so, we need to construct a new definition from this, and mine would be:
#groups-rings-fields
slender estuary
A ring is a set that satisfies all the group axioms for addition. It is assosiative under multiplication, and multiplication distribute over addition.
Multiplication forming a group is an optional thing, and not a requirement.
delicate orchid
if you want a consicse definition:
Let R be a set, then (R, +, *) is a ring if and only if (R, +) is an abelian group, and for all a,b,c in R, we have (ab)c = a(bc) and a(b+c) = ab+ac
delicate orchid
A ring is a set that satisfies all the group axioms for addition. It is assosiat...
yeah exactly
slender estuary
what is (R,+,*)
delicate orchid
I'd personally also require a "1" but some people don't
delicate orchid
what is (R,+,*)
The set R equipped with the operations + and *
slender estuary
So, in actuality, it is actually a set equiped with 2 operations.
+, *
But forms a group under addition.
delicate orchid
yeah that's what I said
fleet pelican
A second example of a ring is polynomials. Notice x doesn't have an inverse which is a polynomial.
It's very helpful to keep those two examples in mind.
delicate orchid
give it a go
slender estuary
i know that:
You can do:
2x^2 + 4x^2 = 6x^2
a o a = a^2
or
2x+2x^2 = 2x + 2x^2
a o b = z
That's closure, right
delicate orchid
it's a nice example to do because it's not like Z where it's like, completely obvious
slender estuary
2x + 2x^2 is a polynomial, which is in the set of polynomials
delicate orchid
i know that:
You can do:
2x^2 + 4x^2 = 6x^2
a o a = a^2
or
2x+2x^2 = 2x + 2...
you can't prove something by example, lol
slender estuary
oh i cant prove, but i want to demonstrate all the axioms work
delicate orchid
oh ok fair enough
delicate orchid
u can skip associativity lol
slender estuary
lol
sure
commutativity then
same shit
?
So, it forms a group under addition!
Therefore, it's a ring, provided it is equiped with multiplication so that: Multiplication is associative, and distributes over addition
boom a ring
unless, uhm, if multiplication is possible operation, but i dont think there is any multiplictive inverses
im thinking of reciprocal, but, then there might be issues. you wont get 1
delicate orchid
you're right, there aren't any inverses for any polynomial with an "x" in it
slender estuary
except for the expression x^-(a number) (which is NOT a polynomial, and therefore is impossible)
since the element does not exist in the set of polynomails
provided our exponent is not an element of the naturals
slender estuary
you're right, there aren't any inverses for any polynomial with an "x" in it
its clicking together, i now understand.
IT HAS those properties. Doesn't mean it has to be a group under multiplicaton! But it can be, because then it would satisify those two points regardless.
delicate orchid
you can't prove something by example, lol
delicate orchid
its clicking together, i now understand.
IT HAS those properties. Doesn't mean ...
yes exactly! If it IS a group under multiplication then all the relevant properties are satisfied PLUS some bonus ones
slender estuary
and i was so blind to not realise this
these statements, do not have to imply it has to form a group, it just HAS those properties
the set itself
slender estuary
now i can talk about rings in alright detail
delicate orchid
now i can talk about rings in alright detail
remains to be seen
slender estuary
💀 if i dont forgor
delicate orchid
EXERCISE: prove that 0*x = 0 for all x in a ring
EXERCISE (Easier) prove that the 0 element is unique
fleet pelican
A slightly more exotic ring is the rings of n x n matrices. For each n there is such a ring.
slender estuary
i can do the second one
delicate orchid
I'm not sure nesymerp knows matrix multiplication lol
slender estuary
Isn't it:
e = ef = f (if e = f, then there is no unique neutral element)
e = f
sonic coral
EXERCISE: prove that 0*x = 0 for all x in a ring
that was on my homework
delicate orchid
Isn't it:
e = ef = f (if e = f, then there is no unique neutral element)
e =...
yes, it is
say we have two 0s, 0 and 0', then 0 = 0+0' = 0'
delicate orchid
that was on my homework
it's a classic
karmic moat
another common exercise (i think) is proving that the cartesian product of two rings forms a ring
under coordinate-wise addition and multiplication
sonic coral
i still have one i can’t figure out for a boolean ring, (r+s)^2 += r^2 + s^2, it might just be ab multiplying it out and using the boolean property i havnt looked too closely
slender estuary
So:
this sounds like an:
E o B = B
Where E is x, and B is 0.
in that case, the neutral element is E? that makes no sense, a multiplicitive neutral element is 1.
and B is not 1
delicate orchid
i still have one i can’t figure out for a boolean ring, (r+s)^2 += r^2 + s^2, it...
uhh lemme think, that's definitely how you do it in a char 2 ring
sonic coral
what does character mean, i havnt learnt that yet
delicate orchid
characteristic
sonic coral
is it to a ring as order is to a group
delicate orchid
also doing boolean rings before characteristics damn
fleet pelican
what does character mean, i havnt learnt that yet
what is 1+1 in a boolean ring?
delicate orchid
wack
sonic coral
what is 1+1 in a boolean ring?
1
delicate orchid
that sounded a bit too complicated for me
hint: yx = (y+0)x for all y and x in the ring
slender estuary
i don't understand 💀
delicate orchid
i don't understand 💀
yx = (y+0)x = yx+0x
karmic moat
what does character mean, i havnt learnt that yet
characteristic of a ring is how many times you add the multiplicative identity (i.e., 1) to get 0
fleet pelican
one plus one is one? what is the additive inverse of one?
fleet pelican
so what if 1+1=1 then (-1)+1=(-1)+1+1?
delicate orchid
le trivial ring has arrived
fleet pelican
1+1=1 doesn't sound correct
slender estuary
but its 2
delicate orchid
boolean rings are characteristic 2, so 1+1 = 0
fleet pelican
boolean rings are characteristic 2, so 1+1 = 0
this makes more sense
delicate orchid
1+1 = 1 implies the ring is trivial
sonic coral
so addition is mod 2 in a boolean ring?
delicate orchid
*my brain gets bricked*
the set {0, 1} forms a ring under 1+0 = 0+1 =1, 0+0 = 1+1 = 0 and normal multiplication
delicate orchid
yeah why not
slender estuary
i dont understand what's happening
delicate orchid
I can have 1+1+1 = 0 too
slender estuary
💀
delicate orchid
and 1+1+1+1+1 = 0
fleet pelican
I can set 4=0.
slender estuary
Okay hold on
delicate orchid
yeah why not
it satisifes the ring axioms, so it's a ring
delicate orchid
2 = 0
that's the actual thing that's going on here
we're quotienting the integers by the equivalence relation 2k ~ 2k+2 for all k in Z
delicate orchid
so the set is {2,1}
no additive identity
the set is {0,1}
if it wasn't I wouldn't have said {0, 1}
delicate orchid
yes
slender estuary
and 1+1+1 = 3 = 0?!
