Mathematics - 2021-01-30 (page 3 of 3)

Balarka Sen

21:02

If $(f_1, \cdots, f_r, x_{n+1} g - 1)$ is not radical, then $(f_1, \cdots, f_r)$ is not radical in $k[x_1, \cdots, x_n]_g$, I think.

Because localization commutes with taking radicals

Ted Shifrin

@BigSocks You're doing a dangerous example, namely a prime power, to start with. And then you choose the very same prime. You have to pick less coincidental examples to get valid insights. What if I'm in $\Bbb Z_{24}$ and choose $(\bar 6)$? The right place to start is: How do you characterize when $(\bar a) = (\bar b)$ upstairs in $\Bbb Z$?

Sha Vuklia

Maybe I have sth better @Balarka

Isn't it true that

If you quotient by an ideal

and the quotient has no nilpotent elements

Then the ideal is radical?

Which is the case here, since we are localising

Balarka Sen

Something like this, but I have to check facts I don't know off the top of my head, which I didn't want to do. You're probably right that is easier.

Sha Vuklia

I mean, it's almost by def

I think I'm convinced by the entirety of the argument

Balarka Sen

Cool!

BigSocks

21:08

@TedShifrin yeah I guess you're right. You would get $(\overline{6}) = (\overline{2})(\overline{3})$, but it's not really equal to either of them, each of which is an ideal generated by a zero divisor...

Ted Shifrin

So the question, @BigSocks, is going to be (once you answer my question above): If you have, e.g., $(\bar 6) = (\overline{12})\subset \Bbb Z/60$, how do you discover the unit?

BigSocks

wait but $6$ is in $(\overline{6}) \subset \Bbb Z/ 60 \Bbb Z$, but not in $(\overline{12})$

Balarka Sen

@ShaVuklia There is a geometric point here. In general, if $I$ and $J$ are radical ideals, radicality of $I + J$ is saying something like, $V(I)$ and $V(J)$ are transversely intersecting. $Z(f_1(x_1, \cdots, x_n), \cdots, f_r(x_1, \cdots, x_n))$ and $Z(x_{n+1} g(x_1, \cdots, x_n) - 1)$ are always transverse in $\Bbb A^{n+1}$, I think.

BigSocks

they're not equal

Sha Vuklia

I don't know what transverse intersection is (I do know it for smooth manifolds)

Ted Shifrin

21:17

Oh, my bad.

Balarka Sen

You can take that as definition of transversality. For example, $I = (y)$ and $J = (y - x^2)$ in $k[x, y]$ are not transverse, because $I + J$ is not radical, it contains $x^2$ but not $x$.

Geometrically, this happens because $y = 0$ and $y = x^2$ intersect with multiplicities at the origin

Ted Shifrin

Yes, of course you're right. Let me try again. How 'bout $(\overline{42})$?

Sha Vuklia

Ok, and what's the definition of transverse intersection geometrically? That the intersection points are isolated?

Or maybe I should just keep a note of this, and when I see the defs spelled out somewhere sometime, I'll remember you mentioned this x)

Ted Shifrin

I wanted an example where $-1$ was not the obvious unit.

Balarka Sen

More than that, the intersection points shouldn't be fuzzy. $V(I + J) = V(I) \cap V(J)$ is the variety $y = 0$ and $y = x^2$, which is the origin. But the ideal $I + J = (x^2, y)$, smaller than $(x, y)$, the defining ideal of the origin.

The variety corresponding to $I + J$ is a fuzzy origin, with a second order fuzz in the $x$-direction

Ted Shifrin

21:19

@ShaVuklia No. $y=x^2$ and $y=0$ intersect at only one point, but most definitely not transversely.

LOL, ooops.

I should know better than to try to beat Balarka to a punch.

Sha Vuklia

We haven't treated multiplicity yet I believe (or maybe we did, but I'm still catching up on the material), but I guess this will come around eventually

Balarka Sen

:)

Yeah, no worries. I think it's a sophisticated point even I don't fully get.

