yes
#groups-rings-fields
fickle brook
solar vessel
thanks ann
chilly ocean
Z_12 = Z mod 12
fickle brook
congratulations that's the most obtuse way to write $e^{i \pi}$ in existence
cloud walrusBOT
Ann:
fickle brook
а еще ты не в том канале постишь @mortal stream
somber rivet
Fk that C/d just looks stupid
mortal stream
fickle brook
если это просто мем, то #chill
если это серьезный вопрос, то в один из каналов #❓how-to-get-help
tranquil creek
I need some guidance on these sorts of equations
I don't understand how c was solved for
Or any of them for that matter
mild laurel
Do you understand why the lambdas exist?
mild laurel
They're not eigenvalues
tranquil creek
Ya I think they are used to denote sonething for mod 11 but I'm lost on how its solved
fossil vapor
for a relaxing Saturday evening, here's some 20's German University Life reflections
The man is most famous for this book, putting together innovations from Noether, Artin, etc.:
https://en.wikipedia.org/wiki/Moderne_Algebra
late sedge
now that ive finished abstract algebra i will be retiring from this channel
thank you all for your help
you guys have been great
onto Complex Anal
tranquil creek
Gl
wind cypress
I will start studying abstract algebra on 7th of January. Topic at the first course/module is rings and fields.
late sedge
strange you usually do groups before rings
fading wagon
I don't find it too strange, dealing with one operation might be easier than dealing with 2
jade sky
@fading wagon Eats, shoots, and leaves
fading wagon
@jade sky ??
jade sky
@fading wagon I think @late sedge meant to include a period after "strange", and it seems (from your response) that you read it instead as "strange that you usually do groups before rings"
late sedge
^
fading wagon
oh that explains it
late sedge
yea exactly @jade sky
golden pasture
does anyone happen to know of a construction of $R$ such that $R^m\cong R^n$ and $m\neq n$ for some $m,n$?
cloud walrusBOT
Ariana:
mild laurel
Are you looking for a ring R
golden pasture
yea
mild laurel
Just take countably many infinite direct product of Z
golden pasture
right oops
woven delta
Smh
golden pasture
brainlet moment
chilly ocean
How do I show 3 isn't prime in Z[sqrt(-5)]?
golden pasture
define a norm?
chilly ocean
sorry, isn't prime
golden pasture
try factorizing 6 in 2 ways
chilly ocean
ok, but is there a different way? Or why did you choose 6?
golden pasture
6 is the easiest option /shrug
chilly ocean
Is that sth you see after you do enough problems because of norm?
golden pasture
its like a really common example on why Z[sqrt(-5)] isnt a ufd
chilly ocean
k thx
golden pasture
3 is irreducible isnt it?
mild laurel
Oh yeah I'm dumb
dawn kiln
try playing around with f(x) = 1/x
thats not the isomorphism, but you may be able to tweak it to get the correct one @chilly ocean
and then remember that the check for f being an isomorphism (after you know its bijective) is f(a)f(b)=f(ab)
chilly ocean
I double checked and x + 1 is an isomorphism
dawn kiln
i might be being stupid here
but f(a)*f(b) = (a+1)*(b+1) = a + 1 + b + 1 + (a+1)(b+1)
= a + b + 2 + ab + a + b + 1
= ab + 2a + 2b + 3
whereas f(a*b) = f(a + b + ab) = a + b + ab + 1
@chilly ocean
warped bay
(a+1)*(b+1) = a + 1 + b + 1 + (a+1)(b+1)
🤔
it's a bijection, you just need to show that it's a homomorphism
i.e. f(a)f(b) = f(a*b)
make sure you did your foil right
also the operation * in G is different from the multiplication operation in R* [the set of nonzero real numbers]
that's what might be confusing you
cloud walrusBOT
fields!:
chilly ocean
If I have proved phi is a homomorphism, and have constructed f st f(phi(x)) = x do I need to prove f is a homomorphism separately?
warped bay
what are you trying to prove in the end? it's easy to prove that f is a homomorphism
chilly ocean
Phi is an isomorphism
warped bay
i don't see why you need f for that
chilly ocean
You need phi to be a bijection and the easiest way is to construct an inverse
warped bay
so you haven't proven yet that phi is a bijection
chilly ocean
If f the inverse of phi exists then f is a bijection
It's the quick way to show a set function is bijective
warped bay
and yes, you need to prove it separately
chilly ocean
Let's consider phi : A -> B just as a function between sets no group structure.
Let's show that a function f between sets f:B ->A exists such that f(phi(x)) = x
Then, just as a function between sets, phi is bijective
Now separately we throw away f and show phi is a homomorphism.
Because the definition of an isomorphism is a homomorphism that is also bijective as a set function, phi is an isomorphism.
warped bay
wait
you didn't even show that there exists a function f
in the end, i prefer showing that phi is bijective
chilly ocean
In my story I constructed such a set function f.
warped bay
that works i guess
chilly ocean
Thanks I needed to double check
lavish sigil
I'm trying to solve the following. Does anybody have any hints? I'm not sure to use the hint given in the book to be honest.
I mean, sure, assuming that $1/2\in A$, we have $(y+\frac{a}{2})^2-\frac{a^2}{4}+b=(z+\frac{c}{2})^2-\frac{c^2}{4}+d$. Not sure how that helps me though.
cloud walrusBOT
gustavn64:
jade sky
urban acorn
@chilly ocean f(phi(x)) = x doesn't guarentee a bijection, because it's just a one sided inverse
a left-inverse only guarentees injectivity
if you can show that phi(f(x)) = x also holds then it's a complete inverse
which guarentees bijectivity
lavish sigil
Rigorously, how do you prove that for an integral domain $A$ and a nonzero element $f\in A$, the ring $A[f^{-1}]$ is isomorphic to $A[x]/(xf-1)$, where $x$ is an indeterminate?
cloud walrusBOT
gustavn64:
lavish sigil
My guess is that I should look at the map $A[x]\to A[f^{-1}]$ which sends $x$ to $f^{-1}$, and prove that its kernel is $(xf-1)$, but this is precisely where I am stuck.
cloud walrusBOT
gustavn64:
chilly ocean
You could just use the universal property of localization on $A\mapsto A[x]/(xf-1)$ to induce the unique homomorphism of the localization at f which is trivially surjective and is also easy to see that the kernel is 0
cloud walrusBOT
gabe:
chilly ocean
Or galaxy brain proof is that quotienting out by xf-1 is setting it to 0 and therefore x = 1/f and by handwave we have our isomorphism
chilly ocean
Hi i have the following question:
I want to implement a bignum program and want to do it formally correct. For that I want to declare a few algebraic structures which I am not sure if they are mathematically correct:
Let $(\mbb{Z}^{\infty}, +, \cdot)$ be the countable infinite dimensional standardmodule over the ring $(\mbb Z, +, \cdot)$ such that the following requirements are met:
$\$1: $\forall (s, v_1,...,v_n,v_{n+1},..)\in \mbb{Z}^{\infty} : v_{n+1} \neq 0 \implies v_n = 2^{64} -1\$
2: $\forall (s, v_1,...) \in \mbb{Z}^\infty : s = 1 \vee s=0 \vee s=-1\$
3: $\phi : \mbb{Z}^\infty \rightarrow \mbb Z , (s,v1,..) \mapsto s\cdot (v_i)_{i\in\mbb N \setminus {0}}$ isomorphism
maiden ocean
The exercise in this book is asking me to prove that given a monoid A the set $A_L$ of left translations $a_L \colon x \to ax$ is the centralizer of the set of right translations (same thing but with xa) in the monoid of transformations of A but this seems flat out false? the transformation x -> $x^1, x1$, etc are not in $A_L$ but are clearly in the centralizer
cloud walrusBOT
hegel:
woven delta
Hegels
mild laurel
Rip
woven delta
Lol
chilly ocean
Now it's good
woven delta
@maiden ocean can you describe your transformation?
maiden ocean
wdym
woven delta
What do you mean x^1, x1
maiden ocean
as in, the translation from x to x1
chilly ocean
I'll ask later
woven delta
Don't the left and right multiplications have the same identity?
chilly ocean
Yeah
cloud walrusBOT
hegel:
chilly ocean
Is some element in Z[n] a unit iff its norm is 1?
