User qing liu - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T02:54:24Z http://mathoverflow.net/feeds/user/3485 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16745/when-is-brx-h2x-g-m/16749#16749 Answer by Qing Liu for When is Br(X) = H^2(X,G_m)? Qing Liu 2010-03-01T09:52:32Z 2013-04-25T10:41:12Z <p>When $X$ is quasi-projective over an affine scheme (or more generally if $X$ has an ample [<b>EDIT:</b> invertible] sheaf), then its Brauer group is isomorphic to the <b>torsion part</b> of $H^2(X, {\mathbb G}_m)$. This is an unpublished result of Gabber, and J. de Jong <a href="http://www.math.columbia.edu/~dejong/papers/2-gabber.pdf" rel="nofollow">wrote down a different proof</a>.</p> http://mathoverflow.net/questions/127908/reduction-types-of-elliptic-curves/128617#128617 Answer by Qing Liu for reduction types of elliptic curves Qing Liu 2013-04-24T13:52:45Z 2013-04-24T17:13:32Z <p>What you said is <i>almost</i> correct. </p> <p>First the list of types is complete if the residue field is algebraically closed. Over a perfect residue field, there are some more types, easy to handle (see nosr's comments). Essentially you identify irreducible components in the special fiber via possible automorphisms of the special fiber. </p> <p>Over an imperfect residue field of characteristic 2 or 3 (see again nosr's comment), there are also some more types, decribed in Szydlo's thesis (J. Number Theory, 2004). See also JSE's answer at <a href="http://mathoverflow.net/questions/10607" rel="nofollow">this question</a>. </p> <p>In residue characteristic 2, it is not true that type $I_n^*$ implies potentially multiplicative reduction, but the converse is true (potentially multiplicative reduction $I_m$ implies type $I_n^*$ for some $n$). There is a paper of Dino Lorenzini (in Pure Appli. Maths. Q., special issue in honor of Tate) where among other results, he gives the relation between $n$ and $m$ when the curve has potentially multiplicative reduction $I_n$. The relation involves the different of the minimal extension realizing the good reduction when $K$ is henseilan with algebraically closed $k$. </p> <p><s>The relation between II and II$^{\star}$, when $p\ne 2, 3$, is that II is obtained by a quadratic twist of II$^\star$, similarly for IV and IV$^\star$ if I remember correctly</s>. You can easily check on some examples using pari/gp. <b>See Will's comments. </b></p> http://mathoverflow.net/questions/127207/scheme-of-generalizations/127344#127344 Answer by Qing Liu for scheme of generalizations Qing Liu 2013-04-12T11:07:24Z 2013-04-12T11:07:24Z <p>Suppose for simplicity that $\mathcal O$ is a henselian (e.g., complete) DVR (otherwise, a point of $X_\eta$ may have several specializations in $X_s$) with field of fractions $K$. Consider $X=\mathbb P^1_S$ parametriezd by a rational function $t$, and let $Z$ be the single point $t=0$ in the closed fiber. The the set $U$ is the open disc $|t|&lt;1$ in the generic fiber $\mathbb P^1_K$ (in more algebraic terms, they correspond to maximal ideals $P(t)K[t]$ with $P(t)\in \mathcal O[t]$ monic and $P(0)\in \mathfrak m$ the maximal ideal of $\mathcal O$). So it is not a scheme, but a rigid analytic subspace.</p> http://mathoverflow.net/questions/122819/the-locus-where-a-sheaf-is-supported-in-a-certain-dimension/122873#122873 Answer by Qing Liu for The locus where a sheaf is supported in a certain dimension Qing Liu 2013-02-25T10:36:49Z 2013-02-25T10:36:49Z <p>Maybe I misunderstand something, but by Nakayama, the support of $E_t$ is just the support of $E$ intersected with $X_t$, and the support of $E$ is closed in $X$. Let $Z$ be the suppor of $E$. By Chevalley's semi-continuity theorem (EGA, IV.13.1.3), the set of $x\in X$ such that $\dim_x Z_t\le r$ ($x\mapsto t$) is open, so the set of $t\in T$ such that $\dim Z_t\le r$ is constructible in $T$. </p> <p>This is true for any morphism of finite type $X\to T$ over a noetherian scheme $T$. </p> http://mathoverflow.net/questions/114429/how-does-one-make-sense-of-the-mathbfc-p-points-of-a-rigid-analytic-space-ov/122531#122531 Answer by Qing Liu for How does one make sense of the $\mathbf{C}_p$-points of a rigid analytic space over $\mathbf{Q}_p$? Qing Liu 2013-02-21T12:46:24Z 2013-02-21T12:46:24Z <p>I think in Coleman-Mazur, it is just to be taken in the sense of $\mathbb C_p$-points of $X_{\mathbb C_p}$. This is compatible with the more abstract definition you propose. Indeed, as $\mathrm{Sp}(\mathbb C_p)$ is just one point, it is enough to work with an affinoide space $X$ associated to an affinoide algebra $A$ over $\mathbb Q_p$. Then it is clear that the continuous homomorphisms from $A$ to $\mathbb C_p$ coincide with continuous homomorphisms from $A\widehat{\otimes}_{\mathbb Q_p} \mathbb C_p$ to $\mathbb C_p$.</p> http://mathoverflow.net/questions/122227/divisor-class-group-on-blowup-of-nodal-surface/122296#122296 Answer by Qing Liu for Divisor class group on blowup of nodal surface Qing Liu 2013-02-19T12:42:06Z 2013-02-19T12:42:06Z <p>This is a partial answer to the part concerning Weil divisors. </p> <p>I prefer to use the terminology of cycles of codimension $1$ instead of Weil divisors. The group of Weil divisors modulo rational equivalence is the Chow group $A^1$. </p> <p>Let $f : \tilde{S}\to S$ be any proper birational morphism of integral normal algebraic varieties (integral normal schemes of finite type over a field) of dimension $\ge 2$. Let $E_1, \dots, E_n$ be the irreducible components of codimension $1$ (in $\tilde{S}$) of the exceptional locus of $f$. </p> <p>As $f$ is proper, there is a canonical pushforward map $f_*: A^1(\tilde{S})\to A^1(S)$ (see Fulton, Intersection Theory, Chapter 1) and it is a group homomorphism. As $S$ is normal, $f$ is an isomorphism outside of a codimension $\ge 2$ closed subset of $S$. In particular, $f_*$ is surjective because for any irreducible cycle $\Gamma$ of codimension $1$ in $S$, $f_*$ maps the class of the strict transform $\tilde{\Gamma}$ to the class of $\Gamma$. However, in general $f_*$ doesn't have a section. </p> <p>By construction, the kernel of $f_*$ is generated by the classes of the $E_i$. Let us show the map $\oplus_{1\le i\le n} [E_i]\mathbb Z \to A^1(\tilde{S})$ is injective. Let $g\in k(\tilde{S})$ be a rational function with divisor <code>$\mathrm{div}_{\tilde{S}}(g)$</code> supported in <code>$\cup_i {E_i}$</code>. Then the divisor <code>$\mathrm{div}_{S}(g)$</code> of $g$ as a rational function on $S$ is $0$. As $S$ is normal, this forces $g$ to be a unit in the ring of regular functions $O(S)$, hence $g$ is a unit in $O(\tilde{S})$ and <code>$\mathrm{div}_{\tilde{S}}(g)=0$</code>. Conclusion, we have an exact sequence $$0\to \oplus_{1\le i\le n} [E_i]\mathbb Z \to A^1(\tilde{S}) \to A^1(S) \to 0.