Unless we reach for some specific machinery ig
#groups-rings-fields
coral spindle
delicate orchid
I don't believe in infinitely generated objects - how do you know it's infinitely generated? just because you got bored halfway through writing down all of the generators?
coral spindle
Perhaps there’s a homological condition, idk
rocky cloak
I guess the boring condition would be wheter it homs to Prod_p Z/p
But maybe one could work from that somehow
rocky cloak
Well, all simple modules
coral spindle
Right, overkill
delicate orchid
why did you have to make it into ring theory....
rocky cloak
Underkill I think
delicate orchid
STOP NERD REACTING ME
rocky cloak
Like having a nontrivial Hom to a quotient of the ring shouldn't be enough
coral spindle
No :)
coral spindle
Like having a nontrivial Hom to a quotient of the ring shouldn't be enough
Surely it is? We’re just looking at a composition factor on top that way
winter shore
In an euclidean domain, if there is a nonzero $a$ such that $N(a)=N(ab)$, does that imply $b$ is a unit? ($b$ is also nonzero)
cloud walrusBOT
Shiranai
rocky cloak
Surely it is? We’re just looking at a composition factor on top that way
Like take the example again with the Prüfer group with zero multiplication, and freely adjoin unit. Then the Prüfer group is an ideal, so maps to the ring, but doesn't have any maximal submodule
The problem is that mapping to the ring doesn't mean mapping to the top
wooden fulcrum
Is this supposed to be obvious
delicate orchid
both sides are tautological
every group has a maximal subgroup (the group itself) and every group has a quotient (G/G)
wooden fulcrum
Oh i thought it meant proper maximal subgroup
delicate orchid
if I did the only counter example would be the trivial group
wooden fulcrum
To which part?
lapis latch
who doesnt require maximal things to be proper
rocky cloak
People who are memeing
delicate orchid
the top of the lattice is the top. Ain't my problem
delicate orchid
The problem is that mapping to the ring doesn't mean mapping to the top
first time in history the top of a lattice has been referred to twice independently within the same year
rocky cloak
I'm talking about the top of a module bro, get out of here with that lattice bs
south patrol
Gm
delicate orchid
I'm talking about the top of a module bro, get out of here with that lattice bs
what's even the difference, pal
rocky cloak
Modules don't frighten me
south patrol
tbf i think usually one says maximal subgroup to mean proper right ye
south patrol
Modules don't frighten me
Ok do my work
rocky cloak
Because I know all modules are finite dimensional
south patrol
I'm stuck on some interesting group theory q but I will not ask because then someone will do my diss for me lmfao
south patrol
Because I know all modules are finite dimensional
everything is a filtered colimit of them and everything preserves filtered colimits
so agreed
south patrol
Look at this guy, wew. Doesn’t even know day from night
lol
rocky cloak
In an euclidean domain, if there is a nonzero $a$ such that $N(a)=N(ab)$, does t...
Is your norm defined to be multiplicative?
delicate orchid
Because I know all modules are finite dimensional
since all modules are k-algebras as well, instead of writing top just write M/J(M)
oh nvm you're actually helping someone I'll stop shit posting
winter shore
Is your norm defined to be multiplicative?
no, but you can asume Na <= N(ab)
wlog
actually I just realized I don't need this yay
south patrol
since all modules are k-algebras as well, instead of writing top just write M/J(...
lol
winter shore
if it's false, we'd have to get a counterexample in a E.D. with nonmultiplicative norm
which I think there's only one known
lol
rocky cloak
Now I'm curious, what example is that
winter shore
Maybe there's more, I just know this https://arxiv.org/abs/1703.02631
rocky cloak
This seems like a euclidian domain that doesn't admit a multiplicative norm though. Which I guess is much stronger than admitting a norm that isn't multiplicative
random pasture
rocky cloak
Try computing (ai + bj + ck)^2 and take it from there
south patrol
doesn't the hint just spoil the entire problem lol
delicate orchid
evidently not
random pasture
oh shoot i multiplied them wrong
i forgot multiplying backwards negates them
so $(ai+bj+ck)^2 = -a^2-b^2-c^2$?
cloud walrusBOT
x496
south patrol
yh
south patrol
evidently not
sorry i mean it spoils the interesting stff
vivid tiger
this is true for square roots in the quaternions
hidden wind
I'm talking about the top of a module bro, get out of here with that lattice bs
what do you have against lattices
dull ginkgo
... just abel goop ;-;
vivid tiger
first time in history the top of a lattice has been referred to twice independen...
wew watch this:
true
celest furnace
I think I accidentally swallowed some soap
R.I.P. gotta call posion control now
vivid tiger
I think I accidentally swallowed some soap
was it that shocking?
i better not mention false.
void cosmos
rocky cloak
what do you have against lattices
The word has too many unrelated meanings. I can think of at least 3
velvet steeple
hidden wind
The word has too many unrelated meanings. I can think of at least 3
good point
glad osprey
Phi is a homomorphism from G to G'. In part 4, it says K' is a subgroup of G' intersect phi[G], but this is just equal to phi[G]. I think it should be K' is a subgroup of G', right?
hollow tartan
Is it implying that G' is 'bigger' right? Also is it saying that K' is a subgroup of G' intersecting phi(G), not K' is intersecting phi(G)?
coral spindle
Phi is a homomorphism from G to G'. In part 4, it says K' is a subgroup of G' in...
I imagine they're trying to avoid conceptual problems with defining phi^-1[K] where K contains elements outside the image.
But yes, if you're happy with that, it's fine to say that K' is a subgroup of G'.
glad osprey
I imagine they're trying to avoid conceptual problems with defining phi^-1[K] wh...
I thought it was common to accept elements outside the image using this notation, but I get it. If that was their intention I wish they would just write phi[G] though
tribal niche
I'm reading a proof that if R is a UFD and f,g are primitive in R[X] then so is fg. It begins as follows:
If fg is not primitive then c(fg) is not a unit, so since R is a UFD, we can find some irreducible p in R with p | c(fg). [Here, c is the content of a polynomial, i.e. a gcd of its coefficients]
By assumption, c(f) and c(g) are units, so p ∤ c(f) and p ∤ c(g). Then...
i'm probably being an idiot, but how does the bolded part follow?
Oh nvm I am an idiot. Got it
dull ginkgo
Here’s a fun exercise if you’re bored. have fun
south patrol
||First consider prime power n. Can't have n = p^k for any p > 2 (except the case p=3 and k=1) since (Z/p^k)^x is cyclic of order (p-1) p^{k-1}. For n = 2^k, note (Z/2^k)^x = Z/(2^{k-2}) x Z/2, so we need k <= 3. So using CRT largest should be taking n = 24, with units (Z/2)^3||
But I think I know another way maybe
Eh nah i think you need to know about Z/2^k being special ofc
hollow tartan
Here’s a fun exercise if you’re bored. have fun
a^2=e (mod n) is this what you mean? if so I wish there was a way to not forget how many "loops" ,say 12 in Z/5Z is 2, has made in order to generalize it, for this case it made 2 loops
dull ginkgo
||First consider prime power n. Can't have n = p^k for any p > 2 (except the cas...
yeppers :D
that's the route I used
hollow mica
Wait what’s the question asking I don’t understand
dull ginkgo
I used CRT, then analyzed powers
south patrol
Well, just the fact that the method of proof is literally computing (Z/n)^x for all n
is kinda sad lmao
dire siren
this problem was asked in the Romanian National Olympiad in 1987
south patrol
Yes
hollow mica
whatt
rustic crown
I was trying to do another way but didn't really work lol
yea looks like fermat primes would be annoying to deal with
south patrol
Well all a in (Z/nZ)^x
hollow mica
ohh
south patrol
has a^2 = 1
dire siren
are the elements a with a^2=1 really called involutions? I only heard this term used for functions
south patrol
But then a finite group where every element has order 2 is isomorphic to (Z/2)^k for some k
south patrol
are the elements a with a^2=1 really called involutions? I only heard this term ...
