m,n) = 1, which of course has got to be necessary somewhere! i am dumb), but at the end you wind up with this complciated sum (the big 2-line inlined equation), note that it's in (x^m) - (y^n), and then conclude that the kernel is equal to (rather than contained in) (x^m) - (y^n)
restive kiln
I concluded what I did because we end up with something which appears to be some thing generated by $$x^m$$ beign summed with something generated by $$y^n$$
mellow vaporBOT
restive kiln
I did try separating out the sums of the coefficients and stuff, but the closest thing I was left with was $$x^{mu}-y^{nu}$$, which isn't generally divisible by $$x^m-y^n$$
mellow vaporBOT
restive kiln
Okay, I think your counterexample is valid
cerulean rune
yeah; i think you should be able to show that your ugly polynomial is divisible by (x^m - y^n), but i am still struggling with the algebra like i do
restive kiln
I spent most of today trying to do that
I'm really not sure how
Okay, so I know that $$\sum a_{ij}=0$$
mellow vaporBOT
restive kiln
I should be able to use this to conclude that $$\sum a_{(k+um,j) }+ \sum a_{(k,un+j)}=0$$
mellow vaporBOT
restive kiln
So we might as well call one a and the other negative a
move them over to the side
and completely factoring, we get: $$ax^ky^j\sum_{u\in U}x^{um}-y^{un}$$
mellow vaporBOT
cerulean rune
sorry, i don't understand what your sums are over in $$\sum a_{(k+um,j) }+ \sum a_{(k,un+j)}=0$$
mellow vaporBOT
restive kiln
The sums are over $$u\in U$$
mellow vaporBOT
cerulean rune
and k and j have been fixed?
cerulean rune
i don't think i agree with this equation
mellow vaporBOT
restive kiln
What's wrong?
cerulean rune
i don't understand what k and j have been fixed to be
like, something kinda weird is going on
usually (i,j) are paired, or (k,l), in your notation
you could fix a pair (i,j) such that ni+mj = d
but in terms of your u you have the relationships (i-k)=um and (l-j)=un, so if you are trying to "fix" a j and a k simultaneously -- i think i may just be missing something dumn
restive kiln
The idea I had in mind was that $$i=um+k$$ and $$l=un+j$$
So if you fix them, you're able to generate every other value by varying u
mellow vaporBOT
cerulean rune
well, if i varies, generally j has to vary as well so as to maintain the relationship ni + mj = d
but if you fix i and l, then since fixing i also determines j, you've just also fixed u
restive kiln
Ah, you're right
I'm silly
The issue I get when I fix i and j though is I get something that seems ugly and hard to work with
restive kiln
cause you have a difference in the exponent
cerulean rune
ugh, okay, my algebra worked out but it was unpleasant
restive kiln
I think I made some progress
cerulean rune
oh good
restive kiln
since i-um needs to be positive, we can pull out $$x^{i-um}$$
mellow vaporBOT
restive kiln
So then we'll have $$ax^{i-um}y^j(\sum\limits_{u\in U}x^{um}-y^{un})$$
mellow vaporBOT
cerulean rune
i do not think you get to pull out a term with u in it from a sum over u :\
cerulean rune
the thing i did was kind of unpleasant but i can outline it if you'd like
restive kiln
I might ask in a few. I want to see if I can get it myself
cerulean rune
ok gl :D
restive kiln
Does your solution make use of the observation that since i is fixed, i-um>0 puts a cap on high the values of u can be?
cerulean rune
not like, clever use of this fact, no, but i am writing down a sum over u and i want it to be finite, otherwise it doesn't make sense
restive kiln
Okay, I think I have it now
So there's a cap on the value of u such that i-um>0, call this u'
Then we can write $$ay^j(x^i-\sum\limits_{u=0}^{u'}x^{i-um}y^{un})=ax^{i-u'm}y^{j}\bigg(x^{u'm}-x^{(u'-1)m}y^{n}-\dots-y^{u'n}\bigg)$$
mellow vaporBOT
restive kiln
Which I think is just equivalent to $$ax^{i-u'm}y^j\bigg(x^m-y^n\bigg)^{u'}$$
mellow vaporBOT
restive kiln
oh wait, I don't think that's correct
Because the negative signs would alternate if it was just $$(x^m-y^n)^{u'}$$
mellow vaporBOT
cerulean rune
so given that you haven't used the fact that the sum of your coefficients a_{i+um, j-un} is zero, it's very unlikely that this works
(that was our translation of the condition that this element be in the kernel after all...)
restive kiln
but I did make use of that fact
I concluded that $$a=a_{ij}=-\sum\limits_{u=0}^{u'}a_{(i-um,j+un)}$$
mellow vaporBOT
restive kiln
so I factored out the a and $$y^j$$ which gives me $$ay^j(x^i-\sum\limits_{u=0}^{u'}x^{i-um}y^{un})$$
mellow vaporBOT
cerulean rune
hrm, maybe i am lost. a is zero? i don't see how it factors out of your sum
restive kiln
it's not zero
the sums of the coefficients are zero
which means $$a_{ij}+\sum_{u=0}^{u'}a_{(i-um,j+un)}=0$$
mellow vaporBOT
restive kiln
which means one is equal to the negative of the other
restive kiln
so then we have $$a_{ij}x^iy^j+\sum_{u=1}^{u'}a_{(i-um,j+un)}x^{i-um}y^{j+un}=ax^iy^j+\sum\limits$$
mellow vaporBOT
cerulean rune
i'm fine with that yeah
i am not sure how it is helpful but it does not seem wrong :P
cerulean rune
haha yes
restive kiln
Anyway, I can get it down to this form: $$ay^j(x^i-\sum\limits_{u=0}^{u'}x^{i-um}y^{un})=ax^{i-u'm}y^{j}\bigg(x^{u'm}-x^{(u'-1)m}y^{n}-\dots-y^{u'n}\bigg)$$
mellow vaporBOT
restive kiln
I need to replace all these u=0 with u=1
cerulean rune
i do not agree that the thing you wrote above is equal to $$ay^j(x^i-\sum\limits_{u=0}^{u'}x^{i-um}y^{un})$$
mellow vaporBOT
cerulean rune
"above" meaning hold on
let me be clearer
$$a_{ij}x^iy^j+\sum_{u=1}^{u'}a_{(i-um,j+un)}x^{i-um}y^{j+un}$$ and $$ay^j(x^i-\sum\limits_{u=0}^{u'}x^{i-um}y^{un})$$ do not appear to be equal to me in general
friggin discord formatting
mellow vaporBOT
cerulean rune
you can't factor that sum of a_{blah} out of the right hand term
restive kiln
:<
Ah I see
hmmm
Okay, so we have $$x^{i-u'm}y^j(a_{ij}x^{u'm}+a_{(i-m,j+n)}x^{(u'-1)m}y^{n}+a_{(i-2m,j+2n)}x^{(u'-2)2m}y^{2n}+\dots + a_{(i-u'm,j+u'n)}y^{u'n}).$$
mellow vaporBOT
restive kiln
How do we get rid of the coefficients, or at least get ourselves to a place where we can ignore them?