$$ </p> http://mathoverflow.net/questions/121880/sections-of-the-cotangent-bundle-of-elliptic-surfaces/122087#122087 Answer by Qing Liu for sections of the cotangent bundle of elliptic surfaces Qing Liu 2013-02-17T17:41:03Z 2013-02-17T17:41:03Z <p>This is a comment starting from a slightly more general context. Most of the following material can be found in <a href="http://www.math.u-bordeaux1.fr/~qliu/articles/conduct-ineg.ps" rel="nofollow"> a paper of T. Saito and me</a> (but most dealing with the positive characterisitc case).</p> <p>Let $f : X\to C$ be a flat morphism of smooth (geometrically connected) projective varieties over a field $k$ of characteristic $0$. Consider the canonical exact sequence $$0 \to f^*\Omega_{C/k}\to \Omega_{X/k}\to \Omega_{X/C} \to 0.$$ We have $f_{*}\mathcal O_X=\mathcal O_C$. Taking $f_*$ we get an exact sequence of sheaves on $C$: $$0 \to \Omega_{C/k}\to f_{*}\Omega_{X/k} \to f_{*}\Omega_{X/C}\stackrel{\theta}{\to} R^1f_{*}(\mathcal O_X)\otimes\Omega_{C/k}.$$ We have $R^1f_{*}(\mathcal O_X)\simeq \omega_{X/C}^{\vee}$ (in characteristic $0$). Let $T=\Omega_{X/C, \rm{tors}}$ (torsion as $\mathcal O_X$-module). A local analysis shows easily that $T$ is an invertible sheaf over the verticla divisor $D:=\sum_{s\in C} D_s$, where <code>$D_s=X_s-(X_s)_{\mathrm{red}}$</code> (here again we need $k$ of characteristic $0$) and we have an exact sequence $$0 \to T\to \Omega_{X/C}\to \omega_{X/C}(-D) \to S \to 0$$<br> with $S$ of finite length. Thus the $\mathcal O_C$-torsion part of $f_*\Omega_{X/C}$ is exactly $f_*T$. </p> <p>At the generic fiber $\theta$ is the Kodaira-Spencer map. It is non-trivial when $f$ is non-isotrivial, and it is injective if moreover the generic fiber has genus $1$. So under these hypothesis, we have $$0 \to \Omega_{C/k}\to f_{*}\Omega_{X/k} \to f_{*}T\to 0.$$ Therefore the canonical map $H^0(C, \Omega_{C/k})\to H^0(X, \Omega_{X/k})$ is an isomorphism if<br> $$H^0(X, \Omega_{X/C, \mathrm{tors}})=H^0(C, f_*T)=0.$$ In a small neighborhood of a non-multiple fiber $X_s$, one can show that $H^0(f_*T)=0$. Otherwise (especially when $X_s$ is irreducible but not reduced) I don't know. </p> http://mathoverflow.net/questions/121297/what-kind-of-subset-is-specr-p-in-specr/121452#121452 Answer by Qing Liu for What kind of subset is Spec(R_P) in Spec(R)? Qing Liu 2013-02-11T08:52:15Z 2013-02-11T08:52:15Z <p>As already said, in general the image is not open and even not constructible. But it is pro-constructible (that is, locally an intersection of locally constructible subsets, see EGA IV.1.9.4, and 1.9.5(ix) because $X:=\mathrm{Spec}(R)$ is affine hence quasi-compact). </p> <p>Denote this image by $S$. This is a subset of $X$ having exactly one closed point and which is stable by generization as pointed out by Martin. I claim that these two properties characterize all possibles images $S$ when $X$ is noetherian (this hypothesis could be weakened if we notice that $S$ is always quasi-compact, but I don't know how exactly). Here $X$ needs not be affine. </p> <p>Indeed, let $s$ be the unique closed point of $S$. Let us show $S$ is the intersection of all open neighborhoods of $s$ in $X$ (then it is easy to show $S$ is the image of $\mathrm{Spec}(O_{X,s})$). Let $U$ be any open neighborhood of $s$ in $X$. Then $S\cap (X\setminus U)$ is noetherian hence admits a closed point if non-empty. But this would be a closed point of $S$ different from $s$. So $S\cap (X\setminus U)=\emptyset$ and $S\subseteq U$. Conversely, any point $x\in X$ in the intersecion of all $U\ni s$ is a generization of $s$ (otherwise the complementary of <code>$\overline{\{ x\}}$</code> is an open neighborhood of $s$ not containing $x$). </p> http://mathoverflow.net/questions/117670/finite-extension-of-projective-space/117739#117739 Answer by Qing Liu for Finite extension of projective space Qing Liu 2012-12-31T16:19:46Z 2012-12-31T16:19:46Z <p>Let $X$ be any projective scheme of dimension $n$ over an arbitrary field $k$. Then there exists a finite surjective morphism from $X$ to $\mathbb P^n_k$. </p> <p>Embed $X$ in some $P=\mathbb P^N_k$. By the homogeneous prime avoidance lemma, there exists a hypersurface $H_0$ of $P$ which doesn't contain any generic point of $X$. Then $\dim (X\cap H_0)=n-1$. Similarly there exists a hypersurface $H_1$ of $P$ which doesn't contain any generic point $X\cap H_0$. We have $\dim (X\cap H_0\cap H_1)=n-2$. Repeating the argument we find $n+1$ hypersurfaces $H_0, \dots, H_n$ such that $$X\cap H_0\cap \dots \cap H_n=\emptyset.$$ Each $H_i$ is defined by a homogeneous polynomial $F_i$. Remplacing $F_i$ with some positive power (this doesn't change the property of the intersection being empty), we can suppose they all have the same degree $d$. These $n+1$ sections of $O_P(d)$ don't have commun zeros in $X$, so they define a morphism $f: X\to \mathbb P^n_k$. </p> <p>It remains to show $f$ is finite. I just repeat the argument from Lemma 3 in Kedlaya: <i>More étale covers of affine spaces in positive characteristics</i>, J. Alg. Geometry (2005): let $z\in \mathbb P^n_k$. There exists $i\le n$ such that $z\in P\setminus H_i$. So the fiber $X_z$ is projective over $k(z)$ and also affine because it is closed in the affine scheme $P\setminus H_i$. This implies that $X_z$ is finite. So $f$ is quasi-finite and projective, so it is finite. It is surjective because $\dim X=n$. </p> http://mathoverflow.net/questions/112668/minimal-number-of-generators-for-an/112723#112723 Answer by Qing Liu for Minimal number of generators for $A^n$ Qing Liu 2012-11-17T22:03:21Z 2012-11-18T22:51:49Z <p>You already completely solved the question over fields, noetherian artinian rings and $\mathbb Z$. </p> <p>Let $A$ be any commutative unitary ring. Then the maximum $f_A(n)$ of all $e_k(n)$ when $k$ runs the residue fields of $A$ (at maximal ideals) satisfies clearly $e_A(n)\ge f_A(n)$ by your (4). </p> <blockquote> <p>Suppose $A$ is noetherian of dimension $d$, then <code>$$f_A(n) \le e_A(n) \le \max \{ d+1, f_A(n)\}.