A unit needs to have its norm = 1 right?
mild laurel
Prove this
chilly ocean
ok yeah I think I got it, cause norm of any element needs to be an integer in this ring, and if ab=1 then a and b are both units and N(1)=N(a)N(b) which menas both need to be 1
is that right?
mild laurel
Yes
potent birch
would there be an algorithm to solve linear systems of equations over an arbitrary ring? i was thinking about when i saw a post online when someone asked if they could solve a system of linear equations where the variables and coefficients are both square matrices.
there was some stuff i saw online for solving single linear equations over arbitrary rings so i could try reading those
mild laurel
Gauss Jordan requires inverses
Smith normal form only requires that your ring is a PID
potent birch
if you assume the matrices are nonsingular gauss jordan would still apply?
mild laurel
If I recall, in general there's not a ton you can do
topaz solar
Can def do some things, but rings are a lot less well behaved compared to fields
since
factorization
ideals
zero divisors
inverses don't exist
(or at least in general, of course)
potent birch
yeah, but for the specific case of equations where the equation is over nonsingular square matrices, it seems like there should at least be some way to see if there isn't a solution. I get that it isn't a PID because 0 divisors.
topaz solar
I mean, I don't know a way, but that doesn't mean a way doesn't exist
There might be something online about this/some article
OR, of course, working on it some yourself in your free time
(though I have no idea how hard this would be to do, so take that with a grain of salt)
ivory dust
What is generally covered in an undergraduate intro (abstract) algebra course? I know the definitions for a group, ring, field, and algebra as well as their motivation and have some basic intuition for them, but I imagine that I'm nowhere close to what I would have from completing a proper course.
So basically I know I'm missing a lot, I just don't know exactly what. Is it intuition from solving lots of problems? Knowing and being comfortable with relevant theorems? General mathematical maturity?
dawn kiln
@ivory dust how much algebra have you covered?
have you done anything like lagrange's theorem, symmetric groups, isomorphisms, homomorphisms, isomorphism theorems etc
delicate bloom
if you feel like you need to go through a course, maybe find a course on youtube or something and go through it with whatever textbook and assignments they give
dawn kiln
this is what's covered in cambridge's course
that would be an "upper bound" for what you would reasonably expect to cover in an intro course
ivory dust
Nothing really. I've just picked up on the basic definitions and intuitions since a lot of things reference them. Of what you mentioned, I know lagranges theorem, but not a proof. I know isomorphisms, and I've seen the definition of symmetry groups, but never really worked with them
Idk about homomorphisms but I imagine they're related to isomorphisms
dawn kiln
they're isomorphisms that aren't bijective
i'd say it would be reasonable to cover the first 6 chapters in this book
https://homepages.warwick.ac.uk/~maseap/teaching/aa/aanotes.pdf these notes are also pretty fun to read through, and good if you're self teaching @ivory dust
ivory dust
I'm not sure yet if I'll self teach or not. I have a long list of things I want to teach myself, and it really comes down to motivation and need
dawn kiln
as someone who has self-taught, i can recommend abstract algebra
its very satisfying to work through, and has plenty of cool results
ivory dust
Like I need to teach myself Java and E&M so I can test into more advanced courses in the fall
I might add Abstract algebra to the list idk. It doesn't seem that hard especially since I'm a lot more comfortable with the algebra/discrete side of math
Continuous things are scary
ivory dust
Idk yet
I know I really like math. Honestly I'm interested in physics for the same reason most people are: QM and relativity. I just want to learn the actual math. Idk how far I'll get though
dawn kiln
fair
i was tempted to try and learn GR over summer, then i realised that it wasn't happening with my level of maths lmao
esp given that im doing a comp sci degree lol
ivory dust
GR seems terrifying given what I've heard
dawn kiln
yep
compared with special rel which has fairly simple maths from what ive seen
GR seems like an entirely new playing field
ivory dust
Like special relativity just takes highschool algebra 2 and maybe a little linear
dawn kiln
anyways, with algebra, defo have a look at the warwick notes i linked
they're very readable, and will probably give you enough of a taste to see if you want to continue with it
ivory dust
Yeah I added it to my long list of "math texts I should read"
dawn kiln
👍
the other one is better if you are seriously getting into it
but worse as a taster
ivory dust
I downloaded both
potent birch
@ivory dust even if you like discrete things more, discrete and continuous structures are very often linked, one example being the prime number theorem.
ivory dust
Yeah I know I need to get more comfortable with continuous things
I imagine taking both real and complex analysis next semester will help with that
potent birch
i was going to try to learn about algebraic geometry, but when i started reading a free book about it, it got into the connections to differential geometry/complex analysis right away, so i realized i needed to learn about those first
or at least complex analysis
ivory dust
Let's move this to #math-discussion
potent birch
ok
lavish sigil
I'm trying to prove the following statement: let $f\in A:=k[x]$, where $k$ is a field, and let $B=k[x,y]/(y^2-f)$. Then $B$ is an integral domain if and only if $f$ has a square root in $A$. So this is equivalent to proving that $(y^2-f)$ is prime in $k[x,y]$ if and only if $f$ has a square root, and one direction is pretty easy: if $f$ has a square root $g\in A$, then $(y+g)(y-g)=y^2-f\in (y^2-f)$, despite none of the factors being in the ideal. But what about the other direction?
cloud walrusBOT
gustavn64:
lavish sigil
One idea I have is that if $gh\in (y^2-f)$, where $g,h\in k[x][y]\subseteq k(x)[y]$, we can write $gh=p\cdot (y^2-f)$ with $p\in k[x,y]$ and then the relation $\frac{g}{p}\cdot h=y^2-f$ holds in $k(x)[y]$.
cloud walrusBOT
gustavn64:
lavish sigil
But I'm not sure how to deduce a relation in $k[x,y]$ from this
cloud walrusBOT
gustavn64:
lavish sigil
oops, that's not true really, it should be in $k(x,y)$
cloud walrusBOT
gustavn64:
chilly ocean
$k[x,y]$ is a UFD, so $k[x,y]/(y^2 - f)$ is a domain iff $y^2 - f$ is a prime in $k[x,y]$ iff $y^2 - f$ is irreducible in $k[x,y]$. So if $k[x,y]/(y^2 - f)$ were not a domain, then $y^2 - f$ is reducible, i.e. there exists $a,b \in k[x,y]$ such that $y^2 - f = a(x,y)b(x,y)$ (so use the UFDness to kill that $p$ factor above) with $a(x,y),b(x,y)$ both non-units in $k[x,y]$. There are two choices: degree _y of $a$ = 0 and deg_y $b$ = 2 or deg_y $a$ = 1 = deg_y $b$. The former is ruled out because $y^2 - f$ has unit coefficient for $y^2$. In the latter case, say $a(x,y) = a_0(x) + a_1(x)y$ with $a_1(x) != 0$ in $k[x]$.then setting $y = -a_0 / a_1$ you have $f = (a_0/a_1)^2$, or $a_1^2 * f = a_0^2$. Use UFDness of $k[x]$ this time to deduce that $f = g^2$ for some $g \in k[x]$.