Ted Shifrin

You want to cut down the dimension of the coordinate ring by $1$ with each one. Something about length of ideals.

Balarka Sen

That should be right yeah

Ted Shifrin

Look at $k[x,y]/(y) \cong k[x]$. Then look at $k[x]/(x^2)$ Dimension 2 as a vector space. Not good.

I can't believe I'm mumbling algebra.

Sha Vuklia

21:22

Loll, Ted

Balarka Sen

I need to write my percolations talk tomorrow!

I'm screwed

Ted Shifrin

Stop playing in here.

I need to take my walk, actually, while @BigSocks tries to figure out number theory. :)

Balarka Sen

Yup, I should be done in half an hour though

Bye for half an hour

Ted Shifrin

LOL, bye.

@Sha: In commutative algebra, there is a notion of a regular sequence. That's what we're talking about.

schn

What does the author mean from the sentence starting with “In this case...”? What is $f(x_i)$ and why is it independent from the original $f(x_i)$?

Ted Shifrin

21:26

It's a typo. $x_i$ should be $t_i$.

You recover the average of the limiting values, so if those limiting values are different (i.e., the function isn't continuous), you never have any idea what the value at the point is.

Sha Vuklia

@TedShifrin Right okay. I'm still recovering from this proof that finally came together x) I'll keep the mental notes, thanks

Ted Shifrin

Integrals never notice individual values, anyhow.

OK, @Sha.

Sha Vuklia

@TedShifrin in discrete spaces they do

lol sry

I'll let myself out

Ted Shifrin

Bubye. I'm going for my walk.

Astyx

have fun

Sha Vuklia

21:28

Enjoy the walk

Ted Shifrin

Hi, bye, Astyx.

Astyx

hi bye

Thorgott

revolting notation

why would you use the same notation for two different things literally at the same time as trying to make the point that these things can be different

schn

@TedShifrin Thanks for the reply. @Thorgott you mean the $f(t_i+0)$ and $f(t_i-0)$?

Thorgott

no, I mean writing $f$ for the inverse FT of the FT of $f$

the author is saying that $f(x_i)$ isn't the same thing as $f(x_i)$

and the notation is just as stupid as this reads

BigSocks

21:34

@TedShifrin I took a shower but now I am back to this

Sha Vuklia

@BalarkaSen That's not a strong example imo, because he's old and it's kind of a modern view

schn

But the author does use different notation, it is $\hat{f}$ for the FT and $f$ for the inverse one, no @Thorgott?

Thorgott

yes, but $f$ is also used for the original function

the rest of the paragraph is also pretty wrong, but I guess it's not supposed to be a mathematically accurate text

Mike Miller

@ShaVuklia Right which proves that what "rational" means is socially conditioned at best

schn

@Thorgott on Wikipedia they also use $f$ and $\hat{f}$, which original function do you mean?

Thorgott

21:42

f

schn

the one inside the integral? :)

Sha Vuklia

@MikeMiller Oh, ye, I wasn't entirely following the discussion, but I can get behind that

Thorgott

the one we start with

schn

@Thorgott Do you understand what is meant with the sentence "In this case..." ?What is $f(x_i)$ and why is it expressed as an average?

Is it simply some inverse transformed function?

where $t=x_i$

Thorgott

as Ted said, $x_i$ is supposed to be $t_i$ and it is the value of the integral on the right of $(2)$ at the point $t=t_i$

it is not the same thing as the value of the function $f$ we started with at the point $t_i$

which is why I've been saying that the notation is utterly abysmal

schn

21:47

So there should be no $x_i$ at all?

Kind of makes sense, totally undefined

Thorgott

the $x_i$ is a typo

schn

Cool

Then how does it follow from (2) that $f(t_i)=(f(t_i+0)+f(t_i-0))/2$?

