User laurent moret-bailly - MathOverflow

First order decidability of rings vs Diophantine decidability

2013-05-06T07:03:55Z

Are there known (preferably ``concrete'') examples of a ring $R$ (commutative, with 1) such that:

$\bullet$ the first order theory of $R$ is undecidable, but
$\bullet$ the positive existential (= Diophantine) theory of $R$ is decidable?

The Diophantine theory consists of formulas of the form $\exists x S(x)$ where $x$ is an $n$-tuple of variables and $S$ denotes a finite system of polynomial equations, with coefficients in (some subring of) $R$.

Answer by Laurent Moret-Bailly for Example of codim 1 regular embedding that is not an effective Cartier divisor?

2013-05-01T13:56:17Z

Let $(A,m,k)$ be a discrete valuation ring. Consider the subring $R\subset A^\mathbb{N}$ of converging sequences (for the discrete topology on $A$), i.e. ultimately constant sequences. We have a morphism $\varphi:R\to A$ sending any sequence to its limit. Put $\mathfrak{p}=\varphi^{-1}(m)$: this is the maximal ideal of sequences with noninvertible limit.

Now it is easy to check that $\varphi$ induces an isomorphism $R_\mathfrak{p}\cong A$. In particular, $\mathfrak{p}$ becomes principal in $R_\mathfrak{p}$, and of course also in $R_\mathfrak{q}$ for all $\mathfrak{q}\neq\mathfrak{p}$ because then $\mathfrak{p}R_\mathfrak{q}=R_\mathfrak{q}$. So $V(\mathfrak{p})\subset\mathrm{Spec}\,(R)$ is a regular embedding of codimension 1.

On the other hand, $\mathfrak{p}$ is not finitely generated. Indeed, if $J\subset\mathfrak{p}$ is a finitely generated ideal, there is an $N\in\mathbb{N}$ such that for each $u:\mathbb{N}\to A$ in $J$ and each $n>N$ we have $u(n)\in m$. Clearly there is no such $N$ for $\mathfrak{p}$.

Answer by Laurent Moret-Bailly for morphism between two elliptic curves over a local field

2013-05-01T07:39:15Z

I assume that by "good models" you mean models $\mathscr{X}$, $\mathscr{Y}$ which are elliptic $R$-curves. These are pointed curves of genus 1, and in particular stable curves. This implies that the sheaf $$I:=\underline{\rm Isom}_{\text{$R$-ell. curves}}(\mathscr{X},\mathscr{Y})$$ is a scheme, finite and unramified over $R$. With your assumptions on $R$ (strictly henselian is enough), $I$ is then a finite disjoint sum of copies of closed subschemes of $R$. In particular, for every extension $L$ of $K$ (finite or not), we have $I(R)=I(L)$, which gives the result.

Answer by Laurent Moret-Bailly for Dimension of formal fiber

2013-04-16T19:15:53Z

For question 1, my geometric explanation is this: you can find an irreducible "analytic curve" $C$, i.e. a 1-dimensional closed subscheme of $X$, which is "as transcendental as possible", meaning that it is not contained in any algebraic hypersurface (i.e. does not map to a hypersurface in $Y$). Now if $p$ is the generic point of $C$, this means that $p$ maps to $y$, hence it is a point of $X_y$ and so is any generization of $p$, which proves that $\dim X_y≥d-1$ because $p$ has codimension $d-1$ in $X$. And of course $\dim X_y<d=\dim X$ , so $\dim X_y=d-1$.

To construct $C$, just note that $\mathbb{C}[[t]]$ has infinite transcendence degree over $\mathbb{C}$, so you can find $f_1,\dots,f_d\in \mathbb{C}[[t]]$ with zero constant term (and nonzero degree 1 term, if you wish) which are algebraically independent over $\mathbb{C}$. Now take for $C$ the image of the closed immersion $\mathrm{Spec}\,\mathbb{C}[[t]]\to X$ corresponding to $\varphi:\mathbb{C}[[x_1,\dots,x_d]]\to \mathbb{C}[[t]]$ sending $x_i$ to $f_i$. (Geometrically, it is the "formal curve parametrized by $f_1,\dots,f_d$").

Remark: I don't have Matsumura's book here, but I wouldn't be surprised if this were essentially his proof!

Answer by Laurent Moret-Bailly for Open idempotents in modules over a local ring

2013-03-06T13:13:23Z

The answer is no in general. It is yes if $R$ is $\mathfrak{m}$-adically separated (e.g. noetherian), where $\mathfrak{m}$ is the maximal ideal.

