i satisfied everything else but the existence of an inverse
#groups-rings-fields
cinder bone
solemn rain
np
cinder bone
so wait
how does this tell me i have an inverse that gives me the identity matrix specifically
rather than just any matrix with determinant of 1
solemn rain
the inverse must have det 1
for it to be in S
you showed S has the inverse
as the inverse is a 2x2 matrix over R with det 1
cinder bone
i guess first i should state that an inverse exists under matrix multiplication in general
and then i can say its an element of s
cinder bone
thanksπ
would i also be able to use this fact to determine if s possesses the commutative property?
cinder bone
nvm i found a sick counterexample
thorn delta
so we have this theorem
and this definition
so is there no special name for whenever a group G is a direct sum (weak direct product) of not necessarily normal subgroups not satisfying (i) or (ii)?
solemn rain
strong
the special case for this is when these normal subgroups are sylows
you call them finite abelian
lol
thorn delta
It seems weird that there aren't many (if any) required conditions to have a product of subgroups iso to the group
solemn rain
the only thing i remember
and im sure of
is that G is nilpotent is equivalent to G is a direct product of its sylows ( the theorem you stated when the normal subgroups are sylows )
G is a finite group
i think
prime gale
Yes.
G=Z*Z. N1=Z*1, N2=Z*1. Then G isomorphic N1*N2. But (i) and (ii) false.
(i) is false because N1=N2 so <N1,N2> != G. (ii) false because N1 intersect N2=N1.
Technically, I mean Z*{1}. But people write the trivial group as 1 all the time.
prime gale
Just do what the exercise says? Start with the exact sequence, nZ -> Z -> Z/n -> 0. Apply Hom. Get result.
slate forum
I didn't see what sequence to use...
it says left exactness though
Unless it's the other functor we're using
prime gale
Hom(.,G) is contravariant.
slate forum
ohh
I don't see how that gives the result... though, 0->Hom(Z/n,G) -> Hom(Z,G)-> Hom(nZ,G)
prime gale
What is Hom(Z,G)?
prime gale
And what is Hom(nZ,G)?
prime gale
nZ is the same group as Z.
prime gale
I could have written my exact sequence as Z -> Z -> Z/n -> 0, where first map is multiplication by n.
But it is easier to write it as nZ -> Z -> Z/n -> 0.
prime gale
And what is map from G->G in that left exact sequence?
prime gale
What is map G=Hom(Z,G) -> Hom(nZ,G) = G?
slate forum
The first identification considers g as the function that sends 1 to G
prime gale
Go back to definitions of what is going on here.
slate forum
Hom(Z,G)->Hom(nZ,G) is the pullback by multiplication by n
slate forum
?
prime gale
If you call composition pullback, then it does. But I've never heard anybody call composition pullback.
prime gale
If I have map f:X->Y, and g is element of Hom(Y,Z), then the map Hom(Y,Z)->Hom(X,Z) takes g to (g o f)
slate forum
Yeah,that's f^*g
prime gale
I just call that f upper star. I call g o f composition.
slate forum
Oh, alrighty then
prime gale
Anyway, calling it whatever, what do you get as your map from Hom(Z,G) -> Hom(nZ,G)?
prime gale
Yes. And after identifying Hom(Z,G)=G and Hom(nZ,G)=G, what is map G->G?
prime gale
Yes. And what is kernel of that map>
slate forum
G[n]
prime gale
Yes. And with left exact sequence, what is kernel?
slate forum
the kernel of a sequence?
uhhh
The kernel of the map we're talking about is the image of Hom(Z/n,G) under the pullback
prime gale
Huh?
We have left exact sequence 0 -> Hom(Z/n,G) -> Hom(Z,G) -> Hom(nZ,G). We have identified second second and third groups as G and map as n. What is kernel of the rightmost map of the left exact sequence I just wrote?
prime gale
If I have a left exact sequence 0 -> A -> B -> C, what is the kernel of the map B->C?
slate forum
The image of A
prime gale
Well, most people would just say "A".
slate forum
But you want it up to isomorphism?
oh ok
fine "a"
"A"
I'm just a stickler for that I guess
prime gale
They are not just isomorphic. They are naturally isomorphic. Big difference. When they are naturally isomorphic they are more or less equal. It is more accurate to call naturally isomorphic things as equal than as isomorphic.
slate forum
naturally isomorphic? I only understand that in the context of a natural transformation of functors
prime gale
Yes. Naturally isomorphic means invertable natural transformation.
Anyway, that's a digression.
prime gale
The two functors are the Hom(.,G) and Hom(.,G).
slate forum
But in terms of independence of basis or natural transformations
I thought you said when you have 0->A->B->C that A is naturally isomorphic to that kernel
prime gale
Okay. In that case you have functor taking LES to groups. First functor is the first element of LES. Second functor is second element of LES.
I guess if you want it to be kernel second functor takes LES to kernel of right most map.
In any case, you have your proof using left exactness?
prime gale
Usually you wouldn't mention all these gory details. Just write down the sequences. Say the map is multiplication by n, and conclude that Hom(Z/n,G) = G[n].
slate forum
What is LES?
prime gale
Category of left exact sequences.
slate forum
oh, was think long exact
I'm not very good at seeing what the map between the identifications is
prime gale
I don't think it always means anything in particular. Without context, it can mean anything.
Just we were talking about left exact sequences, so I figured I could abreviate.
slate forum
When we have R a commutative ring w/ id and M and R module, then Hom_R(R,M) is iso to M? I don't think that quite works, does it?
prime gale
You just have to carefully and slowly chase the stuff around.
slate forum
It just happens for Z modules here it seems
prime gale
Again, I would say equal, not isomorphic. They are more than just isomorphic. They are naturally isomorphic.
slate forum
This works for arbitrary R modules?
prime gale
The distinction is really important in some contexts.
slate forum
I don't see how to get every element of M from a homomorphim from R to M
prime gale
Pick element m in M. The corresponding map in Hom(R,M) is map that takes 1 to m.
slate forum
ok, yeah, never mind, it works
prime gale
Pick map, f, in Hom(R,M). Corresponding element if f(1).
slate forum
That's weird, then we have Hom(R,M) and Hom(M,R) iso to M?
prime gale
Hom(M,R) is something else.
slate forum
Oh, I thought modules would be iso to their duals
prime gale
No way!
prime gale
This is partly where the difference between isomorphic and naturally isomorphic comes up. Modules are (more or less) vector bundles.
prime gale
Vector bundles are a whole lot simpler than modules. A module is like a (potentially) singular vector bundle.
slate forum
I thought a vector bundle like assigns to each point in a set a vector space and then it locally looks like the cartesian product of neighborhoods of the space and the vector space
prime gale
Yes. So you can thing of a vector bundle as a module over the ring of functions on your space.