Oh
Do you just apply polynomial long division and find that the coefficients disappear and the polynomial divides?
cerulean rune
i think that should work in this case (presuming you know what you mean by "polynomial long division" when there are two variables, but this is a simple case of that)
restive kiln
I'm guessing you did something else
cerulean rune
i found that got a little unpleasant when i went to do it, so i made some algebraic simplifications that made things a bit more obvious to me, yeah
restive kiln
Okay
If you wouldn't mind sharing those manipulations, I'd prefer to use them than prove that polynomial division works.
cerulean rune
sure -- gonna take me a moment to translate what i have written down because my u is negative what yours is
so you have $$\sum_{u = 0}^{u'} a_{i-um, j+um} x^{i-um} y^{j+un}$$ which you want to show is divisible by $$(x^m - y_n)$$
mellow vaporBOT
cerulean rune
set $$b_u = a_{i-um, j+um}$$; the sum above becomes (carrying out the computation in $$R[x, x^{-1}, y]$$):
$$x^i y^j \sum_{u = 0}^{u'} b_u x^{-um} y^{un} = x^i y^j \sum_{u=0} b_u (x^{-m} y^{n})^u$$
mellow vaporBOT
cerulean rune
I set $$\tau = x^{-m} y^n$$; the sum on the right becomes $$\sum_{u=0}^{u'} b_u \tau^u$$, and the condition we have is that $$\sum b_u = 0$$; it's easy to do the long division with $$\tau$$ to see that $$(1 - \tau)$$ divides this sum
mellow vaporBOT
cerulean rune
then you just unwrap everything and see that the bad stuff in the denominator all gets cancelled. I think this is what you wind up doing, but you drag some factors of $$x^{u'm}$$ around everywhere to make sure you're always in $$R[x,y]$$, which is fine, just winds up being a bit more notationally painful
mellow vaporBOT
cerulean rune
("bad stuff in the denominator" = factors of x^{-m})
(if you're uncomfortable with working in $$R[x,x^{-1}, y]$$, you can at least save yourself some headache by setting $$\tilde{x} = x^m, \tilde{y} = y^n$$ or something like that, and defining the $$b_u$$ as above, so as to get rid of the $$m$$'s and $$n$$'s plaguing everything)
mellow vaporBOT
restive kiln
Wow! Thanks
restive kiln
I'm not sure I understand why you divide by $$1-\tau$$.
mellow vaporBOT
restive kiln
Because $$1-\tau=\frac{x^m-y^n}{x^m}$$
mellow vaporBOT
restive kiln
Which ends up being equivalent to multiplying everything by $$x^m$$ and dividing by $$x^m-y^n$$
mellow vaporBOT
restive kiln
But I guess since $$x^m$$ isn't divisible by $$x^m-y^n$$, this confirms that the polynomial is divisible by $$x^m-y^n$$
mellow vaporBOT
restive kiln
is that the reasoning you had in mind?
cerulean rune
i am not really sure whether you are describing the thing i had in mind or not
i was sort of picturing writing $$\sum_{u=0}^{u'} b_u \tau^u = (1-\tau) \sum_{u=0}^{u' - 1} c_u \tau^u$$, and then plugging that back in to our original thing
mellow vaporBOT
cerulean rune
there's a bunch of x^{-m} in the denominator of the thing once you re-substitute, but they get cancelled by stuff in the x^i term (this is exactly what you were doing earlier i think, is explicitly putting in that x^{u'm} part of the x^i term onto everything in the sum)
restive kiln
@cerulean rune Thank you for your assistance and patience. I ended up just going with my original strategy, but I genuinely appreciated the sanity checks.
cerulean rune
all right! was fun to work on
dusky oxide
Hey, sorry if I am breaking any rules, I tried posting this in the help channels with no luck. I was wondering if someone could quickly look over my proof of Cayley's theorem. This is what I have: Let G be a group. Then let G act on itself by left multiplication. Then there is an associated permutation representation phi: G -> S_G. Then ker(phi) = {g in G : g h = h, for all h in G} = {g in G : g = 1} = {1}. Then by the first isomorphism theorem, G/{1} = G is isomorphic to the image of phi in S_G.
bleak abyss
Is your statement of Cayley theorem that finite groups embed into symmetric groups?
dusky oxide
@bleak abyss It is that every group is isomorphic to a subgroup of some symmetric group
bleak abyss
Your argument works
dusky oxide
@bleak abyss Ah, thank you. The argument presented in the text (Dummit & Foote) is quite long winded, creating several propositions before giving the proof. To me it looked like the approach in the text was largely pedagogical rather than practical, I just wanted to make sure haha.
lavish fossil
how do we show that any two fields of order p are isomorphic?
quiet cave
Take the map that sends one generator to the other one and its powers to the powers of the other one
That should be your isomorphism
chilly ocean
R be a ring with unity such that its subrings (not necessarily with unity) satisfies ascending and descending chain condition ; then is R finite?
can someone help ? i have a midterm tomorrow : /
inner acorn
I think this goes here?