To be honest I wouldn't use it for groups lol
hollow mica
idempotents is more standard I think
dire siren
that's x^2=x
hollow tartan
idempotents makes more sense
hollow mica
oh
hollow tartan
oh lol
rustic crown
but yea, i've heard of involutary matrices
south patrol
idempotent is different
oh lol
already been said
i think involution is unambiguous though really
hollow tartan
but the word itself says idem - potent, identity-exponent right?
rustic crown
i really like the word "tridempotent" 
dire siren
it's weird actually that nilpotent means x^k=0 for some k, while idempotent means x^2=x
rustic crown
unipotent would make more sense to you ig
(they're commonly defined either as x^k = 1 or x = 1 + nilpotent)
hollow tartan
unipotent would make more sense to you ig
exactly
dire siren
is there a name for x^k=x, for some k?
hollow tartan
what about x^x=e lol that could be a nice exercise
dull ginkgo
this problem was asked in the Romanian National Olympiad in 1987
oh wait really
dire siren
yeah
dull ginkgo
huh
Yeah I used CRT to show n must be a product of some power of 2 and some power of 3
then it's some small calculations to show max power of 2 is 8, max power of 3 is 3
so it's 24
dire siren
I have a proof without CRT
we notice that phi(n) is a power a 2
therefore 7 does not divine n
therefore 7 is in (Z/nZ)*
so 7^2=1, i.e 48=0, i.e n | 48
n=48 doesn't work because 5^2=/=1
but n=24 works (I won't write the details)
south patrol
Nice
languid trellis
Does anyone have any ideas for this exercise?
would appreciate a nudge in the right direciton
dull ginkgo
I'm also on this, i had a biblical brain fart with the G = that union
hollow mica
Does anyone have any ideas for this exercise?
Each gHg^{-1} is a subgroup of G
languid trellis
Each gHg^{-1} is a subgroup of G
so suppose G = \bigcup gHg^-1. Then G is the union of subgroups, which implies that one of these subgroups is G itself.
gGg^-1 = H
implies G =H
so G is not a proper subgrou
coral spindle
so suppose G = \bigcup gHg^-1. Then G is the union of subgroups, which implies t...
Unless there's something I'm missing, this doesn't follow. Every group is the union of its cyclic subgroups, for example.
languid trellis
no you're right
coral spindle
My hint: if $h \in H$ then note that $(gh)H(gh)^{-1} = gHg^{-1}$, so you can bound the number of distinct conjugates by a certain number, and they all intersect at $\set{e}$ at least.
cloud walrusBOT
Boytjie
coral spindle
I do need to double-check this hint but I believe it solves the problem.
dull ginkgo
aHa^-1 = bHb^-1= > ab^-1 in N(H)
H = N(H)
but that's not the contradiction, unless i can force N(H) = G
coral spindle
I think you're barking up the wrong tree
languid trellis
My hint: if $h \in H$ then note that $(gh)H(gh)^{-1} = gHg^{-1}$, so you can bou...
distinct conjugates share the identity. Also, |gHg^-1| = |H|. So, at most, the number of distinct conjugates ig [G:H]
coral spindle
So now how many elements are in the union? Try getting an upper bound.
languid trellis
we also have (gh)H(gh)^-1 = gHg^-1. So, At most |G ⧵ H| (set difference) distinct elements in the union. For G = gHg^-1, we need [G:H]. So it suffices to show that |G ⧵ H| < [G:H].
coral spindle
Hm I don't see how you got there
I was thinking more of an overcounting argument :)
languid trellis
Hm I don't see how you got there
If we have [G:H] distinct conjugates of G then G = \cup gHg^-1
coral spindle
No...
Remember, they intersect nontrivially as sets...
They are distinct, not disjoint!
languid trellis
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
coral spindle
Now try
languid trellis
we need to get rid of some stuff
dull ginkgo
but that's not the contradiction, unless i can force N(H) = G
$G = \coprod_{S \in \mathrm{Orb}(H)}{S}$
coral spindle
and write down the upper bound you get
cloud walrusBOT
Mizalign #1 simp
coral spindle
??
dull ginkgo
that's just orbit stabilizer nvm
dull ginkgo
Orb(H) is the set of conjugates of H
coral spindle
I know.
coral spindle
I'd like swiftee to do the thinking here :)
hardy cedar
coral spindle
The amazing thing is you can read and learn new words!
dull ginkgo
$|G| = [G : N(H)]| |N(H)| < [G : N(H)] |H|$ because of that union NOT being disjoint, contradicting $H \subseteq N(H)$
coral spindle
This is not especially obscure math
cloud walrusBOT
Mizalign #1 simp
dull ginkgo
... how
coral spindle
No well, in any case I simply can't see how you've gotten here
coral spindle
Again, I would really like to see swifteee's thinking rather than yours here.
coral spindle
**Mizalign \#1 simp**
Yeah this is just false, I don't know what to say.
south patrol
Does anyone have any ideas for this exercise?
oh lol which book is this from btw
cause i remember discussing this before
d+f?
maybe lang even
languid trellis
No...
I've tried to type this message 3 times now. When are two conjugates distinct? per your hint, we have (gh)H(gh)^-1 = gHg^-1, so, intuitively, if g,g' \in G \ H, then we have gHg^-1 \neq g'Hg'^-1, as if they were equal, there would exist an h such that gh = g'. i.e. h = gg'^-1, which implies that g, g'^-1 in H (I think?)
coral spindle
OK so this is not the way.
You will not be able to determine in a general case when the conjugates are distinct.
Please just try my hint! We have at most |G:H| subgroups, each of size |H|, but we overcount by some amount. What is the most obvious adjustment to the upper bound we could make?
Miz if you spoil this for swift I'll be quite upset.
languid trellis
|G:H| |H| - |G:H| (for the number of elements in the union)
coral spindle
Yes exactly
languid trellis
so [G:H] |H| - [G:H] + 1
dull ginkgo
G must be partitioned into the cosets of N(H), the normalizer of H, so G is partitioned into [G : N(H)] distinct sets
G is the union over all the conjugates of H, aka the orbit of H. There are [G : N(H)] UNIQUE conjugates. HOWEVER, each gHg^-1 contains the identity due to it being an image of an automorphism, thus the union is NOT A DISJOINT UNION.
By orbit-stabilizer, each coset gN(H) corresponds to a conjugate gHg^-1. This is the contradiction of the fact that the gN(H) partition G... but the gHg^-1 DON'T
coral spindle
so [G:H] |H| - [G:H] + 1
That's right! Now what's a simplification you can make here...