For a counterexample, assume $R$ is a valuation ring (not a field) and $\mathfrak{m}=\mathfrak{m}^2$. Take $F=\mathfrak{m}$, $e=$ the inclusion in $R$. Now $e\otimes \mathfrak{m}$ is injective by flatness (over a valuation ring, flat=torsion free) and surjective because its image is $\mathfrak{m}^2$.

Now assume $R$ separated, let $k$ be the residue field, and put $\overline{F}=F\otimes k$. Then $e\otimes k$ is a linear form $\overline{F}\to k$ inducing an isomorphism $\overline{F}\otimes\overline{F}\to \overline{F}$, which immediately implies $\overline{F}=0$ or $\overline{F}=k$ (and $e\otimes k$ is an isomorphism in the second case).
If $\overline{F}=0$, then $F=\mathfrak{m}F$, hence $F=\mathfrak{m}^nF$ for all $n$, hence $e(F)\subset \mathfrak{m}^n$ for all $n$. By our assumption this implies $e=0$, and finally $F=0$.
In the other case, $e$ is surjective (its image is not contained in $\mathfrak{m}$), hence $F= R\oplus\ker(e)$. Tensoring with $F$ we get that $F\otimes \ker(e)$ must be zero, hence $\ker(e)=0$ since $R$ is a direct summand of $F$. Hence $e$ is an isomorphism. (In this case the assumption on $R$ is not used).

Answer by Laurent Moret-Bailly for Which schemes can be presented as limits of smooth varieties?

2013-03-06T08:15:35Z

Here is an answer for the affine case. Assume $f:A\to B$ is a homomorphism of Noetherian rings. Then $B$ is a filtered colimit of smooth (finitely generated) $A$-algebras iff $f$ is regular (flat with geometrically regular fibers). This is due to Popescu and Spivakovsky; see for instance Teissier's Bourbaki talk
http://www.math.jussieu.fr/~teissier/documents/Approx.BBk.pdf
and the references therein (the above result is Thm. 1.1).
If $A=k$ is a field, this says that $B$ is a colimit of smooth $k$-algebras iff it is geometrically regular over $k$. If $k$ is perfect (e.g. the prime field!) you can remove "geometrically". The field case might possibly be simpler than the general case.

Answer by Laurent Moret-Bailly for Artin approximation theorems over non-regular rings/non-Noetherian rings

2013-02-28T20:37:25Z

Concerning (2), here are some references:

For certain subrings of $R[[T_1,\dots,T_N]]$ where $R$ is a complete valuation ring of rank 1, see:
H. Schoutens: Approximation properties for some non-Noetherian local rings. Pac. J. Math. 131(2), 331–359 (1988).

For any henselian valuation ring, with fraction field $K$, such that the completion $\widehat{K}$ is separable over $K$, see my paper:
An extension of Greenberg’s theorem to general valuation rings, Manuscripta Math. 139, 153–166 (2012).

The case of a henselian valuation ring of rank 1 is already mentioned in Elkik's thesis:
R. Elkik: Solutions d’équations à coefficients dans un anneau hensélien. Ann. Sci. École Norm. Sup. (4) 6, 553–603 (1974) (see Remarque 2, p. 587),
and treated in more detail in chapter 1 of
A. Abbes: Éléments de géométrie rigide I. Progress in Mathematics. Birkhäuser, Boston (2011).

For rings of differentiable functions, perhaps you should look at Tougeron's papers.

Answer by Laurent Moret-Bailly for Are noetherian schemes generically Jacobson?

2013-02-05T17:13:04Z

Here is a counterexample. Let $k$ be a field, $R=k[x,y]$. Choose a set $\Sigma$ of closed points of $\mathbb{A}^2_k=\mathrm{Spec}(R)$ such that:

(1) $\Sigma$ is Zariski-dense,
(2) for every $s\in\Sigma$ there is a curve $C$ containing $s$ such that $C\cap\Sigma$ is finite.

EDIT: (For instance, if $\mathrm{char}\,k=0$ , take $\Sigma=\mathbb{Q}^2$ , and for $s=(a,b)\in\Sigma$ take $C$ defined by $(x-a)^2+(y-b)^2=0$.)

Now let $R_1$ be the localization of $R$ at $\Sigma$, and $X=\mathrm{Spec}\,R_1$ . Let $\emptyset\neq U\subset X_1$ be open. By (1), $U$ contains some $s\in \Sigma$, and by (2), $U$ has a one-dimensional closed subscheme with only finitely many closed points. Hence $U$ is not Jacobson.

Answer by Laurent Moret-Bailly for purity for finite flat group schemes

2013-02-05T08:04:30Z

It is true if $X$ is regular. This is stated (as Lemme 2) in my CRAS 1985 note on purity for families of curves:
http://gallica.bnf.fr/ark:/12148/bpt6k5495813c/f45.image
where, unfortumately, the proof is missing (I only claim that it extends Auslander's proof of the étale case).