*think.
prime gale
No. Fixes them. But it moves vectors.
You have a vector space for each point, right? You have a function on your space. Multiply the vector at that point by the value of your function at that point.
slate forum
It doesn't do anything to the fibers themselves though?
Just moves around vectors in each fiber?
prime gale
Yes.
So you see that every vector bundle is a module. Turns out with some reasonable assumptions they are not that different.
slate forum
weird
Seems weird because vector bundles are a bunch of special modules stitched together
prime gale
One mistake I made in beginning was to prove stuff the hard way the way you initially proved that Hom(Z/n,G) = G[n]. It took a long time before I realized that I was just doing everything the hard way and the stuff is actually really easy.
slate forum
I didn't think it was that bad to prove it directly
prime gale
I wish the grader just knocked me on the head whenever I proved something the hard way.
slate forum
Though it wasn't the intended method I'm sure
prime gale
Don't worry. The hard way will get a lot harder.
slate forum
I'm sure, I'm a grad student...
prime gale
You have to be smart to be a grad student.
slate forum
Eh... I get the feeling that's not always the case sometimes
I often feel like I got lucky with my quals
prime gale
I failed all my classes and dropped out. So now I am a drummer.
slate forum
just undergrad?
prime gale
Yes.
prime gale
Only the easy stuff. I don't know any advanced stuff.
slate forum
vector bundles seem advanced to me
prime gale
Well, I don't know much about vector bundles.
I really need to calm down about equal vs isomorphic. I guess I am stickler too. Most people don't really care.
slate forum
Alright... I'm trying to show that left split-exact implies right split-exact
I have that 0->A-i-> B -p-> C -> 0, and I have a map q:B -> A such that qi=1_A.
I need a map j: C -> B such that pj=id_C, my guess was to use a preimage of c in C as j(c), but this is not well-defined
At least, I don't think it is
I've been trying to see what happens if p(b)=p(b')=c, then b-b' is in the kernel of p, so b-b' = i(a) for some a in A
I can apply q to both sides to get a=q(b-b')
I don't really see where I can go from there though
prime gale
Use j(a) = p^-1(a)-i(k(p^-1(a))), where k is your left splitting. I haven't checked it, but it seems like something like this has got to work.
slate forum
p^-1(a)? wut
prime gale
Pick any element of preimage.
slate forum
p goes from B to C
prime gale
p^-1(a) = any element of preimage of a. Use my j and try to show it is well defined.
prime gale
Of course, I have two p^-1(a) there--you better pick the same one both times.
slate forum
Doesn't look like it's quite working, if p(b)=p(b')=c, then I need to show b-i(q(b))=b'-i(q(b'))
right?
prime gale
Or should I just give you my proof that it works?
prime gale
Start with b1 in p^-1(c), and b2 also in p^-1(c). Then b2 = b1+i(a) for some a in A. Then show my j is well defined.
slate forum
hmm b2-i(q(b2))=j(b2)=j(b1+i(a))=b1+i(a)-i(q(b1+i(a)))=b1+i(a)-i(q(b1))-i(a)=b1-i(q(b1))=j(b1)
prime gale
j isn't defined on b1 or b2. domain of j is C. b1 and b2 live in B.
prime gale
So you've convinced yourself that j is well defined? Only thing left is to prove that p o j is 1 and j is linear.
prime gale
Want to know my thought process to get it?
prime gale
You are starting with something in C. Only thing you can possibly do is apply p^-1 and pick something there, right?
slate forum
Yeah
prime gale
But the problem is that that isn't well defined.
slate forum
exactly
prime gale
Two things might differ by something in image of i. So we need to cancel that somehow. Well, we just have to use what we are given. We are given map k:B->A, right? Nowhere to send the thing other than through k, right? Then once it is in A, nowhere to send it other than through i.
So we subtract off that stuff and hope that it is well defined. And I guess I got lucky and it worked.
slate forum
Yeah... guess there's not really much of a choice
prime gale
That is, we are trying to cancel the "undefinedness". That is, the part that is coming from A.
Given the data, not much choice, right?
prime gale
np.
timid hull
I was wondering if we have any notion of "projective algebra" ? I mean we have free modules over a ring say R and projective modules are generalisation of them and stuff , Similarly we also have a notion of free algebra over a ring (its nothing but a non-commutative polynomial ring with variables indexed by some index set) so thats why I was curious if there is something called "projective algebra" ... Also it seems that free algebra is also a notion which isn't much popular ...
prime gale
We have finitely generated free commutative algebra over a ring, which is a super duper popular concept. You can have scheme that is projective over a ring.
(Scheme is just a bunch of rings glued together.)
If R is a graded ring, then Proj(R) of is projective.
haughty portal
How does one undergo an orthogonal transformation to simplify a dielectric tensor to it's diagonal components only?
particularly the transformation resulting in the principal-axis system?
prime gale
Are you wondering why you can make a symmetric matrix into a diagonal matrix by multiplying it by an orthogonal matrix?
timid hull
If R is a graded ring, then Proj(R) of is projective.
@prime gale I think u misunderstood my ques , By "projective algebra" I meant something analog to a projective module .
prime gale
You can have an algebra over a ring that is projective as a module.