You may know this particular result by a name
south patrol
lol
languid trellis
That's right! Now what's a simplification you can make here...
|G| - [G:H] + 1 = 1, which implies [G:H] = 1, which implies G = H, a contradiction
south patrol
Miz if you spoil this for swift I'll be quite upset.
posts right after boytjie saying this
mfw
coral spindle
Yup, to say the least I am not happy.
coral spindle
|G| - [G:H] + 1 = 1, which implies [G:H] = 1, which implies G = H, a contradicti...
Yes! I imagine you meant = |G| but this is a tiny typo
Cool huh?
dull ginkgo
I wasn't reading the chat, just typing... sorry fam
trying to parse because this problem fried my brain
languid trellis
i didn't read it anyway. miz is doing something with normalisersand orbits
coral spindle
Miz overcomplicated it a lot
dull ginkgo
Miz overcomplicated it a lot
I realized my original proof was wrong, and tried to rework it from that framework
coral spindle
But yeah, you got the argument. Well done!
languid trellis
Yes! I imagine you meant = |G| but this is a tiny typo
yes, sorry haha. But yeah this counting argument is way simpler
It also seems quite weird for maximal subgroups
coral spindle
Yes it's kinda hard to picture for me
dull ginkgo
G must be partitioned into the cosets of N(H), the normalizer of H, so G is part...
wait hold a fucking second does this hold for G, H infinite?
coral spindle
It's always pleasing when the simplest possible bound works
languid trellis
oh lol which book is this from btw
this is from jacobson Basic Algebra I btw
south patrol
wait hold a fucking second does this hold for G, H infinite?
Apparently there are counterexamples in general rigt
languid trellis
It's always pleasing when the simplest possible bound works
for sure, this set of exercises has been an absolute misery so far so, once again, I appreciate your time
coral spindle
wait hold a fucking second does this hold for G, H infinite?
It does not.
dull ginkgo
It does not.
oh okay
coral spindle
It's your chance to be creative and produce a counterexample
south patrol
To me it is slightly disappointing that this problem seems to require just a counting argument - it'd be nice if we could somehow construct an element missing lol
But not very surprising either I suppose
languid trellis
i'm talking a 20 minute walk break now lols
south patrol
nice
coral spindle
To me it is slightly disappointing that this problem seems to require just a cou...
In my head this basically comes down to how difficult conjugacy of subgroups is. Maybe I'm just wrong, but this is why I said that trying to find a condition for when the conjugates of H are distinct isn't the way.
dull ginkgo
It's your chance to be creative and produce a counterexample
Z in the infinite dihedral group
?
wait no
Z/2Z in the free product of Z/2Z with itself?
not even that
I can't think of a counterexample
I'd need to find an infinite nonabelian group G with proper subgroup H such that every element in G is a conjugate of an element of H
languid trellis
I have a suspicion that this doesn't hold when we pick some subgroups of GL(n, R)
south patrol
oh that's very smart swiftee
languid trellis
Just because there exists the idea of 'matrix similarity'
south patrol
you could do ||upper triangular matrices inside of all complex matrices||
south patrol
I was trying to cook up a topological example
As in like think in terms of graphs and stuff
coral spindle
I had a more combinatorial example in mind: ||the group of permutations on N with finite support||
edit: forgot to say ||the subgroup will be permutations whose support doesn't contain e.g. 1||
languid trellis
oh that's very smart swiftee
I don't know enough lin alg to say anything more of substance on this matter
dull ginkgo
G is the free product of Z/2Z with itself. Subgroup H is the subgroup of elements such that the length of the word is even.
Any element not in H is an alternating word of odd length like ababa, which must be symmetric :)
south patrol
well thinking of lin alg was the good bit
dull ginkgo
doesn't conjugacy preserve parity of # of letters
yeah I realized that
south patrol
ah ok
fading field
you won’t find an example with index 2
dull ginkgo
I am trying to see if I can come up with a simple example
south patrol
I had a more combinatorial example in mind: ||the group of permutations on N wit...
ahem
lol
dull ginkgo
yeah I just chekced that
fading field
lol
south patrol
Oh no i mean miz posted the same thing and i just pointed out you had the same one (but accidentally pung'd you, my apologies)
coral spindle
Oh lol
dull ginkgo
Yeah S_N in S_\infty lol
fading field
S_n won’t do it
south patrol
$\mathrm{colim}_n \mathrm{Sym}({1,\dots,n}) \to \mathrm{Sym}(\mathrm{colim}_n {1,\dots,n})$
cloud walrusBOT
Süßkartoffel
south patrol
😼
fading field
conjugation of S_n with a member of S_infty just gets you a guy with at most n guys being acted on miz
dull ginkgo
I meant to describe it differently
languid trellis
What even is S_infty
fading field
What even is S_infty
Aut_set(Z)
fading field
uh
languid trellis
All rearrangements of a sequence? Seems horrid
south patrol
You want like
$S_1 \hookrightarrow S_2 \hookrightarrow S_3$ etc to give you $S_\infty$
cloud walrusBOT
Süßkartoffel
fading field
yes
dull ginkgo
COLIMIT JUMPSCARE
south patrol
not sure how itis ajumpscare but ye lol
dull ginkgo
i just like saying [object] JUMPSCARE
south patrol
miz saying jumpscare jumpscare
languid trellis
not sure how itis ajumpscare but ye lol
Cat theory is scary
fading field
i just like saying [object] JUMPSCARE
you should be working on a group that is a union of conjugacy classes jumpscare
dull ginkgo
Cat theory is scary
nah fam they are just supremum and infima bro
ignoring a lot of structure....
scary
coral spindle
I thought S_oo should be finite support
Yeah this is why I was careful to describe my example in words lol
dull ginkgo
you should be working on a group that is a union of conjugacy classes jumpscare
you killed my mojo just because someone posted an example prior that I didn't see no sad
fading field
i didn’t do that, you killed your own mojo
fading field
right after you find a suitable group, i assume
languid trellis
you could do ||upper triangular matrices inside of all complex matrices||
Wait this is just saying that there's a basis in C_n such that the matrix is upper triangular, which is supposedly I theorem I should know 
south patrol
Yes this is uh
Schur's theorem I guess
Every matrix over an algebraically closed field is triangularisable
south patrol
You need the minimal polynomial to be a product of linear factors
dull ginkgo
Might try to formalize the absurd bullshittery to prove that no group of order 24 is simple
south patrol
But not necessarily distinct
fading field
Might try to formalize the absurd bullshittery to prove that no group of order 2...
what do you mean by absurd bullshittery
south patrol
And then diagonalisability is equivalent to the min poly being a product of distinct linear factors
dull ginkgo
what do you mean by absurd bullshittery
Sylow and a small proof by exhaustion
it's bullshit because it's not trivial, you see
coral spindle
You need the minimal polynomial to be a product of linear factors
Maybe I'm not understanding what you're saying, but I think it's much simpler? We simply are guaranteed an eigenvector by algebraic closure, and we begin building a basis with that, quotienting as we go along to get an inductive argument. Am I missing something?
south patrol
Groups of order 24 are simple as 24 only has two prime factors, boom
south patrol
Maybe I'm not understanding what you're saying, but I think it's much simpler? W...