chilly ocean
actually in the last step, the factorization has to be of the form $y^2 - f(x) = (a_0 + y)(a_1 + y)$ for some $a_0, a_1 \in k[x]$ by comparing the coefficient of $y^2$, so now set $y = -a_0$ and you obtain (-a_0)^2 - f(x) = 0$, or $f(x) = (a_0)^2$ as desired.
cloud walrusBOT
hochs:
Compile Error! Click the 
reaction for details. (You may edit your message)
chilly ocean
so that should be slightly quicker
vestal snow
I got an A in Abstract Algebra
The professor was really good
50% of the grade was hw, 25% was take home final, and 25% was midterms
I was really discouraged from taking by the department it at the beginning of semester but I'm so glad I did
somber rivet
I can't wait to do it 🙂
vestal snow
Yeah man
I still have the second sequence I'm gonna take next semester
That's gonna go over rings, fields, vector spaces, modules, and galios theory
Though I'm kinda worried for next semester because I'm taking linear algebra, abstract Algebra, and topology
somber rivet
Oof i guess if you study hard enough, you'll be fine
I can't wait to do all that stuff, it sounds very cool
sharp sonnet
Abstract algebra and topology are like the best undergrad classes you can take
altho it's very weird not to take linear algebra earlier than those two 🤔
eager willow
depends on how ready you are to jump into abstract notions
topaz solar
weird but not the strangest
eager willow
some people feel ready after taking discrete math and learning sets, which is technically all you need to have a background in for AA and topology
linear algebra is just nice to see that you can put structure onto things that aren't really numbers anymore
sharp sonnet
linear algebra is a nice source of examples for abstract algebra
eager willow
yeah
urban acorn
when i learned linear algebra it was the "highest" math i knew
so i really liked it, and being able to put structure into things that aren't numbers was probably a big part of it
vestal snow
Yeah I could've taken linear algebra instead of abstract Algebra, but my advisor recommended that I have a better chance of getting research if I did Abstract Algebra
And by research I mean research in math
golden pasture
one could do abstract algebra without much lin alg
vestal snow
Yeah iirc you don't need any until you get to modules and vector spaces
jade sky
And even then you don't need to know linear algebra as a prerequisite. You'll be learning (a more general version of) it
coarse storm
Having a good understanding on matrix manipulation from linear algebra would give a good mental model though.
As examples to fall back on.
vestal snow
Is the linear algebra covered in Abstract Algebra more about vector spaces in general rather than vectors within those spaces?
jade sky
What are you thinking of when you say "about vectors within those spaces"?
Linear algebra is always about vector spaces (including subspaces and relations between them) and linear transformations between vector spaces
vestal snow
Like abstract Algebra would focus more on properties about vector spaces themselves, instead of focusing on things like Eigenvalues, orthogonality etc
jade sky
The latter is an more or less an inseparable part of linear algebra (inasmuch as linear transformations and operators are). But I guess if you're primarily focusing on those, then it's more appropriately called matrix theory or operator theory
vestal snow
What did you mean when you said a more general version of linear algebra
eager willow
I'd say that linear algebra is just a kind of abstract algebra. You can ask abstract algebra questions about vector spaces, linear transformations, isomorphisms, things like that and it's pretty clearly still just another flavor of abstract algebra, even though it's still a proper subject of linear algebra. But specific notions like eigenvalues and orthogonality are not going to be covered in abstract algebra, but perhaps just referenced instead.
chilly ocean
Orthogonality requires an inner product
jade sky
https://discordapp.com/channels/268882317391429632/496784958430380033/659451142614417470
Oh I was referring to this ↑
I meant you wouldn't technically need to know linear algebra beforehand in order to study modules and vector space. Modules are generalisations of vector spaces, and you'll be studying them in a different way than you would normally study linear algebra (because things are a lot simpler in the latter case). But many of results from module theory imply results you usually prove in vector spaces as well
chilly ocean
Eigenvalues are still mentioned in abstract algebra
Modules over a PID are generally nice
Especially if finitely generated
coarse storm
Linear algebra is usually about solving linear systems. It is a basic tool. Abstract algebra then comes along and says, "hey... linear algebra, you are making too many assumptions! I can strip you down to a bare minimum and still get the same result!"
chilly ocean
how lewd
coarse storm
😅
chilly ocean
I'll try using that line at a bar / pub next time
Linear algebra (at the right level of generality) is just the study of abelian categories
coarse storm
I am sure that clears things up a lot!
chilly ocean
The techniques you use with modules (resolutions, tensor product, hom) make sense in a general context
wind steeple
Hi everyone, can smb gives me a geometrical intuition for ideals in polynomials ring ? Actually, I really don't know how to deal with it, for instance I have to find a generating set for the ideal defined by {P in R[X,Y] st for all t in R, P(sin(t),cos(t)) = 0}. I have the intuition that P is cancelled on the circle, so I can consider the polynomial X²+Y²-1. But I don't know how to use it, I can look at the quotient ring R[X,Y]/(X²+Y²-1) but it isn't the circle, cl(X) and cl(Y) are coordinates for the circle but I don't see how would it be the whole circle.
glossy otter
a prison contained n amount of prisoners, if
2/7 of the prisoners are women
4/7 of the prisoners like the colo(u)r blue
there is 79% more men that like the colo(u)r blue then women.
can you find out how much men are in the prison.
(this question is just a joke i just made it up on the spot so it doesn't make sense)
just ignore my question if you want
since it's impossible (probably)
idk
also find out what n equals
now it's impossible
wind steeple
Idk why but I think that the ideal I've defined is generated by X²+Y²-1. If my quotient was the circle, I would have P=0 in this quotient, so P is in (X²+Y²-1).
But that's false, this isn't the circle x)
hot lake
are you expecting an ideal of R[X,Y] to literally be a subset of R² ?
wind steeple
no
hot lake
or a quotient ring to be a subset of R² ?
wind steeple
But I see R[X,Y]/(X²+Y²-1) as a ring containing reals where there's a curve like a circle
hot lake
I see R[x,y]/(x²+y²-1) as the ring of algebraic functions on the circle
wind steeple
what's an algebraic function ?
ooh
ok
just a polynomial function defined on the circle
hot lake
it's just a word I put so I could conveniently say that things like exponentials or discontinuous functions aren't algebraic functions
wind steeple
typically, it's ${P_{|S¹} s.t. P \in R[X,Y]}$ ?
cloud walrusBOT
Zak:
hot lake
I suppose there are bijections between the two, yes
wind steeple
assuming there's a good bijection which respects algebraic equations, could I prove with this that my ideal is generated by X²+Y²-1 ?
hot lake
well you have to prove that if P vanishes on the circle then it's a multiple of X²+Y²-1
wind steeple
that's what I'm trying to prove
by showing that the class of P is zero in R[X,Y]/(X²+Y²-1)
hot lake
and you could, I dunno, pick such a P and then look at a representative of its class of the form f(x) + yg(x)
and try to prove that f and g are both 0 ?
wind steeple
don't you mean f(x) + yg(x,y) ?
