User anton geraschenko - MathOverflow

What are some reasonable-sounding statements that are independent of ZFC?

2009-10-22T19:26:48Z

Every now and then, somebody will tell me about a question. When I start thinking about it, they say, "actually, it's undecidable in ZFC."

For example, suppose A is an abelian group such that every short exact sequence of abelian groups 0→ℤ→B→A→0 splits. Does it follow that A is free? This is known as Whitehead's Problem, and it's undecidable in ZFC.

What are some other statements that aren't directly set-theoretic, and you'd think that playing with them for a week would produce a proof or counterexample, but they turn out to be undecidable? One answer per post, please, and include a reference if possible.

When is fiber dimension upper semi-continuous?

2009-10-08T05:09:36Z

Suppose f:X→Y is a morphism of schemes. We can define a function on the underlying topological space Y by sending y∈Y to the dimension of the fiber of f over y. When is this function upper semi-continuous?

I have a "concrete" application in mind. If an algebraic group G acts on a scheme X, I'm pretty sure the dimensions of the stabilizers is upper semi-continuous (i.e. it can jump up on closed subschemes), but I don't know a proof. The stabilizers of points are the fibers of the map Stab→X in the following cartesean square:

Stab ---> GxX
  |         |
  |  cart   |
  v         v
  X ----> XxX

The map GxX→XxX is given by (g,x)→(gx,x), and the map X→XxX is the diagonal x→(x,x). So it would be nice to have a condition satisfied by GxX→XxX which guarantees upper semi-continuity of the fiber dimension.

Examples of great mathematical writing

2009-10-12T17:12:16Z

This question is basically from Ravi Vakil's web page, but modified for Math Overflow.

How do I write mathematics well? Learning by example is more helpful than being told what to do, so let's try to name as many examples of "great writing" as possible. Asking for "the best article you've read" isn't reasonable or helpful. Instead, ask yourself the question "what is a great article?", and implicitly, "what makes it great?"

If you think of a piece of mathematical writing you think is "great", check if it's already on the list. If it is, vote it up. If not, add it, with an explanation of why you think it's great. This question is "Community Wiki", which means that the question (and all answers) generate no reputation for the person who posted it. It also means that once you have 100 reputation, you can edit the posts (e.g. add a blurb that doesn't fit in a comment about why a piece of writing is great). Remember that each answer should be about a single piece of "great writing", and please restrict yourself to posting one answer per day.

I refuse to give criteria for greatness; that's your job. But please don't propose writing that has a major flaw unless it is outweighed by some other truly outstanding qualities. In particular, "great writing" is not the same as "proof of a great theorem". You are not allowed to recommend anything by yourself, because you're such a great writer that it just wouldn't be fair.

Not acceptable reasons:

This paper is really very good.
This book is the only book covering this material in a reasonable way.
This is the best article on this subject.

Acceptable reasons:

This paper changed my life.
This book inspired me to become a topologist. (Ideally in this case it should be a book in topology, not in real analysis...)
Anyone in my field who hasn't read this paper has led an impoverished existence.
I wish someone had told me about this paper when I was younger.

What do epimorphisms of (commutative) rings look like?

2009-10-05T05:33:42Z

(Background: In any category, an epimorphism is a morphism $f:X\to Y$ which is "surjective" in the following sense: for any two morphisms $g,h:Y\to Z$, if $g\circ f=h\circ f$, then $g=h$. Roughly, "any two functions on $Y$ that agree on the image of $X$ must agree." Even in categories where you have underlying sets, epimorphisms are not the same as surjections; for example, in the category of Hausdorff topological spaces, $f$ is an epimorphism if its image is dense.)

What do epimorphisms of (say commutative) rings look like? It's easy to verify that for any ideal $I$ in a ring $A$, the quotient map $A\to A/I$ is an epimorphism. It's also not hard to see that if $S\subset A$ is a multiplicative subset, then the localization $A\to S^{-1}A$ is an epimorphism. Here's a proof to whet your appetite.

If $g,h:S^{-1}A\to B$ are two homomorphisms that agree on $A$, then for any element $s^{-1}a\in S^{-1}A$, we have
$$g(s^{-1}a)=g(s)^{-1}g(a)=h(s)^{-1}h(a)=h(s^{-1}a)$$

Also, if $A\to B_i$ is a finite collection of epimorphisms, where the $B_i$ have disjoint support as $A$-modules, then $A\to\prod B_i$ is an epimorphism.

Is every epimorphism of rings some product of combinations of quotients and localizations? To put it another way, suppose $f: A\to B$ is an epimorphism of rings with no kernel which sends non-units to non-units and such that $B$ has no idempotents. Must $f$ be an isomorphism?

Does smoothness descend along flat morphisms?

2012-02-22T01:06:16Z

Suppose $f:X\to Y$ is a flat morphism of schemes. If $X$ is smooth at $x$, must $Y$ be smooth at $f(x)$?

If $f$ is locally finitely presented, then it is open (using EGA IV 1.10.4), so after replacing $Y$ by $f(X)$, we can assume $f$ is faithfully flat. I'd be happy to understand even the case where $X$ and $Y$ are local:

Suppose $R$ and $S$ are local rings and $R\to S$ is a local homomorphism with $S$ (faithfully) flat over $R$. If $S$ is regular, must $R$ be regular?