Yeah no that is correct, I'm just saying this stronger statement is slightly nicer
coral spindle
Fair enough
south patrol
Like expressing in terms of min poly rather than the field
languid trellis
I think I'll start reading the section on sylow theorems to tonight 
fading field
Sylow and a small proof by exhaustion
sorry but the sylow theorems are not bullshit, and also they are sort of trivial, have a look at a book other than jacobson for a moment (humor me), go read Aluffi chapter 2 section 9 on orbit-stabilizer stuff, and then chapter 4 the section on sylow theory
and see if you still think they’re bullshit
dull ginkgo
sorry but the sylow theorems are not bullshit, and also they are sort of trivial...
this is not an exercise in jacobson and I came up with the exercise to challenge myself ergo I actually like it. I just like calling it bullshit
fading field
i promise it will be more clear
coral spindle
Yeah no that is correct, I'm just saying this stronger statement is slightly nic...
Of course, in general, any smooth connected L.A.G. over an ACF is covered by the conjugates of the Borel subgroup. So I suppose we could generalise it further, presumably with this observation
south patrol
Based
Yeah
Iag being linear algebraic group i assume
isn't there more generally a statement about every element being conjugate to an element of the maximal torus
so that's for compact connected liegroups
i guess the issue is existence lol
coral spindle
isn't there more generally a statement about every element being conjugate to an...
Any semisimple element!
south patrol
oh nvm
coral spindle
Otherwise, every matrix would be diagonalisable!
coral spindle
Right right
delicate orchid
Otherwise, every matrix would be diagonalisable!
I've got good news
dull ginkgo
Let |G| = 24 = 2^3 * 3, assume G is simple
a = |Syl_3(G)| = 1 mod 3 and a | 8 => a = 1 or a = 4.
a cannot equal 1 because the only Sylow-3 subgroup would be normal, thus a = 4. Because every sylow-3 subgroup is cyclic of order 3, they have pairwise trivial intersection, thus the union of the sylow-3 subgroups is 4 * (3 - 1) + 1 = 9.
b = |Syl_2(G)| = 1 mod 2 and a | 3 => b = 1 or a = 3. Ditto, b = 3.
Each sylow-2 subgroup has order 8. The intersection of all of the sylow-2 subgroups must be normal due to conjugate closure, therefore the intersection is trivial.
The pairwise intersection of two sylow-2 subgroups must have order 2 or order 4, and this order is the same for all of the pairwise intersections due to transitivity by conjugation. The intersection of 3 of the sylow subgroups must have order 2 or be trivial, as they cannot be equal to another sylow-2 subgroup. The pairwise intersection order cannot be the same as the triple-intersection order. Since the Sylow-3 subgroups have trivial intersection with Sylow-2 subgroups, that means the union over all the sylow-3 subgroups must have cardinality 24 - 9 + 1 = 16
This leaves a case analysis by inclusion-exclusion:
Pair Inter 4, triple inter 2: 4 * 8 - 6 * 4 + 4 * 2 - 1 = 15
Pair Inter 4, triple inter 1: 4 * 8 - 6 * 4 + 4 * 1 - 1 = 11
Pair inter 2, triple inter 1: 4 * 8 - 6 * 2 + 4 * 1 - 1 = 23
None of these values are 16
I also used a symmetry group argument by letting G act on Syl_2(G) giving a non-mono homomorphism from G to S_3
which is way simpler but I tried to do a more calculation-based one
Frankly you can do a fuckin lot with that shortcut lmao
hollow mica
Let R be a commutative unital ring. I know that if R is a field, then R[x] is a Euclidean domain, but it is also true that for a general ring R Euclidean division works as long as the polynomial you're dividing has leading coefficient that is a unit. Does this mean that the subring of R[x] composed of polynomials with leading coefficient unit are a Euclidean Domain?
coral spindle
Are you sure that's a subring?
celest furnace
Let `R` be a commutative unital ring. I know that if `R` is a field, then `R[x]`...
Yes it is true if S is a field then S[x] is a Euclidean domain.
At the minimum I think you need integral domain so I don’t think the argument works for just general rings R
dull ginkgo
products of units need not be units if they don't commute afaik
wait it's a poly ring
celest furnace
products of units need not be units if they don't commute afaik
R u sure?
celest furnace
if you have uv then v^-1u^-1 is the inverse
I feel like a lot of group theory would break if what you said was true lol
dull ginkgo
if you have uv then v^-1u^-1 is the inverse
Thinking of something else for units for a second
hollow mica
Are you sure that's a subring?
if p and q both have leading coefficient unit then so does their product?
coral spindle
if `p` and `q` both have leading coefficient unit then so does their product?
And what about their sum?
hollow mica
oh shit
dull ginkgo
1 + 1 = 2 isn't a unit in Z
brb
One time I saw a paper use the equivalent def of a unit being an element lying outside of every proper ideal
But like damn
Vibe death
hollow mica
Let `R` be a commutative unital ring. I know that if `R` is a field, then `R[x]`...
I also realized now that the polynomial you're dividing by has to have leading coefficient that is a unit, not the polynomial you're dividing
vivid tiger
products of units need not be units if they don't commute afaik
no the issue is addition
dull ginkgo
no the issue is addition
i corrected myself lol
hollow mica
If f, g, h ∈ R[x] and f is nonzero and has leading coefficient that is a unit, then is it true that fg = fh implies g = h ?
south patrol
yes - equivalently you're asking if f with unit leading coefficient is a zero divisor and it's not too bad to see why that's the case ||given non-zero g, consider top coefficient of fg||
the leading coefficient needn't be a unit either
dire siren
the leading coefficient needn't be a unit either
I think it needs to be a unit
constant polynomials in Z/6Z[x], f=2, g=3, h=0
south patrol
It doesn't need to be a unit
It needs to not be a zero divisor
In fact you can prove that if a polynomial f(x) is a zero divisor, then there's a nonzero constant c such that cf(x) = 0
which is cool
south patrol
Dw
dire siren
yeah, I thought you meant that we can drop the condition and the result will remain true
hollow mica
It makes sense now, thanks
does this look good https://hastebin.com/share/uhubicalej.markdown
dire siren
I think this is mentioned in Dummit and Foote
south patrol
oh yeah fun exercise
dire siren
as far as I know this only works for rings with unity
dire siren
you want to prove commutativity in addition sense, not in multiplicative sense
dull ginkgo
you want to prove commutativity in addition sense, not in multiplicative sense
No
that [x,y] is the same for all x and all y, and then throw 1 into the mix
[x,y] * c = [c*x, c*y]
commutator in the underlying additive group
south patrol
yeah it's just kinda cursed having non-commutative addition
dull ginkgo
what do you mean by no? that's what the problem asks
sorry, i thought you were implying I was doing a different proof
south patrol
Well this isn't the normal proof tbf
dull ginkgo
[x,y] is an "additive commutator"
south patrol
But can't really guess that until you've done it lol
south patrol
ye noice
dire siren
what do you want to do with the commutators tho?
dull ginkgo
show they are 0
dire siren
but why?
dull ginkgo
... All the commutators being 0 imply the additive group is abelian
dire siren
no
dull ginkgo
... why
last spoke
In the context of rings I've known [x,y] to be shorthand for xy-yx
dull ginkgo
I'm not referring to xy - yx
this is a commutator of the additive subgroup
not a commutator of the ring
I've said this like 3 times
south patrol
Lol yeah
dull ginkgo
[x,y] = x y x^-1 y^-1
dire siren
[x,y] is an "additive commutator"
indeed, I missed this message, sorry
south patrol
I thought you had a convo about it lol
dull ginkgo
okie no problemo
just the notation gets a bit fucky when you have lots of operations at once
dire siren
but in my defense, I never encountered any other meaning of [x,y] in ring theory
dull ginkgo
I'm using x * y for multiplication, just xy for the "noncommutative addition" for my sanity
just for this problem
actually might switch that
anyway if the additive commutators are all zero, then coolio we have an abelian group
I'm pretty sure with our properties we can show x * a = x for all a implies x = 0
south patrol
It's a group
dull ginkgo
It's actually because it's the only element outside of the quantification for the multiplication monoid
south patrol
Wait is * now multiplication
south patrol
I'm pretty sure with our properties we can show x * a = x for all a implies x = ...
