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#groups-rings-fields
obsidian sleet
we use x^-1 to find that e is in H but then to check if x^-1 is in H we use e
???
what
oblique river
no, we use x to show that e is in H
and then once we've done that, we know e is in H, so we can use it to show that x^(-1) is in H
obsidian sleet
hmm
chilly ocean
I think you should has yourself what the definition of a subgroup again is
obsidian sleet
ok i will reread that and come back to this
latent anvil
so many 🐺 reacts and so few messages
latent anvil
no i said 🐺
obsidian sleet
ok so we're just going through the group axioms by doing this test but by choosing the right elements in H to do so
cus i see now that the "whenever" part is important
a,b in H => ab^-1 in H
is supposed to be true
for all a,b in H
and since H is nonempty we can choose one element in it and deduce that xx^-1 is in H which is e
hmmmmmmmmmmm
agh this still feels weird
this is why im gonna be getting two passes of this stuff
big thinky
thorn delta
do u have a question?
obsidian sleet
i did
im still thinking on it
what if we changed the rhs of the implication
what if it wasn't ab^-1
in H
chilly ocean
agh this still feels weird
It feels weird at first but if you look at its quite restrictive
obsidian sleet
hmm
chilly ocean
sorry I shouldn't say restrictive
obsidian sleet
it seems like the choice for ab^-1 is the one that ensures that all the group axioms come out of it
hmm maybe i should look at examples of subgroups that Fail this test
chilly ocean
H is a subgroup if it contains the identity, and if its closed under multiplication
thorn delta
no subgroup will fail this test tho 
chilly ocean
no group will fail this test
obsidian sleet
er right
i shoudl say
subsets of a group that are not subgroups
poor phrasing initially :3
chilly ocean
The whole point of this theorem is the following two statements are equivalent:
- H contains e, and H is closed under multiplication
2)a,b in H, implies ab^-1 in H
what does closed under multiplication mean
thorn delta
you also need closed under taking inverses in 1)
chilly ocean
subgroups are automatically closed under taking inverses?
latent anvil
hot take: this theorem is dumb and I think teaching it isn't worth it
thorn delta
we're talking about sufficient conditions to have a subgroup tho lime?
latent anvil
it's a cute trick to for packing together the three conditions (1) identity is in there (2) product is in there (3) inverses are in there
obsidian sleet
closure under multiplication means that if a and b are in H then ab is in H for all a and b
i feel like
this is way too compact
my teacher showed another test
which seemed way more friendly
latent anvil
I would agree metal
I straight up think the a b^-1 thing is not worth thinking about
it's obscuring the real idea of a subgroup
obsidian sleet
hmm
chilly ocean
we're talking about sufficient conditions to have a subgroup tho lime?
I thought H contains e and H is closed under multiplication implies H is a subgroup?
thorn delta
oh yea, it would work for finite groups
chilly ocean
ah gg
golden pasture
hot take: this theorem is dumb and I think teaching it isn't worth it
it looks like an exercise lolz
golden pasture
oh
chilly ocean
I think its a good exercises
chilly ocean
and its a hand trick for doing exercsises
golden pasture
i mean is good way to test if students know what subgroup is ig?
latent anvil
I don't think it's important for actually checking that something is a subgroup
it's usually cleaner to do the conditions separately
golden pasture
(it isnt cuz it's so unnatural)
viscid pewter
agreed
oblique river
coming in late but I think it's a reasonable exercise in order to get students practicing working with group axioms, but it shouldn't be presented as a theorem or anything which is ever used at all
in general I think this is something not done very well in math classes, especially intro ones
from calc to group theory
making clear the difference between what is an important theorem and what is just an exercise meant to help you practice certain skills or applying certain definitions
chilly ocean
making clear the difference between what is an important theorem and what is jus...
This is good point I’ve never thought about
chilly ocean
About modules take A=k[x], atyiah macdonald says that in this case an A-module is a k-vector space with a linear transformation
how do I see this?
To give a module structure is the same thing as giving a ring morphism from A to the endomorphisms of my group M
I think to give a ring morphism from k[x] we need only say were x is sent
vestal snow
I think you just look at the restricted action by $k$
Since A contains k
Oh wait were you asking the other way around?
chilly ocean
Sorry i thought the second definition was easier
vestal snow
Like how to make a k vector space into an A module using some linear transformation?
chilly ocean
no im asking how to see that a k[x]-module is a vector space with a linear transformation
but i think what you said makes sense
If have a module structure, then we can restrict to a k-module structure
this is a vector space
vestal snow
I don't get the "with a linear transformation" part of your question
chilly ocean
then the linear transfomartion is just saying that the point v is sent to x acting on v
example 4, i am trying to understand what they meant by with a linear transformation but i think i get it now
vestal snow
Oh okay this is slightly different
you take an A module M
and you look at the map "multiplication by x"
Define this map as T_x
chilly ocean
this is the linear transformation
vestal snow
Then this A-module can be identified with a k-vector space (simply restrict the action) and the linear map T_x
chilly ocean
on the vector space (M,\mu restricted to k\times M)
vestal snow
I think what it's trying to say here is that there is a bijection between A-modules and ordered pairs (V, T) for a k vector space V and a A-module transformation T
chilly ocean
yeah thanks very much I see now
vestal snow
Can you send a screenshot of what it says before the example?
latent anvil
I think to give a ring morphism from k[x] we need only say were x is sent
This is a good way to think of it fyi
chilly ocean
I was confused because i was thinking what the heck is (x^2+x)*m is
chilly ocean
Can you send a screenshot of what it says before the example?
clearly by k-vector space we allow infinite dimensions
vestal snow
Yup
chilly ocean
Using the standard definition of a module, the elements in M are the basis vectors?
for a k-module
vestal snow
I don't get what you mean
Like is every element of M a basis for M?
the answer is no as 0 can never be a basis
Also, in general, modules do not need to have a basis
vestal snow
Yes
chilly ocean
Say M is k-module. The dimension of M is the cardinality of the abelian group M
thorn delta
R=reals; R^n is an R vector space but R^n Is not a basis
vestal snow
No that's not how the dimension is defined
The dimension of a k-module (usually called a k-vector space) is the cardinality of a basis set of the k-module
thorn delta
i interpreted this as lime asking whether the set M forms a basis
for M as a k-vector space/module
chilly ocean
the set of the group M yes
chilly ocean
I have the answer now but the question was
Let M be a group of size n. Give M the structure of a k-module. Then there is some way to consider M as a k-vector space.
