is it always?
#groups-rings-fields
mild laurel
mild laurel
Yeah okay, then what else is in the image
mild laurel
Yes
pure crest
the image is also a subspace
mild laurel
Yes
pure crest
so the inverse of f(1) should be in it
mild laurel
@pure crest thought of anything?
pure crest
i still havent been able to figure it out.
mild laurel
well, did you figure out how its even possible to have 5 elements in your image
if f(1), f(2), f(3),... are all in your image?
pure crest
well they must all map to 5 elements in Z_175
mild laurel
maybe think about how its possible for a subgroup of Z_175 to have 5 elements
mild laurel
Those are the same element of Z_175
pure crest
but they are two different elements in Z_245
mild laurel
Sure, but you're thinking about subgroups of Z_175
mild laurel
That's a subgroup yeah
pure crest
im really lost haha. i guess i should do some reviewing
mild laurel
Maybe think if that's the only such subgroup, or if there are more
Or think about how many group homomorphisms from Z_245 to Z_175 have the image as the subgroup that you just described
woven obsidian
In a euclidean domain is the decomposition a=bq+r, with d(r)<d(b) unique?
W.r.t a certain d-function
eager willow
here's article from stackexchange on the conditions for uniqueness https://math.stackexchange.com/questions/2171776/uniqueness-of-the-remainder-and-quotient-in-an-euclidean-domain
summary the decomposition is unique if and only if d(a+b) <= max(d(a), d(b)) for any a,b in the domain
woven obsidian
Thank you
steel owl
Hi, I've been struggling with this for a while now, I'm really confused. From what I understand, to prove that a group is a subgroup of another group you have to show the three rules of identity, closure and inverse
stone fulcrum
Want to start with i?
tribal pasture
Can somebody point out the least and necessary ring theory chapters (preferably in Dummit) I should cover before covering module theory (We are starting modules in my alg class and my previous class didnt introduce rings properly)
zealous mortar
I need a clue about how to approach this: Let $\omega = e^{2\pi i/23}$. Prove that 2 is not a prime in $\mathbb{Z} + \mathbb{Z}\omega+\mathbb{Z}\omega^2+\cdots + \mathbb{Z}\omega^{21}$.
Presumably, I just need to find $a, b$ such that $2 | ab$ but $2 \not| a$ and $2 \not | b$. But, how to find?
cloud walrusBOT
jm:
stone fulcrum
@steel owl
Okay well you want to go over it now?
marble bolt
@tribal pasture Chapter 7,8,9 is what we did in class before modules
marble bolt
Yea, it used concepts from the three previous chapters
zealous mortar
Maybe I should ask in number-theory. Hm.
cloud walrusBOT
nnutannep:
cloud walrusBOT
jm:
zealous mortar
got it.
mild laurel
I mean, if I contains m^v, then the radical of I contains m
bitter mauve
the part where i need to show I is primary
mild laurel
Should be able to do this from the definition of primary, take ab \in I, with b \notin sqrt(I) = m, and you want to show that a \in I
bitter mauve
oh
mild laurel
use the maximality to note that b, m generate the whole ring
bitter mauve
i dont understand
mild laurel
what don't you understand?
bitter mauve
sorry, 1 min
hmm.... i think rather than thinking in the sequence that the author gave, i should just note firstly that sqrt(I) is m and hence prime
and then work backward to show I primary
thanks
bitter mauve
question 4.3 how to say its neither open nor closed?
like its not a finite union of closed sets? idk how to demonstrate that in this question?
@bleak abyss
bleak abyss
I'm busy, please don't ping
mild laurel
I think you don't really need to
That they're just noting that its in general not open nor closed
bitter mauve
hmm... all his questions are like these
so i thought this might be a question itself
valid bridge
scarlet estuary
@mild laurel
valid bridge
i wanted to surprise you
scarlet estuary
to compensate
mild laurel
oh, okay this is relevant lmao
valid bridge
indeed it is
im guessing the proof involves thinking of n as the norm of some complex number a+b*sqrt(-2)
can't seem to figure out what the trick is
mild laurel
hm
I'm not immediately seeing it either
So usually for these or questions
you want to assume one of them doesn't happen, and show the other must always happen
valid bridge
ye
mild laurel
hm
okay
So if p = c^2 + 2d^2
then p = (c + d sqrt(-2))(c - d sqrt(-2))
i.e., p isn't prime anymore in the bigger ring
and you should be able to show the other direction, that if p isn't prime, then p = c^2 + 2d^2
So assume that p can't be written as c^2 + 2d^2 and work to show that p^2 divides n
from this hypothesis, like I was saying, you know that p stays prime
@valid bridge
valid bridge
so ur saying that if p can't be written as c^2+2d^2, then its prime in Z[sqrt(-2)]?
valid bridge
ok after i show that what happens
valid bridge
i through irreducible => prime but prime doesn't imply irreducible
mild laurel
wait
prime => irreducible in any domain
but irreducible => prime in a UFD
@valid bridge
valid bridge
oh
i was under the impression that prime was a stricter condition than irreducible
guess i was wrong
what happens after i prove that p is prime in Z[sqrt(-2)]
mild laurel
think about it buddy
valid bridge
this is why i hate mathematicians
bitter mauve
even i got it, dude.
valid bridge
are you a number theorist dude?