slender estuary
wait give me a moment
delicate orchid
literally the only thing that's different is 2 = 0
fleet pelican
oh so to make that easy
delicate orchid
it's the easiest ring quotient in existence
delicate orchid
yeah you can think of it like that
fleet pelican
even plus even is even
slender estuary
even + odd , odd
odd + odd = even
even + even = even
fleet pelican
odd plus odd is even
fleet pelican
think of it as the remainder when you divide a number by 2
that works too, it's what "modulo" means in computer science as well.
slender estuary
so...
odd + odd + odd
is like
even + odd = odd
fleet pelican
yes
delicate orchid
this is an instance of a quotient ring if you wish to "g"oogle it
slender estuary
what is a qoutient ring
formal ermine
what does it mean for an ideal to divide another one
slender estuary
idk what an ideal is
formal ermine
that question was not targetted at you 
delicate orchid
what does it mean for an ideal to divide another one
What ring are we in lol
slender estuary
"In ring theory, a branch of abstract algebra, a quotient ring, also known as factor ring, difference ring[1] or residue class ring,"
so
thats where mod comes from
residue class ring
delicate orchid
Oh lawd he went on the Wikipedia page
delicate orchid
residue class ring
Nobody has called it this ever
formal ermine
What ring are we in lol
I'm trying to decipher this answer
slender estuary
Nobody has called it this ever
but is it correct
delicate orchid
Sure it’s a ring of residue classes
delicate orchid
I'm trying to decipher this answer
Yeah I got no clue, it’s probably a generalisation of the idea from PIDs that a|b => (b) \subseteq (a) would be my guess
karmic moat
"residue class ring" yuck
formal ermine
bro residue class ring
karmic moat
what's next, residue class space?
elder wave
agile burrow
I'm trying to decipher this answer
You have unique factorization into prime ideals in rings of integers, so saying \mathfrak p divides (p) just means that \mathfrak p appears in this prime decomposition of (p)
chilly radish
Equivalent \frak p \cap Z= pZ
formal ermine
wait but if it's a prime ideal already
chilly radish
the ideal generated by a prime number needn't be a prime ideal
agile burrow
The extension of (p) to O_K won't necessarily be prime
formal ermine
dfoiler says "divides the prime ideal" though?
chilly radish
Generally an ideal I divides an ideal J if theres some J' such that IJ'=J
agile burrow
Yeah, I think the 'prime ideal' in the picture you sent may have just been a typo
formal ermine
okay so to repeat
mathfrak p lies over p if mathfrak p divides the ideal generated by p in the ring of integers of K?
agile burrow
Yes
agile burrow
I'm not entirely sure but this might be specific to the fact that O_K is a Dedekind domain
Like in general we say that q lies over p if the restriction of q to the subring is equal to p, which is what ShiN said
formal ermine
ah ok
formal ermine
I have multiple questions
what is a reduction map
and why does the residue field contain a copy of Fp
delicate orchid
oh look the dudes are back
formal ermine
yes one day later still at 1 am trying to understand this
delicate orchid
and why does the residue field contain a copy of Fp
ok this one's easy, is F_\frak{p} char p?
I think so as it lies over p
formal ermine
can you elaborate
agile burrow
Primes are maximal in rings of integers. If q lies over p, then there's a natural way to view O_k / q as a field extension of Z/pZ
I'm using q in place of mathfrak p
formal ermine
Primes are maximal in rings of integers. If q lies over p, then there's a natura...
and that is?
agile burrow
Just write it out lol
Z embeds into O_K, this embedding takes p into q
So the composition Z -> O_K -> O_K / q factors through Z/pZ
formal ermine
I'm still a bit lost at the lying over
there were like 3 different definitions
which one are we going with
agile burrow
Any of them work, I think the easiest to think about is that a prime q lies over p if q divides p O_K
Anyways, the decomposition group of q is the set of elements of Gal(K/Q) which fix the prime q. Every element in the decomposition group restricts to an automorphism of O_K which fixes Z. Since these automorphisms also fix q, they yield an automorphism of O_K / q over Z/pZ, but you know what the latter is
Or I guess I should phrase it differently - O_K / q is a finite field extension of Z/pZ and you know what the Galois group of that is
But yeah, the above describes a map from the decomposition group of q to the Galois group of O_K / q over Z/pZ
formal ermine
Any of them work, I think the easiest to think about is that a prime q lies over...
what does p O_K here mean
formal ermine
Since these automorphisms also fix q, they yield an automorphism of O_K / q over Z/pZ, but you know what the latter is
can you pls elaborate on this part
agile burrow
Which part is unclear?
formal ermine
the yielding
agile burrow
So we have an automorphism, say f, of O_K and it fixes q. This means that it induces an automorphism of O_K / q by sending x + q to f(x) + q. The fact that it fixes q (set-wise) implies that it is well-defined
agile burrow
Do you understand how we can view O_K / q as a field extension of Z/pZ
agile burrow
So we have a composition Z -> O_K -> O_K / q, right?
formal ermine
yes
delicate orchid
should I draw the square walter
agile burrow
Lol go for it, I'm too lazy to typeset
agile burrow
So we have a composition Z -> O_K -> O_K / q, right?
The kernel of this composition is Z \cap q = p by one of our equivalent definitions of lying over
agile burrow
So first isomorphism theorem gives us a homomorphism Z/pZ -> O_K / q, which is precisely the data of viewing O_K / q as a field extension of Z/p
agile burrow
Well yes, that's the quotient map
formal ermine
yeah ok I can see it now
agile burrow
Cool
delicate orchid
[\begin{tikzcd}
{\mathbb{Z}} & {\mathbb{Z}/p\mathbb{Z}} \
{O_K} & {O_K/q}
\arrow[from=1-1, to=2-1]
\arrow[from=2-1, to=2-2]
\arrow[from=1-1, to=1-2]
\arrow[from=1-2, to=2-2]
\end{tikzcd}]
I think this is the $\text{d}$ude
cloud walrusBOT
Wew
coral shale
@formal ermine interested in my course lecture notes for alg nt?
delicate orchid
this square commutes by univesral property of the quotient (and the fact Z is initial) but mainly because I said so
formal ermine
<@391270706186420224> interested in my course lecture notes for alg nt?
my alg nt course starts next tuesday
agile burrow
what does the "over Z/pZ" part mean
Now the 'over Z/p' part just means that since the decomposition group is a subgroup of Gal(L/K), these automorphisms fix Z
So the associated automorphism of O_K / q fixes Z/p, hence yields an element of the Galois group of this extension of finite fields
formal ermine
yeah ok I can see it now
wait nvm I don't
why is the map from the quotient to our O_K/q injective
agile burrow
It's a field homomorphism
formal ermine
So the associated automorphism of O_K / q fixes Z/p, hence yields an element of ...