Answer by Laurent Moret-Bailly for Are rational varieties simply connected?

2013-02-03T09:23:18Z

Every proper, normal, rational variety over an algebraically closed field is simply connected. Dmitri explains this via the smooth case and resolution of singularities, but this is true in any characteristic: see SGA 1, XI, Cor. 1.2.

EDIT: As Dmitri and Vesselin observe, the "proof" in SGA1 is sketchy, to say the least. One can argue as follows: if $X$ is our variety, there is a birational morphism $U\to X$ where $U=\mathbb{P}^n\smallsetminus Z$ and $Z$ has codimension $\geq2$ in $\mathbb{P}^n$. By the purity theorem, $U$ is simply connected. So any étale covering of $X$ is generically trivial (because its pullback on $U$ is trivial), hence trivial since $X$ is normal.

In fact, this proves that if $X$ and $Y$ (both proper and normal) are birationally equivalent, and $Y$ is regular and simply connected, then $X$ is simply connected. But this does not answer Vesselin's final question.

Answer by Laurent Moret-Bailly for Representability of sheaf of Ext^1 of a Néron model by $\mathbb{G}_m$

2012-11-25T08:32:00Z

First question: no. Assume, to fix ideas, that $R$ is complete with uniformizer $\pi$, $k$ is algebraically closed, and $A$ is an elliptic curve with multiplicative reduction. Denote by $\mathscr{E}$ the Ext sheaf in question. Then the restriction of $A^\circ$ to $S_n:=\mathrm{Spec\,}(R/(\pi^{n+1}))$ is isomorphic to $\mathbb{G}_{m}$, so $\mathscr{E}(S_n)$ is zero for all $n$. If $\mathscr{E}$ were a scheme, this would imply $\mathscr{E}(R)=0$ (the functor of points of a scheme "commutes with completion" for local rings), a contradiction.

Second question: yes, because $W$ is a $\mathbb{G}_m$-torsor over the product, and torsors under affine group schemes are schemes.

Answer by Laurent Moret-Bailly for Over which fields does the Mordell-Weil theorem hold?

2012-10-30T10:28:48Z

[Edited November 2 for brevity]

(1) An extension of the finitely generated case for Q1: let $K_0$ be finitely generated over the prime field, and let $K=K_0((x_i)_{i\in I})$ be a purely transcendental extension of $K_0$. Then $K$ "satisfies Q1". Indeed, any abelian variety $A/K$ is defined over some intermediate $K_1:=K_0((x_i)_{i\in J})$, $J\subset I$ finite. Then $A(K_1)$ is finitely generated, but $A(K)=A(K_1)$ since $K/K_1$ is purely transcendental.

(2) Another "easy" case for Q2: if $K/\mathbb{F}_p$ is infinite algebraic, then for any $A$ the group $A(K)$ is torsion, but must be infinite by Weil's estimates, hence is not finitely generated.

(3) A general result on Q2: Say a field $K$ is fertile if for every smooth irreducible $K$-variety $X$, if $X(K)$ is nonempty, then it is Zariski-dense.
(Pop, who invented the concept, called these fields "large"; others say "ample").

I claim that every fertile field $K$ satisfies Q2. This includes in particular:

(3a) all Henselian valued fields (already mentioned by Pete, but there is no restriction on the rank here, except the valuation must be nontrivial).

(3b) Pseudo-algebraically closed fields (i.e. such that every geometrically irreducible variety has a rational point). This includes example (2) above.

Proof of claim: Let $A$ be an abelian $K$-variety of dimension $g>0$, with origin $e$. We may assume $g\geq2$ (if $g=1$, consider $A\times A$). Let $t_1,\dots,t_g$ be a regular system of parameters at $e$. Consider the rational map $(t_1:\dots:t_g):A\dots\to\mathbb{P}^{g-1}_K$ . It induces a morphism $f:U\smallsetminus\{e\}\to\mathbb{P}^{g-1}_K$ where $U\subset A$ is a neighborhood of $e$. Let $\widetilde{U}$ be the blow-up of $e$ in $U$. By the assumption on $t_1,\dots,t_g$, we get a morphism $\widetilde{f}:\widetilde{U}\to \mathbb{P}^{g-1}_K$ which induces an isomorphism $E\to\mathbb{P}^{g-1}_K$ where $E$ is the exceptional divisor. Moreover, $\widetilde{f}$ is smooth along $E$. Shrinking $U$, we may assume $\widetilde{f}$ smooth.
For every $y\in\mathbb{P}^{g-1}(K)$ , $\widetilde{f}^{-1}(y)$ is a smooth curve with a rational point on $E$. Since $K$ is fertile, $\widetilde{f}^{-1}(y)$ also has rational points on $U\smallsetminus\{e\}$ . Hence $f:U(K)\smallsetminus\{e\}\to\mathbb{P}^{g-1}(K)$ is surjective.
On the other hand, if $A(K)$ were finitely generated there would be a finitely generated subfield of definition $K_0\subset K$ for $A$, $U$ and $f$ such that $A(K)=A(K_0)$, which would imply $f(U(K))\subset\mathbb{P}^{g-1}(K_0)$ . This is a contradiction because $K_0\neq K$ (finitely generated fields are not fertile).