In this
dull ginkgo
welcome to the jungle
south patrol
Or what
south patrol
Then take a = 0
dull ginkgo
FUCK
south patrol
Lol
last spoke
what's (-1)*(b+a)
dire siren
where is the distributivity of multiplication towards addition?
dull ginkgo
where is the distributivity of multiplication towards addition?
M3, accidental poor cropping lo
this sucks
i have 0 fucking clue what to do
or where to start
dire siren
||M_3||
dull ginkgo
i think my route is wrong and this is probably a bad omen for what's to come
I need to at least show 0 * x = 0 first
0 * x = (a a^-1) * x = (a * x) (a^-1 * x)
still stuck
if I can't solve this I probably should not do abstract alg
this is fucking stupid it's like the simplest problem
last spoke
Well you're just trying to compute a+b-(b+a) right
last spoke
I thought you wanted to know whether addition is commutative
last spoke
well do you know anything about -(b+a) from the axioms
dull ginkgo
well do you know anything about -(b+a) from the axioms
besides it exists, nothing
idk how to necessarily show it's -1 * it
because Idk how to show 0 * x = 0
dire siren
think about the definition of 0
dull ginkgo
1 + 0 = 1
x * 1 + x * 0 = x * 1
x * 0 = 0
last spoke
tru
dull ginkgo
0 = 1(-1)
0 * x = (x * 1) (x * -1) = 0
x * -1 = (x * 1)^-1
most of me parsing this is the goddamn notation
-1 * (xy) = (-1 * x) (-1 * y) = x^-1 y^-1 = (xy)^-1 = y^-1 x^-1 <=> xy = yx
holy fuck finally
In my notes I am DEFINITELY using i for the multiplicative identity
before I implode
last spoke
I like e better because i is used with decent frequency to refer to the inclusion homomorphism or other things like that later on
cobalt heath
nobody uses anything but 1 for a ring
I heard some CS people uses o and i to avoid confusing it with “real 0 and 1”
south patrol
Huh lol
cobalt heath
Indeed
dull ginkgo
This might be more readible. Let e be the additive identity, where addition is *, and i be the multiplicative identity, where multiplication is just xy:
Lemma 1: ex = e
Proof: e * i = i => x(e * i) = xe * xi = xe * x = x => xe = e by cancellativity
Lemma 2: xi^-1 = x^-1
Proof: e = i * i^-1 => xe = e = xi * xi^-1 = x * xi^-1 <=> xi^-1 = x^-1
Lemma 3: x * y = y * x
Proof: (x * y)i^-1 = xi^-1 * yi^-1 = x^-1 * y^-1 = (y * x)^-1 = (y * x)^-1 <=> x * y = y * x
holy fuck
dire siren
it would be more readable with + instead of *
cobalt heath
Btw, does a thread become inaccessible once I do not visit there often
dull ginkgo
Btw, does a thread become inaccessible once I do not visit there often
hover over the channel, "more active threads"
should be there
cobalt heath
Ahhh, thanks
dull ginkgo
dire siren
here is another proof:
(1+1)(a+b)=(1+1)*a+(1+1)*b=a+a+b+b
whats the idea behind your use of juxtaposition for the group operation in the ring, Mizalign?
dull ginkgo
Nah y’all right
dull ginkgo
here is another proof:
```(1+1)(a+b)=1*(a+b)+1*(a+b)=a+b+a+b
(1+1)(a+b)=(1+1)*a+...
Wow that’s clean
gotta love the fucking terrifying difficulty jumps lmao
probably not that hard inductively
cobalt heath
Hm I do not recall one has to prove the basic properties of ring
cobalt heath
gotta love the fucking terrifying difficulty jumps lmao
Hahahahahahahaaaa
dull ginkgo
probably just some inductive hypothesis bs
cobalt heath
Yeah, or some combinatorics
dull ginkgo
probably an index shift
vivid tiger
Oh, yes, this one, which is my reason for being mad at anyone that includes abelianness as an axiom.
Also prove the following: if you satisfy all the axioms of a module except for commutativity of addition, then commutativity of addition is implied. Thus one doesn't need to assume that a module is an abelian group under addition.
cobalt heath
Interesting, it follows from bilinearity of scaling map
crystal turtle
Nah y’all right
I don't think I've seen anyone give an x a tail like that before. If thye do, i've only seen it on the \ part, not the / part
dull ginkgo
I don't think I've seen anyone give an `x` a tail like that before. If thye do, ...
You should see when I use chi and x in the same thing lol
crystal turtle
Awful.
dull ginkgo
cobalt heath
Which one is chi?
dull ginkgo
Left
cobalt heath
Interesting
dull ginkgo
Idk where I got that from
Which turned into this really weird habit of using chi as a variable when referring to duals often
vivid tiger
do an x as two curves like )(
vivid tiger
Which turned into this really weird habit of using chi as a variable when referr...
...that is good notation, acruallt
cobalt heath
dull ginkgo
valuation of chi at asterisk
vivid tiger
•
dull ginkgo
gotta love the fucking terrifying difficulty jumps lmao
awful
good enough. proof by suction
cobalt heath
Called it
cobalt heath
Computing hell
dull ginkgo
yep
wasn't too bad though, just my hand regrets it
"basic algebra"
i fear not the content, but writing the work
dull ginkgo
lol you wish it was commutative
it's over a module
immediately goes for wedderburn little theorem as the first exercise
waiter, waiter more pigeonhole problems
last spoke
well wedderburn's theorem is a further, much trickier result
showing that it's at least a division ring is easier
cobalt heath
it's over a module
Oh?
What's a matrix over a module? (Also is + here noncommutative)
dull ginkgo
well wedderburn's theorem is a further, much trickier result
I tried proving that if a ring has the property that there's an n for each r such that r^n = r then it's a commutative
god that shit took forever
Wait that proof is actually in this textbook I think
dull ginkgo
What's a matrix over a module? (Also is + here noncommutative)
literally just... same as a vector space but the scalars need not commute lol
and a lack of a "hardly know er" pun
cobalt heath
Ah, then you can state it as a internal hom module
dull ginkgo
true chat
cobalt heath
In this case, it is End(R^n), even
dull ginkgo
Because this section is critical I'm going to do all the exercises
This is gonna take a while
i just answered the first 4 in my head, that's a good sign
cobalt heath
Woah, how do you easily do exercise 1
dull ginkgo
pigeonhole
last spoke
try left multiplication and then think about it
dull ginkgo
every element has a multiplicative "order"
dull ginkgo
i.e x^n = x for some n
It's actually a proof for semigroups
Every finite cancellative semigroup is a group
cobalt heath
I wish I were less dumb 
vivid tiger
@dull ginkgo whenever your journey takes you to modules, mind pinging me? that might be a fine time to learn the structure theorem for modules over PIDs and Morita equivalence and what injective and projective objects actually look like
dull ginkgo
<@315847860400816129> whenever your journey takes you to modules, mind pinging m...
sure
dull ginkgo
I wish I were less dumb <:devastation:898725325335298088>
it's not immediate
dull ginkgo
Every finite cancellative semigroup is a group
I've done this exercise before, I think in D & F?
it just follows from that
cobalt heath
it's not immediate
Wdym?
dull ginkgo
what's a semi group again?
Set with a binary associative operation
vivid tiger
...no unit?
dull ginkgo
doesn't necessarily needs a unit or inverses
cobalt heath
Semigroup is a group but without inverses and unit
dull ginkgo
monoid has a unit
vivid tiger
right
cobalt heath
You can always adjoin a unit to a semigroup to make it a monoid, though.
dull ginkgo
yeah
vivid tiger
okay so for a finite cancellative semigroup
x^n = x so x x^{n-1} = x so there's a unique x^{n-1}.
dull ginkgo
if a monoid is cancellative too you can make a group out of it
vivid tiger
y x^{n-1} x = yx so y=yx^{n-1}
and likewise for the other way
so x^{n-1} is a unit
dull ginkgo
okay so for a finite cancellative semigroup
x^n = x so x x^{n-1} = x so there's ...
Let o(x) = min{n : x^n = x}
then you have to show x^{o(x) - 1} = y^{o(y) - 1} for all x and y
so that makes it the unit
yeah
cobalt heath
Let o(x) = min{n : x^n = x}
then you have to show x^{o(x) - 1} = y^{o(y) - 1} fo...
I think this follows directly from uniqueness of unit
vivid tiger
idc about actually taking the min. I'm lazy so I'll just take an n by pigeon.
err wait
x^{m1} = x^{m2}
and you can cancel all but n
okay we are back in black
vivid tiger
now for inverses, x * ? = x^{n-1}. well, x^{n-2}.
QED.
unless I fucked up, this is indeed immediate.
dull ginkgo
yeppers
cobalt heath
You can also take any number with x^n = x, not necessarily minimum.
dull ginkgo
fair
vivid tiger
yes absta, be lazy like me
cobalt heath
Being lazy is virtue of mathematicians!
dull ginkgo
0 is the only nilpotent in a domain
if x^n = 0 then x^(n-1) x = 0 so x is a zero divisor and is thus 0 qed
vivid tiger
...oh, sweet
cobalt heath
Reminds me of reduced & irreducible => integral
dull ginkgo
zwz is a zero divisor
(zw)z = z(wz) = 0 so either z is a left zero divisor or z is a right zero divisor because it's an OR
vivid tiger
thus the ring of matrices has zero divisors
dull ginkgo
qed
vivid tiger
>zwz is a zero divisor
(zw)z = z(wz) = 0 so either z is a left zero divisor or ...
what no
1*1*1 is not a zero divisor
fading field
thus the ring of matrices has zero divisors
you didn't need that theorem to show that the ring of matrices has zero divisors
lol
dull ginkgo
1*1\*1 is not a zero divisor
*assume zwz is a zero divisor
cobalt heath
Does any ring of matrices have zero divisor?
dull ginkgo
yes
vivid tiger
except for dim 1
cobalt heath
What about matrix over field of one element
dull ginkgo
1 in the top left, 0 all others
dull ginkgo
~~What about matrix over field of one element~~
technically the ring of one element here
fading field
Does any ring of matrices have zero divisor?
by ring of matrices do you mean subring of M_nxn?
vivid tiger
or rather
O 1
O O
dull ginkgo
CHUBBY
vivid tiger
oh, for a subring, me things probably not
cobalt heath
by ring of matrices do you mean subring of M_nxn?
Or M_nxn itself
cobalt heath
Hmm, can one have a ring of matrices over right of 0=1
vivid tiger
suppose you have a subring that is rank >1 as an abelian group
hollow mica
how come in the wikipedia page of ufd it says every non-zero non-unit can be written as a product of irreducible and a unit ?
vivid tiger
how come in the wikipedia page of ufd it says every non-zero non-unit can be wri...
factorization is up to units
hollow mica
other places seem to not have the unit in the factorization
vivid tiger
5=1*5=-1*-5
fading field
Hmm, can one have a ring of matrices over right of 0=1
can you make this more precise
or like what do you mean
hollow mica
factorization is up to units
oh also I just realized
cobalt heath
I meant M_nxn over the trivial ring, sorry for typo
hollow mica
the wikipedia definition allows you to factor units too
which is only possible if you allow a single unit be in the factorization
vivid tiger
okay unit times unit is unit
so you can allow as many units as you want
factorization is unique up to a unit in a UFD
fading field
I meant M_nxn over the trivial ring, sorry for typo
so like, it just has one element, namely the matrix of all 0's?
dull ginkgo
NExt is to show that if 1 - ab is a unit, so is 1 - ba
cobalt heath
so like, it just has one element, namely the matrix of all 0's?
Yep, so I guess the ring is trivially trivial
fading field
that certainly doesn't have any zero divisors
dull ginkgo
NExt is to show that if 1 - ab is a unit, so is 1 - ba
the big think
fading field
that certainly doesn't have any zero divisors
you can also think of subrings of M_n(R)
for example, k * I for k in Z has no zero divisors since it's isomorphic to Z as a ring
@cobalt heath this is what i had in mind
cobalt heath
for example, k * I for k in Z has no zero divisors since it's isomorphic to Z as...
Quite a valuable example though!
fading field
if you have a matrix ring M_n over a nontrivial ring then you have zero divisors (if that is what you wanted originally)
dull ginkgo
NExt is to show that if 1 - ab is a unit, so is 1 - ba
kinda stuck with this one
c(1 - ab) = (1 - ab)c = 1
abc = cab = c - 1
a = c(1-ab)a = (c - cab)a = ca - caba = ca(1 - ba)
cobalt heath
kinda stuck with this one
Yeah so this is a technical one
dull ginkgo
a = c(1-ab)a = (c - cab)a = ca - caba = ca(1 - ba)
(1 - ab)a = a(1-ba)
dull ginkgo
x(1 + ab) = 1 => x + xab = 1 => x = 1 - xab
bxa = b(1 - abx)a = ba - babxa = ba(1 - bxa)
bxa + (1 - bxa) = 1 = ba(1-bxa) + (1-bxa) = (1 + ba)(1 - bxa)
cobalt heath
Did you do it
dull ginkgo
x(1 + ab) = 1 => x + xab = 1 => x = 1 - xab
bxa = b(1 - abx)a = ba - babxa = ba(...
yes
trying to get a better more understandable proof tho
cobalt heath
Oh wow
dull ginkgo
If c is the inverse of (1 + ab), then (1 - bca) is the inverse of (1 + ba)
serene crag
can all differential equations be expressed as a lie group? and what its the geometric meaning behind it. is the lie group just the manifold that contains all the solutions to the diff eq?
rocky cloak
can all differential equations be expressed as a lie group? and what its the geo...