ahh
agile burrow
I'll remark that I remember it being non-trivial to show that the map from the decomposition group to the Galois group of the extension of finite fields is surjective
formal ermine
you should use zeta for e^(2pi i/3)
pastel cliff
these are prof notes
formal ermine
no idea
delicate orchid
lol
pastel cliff
that's what im asking about lmaoo
delicate orchid
u know I think these two might just be isomorphic....
formal ermine
one of the middle ones should be zeta^2 cbrt(2)
.
pastel cliff
u know I think these two might just be isomorphic....
pastel cliff
ok but it does have four subfields then
coral shale
using Id ...
coral shale
lets call it D3
tribal moss
why is one of the subfields repeated am i missing someething
I think one of them was intended to be the complex conjugate.
white oxide
i get why the equality can hold only if q1(x) - q2(x) = 0 because r2(x) - r1(x) = 0, but why does the equality also hold in the case that r2(x) - r1(x) is less than the degree of g(x)?
this is in the proof of the division algorithm for F[x] for context where F is a field
rotund dragon
i get why the equality can hold only if q1(x) - q2(x) = 0 because r2(x) - r1(x) ...
that is justified in the rest of the paragraph
ah wait i think you're stuck on why deg(r_2(x) - r_1(x)) < deg(g(x)) implies q_1(x) - q_2(x) = 0. sorry i shouldve read more carefully
white oxide
all g
rotund dragon
try writing out an example in R[x]
white oxide
ok bet
how important are noncommutative rings
or rather how interesting are they
to study
karmic moat
iirc noncommutativity has lots of stuff in physics
white oxide
since mrfeynman reacted i trust you
so i understand everything in the proof up to consequently s = n; does it have anything to do with a_n = b_rc_s?
steep mesa
I have a very stupid question
what's the cannonical notation for the function from $\mathbb Z$ to $\mathbb Z / n \mathbb Z$
cloud walrusBOT
TheZachMan
steep mesa
or to a quotient in general I guess
thorn delta
pi
rustic crown
just Z --> Z/nZ is a good notation too, assuming you're only drawing the diagram and not doing some calculus
fleet pelican
pi is typically used for all such maps i.e. quotient projections
kind jacinth
i dont get what this example is saying
fleet pelican
I think you left critical information out of that snippet.
kind jacinth
it starts like this which i already dont get but i should send everything
delicate orchid
more like, what is this an example of
kind jacinth
kind jacinth
more like, what is this an example of
its after this corollary
Corollary 8.4.1 Let F be a field and p(x) ∈ F [x] of degree at most 3. Then
p(x) is irreducible over F if and only if it has no zero in F .
delicate orchid
there appears to be 2 other results between that corollary and that example
kind jacinth
they are different examples (the way they number it is weird) like example 8.4.2, 8.4.3 then 8.4.4
fleet pelican
You really haven't provided enough information to make it easy enough to help you.
delicate orchid
the example is just showing you that the kernel of the evaluation map at some point is the space of polynomials that have a solution at that point
maybe one variable is easier
take the evaluation map f_1 : F[x] -> F mapping x -> 1, then ker(f_1) is the space of polynomials that are 0 at x = 1
so x-1 x^2-1 and x^2-x are in the kernel, for example
kind jacinth
i see.
kind jacinth
so x-1 x^2-1 and x^2-x are in the kernel, for example
and in the example case they are saying all the polynomials are in the kernel since they equal 0?
is x1 with a star different from x1 in the polynomial?
im assuming by x1 with a star they mean a value?
a value that gives 0
delicate orchid
each of the p_1, ..., p_m are in the kernel of phi_(x*_1, ..., x*_n) because by definition (x*_1, ..., x*_n) is a solution to that system of equations
that is to say that p_i(x*_1, ..., x*_n) = 0 for all i
kind jacinth
kind of get it because they equal 0. I am confused about what phi_(x*_1, ..., x*_n) ior just in general what p1(x1,...,xn) is. is it a polynomial of form ax1+bx2+ ... zxn
?
and x*_1, ..., x*_n is what i plug in to get 0 for that polynomial?
delicate orchid
I'm just gonna write x* for x*_1,... x*_n now cause it's getting on my nerves
p_1(x_1,..., x_n) is just some polynomial in n variables
like x_1^2x_2^2+x_3+x_4 or something
and phi just maps x_i to x*_i
which is the same as evaluating the polynomial on x*
for example, lets go back to one variable and our evaluation map f_1: F[x] -> F, x -> 1
lets take a polynomial p = x^2+x+1, then f_1(p) = (1)^2+(1)+1 = 3
your example is just this but with more variables
kind jacinth
ohh ok. and if i have a polynomial j that j(1) = 0, the polynomial will be in my kernel
is that correct?
kind jacinth
but why does it have to be for 1? oh cause its supposed to map identity to zero right?
delicate orchid
lets take a polynomial p = x^2+x+1, then f_1(p) = (1)^2+(1)+1 = 3
if we instead took f_4(p) we'd get (4)^2+4+1 = 21
all we're really doing here is formalising the connection between viewing polynomials as formal sums in a ring and also maps p(x_1, ..., x_n) : F^n -> F
formal ermine
given a normal subgroup, how do we find a homomorphism whose kernel is that group?
delicate orchid
pi : G -> G/N lmfao
formal ermine
kind jacinth
all we're really doing here is formalising the connection between viewing polyno...
ok this makes sense. and by evaluation homomorphism they mean its gonna be all polynomials that give 0 given value z
phi_z
delicate orchid
the kernel of the evaluation homomorphism phi_z is all polynomials that are 0 on z
kind jacinth
*the kernel* of the evaluation homomorphism phi_z is all polynomials that are 0 ...
yes. thanks!
rotund aurora
given a normal subgroup, how do we find a homomorphism whose kernel is that grou...
this is a fun question, because you can motivate quotients of groups by solving this (non trivial) problem
formal ermine
what's a motivation for normal subgroups, if not quotients
delicate orchid
ur literally a galois theory head
why don;t u tell me all about ur normal extenstons
rotund aurora
delicate orchid
why don;t u tell me all about ur normal extenstons
the groups were named after the extensions historically, not the other way around
formal ermine
why don;t u tell me all about ur normal extenstons
rotund aurora
what's a motivation for normal subgroups, if not quotients
But also kernels of group homomorphisms, which is what I meant
(and the usual definition is the "algebraic characterization" of those groups)
kind jacinth
Show that the division algorithm does not hold for Z[x]. Why does it fail?
trying to understand why this doesnt hold using an example.