Is it the case that the dimension of M as a vector space is n.
vestal snow
In general no
Also, the dimension depends of the module depends on how you "give M a k-module structure"
chilly ocean
jesus this alot more subtle than i thought it would be, I was thinking that if we had M=Z/3Z, and k=R. Then M as a k-vector space would be R^3.
chilly ocean
But yeah now I don't even see how you can give Z/3z an R module structure
vestal snow
What you're thinking of is the free k-module over the set M=Z/3Z
chilly ocean
I was just thinking of it as a formal multiplication
vestal snow
which you will encounter in this chapter
when it contructs tensor products
Yeah I think this chapter and integral dependence were my favorites in A-M
chilly ocean
ugh flatness
sturdy marsh
chilly ocean
Sorry for the super late reply, but I don’t quite see y you get a map from the cokernel
I don’t have much familiarity with cokernel aside from the definition
next obsidian
This is the universal property of cokernel
Coker f: M -> M’ has the property that if you have a map M’ -> N such that M -> M’ -> N = 0
Then there’s a unique map coker f -> N such that M’ -> coker f -> N = M’ -> N
This is the definition of cokernel
bleak abyss
next obsidian
Then it’s the universal property of quotients
carmine fossil
How do you show each f_i belongs to some ideal generated by some finite subset of S?
sturdy marsh
each fi is some finite combination of things in S
carmine fossil
each fi is some finite combination of things in S
How do you know that?I think that's my question
sturdy marsh
that's how generation works, you don't have infinite sums
like f_i = a_1s_1 + .... + a_ks_k
s_j in S
a_j in the ring
carmine fossil
So,when we say an ideal generated by an infinite set{f_1,f_2...},we mean
elements of form g_1 f_1 + g_2 f_2... Where only finite number of g_i's are nonzero?
golden pasture
I think flatness is pretty cool
annoyed by the name tho so geometric but so ungeometric
chilly ocean
I think flatness is pretty cool
Do you have any good intuition about the property
vestal snow
Hmm I usually just think of it how I think of normal subgroups
It's how one would want a module to behave, but not every module behaves that way
raven fern
are these quantifier-free axioms for boolean algebra complete? if not, is there a complete quantifier-free extension of it?
cloud walrusBOT
katia
sturdy marsh
raven fern
I know of an extension of it adding the following axiom, but the catch is that it uses existential quantification
cloud walrusBOT
katia
raven fern
Huntington says in Sets of Independent Postulates for the Algebra of Logic that this axiom and axioms 7 and 8 are independent from axioms 1-6 and 9, but not that this axiom is independent from 1-9, so it's possible 1-9 are complete, though I'm not sure
cloud walrusBOT
Moth | not male
maiden ocean
@unique juniper
unique juniper
It follows that we can now write I as a product of m transpositions in which the
first transposition to be applied fixes n (this was proved under the assumption
that τm(n) ̸= n, and I is already in this form if τm(n) = n).
i dont get that part
carmine fossil
That just seems like a unnecessarily complicated proof
identity is even,therefore it's a product of even number of transpositions
rustic crown
no... so the point of the the lemma is to define the notion of sign of a permutation
carmine fossil
nvm
unique juniper
is there a better proof?
rustic crown
yep
carmine fossil
identity is even,therefore it's a product of even number of transpositions
This
Which book is this?
unique juniper
Algebra and geometry by Alan Beardon
rustic crown
do you want the better proof or want to understand this proof?
i haven't read it... but the idea looks like pushing the element "n" towards the very left until its gone
unique juniper
could you link a better proof?
rustic crown
i like this one, the group S_n acts on Z[x1,..., xn] by permuting the xi
Now consider the polynomial D = prod_{i<j} (x_i - x_j)
rustic crown
What we're trying to do is push the "n" towards the left as much as possible
unique juniper
ye
rustic crown
if τ_m(n) = n, then τ_m doesn't "contain" n and so we can keep moving leftward from τ_{m-1} and so on
but in case τ_m(n) = a, then we need to work a little to see how to get another product for identity but which doesn't have a "n" in its last transposition.
this is handled by the induction, since m>=2, we can divide it according τ_{m-1} in the 4 cases.
if you look carefully, we replaced the last 2 permutations in each case with 2 other permutations such that n doesn't appear in the last transposition.
so by this whole argument, we say that we can change the product in a way such that the last transposition didn't move n.
if you continue this argument, then you can assume that τ_2 also doesn't move n.
but since the whole product was supposed to be identity, τ_1 must also fix n.
all in all, you completely removed n from the product. and now it only has elements in {1, 2, ..., n-1}, and so we may use induction hypothesis
unique juniper
ugh
i dont get it
"It follows that we can now write I as a product of m transpositions in which the
first transposition to be applied fixes n"
rustic crown
that "first" refers to τ_m...
rustic crown
didn't get what you mean...
the product of these 2 transpositions is identity... so it does fix n...
unique juniper
lemme try it with n = 3 to see if i can see it
Ok
So
I get the idea of pushing it all the way to T_1
all the transpositions fix n
rustic crown
nice
since all fix n, you can just look at the product over the permutation of {1, 2, ..., n-1} and so you'd be done!
unique juniper
nice thanks
rustic crown
gray charm
Let F be a field, and K a field extension of F. Suppose a, b \in K are algebraic over F with degrees m and n, where m and n are coprime. Prove F(a,b) is of degree mn over F and F(a) \cap F(b) = F.
For the first statement to prove, I am thinking of using [F(a,b):F] = [F(a,b):F(a)][F(a):F], so I try to show [F(a,b):F(a)]=[F(b):F]. I let p(x) be the min polynomial of b over F, and want to show that p(x) is also the min polynomial of b over F(a). Am I on the right track?
subtle current
[F(a,b):F] = [F(a,b):F(a)][F(a):F] = [F(a,b):F(b)][F(b):F]
So [F(a,b):F] is divisible by both m and n and hence by mn
But [F(a,b):F(b)] = [F(b)(a):F(b)] <= [F(a):F]
So [F(a,b):F] = [F(a,b):F(b)][F(b):F] <= [F(a):F][F(b):F] = mn
So it should be equal to mn
subtle current
Let F be a field, and K a field extension of F. Suppose a, b \in K are algebraic...