Note that I'm not asking if smoothness is "flat local"; there are certainly flat morphisms from singular things to smooth things (e.g. $k[x,y]/(x^2-y^2)$ is flat over $k[x]$). The question is whether there are flat morphisms from smooth schemes which hit singular points.

Is every functor a composition of adjoint functors?

2009-10-12T03:51:02Z

My understanding of Ben's answer to this question is that even though associated graded is not an adjoint functor, it's not too bad because it is a composition of a right adjoint and a left adjoint.

But are such functors really "not that bad"? In particular, is it true that any functor be written as the composition of a right adjoint and a left adjoint?

Answer by Anton Geraschenko for A Nisnevich cover which is not Zariski

2012-07-27T05:24:19Z

My standard example is an $n$-gon of $\mathbb P^1$'s covering the nodal cubic. For some reason, I especially like the case $n=2$.

Here's the affine version, which always takes me a bit to work out. It's two parabolas joined at two points covering the nodal cubic: $$ \def\spec{\mathrm{Spec\,}} \spec k[s,t]/(t^2-(s^2-1)^2) \to \spec k[x,y]/(y^2-x^2(x+1)) $$ given by $x\mapsto (s^2-1)$ and $y\mapsto st$.

This map is étale. To see it is Nisnevich, you have to check that the residue field extensions of the generic point are isomorphisms (this is where the squaring map on $\mathbb A^1-0$ fails to be Nisnevich). You can see that by restricting to the two components, giving the normalization maps $$ k[x,y]/(y^2-x^2(x+1)) \to k[s,t]/(t-(s^2-1)) \cong k[s] $$ given by $(x,y)\mapsto (t,st)$ (so $s=y/x$), and $$ k[x,y]/(y^2-x^2(x+1)) \to k[s,t]/(t+(s^2-1)) \cong k[s] $$ given by $(x,y)\mapsto (-t,st)$ (so $s=-y/x$).

Can a non-trivial action of a connected group on a reduced scheme be trivial on a dense open?

2012-07-04T19:06:01Z

It is well-known that if a reduced algebraic group $G$ acts on a separated reduced scheme $X$, and $G$ acts trivially on a dense open subscheme $U\subseteq X$, then the action is trivial.

If $X$ is non-reduced, the standard counterexample is $\def\AA{\mathbb A}$ $\AA^1$ with an embedded point, with the group acting non-trivially on the embedded point. If $X$ is non-separated, the standard counterexample is $\AA^1$ with a doubled origin with $G$ swapping the two origins. My question is whether the separated hypothesis can be removed if $G$ is assumed to be connected.

Suppose $G$ is a connected group scheme acting on a reduced scheme $X$, and that $G$ acts trivially on a dense open subscheme $U\subseteq X$. Must the action be trivial?

For reference, the basic argument for the original version is that the graphs of the two morphisms $G\times X\to X$ (projection and action) are closed subschemes of $G\times X\times X$ (since the graphs are pullbacks of the diagonal of $X$, which is assumed separated) which share a common dense open (the image of $G\times U$). Since $G\times X\times X$ is reduced, this means that the two graphs agree, so the action and projection maps are the same.

^†~~For finite-characteristic people, I actually assume $G$ is reduced, but this is fine since the action on a reduced scheme must factor through the reduction of $G$ anyway.~~

Can continuity of a function be checked by restricting to smooth curves?

2012-06-08T01:06:01Z

Well-known example: Consider the function $$f(x,y)=\left\{\begin{array}{c} \frac{x^2y}{x^4+y^2} & \text{if }(x,y)\neq(0,0)\\ 0 & \text{if }(x,y)=(0,0) \end{array}\right.$$ When restricted to any straight line through the origin, this function is continuous. However, if we approach the origin along the parabola $y=x^2$, we get a limit of $\frac 12$, so $f$ is actually discontinuous. The question is whether smooth curves can always ferret out discontinuity in this way.

Does there exist a function $f:\mathbb R^2\to \mathbb R$ which is discontinuous at a point $x$, but is continuous at $x$ when restricted to any smooth curve?

Discontinuity is witnessed by a sequence $\{x_i\}$ converging to $x$ so that $\{f(x_i)\}$ does not converge to $f(x)$. Since we could take $f$ to be the characteristic function on $\{x_i\}$ , the above question is equivalent to the following.

Given a convergent sequence $\{x_i\}$ in $\mathbb R^2$, must there be a smooth curve passing through infinitely many of the $x_i$?

Can a vector space over an infinite field be a finite union of proper subspaces?

2009-09-29T14:22:21Z

Can a (possibly infinite-dimensional) vector space ever be a finite union of proper subspaces?

If the ground field is finite, then any finite-dimensional vector space is finite as a set, so there are a finite number of 1-dimensional subspaces, and it is the union of those. So let's assume the ground field is infinite.

Answer by Anton Geraschenko for A left adjoint for the evaluation functor \Gamma(, U)

2012-03-13T21:42:15Z

It's not hard to give an explicit description of the left adjoint to $\def\O{\mathcal O}\def\mod{\textrm{-mod}}\Gamma(U,-):\O_X\mod\to\O_X(U)\mod$ by stringing together left adjoints. Perhaps a slight modification will do what you want.