Thorgott

that's a version of the Fourier inversion theorem

it's a non-trivial claim

copper.hat

It is not hard to see that the behavior at any specific point is irrelevant and that only local behaviour matters. If $f$ has limits from either side then the transform has no 'choice' but to converge to the average.

Proving it is more work, of course.

schn

@Thorgott So there is another version of the Fourier theorem that states that when evaluating $f(t)$ (equation (2) in the screenshot) at a discontinous point then one gets the average limiting values of $f$ around that point?

Thorgott

21:56

yes

schn

@copper.hat But one has an integral, i.e. a continuous sum, why would it suddenly turn into a discrete sum with only two terms in it, the limiting values around the discontinuous point?

Astyx

@ShaVuklia Just do scheme theory and that's by definition of $O_X(U)$

copper.hat

@schn if the function was zero except, say, one at $t=1$ what would you expect the Fourier transform to be? zero of course. It is just averaging the nearby values.

Sha Vuklia

@Astyx Lol, my alggeo2 (=scheme thy) classes start in ~1 week

Astyx

Have fun

Sha Vuklia

22:01

x)

schn

@copper.hat I would it expect to be zero because of the definition of an integral (thinking in terms of area under the graph of the integrand).

Sorry

copper.hat

yes. so the value at a particular point does not matter as such.

what matters is how the function behaves locally.

Sha Vuklia

oi, Ted, that was a quick walk:o

or maybe I lost sense of time

schn

what happens when one has several discontinuous points? @copper.hat

Ted Shifrin

@schn But we shouldn't use the letter $f$ on the left. It should be $\tilde f$, where $\tilde f$ is the inverse transform of the transform.

It was over a half hour, I think, @Sha.

Sha Vuklia

22:08

Yea, I defo lost sense of time

Ted Shifrin

Supposedly, 4308 steps.

copper.hat

@schn you need to look at the function $\tilde{f}(t^*)={1 \over 2} (\lim_{t \uparrow t^*} f(t) + \lim_{t \downarrow t^*} f(t) )$.

Ted Shifrin

It just shows to go, @Sha, that you really never miss me :P

Sha Vuklia

Loll, so savage

schn

@copper.hat right, one only ever evaluates at one point

Thorgott

22:10

@TedShifrin yeah, that's what I've been complaining about

Ted Shifrin

Yup, we agree again. Damn.

copper.hat

@schn the inverse fourier transform just 'sees' $\tilde{f}$ not $f$ as such.

Ted Shifrin

Of course, I earlier suggested that $\gcd(6,60)=\gcd(12,60)$, so who am I to complain?

copper.hat

you are permitted to make minor errors :-)

perfection is overrated.

schn

@Thorgott @TedShifrin I don't understand why using $f$ is a big issue? :)

Ted Shifrin

22:11

Will you say that if I tell people that the derivative of $f(x)=x^3$ is $3x$?

Thorgott

because the point of that paragraph is literally that the inverse FT of the FT of $f$ need not be the same thing as $f$

Ted Shifrin

@schn: It's amazing that when $f$ is continuous, that $\tilde f = f$. Otherwise, you shouldn't be surprised that discontinuities get "modified."

The fascinating thing you should look up is the Gibbs phenomenon.

Thorgott

my inner pedant remarks that continuity does not suffice

copper.hat

@Thorgott some degree of smoothness required...

@schn If $f$ has a step discontinuity at say $t^*$ then you can write $f$ locally as a continuous function plus a step function. So, look at the Fourier inverse of the Fourier transform of a delayed step function to see where the ${1 \over 2}$ comes from.

Ted Shifrin

Smoothness gets uniform convergence of the Fourier series. I haven't thought about the Fourier transform in a while.

copper.hat

22:18

Lipschitz is good enough for the local average result above.

schn

So one has a function $f$ and then applies the FT to get $\hat{f}$, then the inverse FT of $\hat{f}$ may not equal $f$? So it would have been clearer to have a separate notation for the inverse of $\hat{f}$?

copper.hat

It is essentially equal to $f$.