I think we can continue with this approach as well
But I don't see an easy way to prove that minimal polynomial for b is same over F(a) as well
rustic crown
yea, it would be more or less the same argument... say a smaller degree polynomial q(x) was the minimal polynomial for b over F(a). Then the [F(a,b):F(a)] = deg(q) => [F(a, b):F] = deg(q) * [F(a):F] = [F(a, b):F(b)]* deg(p).
deg(q) * m = [F(a, b):F(b)] * n
n divides the product on left, but is coprime to m,
hence n = deg(p) divides deg(q)
deg(q) >= deg(p)
this contradicts q had smaller degree.
obsidian path
can someone guide me through showing addition and scalar mult. are well-defined?
carmine fossil
Addition being well defined is just W being a abelian(and hence normal) subgroup
obsidian path
makes sense
carmine fossil
For scalar multiplication, Let's say a+W=b+W
Then a-b is in W. Since W is a vector space ca-cb will be in W
Implying ca+W=cb+W
obsidian path
so a and b are any random elements in V?
rustic crown
yep
carmine fossil
Yes
obsidian path
"Since W is a vector space ca-cb will be in W"
what axiom does this follow? (I assume its from an axiom for vector spaces)
obsidian path
Addition being well defined is just W being a abelian(and hence normal) subgroup
could you elaborate?
carmine fossil
Are you familiar with normal subgroups?
obsidian path
I should be, but I'm not very comfortable with them yet
left and right cosets coincide
carmine fossil
Yes
obsidian path
oh, then ofc addition follows
carmine fossil
I mean W is normal because the abelian group V is abelian
unique juniper
"It is clear that f maps even per-
mutations to odd permutations, and odd permutations to even permutations"
dont you need to know how many transpositions that P can be represented as tho?
idk how its "clear"
nova plank
Well, you are always adding one transposition to it because you are composing with σ
So if there is an even number and you add one, then there is an odd number
And vice versa
@unique juniper
unique juniper
ok
nova plank
Do you understand?
unique juniper
ok i misread, got it
nova plank
Okay
unique juniper
ty
gray charm
But [F(a,b):F(b)] = [F(b)(a):F(b)] <= [F(a):F]
So [F(a,b):F] = [F(a,b):F(b)][F(b...
Thanks for the response. How do you know that [F(b)(a):F(b)] <= [F(a):F]?
gray charm
yea, it would be more or less the same argument... say a smaller degree polynomi...
Cool, thanks
latent anvil
Thanks for the response. How do you know that [F(b)(a):F(b)] <= [F(a):F]?
[F(b)(a) : F(b)] is the degree of the minimal polynomial of a over F(b) and [F(a) : F] is the degree of the minimal polynomial of a over F. If m(x) is the minimal polynomial of a over F then m(x) is a polynomial with coefficients in F(b) such that m(a) = 0, and thus the minimal polynomial of a over F(b) divides m(x) in F(b)[x]. In particular the degree of the minimal polynomial of a over F(b) is at most than the degree of m(x)
devout crow
let $X \subseteq \bR$. if $A$ and $B$ are finite subsets of $X$, then the statement $$\sum_{a \in A}2^a = \sum_{b \in B} 2^b \iff A = B$$ is true when $X = \bN$ or $X = \bZ$ - this follows from uniqueness of finite binary representations. however, this statement is false for $X = \bR$ because, for example, $2^0 + 2^2 = 5 = 2^1 + 2^{\log_2{3}}$.
my question: what about the rationals? does the statement hold for $X = \bQ$ ?
cloud walrusBOT
xdres
delicate bloom
yeah, I think we can put ourselves in a finite galois extension of $\bQ$ that contains the roots of 2, and if we have two inequivalent representations then we can pick an element and look at the automorphism that changes this and keeps the other elements fixed to produce a contradiction.
cloud walrusBOT
Merosity
delicate bloom
but might be some details to sort out, I don't think we can quite guarantee we will be able to fix all the other elements every time perfectly
chilly ocean
Soft question
For people who are good at algebra
How do you conceptualize algebra in your head/mind’s eye
For geometry you can have this strong visual intuition or do tricks like picturing a familiar manifold and making sure you don’t use any of the specific properties of that particular manifold
solemn rain
i am not good at algebra but i will answer it anyways cuz im bored
u dont
its abstract
maybe at best u can have an idea or two or like motivation behind definitions
but thats it i think
u just work with defs thems
thats just my bad opinion
chilly ocean
I don’t even strictly mean conceptualizations that give you intuition
solemn rain
what do u mean then
chilly ocean
For example one way of keeping track of stuff if R is a ring with I and ideal contained in J
You can picture those three bits of information as the one image of the set R contains J contains I
But I would like to hear how some of the large brains conceptualize algebra
solemn rain
yea 0 large brain for me but like
i mean
this is just information that u cna think of as you like
but for example
how would u 'conceptualize' key theorems
or like how do u recall them
?
for u?>
chilly ocean
this is just information that u cna think of as you like
Very high information sentence
solemn rain
haha
solemn rain
how would u 'conceptualize' key theorems
.
solemn rain
ig
chilly ocean
Lagrange’s theorem I read as text in a book. When I recall the theorem I don’t think recall the sentence ‘if G is a group and H is a subgroup then the order of G is divisible by the order of H’
solemn rain
what do u do then
chilly ocean
I get an abstract picture of a group H fitting in G n times
I mean you can have plenty of visual intuition for topology even though a general topology space is not a possible physical space
solemn rain
lmao i am the opposite way
im learning about topology now
and i just cant do shit haha
delicate bloom
I imagine the group represented by matrices in GL_n(R) and are the symmetry group of some shape in R^n or find some kind of specific geometric picture for the case at hand
if it's more than just some silly problem that I care about, I have some picture in my mind
thorn delta
A common theme for me is subconsciously thinking in terms of examples.
solemn rain
yea i think examples are the way to go
thorn delta
Rn in topology. Polynomials or integers in algebra etc...
delicate bloom
I imagine a queue of isomorphism theorems, and standing first in line - boom
thorn delta
I imagine it as like, the more of the kernel you quotient out, the more injective the original becomes lol
noble saddle
The first isomorphism theorem is like saying, man I wish this map were injective. But it can't be, because some nonzero shit maps to zero!
So you just pretend everything that maps to 0 is 0
I don't think geometry is really the best way to see it
chilly ocean
first iso theorem
chilly ocean
I don't think geometry is really the best way to see it
I don’t really mean a geometric intuition
But a pictorial representation for touring the information
If I’m doing a problem I don’t want to be thinking of some sentence
noble saddle
But, to be clear, lots of geometry is super visual! I don't understand any commutative algebra other than what I can translate into geometry, for example
lol geometry. I mean algebra
chilly ocean
Yeah so I know there are these nice geometric intuitions with spectrum of rings and intersections of ideals
delicate bloom
yeah, I just think of the first iso thm as like we can lift the lid on an element of the group and peek in side to see all other elements that it was a representative for
idk if that's pictorial enough
noble saddle
but like, even at a much simpler level than spectra
You can translate intuition about polynomial rings over C into decent understanding of rings as a whole. Even if schemes give a better picture in general
chilly ocean
Okay say this question here
Look it’s not to hard to just bang out the definitions
But can you try think about what way you think about the problem when you read it
sturdy marsh
quotient groups can be thought of as fiber bundles
noble saddle
When I read that, I say a ring where every function is supported on one "point"
delicate bloom
x^n=x makes me try to think about a projection I suppose
sturdy marsh
and you have a section precisely when the fiber bundle is trivial
noble saddle
Also every prime ideal is maximal literally means 0 dimensionally, which also is what you wany
*want
chilly ocean
Okay
So in general I should think about an arbitrary ring being a ring of functions
noble saddle
These are really good questions by the way! Yeah, that's what Grothendieck tells us, and it tends to work out more often than it doesn;t
chilly ocean
My topology lecturer was big on talking about ‘meta-cognition’
Now we say a ring of functions
Do you just think about functions form R to R and try not to use and properties of those
noble saddle
I pretend to think about C to C, because real numbers tend to obscure stuff now and again, but basically yeah
chilly ocean
And these functions supported on a point because x^n=x
Ah that’s why we needed complex
noble saddle
Yeah in some sense this couldn't happen otherwise. Probably you can prove that under reasonable geometric assumptions, these are euivalent, but it's been a hot minute for me
*equivalent
noble saddle
that squid still turns me on to this day
chilly ocean
actually could someone help talk me through this picture
The furry-ness is related to taking some closure
golden pasture
the furry stuff are generic points
noble saddle
All algebraic geometry should be thought of as over a field
So over Z, you have a family of different varieties
chilly ocean
furry-ness///
noble saddle
and some italian at some point said "that's a mean-a generic point"
golden pasture
you get the usual affine line Z
golden pasture
https://i.stack.imgur.com/2e9Cd.jpg
so you see the affine line there
then you "extend" those primes to like fibers
noble saddle
and usually if you just plug in the variable, you get a "generic point" (x,y,...)
golden pasture
and the prime at infinity of Z corresponds to like
(x)
(x^2+1)
type of primes
noble saddle
For me, algebraic geometry made sense when I understood that all AG is fibers over something. And GRothendieck's magic fan bullshit about schemes over Z meant every variety ever is a fiber of something over Spec Z just felt right
chilly ocean
and so the space is discrete
Are the supports of functions related to the topology of the spectrum
golden pasture
im at the "fibers over something" part of my ag path rn 
delicate bloom
http://www.neverendingbooks.org/mumfords-treasure-map
nice this is pretty readable
noble saddle
Are the supports of functions related to the topology of the spectrum
They are what is called associated points, which is admittedly confusing as shit
noble saddle
im at the "fibers over something" part of my ag path rn <:smoL_nozoomi:719535199...
gl took me years to have somebody finally tell me how simple that whole idea is
chilly ocean
So an associated point need not be a point
noble saddle
So an associated point need not be a point
It's a point in the cheeky "points need not be closed" sense, yeah
chilly ocean
Okay
Wild guess
Being a local ring means
All the functions have support contained in some compact set
noble saddle
Vakil's book has a really wonderful and readable chapter on (well I guess everything) but especially on associated points and fuzz
Yeahhh sort of
but you have to remember that everything in AG is compact
because I don't fuck with non projective stuff
local rings actually predate rings in like "functions that don't have poles at this exact point"
chilly ocean
local rings actually predate rings in like "functions that don't have poles at t...
Sorry I didn’t get this at all
prisma ibex
whose closure is some subscheme
like think about the case of Spec(Z)
you have closed points (p) for p prime and also a generic point (0) whose closure is all of Spec(Z); this is drawn as some "fuzz" at infinity
noble saddle
Local rings come from complex analysis, originally. They are like "power series" which converge at a point
prisma ibex
it's a point that is "everywhere but nowhere in particular"
prisma ibex
in the usual topological sense
golden pasture
intersection of all closed sets?
prisma ibex
these are points in the Zariski topology
prisma ibex
a lot of points in the Zariski topology don't behave like points in the usual sense: in particular some points in the Zariski topology are not closed
golden pasture
seems like you care more about maps to other things instead of itself
noble saddle
Ehh the Zariski topology is not fundamental
prisma ibex
it's usually kind of hard to think of the Zariski topology as a topological space in the usual sense, too many of the usual intuitions about topological spaces fail
golden pasture
which makes sense cuz spec k is a very boring thing
chilly ocean
I mean generally you just think of the Zariski topology as a bookkeeping device
Really?
noble saddle
The Zariski topology was invented because it makes sense in not the complex numbers
But then we got sad because complex numbers have a decent topology
and invented the étale topology
golden pasture
haha yea
golden pasture
seems like you care more about maps to other things instead of itself
etale topology feels more in this vibes
prisma ibex
well the Zariski topology was invented because it allows you to endow Spec(R) with the structure of a topological space such that the topology corresponds to certain commutative algebraic properties of R
and it constructs Spec(R) as a "universal" topological space with commutative ring of functions R in some suitable sense
noble saddle
But it goes back to..well Zariski, but especially Weil who were not concerned with Spec(R)
prisma ibex
I'm pretty sure Weil was interested in Spec(R)? 🤔
noble saddle
They just wanted finite fields
prisma ibex
no? Weil was concerned with more than that?
chilly ocean
Is there a good intuition in this ring of functions picture for ideals
noble saddle
Weil was concerned with non projective things for sure
but non-fields? not sure honestly
prisma ibex
I mean Weil had a good understanding of the whole number fields/function fields picture
noble saddle
Yes! ideals are subvarieties
prisma ibex
among those certainly includes things like Spec(Z) and many more
noble saddle
I guess, but there is still no sense where that is a literal 1-1 translation
prisma ibex
Maybe not, but Weil is certainly talking about the same objects here
noble saddle
Very fair point. I have in mind a very particular set of papers when I think Weil
So I will concede
prisma ibex
I mean I'm thinking of his letters to Simone Weil in which he communicates this function field/number field analogy
it also shows up in some other of Weil's papers
it was definitely pre-Grothendieck but the same ideas were there
noble saddle
That whole time period is a clusterfuck of "who really was thinking what"
and it's so cool to read about
prisma ibex
yea
chilly ocean
Yes! ideals are subvarieties
Sus variety of ? The spectrum?
noble saddle
Wait, so is Weil the first to write about the function field-number field connection?
prisma ibex
more or less yea
chilly ocean
Is it true that when we at a ring can we viewed as a ring of functions, we actually mean the ring of functions over its spectrum?
noble saddle
Sus variety of ? The spectrum?
Technically, yes. But remember a scheme is a family of a varieties in the classical sense, so that's what we want
noble saddle
Is it true that when we at a ring can we viewed as a ring of functions, we actua...
literally yes
But literally in a sad way where we constructed schemes to make it true
chilly ocean
Technically, yes. But remember a scheme is a family of a varieties in the classi...
I only know the definition of a variety to be a k-space isomorphic to a closed subset of affine space
I have not seen schemes yet
noble saddle
eh, don't worry about schemes if you haven;t seen them. They just extend varieties to work in more situations that geometers want
The intuition is variety
noble saddle
If "large brain" = "pissed away my life getting a PhD" yes
for variety, don't even worry about affine opens
chilly ocean
In the manifold setting this is lovely picture
noble saddle
just think "zeros of polynomials"
chilly ocean
All the nice small disks
noble saddle
we glue because we like projective things
because again
I don't fuck with non-projective
chilly ocean
Is it a good exercise to try rephrase prime maximal radical ideals
In terms of what they look like for a ring of functions
noble saddle
you spit out a lot of words in a row, and I'm not sure what you're looking for
Oh like, prime, maximal, and radical separately?
Yes 100%
chilly ocean
Sorry yes
noble saddle
And read Atiyah-MacDonald for the answer if you haven;t
chilly ocean
And read Atiyah-MacDonald for the answer if you haven;t
Where is the answer in atiyah?
noble saddle
Spread out through the exercises in every chapter
Atiyah and Macdonald aren't your firend
but it's all there
chilly ocean
We can’t have any nilpotent elements as a function from C to C
noble saddle
If you like "learning" or whatever you can find this shit in Vakil's AG notes
we sure can
think about x^2
it's not zero
but it basically is to a physcicist
chilly ocean
think about x^2
I had a strong fear that I fundamentally misunderstood all my training so far
noble saddle
lol algebraic geometry is one of those things that makes no sense until someone who already knows it tells you all the secrets
noble saddle
as is tradition
chilly ocean
lol algebraic geometry is one of those things that makes no sense until someone ...
Do you have the secret sauce
noble saddle
yes, talk to anyone in your department because they know better than I ever learned
I still remember talking to my advisor about some "basic" shit we didn't know the answer to
and he calls over the geometer next door
who already has his coat on and uber called
and answers the question is 30 seconds
golden pasture
so
anyone wants to invite me to their department
my department rn consists of me, my cat, my piano
noble saddle
I'm always happy to talk algebraic geometry, but you have to be willing to deal with someone between piano and cat in intelligence
ivory dust
yall
so im trying to find centralizer G(A) with G = GL(2,R) ie grp of 2x2 matrices w real entries, and A = [ (2,1), (0,1) ]
and im stuck
i let B be arbitrary part of GL(2,R) i wrote it as [ (a,b), (c,d)]
and did AB = BA but my equation is kinda basic*
i get c = 0, b+d = a
and thats it
well also ad - bc =/= 0, so ad =/= 0 so a = b + d =/= 0, and d =/= 0
next obsidian
Fuck divisors
Also shoutout AG secret sauce, the secret sauce you need to be told is actually “think about it for 12 hours a day for a few months and it’ll make some sense”
carmine fossil
i get c = 0, b+d = a
That's enough to conclude the centraliser is made up of elements of form [(a,b),(0,a-b)]
a,b are free variables restricted by those conditions
celest brook
everything is matricies
golden pasture
latent anvil
i am currently
matricesing
idk this might be better suited for top geo
but it's also about rational canonical form
so maybe linear algebra
so ill post it here instead :^)
I'm trying to show that if $J : E \to E$ is an iso of real bundles satisfying $J^2 = -Id$ then $J$ can be locally represented as
$J_0 = \begin{bmatrix} j & 0 & \ldots & 0 \ 0 & j & \ldots & 0 \ \vdots & \vdots & \ddots & \vdots \ 0 & 0 & \ldots & j \end{bmatrix}$
cloud walrusBOT
Shamrock -> E -> B
latent anvil
where $j = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$ whoops
cloud walrusBOT
Shamrock -> E -> B
latent anvil
so the linear algebra version of this is that $J^2 + 1 = 0$, so $J$ has minimal polynomial $m(x) = x^2 + 1$, so $J$ has all invariants factors $m$, boom apply rational canonical form
cloud walrusBOT
Shamrock -> E -> B
latent anvil
but this is not constructive and so it's hard to bring it over to continuous bundle land
or at least it's not explicit
hi banana
vestal snow
Hello
latent anvil
is your username a reference to this? https://www.youtube.com/watch?v=k4f9m4OYkCY
2011 brony deep cut that I still remember lmfao
vestal snow
Haha no, it's a reference to something I said when I was in kindergarten to some girl from my class
latent anvil
ah gotcha
vestal snow
Let $R_1$ and $R_2$ be rings with coprime characteristic. Let $\varphi$ be an automorphism of $R_1\times R_2$. Prove that $\varphi (1,0)=(1,r_2)$ for some $r_2\in R_2$.
cloud walrusBOT
Have a Banana, Bitch
vestal snow
1 goes to 1 in my ring homomorphisms btw
latent anvil
I seem to have shown $r_2 = 0$
cloud walrusBOT
Shamrock -> E -> B
latent anvil
let $p$ be the characteristic of $R_1$ and $q$ the characteristic of $R_2$. write $ap + bq = 1$. Then $p \varphi(1,0) = \varphi(p, 0) = \varphi(0,0) = 0$, so $\varphi(1,0) = (ap+bq) \varphi(1, 0) = b q\varphi(1,0)$ if we write $\varphi(1, 0) = (x, y)$ then $(x, y) = (b q x, b q y) = (b q x, 0)$
cloud walrusBOT
Shamrock -> E -> B
latent anvil
bezout's feels right here regardless
vestal snow
Hmm I have been trying this for the past 20 minutes and I still can't do it
vestal snow
no I think its correct
latent anvil
hmm
weirdly stated then
ah I think maybe I see
so $\varphi(1,0) = (bq x, 0)$ and $\varphi(0,1) = (0, ap y)$
yeah?
cloud walrusBOT
Shamrock -> E -> B
vestal snow
Yup
cloud walrusBOT
Shamrock -> E -> B
vestal snow
I see
latent anvil
second equality is $\varphi(1,1) = \varphi((1,0) + (0,1)) = \varphi(1,0) + \varphi(0,1) = (bqx, 0) + (0, apy) = (bqx, apy)$
cloud walrusBOT
Shamrock -> E -> B
latent anvil
i mean it looks right but it's stronger than your statement
and doesn't use that it's an iso
so idk
latent anvil
ah okay
vestal snow
I just reduced it to this
latent anvil
then im comfy with it
you use a similar trick to show $\mathrm{Aut}(G \times H) \cong \mathrm{Aut}(G) \times \mathrm{Aut}(H)$ when $|G|$ and $|H|$ are coprime
cloud walrusBOT
Shamrock -> E -> B
latent anvil
which is a fun problem
vestal snow
@latent anvil I think there might be a problem with this proof
latent anvil
oh dear
vestal snow
Because both (1,1) and (1,0) are mapping to the same thing
latent anvil
How so?
(1,1) maps to (1, 1) = (bq x, ap y) and (1, 0) maps to (1, 0) = (bq x, 0)
Oh I think I used y to mean two different things at one point
Proof should still go through though
Also I think I was being a fucking idiot with the bundles thing
omfg this is obvious
vestal snow
Proof should still go through though
Let $\phi(1,0)=(x,y)$. We get (ap+bq)\phi(1,0)=\phi((ap+bq)(1,0))=\phi(bq,0)=bq\phi(1,0)=(bqx,bqy)=(bqx,0)$. Similarly, we get $\phi(0,1)=(0,apy).$
cloud walrusBOT
Have a Banana, Bitch
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
vestal snow
Ah! I did the same thing lol
latent anvil
update: I was not an iditio
it ended up being a variant of maschke's theorem, which is kind of cool
i mean i was a little bit of an idiot but still
coarse forge
Hi, I’m starting a course on abstract algebra in a few weeks. What are some textbooks that would be useful as an additional resource?
coarse forge
Oooh I’ll look into them ty!
coarse forge
Funnily enough, the course outline did recommend the Jacobson textbook so I shall look into that! Tyyy
nova plank
Jacobson is strictly better than dnf<:smug:705259507424428065>
That's a low bar
golden pasture
old lava
so many haters
paper flint
G a l l i a n
carmine fossil
Gallian is for kids
old lava
gallian is actually garbage
paper flint
old lava
before I thought it was passable
but now that I actually looked through it a bit
because of my course
actually just use wikipedia page for abstract algebra
over gallian
gallian is to abstract algebra as stewart is to calculus
paper flint
solemn rain
wow
yea the book is like abit not in depth but not to that level
the text is only lacking like group actions ig ( which is obviously very important but still )
its not a bad book 😦
carmine fossil
an algebra book without group actions is a bad book
chilly ocean
AA book with no group actions 
yikes
imagine leaving out the reason anyone actually cares about groups
chilly ocean
I’ve been trying to think about rings in term of rings of functions
If we imagine the functions are over some field or ring that does not have zero divisors
Then a a ring of functions being an integral domain
Is implied by all the functions having finitely many zeros
And I think it’s also implied if every function is non zero on a dense set
We could think of the ring of Z\oplus Z as being the ring of functions generated by Z-combinations of these
oblique river
Is implied by all the functions having finitely many zeros
This only works if the field or ring is infinite, otherwise you're in trouble
chilly ocean
Ah good point
oblique river
there's a nice example of "two elements a, b in a ring which are multiples of each other but not associates" with rings of functions that I've always liked
thinking about rings as rings of functions is not a bad way to get intuition about things
as long as you are careful to not rely too heavily on properties of, say, continuous functions on the real line, to try to make general statements about commutative rings
chilly ocean
as long as you are careful to not rely too heavily on properties of, say, contin...
Yeah
I’m trying very hard at the moment to think about the way I am thinking about maths
With algebra it just feels like I am pushing formal statements about
But this ring of functions picture is really helping
oblique river
yeah I've always found ring theory hard for that reason. I think groups are nice because you can visualize them as symmetries of objects, and with fields I always think of extensions of Q and galois theory which feels concrete to me
chilly ocean
as long as you are careful to not rely too heavily on properties of, say, contin...
In geometry I feel very comfortable picturing a 2 dim manifold and not using any of the properties of being 2d when I reason
But if I think an example of a ring I only have a feel for the specific properties of that ring
oblique river
but with rings I think it's easy to just push statements around and not really feel like you're not really dealing with any particular object
chilly ocean
Yeah
But for me personally this tends to lead to a problem where
I can’t “chunk” all the formal statements together well
And eventually the tower of what I’m learning because too unstable
I always have two go back and check what the statements of the kind like the intersection of prime ideals is prime , or something about an intersection being contained in a prime and such and such is coprime
sturdy marsh
the intersection of prime ideals may not be prime
take a product of fields FxF
0 X F and F X 0 are prime
but their intersection is (0,0)
which isnt prime
chilly ocean
the intersection of prime ideals may not be prime
Proof of my my confusion
sturdy marsh
yeah idk if there's a way to clearly organize all the concepts and interpretations of the stuff you learn during your first go at it
over a longer period of time, it sorta happens automatically?
like say you're learning comm alg right now, if you do AG later you will go back and look at some comm alg stuff that you forgot
and that second pass at it should normally help you digest it completely
idk if it is feasible to ask for a crystal clear picture of everything you learn during your first pass at the subject
also, reading MO soft-question threads that ask for motivation/intuition/interpretations of stuff can help
chilly ocean
idk if it is feasible to ask for a crystal clear picture of everything you learn...
Yeah this is very true
devout crow
yeah, I think we can put ourselves in a finite galois extension of $\bQ$ that co...
I know this is a late response, but how can the extension be finite if there are infinitely many roots of 2?
delicate bloom
the sums are finite, so we only need to look at a specific finite extension if there is a supposed counterexample
so like for instance suppose you give me 2 different representations that are equivalent x, y then we have x=y in some finite extension of Q, with a linear combination of these basis vectors of roots of 2.
then you have x-y=0 which is a nontrivial linear combination of basis elements equal to 0, contradicting that it's a basis
unique juniper
Show that every element of order 14 in $S_{10}$ is odd
cloud walrusBOT
Yes
viscid pewter
so when can an element of Sn ever have order 14
unique juniper
just need to check my reasoning for this one
viscid pewter
go on
unique juniper
if we supposed P = p_1 ... p_m
The LCM of the lengths of those permutations would be 14
And you need an even and and odd number to get 14
Cycle of length 2 has a sign of -1 (odd)
if that makes any sense lmao
viscid pewter
yeah i think that's it
unique juniper
sweet
obsidian path
can someone run me down on how to do this?
I know we suggest a basis and prove its linearly independent + spans the quotient space
I'm stuck on proving it spans
cyan marten
By the division, we have q(x) = a(x)p(x) + r(x), with deg r < deg p = n. So the images of 1, x, x^2, ..., x^(n-1) do span F[x]/(p).
chilly ocean
Show that every element of order 14 in $S_{10}$ is odd
Is it not true in general that ever permutation of odd order is even and vice versa?
old lava
no, that is not true in general
(1 2)(3 4) is of even order and also is even, while (1 2 3 4) is of even order but is odd
old lava
Is there such thing as a cyclic group generated as follows? $G = {g^q: q\in \mathbb{Q}}$
I don't think so, having rational exponents is only like well defined if you have like a divisible group. But, any finite abelian group is not divisible. Any infinite cyclic group can be written as a direct product with \bZ in it, and the integers aren't divisible, and a group that's a direct product is divisible iff every factor in the direct product is divisible. So, like no cyclic group is divisible, so like the rational exponents wouldn't make much sense unless you define it to be something else
@main quiver
main quiver
Oh crap thanks for that lol
viscid pewter
Is it not true in general that ever permutation of odd order is even and vice ve...
i think it's true for n-cycles
unique juniper
um yes
scarlet estuary
if so then every integer can be written as 1 + 1 + 1 + ... + 1
or -1 - 1 - ... - 1
for enough 1s
which is what 1^k means in this group
unique juniper
i see
scarlet estuary
remember that exponents in group theory repreesnt repeated application of the group operation
in this case +
NOT integer multiplication
unique juniper
yeah thats where i got mixed up at :D
chilly ocean
Every ring is the ring of functions on its spectrum
Can some direct me to where to read about this
wind steeple
It's simple, if you have A a ring, then for every f in A you can define a function tilde f from Spec(A) to the sum of A/p for p in Spec(A) by defining (tilde f)(p) = f mod p
latent anvil
@chilly ocean what does "ring of functions on the spectrum" mean to you?
vital quail
ring of ace functions
shut halo
Hey does this definition of what a module is make sense to anyone?
Previously, we discussed how elements of a group act on each other, and we also talked about how elements of a group act on some other object or set of objects (like three painted eggs). We now generalize this notion to a set of $q$ abstract objects a group can act on, denoted $M={m_0,m_1,m_2,\dots,m_{q-1}}$. Just as before, we build a vector space, similar to the one above used in building an algebra. The orthonormal vectors here will be
\begin{equation}
m_0\mapsto|m_0\rangle,\quad m_1\mapsto|m_1\rangle,\quad\dots\quad m_{q-1}\mapsto|m_{q-1}\rangle\tag{3.48}
\end{equation}This allows us to understand the following definition. The set
\begin{equation}
\mathbb{R}\mathbf{M}=\left{\left.\sum_{i=0}^{q-1}a_i|m_i\rangle,\right|,a_i\in\mathbb{R},\forall i\right}\tag{3.49}
\end{equation}is called Module of $M$ (we don't use square brackets here to distinguish modules from algebras).
golden pasture
this is just a free R vector space
shut halo
@golden pasture what do you mean by "free"
golden pasture
oop idt the word free is needed lol
it's just R^q as a vector space
it isn't how we define modules
shut halo
Ok so in general a vector space is a set with some properties, but here it seems that we are looking at the set of all possible linear combinations of some elements
golden pasture
which is exactly what a vector space of dimension q is
chilly ocean
There are many ways to do this problem and it’s easier to just define a subjection from Z and quotient by the kernel but I would like to use the universal property of the tensor product to do this question to get used to working with the universal property
If Z/nZ tensor Z/mZ is Z/gcd(n,m) Z
We can define a map from Z/nZ times Z/mZ to Z/gcd(n,n) Z by
([a],[b]) maps to [ab] where [x] represents the respective class in the quotient, there are some well definedness stuff to check but that’s not too bad
Next we need that this is bilinear, and by that we mean linearity wrt scalars in Z since we view everything as a Z module
This is also fine
This is where I get confused
chilly ocean
If Z/nZ tensor Z/mZ is Z/gcd(n,m) Z
Sorry this is the problem
My English is not very good
latent anvil
No worries
chilly ocean
I am a native hiberno-English speaker but often this leads to problems
My confusion is
latent anvil
What area is hiberno from?
(just curious)
I'm assuming that's a language
Also sorry for interrupting your question again lol
chilly ocean
To use the universal property of the tensor product to show that Z/gcd(m,n)Z is indeed the tensor product
We need to show that if P is a Z module
And there is a bilinear map from Z/n times Z/m it factors over the map we defined above
chilly ocean
What area is hiberno from?
Oh lmao, hiberno-English is just the dialect spoken in Ireland
latent anvil
Ah gotcha lol
chilly ocean
It’s just English but I mumble
latent anvil
Oh lmfao I thought for a sec you grew up in a gaeltacht
chilly ocean
Very funny, I was at a summer school in Italy once, lots of different nationalities at it, Italian spanish Costa Rican American
We were drinking one night about a week into the school
And one of the guys says
‘I’m sorry lime I didn’t understand anything you just said’
latent anvil
Lmfaoooo
chilly ocean
Someone else jumped in
‘Me too! I thought I just had bad english and couldn’t understand him because of that’
‘I usually just try and guess from the context and tone of voice what was said’
Was a very funny night
latent anvil
my parents have been watching reeling in the years recently and it's a very different experience for then than it is for me
Partially because they have cultural context, but also because they can understand the accents 
chilly ocean
Oh you should follow this page if you some Irish fun
Just a bunch of funny moments from reeling in the years
latent anvil
Okay sorry for interrupting your question
tensor product
you want to factor through the map Z/nZ × Z/mZ -> Z/gcd(n, m)Z
chilly ocean
Oh yes
chilly ocean
you want to factor through the map Z/nZ × Z/mZ -> Z/gcd(n, m)Z
Call this map b
Call the map from Z/n x Z/m to an arbitrary Z- module P by p
I want to show that p= i composed with b
latent anvil
for a unique i : Z/gcd(n, m)Z -> P
chilly ocean
Oh I thought it was the other way around?
latent anvil
uhh
chilly ocean
Since b has image Z/gcd and p has image P
chilly ocean
We also have to show that this i is unique?
chilly ocean
Rip to a G
chilly ocean
I am very confused about how to tackle this
The map p and Z-module P are pretty arbitary
all we know is it’s bilinear
latent anvil
yup
So what's this factorization saying really
It says that p doesn't depend on the value of x, y, only on their product
chilly ocean
Ah so
The map i should send [ab] to p(ab,1)
And then bilinear of p makes it work nicely
latent anvil
Makes sense to me
latent anvil
Just be careful about products of residue classes vs scalar multiplication and all
when invoking bilinearity
chilly ocean
Wait
Take Z/5 times Z/3
(2,1) should equal (7,10)
But the first gets mapped to 2 mod 15
The second gets mapped
Oops lmao
latent anvil
lol
chilly ocean
is it true that the gcd of m,n is the smallest number you can write as am+bn with a b nonzero?
delicate bloom
suppose you could write am+bn = k with k not divisible by gcd(m,n)
but the left side is divisible by gcd(m,n)
so k is divisible by gcd(m,n)
contradiction
latent anvil
(and the other direction is bezout's identity)
uncut girder
What's a reductive group? Over an algebraically closed field it's just a Dynkin diagram 👏
chilly ocean
Quick sanity check
M an A module, a an ideal of A that annihilates M, then M is an A/a module
M as an A module is isomorphic to M as an A/a module?
latent anvil
yup
Err
Not exactly
Isomorphic as what?
If you extend and restrict scalars of M along the map A -> A/a you get an isomorphic A-module
(in this case)
(and the extension of scalars is iso to the A/a-module M you're talking about)
scarlet estuary
First sentence: taking a union with an empty set doesnt change the original set
so if A_n is empty we can "pretend it isnt there"
like adding 0 or multiplying by 1
Second sentence: See "the above remark"
Third sentence: The only case you have left to consider, then, is to consider the case where there are infinitely many nonempty A_n sets
now we're "pretending" that all the A_n are nonempty
since again, if one is empty
we can just ignore it
so we're assuming that the A_n are already nonempty
(which is what the yellow highlighted part means)
(also i just realized this wasnt #proofs-and-logic )
(it should really be there)
leaden finch
okie, ty
snow flint
sideurk moment, but am i missing smth or do these as they are not show its a subring
scarlet estuary
sideurk moment
(slightly different from yours but equivalent)
might be a good idea to prove for yourself that those 4 properties imply a given set is a subring!
its a good exercise
though kinda easy
most of the proof boils down to "we inherit this behaviour from the parent ring"
snow flint
i thought subring test involved showing identity was in S hm
placid solstice
subring test i just learned in class involved showing closure over a - b and ab^{-1} on a subgroup (perhaps I'm misremembering)
scarlet estuary
sideurk moment
if you require the unities to agree then
you need an extra criteria
to say their unities agree
not all definitions of subring require this however
and if yours doesnt you're still okay
idk what exact definition youre using
but based on the screenshot i assumed your def doesnt require this
snow flint
well actually we didnt get a def lol
next obsidian
Joe papa
snow flint
ok archsys alt acc
chilly ocean
this is showing that A[x]/p[x] is isomorphic to (A/p)[x]
I don't understand why A to A/p being surjective implies the kernel is a superset of p[x]?
golden pasture
everything in p[x] vanishes
chilly ocean
it seems to me its easier to say in one step that the kernel is precisely p[x]
the image of f is non zero precisely when all its coefficents are zero, which only happens if all its coefficents are in p, which is p[x]
golden pasture
yesh
leaden finch
@scarlet estuary is there an easier proof for that?
golden pasture
just construct an injection from the direct sum to the natural numbers
by using unique factorization of primes
hence it is countable
leaden finch
can we use the union defintion for the proof?
cyan marten
Is every abelian group the additive group of some ring? Here I am assuming the existence of unity (so letting ab = 0 doesn't work).
It is easy if G is finitely generated, for example, because we can use the structured theorem.
carmine fossil
Pretty sure this will have smt to do with choice
cyan marten
Structure*
golden pasture
well you know the classification of abelian groups
soooo
the classification gives a ring structure
cyan marten
Pretty sure this will have smt to do with choice
The axiom of choice is true.
rustic crown
What about infinite direct sum of Z?
rustic crown
I'm not saying infinite product
cyan marten
well you know the classification of abelian groups
We do? I think we only know the finitely generated ones.
cyan marten
I'm not saying infinite product
It still works.
golden pasture
hm time to think of cursed nonfinitely generated ahelian grous
cyan marten
Maybe the Prüfer p-group is a counterexample because it is Artinian but not Noetherian as a Z-module.
rustic crown
What about infinite direct sum of Z?
I feel like any ring structure will force the identity to be (1, 1, 1,...) Which is not there in the group
carmine fossil
What about rng?
rustic crown
I feel like any ring structure will force the identity to be (1, 1, 1,...) Which...
lol Z[x]
cyan marten
Maybe the Prüfer p-group is a counterexample because it is Artinian but not Noet...
Wait, it's actually the direct limit of Z/p^nZ so can be given a ring structure.
golden pasture
Q/Z is a very funny group
carmine fossil
Isn't that just floor for rational numbers
golden pasture
Isn't that just floor for rational numbers
as in?
cyan marten
What about rng?
You can define multiplication by ab = 0 for all a, b. But we can do that, then make this rng into a ring (there's a canonical way to do this, I think)..
cyan marten
Wait... Can't we just take the group ring?
golden pasture
you want to preserve the additive group tho
cyan marten
For a group G, the additive group of ZG is the direct sum of |N| copies of G.
wind parrot
Abelian groups where every element has finite order, but there is no upper bound on the order cannot be made into a ring with unity; namely if 1 was your unit and it has say, order n, then for any x you have n*x = (n*1)x =0
golden pasture
mm makes sense
wind parrot
Say every element has order dividing n, so this goes wrong if there is no upper bound on the order
namely Q/Z is indeed a counterexample
wind parrot
I remember seeing this on MSE I think, lemme see if I can find the post
wind parrot
wait isnt the prufer p group as well
Yes
Here is was, I think the top answer is the same argument as I had https://math.stackexchange.com/questions/93409/does-every-abelian-group-admit-a-ring-structure
cyan marten
Abelian groups where every element has finite order, but there is no upper bound...
Nice!!
wind parrot
I only know because I thought about this once myself before and then I googled and found that MSE answer
Its a nice and clever little argument
golden pasture
all i know is that just check with groups like Qp/Zp Q/Z μ_p^\infty and if they work it's probably true
cyan marten
Wait, it's actually the direct limit of Z/p^nZ so can be given a ring structure.
What went wrong here?
crystal lion
all i know is that just check with groups like Qp/Zp Q/Z μ_p^\infty and if they ...
Pretty much 
cyan marten
Its a nice and clever little argument
Quite so
golden pasture
the additive group is prob different if taken as a direct limit of rings id guess?
wind parrot
What went wrong here?
Ah I think I see it, the maps used to define the direct limit are multiplication by p, Z/p^n -> Z/p^(n+1), but these are not ring homomorphisms
cyan marten
Oh, for the same reason Q/Z doesn't work (Z is not an ideal)..
golden pasture
ye
cyan marten
Is there a nice way to think about the quotient of P/S, where P is the countable direct product of some (group, ring, module) with itself, and S the direct sum?
golden pasture
golden pasture
that sounds very cursed
like say simplest example
direct product of F_2
and direct sum of it
direct sum isnt an ideal so we ignore the ring case
yea i have no idea what kinda group it makes lol
cyan marten
Wait, why isn't the direct sum an ideal? A typical element in S is (a, b, c,..., 0,0,0,0,0,...) so when we multiply by an element in P we get an element in S.
cyan marten
Which is nice
It has the same "two functions are equal if they agree except on a set of measure 0" spirit..
crystal lion
Wait, why isn't the direct sum an ideal? A typical element in S is (a, b, c,...,...
I love your index free notation lol
cyan marten
Suppose A < B < C are rings, where C is fg as an A-algebra and as a B-module. I thought B must be fg as an A-algebra, but apparently this isn't true.
Proposition 7.8 of Atiyah-MacDonald says that B is fg as an A-algebra if A is Noetherian.
cyan marten
Suppose A < B < C are rings, where C is fg as an A-algebra and as a B-module. I ...
(I obviously didn't think of this situation before reading AM)
cyan marten
Proposition 7.8 of Atiyah-MacDonald says that B is fg as an A-algebra if A is No...
I don't understand why, and can't think of an example where the Noetherian condition is necessary. Any help would be appreciated.
carmine fossil
If u1!=v1 multiply both sidea with a_1 and rearrange