hot lake
no
wind steeple
oh
hot lake
that wouldn't be any helpful
wind steeple
ok that's a good representative ok
because Y² = -X²+1
oh ok, bc we have f(sin(t)) - cos(t)g(sin(t)) = 0 for all t, we have f(sin(t))² = (1-sin(t)^2)g(sin(t))², thus f² = (1-X²)g², f and g are necessarly zero bcp (1-X²) isn't a square. I think this works. Thank you.
chilly ocean
@wind steeple the ring you define is the ring of functions on the circle
As a subscheme of the plane
wind steeple
yup ok
chilly ocean
The nullstellensatz doesnt hold over R
wind steeple
and if I define R[X,Y]/(X²+Y²+1) ?
chilly ocean
The second ring is still the ring of functions on some subscheme of the plane
Plane as in R^2
Except this scheme has no R-points
wind steeple
idk what's a scheme x)
chilly ocean
It should have many C-points though
wind steeple
if the field is algebraic closed, this is exactly function over the curve with "the field"-points ?
chilly ocean
it is the function field of P^0. it can't be a "curve" over k because the function field of "curve" would have transcendental degree 1 over the base field
wind steeple
of what are you talking about ?
chilly ocean
@wind steeple of "if the field is algebraic closed, this is exactly function over the curve with "the field"-points ?"
"this" refers to the field?
wind steeple
if k is the field, I refered to k[X,Y]/(X²+Y²+1) by "this"
chilly ocean
then sure, the field of fractions of that is going to be the "function field" of the curve defined by x^2 + y^2 - 1.
x^2 + y^2 + 1
wind steeple
we were talking about for instance, R[X,Y]/(X²+Y²-1) which can be seen geometrically as polynomial functions on the circle. But R[X,Y]/(X²+Y²+1) has no geometrical point of view bcp there's no curves on R² defined by X²+Y²+1 = 0.
I was asking if this intuition is always present when the field is algebraic closed
why do you consider the field of fractions ?
chilly ocean
for birational classifications.
generally speaking it's easier to classify varieties up to birational maps (defined on some open dense subsets rather than the whole thing) and that's classified by looking at the function fields or the "field of fractions" of varieties
@wind steeple : Yes, one nice feature of going to algebraic closure is that you have nice intersection theory theorems (bezout's and its generalizations). one sacrifices a bit of intuition for that (as you have noted)
wind steeple
uh ok thanks. I don't really know about birational classifications and all the stuff aroud that. I'll probably read more about this once I finish reading my books x)
chilly ocean
btw $k[x,y]/(x^2 + y^2 + 1) \cong k[x,y]/(x^2 + y^2 - 1)$ for $k = \mathbb{C}$ but not for $k = \mathbb{R}$.
cloud walrusBOT
hochs:
wind steeple
What's your isomorphism ? P -> P(ix,iy) ?
chilly ocean
You can show that any smooth conic is isomorphic to $\mathbb{P}^1$
cloud walrusBOT
abstract nonsense:
chilly ocean
@wind steeple : That works, though there are other things that can work too.
tranquil creek
Anything I should practice and get used to before dealing with algebra? I find mod addition and multiplication needs to be brushed up. Also proofs
tribal pasture
Set theory
solar vessel
Set theory
potent lynx
Set theory
warped bay
matrices, depends on your course content
dawn kiln
Have you tried using homotopy type theory?
Edit: mb, I meant set theory @tranquil creek
tranquil creek
Any branch of set theory? How much of the basics should I know for an introduction to groups? A list of topics in it would help.
And thankyou already, I think you guys have pointed out a great weakness I have been aware of but havnt been able to address
urban acorn
personally, my knowledge in set theory isn't that deep but i get alone just fine with abstract algebra
chilly ocean
Set theory
coarse storm
Just the basic definitions. What defines a set. Order of a set. Bijection. May be cartesian products?
tribal pasture
And intersections unions, equivalence classes
tribal pasture
Although do practice on some element chasing proofs.
coarse storm
I think.. most of the things are simple enough to be covered in the course itself.
tribal pasture
They are simple once understood. But for a newcomer, the shift and the mindset to settle in requires some time
tribal pasture
thancc for support incompleteness guy
urban acorn
lmao
coarse storm
It does take time to adapt your thinking, yes. I still don't see why that cannot be done why learning group theory. It even provides a way to use those concepts from set theory.
tribal pasture
I mean it would just make the course more obscure. Instead of trying to connect say a symmetric group with something more material for more intuition, one would be struggling with what a set is. And even if one can recite the definition of a set by heart, the set-theoretic proof techniques should be studied in their own and requires some time.
coarse storm
I don't think your examples really help your point, but I think I see what you want to say about proof.
tranquil creek
I'll consider this dialogue in preparing for my upcoming course, ty
coarse storm
Well, I said what I did mostly because that was how I learnt group theory. I did struggle through it, but I have always just assumed it was part of the learning process.
Perhaps I would have had an easier time if there was a separate module.
chilly ocean
Guys I need help, I think my book made a mistake in one solution: How many divisors does $z=-3\left(1-i\right)^3\left(2+i\right)\left(2-i\right)^3$ have? (Working in Z[i], these are all irreducible already, just need to find the combinations)
I got 2^7, the answer seems to be 2^8
cloud walrusBOT
Godel:
chilly ocean
my solution was: divisors are all subsets of a set -3, (1-i), (1-i)^2, (2+i), (2-1), (2-1)^2 (6 elements) and then multiply by 2 (can be + or -)
wind steeple
multiply by 4
in Z[i], 1,-1,i and -i are invertible
your subset representation seems to work but it is not natural: hou would you do for,for instance, (1+i)^4 ?
chilly ocean
ohhh yeah true
wind steeple
The good way to count divisors with the factorized form is tu factor powers : the number of divisor here is (1+1)*(3+1)*(1+1)*(3+1)*4
chilly ocean
and (1+i)^4 doesnt divide
so it works fine
the subset representation
(1-i)^3 would be (1-i( and (1-i)^2
wind steeple
yes but you wouldn't do this representation if you were needed to sudy the number of divisors of (1+i)^4
chilly ocean
And how did you get +1 in each of these terms?
wind steeple
you can see divisors as elements of {0,...,v3(n)}x{0,...,v(1+i)(n)}x...
wind steeple
bc each prime divisor of a divisor of your number is a prime factor of your number
and the power must be less than the powers of your number
chilly ocean
yep
wind steeple
👍
urban acorn
okay what the fuck i am so confused
I assume $C_{13}$ is the cyclic group of order 13.
cloud walrusBOT
Intel:
urban acorn
But that would mean $C_{13} \cross C_{13}$ is of order 169
cloud walrusBOT
Intel:
urban acorn
which clearly does not divide $|S_{27}| = 27!$
cloud walrusBOT
Intel:
urban acorn
27 factorial is divisible by 13 exactly once, right?
wind steeple
13*26 = 13*13*2 divides 27!
urban acorn
i am, actually
chilly ocean
why doesnt 2-2i sqrt3 have unique factorization in Z[sqrt(-3)]
obviously one way is 2(1-i sqrt3)
whats the second one
chilly ocean
yeah its 16
only way is 4 * 4
maybe I forgot 1 case of +/- sign
but its gotta be sth +/- 1 +/- isqrt3 times the same thing
?
wind steeple
-(1+isqrt(3))^2
chilly ocean
btw, if I want to show that there is unique factorization do I have to like check all the possibilites in +/- signs?