For example, if R is a ring, then R is an algebra over R that is projective as a module. Or, for a less trivial example, R[x], is an algebra over R that is projective as a module.
restive shard
Hi. Is there some sort of book talking about applications of abstract algebra in analysis?
smoky cypress
For a commutative ring suppose $x^2=x$ for all $x$, then any finitely generated ideal is principle. For this I noticed that $$(x+1)^2=x+1\implies x^2+2x+1=x+1\implies 2x=0$$ so $$ax+by=(ax+by)(x+y+xy)$$ and so the ideals $$(x,y)=(x+y+xy)$$ and the result follows from induction right?
cloud walrusBOT
Whoever:
solemn rain
yea
smoky cypress
Nice to know
solemn rain
yea cool
smoky cypress
I hate algebra
solemn rain
why
smoky cypress
Because no intuition whatsoever
Like for this problem
I have no idea what a ring where x^2=x will look at
I just came up with the solution by looking at (Z/(2))^N
Ok so like
In analysis you can sort of picture the functions and come up with the inequalities
But you canβt do that with algebra
oblique river
"free" is equivalent to "has a basis"
oblique river
I don't know how you went from "sum r_x*x in IF" to "each r_X in I"
(though I think that's true)
actually, wait, I don't think it even matters
I mean, it does matter, but F being free over R is sufficient but not necessary for F/IF being free over R/I
so I guess what I'm saying is maybe it's okay if you don't use the full strength of that hypothesis
(for example: R = Z, I = (2), F = Z/2Z)
I think the "lemma" to prove is that X being a basis of F over R implies that (sum r_i x_i is in IF <==> each r_i in I)
although that latter condition can also hold in more general settings
but it's really that latter condition which let you conclude that the cosets of X are a basis of F/IF
slate forum
I thought that that
err
that's what it means to be an element of IF
all the r_x's gotta be in I
solemn rain
@smoky cypress yea fuck that
fuck math
somepeople get off by abstraction tho
take a minute break
and think of how abstract your thinking rn
its overwhelming
and cool
compare it to like other sciences or any other mindful shit
its boner raging
so col
not for me tho fuck that shit
oblique river
@slate forum no, being in IF just means you look like a*x for an element a in I and x in F
but when F is a free module, then it is equivalent to that condition
smoky cypress
@solemn rain itβs not about the abstractness of math. And yes I actually agree on that aspect: I really like how abstract I am thinking about some concepts. But the thing is that for algebra, most of the time Iβm just trying random things. Sometimes it works sometimes it doesnβt
solemn rain
cuz it sucks ass
like you have nothing to think about
u just brute force your waya
the cool shit is about the theory itself
not the problems
smoky cypress
Lmao
slate forum
Oh, gotcha
smoky cypress
I guess I can agree with that
slate forum
@oblique river
oblique river
every element in F has a unique representation as a sum of r_x*x, so every element in IF has a unique representation as a*(sum r_x*x) = sum (a*r_x)*x
slate forum
a*?
oblique river
a in I
slate forum
oh
oblique river
the uniqueness is important here
or else if you had some kind of linear dependence among the elements of X
you could just add that to the expression above
and now your coefficients wouldn't be in I anymore
slate forum
I don't understand, it only gives a representation of it by multiplying by things in R, not by things in I
oblique river
for example suppose you were being silly and included x and 2x in X (so it's no longer linearly independent). then something like (a+2)x - 1*(2x) would be in IF (again a in I)
but neither a+2 nor 1 are in I
I'm not sure what you mean
slate forum
The freeness
oblique river
elements of IF are exactly "elements of I" * "elements of F"
slate forum
Yeah
oblique river
elements of F all look like sum(r_x * x) over x in X
slate forum
Ok, so I can write sum(r_x x)=sum(r'_x x) say
oblique river
but there's no reason for the coefficients to be in I
oblique river
because of the uniqueness
that's what I was saying
if there's no uniqueness
then you can ahve r_x and r'_x be distinct
slate forum
Hang on, so if I have sum(r_x x) in IF, then by assumption it is of the form i* sum(r'_x x)
So then I get r_x = ir'_x
so then it works?
oblique river
well, I guess actually it's a sum of things like i*sum(r'_x x)
take something in IF. It is of the form sum_{i=1}^n a_i * f_i where a_i in I and f_i in F
now write f_i = sum_{x in X} r_{i, x} x
slate forum
oh christ, messy
oblique river
now change the order of summation. our element of IF is of the form $\sum_{x \in X} \left[ \left(\sum_{i=1}^n a_i r_{i,x}\right) x\right]$
cloud walrusBOT
Buncho Bananas:
oblique river
the coefficient on each x is in I
because all of the a_i are
and now because X is a basis, this representation is unique
i.e. an element f in F lives inside IF if and only if f = sum r_x x where each r_x is in I
if we didn't have a basis, then there could be other representations of f
which didn't have all their coefficients in I
I gotta go now, gl working this out
slate forum
thanks!
maiden ocean
hm
cloud walrusBOT
Moth:
maiden ocean
They should be, right?
maiden ocean
np
solemn rain
idk about free products but this might help you https://math.stackexchange.com/questions/1191542/subgroups-of-free-products-of-cyclic-groups
maiden ocean
i saw that yea
solemn rain
is this related to topology ?
woven delta
I think you can do like 010 and 1
maiden ocean
yeah
woven delta
And that should generate a copy of Z_2*Z_2
maiden ocean
yea but thats not proper
woven delta
Why is that not proper?
maiden ocean
up to iso
woven delta
Am I being silly
What
I'm not sure what you mean
Also how do you get Z?
That seems unlikely
maiden ocean
subgroup generated by ab
woven delta
Oh ok
maiden ocean
(ab)(ab)(ab)(ab)...
woven delta
Makes sense
Why isn't Z_2*Z_2 in your list though?
I gave a proper copy of Z_2*Z_2 in itself
uncut girder
Haha infinite groups go brr
woven delta
(I think at least)
maiden ocean
im thinking abt this like up to iso only basically so like
yea ig technically proper is the wrong term here
is there a term for this?
woven delta
No
maiden ocean
_<
woven delta
It seems like a weird notion
uncut girder
Tfw you're isomorphic to a proper subgroup of yourself
woven delta
Yeah it's a nice property
Also it's crazy when you have non isomorphic groups A and B with injections of A into B and B into A
ocean magnet
too much freedom for me
woven delta
Being isomorphic to a proper substructure of yourself happens a lot in Boolean Algebras land lol
Also there are universal groups sometimes for certain properties
Which have lots of different embeddings into themselves
(universal for property P meaning they have property P and every group with property P embeds into them)
uncut girder
Example?
woven delta
Another example is every countable Boolean algebra is a sub algebra of the rational interval algebra (the Boolean Algebra of all sets of the form (a, b] where a and b are rationals or infinity)
chilly ocean
@solemn rain yo homeboy teach me group theory
fickle brook
aight so i need a reality check here
- if K is a field, K[x] is a PID.