Thorgott

there's a lot more issues at play

copper.hat

I mean that in an ae. sort of way not a demanding way :-).

Thorgott

the plain integral given there does, even for very nice functions, not converge in general

usually you have to "massage" it a bit, either by introducing a convergence factor or interpreting it as Cauchy principal value

copper.hat

22:19

@Thorgott I think you are focusing on technicalities that are clouding the issue?

Ted Shifrin

It's bizarre to think about the Fourier transform before learning some Fourier series, IMHO.

copper.hat

I agree.

But I am surprised that, when taught, there is not more made of the connection.

Ted Shifrin

I'm thinking back to the year-long applied math course I taught in 1986. So much great stuff.

copper.hat

I know that its old stuff, but I love the Fourier transform.

Sort of linear algebra on steroids.

schn

@copper.hat But why is the inverse FT of $\hat{f}$ "essentially equal to $f$"? As discussed, isn't this only true if $f$ has no discontinuities?

copper.hat

22:22

(I'm thinking of the Plancheral transform really.)

@schn If $f$ is smooth except for a finite number of discontinuities per unit interval then the inverse transform will equal $f$ everywhere except at the discontinuity points.

schn

@copper.hat Clear :)

copper.hat

That is essential enough for most applications.

Ted Shifrin

Aside from a bunch of transforms, discrete and continuous, and some complex analysis and PDE, we also did the method of stationary phase and some asymptotics. Such cool stuff.

copper.hat

I like the insight into physical phenomena.

I am supposed to be an engineer after all.

Ted Shifrin

The other fascinating thing I hadn't learned as an undergraduate was the Huygens principle, and how the wave equation is so different in even and odd dimensions. So fascinating.

copper.hat

22:26

That is something I found/find frustrating about pdes.

Ted Shifrin

Group velocity and Kelvin angle also fascinating.

copper.hat

The behaviours are so vastly different for different pdes.

Ah, the boating angle :-)

Ted Shifrin

Well, that's to be expected. Elliptic is the golden baby; the others are brats.

copper.hat

I am much more comfortable with odes.

It is astounding to me that people can design systems governed by Maxwell's equations.

schn

@Thorgott In that sentence I mentioned earlier, i.e. "$f(t_i)=(f(t_i+0)+f(t_i-0))/2$ independently of the original value $f(t_i)$", how would you have changed it if you could only change the $f$'s? That'll make it clearer I think.

Ted Shifrin

22:29

Well, they are pretty fundamental :)

copper.hat

Antennae, microwave design is such a black art.

In chip design it used to be there were digital designers, analog designers and microwave designers in increasing order of 'black art'ness.

"Can one hear the shape of a drum?"

Thorgott

@copper.hat The integral defining the inverse FT need not be convergent. In fact, if $f$ has at least one discontinuity, it will never converge. You absolutely have to interpret it as a principal value for this statement to be true. This is an integral point, not a technicality.

BigSocks

So I get $(\overline{6}) = (\overline{12})$ in ideals $(42), (18), (30)$, but they are not equal in ideals like $(24), (60), (48)$. The difference seems to be that both divide the latter ones but not the first ones (even thought they both share common factors with the first ones). So I am going to guess that $(a) = (b)$ whenever $(\overline{a}) = (\overline{b})$ in every quotient of your PID

@TedShifrin

Thorgott

@schn the value of the inverse Fourier transform of the Fourier transform of $f$ at the point $t_i$ is $(f(t_i+)+f(t_i-))/2$

Ted Shifrin

You're writing "in ideals" when you mean modding out by ...

copper.hat

22:35

@Thorgott that's a little strong, it depends on what you mean by converge. Plancheral would disagree.

Ted Shifrin

So what is the inverse image in $\Bbb Z$ of the ideal $(\bar a)\subset\Bbb Z/n$?

BigSocks

@TedShifrin yeah

schn

@Thorgott Thanks!

Ted Shifrin

@BigSocks: Remember, of course, that $\overline{a+kn} = \bar a$.

BigSocks

@TedShifrin should be $(a)$

Ted Shifrin

22:37

What did I just finish writing?

Thorgott

I don't follow. Plancherel tells us that $\hat{f}$ is in $L^2$ if $f$ is in $L^1\cap L^2$. $\hat{f}$ does, in general, not lie in $L^1$, regardless (which is what I mean by convergence).

BigSocks

the equivalence class of $a + kn$ in $\Bbb Z /n \Bbb Z$ is the same as the equivalence class of $a$

copper.hat

@Thorgott I presume you are talking about pointwise convergence?

Ted Shifrin

So how can the preimage of the ideal depend on which representative I pick?

BigSocks

oh I see what you're getting at

can't be $(a)$

Thorgott

22:40

no, convergence of the integral $\int|\hat{f}|$, i.e. integrability of $\hat{f}$

copper.hat

the Fourier transform is an isometry of $L^2$.

i see

BigSocks

well I guess it has to be the "most reduced version" of $a$

Ted Shifrin

You need to understand this fully.

BigSocks

I do yes

Ted Shifrin

In general, if you map $R\to R/I$, what is the preimage in $R$ of the ideal $\bar J$ ?

BigSocks

22:42

I don't understand. preimage of something already in the source?

oh ok

Ted Shifrin

Bad English. The bar should have clued you.

So, to be slightly more concrete, assume $\bar J = \pi(J)$, with $\pi$ the projection.

You can stick with $\Bbb Z$, but I want you to think about it clearly.

schn

@copper.hat Previously you said that if the function is zero everywhere except at $t=1$, the Fourier transform would simply be zero. Suppose the function isn't zero everywhere, why would one still get the average limiting value around $t$ then?

copper.hat

@Thorgott ignore my earlier remarks.

@schn with that function $\lim_{t \to t^*} f(t) = 0$ everywhere.

@schn what book did the red boxed statement come from?

Ted Shifrin

@BigSocks: I know you're getting overwhelmed, but I finally have a good numerical example for you to explore. In $\Bbb Z/120$, consider $\bar a = \overline{18}$ and $\bar b = \overline{42}$.

I wonder if @Thor has figured out a proof that in any quotient of a PID same ideal implies associates. :)

BigSocks

@TedShifrin brain broke. $\pi^{-1}(J)$? hmm

Thorgott

22:53

I haven't thought about it, let's see

Ted Shifrin

But what is $\pi^{-1}(J)$? Do a concrete example. Write out $\Bbb Z\to\Bbb Z/12$ and figure out the preimage of $(8)$. Then prove it in general.

@Thor: You will probably have a better proof than I came up with. It's a little subtle.

schn

@copper.hat They are lecture notes, not belonging to a book. Why does $\lim_{t \to t^*} f(t) = 0$? Do you mean the right-hand and left-hand limits at $t$ do not exist?

Ted Shifrin

Actually, I've just been rethinking the $\Bbb Z$ situation, not the general one. But I know there's a surprising hook.

copper.hat

@schn If a function is zero except at a single point then the limit is zero everywhere.

BigSocks

@TedShifrin there's a twist now? amazing

@TedShifrin writing down numbers now

copper.hat

22:55

@schn I found a similar statement in web.stanford.edu/~hca/c276autumn2009/NA1_Fourier.pdf

schn

@copper.hat Object not found. By the way, it is a class in electrodynamics, using Jackson's book. Have you read that book?

Ted Shifrin

Last time I taught algebra, I actually assigned this problem to the graduate and Honors students in the course.

BigSocks

oh wow ok. I thought I was choking at a baby problem

Ted Shifrin

Well, you haven't gotten to the main part of the problem yet.

BigSocks

quietly goes back to writing down numbers

Ted Shifrin

22:57

You definitely need to understand what the inverse image or preimage of the ideal looks like.

copper.hat

@schn Try again, there was an extra / at the end which messed things up.

schn

@copper.hat Works

copper.hat

Jackson, yes, a lifetime ago :-(

I think @Thorgott's point is that the inverse transform needs to be in a $\lim_{T\to \infty} \int_{-T}^T$ sense.

Ted Shifrin

I distracted Thor for you with algebra.

Thorgott

yeah, that is my point

this is important already for the arguably most important example in Fourier theory: rectangular functions and sinc functions are dual via the Fourier transform. but: sinc functions are not integrable; however their principal values make perfect sense.

copper.hat

23:01

@Thorgott sry, i relearn this point every few months.

Ted Shifrin

@Thor just wants to show us that he knows a lot of analysis, not just algebra :)

copper.hat

certainly brought me back to attention.

even with Fourier series there is a similar point with symmetric partial sums

Thorgott

@TedShifrin I wish

Ted Shifrin

I was being sincere, actually.

So, if you didn't see it, a slightly more interesting example of elements generating the same ideal that are not "obviously" associates are $\overline{18}$ and $\overline{42}$ in $\Bbb Z/120$.

BigSocks

@TedShifrin wait is it $J \cap I$

Ted Shifrin

23:08

That can't be right, can it?

It has to contain all of $J$.

BigSocks

dang

Ted Shifrin

Writing down numbers in my example should give it away.

(The easy example, not mod 120.)

BigSocks

I was thinking because the other should be $(4)$ I think, and I thought maybe it was because $4 = gcd(8,12)$ and I think there was a fact $(a) \cap (b) = (gcd(a,b))$ for ideals in the integers

Ted Shifrin

Get that fact correct.

schn

@copper.hat in the linked paper, what does it mean for integrals not converging separately at the upper and lower limits?

simply that $F(b)$ or $F(a)$ do not exist in $\int_a^b f(x)dx=F(b)-F(a)$?

BigSocks

23:11

$(a) + (b) = (gcd(a,b))$ @TedShifrin yep

not $\cap$

Ted Shifrin

Right, it's all about gcd's. And adding the ideals should make sense here if you write out a few numbers.

BigSocks

so should it be $I + J$?

Ted Shifrin

Yes.

Fargle

@TedShifrin What do you mean not "obviously"? Clearly $42 \cdot 109 = 38 \cdot 120 + 18$. /s

BigSocks

woah nice

Ted Shifrin

23:13

LOL. Yes, that was obvious.

So give me a proof in general :)

Fargle

(I lucked into it because $42 \cdot 11 = 462 \equiv -18$ mod $120$.)

BigSocks

ok I wrote down $18,36,54,72,90,108,6,24,42,60,78,96,114,12,30,48,66,84,102$

that is for $\overline{18}$

Ted Shifrin

Well, if you use what we just "discovered," you should know that the ideal upstairs in $\Bbb Z$ will be $(6)$.

BigSocks

and $42$ shows up in the list so I am gunna guess this is the same list as the one for $\overline{42}$ but in a different order

Ted Shifrin

Again, what's $\gcd(42,120)$?

BigSocks

23:16

$6$

Fargle

@BigSocks That doesn't quite work as a guess: $36$ is in the list too, but generates a proper subideal.

Ted Shifrin

So the same gcd suggests that the ideals are in fact equal upstairs, hence downstairs. Now how to find multiples to show associates.

BigSocks

@Fargle this is true, I should not be so quick to guess

but $gcd(18,120)$ is also $6$

Ted Shifrin

Yes, the ideals are equal iff the gcds are equal.

BigSocks

and $120/6$ is $20$ which is the amount of numbers in the list and it would be really wild to have 2 different lists that contain 2 of the same number but 18 other different numbers

copper.hat

23:18

@schn I think you can omit the word separately from that sentence.

Ted Shifrin

Now how to proceed?

BigSocks

Hmm well $(a) = (b)$ in a given quotient ring of $Z$ whenever they have the same gcd with respect to the ideal you quotient by

should I prove that somehow?

