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9:01 PM
@EM4 indeed
 
EM4
yay thank you!
I was having doubts.
 
@user2103480 lol
 
@user2103480 Perhaps he likes impartial fractions.
 
@robjohn I hope so, professors should be an impartial faction
 
EM4
9:10 PM
do you mean partial fractions.
 
we mean jokes
 
Yeah, we fall numerators over denominators around here.
But only proper fractions. We run a clean room.
I walked into a bar and asked for a mixed fraction, and they just stared at me.
It was definitely not an open bar; there were people in a neighborhood of 21 that they would not serve.
 
is this turning into a film noir
 
9:49 PM
So if we have an exact sequence of $A$-modules $0 \to M' \to M \to M'' \to 0$, and $M$ is finitely generated as an $A$-module, apparently we have $M''$ is also finitely generated as an $A$-module. The map $g : M \to M''$ takes the generators of $M$ to the would-be generators of $M''$.

My questions:
$1.$ Some of these generators could be mapped to zero right? What prevents all of them from going to zero? probably exactness somehow I bet.
$2.$ What are some examples of $M$ being finitely generated and then $M'$ not being finitely generated?
 
10:03 PM
2. A non-noetherian ring has ideals that are not finitely generated. A minimal example is $k[x_1,\dots,x_n,\dots]$ and the ideal $(x_1,\dots,x_n,\dots)$.
A more interesting example may be a valuation ring of dimension $>1$. Any finitely generated ideal is principal, but it must have ideals that are not finitely generated.
1. The fact that things map to $0$ isn't really a problem.
 
$1.$ But say the generators for the middle one are $\{a_1, ..., a_n \}$. apparently the generators for $M''$ are $\{ g(a_1) , ... , g(a_n) \}$, but what if they all get sent to $0$? Can that even happen?
and for the valuation example what is an example of a not finitely generated ideal of a valuation ring of dimension $>1$
 
@robjohn I see robjohn has had his cocktails early this afternoon :D
 
10:19 PM
@BigSocks The maximal ideal cannot be principal in a valuation ring.
For 1. it may be helpful to remind yourself of what properties $g$ must have.
 
oh right the kernel must be the image of $f$...
can't just be the whole thing
all of $M$ that is
 
If I gave you $M'\subset M$, there is up to isomorphism, just one choice of $M''$.
 
and I have to learn more about valuation rings I think, but you have already been helpful. I will write this down by the proposition to remember
 
@TedShifrin somehow, fractions inspired some d-bar room humor.
 
So you're merging into elliptic PDE?
 
10:24 PM
@TedShifrin you're getting too complex for me. I'm boxed in at a lower level.
 
it's all fun and games until 0 shows up
 
Yes, we have hit bottom and run into the boundary.
I guess we need a Neu mann.
 
@TedShifrin oh, you haven't seen anything near the bottom ;-)
 
@KarlKroningfeld hmm I am thinking but I can't seem to figure out why.
 
@BigSocks so you can be!
you think, therefore UR
 
10:28 PM
probably has to do with $f$ having to be injective
hahahaha
feels good to be able to be
at times, infrequently, spread out
etale being
 
@user2103480 Speaking of partial fractions, there's a nice elementary proof that any (finite-dimensional) linear map whose minimal polynomial has distinct roots is diagonalizable ... using partial fractions, nothing fancy.
 
@BigSocks I rarely think about the ring structure of valuation rings. The maximal ideal can be principal, as Matsumura tells his poor readers, but then it is the height $1$ prime ideal that is not finitely generated. Defies my Noetherian upbringing.
 
@TedShifrin cool! do you have a reference?
 
@BigSocks What is one short exact sequence that starts with $0\to M'\to M$?
 
No, I discovered it while writing exercises for my linear algebra text. :) So it's a problem in there.
 
10:33 PM
Will $M''$ always be something like $M/M'$?
 
It doubtless is in print in some textbook, but I didn't learn it that way.
 
oooh, neat :)
cool when the easiest guess is the right answer
 
I can give you the main idea if you're interested, @user2103480.
 
and I do not know Matsumura - you are referencing his "Commutative Ring Theory"?
@KarlKroningfeld
 
10:34 PM
That'd be nice
 
something-something $(A-\lambda I)^{-1}$?
 
@BigSocks Right, he gives the construction on p.79
 
time to take a peek :)
 
@TedShifrin oh, go ahead. I'm not sure I've seen it.
 
@Semiclassical the right answer for everything involving eigenvectors
 
10:37 PM
So suppose $A$ satisfies $f(A)=0$ for $f(t)=(t-\lambda_1)\dots (t-\lambda_k)$. Write the partial fraction decomposition for $1/f(t)$. When you clear the denominator and substitute $A$, you get $I = \sum p_j(A)$ where $(A-\lambda_j I)p_j(A)=0$. This gives a decomposition of your vector space into eigenspaces.
 
hmm, i think that counts
 
I "discovered" this because long ago I had written exercises to show that if $A^2=I$ or if $A^2=A$, then $A$ is automatically diagonalizable.
Those exercises are totally standard in every (good) linear algebra book.
 
@TedShifrin neat
 
Of course, we all learn the advanced proofs once we study minimal polynomials and canonical forms.
 