Division algorithm formula is f (x) = g(x)q(x) + r(x),
lets take f(x) = 3x^2 and g(x) = 2x
this would result in q(x) = 3x/2 and r(x) = 0
why would this not be euclidean given Z[x] but apparently is an euclidean given Q[x]?
formal ermine
3x/2 is not in Z[x]
kind jacinth
lol right
formal ermine
R[x] is an euclidean ring iff R is a field
kind jacinth
so then the proof would be something along the lines that we could get a fraction which isnt in Z[x]?
formal ermine
yeah it's literally your counter example lol
terse rose
I’m a bit confused on this proof
How does it know that there are phi(d) elements of order d in G_d
kind jacinth
Give two different ring homomorphisms from R to R[x]
can I say Z_2 -> Z_2[x] for polynomials of 1 or 0 degree?
not 0 degree actually, just 1 degree
lethal dune
kind jacinth
yeah idk what im doing honestly
back to the basics. I need R to be an integral domain (which i think is fine in the case above) and it needs to be homomorphic
lapis trail
Trying to prove a polynomial Eisenstein irreducible by way of a change of variable. Is there a good strategy for determining what the change of variable should be?
Wait I might have found one
kind jacinth
so they are mapping z->alpha*z + beta but im confused on how this works
so say I want to do theta(2) this would be alpha*2 + beta but how do i know what the value of alpha and beta is? are they all values where (alpha*2 + beta) mod 6 = 2?
also, I guess proving commutative is straight forward? (just show theta(a) + theta(b) = theta(b) + theta(a) ?)
white oxide
so i understand everything in the proof up to consequently s = n; does it have a...
can somebody help me here pls?
coral shale
also, I guess proving commutative is straight forward? (just show theta(a) + the...
no?
minor point - the symbol is phi in the image
main point - the elements are functions (this is the main point to understand - the rest is down to set builder notation).
Its not just a single phi thats the function, there are many different ones.
Think to yourself what the group operation is
and then think what the commutativity statement should be.
kind jacinth
Think to yourself what the group operation is
composition in that case right?
ill try to make the commutative statement for this
coral shale
yes
kind jacinth
f = alpha1z + beta g = alpha2z+beta
white pendant
Hey ya'll I'm gonna ask a question if thats alright
blissful crystal
formal ermine
Hey ya'll I'm gonna ask a question if thats alright
why shouldn't it
white pendant
Thats a mighty fine question
Anywho
I'm doing proofs about finite abelian groups
and I there's this proof that I'm like 85% the way there on
So the proof is to show that G is a p-group -> G = G[P^n] for some n
And showing that G[p^n] in G is really easy
But I'm having trouble hammering out the details in the other direciton
My plan was to say that for some a in G the order of a is P^k for some k
And a would be in G[P^k] cause P^k(a) is 0
But there's probably some other elements in G with bigger orders, so G != G[P^k]
But consider some b in G st. |b| = P^m st. n<m
P^m(a) would still be 0, so if we can find some element in G with maximum order then its all good and G = G[P^l] where P^l is the maximum order
formal ermine
what does the notation G[P^k] mean
white pendant
Oh sorry, G[n] = {g : ng = 0}
formal ermine
all elements of order n?
formal ermine
what are you trying to show to begin with? prime power order iff p group for finite groups?
white pendant
That G is a P-group implies that G = G[P^n] for some n
formal ermine
which definition of p group are we going with
white pendant
Oh whoops
there we go
I feel like thinking about this now, it feels like this follows from the definition
Also misunderstood the question, a P-group in my class is the following
formal ermine
essentially like you already said the order of an element is p^k for some k <= n
white pendant
Yeah!
formal ermine
ignore that I had a brainfart
white pendant
All good! thanks for talking btw
white pendant
essentially like you already said the order of an element is p^k for some k <= n
Wait so here are you talking about the definition of a p-group?
formal ermine
my definition for finite groups is p group iff order is p^n
white pendant
A h
Oh and the order of any individual element has to be equal to or less than the order of the group
white pendant
SO REAL
formal ermine
explain why if |a| = p^k then a^(p^(k + c)) = 1 and you're done
formal ermine
Ah but I can't use that fact without proving it 😭
have you not proven lagrange in class yet lol?
that's like one if the first things you usually do
white pendant
No we've proven Lagrange, its just that proving that |G| = P^k is part of the assignment
ah
formal ermine
can you share the assignment
formal ermine
I'm confused on what you're supposed to do lmao
formal ermine
can you send a screenshot of the definition of a p group
white pendant
yeah!
formal ermine
these are usually the definitions of a p group lol
white pendant
Excuse my professors wack handwriting
But basically every element has an order of a power of P
formal ermine
ah ok
white pendant
Ye
formal ermine
that sounds ok. I would personally do b -> a -> c -> b
white pendant
Oh just easier?
formal ermine
no idea just the one I would do
formal ermine
alright
white pendant
So one more question
tell me if I'm going off the deep end, but the idea is just to say that there is an element with a max order, cause the set of order is bounded above by the order of the group and to show that any old element in G, say a, would be 0 if a*(max order)
And that would show that a would be in G[P^n] and thus G would be in G[P^n]
formal ermine
a -> b right?
white pendant
yeah
formal ermine
I honestly am not sure what you're trying
white pendant
So
I'm trying to end up with G = G[P^n] right?
So given that G is a P-group
I show that G[P^n] is in G
and that G is in G[P^n]
formal ermine
given an element g with |g| = p^k (i.e. g^(p^k) = 1), k <= n, you need to show that g^(p^n) = 1
white pendant
Wait my class switched around the notation for abelian groups
so instead of e or 1 we use 0
formal ermine
same thing
formal ermine
ye
formal ermine
given an element g with |g| = p^k (i.e. g^(p^k) = 1), k <= n, you need to show t...
you know that n = k + c for some c
so just do some basic algebra
white pendant
Wait, so what is k here?
formal ermine
some number
formal ermine
n here is our k
white pendant
Okay, so its not special or nothing
OH okay, so that's pretty easy
Yeah just that P^n -> P^(k+c)
g^(P^k)(P^c) -> 1^P^c -> 1 yeah?
formal ermine
yes exactly
white pendant
Okay, so theres some element with order P^n
and that order, P^n is the biggest order of any element in G, right?
formal ermine
you've essentially shown that given any element you have that it's in G[p^n]
formal ermine
yeah we just choose the largest k as our n
white pendant
Oh okay, see thats what I was going on about with the max order thing
But also
like lets say we have k=3 and n=2 couldn't we just split it up into P^(2) -> P^(3+-1) -> P^(3)*P^(-1)
just as a bunch of inverses
Or no
Ignore that
Thank you very much for talking with me about this, I appreciate it yo!
kind jacinth
just to make sure as no answers are given,
the orbits are O_0/O_3 = {0,3} and O_1,2,4,5 = {1,2,4,5}
is the stabilizer going to be {1,2,4,5}?
or how would i find the stabilizer for this actually?
actually is the stabilizer {Z, 5Z+3}
lapis trail
Trying to find the inverse of a+b sqrt(2) + c sqrt(4). proving to be a little difficult. is there a trick? is it obvious and I'm just not seeing it?
by the way
what z is it such that z(a+bi)z^(-1)=a-bi?
formal ermine
Trying to find the inverse of a+b sqrt(2) + c sqrt(4). proving to be a little di...
what structure are we in
formal ermine
field?