For $j:U\hookrightarrow X$ an open immersion, the restriction functor $j^*:\O_X\mod\to \O_U\mod$ has a left adjoint $j_!$. The pushforward functor $f_*:\O_U\mod\to \O_{Spec(\O_U(U))}\mod$ has the left adjoint $f^*$. The functor $\Gamma(Spec A,-):\O_{Spec(A)}\mod\to A\mod$ has the left adjoint taking an $A$-module $M$ to the quasi-coherent sheaf $\widetilde M$. Stringing these together, we get that the functor sending an $\O_X(U)$-module $M$ to $j_!(f^*\widetilde M)$ is left adjoint to $\Gamma(U,-):\O_X\mod\to \O_X(U)\mod$.

I'm not sure how to modify $j_!$ to get a left adjoint to $j^*:QCoh(X)\to QCoh(U)$. If there is such a thing, it finishes the job.

Is there always a toric isomorphism between isomorphic toric varieties?

2011-06-22T23:36:06Z

Suppose two toric varieties are isomorphic as abstract varieties. Does it follow that there exists a toric isomorphism between them?

Edit: the comments below lead me to believe that I'm using the terms "toric variety" and "toric morphism" in a non-standard way, so let me clarify. Here are the definitions I have in mind.

Definition: A toric variety is a normal variety $X$ together with (1) a dense open subvariety $T\subseteq X$, and (2) a group structure on $T$ making it a torus, such that the action of $T$ on itself extends to an action on $X$. (Note: I do not include the data of an isomorphism between $T$ and $\mathbb G_m^{\dim X}$, but only require that such an isomorphism exists.)

Definition: If $(X,T_X)$ and $(Y,T_Y)$ are toric varieties (group structures on $T_X$ and $T_Y$ implicit), a toric morphism between them is a morphism $f:X\to Y$ which restricts to a group homomorphism $f|_{T_X}:T_X\to T_Y$.

Interesting applications of the Pigeon-hole Principle

2009-11-05T18:13:11Z

I'm a little late in realizing it, but today is Pigeon-hole Day. Festivities include thinking about awesome applications of the Pigeon-hole Principle. So let's come up with some. As always with these kinds of questions, please only post one answer per post so that it's easy for people to vote on them.

Allow me to start with an example:

Brouwer's fixed point theorem can be proved with the Pigeon-hole Principle via Sperner's lemma. There's a proof in Proofs from The Book (unfortunately, the google books preview is missing page 148)

By the way, if you happen to be in Evans at Berkeley today, come play musical chairs at tea!

Answer by Anton Geraschenko for Quotients of Schemes by Free Group Actions

2009-10-21T03:29:21Z

Hironaka's example of a proper non-projective 3-fold has a $\mathbb Z/2$ action which is free (actually, there's one fixed point, but you can just throw it out) whose quotient is not a scheme. (see an appendix of Hartshorne (page 443) or Shafarevich (page 75))

In general, the quotient of a scheme by a free action of a group is an algebraic space. If this group is finite, this is just because the action induces an etale equivalence relation (this is where you use freeness), and any quotient of a scheme by an etale equivalence relation is an algebraic space. For infinite groups, the quotient will be an algebraic stack fibered in sets, so it will be an algebraic space (c.f. this question).

You are then reduced to the question, when is an algebraic space a scheme?

Here's one situation where you get a scheme quotient. If you have a quasi-projective variety $X$ with an action of a connected reductive group $G$, then there is a geometric quotient $X/G$ if there exists a line bundle $L$ (with $G$-action compatible with the $G$-action on $X$) such that every point of $X$ is stable with respect to $L$. That is, if for every point $x\in X$, there is an invariant section $s$ of some tensor power of $L$ such that $X_s$ (the non-vanishing locus of $s$) is an open affine neighborhood of $x$ in which every $G$-orbit is closed. This is Theorem 1.10 in Geometric Invariant Theory. I don't know if knowing that the action of $G$ is free helps in finding such an $L$.

Answer by Anton Geraschenko for 'Ampleness' of a big line bundle

2011-12-02T14:05:35Z

Here is a simple counterexample (of the form Zhengyu Hu suggested). Take $X$ to be the blowup of a point in $\mathbb P^2$, $M$ the pullback of $\mathcal O(1)$ under the blowup map, and $F$ the line bundle associated to the exceptional divisor $E$. Sections of $F\otimes M^n$ are rational functions which may have a pole of order 1 along $E$ and a pole of order up to $n$ along a line not meeting $E$. However, any rational function $f$ actually having a pole along $E$ (i.e. generating $F\otimes M^n$ at the points of $E$) must have a pole along some other divisor passing through $E$ (since $X$ has the same rational functions as $\mathbb P^2$).

Answer by Anton Geraschenko for Best Algebraic Geometry text book? (other than Hartshorne)

2009-10-26T00:38:17Z

Perhaps this is cliché, but I recommend EGA (links to full texts: I, II, III(1), III(2), IV(1), IV(2), IV(3), IV(4)).

I know it's a scary 1800 pages of French, but

It's really easy French. I would describe myself as not knowing any French, but I can read EGA without too much trouble.
It's extremely clear. The proofs are usually very short because the results are very well organized.
It's the canonical reference for algebraic geometry. I assure you it is not 1800 pages of fluff.