- quotients of PIDs are themselves PIDs.
fickle brook
do you?
take a PID R, take a quotient R/I, take an ideal J in R/I, lift it back up to R as J+I, get J+I = <g>, then g+I β R/I will generate J
no?
nova plank
Otherwise the quotient isn't an integral domain
fickle brook
hm wait
i don't need the zero product property
for what im doing that is
the property of all ideals being principal still carries over to quotients tho
right?
nova plank
I think every ideal in the quotient is still principal, it's just not necessarily an integral domain
Yes
golden pasture
more generally yes cuz all ideals in A/I are in bijective correspondence to those in A that contains I
unique juniper
im bit a confused with the last statement
if h is the inverse of g
why would h'*g also equal e ?
unique juniper
i see thanks
smoky cypress
Does the unique maximal ideal of a local ring necessarily contain all nonunits
smoky cypress
Is the Zariski topology of Z, C[x], and R[x] just the finite complement topology on the set of principal ideals generated by the prime elements
olive mirage
on the variety yes, on the scheme no.
(the difference is whether you include the ideal (0) in the topology, which is a weird point, usually called a "fuzzy" point)
I think if you want to think about the Zariski topology, it is best to hop up to two variables
though even Spec R[x] has some interesting geometry going on where you can think of it as being connected to the upper half plane.
olive mirage
roughly speaking, the variety corresponds to the set of maximal ideals
and the scheme corresponds to the set of prime ideals
smoky cypress
Ughhh ok
olive mirage
working with maximal ideals is more intuitive, because the correspondence to geometry is more clear
but practically speaking, you gain a lot from also including the primes
smoky cypress
I have no idea yet how to think of them geometrically
olive mirage
but if I recall correctly Atiyah Macdonald tries to talk about commutative algebra without talking about geometry which is, well, not how I think about it.
golden pasture
AM doesnt really focus on the geometry part, it does introduce varieties and really goes into detail with spec but doesnt talk about the geometry
it's a lot easier once you start adding in the geometry
olive mirage
Well, every maximal ideal in C[x] corresponds to (x-a) where a is a point in the complex plane. So the idea is that maximal ideals give you a way to algebraically capture the idea of the complex plane.
golden pasture
Zariski topology of Z, C[x], and R[x]
i think this was referring to a qn in AM chap 1 lel
olive mirage
I accidentally ruined someone's undergraduate research project by pointing out their result was an exercise in chapter one of AM
I see, then the answer is: it is not quite the finite complement topology because (0) is weird.
golden pasture
ooft
yea we're looking at the scheme
wait how would undergrad research project be a exercise in chap 1 of am if it's on com alg wouldnt they read smt like am
olive mirage
haha I don't think the question read like a CA question. It was more a commutative ring theory question.
in any event, perhaps as karma, the next project taht my thesis advisor assigned to me I completed only to find out it had been published in 1979
golden pasture
oh yikes
olive mirage
I think probably the methodology was so different I should have just published it anyway, but eh π
smoky cypress
Eh I donβt quite feel like reading other books
Am seems short enough to commit
Other books seem way too long to finish
golden pasture
seems short enuf to finish
i found reading about ag alongside am helped make a lot of things clearer
smoky cypress
Not short enough to finish lol
Short enough for me to think it is worth it to commit to the book
And that I won't give up mid way
toxic zephyr
so this is in my abstract algebra textbook, and i just want to clarify. it's in a section which defines the determinant, so it's abstracting things a bit. i understand the perspective that the determinant is a function which accepts n^2 arguments and gives one output. However, in the general context of linear algebra, when we explicitly say det(A) where A is an nxn matrix, is it not a function from M_nxn to R? of course M_nxn and R^nxn are isomorphic but is it not normally considered a function with a domain of matrices?
https://puu.sh/GuGly/649d683ea6.png
chilly ocean
nvm that doesnt answer your question
sharp sonnet
$\bR^{n\times n}$ is the set of real nxn matrices
cloud walrusBOT
LochverstΓ€rker:
sharp sonnet
which you apparently want to denote M_nxn(R) or something
toxic zephyr
well thats the notation i learned so sue me i guess π
sharp sonnet
it's fine, but both is common
quaint ivy
you can think of matrices as the ring $\bR^{n^2}$ with the usual sum and a funky "multiplication" if it helps. anyhow it's just a matter of notation
cloud walrusBOT
bastian.uwu:
thorn delta
ok, im a bit confused about terminology. When we say that G is the internal direct product of H and K, we can assume that H and K are normal (as seen in the problem below).
But in general, G could be the direct product of not-normal subgroups. If H, K are subgroups of G which aren't normal and HxK = G, do we not call it an internal direct product? Lookin for some clarification here lol
prime gale
G could happen to be isomorphic to product of non normal subgroups. But when G is the internal direct product of H and K, then it is isomorphic to H*K via a very specific map. That map doesn't work if H or K is not normal.
thorn delta
basically, when we aren't talking about H,K being normal and HxK being iso to G with the specific map, then we just call it a regular ol' direct product (no "internal")?
prime gale
Yes. Of course H is going to be normal in H*K and K is going to be normal in H*K.
thorn delta
yea, i understand that. Idk, the terminology "internal" just makes me think that G is the product of its subgroups, not necessarily normal.
prime gale
Easy exercise: Give example of group, G, with subgroups H and K, so that G isomorphic (as groups) to H*K. But H is not normal.
cinder bone
can someone assist me with d)
c) the identity element is the identity matrix no problem
ocean magnet
sure, it's invertible if the determinant is nonzero
ocean magnet
yup, sum of squares is always positive as long as both aren't 0
cinder bone
ill just look up a proof for "invertible if the determinant is nonzero"
thanks appreciate it
ah okay, so since the determinant of the identity matrix is 1
and det(AB)=det(A)*det(B)
det(A) and det(B) must be nonzero
perfect
thorn delta
@prime gale So, suppose we have H,K < G and HxK is iso to G as a direct product. Then Hx{e} and {e}xK are normal in G, and intersect trivially. Let f be an isomorphism from HxK to G. Then you could say that G is the internal direct product of f(Hx{e}) and f({e}xK), right?
i.e. if you have a direct product of subgroups, you can do what I just did to get an internal direct product of subgroups. Albeit, it sometimes could be that f(Hx{e}) = G and f({e}xK) = {e} which isn't very interesting.
thorn delta
For b), since $H$ is a direct factor of $G$, there is a normal subgroup $K$ such that $H \times K = G$. I thought maybe I could extend a homomorphism $f: H \to G$ by defining $\bar f (g) = \bar f (hk) = f(h) k$, but I am stuck proving this is a homomorphism. Let $a = hk \in G$ and $b = h'k' \in G$. Then $$\bar f (ab) = \bar f (hkh'k') = \bar f (hh'kk') = f(hh') kk' = f(h)f(h')k k'.$$
but this is as far as I can get.
cloud walrusBOT
kxrider:
prime gale
Why not define fline(hk) = f(h)?
