206

I would have preferred not to comment seriously on Mochizuki's work before much more thought had gone into the very basics, but judging from the internet activity, there appears to be much interest in this subject, especially from young people. It would obviously be very nice if they were to engage with this circle of ideas, regardless of the eventual ...

175

I'll take a stab at answering this controversial question in a way that might satisfy the OP and benefit the mathematical community. I also want to give some opinions that contrast with or at least complement grp. Like others, I must give the caveats: I do not understand Mochizuki's claimed proof, his other work, and I make no claims about the veracity of ...

107

Wrong!
Here is Bourbaki document on algebraic geometry, taken from the now available Master's Archives: click on Autres rédactions, then on Chap.I Théorie globale élémentaire (91 p.)
This preliminary draft was apparently written (according to a penciled annotation on the first page) for Bourbaki in 1954 by Samuel, a distinguished algebraic geometer and ...

answered Mar 20 '15 at 9:06

Georges Elencwajg

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86

Last revision: 10/20. (Probably the last for at least some time to come: until Mochizuki uploads his revisions of IUTT-III and IUTT-IV. My apology for the multiple revisions. )
Completely rewritten. (9/26)
It seems indeed that nothing like Theorem 1.10 from Mochizuki's IUTT-IV could hold.
Here is an infinite set of counterexamples, assuming for ...

81

September 2018: There has been a back-and-forth in 2018 between Shinichi Mochizuki and Yuichiro Hoshi (MoHo) in Kyoto, and Peter Scholze and Jakob Stix (ScSt) in Germany, with ScSt spending a week in Kyoto in March 2018 to confer with MoHo.
ScSt have released a report saying they believe there is a gap in the proof of Corollary 3.12 in IUTT-3, and ...

80

The answer is negative. Suppose for contradiction that $S$ is such a surface, and let me first assume that it is smooth and projective.
Fix $g\geq 24$. Then the coarse moduli space of genus $g$ curves $M_g$ is of general type (this is due to Harris, Mumford and Eisenbud, see for instance [The Kodaira dimension of the moduli space of curves of genus $\geq 23$...

answered Oct 27 '15 at 20:08

Olivier Benoist

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77

1 Easy
Proposition Let $f:X\to Y$ be a continuous map of topological spaces, $\mathscr F$ a sheaf of abelian groups on $X$ such that $R^jf_*\mathscr F=0$ for $j>0$. Then for all $i\geq 0$ there exists a natural isomorphism
$$
H^i(Y, f_*\mathscr F)\simeq H^i(X,\mathscr F)
$$
Proof
Apply the composition rule for the derived functors of $G=\Gamma(Y, \_ )$ ...

67

Classically, Grothendieck's motives are only the pure motives, meaning abelian-ish things which capture the (Weil-cohomology-style) $H^i$ of smooth, projective varieties. To see the relationship with motivic cohomology, one should extend the notion of motive so that non-pure (i.e. "mixed") motives are allowed, these mixed motives being abelian-ish things ...

66

I propose the following plan, assuming a basic background in scheme theory and algebraic topology. I assume that you are interested in derived algebraic geometry from the point of view of applications in algebraic geometry. (If you are interested in applications to topology, you should replace part 2) of the plan by Lurie's Higher algebra.) The plan is ...

66

I have it. Mazur gave me a xerox copy off his shelf when I asked him (in grad school) if a copy exists. It's 56 pages and the first sentence is: L'objet de ce rapport est de construire la série L p-adique de Kubota-Leopold et d'établir quelques propriétés fondamentales. It was in my office and I was going to try scanning it this evening, but literally as ...

65

There is no such bijection.
To see this, imagine four circles all tangent to some line at some point $p$, but all of different radii, so that any two of them intersect only at the point $p$. (E.g., any four circles from this picture.) Under your hypothetical bijection, these four circles would map to four squares, any two of which have exactly one point in ...

57

First, recall the step's of Weil's proof, other than defining the surface:
Develop an intersection theory of curves on surfaces.
Show that the intersection of two specially chosen curves is equal to a coefficient of the zeta function.
Prove the Hodge index theorem using the Riemann-Roch theorem for surfaces (or is there another proof?).
By playing around ...

57

I'm a topologist, and so this answer is going to specifically be about analogies with ordinary topology. I like to think of different Grothendieck topologies as corresponding to different "allowable" ways to build up a space using a quotient topology.
The Zariski topology is like recognizing that a space $X$ can be built up from a collection of open subsets ...

answered Jan 29 '15 at 15:01

Tyler Lawson

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55

This is basically the same as roy smith's excellent comment, but I'd like to put a slightly different spin on it.
A normal variety is a variety that has no undue gluing of subvarieties or tangent spaces.
Let me explain what I mean by gluing.
Given a variety $X$, a closed sub-scheme $Y \subseteq X$ and a finite (even surjective) map $Y \to Z$, you can glue $...