In the process of deciding what I want to write about for my undergraduate project, and the Cayley-Dickson construction caught my eye (namely how after each application, the algebras lose a property). Was wondering if anyone had any reading (websites, books, videos) recommendations?
inner acorn
<@&286206848099549185>
patent horizon
better check the advanced mathematics server
inner acorn
there's an advanced mathematics server? o.o
patent horizon
yeaah, go to #old-network
inner acorn
ahhh, thankyou
bleak abyss
@delicate chasm do you have a decent reference at all for rep theory in the style of "semisimple k[G]-modules"?
delicate chasm
Hmmm
I learnt it for group cohomology
I'll have to check and get back to you, I'm fairly random about where I learn things but there were a few main references.
cloud crow
Do you know Serre's linear representations of finite groups?
delicate chasm
The second part of that is quite good
idle halo
can someone help me?
twin oriole
perhaps it wants you to input (4,4,9) since the eigenvalue 4 has multiplicity 2?
idle halo
i tried that its wrong also 😦
snow dew
Anyone has exercises on general arythmetic?
charred lance
- Show that t ∗ (x, y) := (e^tx, e^{−t}y) defines an action of the group (R, +, 0) on the set R^2 . What are the orbits and stabilisers of this action?
how do i find the orbits
or what does it mean to find the orbits i guess
If we let $$x=(x_1,x_2)$$ then $$G_x={(e^tx_1,e^{-t}x_2)|t\in\mathbb{R}}$$
mellow vaporBOT
charred lance
is that what it means to find the orbits?
fierce ferry
It means to find the image of the action for each element in the group
So fix a $$t\in \mathbb{R}$$ and look at $$(e^tx,e^{-t}y)$$. I would draw some pictures
mellow vaporBOT
charred lance
Thanks
charred lance
if $$v\in gGx g^-1$$ what does that mean?
mellow vaporBOT
charred lance
where Gx is an orbit
snow dew
The conjugates of every element of Gx
thorny slate
if you have a two sided action then it's just what it says, act on the left by g and on the right by g^-1
sick acorn
Let f:Z->Z be a homomorphism. Show that there exists a k in Z s.t. f(a) = ka for all a in Z. (Let k = f(1).)
Any advice? I should be able to do the other ones after figuring out how to approach existence here...
sick acorn
Ahhh okay that's what I was thinking. Need to see what to do about that more. I've been trying to figure out f(-1) * f(a) = a. (because of how f(1)^-1 = f(1^-1) = f(-1)
sick acorn
...Oops okay I see
earnest valley
Group or ring ?
snow dew
Oh okay, then you can add a hypothesis to go to R. There are many you can use, can you tell me two examples of hypothesis that will allow you to prove it on R?
@sick acorn
Hypothesis of f i mean
steep quest
Does anyone have any recommendations for notes/texts on 1) semigroup theory or 2) combinatorial group theory?
fierce ferry
@steep quest Howie's Fundamentals of Semigroup Theory
errant drum
I have a question that says M2(R) where R is a ring and M2 are 2x2 matrices?
Does it mean 2x2 matrices whose entries are from the ring R?
ocean marsh
Yes
stone fulcrum
Let x be an element that is in every maximal ideal of a ring A
Let y be an element of A.
Then xy is also a member of every maximal ideal.
1 - xy is not a member of any maximal ideal.
I'm stuck on that last step. I don't understand why 1 - xy isn't a member of any maximal ideal
... I'm also iffy on xy being a member of every maximal ideal.
cerulean rune
For the latter step: If a and 1 - a are both in an ideal, then so is 1 (and hence your ideal is just the whole ring A).
For the former step: this is just part of the definition of (right) ideal (if x is in an ideal so is xy for every y \in A).
delicate chasm
Proving the alternative definition of the Jacobson radical
stone fulcrum
Indeed.
x ∈ J(A) iff 1 - xy is a unit
delicate chasm
Ok, so, let $x$ be an element in every maximal ideal.\
Let $\mfk{m}$ be a maximal ideal of $A$.\
Since $x\in \mfk{m}$ we have $(x) \subseteq \mfk{m}$ and so $xy \in \mfk{m}$
stone fulcrum
@cerulean rune
I agree that xy is in an ideal. But I'm iffy on xy being in every maximal ideal
cloud walrusBOT
Puerøsola:
delicate chasm
Ideals are closed under multiplication action from the ring
If $x$ is in an ideal, $xy$ must be in the same ideal for any $y$ in the ring
cloud walrusBOT
Puerøsola:
stone fulcrum
Oh yeah, duh that makes sense
delicate chasm
👍
For the last step, what would happen if $1-xy$ \emph{was} in the ideal as well?
cloud walrusBOT
Puerøsola:
stone fulcrum
Thinking through it. I'm sure it's right in front of me, lel
delicate chasm
Haha, things can be obvious but at the same time not obvious in this subject :)
What do we know?
stone fulcrum
xy is in every maximal ideal.
delicate chasm
Yep
stone fulcrum
This is enough, alone, to imply that 1 - xy isn't in any of them
delicate chasm
And for the sake of contradiction, we are assuming we have a maximal ideal where $1-xy$ is hanging around as well
cloud walrusBOT
Puerøsola:
stone fulcrum
Assume 1 - xy is in a maximal ideal. Then we can prove xy isn't in one of them.
... But how? Lel
delicate chasm
Assume $1-xy$ and $xy$ are in the ideal
cloud walrusBOT
Puerøsola:
delicate chasm
Ideals are defined by being closed under multiplication from the ring, but also the additive group structure
stone fulcrum
Are they? Oh wow I think I forgot that
delicate chasm
That one is important to remember! They are closed under addition and they contain the 0
stone fulcrum
Then 1 must be in the ideal. But that's silly, since 1 is a unit
stone fulcrum
Makes sense! Thanks for reminding me of that
gentle pendant
lol being closed under addition is like the majority of the definition, defs don't forget that
delicate chasm
So, to write it out fully:\
Assume that $x$ is in every maximal ideal.\
By closure of maximal ideals under ring action, $xy$ is in every maximal ideal for every $y$ in the ring.\
Hence $1-xy$ cannot be in any maximal idea, because otherwise $1-xy$ and $xy$ are in the ideal, i.e. $1$ is in there, and the ideal must be the ring itself (and hence not maximal)\
Since every non-unit is contained in at least one maximal ideal, we have that $1-xy$ must be a unit for every $y$.
cloud walrusBOT
Puerøsola:
stone fulcrum
Makes much more sense!
gentle pendant
🐱 👍
delicate chasm
It's amazing, the exercises teach you a bit of Algebraic Geometry as well
gentle pendant
yeah the course had a view towards algebraic geometry
wasn't directly tied to that book
delicate chasm
Can be a bit hard to get through it though, and there are so many exercises
delicate chasm
My friend and I went through a large chunk of it to strengthen our CA last year, by our selves
I also like it :)
Most of the learning happens there
gentle pendant
start to get above that level and exercises in books becomes kinda rare
delicate chasm
Aye, I think the exercises really add to the learning though
gentle pendant
definitely but you don't have the time when writing research monographs
delicate chasm
I still sometimes use a CA result and go oh, that was in an excercise from A and M! xD
gentle pendant
you trust the reader to ask his own questions and answer them
delicate chasm
Aye
gentle pendant
yeah I can imagine
delicate chasm
Most undergrad and early grad level readers don't know how to ask the right questions though
gentle pendant
yeah
delicate chasm
I still don't, and I've been through a masters and an honours, not that they count for much
The idea of research questions is even scarier
gentle pendant
haha yep
stone fulcrum
I did abstract algebra a while ago from Fraleigh, and I did well with it. Unfortunately I've forgotten some, revised some, and am learning commutative algebra now.
delicate chasm
Especially since from time to time I will think something is true which isn't :P How can I set my sights on something that noone knows is true or not, in the vast unknowable scope of the unknown xD
gentle pendant
terrifying trying to prove X is true and then several months later realise it might be false
(and often walk away without being able to know either way)
delicate chasm
I did that in my honours thesis xD I was working with my supervisor on a result and three months in I had a meeting where he said "I don't think this is possible to do with the methods we have now, I'll have to think about it"
In the end I proved a very special case and left the full result as a conjecture xD
errant drum
Can I ask a hw question here?
delicate chasm
Is it abstract algebra related?
errant drum
Yeah
delicate chasm
Fire away then!!
errant drum
So Let M2(R) be the set of 2x2 matrices with entries from R. The question was to prove that M2(R) is commutative iff ab=0 for all a,b in R.
It's pretty easy to prove that if ab=0 then the matrices commute
I tried doing it the other way but I just have a bunch of equations in 8 variables which I don't think is the right way to do it
I said let A be a matrix with entries a1 a2 a3 and let B be a Matrix with entries b1,b2b3 and b4
Then I did AB and BA and tried to equate the entries but it doesn't really look like I'll get anywhere
R is a ring btw
delicate chasm
Hmm
I feel like you need to set up specialised matrices
which only commute when ab=0
errant drum
Oh of course
delicate chasm
Not necessarily trivial to construct the matrices though
Is R assumed commutative?
errant drum
No
delicate chasm
Heh, that's gnarly
I'm not used to non-commutative algebra
It has unit though I hope?
errant drum
I was thinking if it wasn't commutative then the matrix (a,0,0,0)(b,0,0,0)
That would work
But R being commutative kind of foils that
Also I haven't gotten on to units in my course
delicate chasm
Actually
errant drum
Could I divide it in to two cases?
delicate chasm
Those matrices $\bmqty{a & 0\ 0 & 0} \bmqty{b & 0 \ 0 & 0 }$ prove commutativity of $R$ I guess
cloud walrusBOT
Puerøsola:
errant drum
Oh yeah
delicate chasm
We don't know char 0 either do we
errant drum
Unfortunately not
delicate chasm
Rip
delicate chasm
I can get it to say $ab = -ab$ lol
cloud walrusBOT
Puerøsola:
north tide
I think (a, b; 0, 0) and (a, 0; b, 0) work
delicate chasm
Ok, so taking your general situation from earlier, you had
$$
\bmqty{a_1 & a_2\ a_3 & a_4} \bmqty{b_1 & b_2\ b_3 & b_4} = \bmqty{b_1 & b_2\ b_3 & b_4} \bmqty{a_1 & a_2\ a_3 & a_4}
$$
We just look at the first term of the equivalence,
$$
a_1b_1 + a_2 b_3 = b_1a_1 + b_2a_3
$$
Hence, since we showed commutativity,
$$
a_2 b_3 = b_2 a_3.
$$
So anything which makes $a_2 b_3 = ab$ and $b_2a_3 = 0$ will prove the result.
mellow vaporBOT
Rendering failed. Check your code. You can edit your existing message if needed.
cloud walrusBOT
Puerøsola:
Ok, so taking your general situation from earlier, you had
$$
\bmqty{a_1 & a_2\\ a_3 & a_4} \bmqty{b_1 & b_2\\ b_3 & b_4} = \bmqty{b_1 & b_2\\ b_3 & b_4} \bmqty{a_1 & a_2\\ a_3 & a_4}
$$
We just look at the first term of the equivalence,
$$
a_1b_1 + a_2 b_3 = b_1a_1 + b_2a_3
$$
Hence, since we showed commutativity,
$$
a_2 b_3 = b_2 a_3.
$$
So anything which makes $a_2 b_3 = ab$ and $b_2a_3 = 0$ will prove the result.
For instance we can choose $\bmqty{0 & 0\ b & 0} \bmqty{0 & a\ 0 & 0}$
cloud walrusBOT
Puerøsola:
errant drum
Yeah I was just going to say I think I found one
delicate chasm
Nice :)
errant drum
Thank you so much for your help
delicate chasm
No worries!
steep swallow
oh this is kinda neat
notice that sin(3pi-x) = sin(x)
so y=(3pi-x)sin(x) = (3pi-x)sin(3pi-x)
and also notice that it's an even function
so y = (x - 3pi)sin(x-3pi)
scarlet coyote
thank you!!
steep swallow
np
drowsy meteor
Given a set of all elements U in ring R and there being unity 1, how would someone prove U is closed under multiplication?
delicate chasm
We need a bit more info than that
Generally, take two arbitrary elements of U, apply the multiplication rule, then apply the test for membership of U
All I can really say without knowing more 🤷
drowsy meteor
Whoops I think I confused unity with multiplicative inverse
Given a set of all elements U in ring R which have a multiplicative inverse, how would someone prove U is closed under multiplication?*
errant drum
How many [3,3] cycles are in S6.
As in cycles of the form (123)(456). I know it's 5! divided by something but I'm not sure what?
fierce ferry
@errant drum there should be $$\frac{6!}{(2!) (3^2)}$$
mellow vaporBOT
fierce ferry
@drowsy meteor Take two elements $$a,b\in U$$ and show that $$(ab)$$ has a multiplicative inverse. Since $$a,b\in U$$ you have $$a^{-1},b^{-1}$$ existing and now just state what $$(ab)^{-1}$$ has to be.
mellow vaporBOT
drowsy meteor
Thanks so much!
errant drum
@fierce ferry
Whaaa? Do you think you could explain why?
fierce ferry
@errant drum Sure! The $$6!$$ is for the actual permutation since we have six slots to pick. The $$2!$$ comes from removing the double count created by $$(abc)(def) = (def)(abc)$$ since they are disjoint. Finally, the $$3^2$$ comes from having 3 choices for each of the first slot (or any fixed slot) of the 3-cycle, e.g. $$(abc)=(bca)=(cab)$$.
mellow vaporBOT
errant drum
@fierce ferry Ty that helps a lot. I struggle a lot with combinations and permutations
fierce ferry
keep at it...once you master those the next thing will give you issues...and the cycle continues!
wide moth
guys help me what is the answer to 1+1
snow dew
0
wide moth
thank
inner acorn
Hmm, how would I go about showing p(x) = 3x^6 - 125, is irreducible over the rationals?
Eisenstein's Criterion doesn't apply, nor can I immediately find any transformations p(ax + b) which satisfy Eisenstein's Criterion :c I don't want to check all the cases where p(x) = f(x)g(x) unless necessary o:
I posted in Abstract algebra vs Algebra, since it's a problem in Galois theory - but I'll move to Algebra if necessary :DD
thorny slate
you can write x = a/b coprime
and look at 3a^6 - 125b^6 on Z
or something
honestly you might have to check p = fg
:(
inner acorn
Hmm, how would I go about showing p(x) = 3x^6 - 125, is irreducible over the rationals?
<@&286206848099549185>
north tide
If you can use Gauss' lemma, then you can try to prove p(x) is irreducible over Z (integers)
inner acorn
I was trying to do that indirectly with Eisenstein's Criterion n.n;;
Otherwise the only way I'd know how to do it that way is by considering loads of cases.. qwq
stone fulcrum
What are its roots? They aren't hard to find
@inner acorn
Or, rational roots theorem.
inner acorn
@stone fulcrum I'm aware it has 2 real roots (not rational) & 4 complex roots, however what I'm struggling to show is that it's irreducible.. :c
And although C[x] is unique factorisation domain, and moreover Q[x] a subset of C[x], can't immediately see how I'd show 3x^6 - 125 = f(x)g(x) is a contradiction for f,g non units.
I know it's irreducible, I just don't know how to show it is
Also, despite having not proved the rational root theorem in lectures (doesn't seem difficult to show), that would only deduce f or g don't have degree 1 or 5
thorny slate
maybe factor it over R
and show the products of factors aren't in Q[x]
I guess this would amount to almost computing the galois group of the extensions kinda...
but it shouldn't be hard for this concrete case
inner acorn
I mean, I'd have to show the irreducible bit of the factorisation over R is indeed irreducible.. which I think would require basically the same work as if I were to just check over C
but checking all products of the factorisation over C... is like 2^6 - 7 multiplications x-x
thorny slate
no need to check the C ones, complex roots come in pairs
so those have to be together
so it's like 2^4 or something
you can check for the degree of each root over Q and simplify some arguments
etc
inner acorn
I think 3x^6 - 125 = (ax +b)(cx + d)(ex^2 + fx + g)(hx^2 + ix + j) has this kind of factorisation over R
thorny slate
yes
inner acorn
which would be 2^4 - 5 checks or something like that
thorny slate
in principle, yes, but you can check the degree of the roots -b/a and -d/c for example
oh, well, the degree of all of them over Q is 6
that doesn't help much lol
yeah just check
should be faster from how the roots look
you can for example check subsets of the factors including (ax+b)
and multiply those
since some factor will have ax+b
inner acorn
....I really feel like there's a trick I'm missing
thorny slate
probably
inner acorn
because breaking the problem into 11-12 cases, and arguing each one seems silly for 2 marks
inner acorn
The problem I'm actually tackling is, find the minimum polynomial of sqrt(5)/sixth_root(3) over the rationals
thorny slate
well its square is essentially cube_root(3)
and it's not in Q[cube_root(3)]
so it has degree 6
inner acorn
Hence why I used x^6 - 125/3
which I changed to 3x^6 - 125 so it's a polynomial over Z
thorny slate
do you follow
inner acorn
yes, I follow what you said
thorny slate
okay so your polynomial is the minimal one
but yeah I guess transforming it into that makes it look elementary
when it's not
inner acorn
I don't think I can use that argument n.n;;
thorny slate
how come?