I think the usual connection is that lie groups are the symmetry groups of differential equations.
still dew
For this problem
This is the intended solution
I don't understand the solution
Is the sum |Z_g| = |G| thing true
hardy scaffold
there are |C_g| elements in the conjugacy class of g, so that follows from the previous statement
south patrol
For this problem
Nice
hardy scaffold
once you show that |Z_h|=|Z_g| for every h in the conjugacy class of g
still dew
Also what happened in the last line
hardy scaffold
then you sum over the conjugacy classes to get c|G| on the right. on the left you sum over the conjugacy classes and then over the elements in the conjugacy class, which gives you the sum over every element in G
still dew
there are |C_g| elements in the conjugacy class of g, so that follows from the p...
Okay but how is sum |Z_g| over the conjugacy classes equal |G| like Z_g 's aren't even disjoint they have e in common
hardy scaffold
Okay but how is sum |Z_g| over the conjugacy classes equal |G| like Z_g 's aren'...
you have |Z_g||C_g| = |G|, you just replace the |C_g| by summing over the elements of C_g
generally you won't have that the union of Z_h for h in the conjugacy class of g is equal to G
still dew
Oh so the conjugacy class remains fixed
We sum over the elements of the conjugacy class?
hardy scaffold
yes
still dew
Oh so it's basically a double sum
Thank you I was confused with the notation
It's clear to me now
still dew
I was able to get through the first one isn't it just orbit stabilizer with the fact that elements of H fix H under conjugation with however I am not able to get the second one
delicate orchid
if every element of G was in a subgroup of that form that means that every g in G is conjugate to some element of H, see if you can show that H must be all of G from this
barren sierra
I guess there's an implicit assumption that H ≠ G
still dew
if every element of G was in a subgroup of that form that means that every g in ...
Yes I tried this I am not able to bound this
serene dune
need help
dire siren
is a) true?
serene dune
ans is b,d
dire siren
oh, so you had to pick the true statements
serene dune
oh yeah
dire siren
so I got that (a) is false]
because F* is cyclic, and if you take a generator a
then 1, a^17, a^(2*17), ..., a^(14*17) are distinct, and they are roots for x^15-1, so these are all the roots
(b) is true because 63 does not divide 255
serene dune
oh i was tryna thinking how 256 can be cyclic
dire siren
the multiplicative group of any finite field is cyclic
serene dune
thats some new information for me
dire siren
but I think (a) can be solved without this fact
since F* has 255 elements, then there is an element a of order 3 and an element b of order 5
so ab has order 15
and then any element in the subgroup generated by <ab> is a root of that polynomial
for c) you can prove that a and a^2 are roots, where a is an element of order 3
and for d) you can use the fact that F* has 255 elements, so t^255=1 for any t=/=0
serene dune
awight, much thanks
serene crag
I think the usual connection is that lie groups are the symmetry groups of diffe...
i thought the symmetries are in the lie algebra
or am i misunderstanding something
chilly ocean
elements of the lie algebra are infinitesimal symmetries
serene crag
elements of the lie algebra are infinitesimal symmetries
hm so how do i find these symmetries? i read something about using the infinitesimal generator?
chilly ocean
elements of the lie algebra are infinitesimal symmetries
Is this related to galois theory of inseparable field extensions?
teal wind
Could someone help me with this problem?
last spoke
well it asks you to do something pretty specific so have you tried to set up a group table yet?
dull ginkgo
More of an analysis esque question but here we go:
fg = 0 in \Gamma implies that for each x in [0,1], f(x)g(x) = 0
So for each x, f(x) = 0, or g(x) = 0 as R is a domain
next obsidian
Okay?
dull ginkgo
I am trying to work my way through it
next obsidian
I see
dull ginkgo
but neither f nor g is everywhere-0 on [0,1]
next obsidian
I don’t really wanna say anything
Because it’s hard to say anything without just solving it basically
You need to use continuity
That’s all you can really say
dull ginkgo
Is it easier to use the preimage-open or epsilonic def of continuity here
dull ginkgo
then it's not 0 everywhere?
dull ginkgo
There is a neighborhood of x that isn't 0 under f
dull ginkgo
g is 0 at x
dull ginkgo
There is a neighborhood of x that isn't 0 under f
wait
next obsidian
Okay so if there’s no open nbhd of f where f is constantly 0 then?
next obsidian
then that neighborhood in g must be 0
Okay but this still isn’t quite what we want
dull ginkgo
how is that not what we want
dull ginkgo
next obsidian
Because look
You can’t patch together all these opens
To show g is constantly 0 using this sort of argument
But if f isn’t 0 on any interval
It is not 0 at a bunch of dots near every point
Which means..
dull ginkgo
not continuous?
next obsidian
Actually not necessarily haha
Okay maybe we can just work more directly
Let eps > 0
vivid tiger
what's the problem?
dull ginkgo
I thought the neighborhood argument would be just fine/
next obsidian
Omay I mean
Try to write it down
And I’m pretty sure you’re gonna struggle to make it come togehter
You can’t argue that g is constantly 0 on a bunch of nbhds and get it to work
You need to argue that there’s a bunch of points where g is 0
dull ginkgo
it's showing that a neighborhood exists?
next obsidian
Enough to the point that any nbhd of x contains at least one such point
And then continuity finishes it
vivid tiger
if fg is zero then one of f or g is zero at every point
dull ginkgo
if fg is zero then one of f or g is zero at every point
yeah
vivid tiger
but if you have two subsets that union to the whole thing
perhaps one of them must be dense?
is that true?
A,B \subset \R, is it true that if A \cup B is \R then one of the two must be dense?
cobalt heath
if fg is zero then one of f or g is zero at every point
Oh is this alg geo
next obsidian
No
cobalt heath
Hmm
cobalt heath
A,B \subset \R, is it true that if A \cup B is \R then one of the two must be de...
Depends on spaces
vivid tiger
ℝ.
cobalt heath
Ah, need not dense then
vivid tiger
Actually [0,1].
dull ginkgo
Here's what I was going to show:
Lets say f in C([0,1]) is a zero divisor. Then there exists a function g(x) that's not 0 everywhere such that for each x in [0,1] f(x)g(x) = 0. Then that implies that for each x, f(x) = 0 or g(x) = 0.
Analyzing g(x), because g(x) is continuous, there must be a point x_0 in (0,1) that is not 0. By continuity, there is an open neighborhood of x_0, S, that is also not 0. By R being a domain, that means the image of that open neighborhood must be 0 under f.
next obsidian
Yeah xela your statement is
You aren’t getting st the right thing
Because what you want to be true isn’t you can just halve the interval
And neither is dense
vivid tiger
oh duh
dull ginkgo
You aren’t getting st the right thing
are you talking to Xela or me?
next obsidian
Here's what I was going to show:
Lets say f in C([0,1]) is a zero divisor. Then...
You’re never going to finish your proof this way
You cannot work with neighborhoods like this
dull ginkgo
You’re never going to finish your proof this way
how so?
next obsidian
You never get control over specific points
I’m gonna try to explain this
You need to show f (or g whatever they can be swapped) is 0 at every point x
dull ginkgo
why?
next obsidian
Say f is not constantly zero on any interval
vivid tiger
if fg = 0 at some point and in the nghb of that point
then one of f,g is zero at that point
next obsidian
You’re gonna pick an interval I
dull ginkgo
why do I need to show f is 0 at every point x?
vivid tiger
take the one that's potentially not zero
dull ginkgo
I am so confused
vivid tiger
take a ball such that in that ball f is not zero and this ball is also in the earlier ball
next obsidian
Xela you aren’t really helping when two people are saying different things
Oky look miz
How would you try to write the proof
Let’s say f is not constantly zero on any interval
vivid tiger
so then in this ball g is zero
next obsidian
You’re trying to show g = 0 right?
dull ginkgo
?????>
vivid tiger
and then I get stuck.
next obsidian
You can’t possibly make a proof work showing g is zero in balls
Okay miz
You pick either me or xela
dull ginkgo
I am doing first that f being a 0 divisor implies that the preimage of 0 under f contains an open interval
dull ginkgo
Here's what I was going to show:
Lets say f in C([0,1]) is a zero divisor. Then...