I think your way works too, but it is very slightly more complicated.
thorn delta
Why not define fline(hk) = f(h)?
ugh, cuz im dumb i guess lol.
I don't think my way works without more assumptions, like $\operatorname {Im} f \cap N = {e}$.
prime gale
Your way works.
cloud walrusBOT
kxrider:
thorn delta
how?
prime gale
You have that string of equalities. Just keep going until you get fline(a)fline(b).
thorn delta
yeah, I'm not seeing it. I don't see how we can commute elements of K with elements of the image of f.
prime gale
You are right. You can't. I was wrong. Your way doesn't work.
thorn delta
ah okay, that makes more sense 
thanks
prime gale
Do you have example for second part of b?
thorn delta
i mean, using the homomorphism you defined, its clear that im(f) = im(f bar), so f must be an automorphism for f bar to be an automorphism
prime gale
Huh? f goes from H to G. No way can it be an automorphism. Groups are different. Besides, maybe there is a different way to extend f to an automorphism.
thorn delta
yea...
prime gale
There is a really trivial abelian example.
So don't bother thinking about weird groups or complicated things.
thorn delta
if you have $\bZ_2 \oplus \bZ_4$ then I don't think the monomorphism $f: \bZ_2 \oplus {0} \to \bZ_2 \oplus \bZ_4$ defined by $f(n(1, 0)) = n(0, 2)$ can be extended to an automorphism.
cloud walrusBOT
kxrider:
prime gale
That works.
thorn delta
was that the example you were thinking of?
prime gale
I was thinking of an easier example.
thorn delta
hmm, i wonder. Klein 4 wouldn't work
prime gale
Well, my example is bigger, but technically easier.
H=Z. G=Z*1. 1->2. It is mono. Obviously can't extend to automorphism, because there is nothing to extend.
For your example, you would have to make some argument about why it can't extend.
thorn delta
ah yeah of course. For mine, a map is determined by the action on (1,0) and (0,1). Since (1, 0) maps to (0,2), (0,1) maps to (1, x) for some x. No matter what, (0,2) is the image of some multiple of (0,1)
much too complicated counter example tho lol
prime gale
Your example is good. I like it.
thorn delta
thanks. Actually, if (0,1) maps to (1, x), then (0,2) is not necessarily the image of some multiple of (0,1). If x = 2, you get a different case, which fails to extend to an automorphism in a similar way.
prime gale
My example is so simple that it is stupid. Probably doesn't illustrate what author had in mind.
prime gale
Just curious. Why are these questions here instead of the "math help" section? Not saying that they belong in "math help". Just wondering.
slate forum
How do I show that 0->B' -i-> B -p-> B''-> 0 is exact if 0 -> Hom(B'',M) -p^*->Hom(B,M) -i^*-> Hom(B',M) -> 0 is exact, in particular, how do I show i is injective
I think they the questions down there are for more simple things
prime gale
Well, it doesn't have to be exact.
For example, pick 0->B'->B->B''->0 not exact. Pick M=0. Then second sequence is exact, but first is not.
scarlet estuary
Why are these questions here instead of the "math help" section?
people qualified to help with advanced mathematics generally dont check the math help section
prime gale
I see. Assume second is exact for all M. I assume B' are modules over commutative ring with unit?
scarlet estuary
as 99% of it ranges from grade school level to intro calculus/linear algebra
so it's better for everyone involved to just put the questions here
prime gale
Well, to show i is injective, I would let M=B, and look at i as element of Hom(B',B). Haven't thought it through, but something like that has got to work.
slate forum
Hmmm
given that, then since i^* is surjective, i=i^*f for some f:B -> B
So i = f(i)... weird
prime gale
No. I don't think my idea works.
prime gale
Pick M=B'. And identity map in Hom(B',B'). Then since map is surjective, there is something in Hom(B,B') that composes with i to be identity. If I is not injective, no way it can have inverse.
slate forum
oooh
prime gale
Anyway, it looks like I have just proved original sequence is split, so everything is trivial.
slate forum
Oh, that's what the main question was
prove that if i^* is surjective, then the sequence is short exact and splits
prime gale
Oops. Didn't mean to spoil the problem.
slate forum
it's ok
I was stuck for a bit, I was going to try and find a map to show it was right-split instead, but that's a lot harder
Does split imply exact?
prime gale
I just tried putting random things in for M until something worked.
slate forum
It doesn't seem like it to me
prime gale
Usually people don't talk about split sequences unless they are exact. But once you have an inverse for i, proving exactness is probably trivial.
Not every sequence in which first map has inverse is going to be exact. But it'll probably make proving exactness given your other conditions easy.
slate forum
aight thanks!
prime gale
Welcome.
prime gale
Are you assuming A is noetherian?
smoky cypress
I was thinking maybe I could mimic the proof that closed subsets of a compact space are compact, but then X_f is not closed
No
Only commutative and unital
prime gale
Well, if you already proved v, then doesn't that imply vi, with A_(f)?
smoky cypress
How?
prime gale
Doesn't your picture define X_f?
smoky cypress
Yeah I was gonna ask what was A_f but mistyped it as X_f and then realized you meant X_f
prime gale
X_f = Spec A_(f). So just apply v to A_(f).
prime gale
A_(f) = A localized at multiplicative set generated by f.
If you already proved v, you can use v to prove vi. Not circular.
smoky cypress
I have no idea why that means
A_(f) = A localized at multiplicative set generated by f.
prime gale
But probably you shouldn't use the fact that X_f = Spec A_(f). Since apparently book didn't define it yet.
Anyway, you know what basic open sets on X_f are, right?
prime gale
Fix f. vi asks you to prove X_f is quasicompact. What are basic open sets on X_f?
prime gale
Yes. Which is?
prime gale
You know what V(f) is, right? You can throw in a few basic open sets to cover V(f) too, right?
prime gale
So now you have open cover of X, right?