lapis trail
Splitting field of t^3-2
formal ermine
do you want the inverse expressed as a linear combination of 1 cbrt2 cbrt4?
lapis trail
Yeah that's what I'm after
formal ermine
you can like
express the inverse as a linear combination of unknowns
then compare coefficients
because they will always lie in Q
formal ermine
yeah
but you know 3 of them
so it's just 3 equations in 3 unknowns
I get these solutions
kind jacinth
actually polynomials that accept as outpit both 0 and 1 wont count as thats just the whole Z_2[x]... i am not sure how to do this then
formal ermine
the simple way would be to find the image of phi then use first iso and argue that the image is not a field
white pendant
Illuminator you are a benevolent god
Given that G is cyclic, I'm trying to prove that G is isomorphic to Zp1^n1 X ... X Zpk^nk
And I have very few Ideas
I guess part of the proof is just coming up with a homomorphism
delicate orchid
guessing you can't just structure theorem bash this one
white pendant
Are you responding to me?
delicate orchid
yus
white pendant
Whats structure theorem?
delicate orchid
I'd try first constructing an isomorphism between Z_{p_1p_2} and Z_{p_1}xZ_{p_2}
white pendant
Real, real
I think there's some lectuer notes about that too, I shall investigate
Thank you very much!
Awww ya'll have the best reaction dudes
kind jacinth
the simple way would be to find the image of phi then use first iso and argue th...
i mean isnt the image of phi just
1 0
0 1
and
1 1
1 1 ?
formal ermine
and its combinations
kind jacinth
why and its combinations? because it hsa to be closed?
kind jacinth
ahh ok. in that case i get all of S besides the one where its all 0
formal ermine
are you sure
kind jacinth
i know im getting
0 1
1 0
by adding the theta(0) + theta(1)
and the same element 1 1
1 1
by multiplying
formal ermine
what's phi(0)?
formal ermine
why?
kind jacinth
because phi(x) is defined as 1 1
1 1?
formal ermine
yeah but phi(0) = phi(x + x)?
kind jacinth
what is the x supposed to represent cause i am a bit confused. its integers in Z_2 right?
formal ermine
what?
kind jacinth
what is x in phi(x)?
formal ermine
Z_2[x] is the polynomial ring with coefficients in Z_2
kind jacinth
Z_2[x] is the polynomial ring with coefficients in Z_2
yes
kind jacinth
Z_2[x] is the polynomial ring with coefficients in Z_2
this one i get it. and polynomial rings that would give as output "1" such as x^2+x+1 (because no matter the x value id always get 1) belong in the kernel set right?
formal ermine
What
kind jacinth
x^2 +x+1 is one example of a polynomial in Z_2[x] right?
formal ermine
yes
kind jacinth
what does the "1" in phi(1) represent? and the "x" in phi(x)
kind jacinth
yes
but phi is defined to be from Z_2[x] -> S
so its a polynomial that gets mapped to S
kind jacinth
x is a polynomial in Z_2[x]
so by phi(x) it maps the polynomials to (1 1
1 1)?
kind jacinth
the simple way would be to find the image of phi then use first iso and argue th...
I will review the topic then reattempt it with the hint u provided. thanks.
white oxide
so by phi(x) it maps the polynomials to (1 1
...
sorry what is S in this case
a field containing Z_2?
formal ermine
.
white oxide
ah
can i get a hint for this question? how do I know how many elements F[x] has, isn't it infinite?
<@&286206848099549185>
icy cedar
Can we see exercise 29
white oxide
my fault forgot to include
void cosmos
can i get a hint for this question? how do I know how many elements F[x] has, is...
no
white oxide
wait but like
a0
a0 + a1x
a0 + a1x + a2x^2
....... so on and so forth
as long as all the ajs are in F that's fine right
oh wait
i suppose the definition of polynomials calls for a finite amount of the ajs to be nonzero
ah
void cosmos
would this proof be an example of diagram chasing?
oblique river
Yep
void cosmos
that was actually cool
i smiled when he went from the right commutative diagram to the left by exactness
am i stupid?
barren sierra
Nope
Some diagram chases are pretty
Like snake lemma is another good one
Today in class we were introduced to the exterior algebra
It's this tensor algebra with some extra relations
Why do we care about it tho
Like cool it's a thing but I don't get why we are about it (or the tensor algebra for that matter)
void cosmos
its a bit important in diff geo
void cosmos
Like snake lemma is another good one
i actually saw it in a movie believe it or not
like literally the snake lemma in a movie it was a super old and bad one
and i was just super surprised cuz like most of the time the math shown is weird calculus triple integral whatever to scare the viewer
but it was just some hot chick proving the snake lemma and there was this douchebag of a student ig
lmfao
prisma ibex
Like cool it's a thing but I don't get why we are about it (or the tensor algebr...
have you heard the universal property that the tensor algebra satisfies?
stone hamlet
It's this tensor algebra with some extra relations
(as an example for why they're great, from what i remember) the determinant of a matrix, i.e. volume deformations of linear transformations on the unit cube, can be defined as the exterior product of the columns of the matrix. this machinery carries forward when you consider the "global deformation of a differentiable map" -- look at the map's deformation locally (derivatives are linear transformations!) and then add'em up (with the right definition of "add")
void cosmos
https://en.wikipedia.org/wiki/It%27s_My_Turn_(film)
let me see
yupp its this one
wow u know ur movies
it was so bad lmfao
yea now that i know more about like homological algebra stuff , this shit was really accurate
tribal moss
It's a common piece of mathematical trivia.
void cosmos
is she a mathematician
tribal moss
They probably had an actual mathematician advising on the screenplay.
void cosmos
yea but its too accuarte in a weird way do u get me?
like its too accurate that its actually was useless to be that accurate
i looked it up again now on youtube and its like taken from a class ig
but hey i dont understand the kids question?
wtf does he mean not well defined? the elements?
agile burrow
the connecting map
void cosmos
arent the maps just layed out?
agile burrow
You have to check that it's well-defined
void cosmos
like its assumed
like arent those diagrams like meant to say that "given that those maps are epimorphisms and such and ssuch then the following holds?"
agile burrow
The definition of the connecting map involves choosing certain elements, you need to show that different choices lead to the same map in the end
prisma ibex
yup
void cosmos
idk the snake lemma yet
agile burrow
Yeah, on the level of two short exact sequences, the codomain of the connecting map is the cokernel of some map
void cosmos
but its just a rule of thumb for me that whenever u have equivalence classes shit goes south
are there other movies
where like math is cool like that
stricly algebra cuz we are in #groups-rings-fields lmfao
fast stratus
How to prove this
agile pine
what have you tried
delicate orchid
How to prove this
I don’t think this is true. Take G = M = N = Z_2, and let f, h be the trivial automorphism of Z_2. Then the image of phi is {(0,0), (1, 1)}, which is not onto Z_2 x Z_2. But clearly each of the automorphisms are onto, so this is a counter example?