$$</code></p> </blockquote> <p>Proof. Let <code>$m=\max \{ d+1, f_A(n)\}$</code>. We want to show that the affine space $\mathbb A^m$ over $A$ contains $n$ disjoint sections. Let $r\le n-1$ be such that $\mathbb A^m$ contains $r$ disjoint sections. We are going to show that $\mathbb A^m$ contains one more section disjoint from the previous one. This will prove the claim.</p> <p>Let $T$ be the union of $r$ sections. For every residue field $k$ of $A$, $\mathbb A^m_k$ contains at least $r+1$ rational points. In particular, $T$ doesn't contain $\mathbb A^m_k(k)$. By hypothesis, we also have $\dim T=\dim A&lt; m$. By Proposition 1.10 of <a href="http://www.math.u-bordeaux1.fr/~qliu/articles/hypersurfaces.pdf" rel="nofollow">this preprint</a>, there is a section in $\mathbb A^m$ disjoint from $T$ and we are done.</p> <p><b>Edit</b> (Remove generalization to non-noetherian rings). </p> <p><b>Remark</b>. Let $A$ be any finite dimensional noetherian ring. If $A$ has a finite residue field, there exists $q$ such that $e_A(n)$ coincides asymptotically with $\lceil \log_q n \rceil$. It is enough to take for $q$ the smallest cardinality of the finite residue fields of $A$. If all residue fields of $A$ are infinite, then $e_A(n)$ is bounded hence asymptotically constant (because it is increasing). It would be interesting to decide whether these properties hold without noetherian and finite-dimensional hypothesis.</p> http://mathoverflow.net/questions/112781/about-the-definition-of-flat-morphism-flat-sheaf/112782#112782 Answer by Qing Liu for About the Definition of Flat Morphism (Flat Sheaf) Qing Liu 2012-11-18T20:20:19Z 2012-11-18T20:20:19Z <p>The answer to the first part of your question is no. See <a href="http://mathoverflow.net/questions/65267/" rel="nofollow">this thread</a>. </p> <p>However, in the case of finite (or more generally affine) morphism, $\mathcal F$ is flat over $Y$ if and only if $f_*\mathcal F$ is flat over $Y$. This is because is $\phi: A\to B$ is a ring homomorphism and if $M$ is a $B$-module, then $M$ if flat over $A$ if and only if for any prime ideal $\mathfrak q$ of $B$, $M_\mathfrak q$ is flat over $A_{\mathfrak p}$ where $\mathfrak p=\phi^{-1}(\mathfrak q)$. </p> http://mathoverflow.net/questions/112725/genus-computation/112727#112727 Answer by Qing Liu for Genus computation Qing Liu 2012-11-17T22:19:01Z 2012-11-17T22:19:01Z <p>I would use the double cover $C\to \mathbb P^1$ induced by the rational function $x$ and compute the genus with Riemann-Hurwitz as you do. </p> <p>Over the finite part $\mathrm{Spec} k[x]$, the cover is given by $k[x][y]/(y^2-f(x))$. Over the part containning $x=+\infty$, the cover is given by $k[1/x][y/x^d]$, where $d=[(\deg f+1)/2]$, with the equation $$(\frac{y}{x^d})^2=\frac{f(x)}{x^{2d}}\in k[1/x].$$ </p> http://mathoverflow.net/questions/112085/finitely-affine-morphisms-cohomological-dimension-of-schemes/112325#112325 Answer by Qing Liu for Finitely-affine morphisms; cohomological dimension of schemes Qing Liu 2012-11-13T23:53:02Z 2012-11-14T08:43:57Z <p>"This is too long to be a comment". </p> <p>A way to construct examples $X\to U$ with a given $n$ is to take a (quasi-)projective scheme $X\to U$ whose fibers have dimension $\le n-1$. Indeed, let $s\in U$. There are $n$ hypersurfaces in $X_s$ with empty intersection, so $X_s$ is covered by the $n$ complements of hypersurfaces (more care are needed if $X_s$ is not projective, anyway, the exercise in Hartshorne you refered to is done this way). Now lift these hypersurfaces to hypersurfaces over $O_{U,s}$, then the same arguments show that $X\times_U \mathrm{Spec}(O_{U,s})$ is covered by $n$ affine open subsets. By standard arguments, this will hold above an affine open neighborhood of $s$. So $X\to U$ satisfies your hypothesis. </p> <p>Now in a very special case, $X$ is actually covered by $n$ affine open subsets. Namely, if $U$ is the spectrum of a ring of integers or if $U$ is a regular curve over a finite field, then any projective $X\to U$ with fiber dimensions $\le n-1$ admits a finite morphism $\pi: X\to \mathbb P^{n-1}_U$. This is proved independently in a preprint of Chinburg, Moret-Bailly, Pappas, Martin Taylor, and a preprint of Gabber, Lorenzini and myself. As $\mathbb P^{n-1}_U$ is covered by $n$ affine open subsets, the same is true for $X$. </p> <p><b>Edit</b> It would be interesting to see whether any projective curve over a Dedekind domain can be covered by two (or more, but absolutely bounded number of) affine open subsets. </p> http://mathoverflow.net/questions/112244/uniform-bound-of-the-number-of-generators-of-prime-ideals/112328#112328 Answer by Qing Liu for uniform bound of the number of generators of prime ideals Qing Liu 2012-11-14T00:08:54Z 2012-11-14T00:08:54Z <p>The answer to Question 3 is yes if $R$ is excellent (it is enough that the normalization of $R$ is finite over $R$). Indeed the normal locus of $\mathrm{Spec}(R)$ is then open, so there are only finitely many $\mathfrak p$ of height $1$ with non-normal $R_{\mathfrak p}$. The other prime ideals are either $0, \mathfrak m$ or height $1$ with normal (hence regular) $R_{\mathfrak p}$.</p> http://mathoverflow.net/questions/112270/smooth-morphism-on-schemes/112324#112324 Answer by Qing Liu for smooth morphism on schemes Qing Liu 2012-11-13T23:20:19Z 2012-11-13T23:20:19Z <p>If you just want $S'$ with the property that $C\to S'$ has connected fibers, then $S'=S$ suits perfectly as said Matthieu in the comments. </p> <p>In general, let $f : C\to S$ be a proper surjective morphism of noetherian schemes. Usually, the Stein factorization refers to $S'=\mathrm{Spec}(f_*O_X)$ (it has the advantage to be canonical). This $S'$ is not always isomorphic to $S$: consider $S'\to S$ the normalization morphism of an integral curve with a cusp over $\mathbb C$ and $C=S'\times_{\mathbb C} \mathbb P^1_{\mathbb C}$. However it is true that $S'\to S$ is an isomorphism on an open subset under mild hypothesis. </p> <blockquote> <p>(a) If $S$ is reduced and the fiber of $f$ at a generic point $\xi$ of $S$ is geometrically reduced and geometrically connected, then $S'\to S$ is an isomorphism above an open neighborhood of $\xi$. </p> </blockquote> <p>Proof: As the question is local on $S$, we can suppose $S=\mathrm{Spec}(R)$ is affine. As $S'\to S$ is finite, $S'=\mathrm{Spec}(R')$. Because $S$ is reduced at $\xi$, $O_{S,\xi}$ is equal to the residue field $k(\xi)$ of $S$ at $\xi$. So the canonical map $R\to O_{S,\xi}=k(\xi)$ is a localization hence flat and we have<br> $$R'\otimes_R O_{S,\xi}= H^0(C, O_C)\otimes_R k(\xi)=H^0(C_{\xi}, O_{C_{\xi}})=k(\xi)=O_{S,\xi}.