Sorry--I don't think my line of reasoning here works.
smoky cypress
Yeah I was going to do what you did, and this is what I meant to mimic the proof that closed subsets of compact set is compact
and it doesn't work
Oh ok
Let me see
prime gale
I think so. Just do the same thing for A_(f).
Isn't that what you are talking about?
vi is just v, except using A_(f) instead of A.
Huh?
If you prove v for all rings, you have in particular proved it for A_(f).
I am saying that if you accept Spec A_(f) = X_f, that vi is same as v.
slate forum
Just so you know
This exercise is from Atiyah-Macdonald
The person likely knows no AG
smoky cypress
I don't
prime gale
Right. So v proves affine scheme is quasi compact. So to prove vi, if you know that X_f is Spec A_(f), then you are done.
slate forum
I'm in AG and I don't even know what a scheme is yet
slate forum
is it true that the intersection of all nonzero ideals is a subset of the nilpotent elements?
golden pasture
i mean AM is a pretty popular bookπ€
prime gale
I don't claim ever to have read any of AM at all.
slate forum
And even those I couldn't do all of them on my own :L
prime gale
Actually, I've never read a bit of AM.
prime gale
Anyway, does OP know about localization?
Okay. I would have done it using localization, but that's cheating. So, namirin, please continue and ignore my digression.
Helix--you have same initials as Hentai Kamen. So does that mean you have read Hentai Kamen?
carmine fossil
How is helix even related to that?
slate forum
This confuses me, I don't understand how to show the intersection of all nonzero ideals is a subset of the nilpotent elements, if R is a domain, then it has no nonzero nilpotent elements... This seems weird
prime gale
Didn't namirin just tell you how to do that?
golden pasture
@smoky cypress
For vi) Suppose you have a covering with some X-V(a_i) for ideals a_i
Show that this implies that v(some massive ideal)=V((f)), and hence f is a finite sum of elements in the ideals which means V(f)=V(finite sum of a_i)
slate forum
Yes, but it may not be a nonzero ideal
I mean
it isn't a nonzero one
because R is an integral domain
slate forum
yeah
prime gale
I see. small gap in namirin's reasoning.
slate forum
I was just trying to understand
How it actually makes sense to say it's a subset if the nilradical here is literally zero
prime gale
So actually it isn't true.
slate forum
right? bizarre
prime gale
But that doesn't mean statement in picture is false.
slate forum
I mean, yeah... it's weird how an intersection of nonzero things can be zero still
prime gale
Well, in Z/2, intersection of all non zero ideals is not contained in set of nilpotents.
For any field, for that matter.
Come on.
prime gale
Just do what Jane sasid.
slate forum
By the way, is V the "same" as Z?
slate forum
Can I realize every topological space as the spectrum of a ring with the zariski topology?
I'm guessing no
prime gale
Well, didn't we just decide spectrum of ring is quasicompact? Is every topological space quasicompact?
slate forum
guess not? If quasicompact is the same as compact as I usually see it
I don't understand, isn't (ii) literally the definition of (i)?
prime gale
If you are asking if (i) => (ii) is trivial, then you are right. It is.
slate forum
Oh ok, cool
ii to iii I actually gotta do something though, and iii to i I think I do as well.
prime gale
(ii) => (iii) you don't do much. (iii) => (i), I think definition of coproduct works.
smoky cypress
Ok nvm I still don't know how to do it
and @golden pasture I'm not sure what you meant
slate forum
test $\phi: R\rightarrow S$
cloud walrusBOT
HelixKirby:
prime gale
I don't think what Jane said is quite true though. It might be that some power of f is a finite sum...
But Jane's argument still works because X\V(f^n) = X\V(f) for n>=1.
prime gale
Anyway, @smoky cypress do you still want help with vi?
golden pasture
ok so right lets suppose that $X_f$ is covered by $X-V(\mathfrak a_i)$
cloud walrusBOT
ariana:
golden pasture
what does this tell you about $V(f)$ and $V(\text{something about }\mathfrak a_i\text s)$
cloud walrusBOT
ariana:
prime gale
I don't think Whoever is here anymore.
golden pasture
still in serverΒ―_(γ)_/Β―
this implies that $\sqrt f$ is the radical of a finite sum of $\mathfrak a_i$
cloud walrusBOT
ariana:
golden pasture
(why?)
smoky cypress
I don't see what this says about V(f) and V(some a's)
prime gale
Do you see what it says about V(f) and V(all the 'a's (not just some))?
Does one contain the other?
smoky cypress
Well I can show that V(all of a's) is a subset of V(f)
prime gale
That's all you can show. And it is enough. So what does that tell you about <f> and <all the 'a's>?
smoky cypress
I don't see how I can deduce anything
prime gale
I don't know context and have not read book, so forgive me if I use stuff I am not allowed to. Do you know what radical of an ideal is?
prime gale
Do you know relationship between V(I) and I_V(I)?
smoky cypress
What's I_V(I)
prime gale
Ideal of things vanishing on V(I). That is, intersection of all prime ideals containing I.
smoky cypress
Ideal of things vanishing on V(I).
Not sure what you mean
But I do know the radical is the intersection of all prime ideals containing I
prime gale
Okay. Ignore that part. Just look at second sentense.
Okay. Since you know that V(all of the 'a's) is subset of V(f), what does that tell you about intersection of prime ideas containing f or all of the 'a's?
smoky cypress
So the radical of f is a subset of the radical of ideal generated by a's
prime gale
Yes. So f^n is in <a> for some n, right?
prime gale
Can you finish proof?
prime gale
Great! Sorry about fighting with the others.
prime gale
Well, it was quite a distraction for you. So I am sorry.
smoky cypress
I think it's good to just see people discussing about things
Eh it's fine
gabe I think algebra sucks
prime gale
Just because we sucked this time doesn't mean Algebra sucks.
smoky cypress
Lmao I didn't think algebra sucks because of this time
It's just that
When I'm doing analysis, it just feels way more intuitive
Like I draw a picture and usually I know what's going on and what I should do
But algebra? I just try random stuff and hope for the best
prime gale
That's funny. The reason we all sucked this time is we let intuition get ahead. That why all the false statements.
smoky cypress
Hmmmm well that's you guys. I don't have any intuition for these stuff
smoky cypress
Why does this necessarily mean that zero divisors is a union of prime ideals
It doesnβt say anything about any zero divisor is in a prime ideal of zero divisors
prime gale
I think you need more than just what it says. For example, if every element is contained in a maximal element, then you could conclude conclusion.