T H I S
vivid tiger
don't pick me i was trying to figure it out
dull ginkgo
f(x) is not 0 everywhere
f being a 0 divisor implies there is a g that is NOT 0 EVERYWHERE such that f(x)g(x) = 0 everywhere
next obsidian
Analyzing g(x), because g(x) is continuous, there must be a point x_0 in (0,1) that is not 0. By continuity, there is an open neighborhood of x_0, S, that is also not 0. By R being a domain, that means the image of that open neighborhood must be 0 under f.
Okay so this sentence
What does x_0 in (0,1) that is not 0 mean
That’s clear because it is in (0,1)
dull ginkgo
sorry:
g(x_0) not 0\
Analyzing g(x), because g(x) is continuous, there must be a point x_0 in (0,1) such that g(x_0) is not 0. By continuity, there is an open neighborhood of x_0, S, that is also not 0 under g. By R being a domain, that means the image of that open neighborhood must be 0 under f.
next obsidian
Okay
vivid tiger
wait, this is indeed what I showed, I just didn't realize we were proving that zero divisors only if the 0-set has an open neighborhood in it.
next obsidian
Yeah that works
vivid tiger
i def thought that we were trying to show that there were no zero divisors
next obsidian
Shit
dull ginkgo
what
next obsidian
I thought you were trying to like
Pick a bunch of points f isn’t 0 at
extend that to a nbhd of each
hollow mica
|| g is nonzero somewhere --> g is nonzero on an open interval by continuity --> f is zero on this open interval ||
vivid tiger
Truth has stepped.
next obsidian
This won’t work because you can’t show that g is 0 around specific points, only in smaller intervals in each interval
And you could somehow miss a bunch of shit
Anyway what I was initially trying to have you argue was
In any eps nbhd of any point x, there’s a point where f is not zero
Aka g is zero in any nbhd of x
So g is 0 on a dense set and thus 0 by continuity
dull ginkgo
The other way is even easier
next obsidian
Yeah
dull ginkgo
if f is 0 on (a,b)
cobalt heath
So for each x, f(x) = 0, or g(x) = 0 as R is a domain
Oh, was this the starting point
dull ginkgo
then consider a bump function on (-1,1) and shift by ((b - a)x + a + b)/2
next obsidian
Yeah
vivid tiger
the idempotents are such that at each point f(x)=1 or 0. thus it takes on a constant value of either 1 or 0 on each connected component of the domain.
the nilpotents must be 0.
the invertible ones are not zero
cobalt heath
The units of this is the most interesting imo
dull ginkgo
Yeah
Yeah if $b(x)$ is a bump function on $[-1,1]$, then $b\left(\frac{2x - b - a}{b - a}\right)$ works if $f$ is 0 on $(a,b) \subset [0,1]$
cloud walrusBOT
Mizalign #1 simp
next obsidian
The units of this is the most interesting imo
Isn’t it literally just functions which are nowhere vanishing?
dull ginkgo
Isn’t it literally just functions which are nowhere vanishing?
yeah
cobalt heath
Isn’t it literally just functions which are nowhere vanishing?
I am sorry that trivial stuff interests me
dull ginkgo
nowhere vanishing or strictly positive/negative
next obsidian
IMO personally a functions that’s strictly negative shouldn’t be a unit
dull ginkgo
There are no nilpotents, and the only idempotents are 0,1
next obsidian
I don’t neee that kinda energy in my life
vivid tiger
0 is nilpotent
vivid tiger
generalize to topological spaces
dull ginkgo
0 is nilpotent
NONTRIVIAL
dull ginkgo
generalize to topological spaces
you need an algebraic structure dingus
vivid tiger
dingus
next obsidian
~~you need an algebraic structure dingus~~
No you don’t
vivid tiger
you are the dingus
hollow mica
if you don't include zero as a nilpotent then they don't form an ideal
cobalt heath
Hmm, maybe complement of zero set of continuous functions form subbasis
dull ginkgo
or S -> R
next obsidian
Maps into R get the algebra structure from R
vivid tiger
T->R
C(T,R) form a ring
connected component stuff because I personally like the idea of being constant on a connected component.
dull ginkgo
fuck I can't use my bump function argument like before
cobalt heath
IIRC you can recover manifold structure from the ring of continuous functions C(M, R)
vivid tiger
IIRC you can recover manifold structure from the ring of continuous functions C(...
because Morse and stuff?
dull ginkgo
BEHOLD: URYSOHN
cobalt heath
because Morse and stuff?
I don't think it's Morse
cobalt heath
After all, continuous functions need not be morse, also I did not specify differential manifolds
cobalt heath
NEEDS TO BE NORMAL
Just assume metrizable or being a manifold
Other spaces are for algebraic geometrists
(It isn't butt)
dull ginkgo
i was gonna invoke urysohns and call it a day
next obsidian
You want like paracompact Hausdorff or something
dull ginkgo
you have to consider the closure
dull ginkgo
You want like paracompact Hausdorff or something
Normal hausdorff
cobalt heath
because Morse and stuff?
Do you know of ring spectrum (Spec A) stuff
next obsidian
Paracompact and Hausdorff => normal or something
dull ginkgo
Two disjoint closed sets have disjoint open sets
next obsidian
Idk
teal wind
well it asks you to do something pretty specific so have you tried to set up a g...
Could I show you what I've tried?
dull ginkgo
probably lol
next obsidian
Idk
vivid tiger
Do you know of ring spectrum (Spec A) stuff
i know of the definition
dull ginkgo
Idk this is a psychotic bastardization of the already psychotic bastardized exercise jacobson gave
cobalt heath
Yeah so, you can perhaps compute what Spec C(M, R) is
vivid tiger
it isn't psychotic bastardized
dull ginkgo
it isn't psychotic bastardized
i know my truth
now I just need to write down the fuckin thing
Hey xela, do you want another jacobson exercise I did to keep ya busy
vivid tiger
Yeah so, you can perhaps compute what `Spec C(M, R)` is
it's something.
what do the prime ideals look like?
cobalt heath
For the purpose, let's only think of maximal ideals
What are ideals of C(M, R)? Can you identify that
dull ginkgo
dull ginkgo
Are there maximalmidelasmother than {f| f(x) = 0}
you can probably show every ideal lies in one of em
next obsidian
you can probably show every ideal lies in one of em
I think maybe you’re right but ugh
vivid tiger
are the only prime ideals here maximal? yup
vivid tiger
so then the zariski topology's closed sets are basically choosing single points in the zero set.
next obsidian
you can probably show every ideal lies in one of em
So if for every x there’s a function f_x not vanishing at x in I
I want to show 1 exists in I
I think you want to take a partition of unity or something and then use that to do an infinite sum
But I don’t actually see how you can argue that remains in I
vivid tiger
that is, for every zero set, the set of zeros under our identification of zero sets with ideals gives closed sets