and just calculate and show only 1 of thsoe works
im asking cuz its very boring
chilly ocean
if I want to check factorization of element x in lets say Z[i] I check its norm and find elements a,b,c,d such that x = (a +bi)(c+di) or N(x) =(a^2 +b^2)(c^2 + d^2)
and I deduce (a^2 + b^2) has to be lets say 9
that gives me few possibilities: a=3 and b=-3 , both positive etc.
there will be few cases of that
golden pasture
yea you prob have to check al of them
chilly ocean
yeah thats what I was thinking, I remember reading some properties but that worked only in gaussian integers, but in any ring probly need to calculate all
wind steeple
I thnik nobody will never ask you to check taht
chilly ocean
And if I find 2 of those factorizations this way would need to check if they are conjugated or whatver the word in eglish for it is
wind steeple
it's very annoying
I would only recall that if N(x) is prime then x is irreductible and x in invertible iff N(x)=1
wind steeple
it's works in all rings
in all number rings
uh maybe you didn't study number rings yet
chilly ocean
OK what if we had 40+40i how many divisors does it have in Z[i]
if Im notmistaken its factorization is (2-i)(2+i)(1+i)* 2^3
answer is 112
2^3 can be reduced further
but its (1-i)^3 (1+i)^4 (2-i)(2+i)
just noticed (1+i)= i(1-i) but still not sure how they got 112
plain sequoia
Has anyone in here by any chance a physical softcover copy of Eisenbud’s commutative algebra book
It’s sale on springer so I can get the softcover for like 10$ to accompany A&M but I heard that the softcover binding for it is terrible as the book is so thick
Does anyone in here have any experience with that?
potent birch
i'd figure as long as you're careful with it, it shouldn't tear apart, though a crease in the binding is probably unavoidable
maiden ocean
So I'm trying to prove that if G1 and G2 are simple groups and G is a proper normal subgroup of G1 x G2, G is isomorphic to G1 or G2
i feel pretty lost
my first intuition was projective map or something but :/
raw moth
consider {h : (h, g2) in G for some g2 in G2} @maiden ocean
oh wait that's probably projective map
it would make sense for it to be at least
maiden ocean
yea
raw moth
coolcool
maiden ocean
also i just checked our mutual servers and we have the same first name 
raw moth
lol nice
had to check that was a server where it was my actual first name but it is
oh wait hmm the problem statement is slightly wrong
G also needs to be non-trivial
solar vessel
first name doxed
raw moth
it doesn't count if you go to uni with me smh
maiden ocean
i definitely do not go to uni with you 
solar vessel
you successfully both got wooshed
maiden ocean
raw moth
I knew what you actually meant :^)
solar vessel
my detection is getting dull sorry
solar vessel
yes that is a word
maiden ocean
no its not
cloud walrusBOT
gustavn64:
solemn hollow
Which page does Dummit and Foote introduce index, I've been looking for 15 mins
urban acorn
i think it's in chapter 3 section 2
chilly ocean
Does an isomorphism between $(\mbb{Z}^2,+)$ and $(\mbb Z, +)$ exist?
cloud walrusBOT
N/𝔄:
chilly ocean
I know a bijection exists since $\mbb Z\sim\mbb N\sim\mbb{N}^2\sim\mbb{Z}^2$
cloud walrusBOT
N/𝔄:
stone fulcrum
Z is cyclic, Z² is not. There's no iso between them
chilly ocean
Alternatively if you tensor both with Q, they they are not isomorphic as vector spaces @chilly ocean
That would make them a module, would it not?
Sorry I'm quite unsure about the word tensor, maybe it's an english thing
Or my course isn't this far yet
@chilly ocean
chilly ocean
Wow an understandable explanation from wikipedia
dry sleet
Usually I just skip down to the definition section, but to each their own.
chilly ocean
Under abstract tensor product?
dry sleet
yes.
chilly ocean
So what does "tensor both with Q" mean?
Q as a vector space over Q?
Also since Z is a ring that would make it a module over Z
Or a group with addition
dry sleet
Z, Q, and Z \otimes Q all satisfy the axioms of many algebraic structures. We usually focus on the highest structure they satisfy and call them that.
chilly ocean
Which one?
dry sleet
They are all vector spaces over Q.
dry sleet
Yeah. I don't actually know a nice argument about those two vector spaces to show they are different, I personally like the cyclic (generated by a single element) argument better.
chilly ocean
Hmm it can probably be shown that they have a basis of differing cardinality
But thanks for your time
woven delta
The arguments in this case for Z modules and Q modules are the same
Tensoring with Q doesn't make it easier or harder
chilly ocean
I meant take Z \otimes Q and Z^2 \otimes Q as vector spaces
Over Q
They have different dimension
Alternatively you can find an endomorphism of Z^2 that is not multiplication by a scalar
woven delta
Z and Z^2 have different rank as Z modules
plain sequoia
Did you end up buying any book @bleak abyss? In that case, which one? 
bleak abyss
Fulton and Harris
plain sequoia
Fulton is the book qbout algebraic curves?
Or sorry
Are they coauthors?
Which book?
Nevermind, it is the representation theory book I guess
still rapids
if i have a polynomial ring what are the conditions on the evaluation map p(x) such that it's 1-1
chilly ocean
It has to be surjective and injective
still rapids
@chilly ocean lol wut
dry sleet
You gotta say a little more about the evaluation map. \phi: R \times \mathbb{R} \rightarrow \mathbb{R} ? this is the normal evaluation. My guess is that you're asking about \phi_{p(x)} : \mathbb{R} \rightarrow \mathbb{R}. I suppose you want some statement like, "all the coefficients of p(x) are positive." You should think about how it can fail to be 1-1 (for instance x^2), and how to avoid that circumstance.
main quiver
Just trying to get second opinions, but how is the harvard course on abstract algebra from the 2000s?
uncut girder
Pretty good
main quiver
e🅱️ ic
golden pasture
https://discordapp.com/channels/268882317391429632/576511723574525962/660200016606003231
anyone have any clue if theres some alternative instead evaluating trashy binoms
fading wagon
this doesn't look like AA, unless those are commutators
wait how does this work multiplying numbers by a and b
are a, b matrices?
golden pasture
a and b are arbitrary elements
oops forgot to mention, yes they are commutators, [a,b]=ab-ba
fading wagon
what arbitrary elements? Do you use a pair (number, group element)?
can I just treat them like matrices?
golden pasture
ah prob wasnt clear a,b are 2 elements of a ring
fading wagon
which ring, any ring?
golden pasture
yup
fading wagon
noncommutative ring huh
$a^{(0)}=a, a^{(1)}=ab-ba, a^{(2)}=(ab-ba)b-b(ab-ba)=abb-bab-bab+bba$
cloud walrusBOT
Element118:
fading wagon
hmm, all these expressions are $b^xab^y$
cloud walrusBOT
Element118:
fading wagon
okay so that's where all of that is coming from
$a^{(k)}=\sum_{i=0}^{i=k}{k \choose i}(-1)^ib^iab^{k-i}$
cloud walrusBOT
Element118:
coarse storm
I guess? { x²∈ℤₚ : x∈ℤₚ } would be well-defined, and you can remove the zero if you want. I am probably missing the point of your question. Are you looking for quadratic residues?
chilly ocean
yes
I know there will be (p-1)/2 of them
and if k is one then p-k too, are there any other interesting properties that help me find them?
coarse storm
Closed under product. Legendre symbols might be of interest.
stone fulcrum
I believe the squares form a subgroup themselves
chilly ocean
yeah @coarse storm im trying to study legendre stuff actually thats why im asking
coarse storm
I don't really see an actual question 😅
stone fulcrum
"Is there an explicit way to write the squares?"