Answer by Laurent Moret-Bailly for An etale version of the van Kampen theorem

2012-10-25T09:10:46Z

To simplify notation, let me write $U_i$ for $V\smallsetminus W_i$, and $U_{12}$ for $U_1\cap U_2=V\smallsetminus(W_1\cup W_2)$.

Fact: The obvious functor $$(\mathrm{Sch}/V)\longrightarrow (\mathrm{Sch}/U_1) \times_{(\mathrm{Sch}/U_{12})} (\mathrm{Sch}/U_2)$$ is an equivalence. In other words, a $V$-scheme $X$ is the same thing as a $U_1$-scheme $X_1$, a $U_2$-scheme $X_2$, and a $U_{12}$-isomorphism of their restrictions to $U_{12}$.
This is probably somewhere in EGA1. [EDIT: all I could find was section 2.4 of EGA1, relying on (4.1.7) of Chapter 0 (glueing of riged spaces).]
However, we are dealing here with finite étale schemes, which happen to be affine over the base, so this boils down to the analogous statement for categories of quasicoherent sheaves, which is essentially trivial (plus the fact that ``finite étale'' is a local condition).

If we describe the categories of finite étale covers in terms of $\pi_1$-sets, the above equivalence says that the diagram of groups $$(*)\qquad\begin{array}{rcl} \pi_1(U_{12},p)=:G_{12}& \longrightarrow &G_1:=\pi_1(U_{1},p)\cr \downarrow && \downarrow\cr \pi_1(U_{2},p)=:G_2& \longrightarrow &G:=\pi_1(V,p) \end{array}$$ is cocartesian. In other words, we get the usual van Kampen statement: the natural map $$\pi_1(U_{1},p)\ast_{\pi_1(U_{12},p)}\pi_1(U_{2},p)\longrightarrow \pi_1(V,p)$$ is an isomorphism. [EDIT: the coproduct is in the profinite category, which perhaps makes it hard to describe in general. See Will Savin's comment.]

What we want to prove is that the map "on the other side" $$G_{12}\longrightarrow G_1 \times_G G_2$$ is surjective, given that all the maps in diagram ( $\ast$ ) are surjective.

Identifying $G_i$ ($i=1,2$) with $G_{12}/N_i$, we see from the universal property of the coproduct that $G=G_{12}/N_{1}N_{2}$. [EDIT: clearly this works also in the profinite category: since $N_1$, $N_2$ are both compact normal subgroups, so is $N_1 N_2$, hence $G_{12}/N_{1}N_{2}$ is profinite].

Take any $(x_1,x_2)\in G_1 \times_G G_2$: thus we have $x_i=g_i N_i$ for some $g_i\in G_{12}$, and the fiber product condition says that $g_1=g_2 n_2 n_1$ for some $n_i\in N_i$ (recall that $N_1 N_2=N_2 N_1$). So, $(x_1,x_2)$ is the image of $g_1 n_{1}^{-1}=g_2 n_2\in G_{12}$. QED

Constructible sets in Hausdorff spaces

2012-10-06T15:42:54Z

In an attempt to write a proof by contradiction, I end up with a space $X$ with the following properties:

(0) $X$ is nonempty,
(1) $X$ is Hausdorff,
(2) $X$ has no isolated points,
(3) every subspace of $X$ is constructible (finite union of locally closed subsets).

Is this indeed a contradiction?

It would suffice to know that any $X$ with properties (1) and (2) has a dense subset with dense complement: such a set cannot be constructible unless $X=\emptyset$. [Edit: there are counterexamples to this, see the comment by Yves Cornulier]

Answer by Laurent Moret-Bailly for Is every algebraic space the quotient of a scheme by a finite group?

2012-02-05T20:58:31Z

As I just remembered, the answer is yes if $X$ is quasiseparated, noetherian and normal: see Laumon and Moret-Bailly, Champs algébriques, (16.6.2). The quesiseparated assumption is needed (see Scott Carnahan's answer). Without the normality condition, I don't have a counterexample but my guess is that there is one.