$$ So the finite morphism $S'\to S$ is an isomorphism when localized at $\xi$. By standard arguments, this isomorphism propagates above an open neighborhood of $\xi$. </p> <blockquote> <p>(b) Suppose $f$ is flat and some geometric fiber $C_{\bar{s}}$ is connected and reduced. Then $S'\to S$ is an isomorphism above an open neighborhood of $s$. </p> </blockquote> <p>Proof. We have $H^0(C_s, O_{C_s})=k(s)$. Then the statement is just EGA III, 7.8.8.</p> <p>As a corollary of (b): </p> <blockquote> <p>(c) If $f$ is flat with reduced and connected geometric fibers, then $S'\to S$ is an isomorphism. </p> </blockquote> http://mathoverflow.net/questions/112217/locally-free-sheaves-trivial-on-fibers-of-a-flat-projective-morphism/112227#112227 Answer by Qing Liu for Locally free sheaves trivial on fibers of a flat projective morphism Qing Liu 2012-11-12T22:02:01Z 2012-11-13T09:05:57Z <p>Let me first show that the canonical morphism $O_Y\to \pi_{\star} O_X$ is an isomorphism under OP's hypothesis <b>plus the assumption</b> that the generic fiber of $X\to Y$ is reduced (normality as in my previous edit is not needed). For any affine open subset $V$ of $Y$, the canonical map $O_Y(V)\to K:=K(Y)$ (field of rational functions) is injective. By flatness of $\pi$, the middle canonical morphism $$(\pi_\star O_X)(V)=\pi_\star O_X\otimes O_Y(V)\to \pi_\star O_X\otimes K=H^0(X_K, O_{X_K})=K$$<br> is injective. So $\pi_\star O_X(V)$ is a finite sub-algebra of $K$ containning $O_Y(V)$. As $Y$ is normal because smooth, $O_Y(V)\to (\pi_\star O_X)(V)$ is an isomorphism. This proves $O_Y\simeq \pi_\star O_X$. </p> <p>As $H^1(X_y, O_{X_y})=0$ for all $y\in Y$, the canonical morphism $\pi_*O_X\otimes k(y) \to H^0(X_y, O_{X_y})$ is an isomorphism for all $y\in Y$ (Mumford, Abelian varieties, p. 53, Cor. 3). Hence $H^0(X_y, O_{X_y})=\mathbb C$ for all $y$. </p> <p>Now let $r$ be the rank of $E$. Then for all $y\in Y$, $H^0(X_y, E_y)\simeq H^0(X_y, O_{X_y})^r\simeq \mathbb C^r$. Hence $\pi_{\star}E$ is locally free of rank $r$ (op. cit., p. 50-51, Cor. 2). This implies that $\pi^*\pi_\star E\simeq E$ as in Sándor Kovács's answer. </p> http://mathoverflow.net/questions/112158/surjective-and-injective-criteria-via-hilbert-polynomials/112160#112160 Answer by Qing Liu for Surjective and injective criteria via Hilbert polynomials Qing Liu 2012-11-12T10:11:49Z 2012-11-12T10:11:49Z <p>No. For the first question, let $X$ be a point and $\mathcal M=\mathbb C^2$ (vector space) or any $X$ and $\mathcal M$ be a vector space of dimension $\ge 2$ supported only at $x$. </p> <p>For the second question, consider a Riemann surface $X$ and $\mathcal F$ a line bundle. Then the Hilbert polynomial of $\mathcal F$ is that of $\mathcal O_X$ plus the degree of $\mathcal F$, but $\mathcal F$ can inject into $\mathcal O_X$ only if its degree is not positive. </p> http://mathoverflow.net/questions/111925/curve-through-a-point-on-a-variety/112087#112087 Answer by Qing Liu for Curve through a point on a variety Qing Liu 2012-11-11T16:44:36Z 2012-11-11T22:12:21Z <p>Your idea is good. Let $X'\to X$ be the blowup along $x$. Then $X'$ is projective, geometrically integral and of dimension $\dim X>1$. Embed $X'$ in some $\mathbb P^n_k$. </p> <p>When $k$ is infinite, by Jouanolou, "Théorèmes de Bertini et applications" (Progress in Maths), Corollaire 6.11 (2)+(3), there exists a hyperplane $H$ such that $H\cap X'$ is geometrically integral. </p> <p>When $k$ is finite, the existence of such a <b>hypersurface</b> $H$ is proved in Poonen "Bertini theorem over finite fields", Ann. Math. (2000), Proposition 2.7. </p> <p>Now the image of $H\cap X'$ in $X$ is a geometrically integral closed subscheme of $X$ passing throught $x$ of dimension $&lt;\dim X$. By induction we find a geometrically integral curve in $X$ passing through $x$. </p> <p><b>Edit</b> In fact through any closed finite subset $Z$ of $X$, it passes a geometrically integral curve in $X$. The proof is the same, we just blowup $X$ along $Z$ instead of $x$. </p> http://mathoverflow.net/questions/111332/projectivity-in-flat-families/111359#111359 Answer by Qing Liu for Projectivity in flat families Qing Liu 2012-11-03T10:29:26Z 2012-11-03T10:29:26Z <p>For question 2: Let $Y\to S$ be a proper morphism with $S$ integral and noetherian (for simplicity) and $Y$ irreducible with non-empty projective generic fiber $Y_K$ (where $K$ is the field of rational functions on $S$). Then there exists a dense open subset $U$ of $S$ such that $Y_U\to U$ is projective: </p> <p>Let $Y_K\to \mathbb P^N_K$ be a closed immersion. Let $Z$ be the scheme-theoretical closure of the image $Y_K$ in $\mathbb P^N_K$. Then we have a birational map $f$ from $Y$ to $Z$. The domain of definition $\Omega$ of $f$ is open and contains the generic fiber $Y_K$. So $\Omega$ contains $Y_V$ for some dense open subset $V$ of $S$ and we have a birational morphism $Y_V\to Z_V$. </p> <p>Applying the same reasonning to this morphism or to the inverse birational map, we see that there exists a dense open subset $U$ of $V$ such that $Y_U\to Z_U$ is an isomorphism. So $Y_U\to U$ is projective. </p> <p>Now for all $s\in U$, $Y_s$ is projective over the residue field $k(s)$ of $S$ at $s$. </p> http://mathoverflow.net/questions/110359/do-all-curves-have-neron-models/110592#110592 Answer by Qing Liu for Do all curves have Néron models Qing Liu 2012-10-24T22:47:20Z 2012-10-24T22:55:50Z <p>Here are some observations. I include the case $g=1$ (even if $X$ has no rational point). Denote by $\hat{\mathcal X}$ the (proper) minimal regular model of $X$ over the $O_K$ and let $\mathcal X$ be the smooth locus of $\hat{\mathcal X}$. </p> <ul> <li><p>(1) If the Néron model exists, it is equal to the smooth locus $\mathcal X$ of the minimal regular model. </p></li> <li><p>(2) If the fibers of $\mathcal X$ over $O_K$ have no rational irreducible component (e.g. if $X$ has good reduction), then $\mathcal X$ is the Néron model of $X$. </p></li> <li><p>(3) (localization) If Néron models exist over DVRs, then they exist over any Dedekind domain. </p></li> <li><p>(4) (base change) Let $R$ be a DVR. Let $R'/R$ be an extension of DVR such that an uniformizing element of $R$ is also an uniformizing element of $R'$ and such that the residue extension is separable (e.g. $R'$ can be the completion of a strict henselization of $R$). If the Néron model exists over $R'$, then the Néron