54

The word modulus (moduli in plural, cf. radius and radii, focus and foci, locus and loci) comes from Latin as a word meaning "small measure" or "unit of measure". This is why the absolute value of a complex number z is sometimes called the modulus of z and why the word is used in physics for Young's modulus. In 1800 Gauss introduced the congruence relation $...

ag.algebraic-geometry nt.number-theory at.algebraic-topology ct.category-theory topological-modular-form

52

The answer to this quite beautiful question is that there does exist a commutative ring $R$ with $R\cong R[X,Y]$ but $R\not\cong R[X]$.
Let $F$ be a field, and take
$$
R=F[x_i,y_i,r_i\ (i\geq 0)]
$$
subject to the relations
$$
\forall\ i\geq 0,\ r_i=x_i y_i(x_i+y_i^2)(x_i+y_i^3)(x_i+y_{i+1}^4)r_{i+1}.
$$
First, note that the relations allow us to remove $...

48

Let me describe a common generalization of Nakayama's lemmas and Burnside's basis theorem which may shed some light here. Let $G$ be a group and $P$ a set of endomorphisms of $G$. A $P$-subgroup will be a subgroup of $G$ which is closed under acting by elements of $P$. We'll call $\mathbb{Fr}_P(G)$ the "$P$-Frattini subgroup of $G$", defined as the ...

47

What I think of as the standard proof also gives a pretty clear picture. Let $K$ be a number field with $\rho_1$, $\rho_2$, ..., $\rho_r: K \to \mathbb{R}$ the real places and $\sigma_1$, $\sigma_2$, ..., $\sigma_s: K \to \mathbb{C}$ representatives with the complex places up to conjugacy. Define $\lambda: \mathcal{O}_K \to \mathbb{R}^{r+s}$ by
$$\lambda(x) ...

answered Aug 2 '18 at 21:28

David E Speyer

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47

NB: This answer is directed to the questions about the real case, not the complex case, which was already treated by Francesco.
In some sense, the reason you are running into these 'problems' is that you are working with the ring of real polynomials rather than the ring of real-valued analytic functions, which is also a UFD. For example, it is not hard to ...

answered May 30 '12 at 17:46

Robert Bryant

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45

Let me wrote a quick introduction to this idea:
1) Locales
I do not know if you are already familiar with the notion of locale that Andrej is referring to in his talk: They are a small variation on the idea of a topological space, where instead of defining a space by giving a set of points together with a collection of "open subsets" stable under arbitrary ...

43

[The answer below is a response to an earlier version of the question that was rather different in certain respects. Minhyong Kim's answer gives excellent insight into ideas that Mochizuki had back in 2000 and that provide essential building blocks for the more recent work. But I still believe that it is too premature for a non-expert to seek insight into ...

43

The answer is in fact no.
A complex variety $X$ can never be a differentiable manifold (not even of class $C^1$) throughout a neighborhood of a singular point.
You can find a proof in Milnor's book "Singular Points of Complex Hypersurfaces", Annals of Mathematics Studies 61, remark at page 13.
Notice that $X$ can be a topological manifold (i.e., a ...

answered May 30 '12 at 12:31

Francesco Polizzi

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41

A very good expository article (in Farsi) on recent work of Maryam Mirzakhani can be found here. (PDF)

41

First of all, let me recommend a book: J. Hubbard, Teichmüller theory, vol. 1.
Let me try to list briefly Teichmüller's own contribution to Teichmüller theory.
Bers's papers of 1960-s are good primary sources. The few papers of Teichmüller himself
that I read are also exciting, but my poor knowledge of German does not allow me to
read all of them.
Perhaps ...

answered Nov 28 '12 at 17:40

Alexandre Eremenko

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41

One usually considers the analogue of Morse theory in algebraic geometry to be the theory of vanishing cycles and Lefschetz pencils.
Because of the nature of algebraic functions, Morse theory must be a little more complicated. A Morse function on a compact manifold lets us build the manifold up step by step, starting with a local minimum from which the ...

40

If you forget about all the layers of abstraction, algebraic geometry is, ultimately (and very roughly speaking), the study of polynomial equations in several variables, and of the geometric objects they define. So in a certain sense, whenever you're doing anything with multivariate polynomials, there's probably some algebraic geometry behind it; and ...

40

Let me expand on Yosemite Sam's comment. Pullbacks are indeed easier to define if you view a sheaf as a local homeomorphism. On the other hand, pushforwards are easier to define if you view a sheaf as a set-valued functor.
Suppose we have a continuous map $f: X \to Y$ of topological spaces.
Given a sheaf $F$ on $Y$, viewed as a local homeomorphism $\...

40

The easiest way (I know) to see that there are no nonconstant holomorphic maps from a complete elliptic curve $E$ to the stack $M_g$ is to observe that such a map $f$ would lift to a holomorphic map of the universal covers $\tilde{f}: {\mathbb C} \to T_g$, where $T_g$ is the Teichmuller space. The latter is a bounded domain in ${\mathbb C}^{3g-3}$, so ...

40

I don't know the history (in particular what role primary decomposition held that is now "obsolete"), but I do know a couple places where primary decomposition appears in various other guises.
I realize this isn't exactly an answer to the question, but perhaps it might illuminate some of the relationship between primary decomposition and localization, and ...

40

Not an answer, but way too long for a comment:
According to Ilyashenko ("Centennial history of Hilbert's 16th problem," http://www.ams.org/journals/bull/2002-39-03/S0273-0979-02-00946-1/S0273-0979-02-00946-1.pdf), the claimed result of Petrovski and Landis was disproved by Ilyashenko and Novikov (pg. 303). A citation to this disproof is not given, ...

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