I've found it quite rewarding to to familiarize myself with the contents of EGA. Many algebraic geometry students are able to say with confidence "that's one of the exercises in Hartshorne, chapter II, section 4." It's even more empowering to have that kind of command over a text like EGA, which covers much more material with fewer unnecessary hypotheses and with greater clarity. I've found this combined table of contents to be useful in this quest. [Edit: The combined table of contents unfortunately seems to be defunct. Here is a web version of Mark Haiman's EGA contents handout.]

Answer by Anton Geraschenko for Clean Proofs of Properties of Projective Space

2011-10-18T06:34:47Z

I think the answer is "probably not." The reason is that projective space has two universal properties which are used to prove different kinds of things about it. One of these is the slick universal property you like, and the other is the clunky one which results in unpleasantries.

Though each universal property implies the other (since it uniquely identifies projective space), it seems unlikely to me that you can effectively do anything if you try to avoid one of them altogether.

One universal property makes it easy to understand maps to projective space:

$$ Hom(T,\mathbb P^n) = \{\mathcal O_T^{n+1}\twoheadrightarrow \mathcal L| \mathcal L\text{ a line bundle}\} $$

Without bending over backwards (i.e.~reproducing the usual theory), I'd be surprised if you could use this universal property to even prove that there are no non-constant regular functions on $\mathbb P^n$.

I expect constructions that naturally pull back along morphisms (e.g. line bundles, regular functions) to behave like morphisms from projective space, so it would be strange if you could attack such constructions with this universal property.

Another universal property makes it easy to understand maps from projective space: $Hom(\mathbb P^n,T)$ is the equalizer of the two restriction maps $Hom(\coprod_{i=0}^n \mathbb A^n,T)\rightrightarrows Hom(\coprod_{i,j}\mathbb A^{n-1}\times (\mathbb A-0),T)$.

I guess this is the one that you don't like, but we're lucky to have it since it actually makes it possible to make sense of projective space having Zariski local properties (e.g. being smooth, $n$-dimensional, etc.), and thereby makes it possible to do geometry on it.

Are "fpqc algebraic spaces" algebraic spaces?

2011-10-09T00:22:37Z

Suppose $F:Sch^\text{op}\to Set$ is a sheaf in the fpqc topology, has quasi-compact representable diagonal, and has an fpqc cover by a scheme. Must $F$ be an algebraic space? That is, must $F$ have an étale cover by a scheme?

Note that it is enough to find an fppf cover of $F$ by a scheme, at which point you know that $F$ is an algebraic stack ("Artin's slice theorem", 10.1 of Laumon-Moret-Bailly) fibered in sets, so it's an algebraic space.

The analogous question, "is every fpqc stack an algebraic stack?" has a negative answer, but the counterexample is "purely stacky".

Wanted: example of a non-algebraic singularity

2011-08-29T06:36:17Z

Given a finitely generated $\def\CC{\mathbb C}\CC$-algebra $R$ and a $\CC$-point (maximal ideal) $p\in Spec(R)$, I define the singularity type of $p\in Spec(R)$ to be the isomorphism class of the completed local ring $\hat R_p$, as a $\CC$-algebra.

Do there exist non-algebraic singularity types? That is, does there exist a complete local ring with residue field $\CC$ which is formally finitely generated (i.e. has a surjection from some $\CC[[x_1,\dots, x_n]]$), but is not the complete local ring of a finitely generated $\CC$-algebra at a maximal ideal?

Googling for "non-algebraic singularity" suggests that the answer is yes, but I can't find a specific example. I would expect that it should be possible to write down a power series in two variables $f(x,y)$ so that $\CC[[x,y]]/f(x,y)$ is non-algebraic.

What is a specific formally finitely generated non-algebraic singularity?

Can cones (toric monoids) be built as colimits of their faces?

2011-04-13T21:23:02Z

Suppose $L$ is a lattice (free abelian group) and $\sigma$ is a (pointed) spanning rational cone in $L\otimes\mathbb Q$. Then $M=L\cap \sigma$ is a monoid with $M^{gp}=L$. A monoid of this form is called a toric monoid. Toric monoids are are precisely the finitely generated, commutative, integral (cancellative), sharp (unit-free), saturated monoids. Let $TMon$ be the category of toric monoids.

A face of a monoid $M$ is a submonoid $F\subseteq M$ so that $a+b\in F$ implies $a,b\in F$. The faces of a toric monoid $\sigma \cap L$ are exactly submonoids of the form $\tau \cap L$, where $\tau$ is a face of the cone $\sigma$.

Suppose $D$ is a diagram of toric monoids in which every morphism is the inclusion of a face, and so that for every pair of monoids $D_i$ and $D_j$ in the diagram, there is a unique maximal common face $D_i\cap D_j$ in the diagram. Let $M$ be the colimit of the diagram in $TMon$. Are the maps $D_i\to M$ inclusions of faces?

My feeling is that this problem should be straightforward, but I've done a good job getting myself confused.