model exists over $R$. </p></li> <li><p>(5) You were right to not include the case $g=0$. The projective line doesn't have Néron model. </p></li> <li><p>(6) Let $Y$ be a smooth scheme over a noetherian regular scheme $S$, let $Z$ be a regular scheme, flat and of finite type over $S$ and let $f: Y\to Z$ be a morphism. Then $f(Y)$ is contained in the smooth locus of $Z/S$. In particular, the canonical map ${\mathcal X}'(O_K)\to X(K)$ is bijective if $\mathcal X'$ is the smooth locus of (any) proper regular model of $X$. </p></li> <li><p>(7) If $g=1$, then $\mathcal X$ is the Néron model of $X$. </p></li> </ul> <p>Proof. Sorry I can't give all details by lack of energy and because it would be pretty unreadable in MO. </p> <p>(1) Let $\mathcal N$ be the Néron model. Embedd it in a proper flat model, solve its singularity without touching to the regular locus (which contains $\mathcal N$). Then we get a proper regular model $\hat{\mathcal N}$ containing $\mathcal N$ as an open subset. The identity on $X$ extends to morphism $\hat{\mathcal N}\to \hat{\mathcal X}$. By (6), this morphism induces a morphism $\mathcal N\to \mathcal X$. Then $\mathcal X$ satisfies the universal Néron mapping property. By the uniqueness of Néron model, we get $\mathcal N\simeq \mathcal X$. </p> <p>(2) Let $\mathcal Y -\to \mathcal X$ be a rational map defined over $K$ with $\mathcal Y$ smooth (regular is enough). The projection $p: \Gamma\to \mathcal Y$ is birational. Let $y\in Y$ and suppose $\Gamma_p$ is not finite. By a theorem of Abhyankar (the base scheme is excellent here, otherwise, localize and pass to the completion and use (4)), the components $E$ of $\Gamma_p$ are uniruled. But $E\to \mathcal X$ is a closed immersion, so $E$ is a rational curve in a close fiber of $\mathcal X$. Contradiction. Thus $p$ is quasi-finite biratonal and surjective. As $\mathcal Y$ is normal, $p$ is an isomorphism by Zariski's Main Theorem and the rational map we consider is actually defined everywhere. So $\mathcal X$ is the Néron model. </p> <p>(3) The curve $X$ has good reduction away from finite many places. Using (2) for good reduction places and by gluing with Néron models over bad reduction places, we get a global Néron model over $O_K$. </p> <p>(4) First the formation of the minimal regular model (and its smooth locus) is compatible with such base change. So if $\mathcal X\otimes R'$ satisfies the universal Néron mapping property over $R'$, then so does $\mathcal X$ over $R$ by faithfully flat descent for the definition domain of rational maps. </p> <p>(5) Fix a model $\mathbb P^1_{O_K}$ of $\mathbb P^1_K$. They are plenty of endomorphisms of the generic fiber which don't extend to $\mathbb P^1_{O_K}$ (e.g. $[x,y]\mapsto [x, py]$). This shows that $\mathbb P^1_K$ has no proper smooth Néron model. The general case can be proved similarly with some extra works. </p> <p>(6) Sketch: Consider $Y\times_S Z\to Y$. It is enough to show that its sections have images in the smooth locus (over $Y$), then use descent of smoothness (easy). The left hand side is regular because it is smooth over the regular scheme $Z$, and the right hand side si regular because it is smooth over the regular scheme $S$. So we can reduced to the case of flat morphism of finite type $W\to Y$ between two regular schemes. Let $y\in Y$ and let $w\in W$ be its image by a section $Y\to W$. Then $O_{W,w}\to O_{Y,y}$ is a surjective map of regular local rings. Its kernel is generated by {$t_1, \dots, t_d$}, a part of a system of coordinates of $O_{W,w}$. So the maximal ideal $m_w$ of $O_{W,w}$ is generated by $t_1, \dots, t_d$ and $m_y$. Thus the maximal ideal of $O_{W_y, w}$ is generated by the images of $t_1, \dots, t_d$. The flatness of $W\to Y$ implies that $W_y$ has dimension $d$ at $w$. So $W_y$ is regular at $w$. It is in fact smooth because $w$ is a rational point of $W_y/k(y)$. </p> <p>Another proof is to use $\Omega^1_{W/Y}$ and the fact that the image of a section is locally complete intersection. </p> <p>Application: if $\hat{\mathcal X'}$ is a proper regular model over $O_K$, by the valuative criterion, $\hat{\mathcal X'}\to X(K)$ is bijective. But we just saw that the LHS is $\mathcal X'(O_K)$. </p> <p>(7) We can work over a DVR $R$. If there exists a smooth $R$-scheme $\mathcal Y$ with non-empty special fiber and a morphism $\mathcal Y_K\to X$, then $\mathcal Y_K$ has a point in an étale extension of $R$. So $X$ has a point in an étale extension. By (4), we can thus suppose $X(K)\ne\emptyset$. So it is an elliptic curve, and Néron showed that $\mathcal X$ is the Néron model. If such $\mathcal Y$ doesn't exists, then $\mathcal X$ trivially satisfies the Néron mapping property. </p> http://mathoverflow.net/questions/109903/embedding-of-a-scheme-into-a-regular-scheme/109982#109982 Answer by Qing Liu for Embedding of a scheme into a regular scheme Qing Liu 2012-10-18T07:20:42Z 2012-10-18T07:40:44Z <p>A noetherian regular scheme is universally catenary (Matsumura, 14.B, 16.D), so any subscheme of a regular scheme is universally catenary. </p> <p>But there are affine noetherian schemes (integral of dimension $2$) which are not universally catenary. See an example of Nagata in Matsumura, 14.E, or a slightly simpler one in EGA IV.5.6.11 (it consists in identifying two points of respective codimension 1 and 2 in an affine regular scheme of dimension 2). </p> http://mathoverflow.net/questions/106785/reduction-of-elliptic-curves/106787#106787 Answer by Qing Liu for reduction of elliptic curves Qing Liu 2012-09-10T08:01:28Z 2012-09-10T08:01:28Z <p>The Néron model of $X$ is the smooth locus of the minimal regular model of $X$ (see Bosch-Lütkebohmert-Raynaud: Néron models, §1.5). The equivalence is then clear using the classification of Kodaira-Néron of the types of reduction of the minimal regular model. </p> <p>Note that if $g(X)\ge 2$, then it is also true that $X$ has semi-stable reduction if and only if its Jacobian has semi-abelian reduction (Deligne-Mumford, based on Raynaud's description of Néron models of Jacobians). </p> <p>If $g(X)=1$ but $X$ doesn't have a rational point, then the statement is no longer true. But one can show that the Jacobian of $X$ has semi-abelian reduction if and only if the type of the reduction of $X$ is a multiple of $I_n$, $n\ge 0$ (use the fact that $X$ covers its Jacobian. The additive reduction case is treated in a paper of Lorenzini, Raynaud and myself in 2004). </p> <p>You don't need the completeness hypothesis on the base DVR.