But easy enough to prove in any case.
chilly ocean
@smoky cypress what book is this
chilly ocean
@solemn rain have you learned Tits groups yet?
cerulean wraith
Hi guys, wanted to ask, does every group have a subgroup?
golden pasture
no, consider the group of one element
otherwise {e} is a subgroup
carmine fossil
Every group is its own subgroup
hot lake
isn't G always a subgroup of G
deft plume
Well, I suppose we could look at that in two ways. Proper and improper subgroups. If improper, then @carmine fossil is right. If proper, then @golden pasture is right - in that vein, I guess we could define G = ({e}, β’) the identity under multiplication. It satisfies associativity, closure, identity, and inverse. Even more so commutativity so I guess abelian. There wouldnβt be a subset of e that wouldnβt contain e, so as a proper subset, one could say e doesnβt have a subgroup.
Aside, I think every group overall usually always has subgroups so long the improper subset is applied
solemn rain
call this map f
img(f) intersect A/P_k is nonzero
for any k in {1,.....,n}
@woven obsidian
i think this is called chinese remainder theorem
woven obsidian
Isn't that trivial
solemn rain
the whole theorem?
woven obsidian
The image of 1 intersects them all at some nonzero elt
solemn rain
yea ig
woven obsidian
Some solution notes seem to imply it should be interpreted as im(Pi) is nonzero
solemn rain
whats im(Pi)
prime gale
The image of 1 intersects them all at some nonzero elt
@woven obsidian That isn't how I interpret statement. I interpret it as for each i, the image contains something that looks like (0,...,a_i,...,0) with a_i!=0.
smoky cypress
@prime gale yes every element is contained in some maximal ideal. The problem is that, can you say that all of the elements of the ideal are zero divisors
@chilly ocean Atiyah Macdonaldβs Commutative Algebra
prime gale
Well, if n>1, then everything in each summand is a zero divisor.
smoky cypress
Ah
smoky cypress
Actually I got it
chilly ocean
@solemn rain yo homeboy
prime gale
yes every element is contained in some maximal ideal. The problem is that, can you say that all of the elements of the ideal are zero divisors
@smoky cypress No way. For example, take K[x,y]/<xy>. This has two minimal primes, <x> and <y>. It also has a maximal prime, <x-1>. x-1 is not a zero divisor.
solemn rain
hey
smoky cypress
Then that was my question lol
The problem said therefore the set of zero-divisors is a union of prime ideals
prime gale
I see. You are talking about that previous problem. Have you solved it?
smoky cypress
Well easy application of zorn's lemma will show that it has a maximal element
prime gale
Yes.
But as you pointed out, simply the existence of a maximal element isn't enough to conclude conclusion.
smoky cypress
But I got it
prime gale
Good.
smoky cypress
If x is a zero divisor then (x) consists of zero divisors only
and it was quite trivial
you don't even need a maximal prime ideal
prime gale
Well, that is true. But I don't follow your reasoning.
I agree. You don't need to say anything about maximal ideals.
prime gale
Keep at it!
smoky cypress
Oh bruh
I must be stupid
The whole exercise is to prove there is a maximal element and it's prime
and I'm not using it
prime gale
I think we might have some miscommunication about "maximal prime ideals". You probably mean "maximal prime ideal in E." I was thinking "maximal ideal."
smoky cypress
I meant maximal ideal in E yeah
fringe bane
Kind of a vague Lie group question here. So, SO(2N+1) acting on R^n always has at least one axis fixed, whereas SO(2N) does not. I remember this relating to a slightly deeper theorem about linear algebra and irreducible representations, and I remember it having a nice proof, but I can't quite remember how/what. Does anyone have a sense of what I'm talking about / where I would find this tidbit in a textbook or wiki page?
opal osprey
So
I was solving some problems
And a little theorem crossed my mind
Idk if this is true yet
But the question is not about how to prove it
It's just that I want to state this in a more generalized way
Basically
For any permutation of these *, Β° operations
Then the square operation is also associative
How could I state this?
<@&286206848099549185>
Sorry for the ping, btw. It's just that I'm a little anxious ngl
stone fulcrum
There's a 15 minute rule on helper pings
stone fulcrum
Anyway, what's a permutation here? It's not clear to me
Just any linear combination and parenthesis combination?
opal osprey
I want to state that proposition in a more generalized way
So I want to know if for any given permutation of the symbols *,Β°
Then the square operation is also associative
For example
This permutation could be
[Β°(a,b)]*[Β°(a,b)]
Or
[*(a,b)]Β°[Β°(a,b)]
That one I wrote on LaTeX is just one particular case
stone fulcrum
"Any linear combination of *(a,b) and Β°(a,b), using * and Β°"
stone fulcrum
It's a bit messy if you allow stuff in *( , ) and I'm not sure if questions like that are well understood
You can't afaik
opal osprey
It's a bit messy if you allow stuff in *( , ) and I'm not sure if questions like that are well understood
What do you mean?
stone fulcrum
Like if you put a *(Β°(a,a),*(b,*(b,a)))
opal osprey
I see, maybe the problem is just silly. I'm just starting at Abstract Algebra and some of the first problems in the Dummit and Foote book is about proving if some given operation is associative or not
And I thought about it and a little bit and tried to prove a neat result
But idk it it's well stated or important at all
Just something that crossed my mind and I thought it would be a nice little problem
prime gale
@opal osprey I don't think your square is associative. For counter example, just look at free (set with two associative operations) generated by {a,b,c}.
Or for a more concrete counterexample, let star be addition on integers and circle be multiplication on integers. Then 1 square (2 square 3) = 930 and (1 square 2) square 3 = 162.
opal osprey
Here is the generalized statement
With a little bit of abuse of notation at the end
But I guess it makes sense
Could it be stated like this?
Or for a more concrete counterexample, let star be addition on integers and circle be multiplication on integers. Then 1 square (2 square 3) = 930 and (1 square 2) square 3 = 162.