No. But there's a lot of tips that help
coarse storm
Oh sorry, I sort of ignored that one, because... well, I felt it was a little broad.
chilly ocean
mb
stone fulcrum
Still a very good question under research
chilly ocean
Like, I'm reading this old intro NT books and its so cool. Wrote in like 1940s, there's a lot of elementary approach, a lot of interesting results that are quite elementary to prove as well
And in my NT class we use a lot of group theory to get some results which is sometimes confusing for me as I'm pretty bad at it
And book's at times funny, cuz many things hes talking about being unproven are in fact now proven. And the biggest prime number known at the time was like 2^3000 or sth
stone fulcrum
Fermat's last theorem I'm sure gets a mention somewhere
chilly ocean
yep, being unproven for few hundred yrs
stone fulcrum
My advice is get better at group theory lol. Elementary number theory can be better learned as abelian group theory instead
Because there's nothing special about the integers at this level
chilly ocean
I mean, I know the basics, but like it gets hard to compute and remember all the facts from group theory. fortunately most of the ones we use are just some trivial facts
'Because there's nothing special about the integers at this level' what do you mean
There's a lot of interesting things you can prove without using group theory
coarse storm
We are looking elementary abelian groups. So they are by definition elementary group theory~!
A bad pun to start the decade!
urban acorn
I've seen that one but it's not the kind of question I'm looking for
it is related to group theory but it's not very group theoretic if you know what I mean
lavish sigil
Why can $k[X,Y]/(XY)$ be considered as a subring of $k[X]\times k[Y]$? Here's exercise 1.3:
cloud walrusBOT
gustavn64:
lavish sigil
chilly ocean
It's not
The exercise asks you to prove that every element has a representative of the given form (easy, just wipe out all mixed monomials)
clear obsidian
Maybe this is a soft question, but is it considered "more standard" if we define a ring as already having unital multiplication? I first read about rings as being only semigroups multiplicatively, but the wiki page states a ring as being defined with unity.
delicate bloom
no, not even the ring of even integers would satisfy that standard
idk maybe I actually do implicitly use that standard you describe now that you mention it, I would just call this an ideal and not a ring, meh idk it's not important to me I guess
shrewd halo
@clear obsidian it’s a mess. I’ve seen “unitary ring” for people that think a ring doesn’t need one and are specifying, I’ve seen “rng” for people that think a ring does need one and want to name the other thing
coarse storm
I think it depends on the field/area you are studying, department convention, or the lecturer's area of research if you are in a course. Same question can also come up for commutativity.
urban acorn
non-associative rings are a thing apparently
dawn kiln
i have literally never seen associativity dropped in a ring
just use a magma or some shit
or a near-ring, whatever the equivalent is for rings
topaz solar
So long as you're consistent and your convention is clear it's fine imo
but if you drop associativity i'm going to throw you into the p i t
topaz solar
yes
chilly ocean
I have this problem about ideals in gaussian integers: Given two elements a,b in Z[i] check if aZ[i] + bZ[i] is a maximal ideal. My question is whats the usual strategy to solve these kind of problems?
I was thinking, but not sure if this is true, that this ideal is isomorphic to GCD(a,b)Z[i], maybe then it would be easier for me, but not sure if thats true
maybe that works only in PIDs?
wind steeple
Z[i] is a PID
chilly ocean
I know, just asking if my reasoning works
and what would be needed for that to work
or if theres another approach
wind steeple
it works
chilly ocean
(my idea was to write is as one ideal and check if Z/gcdZ is a field, not sure if thats the easiest approach)
wind parrot
you could also use that an ideal $(d) \subset \mathbb Z[i]$ is maximal iff $(d)$ is prime and not zero iff $d$ is a prime element, and there is a pretty easy characterization of the primes of $\mathbb Z[i]$
cloud walrusBOT
killer_memestar:
chilly ocean
ok I got gcd of those 2 ideals is (2+i)
smoky briar
gcd works for gcd domains
which is stronger than PID
and slightly weaker than euclidean
chilly ocean
yep
good to see you sir
I have to show that if d is not a square and a+b*sqrt{d} is irreducible in Z[sqrt{d}] then so is a - b*sqrt{d}
any hint?
mild laurel
Prove the contrapositive
mild laurel
What do elements of Z[sqrt(d)] look like
chilly ocean
(a+b *sqrt(d))
for a and b integers
ok I think I got it but not sure, I'll latex it
$x- y \sqrt{g} = \left(a + b \sqrt{g}\right) \left(c + d \sqrt{g} \right) \implies ac +bdg = x, bc + ad = -y \implies x + y\sqrt{g} = \left(-a + b \sqrt{g}\right)\left(-c + d \sqrt{g}\right)$
cloud walrusBOT
Godel:
chilly ocean
but idk there's probably a better way
chilly ocean
How do I calculate the inner automorphisms of a group
you conjugate group elements
That gives me the number of possible inner isomorphisms, but a few of those will be the same.
idk what you mean, you have a fixed element, call it a, and to calculate the inner automorhpism you just do a^-1 x a for all x in G right
Listing all functions of the form g^{-1}xg (there are |G| such functions) might give functions that are defined by distinct g's but nonetheless are the same.
Oh wait I know what you're saying now
no wait I misunderstood your question I think, dont listen to me
chilly ocean
is $I = 3i \mathbb Z \left[i\right]$ a prime ideal?
cloud walrusBOT
Godel:
chilly ocean
in Z[i] ofc
Like I tried to show Z[i]/3i is field/integral domain but not sure how
maybe theres a different way
mild laurel
Think about 3 * i in Z[I]/3i
wind steeple
3i = 3*i
chilly ocean
3i is 0 in this ring
I don't know how to describe cosets of this
maybe then I would be able to find an isomorphism
like I thought that its isomorphic to Z/3Z but I have a feeling it was a faulty logic
bleak abyss
@mild laurel the issue is 3 = 3i(-i)
But there is something to note here, namely that the ideal generated by 3i is the ideal generated by 3
So this ring is just Z[i]/(3). But Z[i] is Z[x]/(x^2+1)
So Z[i]/(3) = Z[x]/(3,x^2+1) = (Z/3[x])/(x^2+1)
So now you just need to find out whether x^2 + 1 is irreducible mod 3 or not. Turns out it is
Just check that 0, 1, and 2 aren't roots
So actually this ideal is prime, in fact maximal
(To be fair in Z[i] all non-zero prime ideals are maximal)
The quotient is gonna be a finite field of order 9
@chilly ocean
chilly ocean
So Z[i]/(3) = Z[x]/(3,x^2+1) = (Z/3[x])/(x^2+1) - can u explain how are those true?
is this just some iso theorems?
I dont see it quite yet, but thx
magic owl
Godel you can reason about them even without high level theorems
magic owl
Its like relations and generators are important
bleak abyss
Well okay I don't really believe in coset reps
magic owl
But the coeffs cant be multiples of 3!