EDIT after Chris' and Jason's comments: in fact the proof in the case of normal algebraic spaces can be made substantially simpler than in the book (which proves a more general result about noetherian Deligne-Mumford stacks). It goes like this:

Assume $X$ noetherian, integral, normal and, to simplify, separated (I am not sure how much this helps). Cover $X$ by étale maps $X_i\to X$ with each $X_i$ integral and affine. There is a dense open subspace $U$ of $X$ which is a scheme and such that each induced map $U_i:=X_i\times_X U\to U$ is finite. Let $V\to U$ be an étale Galois cover, with Galois group $G$, dominating all the $U_i$ 's. Now let $\overline{V}\to X$ (resp. $\overline{X_i}\to X$) be the normalization of $X$ in $V$ (resp. in $X_i$); we have dense open immersions $V\subset \overline{V}$ and $X_i\subset \overline{X_i}$. By functoriality, $G$ acts on $\overline{V}$, with quotient $X$.

I claim that $\overline{V}$ is a scheme. Indeed, for each $i$, $\overline{X_i}$ is also the normalization of $X$ in $U_i$. In particular there is an $X$-morphism $f_i:\overline{V}\to\overline{X_i}$ (deduced from $V\to U_i$) which must be finite surjective (everyone is integral, finite and surjective over $X$). Put $V_i:=f_i^{-1}(X_i)$: this is an open subspace of $\overline{V}$ which is finite over $X_i$, hence an affine scheme. So, the union $W$ of the $V_i$'s is an open subspace of $\overline{V}$ which is a scheme and maps surjectively to $X$ (since $V_i\to X_i$ is surjective), hence $\overline{V}$ is covered by $\{gW\}_{g\in G}$ .

Answer by Laurent Moret-Bailly for Irreducible "family" of relative effective divisors of a smooth morphism

2012-07-11T08:34:55Z

Take $Y=\mathbb{A}^1$ (with coordinate $t$), and $X=\mathbb{P}^2_Y$ with homogeneous coordinates $u$, $v$, $w$. Now let $Z$ be the zero scheme of $(tu, u^2, uv)$. Over any point $y$ where $t\neq0$, $X_y$ is the line $u=0$ in $\mathbb{P}^2$, while $X_0$ is defined by $u^2=uv=0$, hence has an embedded component.

Answer by Laurent Moret-Bailly for is intersection of a curve and a family of curves generically constant as a scheme?

2012-06-24T19:39:05Z

Probable counterexample: take $D=\mathbb{A}^2$ (or $\mathbb{P}^2$, if you prefer), $T=\mathbb{A}^1$, and let $X_t$ be the union of the four lines $x=0$, $y=0$, $x=y$, and $x=ty$. It is well known that $t$ is "almost" determined by the isomorphism class of $D_t$, as the cross-ratio of the four tangents at the origin. ("Almost" refers to permutations of the four lines, but only finitely many $t$'s give rise to isomorphic $D_t$'s). My claim is that this remains true if you replace $D_t$ by a sufficiently big infinitesimal neighbourhood of the origin (presumably the one given by the ideal $(x,y)^5$ works).

Assuming this, you have a counterexample as soon as $E$ contains the neighbourhood in question, i.e. is sufficiently singular at the origin.

Answer by Laurent Moret-Bailly for Decomposition of a proper morphism

2012-06-24T10:02:27Z

By Nagata's theorem there is an open immersion $j:X\hookrightarrow \overline{X}$ where $\overline{X}$ is complete. Then, as ABayer suggests, take $(j,f): \overline{X}\times Y$.

When is a valued field second-countable?

2012-05-15T13:58:41Z

Let $R$ be a valuation ring, with fraction field $K$ and residue field $k$. Denote by $\Gamma=K^{\times}/R^{\times}$ the valuation group (assumed nontrivial).
The valuation $v:K^{\times}\to\Gamma$ decomposes $K^{\times}$ as a disjoint union of nonempty open subsets, indexed by $\Gamma$. Each of these is homeomorphic to $R^{\times}$, which is in turn (using the reduction map to $k$) a disjoint union of nonempty open subsets, indexed by $k^{\times}$.
We conclude that any basis for the topology of $K$ must have cardinality at least $\kappa:=\max(\mathrm{Card}\,\Gamma, \mathrm{Card}\,k)$ .

Question: does there exist a basis of open subets of $K$ with cardinality $\kappa$?

Remarks:
(1) It is true if $v$ is discrete, i.e. $\Gamma\cong\mathbb{Z}$. Proof: take a set $S\subset R$ of representatives of $k$, and a uniformizing parameter $\pi$. Let $X\subset K$ be the set of finite sums $\sum_i s_i\pi^{n_i}$ ($s_i\in S$, $n_i\in\mathbb{Z}$). Then the balls centered on $X$ form a basis.
(2) I am especially intereseted in the case $\kappa=\omega$. Explicitly: if $\Gamma$ and $k$ are countable, does it follow that $K$ is second-countable (or, equivalently, separable)?

Answer by Laurent Moret-Bailly for non-isomorphic stably isomorphic fields

2012-05-14T08:55:07Z

An answer to Q2, generalizing Ralph's comment: "$K$ is algebraically closed" is a sufficient condition. Indeed, you can characterize $K$ inside $K(x_1,\dots,x_n)$ as the set of elements having $m$-th roots for infinitely many integers $m$. More generally, it is enough to assume that for some $m>1$, the $m$-th power map on $K$ is onto. Examples: $K$ perfect of positive characteristic, or $K=\mathbb{R}$.

Answer by Laurent Moret-Bailly for Is reflexivity an open condition?

2012-04-02T14:32:45Z

[EDIT: as Sasha points out, this does not answer the question. Please see it as a comment explaining why $X$ should be proper!]

The answer is no in general if $X$ is not proper: take $X=\mathrm{Spec}\,\mathbb{C}[x]$ , $T=\mathrm{Spec}\,\mathbb{C}[t]$ , and $F=$ the structure sheaf of $Z=\mathrm{Spec}\,(\mathbb{C}[t,x]/(1-tx))$ . Then $T'$ is just the origin.

Variant: if instead you take $T=\mathrm{Spec}\,\mathbb{C}[t,u]$ and $Z=\mathrm{Spec}\,(\mathbb{C}[t,u,x]/(u,1-tx))$ (i.e. the same $Z$ as before, but embedded in 3-space), then $T'$ is the union of the origin and the complement of the $t$-axis, hence not locally closed (but still constructible).

Of course the point here is that $Z$ "goes to infinity" at the origin. I don't have a counterexample where $X$ is proper, but the main problem then is "taking the dual in the fibers", as in Sasha's comment above.

Answer by Laurent Moret-Bailly for finite non-commutative local group schemes

2012-01-05T15:07:47Z

If $\mathrm{char}(k)=p>0$ and $G$ is a $k$-group scheme of finite type, the kernel of the relative frobenius $F_{G/k}:G\to G^{(p)}$ is a finite connected $k$-group scheme. It has the same Lie algebra as $G$, and in particular it is noncommutative if the Lie algebra is nonabelian, e.g. for $G=GL_{n,k}$, $n\geq2$.
If $G$ is smooth over $k$, then $F_{G/k}$ is faithfully flat, so we get an exact sequence (of pointed sets) $G(k)\to G^{(p)}(k)\to H^1(k, \ker(F_{G/k}))$ which may be used to get nontrivial torsors if $k$ is not perfect. For instance, if $G=GL_{n,k}$ as above, we have $G=G^{(p)}$ and $F_{G/k}$ acts on matrices by raising entries to the $p$-th power, so any $g\in GL_{n,k}$ whose entries are not all $p$-th powers gives rise to a nontrivial torsor.

Remark: the example given by anon as a comment to the question may be seen as the special case where $G$ is the natural semidirect product $\mathbb{G}_a \rtimes \mathbb{G}_m$ , i.e. the group of affine automorphisms of the affine line.

Answer by Laurent Moret-Bailly for What are some examples of ingenious, unexpected constructions?

2012-03-16T13:44:09Z

I am very fond of Goodstein's theorem and especially of its proof, using ordinal arithmetic to prove that an integer sequence (which at first sight seems hopelessly increasing) is ultimately zero. See for instance here.

Answer by Laurent Moret-Bailly for Categorical interpretation of quasi-compact quasi-separated schemes

2012-03-04T20:15:30Z

Let me show that if $\Gamma$ preserves filtered colimits, then $X$ is quasicompact. (At the moment I don't know about 'quasiseparated'; but, as Martin points out, I only use the injectivity of $\varinjlim\circ \Gamma \to \Gamma\circ \varinjlim$ for filtered inductive systems).

Assume $X$ is not quasicompact. Then there is a filtered decreasing family $(Y_i)$ of nonempty closed subschemes of $X$, with empty intersection. The structure sheaves $\mathcal{O}_{Y_i}$ form an inductive system with colimit zero, but the unit section of any $Y_{i_0}$ is an element of $\varinjlim_i \Gamma(\mathcal{O}_{Y_i})$ which is nonzero because each $Y_i$ is nonempty.

Answer by Laurent Moret-Bailly for In what sense is a generically submersive morphism of varieties subermersive over singular points?