@prime gale Thanks a lot man
prime gale
I must be misunderstanding something. I don't think that is true.
opal osprey
I was about to think about some counterexamples once I figured how to state that proposition
prime gale
Even easier counter example. Take circle = + and star = + on integers. Then square is not associative.
I see. I missed the "two two" associative. I read it as "two associative". I don't know what a two two associative binary operation is. It might be true for a two two associative binary operation.
prime gale
What does "two two associative" mean? I haven't learned about that.
opal osprey
Ohhh
My bad
Just a typo :(
But yeah, the result is not true
Thanks for the counterexamples
prime gale
I see. I thought maybe two associative meant something different that could make statement true.
opal osprey
Now I wonder if *_1 and *_2 are commutative if it makes any difference
prime gale
Well, + is commutative.
opal osprey
To the behavior of *_3
prime gale
Well, we got some practice checking associativity. So it was all worth while.
opal osprey
I hope this is true now
smoky cypress
Ok I haven't read the entire chat but I feel like you shouldn't write $\star_i(a,b)$ and $a\star_ib$ in the same line
cloud walrusBOT
Whoever:
smoky cypress
Stick to one
So write $(a\star_ib)\star_j(a\star_kb)$ or $\star_j(\star_i(a,b),\star_k(a,b))$
cloud walrusBOT
Whoever:
opal osprey
Ok I haven't read the entire chat but I feel like you shouldn't write $\star_i(a,b)$ and $a\star_ib$ in the same line
Oh I see, I will try to do that more often
But is my result true now?
I am always not so confident about my results even if they are quite trivial so idk
smoky cypress
yeah it is true
prime gale
You don't need starj to be commutative for star3 to be commutative. As long as stari and stark are commutative, and starj is any binary operation, then star3 is commutative. But it is pretty weird to talk about non associative commutative binary operations.
smoky cypress
rock paper scissor is a nonassociative commutative binary operation
which I think is interesting
opal osprey
You don't need starj to be commutative for star3 to be commutative. As long as stari and stark are commutative, and starj is any binary operation, then star3 is commutative. But it is pretty weird to talk about non associative commutative binary operations.
I also noticed that. I always thought that associativity was in some sense a "stronger" statement than commutativity.
Sorry
The other way around
Haha
prime gale
Well, there is plenty of math that deals with non commutative associative binary operations (Groups). And plenty of math that deals with non commutative non associative binary operations (Lie Algebras). But I don't know of any math that deals with commutative non associative binary operations.
prime gale
Doubtful. Just let one binary operation be non (whatever) and the other binary operation be constant.
opal osprey
What can we say about *_j and *_k?
prime gale
Let star1=a-b if a!=0 and b!=0, otherwise 0. Let star2=0. Then for any i,j,k in {1,2}, star3=0. So star3 is commutative. But star1 is not.
So converse is not true.
opal osprey
That's a really good example
I see why it's not true
I will try to prove later if in order for *_3 to commute, then either *_1 or *_2 commute.
That's my last conjecture and then I will stop lmao
I will ask for some hints if I need
Thanks a lot
coarse storm
I think tweaking am's example a bit would give a counter-example.
opal osprey
Let $\star_3 : {1,0} \times {1,0} \rightarrow {1,0}$ be such that $\star_3 (x,y) = \frac{x-y}{x-y}$ for $x \neq y$ and $\star_3 (x,y) = 0$ for $x=y$, with $-$ being the usual subtraction of the integers and the same goes for the division.
\
\
Then, $\star_3 (x,y)$ commutes, but $\star_1 (x,y) = x-y$ and $\star_2 (x,y)= \frac{x}{y}$ don't.
cloud walrusBOT
MisterSystem:
opal osprey
Hope this is right
coarse storm
Do we want it to be commutative?
opal osprey
Nope
It's actually the other way around
We want star_3 to commute
But star_1 and star_2 to not
eager bobcat
My book asks me to prove this:
A nonempty subset $H$ of the group $G$ is a subgroup of $G$ iff for all $a, b \in G$, $ab^{-1} \in G$.
cloud walrusBOT
ratsella:
eager bobcat
But I don't see what subgroups have to do with this ...
eager bobcat
If a and b are in G then b^-1 also must be in G since it is a group and since G is a group it must be closed so ab^-1 is necessarily in G?
Not sure where H comes into it at all
I swear this is a typo
opal osprey
Let x/y for y=0 be 1
Then gg
It really doesn't matter, it could be anything
Except 0
As long as star_2 doesn't commute
Then I can be sure that star_3 commutes but star_1 and star_2 do not
prime gale
@eager bobcat Your book is wrong.
coarse storm
We then have ββ being constant, sending to 1, I think. Should work as an example, yea.
opal osprey
It sends either to 1 or to 0 depending of whether x and y are equal or not.
But it commutes anyway
coarse storm
Your definition of ββ makes it unable to send to 0.
opal osprey
When x=0 and y=1 it sends to 0
But when x=1 and y=0 it sends to 1
Let x/y for y=0 be 1
Here
coarse storm
Yea. Let me rephrase. "Your definition of ββ makes ββ unable to send to 0.
eager bobcat
@prime gale thanks, I suspected that ... I feel like they should have written H where they put G at least once
coarse storm
I feel like I have seen that typo before...
May be it is just a common mistake.
But well, your example does work. Good job!
Uh.. wait..
Binary subtraction is commutative, isn't it?
coarse storm
Hm... so, 0-1 = 0?
prime gale
Damn--this has gotten way too complicated for me.
slate forum
Holy shit, can you imagine learning about parity of permutations by doing this?
smoky cypress
1 is a generator of Z^+ as a monoid
So if you view Z has a monoid under +, then Z^+ with 0 is the submonoid generated by 1
vocal depot
I've got a silly question. I'm just learning cosets and I'm using my books exercises to practice. I came across one that defines a subset $H ={0, \pm 3, \pm 6, \pm 9, \ldots}$.
cloud walrusBOT
HisMajestytheSquid:
vocal depot
Then they want you to find all the left cosets of $H$ in $\mathbb{Z}$.
cloud walrusBOT
HisMajestytheSquid:
vocal depot
So my question is, do I have to add the elements of $\mathbb{Z}$ separately to both the positive and negative elements of H. Or is it sufficient to say things like $1+H={1, 1\pm 3, 1\pm 6, 1\pm 9,\ldots}$.
cloud walrusBOT
HisMajestytheSquid: