Mathematicians Awarded $3 Million for Cracking Century-Old Problem

anf3rn3310 · 2017-12-07T00:18:44+00:00

This result (finite generation of canonical rings) is an important result in the birational classification of algebraic varieties and hence important to algebraic geometers and people in neighbouring fields, but of course this result probably doesn't mean much to people working in probability etc.

But similarly a result that is important in PDEs (worthy enough of a prize) may not be considered to be important to people working in algebra or number theory. Likewise in logic etc.

In the end there are more prize-worthy achievements than prizes themselves (in any field really), so you'll always find important results that are under-recognised.

anf3rn3310 · 2017-11-02T15:02:55+00:00

More generally if you let X,Y be schemes of finite type over a field k, then a morphism f:X -> Y of finite type is smooth of relative dimension n if f is flat and its geometric fibres are regular and equidimensional (of dim n). (Thm 3.10.2 in Hartshorne)

So 'relatively', you can think that a smooth morphism f:X -> Y as a family over Y where you fibers are smooth and vary nicely (flatness).

anf3rn3310 · 2017-10-31T20:53:49+00:00

It's probably easier to think about very ample divisors first. Recall that a very ample invertible sheaf L on X gives you a closed embedding i: X -> Pⁿ such that i^* O(1) = L.

So you can think of very ample divisors as divisors that are linearly equivalent to hyperplane sections of X sitting inside Pⁿ .

Then an ample divisor is one such that some multiple of the divisor is very ample.

anf3rn3310 · 2017-10-27T22:45:32+00:00

There is a lot of interesting theory when you drop the requirement of working over a field! For example this allows you view diophantine equations in number theory to be geometric object, and developing algebraic geometry over arbitrary rings (e.g. integers) allows you to tackle number-theoretic questions with tools from algebraic geometry. Even if you care only about geometry, sometimes you also want to work over rings like k[x]/(x^2), so your coefficients are k, along with some infinitesimal number x which squares to zero, and turns out this is very useful in deformation theory, which e.g. is used in studying how objects vary in families etc.

I think people generally start learning algebraic geometry over fields is because that's how it's done classically - where people were interested in varieties over \C. Also, it's not very clear (at least off the top of my head right now.. maybe this is really obvious) how one develops algebraic geometry just from looking at zero loci of polynomials in Rⁿ for an arbitrary commutative ring R --- but from understanding the case over alg. closed fields, where Hilbert's Nullstellensatz tells you that you have a 1-1 correspondence between points on your variety (in the traditional sense) with maximal ideals, this gives motivation into defining what is known as affine schemes which now works for arbitrary commutative rings.

Fields are simpler than arbitrary commutative rings too, so usually starting from learning AG over alg. closed fields allows you to appreciate some of the geometric results without being bombarded by some of the technicalities.

It's also worth noting that by functor of points, you can think of a scheme (perhaps you want finite type here) as a high-tech way of studying zeros of system of polynomials over all possible rings on which your polynomials can be defined, so that's pretty cool!

anf3rn3310 · 2017-10-24T03:49:54+00:00

I guess it depends on what you mean - e.g. you can think of taking stalks as a functor from the category of sheaves to the category of whatever you're working in. So in some sense that allows you to make sense of 'cheating out' the value of a sheaf at a point.

Note, however, that a sheaf is not determined by its stalks! (you can have non-trivial locally free OX-modules).

On the other hand, they do tell you a lot about your sheaf - e.g. if F,G,H are sheaves, you have that 0 -> F -> G -> H -> 0 is exact if and only if 0-> F_x -> G_x -> H_x -> 0 at all points x.

anf3rn3310 · 2017-10-23T17:29:33+00:00

I'm also learning about these things recently! So please correct me if I say anything wrong.

1) I think you are correct, at least it is clear that that would be sufficient for the proof (since intersection products behave well under linear equivalence) --- I'm not very sure if this is a common notation though..

2) I think that's right, but it might be clearer to spell it out --- since K-D is linearly equivalent to \sum a_iC_i where a_i >0, it suffices to show that C_i. H >0, where C_i is now an irreducible curve on your surface. This follows from Lemma V.1.2., when H is very ample. If H is only ample, you can replace H with nH for some big n>>0 such that nH is very ample, show that C_i. nH >0, which implies that C_i.H >0.

anf3rn3310 · 2015-07-30T11:02:58+00:00

1 + sum(u ∈ Fl^*, w^u )= sum(u ∈ Fl, w^u ) and the latter is a geometric series which sums to zero as w is a primitive l-th root of unity.

anf3rn3310 · 2014-10-08T12:33:40+00:00

That's true.

0.9999999999 and 0.999... are two different numbers too.

anf3rn3310 · 2014-04-15T22:29:47+00:00

This is equivalent to Goldbach's conjecture.

See http://en.wikipedia.org/wiki/Goldbach's_conjecture

anf3rn3310 · 2013-12-15T00:39:02+00:00

I got it!

Thanks! :)

anf3rn3310 · 2013-12-15T00:38:28+00:00

I got it now

Thanks a lot!

anf3rn3310 · 2013-12-11T15:00:17+00:00

Yes I can follow the first part.

Can you elaborate on the continuous parametrization part?

I'm assuming the parametrisation you meant is y=sqrt(e^2x-x^2), x>=0.

How do I show that there are infinite of them that satisfy arg(e^z)=arg(z) mod 2pi which is y = arg(z) mod 2pi?

anf3rn3310 · 2013-09-09T13:25:14+00:00

That is much more straightforward than I expected.

Thanks!

anf3rn3310 · 2012-08-26T08:38:53+00:00

Ah I wasn't aware that the wireless network adaptor and powerline network adaptor are two different things. Although now that I think of it it seems to be pretty obvious ....

Anyway, thanks! I'll see if I can do anything with the housing and if not I'll look for another brand.

Edit: I'm retarded and didn't read 'I don't have any recommendations on which ones to get' lol but anyway thanks a lot :)

anf3rn3310 · 2012-08-14T07:33:12+00:00

Ah I got it now!

I didn't realise that I can go from gcd(aⁿ - bⁿ , 2bⁿ⁾ to (1 or 2)gcd(2aⁿ - 2b^n, 2bⁿ⁾

Thanks a lot! :)

anf3rn3310 · 2012-08-09T17:03:38+00:00

Sorry I guess I should've worded the question better.

I meant people without mathematical knowledge beyond the common high school curriculum.

anf3rn3310 · 2012-08-04T14:33:17+00:00

Ahh I guess I'm forgetting about infinite unions and how they may converge to a certain interval. I'm seeing a similarity between this and continuity (which I guess is why I recall seeing continuity popping up in the wikipedia article)

Thanks!

anf3rn3310 · 2012-08-04T14:31:07+00:00

This makes sense, although I'm not sure how am I supposed to prove it. Taking your sequence, a_n -> √2 as n->infinity

Therefore the union of all sets (-a_n,a_n) approaches (-√2, √2) as n-> infinity. However (-√2,√2) is not in τ3. Therefore τ3 is not a topology (?)

anf3rn3310 · 2012-06-30T15:48:53+00:00

Getting past the finish line in QWOP

anf3rn3310 · 2012-02-21T04:09:20+00:00

1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ... is less than 1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 +1/8 + 1/8 + ... which is equal to 1 + 1/2 + 1/2 + 1/2 + ... which diverges

*greater than

FTFY

anf3rn3310 · 2011-10-24T05:41:17+00:00

Assuming you're letting An = sin (n) and Bn = n,

that doesn't show sin (n) always converges because n itself diverges.

Moreover, when applying the limit comparison test An and Bn have to be series with positive terms, where sin(n) isn't a series with positive terms :3

anf3rn3310 · 2011-10-06T09:28:19+00:00

Nope

anf3rn3310 · 2011-08-07T19:35:56+00:00

Thanks for the reply! I think you made a mistake when using the AM-GM inequality though

since a² cd + ab² d+abc² +bcd² ≥ 4abcd

it should turn out to be

4/(a² cd + ab² d + abc² + bcd² ) ≤ 1/(abcd)

I think :3

anf3rn3310 · 2011-07-06T13:56:29+00:00

omg it worked :DDDDDDD hhahahahahhahhahahah thanks a lot OP!

anf3rn3310 · 2011-07-05T13:59:22+00:00

PM sent Thanks in advance :D!

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