So your choices for coeff are 0,1,2
But thats just the polynomial ring over Z/3
bleak abyss
Come up with a map outta Z[x] such that 3 and x^2 + 1 are in the kernel
chilly ocean
they even cant be multiples of 3
magic owl
Yeah thats my point
bleak abyss
Lmao
magic owl
Youre secretly working mod 3
bleak abyss
You got factorial'd Max
chilly ocean
ok I think I see it now, I'll try to formalize tomorrow
anyways thanks for help guys
magic owl
Polynomials rings are so cute
mild laurel
Yeah whoops my bad
magic owl
V fun ti work with
bleak abyss
Tru
chilly ocean
yeah I like rings
magic owl
They feel like, really nice, all of their data is very explicit
And they represent cool stuff
bleak abyss
Dem universal properties doe
magic owl
Whooo dont get me started
bleak abyss
But yeah actually speaking of cosets
I prob mentioned this here before
But in my algebra class in office hours
Someone had a proof of CRT
bleak abyss
Which was super explicit on cosets
But it worked
And my prof was like holy shit
We had it as a pset problem and we were meant to do it the usual way
R->(R/I)\times (R/J)
I\cap J is in the kernel
But the way it was written was backwards, I guess to provide a slight mental roadblock or something
And the guy was like okay
I'm gonna map (x+I,y+J)
To their intersection
As sets
And I will never get over that for the rest of my life
bleak abyss
Yee
topaz solar
I mean, sometimes I go way more explicit than necessary
like i just quickly tried to throw down CRT and i took up like ~1/3 of a page
*also looked up what the actual statement was because i'm a brainlet :^)
took me way too long and too much looking at the original integer version tbh I gotta step it up
woven delta
Lol not knowing a statement doesn't make you a brainlet
Pretending you knew the statement while not knowing it makes you a brainlet
Tbh I forget the statement of things all the time, but it's fine usually
Because you can just look shit up
topaz solar
not knowing a statement doesn't make you a brainlet
CRT is lowkey kinda common
and i've literally never seen the actual statement afaik
(not counting looking it up a moment ago)
bleak abyss
When I was in high school I didn't know what a ring was lol
golden pasture
smh
chilly ocean
Hey, I have a small doubt
When defining a normal subgroup,
they say that gH = Hg
do they mean that left coset is equal to right coset of the subgroup
or do they mean that each element of H does this?
(in which case wouldn't it just be the centre?)
solemn hollow
left coset equal to right coset
chilly ocean
oh, thanks a lot!!
solemn rain
im bad with permutation groups
slate brook
thanks for letting us know
chilly ocean
how do I prove that if R=S/I then R/J = S/(I+J) where S,R rings, I,J ideals
because I dont believe Z[i]/(3) = Z[x]/(3,x^2+1) = (Z/3[x])/(x^2+1)
worthy kindle
how do you even make sense out of I+J if I is an ideal of S and J is an ideal of (?) S/I ?
chilly ocean
true
chilly ocean
It is almost true though
If $\pi: S \twoheadrightarrow R$ is a surjective morphism of rings, $I = \ker \pi$ and $J$ another ideal of $S$, then $S/(I+J) \simeq R/\pi(J)$
cloud walrusBOT
trex:
chilly ocean
as a corollary, if $K$ is an ideal of $R$, we have $K = \pi(\pi^{-1}(K))$ (this is just restating surjectivity), and therefore $R/J \simeq S/(I+\pi^{-1}(K))$
cloud walrusBOT
trex:
chilly ocean
Basically, the only identity that makes sense also happens to be true here. The reason is very simple: We have a surjective composition $\pi': S \overset{\pi}{\twoheadrightarrow} R \twoheadrightarrow R/\pi(J)$. There remains to prove that $\ker \pi' = I+J$.
We check directly that $I + J \subset \ker \pi'$.
Now, if $x \in \ker \pi'$, then $\pi(x) \in \pi(J)$. This means that there exists $y \in J$ such that $\pi(x) = \pi(y)$. In other words $x - y \in I$. Hence $x \in J + I$
cloud walrusBOT
trex:
chilly ocean
Ok that makes sense!!! Thanks
waxen iron
If the order of a group is prime, can I say it's the only one (excluding isomorphisms)?
chilly ocean
Are $\mathbb R\left[x\right] / \left(x^2+3\right) \mathbb R[x] $ and $ \mathbb R\left[x\right] / \left(x^2+2\right) \mathbb R[x] $ isomoprhic?
cloud walrusBOT
Godel:
waxen iron
Oh, of course, from Lagrange's we know it's cyclic, and if it's cyclic it must be the only one
Many thanks!
fickle brook
@chilly ocean they are both iso to C
chilly ocean
thats what I thought but how do I prove it
Like I thought that elements of each are some polynomials of the form ax+b so there's R^2 possibilites which is isomoprhic to C
but how do I show that the elements of those quotient rings are like the possible remainders after division?
is that obvious enough?
ok thats obvious
lone shard
what groups are $\Delta(N)$ and $\Sigma(N)$? they're in this paper im reading but they dont give a name. from context, theyre discrete groups
cloud walrusBOT
chilly ocean
Not sure here but sometimes $\Delta$ refers to the simple roots of a root system of in a cartan subalgebra of a lie algebra of a lie group
cloud walrusBOT
N/𝔄:
chilly ocean
That's the only connection to groups that comes to mind for me
wind steeple
Whats N ?
urban acorn
Can someone explain to me why use $\pi^{-1}$?
cloud walrusBOT
Intel:
cloud walrusBOT
Intel:
hot lake
they probably want phi(pi) ° phi(sigma) = phi(pi ° sigma)
chilly canyon
Hello,
If I have $[\mathfrak{S}_3:G]=3$,
how to prove that $G\cong\mathfrak{S}_2$ with a simple argument ?
cloud walrusBOT
Matplotlib:
quiet cave
gusty plinth
Is there a way to assign a unique identifier to a member of a group?
I'm specifically looking at permutation groups (Rubik's cube)
solemn rain
what does unique identifier mean
gusty plinth
Attempt at a formal definition:
given a permutation group with cardinality n, can I assign an identifier, x_n, to all elements 1-n such that all x_i are unique?
effectively give an identifier to every element of the group
For example the permutation group of order 3 ex. (1, 2, 3) has 8 elements
Could I uniquely identify and compute this identifier easily for all 8 elements?
And could I generalize this to groups of larger order (specifically, I'm thinking about n=48) and still remain easily computable?
mild laurel
You might be interested in group presentations
gusty plinth
damn, I'm already reaching my limits in understand of group theory lol. Time to burrow down the wikipedia rabbit hole...
This does seem to do what I need I think
I guess I could also just hash the element too?
ex:
the more I fiddle with Rubik's cube related programs the more I wihs I'd paid more attention in Group theory
smoky briar
this isn't a group theory question
you're asking if we can give names to objects
which is a very weird question
chilly ocean
lone shard
@wind steeple a natural number
ah, i found this (in case anyone cared what the answer to my question is) https://arxiv.org/abs/1101.2308
ivory dust
Cycle notation bothers me
Like I much prefer the normal permutation notation
But like omit the top row since it's always just 1234...