Problem 1: What is $M$? Colimits exist in the category of finitely generated commutative monoids, but they're unwieldy. See, for example, the first chapter of William Gillam's notes on log geometry. The inclusion of integral saturated monoids into all commutative monoids has a left adjoint, so the colimit $M$ can be formed by taking the colimit in commutative monoids, and then "integral-and-saturifying" it. In particular, $M$ exists. This description makes it impossible to deal with $M$.

Here is another description which is a bit better, but which I'm still not sure how to handle. Let $L$ be the colimit of the induced diagram of free abelian groups $D^{gp}$. This $L$ is probably free even if you take the limit in the category of abelian groups (rather than free abelian groups). Then $M$ is the image of $\bigoplus D_i\to L$. The problem with this description is that it's hard to keep track of faces once you turn everything into groups.

Problem 2: The fact that every pair of monoids in the diagram has a unique maximal common face in the diagram is necessary, but I'm having trouble making use of it. Here is a counterexample where this condition fails.

Let $\def\N{\mathbb N}M_n\subseteq \N^2$ be the submonoid generated by ${(1,0), (1,1),\dots, (1,n)}$

Let $f_1,g_1:\N\to M_2$ be given by $f_1(1)=(1,0)$ and $g_1(1)=(1,2)$. Let $f_2,g_2:\N\to M_3$ be given by $f_2(1)=(1,0)$ and $g_2(1)=(1,3)$. Consider the diagram

$$\begin{array}{ccc} & \N & \\ {}^{f_1}\swarrow & & \searrow^{f_2}\\ M_2 & & M_3\\ {}_{g_1}\nwarrow & & \nearrow_{g_2}\\ & \N \end{array}$$

In this case, I'm pretty sure the colimit is $M_6$ with the maps $\begin{pmatrix}1&0\\ 0&3\end{pmatrix}:M_2\to M_6$ and $\begin{pmatrix}1&0\\ 0&2\end{pmatrix}:M_3\to M_6$. These maps are not inclusions of faces.

Answer by Anton Geraschenko for Can cones (toric monoids) be built as colimits of their faces?

2011-09-20T06:03:24Z

Yes, assuming that $D$ includes all faces (i.e. if $D_i$ is in $D$, the so are all of its faces). This is Corollary 2.12 of Toric Stacks II: Intrinsic Characterization of Toric Stacks.

The basic solution to both problems is to note that a toric monoid $M$ is determined by its dual $M^\vee = \{\chi\in Hom(M^{gp},\mathbb Z)| \chi$ is non-negative on $M\}$, and that the faces of $M$ are precisely the vanishing sets of elements of $M^\vee$. The advantage of this observation is that by the universal property of the colimit, functions on the colimit are equivalent to compatible systems of functions on the diagram.

Given an element $D_i$ of the diagram, one then inductively shows that

It is possible to extend any linear function on $D_i$ to a compatible system of linear functions on $D$. This shows that the restriction map $Hom(colim(D)^{gp},\mathbb Z)\to Hom(D_i^{gp},\mathbb Z)$ is surjective, from which it follows that $D_i\to colim(D)$ is injective.
The zero function on $D_i$ can be extended to a compatible system of functions on $D$ which are strictly positive away from $D_i$ and its faces. This shows that $D_i$ is in fact a face of $colim(D)$.

Answer by Anton Geraschenko for Sheaf cohomology and injective resolutions

2009-10-19T05:49:51Z

Since everybody else is throwing derived categories at you, let me take another approach and give a more lowbrow explanation of how you might have come up with the idea of using injectives. I'll take for granted that you want to associate to each object (sheaf) $F$ a bunch of abelian groups $H^i(F)$ with $H^0(F)=\Gamma(F)$, and that you want a short exact sequence of objects to yield a long exact sequence in cohomology.

I also want one more assumption, which I hope you find reasonable: if $F$ is an object such that for any short exact sequence $0\to F\to G\to H\to 0$ the sequence $0\to \Gamma(F)\to \Gamma(G)\to \Gamma(H)\to 0$ is exact, then $H^{i}(F)=0$ for $i>0$. This roughly says that $H^{i}$ is zero unless it's forced to be non-zero by a long exact sequence (you might be able to run this argument only using this for $i=1$, but I'm not sure). Note that this implies that injective objects have trivial $H^{i}$ since any short exact sequence with $F$ injective splits.

Now suppose I come across an object $F$ that I'd like to compute the cohomology of. I already know that $H^{0}(F)=\Gamma(F)$, but how can I compute any higher cohomology groups? I can embed $F$ into an injective object $I^{0}$, giving me the exact sequence $0\to F\to I^{0}\to K^{1}\to 0$. The long exact sequence in cohomology gives me the exact sequence $$0\to \Gamma(F)\to \Gamma(I^{0})\to \Gamma(K^{1})\to H^{1}(F)\to 0 = H^1(I^{0})$$

That's pretty good; it tells us that $H^{1}(F)= \Gamma(K^{1})/\mathrm{im}(\Gamma(I^{0}))$, so we've computed $H^{1}(F)$ using only global sections of some other sheaves. We'll come back to this, but let's make some other observations first.

The other thing you learn from the long exact sequence associated to the short exact sequence $0\to F\to I^{0}\to K^{1}\to 0$ is that for $i>0$, you have $$H^{i}(I^{0}) = 0\to H^{i}(K^{1})\to H^{i+1}(F)\to 0 = H^{i+1}(I^{0})$$

This is great! It tells you that $H^{i+1}(F)=H^{i}(K^{1})$. So if you've already figured out how to compute $i$-th cohomology groups, you can compute $(i+1)$-th cohomology groups! So we can proceed by induction to calculate all the cohomology groups of $F$.