10:46 PM
@BigSocks if you have an exact sequence $M\rightarrow M^{\prime\prime}\rightarrow 0$, the first map is surjective. if a surjective map maps a generating set to $0$, what can you say about the codomain of that map?
 
it's just zero?
 
yup
 
oh thank god. I was wondering lol
 
and in that case 0 is a generating set
 
but like... trivially so I guess
but the statement is still true, which is what matters in a way
 
10:59 PM
of course, but the point is that things going to 0 is not an issue
 
right yeah- gunna put this as a side note. thanks
 
Someone cast an undeserved downvote, they commented why they downvoted, but the reason was wrong. I explained this in a comment and even updated the answer because the information in my comment was helpful (though not necessary). Of course, they have not removed their downvote (since I've edited, they can) or commented to the contrary. :-( I guess I should be happy they commented rather than drive-by downvoting.
I'm going to suggest that when someone downvotes, and the question or answer is edited, they are notified of the edit. That might promote people to actually improve their post in response to a downvote.
They know that their edit won't be sent into a vacuum.
 
the irritating thing i came across earlier was someone who put up a question which attracted an answer...and then deleted the question and put it up again. (there was an algebra error in their original question, so they're not identical in text. but it's the same problem...)
 
@Semiclassical They should have simply corrected the question, unless that would invalidate the answer. In which case, they can post again, but they should not have deleted the earlier question.
 
yup
it's been flagged
(but flag is still pending)
 
11:06 PM
I will look
 
@Semiclassical It doesn't appear you flagged it. Do you remember the post?
 
I flagged the new question, let me find it
 
hmmm
 
0
Q: Taylor series remainder term to yield ln4 with error terms less than 2^(-10)

johnsdghI need to derive the taylor series for ln(1+x) about 1 then create an inequality that gives the number of terms that must be taken to yield ln4 with error terms less that $2^{-10}$. I got the taylor series to be $$f(x) = ln(2) + \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(x-1)^n}{n \cdot 2^n}$$ I was wo...

 
11:12 PM
@Semiclassical Yeah, I undeleted it and handled your flag
 
I was looking in the wrong place for the flag
 
i'll edit their question to reflect what they included in the updated version
 
vzn
@Semiclassical hi again, re majorana representation, what about the scenario where its not "mere philosophy" or has implications outside of philosophy?
 
or, hmm. how do we typically deal with the version 2.0 of the problem?
 
11:14 PM
hi chat
 
eh. the trouble with the Bell analogy is that, on the QM side of things, the Bell inequality is pretty trivial. the interesting side of the inequality is showing that "local realism" has specific implications
it's perhaps a failure of imagination to not see any interesting implications of the majorana representation, but i see little point to pretending that I see such when I don't
 
vzn
yes was going in that direction, theres a 100+ page paper on that, managed to track it down again with some trouble wink... my idea is that some kind similar analysis applies to or could be applied to the majorana formulation...
 
one thing I particularly don't see anything philosophically-suggestive of is the 2D Coulomb gas aspect. that's very useful from a visualization purpose, but it's effectively just an implication of going from the unit sphere to the plane via stereographic projection
and the unit sphere, in turn, is basically just a matter of "you can orient a stern-gerlach device to any direction in space"
 
vzn
dont grasp it yet myself, but fundamentally its fluid dynamics. theres a huge trend of papers look at fluid dynamics connections to QM. have you looked into "emergent QM" some?
 
if there's fluid dynamics comparisons, it's the same comparisons that have always been around
 
vzn
11:19 PM
there are also some obscure papers linking bells thm to a hidden variable scheme connected to the bloch sphere.
@Semiclassical thought you had some taste for that (hard to remember)...
 
i have sympathy for the pilot wave stuff, but on the "how to make QM safe for trajectories" line of thinking rather than fluids
(and even that mindset only seems to work for nonrelativistic QM)
regardless, need to go eat dinner
 
vzn
bohm himself was considering fluid dynamics ideas but never wrote them up much, have found it in some paper(s)
did you have any reaction/ opinion on this? think it shows that relegating deep questions of QM foundations to mere "philosophy" is no longer tenable to say the least. Quantum Leaps, Long Assumed to Be Instantaneous, Take Time quantamagazine.org/…
 
Ok so same scenario with the exact sequence $0 \to M' \to M \to M'' \to 0$. Apparently it is really obvious that $M$ being noetherian makes $M'$ noetherian (by definition)
 
@robjohn: Have you noticed a bunch more editing (recently than before) of very old posts? I guess people are trying to earn rep by editing, but I find it very annoying. I've complained to a few of them.
 
Anything to get those sweet sweet points
hi Ted
 
11:33 PM
@BigSocks yes
no need for the sequence
 
just like injectivity from $M'$ to $M$, that's it
o bc it's a submodule right
definition of... noetherian
 
yes
 
all submodules are finitely generated
ok cool cool
 
Hi @Astyx
 
@TedShifrin Are you sure they are edits? I saw what I thought was an edit because it was on the front page, but it turned out to be an auto bump.
 
11:42 PM
Yeah, it said edited an hour ago. Here.
 
@TedShifrin I'm resisting the urge to append an s at "but different second fundamental form"
 
Ugh.
 
it's a plural grrr these objects are different
 
What's crazy is that this is an exercise for a 2nd or 3rd year undergraduate course, and the OP says he's doing "research" as if he's advanced.
Yes, I understand. I don't make cosmetic edits like that unless it's a new post and I'm making a math change.
 
@user2103480 Do you mean they're two different object ?
 
11:54 PM
Rectifying curves on a smooth surface immersed in the Euclidean space
apparently a junior research fellow
 
Very elementary stuff. I don't know what he means by a rectifying curve, to be honest, but I'm not going to pay to find out.
 
yeah doesn't look particularly advanced
 
well, a lot can happen in 3 years
 
fair fair
 
Looks like an undergraduate research project. shrug Just goes to show that most math publications aren't necessarily good.
 
11:58 PM
arxiv.org/pdf/2001.07965.pdf this is the newest stuff. But 4 authors for a ten page paper? Including abstract and references
Still, definitely more advanced
 
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