Unless “onto” doesn’t mean surjective here
Which would be concerning
barren sierra
have you heard the universal property that the tensor algebra satisfies?
I saw this
barren sierra
(as an example for why they're great, from what i remember) the determinant of a...
and I saw this
prisma ibex
but yeah the universal property helps, it says that T(V) is the free algebra on the vector space V, it's universal for algebras with an embedding V->T(V)
it's very useful because a ton of other algebras that show up (usually each with their own useful universal properties) are built out of the tensor algebra
barren sierra
so it's useful because it appears in other constructions essentially?
what about exterior algebras?
stone hamlet
(again from what i remember) the exterior algebra forms the underpinning of integration theory as you move forward
since integration relies on a notion of "local volume", which can be built from the ground by the exterior product
barren sierra
idk what integration theory means in this context
void cosmos
idk what integration theory means in this context
Prob means calc on manifolds
stone hamlet
unfortunately i learned my silly diffgeo/integration from munkres's analysis on manifolds, so i only know the R^n theory
barren sierra
I really should take a diff geo class at some point
stone hamlet
but if you have a k-dim manifold in R^n you gotta have a notion of "k-dimensional volume" -- which is conveniently modeled by alternating tensor fields
yeah i think it would be clarifying when you run into it if you take diffgeo in the future
barren sierra
hm it doesn't conflict with my classes next sem but I'm doing 4 other classes already
all of which I'd rather take
prisma ibex
what about exterior algebras?
this appears loads in differential geometry but also for like Lie algebras or dg-algebras
e.g. if you have a Lie algebra g you can form the Chevalley-Eilenberg algebra CE(g) which is a differential graded algebra whose underlying vector space is the exterior algebra on g and whose differential encodes the Lie bracket of g
Lie algebra cohomology is essentially defined in terms of this gadget, the cohomology of this Chevalley-Eilenberg complex is the Lie algebra cohomology of g with trivial coefficients
barren sierra
damn this makes me wish I took that lie algebra course instead of AG
prisma ibex
Lie theory is more fun with AG anyways 🙂
barren sierra
well this AG course is utter shit
prisma ibex
rip
barren sierra
and the lie algebra course had a really really good prof
but I don't know diff geo + I was like "ooo AG more foundational"
stuck fiber
So how combinatorics oriented is CRT?
trail stump
Is it possible for a joint of two finite groups to not be finite?
lethal dune
free prod
trail stump
I know that there exist such group that contain the two finite group but not finite itself, but since joint is defined as the smallest such group…
lethal dune
what do you mean by "join" here?
trail stump
Join my bad
lethal dune
trail stump
Oh shit😂
lethal dune
do you mean the smallest subgroup containing both?
trail stump
Yes
trail stump
Ok I will look into that
solar shore
im not really too sure as to what's going on here
is that supposed to be a typo? should it be 4|bar k|?
thorn delta
is that supposed to be a typo? should it be 4|bar k|?
It should be 4|H|
solar shore
ah
okay thank you
im still a bit confused as to what this example is saying
is it just showing that it indeed is a subgroup?
(of G)
or wait, is it showing closure in K?
feel like this is a subgroup test im forgetting
thorn delta
Are you taking about the “to see this…” part?
solar shore
yeah
thorn delta
They are showing that K, the union of all the cosets in K bar, is a subgroup of G
solar shore
oh wait, is that by finite subgroup test?
thorn delta
Not sure what that is
They omit justification for why K is closed under taking inverses ig
thorn delta
What they are getting at hopefully is that any subgroup of G/H is just K/H for some K a subgroup of G (containing H)
solar shore
ah i see
tender wharf
gallian has a few typos like that yes
I did remember asking about this a few months ago
broken stirrup
Does anyone have any idea how to use the hint?
G=H*K denotes the free product of the groups
celest shadow
Is it correct for a group G , G^-1 = G ?
broken stirrup
Is it correct for a group G , G^-1 = G ?
Yes
every element has a unique inverse and its in the group. Inverting all elements is a bijection. Consider f: G > G^(-1), f(g)= g^(-1). Then f is a bijection
celest shadow
@broken stirrup Is it possible to have same inverses in a group?
broken stirrup
proof: exercise
Well said
broken stirrup
<@414486932119093248> Is it possible to have same inverses in a group?
An element can be its own inverse but no two elements can have the same inverse
delicate orchid
Let h be an inverse to both a and b. ||Then ha = hb = ah = bh = 1, so specifically we have ha=hb, then a(ha) = a(hb) => (ah)a = (ah)b => a = b||
celest shadow
@delicate orchid I see, let a^-1 = b^-1, a(a^-1) = a(b^-1), 1 = a(b^-1), a=b ?
delicate orchid
<@233993116418441216> I see, let a^-1 = b^-1, a(a^-1) = a(b^-1), 1 = a(b^-1), a=...
Yeah, even faster than mine!
celest shadow
thanks anyway 🙂
kind jacinth
given this definition
Definition: Let (R,+1,∗1) and (S,+2,∗2) be rings with additive identities 0R and 0S respectively. If ϕ is a homomorphism from R to S then the Kernel of ϕ is defined as ker(ϕ)={a∈R:ϕ(a)=0S}
In the exercise below, since the additive identity matrix S is the0 matrix is the kernel going to be the set of all polynomials in Z_2[x] where phi(x) = 0 matrix?
.
delicate orchid
Yes
delicate orchid
and how would i find these polynomials? as I dont know what maps to the matrix 0
Well yeah you have to find what maps to zero. Remember you’re mapping into matrices over Z_2 so you have 1+1 = 0
south patrol
I suggest you compute phi(x^2)
kind jacinth
I suggest you compute phi(x^2)
oh yes this gives 0. So does this mean all polynomials with degree 2 get mapped to 0?
i can also have phi(1+1) which would give 0.
or phi(x+x)
delicate orchid
Yeah but those are already 0 in Z_2[x]
south patrol
oh yes this gives 0. So does this mean all polynomials with degree 2 get mapped ...
No
delicate orchid
So you’re just saying that 0 is in the kernel
south patrol
But it means
kind jacinth
Yeah but those are already 0 in Z_2[x]
ahh i see.
kind jacinth
So finding the kernel should be ok
so if my understanding is correct I want all polynomials x in Z_2[x] such that phi(x^2) = 0.