2012-02-28T20:13:59Z

Negative answer to first question: let $C \subset \mathbb{A}^2$ be the plane curve (in characteristic $\neq 2,\,3$) with equation $y^2=x^3$ , with singular point $Q=(0,0)$ , and let $p:\mathbb{A}^1\to C$ be the normalization morphism $t\mapsto (t^2,t^3)$ . Now take $X=\mathbb{A}^2$ , $Y=C\times \mathbb{A}^1$ , and $f: X\to Y$ given by $f(t,u)=(p(t),tu)$. Finally, take $W=\{Q\}\times \mathbb{A}^1$ , $x=(0,0)$ , $y=(Q,0)$ . The differential of $f$ has rank $2$ whenever $t\neq0$ but is zero at $x$.

The second question can probably be treated similarly.

Answer by Laurent Moret-Bailly for Spreading out of flat morphisms of schemes

2012-02-26T17:34:25Z

I think what you are looking for is in the book, only later: see EGA IV, (11.2.6).

Answer by Laurent Moret-Bailly for Do Disjoint Unions and Fiber Products Commute?

2012-02-24T21:54:52Z

Answer to 2: If $A$, $B$, $C$ are three sets, it is not true in general that $(A\times B)\coprod C=(A\coprod C)\times (B\coprod C)$. Hence the category $(Sets)^{op}$ is a counterexample.

Answer by Laurent Moret-Bailly for Rigidity of the category of schemes

2012-02-18T15:22:20Z

A scheme $X$ is reduced if and only if the natural map $$ \coprod_{x\in X}\mathrm{Spec} \kappa( x )\to X$$ is an epimorphism. So "reduced" is categorical, and so is $X\mapsto X_{red}$.

[EDIT to answer Martin's question:

If $X_{red}\subset X$ is an epimorphism, then it is an isomorphism; this is true for any closed immersion $X_{0}\subset X$, because closed immersions are equalizers. Indeed, let $I$ be the ideal sheaf, and let $p:Y\to X$ be the spectrum of the symmetric algebra $A$ of $I$. Then $p$ has a natural section $s$ deduced from the inclusion $I\subset\mathcal{O}_{x}$, and $X_0$ is the equalizer of $s$ and the zero section.]

Strong specializations (edited after Martin's comments):
Say a point $x\in X$ is a strong specialization of a point $y$ if there is a morphism $T\to X$ where $T$ is a connected two-point scheme, sending the closed point $a$ to $x$ and the generic point $b$ to $y$ (note that $T$ is automatically local, irreducible and one-dimensional).
This notion is categorical: to see this, it remains to distinguish $b$ from $a$ categorically on a scheme $T$ as above. We may assume $T$ reduced, and then there is only one monomorphism $Y\to T$ from a one-point scheme $Y$ with image $b$ and infinitely many with image $a$. (Proof: we have $T=\mathrm{Spec}\,R$ where $R$ is a 1-dimensional local domain with fraction field $K$ and residue field $k$. First, a morphism $Y\to T$ with image $b$ must factor through $\mathrm {Spec}\,(K)$ (which is open), hence must be the inclusion if it is a monomorphism. Second, take some $t\neq0$ in the maximal ideal of $R$: then the closed immersions $\mathrm{Spec}\,(R/t^n)\to\mathrm{Spec}\,R$ ($n\geq1$) are distinct monomorphisms with image $a$.)

As Martin points out, all specializations are strong on a locally noetherian scheme, but probably not in general.

Answer by Laurent Moret-Bailly for is a cartesian square of a group scheme with $\mathbb{G}_a^n$ fibres reduced?

2012-02-15T21:40:41Z

Take $S=\mathrm{Spec}(R)$ where $R$ is, say, a noetherian domain. Let $t\in S$ be nonzero and noninvertible. Let $G$ be the kernel of $f:\mathbb{G}_{a,S}^2\to\mathbb{G}_{a,S}$ sending $(x,y)$ to $t^2\,y$ . For each $s\in S$, the fiber $G_s$ is equal to $\mathbb{G}_{a,s}^2$ if $t$ is zero at $s$, and to $\mathbb{G}_{a,s}\times\{0\}$ otherwise. But $G$ is (in general) not reduced since the function $ty$ is nonzero, with square zero.

Answer by Laurent Moret-Bailly for sections of very ample line bundle

2012-02-12T15:22:38Z

Embedding $X$ in $\mathbb{P}(H^0(L)^*)$, this means that every hyperplane through a certain point $Q$ is tangent to $X$. In other words, $X$ must be a hypersurface, and every hyperplane tangent to $X$ contains $Q$. Conics in characteristic two are an example, and, in fact, the only one in dimension 1 (E. Lluis, Bol. Soc. Mat. Mexicana (2) 7 (1962) 47–56; MR0147479 (26 #4995)). I don't know what is known in higher dimensions.