Concretely, to compute $H^{2}(F)$, you'd have to compute $H^{1}(K^{1})$. How do you do that? You choose an embedding into an injective object $I^{1}$ and consider the long exact sequence associated to the short exact sequence $0\to K^{1}\to I^{1}\to K^{2}\to 0$ and repeat the argument in the third paragraph.

Notice that when you proceed inductively, you construct the injective resolution $$0\to F\to I^{0}\to I^{1}\to I^{2}\to\cdots$$ so that the cokernel of the map $I^{i-1}\to I^{i}$ (which is equal to the kernel of the map $I^{i}\to I^{i+1}$) is $K^{i}$. If you like, you can define $K^{0}=F$. Now by induction you get that $$H^{i}(F) = H^{i-1}(K^{1}) = H^{i-2}(K^{2}) = \cdots = H^{1}(K^{i-1}) = \Gamma(K^{i})/\mathrm{im}(\Gamma(I^{i-1})).$$

Since $\Gamma$ is left exact and the sequence $0\to K^{i}\to I^{i}\to I^{i+1}$ is exact, you have that $\Gamma(K^{i})$ is equal to the kernel of the map $\Gamma(I^{i})\to \Gamma(I^{i+1})$. That is, we've shown that $$H^{i}(F) = \ker[\Gamma(I^{i})\to \Gamma(I^{i+1})]/\mathrm{im}[\Gamma(I^{i-1})\to \Gamma(I^{i})].$$

Whew! That was kind of long, but we've shown that if you make a few reasonable assumptions, some easy observations, and then follow your nose, you come up with injective resolutions as a way to compute cohomology.

Answer by Anton Geraschenko for Is there an example of a formally smooth morphism which is not smooth?

2009-10-08T14:50:17Z

Here's an elementary example. For any field $k$, consider the ring $k[t^q|q\in\mathbb Q_{>0}]$, which I'll abbreviate $k[t^q]$. I claim that the natural quotient $k[t^q]\to k$ given by sending $t^q$ to $0$ is formally smooth but not flat, and therefore not smooth.

First let's show it's formally smooth. Let $A$ be a ring with square-zero ideal $I\subseteq A$, and suppose we have maps $f:k[t^q]\to A$ and $g:k\to A/I$ making the following square commute (I drew it backwards because you're probably thinking of Spec of everything)

$$ \begin{array}{ccc} A/I & \xleftarrow g & k \\ \uparrow & & \uparrow\\ A & \xleftarrow f & k[t^q] \end{array} $$

We'd like to show that there's a map $k\to A$ filling the diagram in. For any $q\in \mathbb Q_{>0}$, note that $f(t^q)\in I$ by commutativity of the square, so $f(t^{2q})\in I^2=0$. But every $q$ is of the form $2q'$ for some $q'$, so we've shown that $f(t^q)=0$ for all $q\in \mathbb Q_{>0}$. So $f$ factors through $k$, as desired.

Now let's show that $k$ is not flat over $k[t^q]$. Consider the exact sequence $$0\to (t)\to k[t^q]\to k[t^q]/(t)\to 0.$$ When you tensor with $k$, you get $$0\to k\to k\to k\to 0,$$ which is obviously not exact. So $k$ is not flat over $k[t^q]$.

Do simplicial toric varieties have "lots" of base point free linear systems?

2011-08-23T18:05:29Z

Question: Let $n$ be a positive integer and let $X$ be a simplicial toric variety. Does every coset of $n\cdot Pic(X)\subseteq Pic(X)$ contain a base point free linear system?

If $X$ is not simplicial, the answer is definitely no, since there exist proper toric varieties with no non-trivial line bundles (for example, Example 4.2.13 of Cox, Little, and Schenck).

Obviously, if $X$ is quasi-projective, every coset of $n\cdot Pic(X)$ contains a very ample (so base point free) line bundle. So the question is only interesting for non-quasi-projective $X$.

I'll rephrase the question in terms of fans. Given a rational ray $\rho$, let $v_\rho$ denote the first lattice point along $\rho$.

Rephrasing: Let $n$ be a positive integer and let $\Sigma$ be a simplicial fan on a lattice $N$. Let $\{a_\rho\}_{\rho\in \Sigma(1)}$ be integers^† associated to the rays of $\Sigma$. Do there exist integers^† $\{b_\rho\}_{\rho\in \Sigma(1)}$ so that for every $\rho$, there is a point $\chi\in N^*$ so that $\chi(v_\rho) + a_\rho+nb_\rho=0$ and $\chi(v_{\rho'})+a_{\rho'}+nb_{\rho'}\ge 0$ for every $\rho'\in \Sigma(1)$?

^† These sets of integers are assumed to be induced by integer-valued piecewise linear functions. That is, they correspond to Cartier divisors.

In other words, given a bunch of integral^† hyperplanes perpendicular to the rays of $\Sigma$, is it possible to move each of them by a multiple of $n$ (^†, sigh) so that each of them touches the polytope they define?