</p> http://mathoverflow.net/questions/106568/cohen-macaulay-sheaves-which-are-not-locally-free/106574#106574 Answer by Qing Liu for Cohen-Macaulay sheaves which are not locally free Qing Liu 2012-09-07T06:43:26Z 2012-09-07T06:53:35Z <p>Take any noetherian local domain $A$ of dimension $1$ and a finitely generated torsion-free $A$-module $M$. Then $\mathrm{depth}_A M=1=\dim A$. If $A$ is not integrally closed, then you have plenty of such modules which are not free (e.g. let $\alpha\in \mathrm{Frac}(A) \setminus A$ be integral over $A$, let $M=A[\alpha]$).</p> <p>Proof of $M$ not being free. If it were free, then it would be of rank $1$ because two elements in $\mathrm{Frac}(A)$ are always $A$-linearly dependent. So $A[\alpha]=\beta A$ and then $\beta=1/a$ with $a\in A$ and $\beta$ is integral over $A$. This easily implies that $a\in A^{\star}$, thus $A[\alpha]=A$). </p> <p>If $X$ is regular, then yes, locally free is equivalent to Cohen-Macaulay (see EGA IV, 6.1.5). In general I don't know a nice criterion ($X$ normal will not be enough). </p> http://mathoverflow.net/questions/101915/irreducible-family-of-relative-effective-divisors-of-a-smooth-morphism/106130#106130 Answer by Qing Liu for Irreducible "family" of relative effective divisors of a smooth morphism Qing Liu 2012-09-01T20:23:25Z 2012-09-01T21:00:53Z <p>In fact both sets are constructible in $Y$. </p> <p>Suppose for simplicity that $X$ is connected. Then the dimension of the (non-empty) fibers of $X\to Y$ is constant (EGA IV.12.1.1(i), and flatness is enough), denote it by $d$. Let $Y'$ be the (integral) image of $Z$ in $Y$. We can replace $X\to Y$ by $X\times_Y Y'\to Y'$ and suppose that $Z\to Y$ is surjective. Let $e$ be the dimension of the generic fiber of $Z\to Y$. </p> <p><b>Special case: $e=d-1$.</b> By Chevalley's theorem (EGA IV.13.1.1), for any $y\in Y$, the irreducible components of $Z_y$ all have dimension $\ge d-1$. By hypothesis, $Z_y\ne X_y$, hence $\dim Z_y\le d-1$. Therefore $\mathrm{Pic}_\pi(Z) = Y$ (and it is closed in $Y$ if $Z$ doesn't dominate $Y$). </p> <p><b>General case.</b> The set of $x\in Z$ such that $\dim_x Z_{\pi(x)}\le d-2$ is open in $Z$ (EGA IV.13.1.3). The complementary of the image by $\pi$ of this open subset is constructible and is your ${Pic}_{\pi}(Z)$. </p> <p>In case $Z\to Y$ is flat, then $\mathrm{Pic}_\pi(Z)$ is closed by openess of $\pi|_Z$.</p> <p>For the second question, if $F$ denotes the set of $x\in Z$ such that all associated components of $Z_{\pi(x)}$ passing through $x$ have dimension $\ge d-1$, then you are considering $\pi(X\setminus F)$. By EGA, IV.9.9.2(iii), $F$ is constructible, so your set is constructible. </p> <p>If $Z\to Y$ is moreover flat, then $F$ is open by EGA, IV.12.1.1(i) (I learn recently this reference from an anonymous referee.), hence your set $\pi(X\setminus F)$ is in fact closed.</p> http://mathoverflow.net/questions/89152/does-smoothness-descend-along-flat-morphisms/106071#106071 Answer by Qing Liu for Does smoothness descend along flat morphisms? Qing Liu 2012-08-31T21:08:20Z 2012-08-31T21:08:20Z <p>The answer to the first question is yes and (probably you known) is a consequence of the second question. </p> <p>Suppose $f : X\to Y$ is a flat morphism of schemes over $S$ with $X$ smooth. Let $y=f(x)$ and let $s$ be the image of $x$ in $S$. By the positive answer to the second question, the geometric fiber $Y_{\bar{s}}$ is regular. So it only remains to see that$Y$ is flat over $S$ at $y$. </p> <p>Consider the homomorphisms of local rings $$O_{S,s} \to O_{Y, y}\to O_{X,x}$$<br> The second is flat by hypothesis, hence faithfully flat (because we deal with local rings), and the composition is flat by the smoothness of $X\to S$. So the first one is flat (easily seen using definition of flatness). </p> http://mathoverflow.net/questions/103686/cartier-divisors-on-singular-curves/105921#105921 Answer by Qing Liu for Cartier divisors on singular curves Qing Liu 2012-08-30T09:53:23Z 2012-08-30T09:53:23Z <p>Here is a part of an email I sent you because I couldn't log in MO. </p> <p>The answer to (3) is yes: there is a Cartier divisor $D$ on $X$ such that the associated Weil divisor ($0$-cycle for a curve) is $[P]$. This idea is just to consider the normalization $f : Y\to X$. Let $Q\in Y$ be a point lying over $P$. Then $[Q]\sim Z$ (rational equivalence) for some $0$-cycle $Z$ on $Y$ with support disjoint from $f^{-1}(P)$ (use e.g. Riemann-Roch). By pushforward, $[P]=f_*[Q] \sim f_*Z$. By definition of rational equivalence, this means that $[P]=f_*Z+[\mathrm{div}(h)]$ for some rational function $h$ on $X$. As the support of $f_*Z$ is disjoint from $P$, this means that in an open neighborhood $V$ of $P$, $[P]$ coincides with $[\mathrm{div}(h)]$. Define a Cartier divisor $D$ by ${D|}_{V}=\mathrm{div}(h)$ and $D|_{X\setminus { P}}=1$. Then $[P]$ equals to (the $0$-cycle associated to) $D$. The idea of taking normalization can be found, say, in a paper of Colliot-Thélène (Inv. Math. 2005, p. 599).</p> <p>This works over any algebraically closed field. If the base field is not necessarily algebraically closed, let $Q_1, \cdots, Q_r$ be the points of $f^{-1}(P)$, let $n$ be the gcd of the $[k(Q_i): k(P)]$. Then the above argument shows that $n[P]$ is associated to a Cartier divisor on $X$. And $n$ is the smallest possible.</p> <p>In higher dimension, as you already known, this conclusion is false. However the following weaker result holds. Let $X$ be a noetherian scheme, let $E$ be a prime cycle. Then for some positive integr $n$, $nE$ is the cycle associated to a Cartier divisor (on an integral closed subscheme of $X$ containing $E$) <b>in an open neighborhood</b> of the generic point $P$ of $E$. The integer $n$ can be controlled by some invariants related to the singularity at $P$ (can take $n=1$ is $P$ is a regular point of $X$, but this is not necessary as we saw in the case of curves). See Theorem 4.5, §5, and Theorem 6.5 in <a href="http://www.math.u-bordeaux1.fr/~qliu/articles/GLL1.pdf" rel="nofollow">this paper</a>.