So for example lets say I have two polynomials "x" and "x^2"
polynomial x is in the kernel because this would just give phi(x^2) = 0.
polynomial x^2 is also in the kernel because phi((x^2)^2) = phi(x^4) = phi(x^2)*phi(x^2) = 0
kind jacinth
ok my bad. But what about what is contained in the kernel set?
south patrol
Wdym
kind jacinth
ker(ϕ)={a∈R:ϕ(a)=0S}
I mean in this set, its a set of what?
but are they like polynomials that are mapped to 0? like something in this image? not exactly this but instead of Q[x] z_2[X]
delicate orchid
I already confirmed they are the polynomials that get mapped to 0
kind jacinth
I already confirmed they are the polynomials that get mapped to 0
so if I for example take an arbitrary polynomial in Z_2[x] how can I check if its in the kernel?
south patrol
The fact x^2 maps to zero allows you to just compute phi of any polynomial
kind jacinth
The fact x^2 maps to zero allows you to just compute phi of any polynomial
why is that the case?
south patrol
Phi(x^n)=0 for any n>=2
wraith cargo
here [h,k] is the commutator?
broken stirrup
here [h,k] is the commutator?
Yes
broken stirrup
here [h,k] is the commutator?
The book uses the following definition: [h,k]= h^(-1)k^(-1)hk
wraith cargo
My idea is that maybe the hint is guiding you towards showing that [H,K] is a subgroup of G
tho hmmmm
wraith cargo
The book uses the following definition: [h,k]= h^(-1)k^(-1)hk
Has the book defined the universal property of free groups?
wraith cargo
OK then my idea would be to show that [H,K] satisfies said property by constructing a unique map [H,K] -> G such that the diagram commutes
broken stirrup
Not about showing the freenes of it over < [h,k] >
delicate orchid
The freeness follows immediately from Nielsen-Schreier
wraith cargo
Ahh the hint yeah that's probably it
broken stirrup
OK then my idea would be to show that [H,K] satisfies said property by construct...
To G?
You didnt mean the same G=H*K
wraith cargo
The freeness follows immediately from Nielsen-Schreier
this isn't true because G need not be free
delicate orchid
Oh yeah it’s a free product of two memers
wraith cargo
tho ngl I had that thought too lol
delicate orchid
OK then my idea would be to show that [H,K] satisfies said property by construct...
This sounds like the way to go then
broken stirrup
this isn't true because G need not be free
I came to reply the same thing
Yes
Free product is not necessarily free
south patrol
Lol
broken stirrup
Free product is not necessarily free
Lol
broken stirrup
Free product is not necessarily free
Sounds funny
broken stirrup
I know the problem is dead, just wanted to share an idea. I think the best way to do it would be showing that every element can be uniquely expressed, and {<[h,k>} already generates [H,K] by definition
@wraith cargo
frigid lark
Hey, there's a really quick fix to replying to dead problems. Microwaved bananas if you know what I'm getting at
wraith cargo
I know the problem is dead, just wanted to share an idea. I think the best way t...
hmmmm this might work but idk how problematic it might be to actually show uniqueness
untold cloud
Hi, guys, for free product G of arbitrary many groups, do we always take finitely many length word?
coral shale
yes
tribal moss
Yes.
coral shale
defn of word.
untold cloud
Thanks!
tribal moss
(It's not necessarily a definition if you define the free product by a universal property. But the outcome is the same).
I don't think there is a reasonable way to make a construction with "infinite words" that even produces a nontrivial group.
You'd get b·(ababab...) = a^-1·(ababab...), so by cancellation b = a^-1 whenever a and b are not from the same group originally.
kind jacinth
Z2[x]/〈p(x)〉
is this just the set of all remainders when I divide a polynomial in Z2[x] by polynomial p(x)?
and the set only contains {0, 1, x, 1+x} given that p(x) = x^2 + x + 1
tribal moss
Z2[x]/〈p(x)〉
is this just the set of all remainders when I divide a polynomial ...
Yes, as a set, but the notation also implies certain addition and (in particular) multiplication operations on that set.
kind jacinth
can someone explain what exactly they are trying to do here?
like i get how the 0, 1, x, 1+x come from
but what does all of this mean
delicate orchid
They’re constructing a new field by appending an element a that satisfies a^2+a+1
kind jacinth
They’re constructing a new field by appending an element a that satisfies a^2+a+...
how is alpha^2 + alpha+1 = 0 now?
delicate orchid
how is alpha^2 + alpha+1 = 0 now?
always has been, alpha is a root of that polynomial by definition
it's like how you can construct the complex numbers by doing R[x]/(x^2+1)
you're appending an element such that that element squared +1 is 0, or equivalently that element squared is -1
kind jacinth
it's like how you can construct the complex numbers by doing R[x]/(x^2+1)
oh so like this is now adding the element x^2 = -1 to R which is i and -i
which is why its isomorphic to the complex C
in the example case the element that would result in x^2 + x + 1 to be 0 isnt contained in my Z_2[x]. So im 'adding' it in Z_2 by taking the quotient
did i get this right?
and how is (1+alpha)^2 = alpha?
agile burrow
in the example case the element that would result in x^2 + x + 1 to be 0 isnt co...
That's right, by taking the quotient Z_2[x] / (x^2 + x + 1), the coset of x becomes a zero of the polynomial x^2 + x + 1
agile burrow
and how is (1+alpha)^2 = alpha?
We have (1 + a)^2 = 1 + 2a+ a^2 = 1 + a^2 (since we are in characteristic 2). By construction, 1 + a + a^2 = 0. Adding a to both sides yields 1 + a^2 = a
kind jacinth
We have (1 + a)^2 = 1 + 2a+ a^2 = 1 + a^2 (since we are in characteristic 2). By...
ahhh i see
thanks!
kind jacinth
We have (1 + a)^2 = 1 + 2a+ a^2 = 1 + a^2 (since we are in characteristic 2). By...
what about (1+a)^3. this gives me (1+a)^2 * (1+a) = a*(1+a) = a^3 but this a^3
nvm got it
its 1
tribal moss
As it had better be, because the multiplicative group of the field has order 3, so g^3=e for every element of the group.
kind jacinth
As it had better be, because the multiplicative group of the field has order 3, ...
how do u know that is the case? i mean I realized it by doing the computations but how would u do that otherwise?
tribal moss
How do I know the multiplicative group has order 3, or how do I know g^3=e for all g?
kind jacinth
How do I know the multiplicative group has order 3, or how do I know g^3=e for a...
how do u know the multiplicative group has order 3
tribal moss
I count the elements:
- 1 (one)
- x (two)
- x+1 (three)
tribal moss
The multiplicative group of a field consists of all of the field's elements except 0.
kind jacinth
The multiplicative group of a field consists of all of the field's elements _exc...
ahhh yes thanks!
delicate orchid
Depends what distribution we’re selecting the order of the abelian groups with I suppose
hot lake
you choose an order in the completion of Z and pretend that it makes sense
that is
your order is divisible by p^n with probability 1/p^n
nevermind that it will almost surely be divisible by infinitely many primes
if you are divisible by p^n and not p^(n+1) there are uh partition(n) ways of making an abelian p-group with it
delicate orchid
Just do a Poisson distribution with lambda = 583 quadrillion
hot lake
and the number of abelian groups is multiplicative
so if you are divisible by p^1 the only p-group of that size is Z/pZ
so that's not really doing anything
so even if your uniformly chosen element of the completion of Z has almost surely infinitely many primes dividing it, they have exponent 1 except finitely many so the "number of abelian groups" is still finite
delicate orchid
Do you mean exponent 0?