bitter mauve
we all are
valid bridge
on this blessed day
mild laurel
did you figure it out
valid bridge
i think so
if p divides n then p divides (a + b sqrt(-2)) or p divides (a - b sqrt(-2))
mild laurel
uh what
mild laurel
why
mild laurel
I mean, they show its a euclidean domain
valid bridge
you just apply norm to both sides right
valid bridge
alright thx for the helping hand
valid bridge
all euclidean domains are UFDs
mild laurel
they probably assume you know this lmao
valid bridge
you might be starting out with a harder textbook
steel owl
Hi, I know this is asking for a lot, but could someone help me work through one of the parts of this question (I have gaps in my knowledge). I really appreciate the helps, thanks in advance. ❤️
quiet cave
What have you already tried
steel owl
So, initially I wasn't sure what S3 is, I know understand what it tangibly represents, (or at least I think so)
Using a triangle to understand the different possible mappings/permutations (I know this is a rookie method)
So, firstly, I'm not sure what the subgroup or set H is referring to, I can see that it's defined but I don't actually know what g^3 = identity means
quiet cave
The permutation group $S_n$ is defined to contain precisely every permutation of a set of n elements, such as ${1, 2, ..., n}$
cloud walrusBOT
Rijinaru:
steel owl
but every third element in the subgroup is the identity element, is that what it means?
quiet cave
In particular, its elements are precisely the functions $\sigma:{1,2,\ldots,n}\to{1,2,\ldots,n}$ which are bijective
cloud walrusBOT
Rijinaru:
quiet cave
Now, it might not be that difficult to see what the identity is of, let's say, S_3
(hint: it's the identity function)
steel owl
yeah
quiet cave
And $g^3$ simply means $g$ composed with itself three times
steel owl
So this is just the function which maps (1,2,3) to (1,2,3)
cloud walrusBOT
Rijinaru:
steel owl
oh I see
quiet cave
So now you know what g^3 = e means
steel owl
yeah
I mean, is it wrong to use the transformations using a triangle to understand the mappings
quiet cave
It's only not wrong when n = 3
steel owl
Because when you deal with S_5 we have 5! permutations...
And I don't actually know the generalised method
We went over function composition earlier in an example with S_3
so if you're sayingthat g^3 = e, then that's referring to a 120 degrees clockwise rotation, right?
upbeat burrow
I’ve got a question about taking tensor products. I have $x = [x_1 x_2 x_3]$ and $y = [y_1 y_2 y_3]^T$. I’m asked to take $x\otimes x, x\otimes y, y\otimes x, y\otimes y$.
cloud walrusBOT
iamtim:
upbeat burrow
Oh my spaces didn’t enter properly in those vectors
I’m not exactly sure how to do these though. Would tensoring x with itself result in a length 9 vector?
And would tensoring x with y result in some sort of matrix?
A matrix of size 3x3
cloud walrusBOT
nnutannep:
bitter mauve
so i just need a hint for that
grave ivy
let P primary in B, A->B, Q contraction of P. then A/Q -> B/P is an inclusion, so every zero divisor in A/Q is nilpotent, so Q is primary
upper pivot
<@&268886789983436800>
covert vector
I gotchu
upper pivot
ty
latent anvil
Lol somehow this pinged me?
I'm very confused
I sometimes get notifications when people @ helpers in here too
latent anvil
No
????
Even weirder
Unrelated, what conditions on A in GL(n, C) guarantee that it's minimal polynomial is x^k - 1 for some k?
If A is finite order, its minimal polynomial will divide that
And all the eigenvalues will be |A|th roots of unity
I was going a representation theory problem and I thought this might be useful, but it seems too specific. I'm still curious though
grave ivy
diagonalizable, has all roots of unity as eigenvalues
grave ivy
yes
latent anvil
Oh "all roots of unity as eigenvalues" not "only roots of unity as eigenvalues"
As in all n-th roots?
latent anvil
Oh I think I'm being dumb. Do all entries along a diagonal matrix need to satisfy any polynomial the matrix satisfies? They do because all the matrix operations happen entrywise, right?
grave ivy
yes the eigenvalues are the roots of the minimal polynomial
latent anvil
Are you saying this is true for a diagonalizable matrix or just any matrix?
grave ivy
any
latent anvil
Oh sorry I did know this
grave ivy
you can see it from the jordan normal form but that's circular
latent anvil
Yeah no I remember how this works now
I was thinking about seeing it from rational canonical form
Because you get a bunch of polynomials which divide eachother
The biggest one is the minimal polynomial
latent anvil
And the product is the characteristic polynomial
Kk
Yeah
I'm not very good at linear algebra lol
clear obsidian
I guess then A^x is just the maximal multiplicative group you can get from A? Because the set on the right side I dont think is equal to Z/nZ (doesn't contain the class of 0 for example)?
fickle brook
A^× is the group of all invertible elements of A
clear obsidian
Thank you!
Is this A^x notation only when A is a group or just when (A, *) is a binary structure and the invertible elements form a group?
mild laurel
It's for rings
clear obsidian
ah rings ok interesting
mild laurel
The group of multiplicative inverses
placid solstice
y'all i'm so confused at this problem
(in Z_2[X])
i literally do not even know how to approach it
I know i want to show taht a solution exists for
X^15 = 1 + (X^4 + X + 1)y
in Z_2[X]
so I guess I could solve for
X^15 + 1 = (X^4 + X + 1)y
but that would involve finding the inverse of (X^4 + X + 1) in Z_2[X]... which I have no clue how to do
wait, fuck it, I can just long divide lol lol lol
mild laurel
Yes you can
old tendon
mild laurel
There are different ways to do it
Like you can note that X^4 = -X - 1
And use this to reduce down x^15
subtle granite
Seems like a very dumb question to ask but what do professors mean when they say something is “Well defined”?
Right now I’m doing some Algebra, proving that if R is a ring and I is an ideal of R, then R/I is a ring, known as the quotient ring with multiplication (I + a)(I + b) = I + ab for any a,b in R.
My professor starts the proof by saying “we need to check that this is well defined”. What does he actually mean when he says this? What do I actually have to show?
oblique river
the problem is that you want to be multiplying cosets
but in order to multiply the cosets, you just chose an element of the cosets
what if you chose different elements? could you get a different answer? that would be bad
you need to make sure that, for example, if I + a = I + a', then choosing a vs. a' isn't going to affect the final answer you get
i.e. (I+a)(I+b) = (I + a')(I+b)
said another way: you need to prove that if I+a = I+a' then I + ab = I + a'b
oblique river
👍
wind steeple
Hey I have a question, In my book we prove a property stating that a noetherian ring has a finite number of minimal prime ideals, but wtf, Z is a noetherian ring and pZ is a minimal prime ideal, then there is infinitely many minimal prime ideal. Where am I wrong ?
upper pivot
a minimal prime ideal is over some other ideal
i.e minimal prime ideals of I are prime ideals containing I that are prime and minimal with respect to inclusion
Z is a PID, so every ideal is (n). can (n) be contained in infinitely many (p)?
wind steeple
for me, a minimal prime ideal is a prime ideal I which for every other prime ideal J subset of I, J=I
and the book shows that for every I, the set of minimal prime ideals containing I is finite
upper pivot
yeah thats what i said basically
the "containing I"
is the important part here
for finiteness
upper pivot
hmm
well actually yeah Since Z is an ID
(0) itself is prime
and since every other ideal contains (0), they are not minimal
upper pivot
np
cursive temple
2k+2(-k) =0 and 2(-k) is also even, so 2(-k) is in E
tribal pasture
Can somebody tell why is R[x] commutative? I mean a0b0 would not necessarily equal b0a0
upper pivot
well the assumption is that R itself is commutative
usually books just assume rings are commutative unless explicitly stated otherwise
tribal pasture
oh I see. Thanks
tribal pasture
One more question, does a unit p(x) in R[x] means there exist a q(x) such that p(x)q(x)= q(x)p(x)=1 for all x ?
stone fulcrum
Yes
"unit" simply means "has a multiplicative inverse"
Units cannot be zero divisors
tribal pasture
I was confused about for all x or just one x
stone fulcrum
O
tribal pasture
But yes youre right going from the defintion of multiplicative inverse, it should be for all x
stone fulcrum
I wouldn't think about x as "a variable ready to accept a number" here. It's best thought of as an algebraic object
But for all x, if you must
tribal pasture
Thank you thank you!
tribal pasture
Wdym? Even my book refers to x as an indeterminate
grave ivy
lol
tribal pasture
Its Dummit tho
grave ivy
the word doesn't mean what you think it does
hot lake
a unit p in R[x] means there exist a q such that pq= qp=1
grave ivy
R[x] is not polynomial functions on R
mild laurel
The point is you shouldn't think about these polynomials as functions
hot lake
especially once you get to its field of fractions R(x)
grave ivy
the map from R[x] to polynomial functions on R is not injective in general
hot lake
yep, for example if R is finite you will have some issues
grave ivy
so don't
mild laurel
So maybe it's better to think about two polynomials as equal if all their coefficients are equal
oblique river
just chiming in to second that
hot lake
yes it's always nice to know the definitions
tribal pasture
the map from R[x] to polynomial functions on R is not injective in general
@grave ivy Why is this not true?
grave ivy
why would it be true?
oblique river
you should NOT confuse a polynomial p(x) with the induced function R --> R
hot lake
if R is finite there is only a finite number of functions R -> R
but R[x] is infinite because it has x, x²,x^3, x^4, ...
tribal pasture
Oh no I meant isnt there a correpondence between an element in R[x] and a function that maps from |R to |R?
grave ivy
read what people are saying
oblique river
yes. a polynomial in R[x] induced a map R --> R
but the statements "two polynomials are equal if they induce the same map" and "two polynomials are equal if they agree on every x-value" are false
(to be clear those two statements I wrote are equivalent to each other)
upper pivot
x^p-x and 0 induce the same map on F_p
but they are not the same poly
as an explicit example
grave ivy
your name gave me cancer
oblique river
to make it even more explicit: Consider R = Z/2Z = {0,1} and the polynomials p(x) = x and q(x) = x^2
they are different polys but induce the same function
oblique river
np
also @upper pivot your name is only true if you consider spec as a topological space but not as a scheme
so I would say that you are being
MISLEADING
upper pivot
smh
upper pivot
idk what schemes are
oblique river
smh
upper pivot
so its fine
grave ivy
if he had a true statement it wouldn't be a joke
oblique river
wait, so do you actually not know about schemes or are you trying to make a joke with a false statement
cuz i'm getting two different vibes here
hot lake
196884 is the only one with a true name
oblique river
oh gotcha. yeah that's a pretty cool fact actually :)
mild laurel
Yes
grave ivy
"all schemes are reduced"
tribal pasture
"All schemes are governmental"
stone fulcrum
Waht is scheme
grave ivy
??????????????
oblique river
here is my spiel about schemes and why you should care about them (just a reddit comment I wrote from a year ago) https://www.reddit.com/r/math/comments/a0dd6j/mathematicians_what_is_your_favourite_algebraic/eah33w3/?context=3
@upper pivot @stone fulcrum
cloud walrusBOT
Buncho Bananas:
oblique river
assuming you know what varieties are
stone fulcrum
Waht ore varieties
stone fulcrum
Okay I think I get the idea when you say that
tribal pasture
Nothing beats this tho https://www.reddit.com/r/math/comments/cuv8me/give_me_any_space_and_ill_tell_you_why_s1_is/
oblique river
that's... not entirely relevant here but thanks lol
grave ivy
??????????????????????????
stone fulcrum
S¹ is pretty lit
grave ivy
god /r/math is awful
stone fulcrum
I thank the math gods for this place often
mild laurel
YOURE ALEPH NOT HOLY SHIT
oblique river
who else would I be? :P
tribal pasture
@oblique river Ya welcome :B
mild laurel
Not a redditor isk
oblique river
haha
I don't really post as much as I used to though cuz I've been real busy lately
but I still lurk
also I didn't realize people actually recognized that name
well I guess I recognize people who post a lot on /r/math
mild laurel
You and functor7 have taught me a lot of math on reddit
oblique river
aw :)
I don't know who functor7 is irl
but I think they're some kind of junior faculty who does NT
I do know user jm691 irl, they occasionally answer NT questions as well
mild laurel
I recognize that name
vital night
upper pivot
hint is bezouts
tribal pasture
One question. Is R[x] just R[<x>] where <x> is a cyclic group of order infinite?
oblique river
no
I'm assuming that by R[<x>] you mean the group ring
x is invertible in R[<x>] but not in R[x]
tribal pasture
Ah so following is R[x] atleast a subring of R[<x>]?
oblique river
yes
tribal pasture
Perf. Thanks!
oblique river
if you're happy talking about "semigroup rings"
i.e. like a group ring but for a semigroup instead of a group
R[x] is the semigroup ring for the semigroup generated by x
(isomorphic to the natural numbers)
tribal pasture
Would the isomorphism still hold if my ring R is infinite?
oblique river
yes
it's not really a deep fact
look at the definition of group ring (or semigroup ring)
and the definition of polynomial ring
they are exactly the same
clear obsidian
so is there a more efficient way of checking for a group being Dedekind than just checking every subgroup?
oblique river
isn't there some classification of dedekind groups?
clear obsidian
I found one if the group order is finite https://projecteuclid.org/download/pdf_1/euclid.nmj/1118801610
then I have to contemplate noncyclic infinite groups (clearly since theres only 1 infinite cyclic group)
oblique river
oh oof infinite groups
https://en.wikipedia.org/wiki/Dedekind_group according to this, there is still a classification of sorts for infinite order
clear obsidian
I see the thing about George Miller there but thats Hamiltonian
clear obsidian
right
so that would only leave out abelian groups which are already classified, I see
oblique river
also i'm talking about the previous paragraph
every Hamiltonian group is a direct product of the form G = Q8 × B × D, where B is an elementary abelian 2-group, and D is a periodic abelian group with all elements of odd order.
idk if you are gonna get better than that
clear obsidian
I saw that, yeah thats true
so for Q8 itself I guess we just let B = D = {e} right?
ah it says 2-group though, so order 2
ack
oblique river
you're doing fine :)
clear obsidian
ty
tribal pasture
Say I have two rings R,S and i have to argue that they are not isomorphic. Would it be fine if I argue that for some ring homomorphism f, f(R) ≠ f(S)?
mild laurel
Uh what's K
tribal pasture
Sorry typo
gentle pendant
If f is a ring hom, what is its domain and target object?
tribal pasture
Okay so I have to prove that 2Z and 3Z are not iso. Now consider the canonical map from these rings to the Z/2Z. We know that the image for 2Z will be 0 (all are even numbers), whereas for 3Z, the image will be 0 and 1.
I now realise that domain is included in the def of homomorphism but surely this must be a valid criteria to identify when rings are not iso.
gentle pendant
(Well earlier you did not clarify that R and S were subrings of some ring, and f had this larger ring as domain. Given two arbitrary rings R and S your question was ill-posed.)
tribal pasture
I see, so the correct criteria would be to identify a larger ring K in which R and S might be subrings and then define a homomorphism from K and then argue that f(R) and f(S) are not equal in cardinality (if possible).
gentle pendant
Well I was just clarifying your question, you should investigate yourself whether this is a valid isomorphism test.
Suppose R and S are isomorphic, and try to prove their image under a hom from the larger ring must be equal or isomorphic or whatever you conjecture.
tribal pasture
I mean I will do that but do you think it holds?
So as to not spend my time proving a false conjecture
gentle pendant
the point is in your attempt to prove it, you might discover things that lead you to realise it is not actually true.
this is a very important part of learning math, asking yourself these questions rather than googlings or asking on discord.
tribal pasture
I just dont have the time 😅 to do all the investigation. This is just prereq reading I am supposed to be doing to actually catch up with module theory
But I understand where you are coming from.
gentle pendant
Okay, then consider the ring ZxZ, with standard operations in each component. If you fix either component to be 0 you get two subrings isomorphic to Z. On the other hand if you apply the homomorphism that just forgets the second component, you get?...
tribal pasture
oh... one gives me the trivial ring and the other Z. True true thanks.
gentle pendant
yep
counterexample hunting is also a good skill to build. I find it kind of fun, it's like you are trying to break things.
in some sense less stressful than trying to carefully build things.
tribal pasture
Yeah I just havent covered product rings yet so this counter didnt jump to mind
Although I strongly "feel" that there should be some condition under which the case I have given (image of 2Z, 3Z to Z/2Z from their canonical maps) is necessitated for iso, like there seems to be an easy generalisation for this case, might require using fundamental groups idk and Thanks for your help!
sharp sonnet
what do you not understand
scarlet estuary
do you know what "congruence modulo 4" means?
its just the set {0, 1, 2, 3} with addition modulo 4
(more formally speaking, its a set of equivalence classes, but blahblah)
so for example, 2 + 3 = 1
since 2 + 3 = 5, and 5 modulo 4 = 1
latent anvil
Is CRT true for rings without identity?
I dont see how to generalize the standard proof because we use coprimeness to get x + y = 1 when proving surjectivity
But I can't think of a counterexample (mostly because who cares about rings without 1)
latent anvil
also, what are the ideals in the ring (without 1) of sequences converging to 0? This if very big so maybe there's no complete classification, but families of ideals are also interesting
gentle pendant
You could talk about the subset of sequences that are o(f(n)) I guess for some f(n)->0
Eg, those that decay at rate 1/n or quicker.
latent anvil
That makes sense
gentle pendant
maybe they are even all essentially of that form, but I am too tired to think about that right now.
actually nah they probably aren't
latent anvil
You can require that terms are zero
gentle pendant
yeah, if you make the ideal defining condition more algebraic than analytic.
then you won't get the same kind of object.
interesting question though.
latent anvil
It's like the only nontrivial ring without 1 I know (also continuous/smooth functions on something with compact support)
But none of these seem like good candidates for a CRT counterexample
gentle pendant
I think I remember CRT being true with or without 1.
have a google, it'll surely be on MSE or something
(or I guess figure it out yourself is better exercise)
but I think it is true
latent anvil
So you have an obvious map φ : R -> R/I × R/J with kernel I cap J
Also I did think about it/google first
But I don't see why φ is surjective
The proof for the 1 case has you take (a+I, b+J), write 1 = x + y for x in I and y in J, and notice that ay + bx is a mod I and b mod J
But you can't pull that trick in the non-1ed case
latent anvil
Cool, ty
gentle pendant
np
charred forge
For some reason, today, I've been thinking about a math project I worked on back in college, related to if you have a committee with n members, each of which has some number of votes, and you specify how many votes are required to pass a particular motion. E.g., {4: 3, 2, 1} is a committee of 3 people where 4 votes are needed to get something done, the first member has 3 votes, the second member has 2 votes, and the third has 1 vote. What I've been trying to figure out today is if there's some algebraic structure around such "vote assignments" for committees of n members. I'm not even sure if I'm using the right language to describe the question...
scarlet estuary
i'm not sure what sort of structure you're looking for
it doesnt seem like theres any relation on your set?
like you could construct sets of "all possible vote distributions to pass a given motion"
but that isnt really structure
at least, not in the algebraic sense - it has some combinatorial connections but i dont think those would be particularly deep
fading wagon
You can get the set of all subsets of committees which can pass the motion
but other than that I don't know what you mean by "vote assignments"
charred forge
I'm trying to figure out how to make statements about assignments that are "adjacent", in the sense that {4:3, 2, 1} is "adjacent" to {5:3,2,1} and {4:3,3,1}
fading wagon
seems like it's just changing one value by 1
but what we know about changing number of votes to pass a motion, is that the new set of subsets of committee members that can pass it is either now a subset or a superset
same idea for making people's votes less or more powerful
scarlet estuary
Define two "committees" (or whatever you call them) to be "adjacent" if they differ by 1 at one "voter" or one "motion", and otherwise agree at every voter and motion.
is that what you mean?
charred forge
Hmm.
scarlet estuary
or do you mean like
have the same set of subsets of voters who can pass the motion?
actually wait nvm then {4:3,2,1} and {5:3,2,1} arent "adjacent"
since {3,1} passes the former but not the latter
charred forge
Maybe what would be more useful is experimenting with a notation like x_y, where x is the number of committee members, and y is the number of votes needed to pass a motion
Sorry, I don't even know what I'm trying to do myself, lol
scarlet estuary
well, when we talk about algebraic structures, we're usually more concerned with the properties and relations of the structure rather than the objects themselves
[this is sometimes called the "grothendieck view", but the idea far predates him]
if you don't have a clear notion for what sort of relationships you're looking at, i'm not sure it'll be that illustrative
charred forge
Ok, let me try this: are there any interesting structures formed out of, say, triplets of non-negative numbers?
scarlet estuary
sure, you can define a sort of "pseudo-vector space" with no additive inverses
i.e. a module over a semiring
IIRC there's some computer science applications or something
and a google search gives this paper which seems fairly interesting https://arxiv.org/pdf/1705.01075.pdf
but it kind of "cheats" by introducing further structure, a negation map
which acts somewhat like "the next best thing to" multiplication by -1 in a semiring, though loses the nice property that a - a = 0 (preventing it from being an actual ring)
charred forge
Ok, thanks, I'm gonna chew on this some more.
scarlet estuary
anyway, i havent seen much study of semimodules for their own sake
but the study of modules is very, very deep
charred forge
What's the difference between a module and semi-module?
scarlet estuary
a semimodule is a module over a semiring
whereas normally, modules are defined over rings
hence a semimodule may not have additive inverses
charred forge
Gotcha
scarlet estuary
a semimodule over $\bR_{\geq 0}$, for example, would not be a module over a ring
cloud walrusBOT
Namington:
scarlet estuary
since there's no additive inverses whatsoever (except for 0)
but every "true" module is also a semimodule
oh, we can also think of this in terms of convex cones, i believe
indeed, this may be a more natural setting since convex cones automatically imply being a subset of a vector space
and it's natural to think of your structure as a subset of R^3
but it's still a bit unclear what you're after specifically, so it's hard to give a straight answer.
still, these are objects that are very well-studied, due to applications to linear programming and to optimization
see the generalized Farkas' lemma
charred forge
Thank you for the interesting discussion, I appreciate it
What would you call R^3?
A vector space, I guess?
worthy geode
This is a vector space, in my opinion.
thorn flint
What's the difference between the Cayley Table and the group/multiplication table?
south coral
None?
thorn flint
Okay, I was not sure if there was any because my book only references multiplication/group tables but when I search online, all the results were about Cayley tables
somber bramble
does every abelian group G have a “basis” B with the following properties:
•every element g ∈ G can be written as a finite sum Σaᵢbᵢ, aᵢ∈ℤ, bᵢ∈B
•if we demand the sum to be in a normal form such that if deg(bᵢ)=n<∞, then aᵢ ∈ {0, …, n-1}, then this representation is unique
what I actually want to prove ultimately is the statement that for any abelian group G, there exists a free resolution of the form
0→F₁→F₀→G→0
and if the above statement holds, then I could choose F₀ = ℤ^B and F₁ = ℤ^Bₑ, where Bₑ ⊆ B is the set of elements with finite order. the second map is given in the obvious way by mapping aᵢ ↦ aᵢbᵢ, and the first does aᵢ ↦ deg(bᵢ)aᵢ
grave ivy
0->G is not what anyone would call a resolution
somber bramble
wdym?
grave ivy
you edited it out. does this work for Q and Q/Z?
somber bramble
yea that was just a typo
so if this worked for ℚ then that would be equivalent to stating that ℚ is free, as every element has infinite rank. according to stackexchange, ℚ is not free
grave ivy
what?
of course Q is not free
im asking if you sanity checked this for the obvious examples
that's the first thing you should do
somber bramble
well I don’t know the obvious examples
grave ivy
what made you think this worked?
somber bramble
not knowing better?
I’ve done relatively little algebra
I figured one could possibly adapt the proof that every vector space has a basis (using zorn’s lemma)
it was not at all obvious to me that ℚ it not free (simply because I had never even heard that statment before, as said, I’ve done very little algebra), but I’ve read a proof now
oh well, back to the drawing board
okay I looked up a proof and god I was overthinking this
wind steeple
Hey, I'm reading Grothendieck's Tohoku paper and I'm in the begining we he states axioms for infinite sums/product. He says that an abelian category satisfying 5) and 5*) is zero because A^(I) is isomorphic to A^I. I can't figure out why 5) and 5*) imply that A^(I) is isomorphic to A^i.
5) is the axiom that states that for every family there's a sum, the sum of any monomorphism is a monomorphism and for any increasing filtrant family (Ai), "sous-truc" (= sub-object) of C, B a subobject of C, then we have (sum Ai)inter B = sum (Ai inter B). 5*) is the dual one, I guess that it states that for an decreasing filtrant family (inter Ai) + B = inter (Ai + B)
thorn flint
Hi, I’m trying to make the group table for D10. Where r is the clockwise rotation 2pi/n and s is the reflection. Is this correct?
solemn rain
yea
thorn flint
My book uses $D_{2n}$. So D10 is just n = 5 right?
cloud walrusBOT
Mac:
solemn rain
yes
worthy geode
Oh, I see.
thorn flint
Yup
solemn rain
cool
try to ask as much as you can here lmfao xD
so i can like try to helpa nd learn as well
and learn*
worthy geode
It is the same notation, but my professor ended up using D5 for D10 and D4 for D8
solemn rain
yea
sometimes tricky af tho
cuz books like
contemporary bstract algebra gaillan
( very famous )
usues D_n
thorn flint
Oh wait is that still correct even though s is supposed to be the reflection through vertex 1 and origin?
Just saw it that it was defined like that in the book
How far are you in the book btw? @solemn rain
solemn rain
im pretty slow cuz school adn stuff
i should be doing chapter 3 now
im considering skipping the part of subgroups generated by subsets
idk why lmlao im lazy
you?
thorn flint
I think my table is wrong if s is defined as the line of symmetry through vertex 1 and origin
solemn rain
y
solemn rain
sr(s)=r^-1
thorn flint
r^-1 is e? or r^4?
fringe nexus
I'm trying to use group theory to show that x^2 = -1 mod p has exactly two solutions when p is 1 mod 4
I can easily show that it has at least one solution, but I'm kinda stuck there
We have that the group of units Z_p^x has order p-1 and is cyclic, so there is some a^(p-1) = 1 mod p. Now a^(p-1)/2 has to be -1, as it generates the whole group, so I have one solution
and (p-1) = 4n so (p-1)/2 is divisible by 2
fringe nexus
yea
sorry I didn't include that part
I think i can do it with the fact that if n| |G| then there are phi(n) elements of order n in G
where we pick our n to be 4
oh wait yea i just needed that part
mild laurel
If you can do it elementarily too, i.e if a^2 = b^2 (mod p), then (a-b)(a+b) = 0 (mod p)
p must divide one of the factors so either a = b (mod p) or a = -b (mod p)
But yeah, you can use the fact that cyclic groups have a unique subgroup of each order dividing the order of the cyclic groups
scarlet estuary
its just giving an example of two matrices that aren't equal.
woven delta
Lmao
stone fulcrum
For matricies A and B, it's usually the case that:
AB ≠ BA
scarlet estuary
that is to say, you can't "swap" the order of matrix multiplication and expect the same answer
even though you can do this for, say, real numbers
stone fulcrum
For any general ring multiplication isn't commutative.
xy ≠ yx
But most rings we work with are commutative
south coral
Something every ring must have, though, is an identity for both operations
And they are just saying that the identity for addition is the zero matrix
stone fulcrum
What a controversial thing to say
stone fulcrum
Not really haha. Some books don't define a mult identity
south coral
At least it has to have zero
stone fulcrum
But 99% of people use one anyway
slate brook
learn abstract algebra before linear algebra
tribal pasture
Why dont we require the ideals to be closed under addition by the elements of the ring just like we do it for multiplication?
Would I be correct in arguing that this is because we want R/I to have both addition and multiplicative structures, but since R is abelian under addition, thus I, being its subgroup, is necessarily normal and since quotient of abelian groups by abelian is also abelian, thus R/I is an abelian group under addition so no more structure on I is required to guarantee addition on R/I?
Laslty, is saying I is an ideal, equivalent to rI =Ir for all r in R (as analogous to the normal group definition in group theory)?
delicate bloom
well
the ideal contains 0
so if you're saying r in R and i in I we require r+i to be in I, we just recover the entire ring
bleak abyss
Yup
thorn flint
I’m trying to solve this question
But I’m stuck on the induction part
I have this
Sorry the question is the bottom image and my answer so far is the top image
Can someone help?
thorn flint
I tried to change the (ab//01)^m to the general expression and multiplied by (ab//01)^1
thorn flint
Didn't know any other ways to make it work for this case
scarlet estuary
sorry, i'm still confused
why can you just multiply by (a b \ 0 1) randomly and preserve equality?
like surely this is (a b \ 0 1)^(m+1) right?
so wouldnt this be (a b \ 0 1)^(m+2)?
thorn flint
Yup
scarlet estuary
so surely equality isnt preserved
thorn flint
Can I multiply both sides of the equality by (ab\01)^-1?
To get (ab\01)^m = (ab\01)^m
scarlet estuary
i mean im not sure why you dont just
expand $\begin{pmatrix}a^m&a^{m-k}b\cdots b\0&1}\end{pmatrix}\begin{pmatrix}a&b\0&1\end{pmatrix}$
cloud walrusBOT
Namington:
Compile Error! Click the 
reaction for details. (You may edit your message)
scarlet estuary
if my mental matrix multiplication is right, this should immediately give you the form required to complete the induction?
cloud walrusBOT
Namington:
thorn flint
I wasn't sure how to write it properly but I wanted to write that k keeps increasing up to m
scarlet estuary
so like, $\sum_{k=0}^{m}a^{k}b$?
thorn flint
from 1
thorn flint
Thanks!
cloud walrusBOT
Namington:
thorn flint
Btw how should I rewrite the expression more clearly?
scarlet estuary
changed the index since using n was a bit confusing
i'd probably write
$\begin{pmatrix}a&b\0&1\end{pmatrix}^n = \begin{pmatrix}a^n&\sum_{k=0}^{n-1}a^kb\0&1\end{pmatrix}$
cloud walrusBOT
Namington:
thorn flint
Haha I thought putting the sum in the matrix would look weird so I avoided that
scarlet estuary
and then, at the end of your induction, your top-right term would be $a^m b + \sum_{k=0}^{m-1}a^kb$
cloud walrusBOT
Namington:
scarlet estuary
which is of course equal to $\sum_{k=0}^{m} a^k b$
cloud walrusBOT
Namington:
scarlet estuary
as desired
but yeah, i understand that the "sum in the matrix" looks a bit weird
as long as its clear what you mean
thorn flint
Thanks alot!
cloud walrusBOT
Mac:
Compile Error! Click the reaction for details. (You may edit your message)
Mac:
thorn flint
Oh nevermind
leaden finch
uhh wym
scarlet estuary
it says R is a ring @fading wagon
aaanyway, have you tried folloiwng the hint?
it lays out the argument fairly explicitly
fading wagon
... yeah, it says R is a ring, but different authors may use "ring" to mean different things
scarlet estuary
i've never seen a ring be nonassociative
and you dont need commutativity for this result to hold
anyway, as a hint
fading wagon
yeah, but still it's just confusing what a ring is
scarlet estuary
you want to show that $ad = 0$ or $da = 0$ for some $d$
cloud walrusBOT
Namington:
scarlet estuary
you know that $(ab)c = 0$ or $c(ab) = 0$
cloud walrusBOT
Namington:
fading wagon
a nonzero d
scarlet estuary
can you use this, along with ring axioms, to get something of the form $a \cdot \text{something} = 0$?
cloud walrusBOT
Namington:
scarlet estuary
sorry, yeah, d must be nonzero
chilly ocean
@scarlet estuary do you know if there is an algorithm for that except trying every combination?
stone fulcrum
There's only one way to apply associativity to
(ab)c = 0
Once you do, you've got the answer
scarlet estuary
an algorithm for what?
stone fulcrum
Well, half of it
chilly ocean
For finding zero divisors, sorry for bursting in here just saw this is on topic
scarlet estuary
in the general case, i dont think so
but if you add any more structure whatsoever
theres generally a way to trivialize it
if you require e.g. polynomial factor rings
then the zero divisors are precisely the multiples of the factors of the polynomial
which gives you a very simple algorithm, naturally
clear obsidian
So if 2 groups are iso then their centers must be iso, correct?
ocean magnet
ye
clear obsidian
So I guess you just have to restrict the iso to the center of its domain
night pier
how do we show that (Zn , + , x ) is a field (? not sure what corps is in english) only when n is a prime number
mild laurel
field is right
look at the definition of a field
see which one Z_n satisfies always
night pier
confused
mild laurel
About what exactly?
night pier
honestly don't get how the bezout theorem connects to this
misty spade
anyone on now?
chilly ocean
how is $\frac{\frac{A}{C}}{\frac{B}{C}} \simeq \frac{A}{B}$ called, for groups $B \vartriangleleft A$, $C \vartriangleleft A$, $C \vartriangleleft B$?
cloud walrusBOT
tet:
upper pivot
you mean 3rd isomorphism theorem for groups?
is the name, if thats what ur looking for
chilly ocean
yes, thanks!
steel owl
Hi, just wanted to clarify something:
Does the group table of a group refer to a table of the results of all combinations of operations between any two elements in a group?
So the group table for a group H={e,a,b,ab} should contain 16 different results? (counting 7 with the identity element)
solemn rain
yea
steel owl
Right, gotcha
I'm confused with part of a question
sec
3 part ii
I don't get a^2 to be equal to b^2
upper pivot
and whats x^2
steel owl
wdym?
steel owl
yea
what does it mean for a group to be to be generated
I've got a brief reference to it in my notes but it isn't an explanation
Does that mean that every (product?) in the group table can be generally described?
sharp sonnet
a generating set for a group is a subset of the group such that any element in the group can be expressed as a finite combination of those elements
combination with respect to the group operation
steel owl
what do you mean by a "finite" combination? nvm I think I get it
so, the part where I'm asked to "deduce" the generating set of the group - how am I meant to "deduce" it, don't I just check all the possible products from the group table and see a pattern?
sharp sonnet
where in the question you posted a pic of is there even a mention of generating sets?
mild laurel
I mean, it tells you what should generate H
steel owl
Sorry for the late response, would appreciate a tag if you're responding
mild laurel
namely, a and b
steel owl
Namely, a and b?
Oh, lol
just saw your message popped up at the same time as I sent mine
didn't see the earlier message
yeah
It tells you, but how are you meant to "prove" that, if you know what I mean?
Like, don't you just see it directly from the group table?
mild laurel
What is the definition of elements generating a gropu
steel owl
that any product of the group table can be represented by a finite combination of the generating elements
So yeah I've shown that ab = ba, a^2 = e and b^2 = e
mild laurel
uh, what do you mean by product of the group table
steel owl
like, the result of the binary operation between any two elements in the group?
I don't know what else to call it
mild laurel
Yeah uh, this isn't the right definition of generate
Or well, its technically equivalent but this isn't the way you should think about it at all
steel owl
Okay, how should I think about it?
mild laurel
We've already told you
"a generating set for a group is a subset of the group such that any element in the group can be expressed as a finite combination of those elements"
steel owl
Okay, do you mind clarifying that a little?
isn't a generating set representing the entire group?
like, why do you refer to it as a subgroup of the group?
assuming you meant subgroup and not subset
steel owl
So, the generating set is a subset of the group representing the entire group
and has nothing to do with the "products" of the elements within the group
But, as you said those products would also lie within the group
mild laurel
I mean, it has things to do with the products of elements
finite combination of those elements is a product of elements
steel owl
yeah I meant that it's not in particular about the products of the elements
but about the elements themselves
though, correct me if I'm wrong, the products of the elements are the elements, since by defn the binary operation involving a group is closed?
mild laurel
sure
every element is the product of itself and the identity so every element is a product in some way
stone fulcrum
@steel owl
Heyo. A generating set is a subset of the group. The idea is that you can create any element in the group using only the generating set. Not sure if it was clear haha.
cobalt pilot
A bit of a basic question, but I'm trying to make sure I understand the way of writing an ideal made up of generators correctly. Let (x,y) be an ideal of R[x,y]. What types of polynomials are in (x,y)? Do all polynomials in (x,y) have to have both x and y elements or is it all polynomials with x elements, all with y elements and all with both?
I found a potential definition. If R is a ring and r,s \in R. Then the ideal generated by (r,s)={rx+sy | x,y \in R }. Is this correct? Also, that seems to to be the same as (r)+(s) in commutative rings.
hot lake
yes
if your ring is not commutative then you talk about left-ideals or right-ideals or double-sided-ideals (not sure about that exact term)
cobalt pilot
ok, thanks. I also assume that this notion can be extended to any number of generators. Also, x\in (x,y) since x=x+0*y since 0 in R, correct?
hot lake
yes
steel owl
@steel owl
Heyo. A generating set is a subset of the group. The idea is that you can create any element in the group using only the generating set. Not sure if it was clear haha.
@stone fulcrum Thank you, that really helped clarify. ❤️
cobalt pilot
thanks again
cobalt pilot
@chilly ocean I used to. I actaully have a degree in math, but it was some years ago and I've forgotten a lot of it. I'm trying to refresh my knowledge in commutative algebra since it was my worst course but also the one I found most interesting.
I work as a high school math teacher atm
golden pasture
^^^
cobalt pilot
I'm Norwegian and to teach high school here you need at least one year on each subject and have one subject that you specialized in, which is somewhat similar to master's degree in that subject. My main subject is math.
steel owl
Ouch really
stone fulcrum
When I look up the salary on Google, I see an average $60k a year. I think that's very good. Is it not trustworthy?
upper pivot
thats teachers who are well into their careers
even then thats a bit high i think
(starting is much less than 20 yrs in for example)
stone fulcrum
I just googled starting wages, and did see $30k. That's a bit sad
chilly ocean
I thnk so
slate brook
im going to cry this is too hard
mild laurel
do you know what FG is?
slate brook
$FG=\Bigg{\sum_{g\in G \text{ finite}}f_gg ~\Bigg|~ g\in G, f_g\in F\Bigg}$
cloud walrusBOT
Publius:
slate brook
this is the defn that i'm provided
so sort of finite linear comb. of g with "scalars" from F...
i think
mild laurel
yeah
and you should know what the operations on this set is
that make it into a ring
brisk granite
Sorry if I'm stupid, but why is this obvious?
mild laurel
What does degree of an extension mean?
mild laurel
@brisk granite
brisk granite
If we treat K as a vector space of F, then the degree is the size of the basis
I hope that makes sense
mild laurel
Yep that's right
and so why must any n + 1 elements of K be linearly dependent over F?
brisk granite
yea, that's my question
woven delta
Do you know linear algebra?
brisk granite
not really, the book gives a quick intro before we get here
woven delta
Lol
latent anvil
Is this for a class?
brisk granite
no
latent anvil
Then stop and learn linear algebra
woven delta
^
mild laurel
You really should learn linear algebra before you study any abstract algebra
woven delta
Is this herstein?
bleak abyss
Read Artin
brisk granite
uh ok
bleak abyss
Artin does linear algebra with algebra
latent anvil
In this specific case, any set of n + 1 vectors in an n dimensional space is linearly dependent
But you should learn linear algebra
harsh kettle
i'm working through artin now (amateur) and struggling after reaching more advanced rotation groups
i was wondering how long it took for the pros to understand it
delicate bloom
like a half a milisec probably, or maybe it was 1/3, that was so long ago now
onyx coyote
Is this a place for algorithms?
bleak abyss
Ugh the rotation groups
Uh in principle yeah it's algorithms are fair game, though I don't know how many people here (as opposed to say, the CS server in #old-network) are familiar with them
harsh kettle
I skipped it when I first read artin
@oblique river so i can proceed without a crystal clear understanding of that ?
real spoke
solemn rain
field theory needs linear algebra?
stone fulcrum
@brisk granite
I agree with "learn linear algebra before abstract"
To answer your question fully, a basis:
- is linearly independent
- spans the space fully
- has a certain amount of vectors, determined by the vector space
That last one is important, called the dimension of the space. If you add another vector to any basis, it won't be a basis anymore. It will no longer be linearly independent.
solemn rain
i been learning group theory from df
and uptill now the only lienar algebra i encounterd
like
high school amtrices basically
does later abstract algebra require linear algebra?
upper pivot
yeah. you definately want to know LA. the linear groups are p important depending on field
solemn rain
but like not too much LA
like the linear groups are like
SL or GL
you only like
want to know whats a determinant
and that det(ab) = det(a)det(b) ig
i dont think knowing vector spaces helped me much
idk im ignorant
upper pivot
thats like basic stuff
you definately want to know more LA for later on
vector spaces are important in stuff like field theory as an examle
gentle pendant
vector spaces are important pretty much everywhere
upper pivot
yeah
stone fulcrum
Once you get into field theory, vector spaces are important
upper pivot
field extensions are vectorspaces over the original field