So like the "shift by 1" permutation in S_4 would be (2341)
Not (1234)
And like "flip 1 and 3 would be (3214) not (13)
It feels much more intuitive to me
For composition of permutations
delicate bloom
you'll grow out of it
delicate bloom
multiplying group elements is just easier, you read through the cycles and write where it lands
plus shorter to write
ivory dust
It certainly isn't easier for me now
Like I keep messing up
But I imagine that'll just change with practice
delicate bloom
I can show some example maybe you don't know how it works or something
ivory dust
I know how it works
Like you just apply the cycles in order
I think I'm tripping myself up because part of my brain wants to be using the other notation
shrewd halo
@ivory dust conjugating a permutation by another is equivalent to performing the permutation on the cycle notation of what’s getting conjugated
Which is awesome
mild laurel
There's a lot of information that you can glean from the cycle notation that isn't immediately obvious from writing it the way you'd want it to
Like the order of the element, its conjugacy class, whether its even or odd etc
ivory dust
Wait so let's say I have (1234) and (13)
And I apply (13) to (1234)
I get (23) (14)
But if I do it your way @shrewd halo I would get (3214)
Which if you "break it down the middle" is equivalent to (23) (14)
But if you don't it's not the same
shrewd halo
If you conjugate 1234 by 13 you do get 3214, as expected
ivory dust
Oh so it's the other way around
shrewd halo
Correct, the thing conjugating is what gets applied to the cycle notation of the thing getting conjugated
ivory dust
And you can't apply it the other way around
Since there is no 3rd or 4th place to reference in (13)
Hm?
shrewd halo
You can apply it the other way around, nothing should change
ivory dust
If you apply (1234) to (13) you get (14)(23)
shrewd halo
Not if you conjugate it
ivory dust
Define conjugate
shrewd halo
g conjugating h would be ghg’
ivory dust
Oh
shrewd halo
Where g’ is g’s inverse
ivory dust
Right
Wait so what if you conjugate (1234) with (13)
Like that's not the same as multiplying by (13) right?
Actually I'm dumb
Nevermind
Yeah that works out
(13)(1234)(13) = (3214)
woven cloak
Sorry if this is the wrong place to ask but I couldn't find anywhere about books, I am taking an abstract algebra class in uni and the text book is trash, nothing is explained, everything is rushed and the teacher is not helping either, I reached a chapter about groups and I am strugelling with it so can anyone recommend any book that explains stuff like groups and rings well?
stone fulcrum
Fraleigh is a rather slow and wordy book, perfect for someone first learning.
But it's a bit surface level and if you're looking for advanced group theory, dummit and foote is good too
@woven cloak
My suggestion is to use the books to be sure you are looking at everything, but use YouTube for what you don't understand
woven cloak
@stone fulcrum Do you think dummit and foote would be good for a first timer?
a friend has it and I can borrow it if I want, I don't need mcuh tbh, my text book has a chapter about groups and another about rings so I don't need many advanced stuff
stone fulcrum
It's thick. I get sidetracked reading it. It may be too in depth for your course
But it's accepted as a good algebra textbook, it has a "reference like" nature to it
woven cloak
Ok, thanks alot!
waxen iron
So I had this to prove " If p and q are to prime numbers such that (pq , (p - 1)(q - 1)) = 1, then there's only one group of order pq, excluding isomorphisms "
I used Sylow's theorem's and... well, I got to "proof" that is obviously wrong. I didn't even use the condition.
I hope it's readable
I got that basically every group whose order is a product of primes is cyclic, which is obviously wrong
My thinking was:
There's only 4 cases possible:
1º 1 (sylow p) subgroup of order p and 1 subgroup of order q
2º 1 subgroup of order p and p subgroups of order q
3º q subgroups of order p and 1 subgroup of order q
4º q subgroups of order p and p subgroups of order q
The 2º,3º amd 4º are not possible since they lead to number of elements bigger then the group itself
And the 1º case implies that there are elements not in any of the subgroups. By Lagrange's theorem they must of order 1, p, q or G
There's only 1 element of order 1, and the Sylow subgroups take care of orders p and q. So all it remains is order G, meaning G is cyclic.
But groups like D5 shut down this reasoning since it has 10 elements and it's order is a product of primes 5*2
waxen iron
For 2º we got 1*p+p *q elements
waxen iron
Oh... I forgot to subtract the identity
That explains it
But just to make sure, the other elements are all different right?
By Sylow's second theorem
mild laurel
I'm not sure what you're saying
waxen iron
The only intersection between different Sylow-p subgroups is the identity
mild laurel
This is true, but it's not by Sylow's second theorem
Or at least, I don't see how Sylow's second theorem would imply this
Especially if you're considering the intersection of a sylow p subgroup and a sylow q subgroup
waxen iron
Right, you're correct, I was getting what the 2º theorem says wrong.
Thanks for pointing that the identity is in all of them, completely forgot!
Makes way more sense now
mild laurel
It's not always true that the only intersection between different Sylow p-subgroups is the identity
But in this case, this fact is true
You should think about why
waxen iron
Will do!
Many thanks again for the help!
ivory dust
@woven cloak I got Jacobson on Daminark's recommendation and I really like it so far
bleak abyss
1, he's learning the very basics
solemn rain
yo sloth king daminkak
what is the book aimed at
like what audience
prerequisits ?
ivory dust
linear algebra and a decent amount of mathematical maturity
ivory dust
it requires linear algebra
actually maybe not requires at least for the first chapter
which is all I've read
but familiarity with the concepts of linear will still help a lot
and they also might be assumed for the later chapters
chilly ocean
it doesnt require linear algebra
mild laurel
uh
did you actually look at the book before you said this?
In the preface it says "is intended to serve as a text for a first course in algebra beyond linear algebra"
ripe basalt
No Artin?
solemn rain
wtf is linear algebra
topaz solar
artin oh boy
bleak abyss
Yeah Artin's the book if you don't know linear algebra
Or if you "know" it but not well
solemn rain
linear algebras so boring do i have to know it lmao
chilly ocean
bruh moment
bleak abyss
Math is linear algebra lol
delicate bloom
if you think linear algebra is boring you probably didn't get far enough to learn about eigenvalues and eigenvectors
ivory dust
Linear algebra was my favorite math class do far
Such a fascinating subject
Also I think you overestimated the amount of linear required for Jacobson @bleak abyss
Like I think your original statement about the amount of linear was correct
But I also don't think you have to know it really well either
Like just having completed an introductory course where you covered more than how to manipulate matrices
At least that's my impression of Jacobson so far. I'm pretty early into it though. I haven't even finished chapter 1
bleak abyss
X requires linear algebra \ne X requires knowing linear algebra from page 3
chilly ocean
call me a nerd but when i was learning math in school algebra was my favorite
maiden ocean
??
bleak abyss
I mean like, I think Jacobson is gonna require more as it goes on
But beyond like, the stuff in linear algebra that it references, there is the problem that it doesn't cover certain things in linear algebra that people should know
ivory dust
Like what
Because "the proof that in an algebraically closed field, all matrices are diagonalizable" wouldn't be covered in a linear class
At least not an intro one
Because they wouldn't even know what a field is
bleak abyss
So I never had a linear algebra "class" per se but I did two different things which both did not assume you knew any linear algebra and demanded you knew it by the end
And both of them covered fields
If you had a class which didn't do that proof then it's a subpar class
I mean like
If nothing nothing else
Prove it over C
solar vessel
but
bleak abyss
Also wait wait wait I just reread your statement and that's wrong
Triangularizable
Not diagonalizable
solar vessel
I was getting worried
bleak abyss
Look I don't read sentences in detail, I just read enough words to get the gist and interpolate based on what makes sense
ivory dust
Oh I misspoke
I meant to say triangularizable
Obviously not all matrices are diagnonalizable
I think you overestimate what's typically covered in an intro linear algebra class
Or what needs to be covered imo
bleak abyss
I mean idk maybe there are a bunch of LA lite classes out there
But they need to be followed up
ivory dust
Obviously
bleak abyss
"Or what needs to be covered" I strongly doubt you have any actual perspective on what needs to be covered
ivory dust
Yes but math departments do
bleak abyss
Not all math departments are good lol
bleak abyss
I'd wager most undergrad math curricula are actually kinda shit
ivory dust
I'd agree