</p> http://mathoverflow.net/questions/105381/henselization-and-completion/105891#105891 Answer by Qing Liu for henselization and completion Qing Liu 2012-08-29T23:13:15Z 2012-08-30T08:58:17Z <p><b>Edit</b> Add noetherian hypothesis because $R$ might no be a subring of $\hat{R}$ otherwise. </p> <p>The answer is no as pointed out by quasi-coherent in the comments. But suppose $R$ is a discrete valuation ring and denote $K=\mathrm{Frac}(R)$, then </p> <blockquote> <p>$R^h=\hat{R}\cap K^{sep}$. </p> </blockquote> <p>There are several equivalent properties defining henselien local rings. I will use two of them: let $(A, m)$ be a local ring and $k=A/m$. Then $A$ is henselian iff </p> <p>(a) for any monic polynomial $f(X)\in A[X]$, any simple root of $\bar{f}(X)\in k[X]$ lifts to root of $f(X)$ in $A$. </p> <p>or equivalently</p> <p>(b) If $A\to A'$ is a local homomorphism, étale and with trivial residue extension, then $A'=A$. </p> <p>One can find a proof in Raynaud: Anneaux locaux henséliens, IV, §3. For discrete valuation rings, one can certainly find easier references. </p> <p>Now let us prove the claim above $R^h$. As $\hat{R}$ is henselian (Hensel Lemma), we have $R\subseteq \hat{R}$. </p> <p>(1) For any algebraic separable extension $F/K$ contained in $\hat{K}$, $B_F=\hat{R}\cap F$ is a DVR with ramification index $1$ over $R$ and trivial residue extension. This is easy. </p> <p>(2) Let $B=B_L$ where $L$ is the separable closure of $K$ in $\hat{K}$. Let us prove that $B$ is henselian using Property (a) above. Let $f(X)\in B[X]$ as in (a). Then any simple root of $\bar{f}(X)\in k[X]$ lifts to a root $\lambda\in \hat{K}$. It is automatically a simple root, so $\lambda$ is separable over $L$, hence $\lambda\in L\cap \hat{R}=B$. </p> <p>(3) Let $R\to R'$ be an extension to a henselian DVR. Let's us prove it factorizes throught $R\to B$, which will show that $B$ is a henselization of $R$. It is enough to prove this factorization for $B_F$ for any finite separable exension $F/K$ contained in $\hat{K}$. As $B_F$ is an étale extension of $R$, $B_F\otimes_R R'$ is étale over $R'$, with a unique maximal ideal above the maximal ideal of $R'$ and trivial residue extension at this maximal ideal (because the quotient $B_F\otimes_R R'/(m')\simeq R'/m'$). By Property (b), this implies that $F\otimes_K K'$ has a direct factor equal to $K'$. Hence $F\subseteq K'$ (the minimal polynomial of a primitive element of $F$ has a root in $K'$) and $B_F\subseteq R'$. </p> <p><b>Remark</b> Parts (1)-(2) work for any local ring $R$ such that $\hat{R}$ is a domain and such that $R\to \hat{R}$ is injective. In particular $\hat{R}\cap K^{sep}$ is henselian. <s>(3) works for any valuation ring</s> But in general, I don't think (3) holds because $B_F$ has no reason to be étale over $R$. </p> http://mathoverflow.net/questions/9436/reduced-reduced-reduced-what-about-connected/16404#16404 Answer by Qing Liu for reduced ⊗ reduced = reduced; what about connected? Qing Liu 2010-02-25T15:33:38Z 2012-08-29T23:19:18Z <p>Over $\mathbb Z$, there are several possibilities. </p> <ul> <li>If both $A, B$ are torsion-free over $\mathbb Z$, then $A\otimes_{\mathbb Z} B$ is connected if and only if its generic fiber $(A\otimes_{\mathbb Z} \mathbb Q)\otimes_{\mathbb Q} (B\otimes_{\mathbb Z} \mathbb Q)$ is connected, and you are reduced to the case of algebras over a field. </li> </ul> <p><b>EDIT</b> the above claim is incorrect. See a partial answer <a href="http://math.stackexchange.com/questions/160853/" rel="nofollow">here</a>.</p> <ul> <li><p>If $A$ (or $B$) have positive characteristic, then $A_{\rm red}$ is an algebra over a product of finite fields, then so is $A\otimes_{\mathbb Z} B$. Again you are reduced to the case of algebras over a field (if at least two finite fields involve really, then the tensor product is not connected. </p></li> <li><p>The remainning case: if $A$ is noetherian, then ${\rm Spec}(A)$ has a flat part and a ''vertical part'': take the ideal $I$ of $A$ consisting in torsion elements. Then $A/I$ is flat and $mI=0$ for some non-zero integer $m$, and we have ${\rm Spec A}=V(I)\cup V(mA)$.<br> The same is true for $A\otimes B$. To have the connectedness, you want [<b>EDIT: it is enough </b> (but far to be necessary) that] the flat and vertical parts to be connected and meet each other. Sorry, I don't have a simple statement. </p></li> </ul> http://mathoverflow.net/questions/83626/is-the-normalisation-of-an-integral-noetherien-dimension-one-ring-a-finite-morphi/83647#83647 Answer by Qing Liu for Is the normalisation of an integral noetherien dimension one ring a finite morphism? Qing Liu 2011-12-16T18:08:23Z 2011-12-16T18:08:23Z <p>Let $R$ be any non-excellent DVR with field of fractions $K$, let $L/K$ be a finite extension such that the normalization $B$ of $R$ in $L$ is not finite over $R$. We have $L=K[a_1,...,a_n]$ for some $a_i\in B$. Consider $A=R[a_1,...,a_n]\subseteq B$. Then $B$ in the integral closure of $A$, but is not finite over $A$ (because $A$ is finite over $R$). </p> http://mathoverflow.net/questions/82232/does-each-finite-morphism-of-curves-have-a-model-whose-minimal-resolution-is-semi/82328#82328 Answer by Qing Liu for Does each finite morphism of curves have a model whose minimal resolution is semi-stable Qing Liu 2011-11-30T23:29:30Z 2011-11-30T23:29:30Z <p>If $X$ has no (potentially) good reduction, then the answer to your question is no. More precisely, there always exists a finite cover $Y\to X$ such that for any finite extension $L/K$, no regular semi-stable model of $Y_L$ dominates a regular semi-stable model of $X_L$. </p> <p>Suppose we are given a finite morphism of semi-stable models $\mathcal Y\to\mathcal X$ with $\mathcal Y$ regular but $\mathcal X$ is singular. I claim that in this case, no semi-stable regular model of $Y$ can dominate a semi-stable regular model of $X$ dominating $\mathcal X$. </p> <p>For simplicity, I will work over a strictly henselian DVR. Let $x_0$ be a singular point of $\mathcal X$ and let $y_0$ be a point of $\mathcal Y$ lying over $x_0$. Then $x_0$ is a double point in its fiber $\mathcal X_s$ and similarly for $y_0$. Let $\mathcal Y'$ be a regular semi-stable model dominating $\mathcal Y$ and dominating a desingularization of $\mathcal X$. As $\mathcal X'\to \mathcal X$ is not an isomorphism above $x_0$, $\mathcal Y'\to\mathcal Y$ is not an isomorphism above $y_0$. So some irreducible component $\Gamma$ of $\mathcal Y'_s$ must be contracted to $y_0$ in $\mathcal Y$. A smooth point $\mathcal Y'$ contained in $\Gamma$ lifts to a section of $\mathcal Y'$. This section is mapped to a section of $\mathcal Y$ passing through $y_0$. But as $\mathcal Y$ is regular, its sections are contained in its smooth locus. Contradiction because $y_0$ is not a smooth point. </p> <p>With