@TedShifrin just saw this

Ted Shifrin

Right. So you somehow need to use that gcd to prove the things are associates.

When you have three numbers with a common gcd, what's a natural thing to do?

BigSocks

I dunno, you could try and write something like the Bezout eq. $ax+by = d$, where $d$ is the $gcd(a,b)$?

I guess maybe since they have the same gcd you could equate those...

Ted Shifrin

Remember we want to get a multiple of $18$ equal to a multiple of $42$ (mod $120$).

Thorgott

23:24

@copper.hat You might know this already, but there's a cool fact known as Fejer's theorem (it's usually only stated for Fourier series, but also holds for Fourier transforms). It says that if $f\in L^1$ and $f$ is continuous (no smoothness assumption whatsoever), then we can look at the Césaro principal value instead of the principal value (which means that we take the limit over successively averaged versions of the integral instead) and this always converges to $f$.
Under some comparatively mild conditions (I think piecewise continuity suffices, but I don't recall), you also get that if $…

BigSocks

huh, maybe Chinese remainder theorem to solve this system?

Ted Shifrin

What system?

BigSocks

oh wait but they're not relatively prime and I think you need that nvm

Sha Vuklia

I want to describe this natural isomorphism. Let $(U,f)\in\mathfrak m_{X,P}$. We know there exists some pol $g$ such that $P\in X\cap D(g)\subset U$. We can restrict $f$ to $X\cap D(g)$, and project it to $\mathfrak n/\mathfrak n^2$, where $\mathfrak n$ is the maximal ideal of $O_X(X\cap D(g))$ corresponding to $P$.

copper.hat

@Thorgott that is interesting. thanks!

Sha Vuklia

23:26

According to a previous result, $\mathfrak n/\mathfrak n^2$ is isomorphic to $\mathfrak m/\mathfrak m^2$, where $\mathfrak m$ is the maximal ideal of $\mathcal O_X(X)$ corresponding to $P$. For $\mathfrak m/\mathfrak m^2$ we have a map to $k$, so we apply this map.

This will work if what I just did is well-defined. So say I consider another principal nbhd, $X\cap D(h)$ with maximal ideal $\mathfrak n’$. I want to show that it sends the projection to ${\mathfrak n’}/{\mathfrak n’}^2$ to the same element in $\mathfrak m/\mathfrak m^2$.

The isomorphism $\mathfrak m/\mathfrak m^2\to\mathfrak n/\mathfrak n^2$ is a matter of restricting a regular function on $X$ to $X\cap D(g)$. Furthermore, we have that $\mathcal O_X(X\cap D(g))\cong\mathcal O_X(X)[g^{-1}]$ and $\mathcal O_X(X\cap D(h))\cong \mathcal O_X(X)[h^{-1}]$. The isomorphism is just given restricting once again. But I don't see how any of this is giving me well-definedness.

idk, if anyone sees it, I'd be happy

Thorgott

@TedShifrin thanks, but "a lot of" is an overstatement :P

Ted Shifrin

Well, you can nail this algebra proof :)

I'm actually not seeing the end of it when things have too many factors in common (like my example).

BigSocks

hmm maybe you divide everything by the gcd and write $18x - 24 y = 0 mod 120$ whenever $3x - 4y = 0 mod 20$

Ted Shifrin

Interestingly, not to ruin @BigSocks's fun, we have $18=42\cdot 9 \pmod{120}$ and $42=18\cdot 9\pmod{120}$.

OK, dividing by the gcd is a good thing. But the Bezout equation isn't relevant, @BigSocks.

So divide everyone by $6$ and try to answer the associate question in $\Bbb Z/20$.

BigSocks

but $(1,9)$ is one of the solutions- like the one you just wrote down?

(same as $(9,1)$ I guess)

For the associates you have to find the solution to that equation when one of the variables is set to $1$ I guess

So $xy = x \vee xy = y$

just realized I wrote the wrong thing- should have been $3x - 7y = 0 mod 20$, but still, those are the solutions

Jack Zimmerman

23:36

Do any of us have any mathematical predictions for the next year?

Thorgott

yeah, I'm thinking about this algebra fact, it's surprisingly non-trivial

might have to get my hands dirty

copper.hat

well, i find hindsight much easier.

BigSocks

thank god I am not looking like a complete chump here

Fargle

My pithy example from earlier makes me now finally realize the punchline.

BigSocks

I still don't see what was special about $18$ and $42$ here

I want to tho

Ted Shifrin

23:40

What's special is that you can't just take the additive inverse like we could in easy examples.

And what else is special is the equation I wrote down (with the $9$'s). Unfortunately, $\bar 9$ is not a unit in $\Bbb Z/120$.

BigSocks

fwiw I was about to write that $9$ was what you had to multiply by

Ted Shifrin

Yes, it was cool that $9$ showed up both times (because $9=9^{-1} \pmod{20}$).

But how does this help us in $\Bbb Z/120$? That's why I worked so hard to come up with this example.

Yes, @BigSocks, the fact that Thor has to think means this isn't trivial :P

Sha Vuklia

Nvm, I think I found an argument for my Q a few msgs above

BigSocks

@TedShifrin well it gives us the $r$ in $a = rb$

although it's not a unit so now I am wondering how to reconcile that

Ted Shifrin

But $9$ isn't a unit.

That's why I had to work hard to make up the example.

BigSocks

23:47

then this was a big hoax!

Ted Shifrin

If it is a unit, we're done and we go have a martini.

BigSocks

@TedShifrin that would be good hmm... but it's not so we delay

Fargle

@BigSocks: If it's $9$ in $\Bbb Z/20$, what is it in $\Bbb Z/120$?

Ted Shifrin

It's still 9.

9 is a perfectly nice unit mod 20.

BigSocks

@TedShifrin was gunna write this

Ted Shifrin

23:49

Is @Fargle trying to confuzle us?

Fargle

No, I have a point.

BigSocks

@TedShifrin oh then is this the trick? we have to reduce everything by the gcd of $a$ and $b$?

Ted Shifrin

That's where we start. But it's not winning the game for us in this case.

Fargle

My point was: must it be 9 upstairs?

BigSocks

I can answer- if you mean $\Bbb Z/120$ the answer is yeah

Ted Shifrin

23:51

No. 9 is certainly not the only solution.

BigSocks

well it's the only solution where the other number is a $1$ I think

Ted Shifrin

There really are two numbers; they just both happened to be 9 this time.

BigSocks

it's like 9 has multiplicity 2

Ted Shifrin

I had a Chinese Remainder Theorem argument sketched in my notes for my problem set, but I'm not seeing how to make it work yet. But Fargle is suggesting we try -11 or 29 or something else. Those would be units mod 120.

So probably a proof will work like this ...

BigSocks

So you do do a Chinese Remainder thing. hmm

Ted Shifrin

23:53

Not what I had in mind.

BigSocks

wait!

But I have been trying so much

don't just give it away now

Fargle

Ted, I do think your misgivings about my idea are valid. I don't know how I'd formalize it in the general case---I'd need to tinker with it more.

Ted Shifrin

Well, I'm trying to figure out the end of the proof I had in mind, and I'm stuck at the moment.

I still say this is a great question :)

BigSocks

I would argue it is better now than before

ok so I was thinking about that one fact. $(a) + (b) = (gcd(a,b))$. so if $a,b$ are relatively prime, $(a) + (b) = (1)$

kinda cool

Ted Shifrin

Relatively prime is NOT an interesting case.

Well, of course, that's what Bezout has told you about gcds for years.

BigSocks

23:58

yes, but when you put parenthesis around everything it looks so nice

(but thank you for making the (kinda obvious now) connection with Bezout for me there)

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$Mathematics$ Mathematics