have you seen field extensions yet?
let w = sqrt(5)/sixth_root(3)
note that w^2 = 5/cube_root(3) so that Q[w^2] = Q[cube_root(3)]
which has degree 3 over Q
inner acorn
yes, but all the examples of finding minimum polynomials have directly checked the irreducibility
thorny slate
now note that Q[w] is a proper extension of Q[w^2]
so it has degree 6 and not 3
well
maybe there's a trick
inner acorn
also, isn't Q[w] the smallest ring containing Q and w
and Q(w) the smallest field containing Q and w
??
thorny slate
Q[w] = Q(w)
inner acorn
true xD
inner acorn
@thorny slate yeah.. I'm still really struggling with this
thorny slate
Which part
inner acorn
I mean, don't know how to implement what you've suggested... so I've been trying to go the route of showing [Q(sqrt(5)/sixth_root(3)) : Q] = 6 and therefore the minimum polynomial has degree 6 etc...
thorny slate
Okay
chilly ocean
x = (5)^(1/2) * 3^(-1/6) ?
thorny slate
So split that with the intermediate extension Q[w^2]
inner acorn
and I've been playing around with Q(sixth_root(3), sqrt(5)) = Q(sqrt(5)/sixth_root(3)) but I don't think I can show that's true.. otherwise I'd use the tower law etc...
I don't know what you mean by that :c
inner acorn
yeah, I guessed it was
hence why I then started playing with Q(cube_root(3), sqrt(5)) = Q(sqrt(5)/sixth_root(3))
thorny slate
All you need to show is that rt5 /srt3 isnt in Q(crt3)
inner acorn
why?
thorny slate
Cuz then Q w is a proper extension of Q w^2
Which has degree 3
So it has to have degree 6
inner acorn
I don't think we've looked into proper extensions in lectures
either that or I've missed something
thorny slate
I just mean it isnt trivial
Like, it's strictly bigger
So the degree is higher than 3
And divides 6
So it is 6
thorny slate
Yeah
inner acorn
okay.. why does Q(w) being a proper extension of Q(w^2) mean the minimum polynomial is 6?
inner acorn
I mean, what does this have to do with the min poly?
thorny slate
The degree of minpoly is the degree of the extension
inner acorn
ohh
so because deg(min_poly) = [Q(w):Q] = [Q(w):Q(w^2)][Q(w^2):Q(w)] > [Q(w^2):Q] = 3
deg(min_poly) = 3k for some k =2,3,4,...
and since I've found a poly with degree 6, deg(min_poly) = 6 ??
thorny slate
yeah
inner acorn
which shows that 3x^6 - 125 is the minimum polynomial of sqrt(5)/sixth_root(3) over Q???
yay
thankyou<3
thorny slate
yw
inner acorn
xD some day I might figure this stuff out
weak sail
idle halo
cann someone help me with this
lone shard
the first three are linearly independent but the fourth is in the span of the others: if you label them a b c d then a - b = d
thorny slate
not really
lone shard
hmm?
bleak abyss
a-b isn't d
b-a gives {{0,1},{0,0}}. If you then add d to that you get {{0,0},{1,0}}. If you add those two matrices up and subtract that from b you get {{1,0},{0,0}}. And then if you take c-b you get the last one
So yeah it spans, because you can generate a basis
(In general there are systematic ways to figure this out but I was just mildly bored and this was quick)
lone shard
oh whoops that’s what i get for trying to do math right after i wake up lmfao
timber bay
Hey guys. I got a question i think im close to figuring out, I just need pointers
I know i can suppose f(x) = f(y) and see f(x-y) = 0
and from this, i can kind of gather that f(1) = 0 by doing f(x-y)f(1/(x-y)) = f((x-y)/(x-y)) = f(1/(x-y))*0 = 0
but i cant get to show that x must equal y
when f(1) =/= 0
timber bay
I think I have all the math I need. Just need to finish the logical steps. I see that if x=y, then we have f(0)=0 and if x=/=y then f(anything)=0
thorny slate
it's fair
it's going to work
but you are basically repeating the proof that in a field, any nonzero ideal is the whole field
you can directly use that fact and get a shorter proof
thorny slate
I see, then you do have to do this explicitly
your idea is exactly what you want, f(1) = 0 implies f = 0
your proof is already complete
if the function is not injective, then f(1) = 0 is what you have proved
this is logically equivalent to f(1) =/= 0 implies function injective
buoyant fox
hi
im reading Abstract Algebra: Theory and Applications (http://abstract.ups.edu/) and i'm having some trouble understanding equivalence relations
I feel like I have 25% of an understanding and i'm not sure how correct that understanding is
chilly ocean
If no one offers to help in 5 minutes, feel free to PM me
buoyant fox
sure thing thanks :)
if i had to articulate my interpretation, it seems like
given the cartesian square of a set X, the subset R which is an equivalence relation on X contains all elements of the cartesian square where... I suppose if you were to attempt to relate(?) one element of X to another element of X:
if the elements have the same value, they are related
if one element has the same value as another element, they are related
if one element has the same value as another element which has the same value as a third element, the first element and the third element are related
thorny slate
yeah that's the idea
consider the following relation on Z^2: two integers (a,b) are congruent if the remainder when divding them by 2 is the same
for example 3 ~ 7 because the remainder when dividing by 2 is 1 in each case
you can check that this is an equivalence relation
thorny slate
use morphisms Z[x] -> Q instead
whose kernel contains f
easier to interpret
then check what the condition means both ways
dense hull
So we can take a homomorphism like: φ(f(x)) = f(q) for a fixed q ∈ Q
The kernel (for a fixed q) is {f: f(q) = 0}
dense hull
Ok, that make sense I think
So if it has a root c, I would just let q = c in my thing above
and that homomorphism works
thorny slate
yes
thorny slate
assume you have a morphism Z[x]/f -> Q
this means you have a morphism Z[x] -> Q with f in the kernel
let c be the image of x under this morphism
you know that f(x) gets mapped to f(c) because morphism and to 0 because it's in the kernel
so f(c) = 0
errant drum
In a quotient group G/N. Can you still generate groups by taking some coset hN and generate a group with its powers?