some extra work, one can show that the same property holds over any finite extension $L/K$ (one has to desingularize first the singular points of $\mathcal Y_{\mathcal O_L}$.). </p> <p>Finally, for any $X$ with potentially bad reduction at a prime $\mathfrak p$ and for any integer $n\ge 2$ prime to $\mathfrak p$, there exists (after enlarging $K$) a cyclic étale cover $Y\to X$ of degree $n$ with the required property on the double points. See §6.3 and especially Prop. 6.6 in <a href="http://www.math.u-bordeaux.fr/~liu/articles/modcove.pdf" rel="nofollow">this paper</a>. </p> http://mathoverflow.net/questions/127908/reduction-types-of-elliptic-curves/128617#128617 Comment by Qing Liu Qing Liu 2013-04-24T17:11:58Z 2013-04-24T17:11:58Z Thanks Will for the correction ! And III$^\star$ is a quadratic twist of III. http://mathoverflow.net/questions/119881/absorbing-ramification-and-factoring-finite-flat-maps/119902#119902 Comment by Qing Liu Qing Liu 2013-02-26T15:19:54Z 2013-02-26T15:19:54Z Each $Y_i$ is given by the subextension $K(Y_i)$ of $K(X)/K(Y)$. As $K(X)$ is finite separable over $K(Y)$, there are only finitely many subextensions. http://mathoverflow.net/questions/119881/absorbing-ramification-and-factoring-finite-flat-maps Comment by Qing Liu Qing Liu 2013-02-26T15:17:18Z 2013-02-26T15:17:18Z (1) If there exists $Z\to Y$ faithfully flat such that $X\times_Y Z\to Z$ is &#233;tale, then by faithfully flat base change, $X\to Y$ is already &#233;tale ! http://mathoverflow.net/questions/122943/one-problem-related-with-singular-elliptic-curves Comment by Qing Liu Qing Liu 2013-02-26T10:22:43Z 2013-02-26T10:22:43Z Please post your question on math.stackexchange.com. It is not appropriate for this forum. By the way, the condition that $a_2$ has no square root in $K$ is useless. http://mathoverflow.net/questions/122819/the-locus-where-a-sheaf-is-supported-in-a-certain-dimension/122873#122873 Comment by Qing Liu Qing Liu 2013-02-25T17:20:59Z 2013-02-25T17:20:59Z Yes because in your special case $X\to T$ is proper. As for the reducedness (in the general case), apply your functor to all $U$ affine open in the (moduli scheme)$_{\mathrm{red}}$. http://mathoverflow.net/questions/122819/the-locus-where-a-sheaf-is-supported-in-a-certain-dimension/122873#122873 Comment by Qing Liu Qing Liu 2013-02-25T12:13:31Z 2013-02-25T12:13:31Z As the fiber dimension doesn't change by base change, if there is a scheme which represents your functor, it will be reduced with underlying space equal to the set of $t\in T$ such that $\dim Z_t\le r$. If the latter is locally closed, then endow it with the structure of a reduced subscheme of $T$ and you get a scheme representing your functor. Otherwise I don't know, probably your functor is then not representable. http://mathoverflow.net/questions/122162/formal-power-series-over-a-henselian-ring Comment by Qing Liu Qing Liu 2013-02-18T16:29:44Z 2013-02-18T16:29:44Z Dear kiseki, you should consider accepting answers you find helpful. http://mathoverflow.net/questions/121880/sections-of-the-cotangent-bundle-of-elliptic-surfaces/122087#122087 Comment by Qing Liu Qing Liu 2013-02-17T23:04:40Z 2013-02-17T23:04:40Z @rvarma: yes, if $\Omega_{X/C}$ is torsion free (equivalently, all fibers of $f$ are reduced), your isomorphism is correct. But this isomorphism holds under a slightly more general situation, it is enough that there is no multiple fibers (this is proved in the paper with T. Saito in any characteristic, but probably there is a simpler proof in characteristic $0$). http://mathoverflow.net/questions/121892/are-f-g-projective-modules-free-over-total-quotient-ring-of-a-reduced-non-noeth Comment by Qing Liu Qing Liu 2013-02-17T17:55:08Z 2013-02-17T17:55:08Z Fred: one can descend the module to a f.g. module $M_n$ over some $k[x_1,..,x_n]$. As $k[x_1,..,x_n,...]=k[x_1,...,x_n][x_{n+1},...]$ is faithfully flat over $k[x_1,..,x_n]$, $M_n$ is flat hence free. So the answer is yes (any number of indeterminates). http://mathoverflow.net/questions/120016/galois-action-on-special-fiber-of-a-stable-model/120037#120037 Comment by Qing Liu Qing Liu 2013-02-15T17:57:56Z 2013-02-15T17:57:56Z It is proved with full details in Mireille Deschamps's talk in Asteristisque vol. 86 (I guess this is [D] in Green-Matignon). http://mathoverflow.net/questions/121598/weierstrass-models-and-canonical-models Comment by Qing Liu Qing Liu 2013-02-15T17:14:07Z 2013-02-15T17:14:07Z continued: does not change when contracting $X$ to the canonical model $W$ (the latter has only rational singularities), this can be found in &quot;N&#233;ron models&quot;, Theorem 9.7/1. http://mathoverflow.net/questions/121598/weierstrass-models-and-canonical-models Comment by Qing Liu Qing Liu 2013-02-15T17:10:54Z 2013-02-15T17:10:54Z @nosr: the isomorphism is explained in S. Bloch: &quot;de Rham cohomology and conductors of curves&quot;, Duke Math. J. 54 (1987), Lemma 1.2(i) when the model $X$ is regular. If we contract $X$ to the canonical model, this does not change the generic fiber, and the $H^1$ of the closed fiber doesn't change either (by explicit computations, e.g. Dolgacev: &quot;On the purity of the degeneration loci of families of curves&quot;. Invent. Math. 8 (1969), Prop. 2.4). This can also be explained (at least when the curve has a rational point) by the fact that the neutral component of the N&#233;ron model of $Jac(X_{K})$... http://mathoverflow.net/questions/121598/weierstrass-models-and-canonical-models Comment by Qing Liu Qing Liu 2013-02-15T17:04:08Z 2013-02-15T17:04:08Z @TomPGR:The canonical model in higher genus is obtained by contracting some $(-2)$-rational curves, and all such configurations of rational curves is classified by M. Artin (see my book, 10.1.53). They all appear in Kodaira-N&#233;ron's classification for elliptic curves. So the computation of zeta function is the same as for elliptic curves. http://mathoverflow.net/questions/118260/do-regular-noetherian-schemes-of-dimension-one-only-have-finitely-many-etale-cove Comment by Qing Liu Qing Liu 2013-01-08T15:07:59Z 2013-01-08T15:07:59Z @Masse: pranavk's counterexample is dimension one and not affine. http://mathoverflow.net/questions/118329/stein-factorization-and-flatness Comment by Qing Liu Qing Liu 2013-01-08T15:03:19Z 2013-01-08T15:03:19Z @LMN: Tong's answer in the MO thread you linked to is a counterexample for you (take his $X_n$ which is proper faithfully flat over $R/\pi^nR$).