Answer by Anton Geraschenko for Bertini theorems for base-point-free linear systems in positive characteristics

2011-08-23T16:32:04Z

I think Corollary 4.3 of Spreafico's Axiomatic theory for transversality and Bertini type theorems does what you want. It says (in the case where the property is taken to be smoothness) that if $f:X\to \mathbb P^n$ is a finite type morphism from a smooth scheme $X$ over any infinite field, and if $f$ is residually separated (i.e. the induced extensions of residue fields are separable), then the pullback of a generic hyperplane is smooth.

Answer by Anton Geraschenko for Finite, Étale Morphism Of Varieties

2011-08-08T16:00:16Z

No, it's not true. Consider the map $x\mapsto x^2$ as a map from $X=\mathbb A^1-{0}$ to itself. The "problem" is that Zariski neighborhoods are too big. Any open subset of $X$ has exactly one irreducible component (in general, an open subset cannot have more components than the ambient space), so there is no hope to get the preimage of an open set to have two components.

However, if you refine your topology to allow "etale open neighborhoods" (i.e. you allow pullback by etale maps $U\to X$, not just open immersions $U\to X$), then the answer is yes. Perhaps the easiest way to prove that is to pull back by $\pi$ itself, after which you can "peel off" the diagonal component of $Y\times_X Y$. Now you have a finite etale map $(Y\times_X Y -\Delta)\to Y$ which has degree one less. Repeat until the degree is $1$, at which point you have an etale map $U\to \cdots \to Y\to X$ so that $Y\times_X U$ is $deg(\pi)$ copies of $U$.

Answer by Anton Geraschenko for Weighted blow up of a Toric Variety

2011-08-05T04:23:51Z

(Essentially reposting Jesus Martinez Garcia and Karl Schwede's comments as an answer)

If you are blowing up a torus-invariant sheaf of ideals, then the Rees algebra (the thing you take Proj of to get the blow-up) has grading by characters of the torus, so the blowup has an action of the torus, so it is toric (since it also contains a dense copy of the torus that acts on it). Thus, if your weighted blow-up is weighted "along the coordinate hyperplanes", it will be toric.

If you blow up a non-invariant sheaf of ideals, you do not get something toric in general. If you blow up a non-invariant point of $\mathbb A^2$, then the blow-up map already cannot be toric, even though both varieties are toric. Blowing up two different points of $\mathbb A^2$ gives a total space which has no toric structure at all: the two (-1)-curves must be torus-invariant and blowing them down gives a toric map to $\mathbb A^2$, but any toric blow-up of $\mathbb A^2$ must have the exceptional locus lying over a single point.

Answer by Anton Geraschenko for Schemes do not form a stack in the etale topology?

2011-08-05T03:45:46Z

I'm pretty sure this example works. I do not include any proof that this is the simplest example, and it may not be, but it's not too complicated.

Let $L_1$ and $L_2$ be two rational curves in $\def\P{\mathbb P}\P^3$ which intersect in two points. A standard example of a proper non-projective variety $X$ is obtained by blowing up $L_1$ and $L_2$, but doing it in one order at one intersection point and in the other order at the other intersection point (I think this example is explained at the end of Hartshorne).

There is an involution $\sigma$ of $\P^3$ which switches the two lines and the two intersection points. Let $U\subseteq \P^3$ be the open locus where $\sigma$ acts freely, and let $Y=U\times_{\P^3}X$. Then $Y/\sigma$ is an algebraic space (over the scheme $U/\sigma$) which is not a scheme. It becomes a scheme after the etale base change $U\to U/\sigma$.

Answer by Anton Geraschenko for When can a finite map be blown up to a flat one?

2011-08-03T19:12:29Z

Since the blowups are proper and $f$ is proper, the "property P argument" shows that $\tilde f$ is proper. A proper quasi-finite morphism is finite (EGA IV 18.12.4), so $\tilde f$ is finite.

This (more or less) reduces to the case when $f$ is finite to begin with, so no blowups are needed. Then the only condition I know to ensure flatness is the one Dave Anderson cited, [Matsumura's Commutative Ring Theory, Theorem 23.1]: if $Y$ is regular and $X$ is Cohen-Macaulay, then $f$ is flat.

Answer by Anton Geraschenko for Maps to projective space determined by a line bundle

2011-07-18T19:20:23Z

This answer is meant as (1) the stacky answer Ben anticipated, and (2) an answer addressing the generalization to higher rank bundles.

First, the stacky interpretation. A choice of a line bundle $\mathcal L$ on a scheme $X$ is equivalent to a morphism to the stack $\def\GG{\mathbb G} B\GG_m$; the $\GG_m$-torsor is the complement of the zero section in the total space $\mathbb V(\mathcal L)$. A choice of a line bundle together with $n$ sections is equivalent to a morphism $f:X\to\def\AA{\mathbb A} [\AA^n/\GG_m]$; the $\GG_m$ torsor is the complement of the zero section of $\mathcal L$, and the $n$ map to $\AA^n$ is given by the $n$ regular functions which are the pullbacks of the sections. The condition that $f$ misses the one stacky point of $[\AA^n/\GG_m]$ (i.e. the point where the $\GG_m$ doesn't act freely)—and therefore factors through the open subscheme $[(\AA^n\smallsetminus 0)/\GG_m]=\mathbb P^{n-1}$ —is precisely the condition that the sections do not all vanish at the same point.

Now the generalization. A rank $k$ vector bundle $\def\E{\mathcal E} \E$ on a scheme $X$ is equivalent to a morphism to the stack $BGL_k$; the $GL_k$-torsor is the sheaf $\def\O{\mathcal O} Isom(\O^k,\E)$. A vector bundle together with a choice of $n$ sections is equivalent to a morphism $f:X\to [(\AA^k)^n/GL_n]$; again, the $GL_k$-torsor is $Isom(\O^k,\E)$, and the $k\cdot n$ regular functions on the torsor are given by the pullback of the $n$ sections of $\E$, and the fact that the pullback of $\E$ is canonically identified with $\O^k$. Regarding $(\AA^k)^n$ as the space of $k\times n$ matrices, the stacky locus of $[(\AA^k)^n/GL_n]$ (i.e. the points with non-trivial stabilizers) is the locus where the rank of the matrix is less than $k$. So the condition that $f$ misses the stacky locus is equivalent to the condition that the $n$ given sections span the fiber of $\E$ at any point. The non-stacky locus is the open subscheme $[${$k\times n$ matrices of rank $k$}$/GL_k]$, the Grassmannian of $k$-planes in $n$-space, $Gr(k,n)$. We therefore have the following interpretation.

A vector bundle $\E$ on a scheme $X$, together with $n$ sections $s_1,\dots, s_n$ which span every fiber is equivalent to a morphism $X\to Gr(k,n)$.

If you don't want to dirty things up by choosing the sections, you can replace $(\AA^k)^n$ by $Hom(\Gamma(\E),\mathbb C^k)$ everywhere. The non-stacky locus is then the space of surjective linear maps. Then it looking at kernels, it looks like you naturally get $Gr(n-k,n)$ instead of (the isomorphic) $G(k,n)$. I'm sure this has something to do with a dualization involved in forming the total space of a locally free sheaf; it still regularly confuses me, so I'll just leave this thread loose.

Comment by Anton Geraschenko

2013-01-02T14:58:01Z

@ZhuangXiaobo: Let's work locally, so that the map is $Spec(S)\to Spec(R)$. By definition, "degree $n$" means that the corresponding ring homomorphism $R\to S$ makes $S$ into a free module of rank $n$ over $R$. If $n=1$, the map is bijective, so an isomorphism of rings.

Comment by Anton Geraschenko

2012-09-11T02:19:07Z

I've CWed.

Comment by Anton Geraschenko

2012-07-28T20:13:38Z

I've moved this to a comment on the original question.

Comment by Anton Geraschenko

2012-07-28T20:13:19Z

Johannas (mathoverflow.net/users/22991) asks (in a now-deleted answer): Contradiction to what? That $G$ is simple? $$ $$ Rashed answers: Yes Contradiction to $G$ is simple.

Comment by Anton Geraschenko

2012-07-27T07:48:14Z

You're absolutely right about only needing some point where the extension is trivial; I guess I've misunderstood the definition all this time. Of course you need to check more than the generic point, but the, but checking the generic point is the interesting part. The calculation of the maps on the two components shows that the residue field extensions away from the node are all isomorphisms (since the map is just the disjoint union of two isomorphisms away from the node), and the residue field extensions at the node are $k\subseteq k$.

Comment by Anton Geraschenko

2012-07-27T07:38:51Z

You're absolutely right. I guess I've misunderstood the definition all this time.

Comment by Anton Geraschenko

2012-07-27T06:07:48Z

I don't understand your example. The squaring map $A^1-0\to A^1$ is never Nisnevich since the extension of residue fields at the generic point is $k(x^2)\hookrightarrow k(x)$, which isn't an isomorphism.

Comment by Anton Geraschenko

2012-07-27T00:16:07Z

FYI: The owner of the question gets notified of comments on his question, so if the original poster is the sole target, a comment is just as noticeable as an answer.

Comment by Anton Geraschenko

2012-07-26T05:22:42Z

I'm moving this to a comment on the question.

Comment by Anton Geraschenko

2012-07-26T05:22:24Z

Nam-Hoon says (in a now-deleted answer): Well, I am an algebraic geometer. You can assume that $X$, $Y$ are projective manifolds over complex number and $f$ is a fat morphism if you wish.

Comment by Anton Geraschenko

2012-07-26T05:15:16Z

I've moved this to a comment on the question.

Comment by Anton Geraschenko

2012-07-26T05:14:50Z

Ed Wolf says (in a now-deleted answer): BTW, I have the exponential generating function for this, but since I'm looking for an answer to 3), I was hoping to find something more direct.

Comment by Anton Geraschenko

2012-07-12T19:35:37Z

This is completely off topic.

Comment by Anton Geraschenko

2012-07-12T19:33:01Z

Just in case there are curious cynics out there: by accepting his own answer, Justin did not recover the bounty (see mathoverflow.net/users/…). It's too bad Peter didn't post his comment as an answer. @Justin: consider editing Peter's comment verbatim into this answer.

Comment by Anton Geraschenko

2012-07-09T18:14:31Z

Unless I'm missing something, this is strictly off-topic, so I'm deleting this post.