thorny slate
yeah those are the cyclic groups
elements of G/N look like hN
so <hN> is just that
delicate chasm
then you take the grothy spectral sequence
cloud junco
Hey, all! Apparently, when I was late to lecture the other day, I missed the beginning of this topic and am not sure where to start... Particularly the latter sequence of symbols (in red)? Thanks in advance! I initially thought about a polar representation but idk what t and p are
thorny slate
looks like translation, rotation, reflection
buoyant fox
I don't understand these examples of equivalence relations
The transitivity proofs aren't making sense to me
How do you get from $\frac{p}{q} \sim \frac{r}{s} \Rightarrow ps = qr$ to
cloud walrusBOT
æshthetic:
buoyant fox
$psu = qru = qst$
cloud walrusBOT
æshthetic:
buoyant fox
Wow this is a nicer tex functionality than mathbot
Also maybe I don't know enough calculus for the second one because I didn't know
$f(x) - g(x) = c_1$
cloud walrusBOT
æshthetic:
jagged gate
you know integration ? @buoyant fox
buoyant fox
Not practically 
theoretically sure
jagged gate
$$f(x) \sim g(x) \iff f'(x)=g'(x)$$ $$f(x) \sim g(x) \iff f'(x)-g'(x)=0$$ $$\int [f(x)-g(x)]\mathrm{d}x = \int 0\mathrm{d}x$$
cloud walrusBOT
emeric75:
buoyant fox
Well this makes sense
jagged gate
$$f(x) \sim g(x)\iff\int [f'(x)-g'(x)]\mathrm{d}x = \int 0\mathrm{d}x$$
(here it comes)
jagged gate
oops i did a typo
buoyant fox
Oh wait
cloud walrusBOT
emeric75:
buoyant fox
So you have that constant of integration right
Is that what c_1 is
Since \int 0dx is 0?
0 + c = c
jagged gate
well you'd have constants of integration for f'(x)-g'(x) also
but it's englobed in c_1
buoyant fox
I didn't make the connection between f(x) - g(x) and integration hm
jagged gate
$$\int [f'(x)-g'(x)]\mathrm{d}x = \int f'(x)\mathrm{d}x - \int g'(x)\mathrm{d}x$$ ie linearity of integration
cloud walrusBOT
emeric75:
buoyant fox
Well this makes sense, but what does it imply?
jagged gate
then an antiderivative of f' is f and an antiderivative of g' is g
so you get $$\int [f'(x)-g'(x)]\mathrm{d}x = f(x) - g(x)$$
cloud walrusBOT
emeric75:
buoyant fox
Ohh
jagged gate
with some constants of integration
buoyant fox
So $$\int [f'(x) - g'(x)]\mathrm{d}x + c_1 = \int 0\mathrm{d}x + c_1 = c_1 = f(x) - g(x)$$?
cloud walrusBOT
æshthetic:
buoyant fox
(or is that how how constants of integration work)
jagged gate
well constants of integration appear after actually integrating
$$f(x)-g(x)+c_{fg} = c_0$$ $$ f(x)-g(x) = c_0 - c_{fg}$$
cloud walrusBOT
emeric75:
jagged gate
then let c_0-c_fg = c_1
jagged gate
the constant of integration from int(0dx)
cloud walrusBOT
æshthetic:
buoyant fox
f(x)-g(x) should be 0 right? 👁️
jagged gate
f(x)-g(x) should be some real constant
buoyant fox
Hmm
I guess that makes enough sense
Can't jump to the conclusion that since f' - g' = 0 that f - g = 0
jagged gate
"f'(x) - g'(x) = 0 that f(x) - g(x) = 0
it's not f(x)-g(x) = 0
c1 is not 0, it's a real constant (i literally copy-pasted that image above from your thing)
buoyant fox
Yeah I see
thorny slate
you are defining an equivalence relation in Z^2
buoyant fox
How do you extrapolate $$p/q \sim r/s \iff ps = qr$$ like that
buoyant fox
👀
cloud walrusBOT
æshthetic:
buoyant fox
@thorny slate I suppose that makes sense, but how do you get to that proof at the bottom?
What're they doing to go from the definition to psu = qru = qst
thorny slate
mutiply both equalities by u and q respectively
thorny slate
this is how rational numbers are defined by the way
thorny slate
well, you have to relate pu with qt
so you have to connect them using the equalities somehow
finite prairie
take em boys to school trigonometry
errant drum
\text {Im doing the final bit where it says suppose 'G is a finite group'}
\text {The quotient group G/N has cardinality m which is coprime to n}
\text {I'll definite HN to be the collection of cosets hN where h is in H}
\text{HN is not a subgroup of becuase HN has size n which is coprime to m (the size of the quotient group G/N}
\text {My thought was to consider } $\langle {hN} \rangle$ \text { and I know it generates a subgroup. My idea is to show it generates the coset N but I'm stuck any clues? }
cloud walrusBOT
Cosmicrays:
errant drum
chilly ocean
This sounds like mostly definitions. Perhaps try Googling.
errant drum
I've done the definitions bit the bit below is where im stuck
that's why i wrote the latex above
chilly ocean
Oh, sorry. On my screen the LaTeX is tiny, but your image is huge
I'm too lazy to brush up on group theory sorry :) Hopefully, someone else will show up
errant drum
Yeah I just realised but the LaTeX was automatic :/
chilly ocean
So all proper subgroups of G are also subgroups of some normal subgroup of G. Interesting
errant drum
Contingent on m&n being coprime I think
chilly ocean
Right. I should've added "under the specified conditions"
I should have said "every normal subgroup", but that's probably not true, hmmm
errant drum
<@&286206848099549185>
covert vector
@errant drum
since |H|=n
for any h \in H, h^n = 1
so that gives that (hN)ⁿ = (hⁿ)N = N
errant drum
@covert vector I'm still stuck could I have one more clue? I mean h is closed under the group operation but I can't do much if h^n = 1.
covert vector
order of hN divides n, because (hN)ⁿ = N
also since hN \in G, whose order is m
then order of hN divides m
so it divides gcd(n,m)=1
errant drum
Ohhhh. Damn I was trying to write an equation with h^n * x1 = x2. Where x1 and x2 are in n.