hot lake
I did mean 1
the sum of 1/p diverges
so you expect to be divisible by infinitely many primes
if you're a uniformly random citizen of the completion of Z
delicate orchid
Right ok
hot lake
however, the sum of 1/p² converges
delicate orchid
I’m not gonna pretend to follow 
chat yo stuff king
hot lake
so you expect that there are finitely primes that divide you twice
tribal moss
The claim is probably that $$\lim_{n\to\infty} \frac{#\text{ abelian groups of order}\le n}{n} = 2.29...$$
cloud walrusBOT
Troposphere
tribal moss
Due to the structure theorem if feels sensible that this limit would exist and have something to do with the zeta function.
formal ermine
delicate orchid
<:catThink:526983414226747393>
This is nifty
fast nova
What is it that troubles you?
void cosmos
just proving commutativity of the diagram thats all
verifying the diagram is commutative
fast nova
It takes a bit of work. I once tried to outline the proof in detail for a friend, I can try and find the pdf I had for it if you are interested?
void cosmos
yea sure
fast nova
Actually it would be probably more fruitful for you to go over the sketch for the proof in Hatcher’s book (p. 147). If that doesn’t clear it up just let me know and I’ll try to help :).
void cosmos
okaay cool
will check that out
its not super hard ig
thank you!
a more important question for me
is the point of a splitting exact sequene is when to realise that its just the exact sequence of the usual direct product with inclusion and natural projection?
south patrol
That's more or less the definition right?
But it means you can deduce the structure of one of the objects from the other two, which is super useful
void cosmos
That's more or less the definition right?
yea i was just making sure haha
ig splitting exact squences are rare then ig
void cosmos
hey how is Q a finitely generated Q-module?
prisma ibex
Use your noggin
void cosmos
omg yea
cuz we can multiply by Q now
but we cant as groups right
so thats why Q is a counterexample ( for finitely generated R module --> fin generated as abelian group)
am i right? sorry im stupid rn
void cosmos
yeaa
okayy got it
tysm
can i do number 9 by some splitting lemma stuff
or is that not the waay to approach it
rank-nullity ig
ig it has to do with like hilbert spaces projections and stuff
so it must be some rank-nullity
or nvm yea
i just did it manually
but it sucks
can it work using splitting lemma tho?
void cosmos
so Ker(f) --> A --> im(f)
and under that is what..
ig the ff=f would help with commutativity
south patrol
given x in A, f(x - f(x)) = 0, so x - f(x) is in ker f and x in ker f + im f
then it is easy to see that this sum is direct
void cosmos
given x in A, f(x - f(x)) = 0, so x - f(x) is in ker f and x in ker f + im f
yea thats what i did
south patrol
Tbf okay yeah sure splitting lemma does apply
void cosmos
then it is easy to see that this sum is direct
yo i dont understand the difference between direct and external?
or external and internal
why is there a difference?
im confused
south patrol
External vs Internal
Basically just a technical thing that doesn't really matter too much lol
void cosmos
yea cool fuck that
south patrol
But i can explain if you want
void cosmos
go ahead
south patrol
so like
I'll work with modules to make it easier/ for definiteness
Given a module M and submodules M' and M'' we can write M = M' (+) M'' if like each m in M can be written uniquely as a sum of an element of M' and an element of M''
But we're interested in like reversing that construction
Namely, if I have modules M' and M'', can I create a module M which M' and M'' embed in such that (identifying M',M'' with their images) M = M' (+) M''
void cosmos
Namely, if I have modules M' and M'', can I create a module M which M' and M'' e...
wouldnt this be M + M'?
south patrol
Yeah pretty much, cutting the module up into parts kinda
void cosmos
m+m' where m in M m in M'?
void cosmos
why cant we say {m+m'| m in M , m' in M'}
south patrol
How can you add elements from different modules?
void cosmos
oh lfao
south patrol
But yes, you can do pairs
The normal cartesian product M' x M'' with standard addition etc
void cosmos
wait those arent submodujles/
south patrol
They aren't because I was saying "reversing" the construction
void cosmos
mb mb
south patrol
Like we know a module can decompose as a direct sum
Can we, conversely, direct sum modules into one thing
And yeah we can by that cartesian product thing here
void cosmos
Like we know a module can decompose as a direct sum
when can we do that?
is it the fundamental theorem?
south patrol
I mean there is no like easy way to detect it
south patrol
np
But yeah for the record uh like
Being able to understand when a module can be split up into smaller ones is important in representation theory and other things
and there are methods to do it in special cases but ye
void cosmos
yea
im preparing for an exam
yo can u help me with the splitting lemma proof for this
i really wanan do it cuz it was the first thing that acme to my mind
came*
for problem 9
nvm i got it
ker(f) --> A --> im(f) --> 0
u have the map f:im(f)--> f satisifes the lemma
so i get the iso for literally free
tf
south patrol
yup
void cosmos
yea thats so op
prime sundial
let R be a commutative ring with identity , e in R be idempotent, and phi:R->R, phi(x) = ex
i want to show ker(phi) is a principal ideal. any hints?
rustic crown
say f = 1-e
prime sundial
i see, thank you
rustic crown
lapis trail
How'd they get b=d=+/- 1 in the example considering F is a field?
celest furnace
Hey guys, this quarter I can get writing credit for my CA class for writing about a mathematician/theorem/advanced topic. My professor seems to REALLY want me to include history (I offered to write about the classification theorem for modules over a PID, and he denied it because it didn't include enough interesting facts/histories about someone), so he sent me back an email about Lasker's smaller theorem on the topic. Problem is, I can't read any of his work, because I am engish, and (I believe) he is german.
TLDR: I was wondering if anyone knows of an interesting result by an english-speaking mathematician (in algebra) which I could understand at the undergrad level (including some reading, at this point I know at least a little module theory)
prime sundial
Problem is, I can't read any of his work, because I am engish, and (I believe) he is german.
did you bring this up to your professor?
also, is there no translations anywhere?
celest furnace
https://zenodo.org/record/2052447 here's the paper, I have tried looking for a translation (with a quick google search) to no avail, and no i haven't emailed him about that yet
already sent him like ~4 emails and I don't want to waste his time 😦
Lasker isn't that well-known (at least, not as much as Noether), so I doubt there will be a translation out there
summer path
Perhaps I'm misunderstanding something, what does history facts of a mathematician have to do with reading their original paper?
prime sundial
How'd they get b=d=+/- 1 in the example considering F is a field?
maybe they're assuming the coefficient field is Z
celest furnace
Perhaps I'm misunderstanding something, what does history facts of a mathematici...
Well... he wants me to include like original good ideas (i.e., what lead to the theorem being proved), and Lasker in particular "also wrote some philosophy that nobody has ever cared about as far as I can tell. "