Comment by Laurent Moret-Bailly

2013-05-21T05:46:57Z

OK if "connected" means "with connected fibers". But I suspect the OP means "connected" literally.

Comment by Laurent Moret-Bailly

2013-05-12T08:29:18Z

@SJR: I agree, but my feeling is also (for the same reason) that examples should be easier to come by.

Comment by Laurent Moret-Bailly

2013-05-09T08:15:00Z

@SJR: Thanks for this nice example. It is in a sense a bit disappointing since a key point here, as was noted in the comments, is the choice of $\mathbb{Z}$ as the subring of coefficients. In fact, if we choose $\mathbb{Q}(t)$ instead, then the existential theory of $\mathbb{R}(t)$ is not e.c: this goes back at least to Denef. What surprises me here is the first order undecidability without extra constants.

Comment by Laurent Moret-Bailly

2013-05-06T10:42:16Z

As SJR notes, I guess equations should have coefficients in some fixed (recursive) subring of $R$ in order for decidability to make sense. Likewise, first order formulas are in the language of rings augmented with constants from such a subring.

Comment by Laurent Moret-Bailly

2013-05-02T09:26:59Z

It is not trivial since there are always quadratic twists (besides, is not a group!). The point here is that any nontrivial twist of $X$ must have bad reduction.

Comment by Laurent Moret-Bailly

2013-04-23T15:41:06Z

With the notation in the definition of $\overline{\mu}$, assume $P$ is a singular point. Then in general there is an $f\in\mathcal{O}_{X,P}$ such that $\mu_f(I)=1$, but none such that $\overline{\mu}_f(I)=1$.

Comment by Laurent Moret-Bailly

2013-04-23T07:03:25Z

There is something wrong with your TeX. The math doesn't show properly: specifically, the math italic alphabetic characters don't show at all.

Comment by Laurent Moret-Bailly

2013-04-22T19:16:54Z

Standard fact in commutative algebra: Let $A$ be a noetherian ring (I am not sure about the minimal assumptions), $J$ an ideal such that $A$ is $J$-adically complete and separated. Let $M$ be an $A$-module. If $M$ is $J$-adically complete and separated and $M/JM$ is finitely generated, then $M$ is finitely generated. Apply this to $A=\mathbb{C}[[f_1,\dots,f_d]]$, $J=(f_1,\dots,f_d)$ and $M=\mathbb{C}[[t]]$. Things are even simpler if, say, $f_1$ is a uniformizer in $\mathbb{C}[[t]]$: then it is easy to see directly that $\mathbb{C}[[f_1]]=\mathbb{C}[[t]]$.

Comment by Laurent Moret-Bailly

2013-04-21T07:17:55Z

@Omprokash Das: it is not true in general that you can embed $k(x)$ into $\mathcal{O}_{X,x}$ or even into the function field of $X$. Example: assume $k$ is not algebraically closed, take $X=\mathbb{P}^n_k$ and take for $x$ a closed point with residue field $\neq k$. Variant: same $X$ ($n\geq2$, $k$ arbitrary), $x$= generic point of a non-unirational subvariety, e.g. a curve of positive genus.

Comment by Laurent Moret-Bailly

2013-04-20T14:42:04Z

I don't understand "over $k(x)$". Clearly $S$ is a two-dimensional local scheme with closed point $\mathrm{Spec}\,k(x)$, but in general it is not a $k(x)$-scheme in any natural way.

Comment by Laurent Moret-Bailly

2013-04-19T16:43:39Z

@Serge: the question is what is meant by "generate the stalk". With your phrasing, it should mean "generate the stalk $E_x$ as an $\mathcal{O}_{X,x}$ -module". And then of course it is true for $E=\mathcal{O}_{X}$ . In condition 1 of the question, "$E_x$" should be understood as the stalk at $x$ tensored with the residue field. An even then, that version of the definition works only if you work with rational points over an algebraically closed field.

Comment by Laurent Moret-Bailly

2013-04-19T12:32:54Z

@Serge: $\mathcal{O}_X$ is always globally generated!

Comment by Laurent Moret-Bailly

2013-04-17T12:16:44Z

And what is the quotient of a set by a subset?

Comment by Laurent Moret-Bailly

2013-04-17T10:55:18Z

I don't understand what the quotients mean.

Comment by Laurent Moret-Bailly

2013-04-06T06:23:58Z

Maybe I misunderstand the question, but what if you start with an extension $L/F$ which is not simple and then take $K=F((t))$?