Ty
covert vector
np
chilly ocean
Could someone explain cosets to me? Just what they are and why them seem so powerful? Specifically why aH = H? Does this mean a left coset of a subgroup operating on the subgroup is equal to the subgroup? As far as I know a coset is just a partition of a subgroup but how is it that you can do all of this fancy stuff with them?
muted shoal
@chilly ocean sorry I can't help, but I'll just link this channel in case you've not seen it : https://www.youtube.com/playlist?list=PLAvgI3H-gclb_Xy7eTIXkkKt3KlV6gk9_
stone fulcrum
@chilly ocean
The important part is that if N is a normal subgroup, then its cosets form a group themselves. This gives you a way to form a smaller group using your group.
chilly ocean
@muted shoal thank you, I'll look into that.
@stone fulcrum So each coset of a normal subgroup is a group? Or if you combine all cosets of the normal subgroup it forms a group? Wouldn't that just be N again since cosets are just partitions?
stone fulcrum
Each coset can be considered a single element of a group
So this gives a way to put elements together such that, if you start in coset A, add an element in coset B, you get an element in coset C, no matter which elements of A, B, C you're working with
In that sense, A + B = C
chilly ocean
Oh okay interesting. That seems to be a group then, since when you perform an operation on an element in a collection of cosets, you stay within the collection of cosets.
stone fulcrum
Yus yus.
Modular arithmetic is the perfect example, since you can create these as quotient groups of Z.
1 + 2 = 3 (mod 5)
6 + 7 = 8 (mod 5)
Note that 1 and 6 are in the same coset
2 and 7 are in the same coset
3 and 8 are in the same coset
So those two lines are both A + B = C where A, B and C are cosets
muted shoal
Sorry to interrupt (hopefully it adds more than detracts) - but @stone fulcrum , when you mention modular arithmetic here, are cosets synonymous with equivalence classes?
chilly ocean
Ah okay that makes sense. I'm not sure I understand operations of cosets onto subgroups though. Particularly if a is a coset of a subgroup H, then aH = H. Is it because you're taking all cosets of H and saying that hey, these evenly partition H, so then the collection of cosets of a subgroup H is the subgroup itself? Maybe its a matter of notation confusion more than anything. I don't understand how a coset of a subgroup is equal to the subgroup itself (if thats what aH = H implies)
muted shoal
Perhaps having something concrete to work with might help :
coset 1 : 0 3 6 9 12 15
coset 2 : 1 4 7 10 13 16
coset 3 : 2 5 8 11 14 17
for mod-3
(obviously they extend further)
Also, I hope someone checks that what I'm saying here is correct
But the only subgroup is the first coset there, in that it's closed under addition
muted shoal
if you take 1 and 4 from coset 2,3 and add them then you get an element in coset 3
they're not closed in the way that coset 1 is
however, whichever you do add will always be in 1,2 or 3
chilly ocean
right i see that now. 1 + 4 is 5. And 5 mod 3 is 2 which is in coset 3
muted shoal
@faint mica these are referred to as equivalence classes if I remember correctly ( when It's mod arithmetic )
so I guess it's a fair bit less general than a group, but I find concrete examples useful
chilly ocean
okay so all of the cosets combined are closed under addition, which makes sense since they partition a larger group
faint mica
@muted shoal I love mod arithemtics but I never wrote anything here, so I guess you wanted to ping @chilly ocean instead~ 😋
chilly ocean
So what does aH = H mean? I think "a" is a coset of a subgroup "H". How does one coset equal to the entire subgroup? I get that all of the cosets combined are equivalent, and that one of the cosets are closed under the operation and therefore form a group
muted shoal
as for the operations onto subgroups, my understanding is pretty fuzzy but I seem to think that if you have c2 from coset-2 (above) and C3 is coset-3, then c2*C3 would just apply the group operation of c2 to all elements in C3
(which in the case of mod-3 would give you C1 )
@faint mica >.< yes, sorry
@chilly ocean note you say a is a coset of a subgroup H, and I'm hoping that the mod-3 example above is correct
by which I mean that could have been written as :
H : 0 3 6 9 12 15
coset 1 of H : 1 4 7 10 13 16
coset 2 of H : 2 5 8 11 14 17
Not sure why I'm looking at this instead of the coursework I'm meant to finish for tomorrow morning : ' )
chilly ocean
I thought H would be all of those numbers
muted shoal
yeah, but ignore thinking of N too much as it's not a group ( Z is a group wrt addition, there's no inverse in N )
chilly ocean
right, thats fine. H can be Z3 or Z mod 3
I thought cosets were partitions of H, so how does "coset 1 of H" contain elements not in H? Maybe I don't know what a coset is then.
faint mica
A coset is not a partition of the subgroup H though. H is a subgroup that lives in a bigger group (call it G): the subgroup H, together with its cosets, form a partition of the group G they live in, not of H itself.
muted shoal
@faint mica can this be made concrete wrt mod-3 there?
chilly ocean
I'm following. Wouldn't H plus all of its cosets contain all of the elements of G?
faint mica
Eeeh... well, let me use @muted shoal's example to explain this.
So... you have Z, the set of all integers, which is our group G. The notation I've seen here is kind of non-standard--but anyway, consider the subset of G made of all multiples of 3 (6, 9, 12, -3, -6, 0, etc.), that is usually called 3Z, but that doesn't matter since we can just call it H. It is easy to see that H is indeed a subgroup, but that's not important.
What's important is how you define the cosets of this subgroup. And you do it this way: