User j&#233;r&#244;me poineau - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T22:24:19Z http://mathoverflow.net/feeds/user/4069 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/131126/why-do-rigid-spaces-have-not-enough-points/131274#131274 Answer by Jérôme Poineau for Why do rigid spaces have "not enough points"? Jérôme Poineau 2013-05-20T21:25:28Z 2013-05-22T20:58:01Z <p>If you are familiar with Berkovich spaces, you can do the following construction. Let $X$ be an affinoid space of positive dimension and pick a point $x$ in $X$ that is not a rigid point. Consider the inclusion map $i\colon x \to X$. Then the sheaf $F = i_*\mathbb{Z}$ does the job. Since the space $X$ is Hausdorff, the point $x$ is closed and the sheaf $F$ has no section on the open set <code>$X\setminus \{x\}$</code>.</p> <p>Let me try to rewrite this example purely in terms of rigid geometry in a simple case: $X$ is the closed unit disc over an algebraically closed field $k$ and $x$ is the point at its boundary (the Gauss point). The previous sheaf may then be described as follows: for an affinoid domain $V$ of $X$, we have $F(V)=0$ if every connected component of $V$ is contained in an open unit disc (with any center) and $F(V)=\mathbb{Z}$ otherwise. </p> http://mathoverflow.net/questions/127622/tame-morphism-from-a-curve-to-mathbbp1 Tame morphism from a curve to $\mathbb{P}^1$ Jérôme Poineau 2013-04-15T14:02:53Z 2013-04-25T09:48:03Z <p>Let $k$ be an algebraically closed field of characteristic $p\ge 0$. Let $C$ be a smooth projective curve over $k$. Is it possible to find a map $C \to \mathbb{P}^1$ that is tamely ramified at every point of $C$, <i>i.e.</i> such that the ramification index at every point of $C$ is prime to $p$?</p> <p>A result of Fulton says that, if $k$ is (algebraically closed) of characteristic $p\ne 2$, then it is possible to find a morphism $C \to \mathbb{P}^1$ that is a simple cover: only double points may appear and at most one in every fiber. (This is theorem 8.1 in "Hurwitz schemes and the irreducibility of moduli of algebraic curves", Ann. of Math. 90, 1969. He says it is classical and dates back to Severi.)</p> <p>Fulton's result gives a positive answer for fields of characteristic $p\ne 2$. But what about characteristic 2? Does the result still hold? I would already be interested in answers in particular cases (elliptic curves for instance).</p> <p>EDIT: I added the hypothesis that the field is algebraically closed in order to focus on what I am really interested in. Still, I would also appreciate comments on how relevant this hypothesis is.</p> http://mathoverflow.net/questions/121187/reference-for-rigid-analytic-gaga/121215#121215 Answer by Jérôme Poineau for Reference for rigid analytic GAGA Jérôme Poineau 2013-02-08T16:26:52Z 2013-02-08T16:26:52Z <p>I am quite surprised by the attribution to Kiehl that you saw. Anyway, I think the result is due to Ursula Köpf (not only over a field $K$ but actually over an affinoid space): "Über eigentliche Familien algebraischer Varietäten über affinoiden Räumen", Schriftenreihe Univ. Münster, 2 Serie, Heft 7 (1974).</p> <p>Brian Conrad gave another proof as an application of his results of relative ampleness in the rigid analytic setting (see "Relative ampleness in rigid geometry", Ann. Inst. Fourier (Grenoble) 56 (2006), n° 4).</p> <p>I also learned a proof from Antoine Ducros in the setting of Berkovich spaces. I wrote in down in an appendix to my paper "Raccord sur les espaces de Berkovich", Algebra &amp; Number Theory 4 (2010), n° 3). It is very close to Serre's proof in the complex analytic setting and probably very close to Köpf's proof too, but I cannot say for sure since I never saw her paper.</p> http://mathoverflow.net/questions/105080/analytic-elements-in-non-archimedean-geometry Analytic elements in non-archimedean geometry Jérôme Poineau 2012-08-20T08:56:57Z 2012-08-23T16:24:52Z <p>Let $(k,|.|)$ be a complete non-archimedean valued field. Let $D$ be the open unit disc over $k$. (Anything I write could be adapted to the case of an open annulus.) The ring $\mathcal{O}(D)$ of analytic functions on $D$ admits an explicit description: it consists of the series $\sum_{n\ge 0} a_n T^n$ such that for any $r&lt;1$, the sequence <code>$(|a_n|r^n)_{n\ge 0}$</code> tends to 0. If such a function is bounded, its uniform norm is precisely $\sup_{n\ge 0}(|a_n|)$.</p> <p>For some purposes, it is convenient to consider smaller rings: the ring of bounded functions or the ring of analytic elements, for instance. Let me recall that this latter ring $\mathcal{H}(D)$ is defined as the completion of $k(T) \cap \mathcal{O}(D)$ (<i>i.e.</i> rational functions with no poles in $D$) for the uniform norm on the disc. Remark that this definition depends on the choice of a coordinate $T$ on $D$. Those functions have nice properties: for example, they only have a finite number of zeros on $D$. We refer to Gilles Christol's <a href="http://www.math.jussieu.fr/~christol/courspdf.pdf" rel="nofollow">book</a> (chapter I) for a more detailed account. </p> <p>Consider the disc $D$ inside the affine line $\mathbf{A}^1$. Let $X$ be an affinoid domain of $\mathbf{A}^1$ that contains $D$. One may check that any function on $X$ restricts to an analytic element on $D$, whatever the choice of the coordinate (because the approximation property actually holds for any function on $X$). </p> <p>The question is the following: if $D$ is embedded inside an arbitrary affinoid space $X$, do functions on $X$ still restrict to analytic elements? </p> <p>The issue is that the choice of a coordinate in the definition of $\mathcal{H}(D)$ makes it difficult to use outside the affine line. I would appreciate any idea that helps recognizing analytic elements on a disc inside an arbitrary curve. </p> http://mathoverflow.net/questions/105080/analytic-elements-in-non-archimedean-geometry/105336#105336 Answer by Jérôme Poineau for Analytic elements in non-archimedean geometry Jérôme Poineau 2012-08-23T16:24:52Z 2012-08-23T16:24:52Z <p>Vladimir Berkovich has indicated to me that the answer is no: on a general curve, functions need not restrict to analytic elements. Let me copy here his argument.</p> <p>The algebra of analytic elements $\mathcal{H}(D)$ can be defined as the completion of the inductive limit of the $k$-affinoid algebras $\mathcal{A}_X$ with respect to the supremum norm on $D$ taken over all affinoid domains $X$ in the affine line that contain $D$. It follows that the spectrum of $\mathcal{H}(D)$ is the closure of $D$, <i>i.e.</i> the union of $D$ with the Gauss point $x$, and the non-Archimedean field $\mathscr{H}'(x)$ associated to $x$ with respect to the Banach algebra $\mathcal{H}(D)$ coincides with the similar field $\mathscr{H}(x)$ of the Gauss point in the affine line. In particular, the residue field of $\mathscr{H}'(x)$ is the field of rational functions on the projective line over the residue field of $k$.</p> <p>Let now $Y$ be the analytification of the generic fiber of a smooth projective curve $Z$ over the ring of integers of $k$. The space $Y$ contains a unique "generic point" $y$, whose image under the reduction map is the generic point of the closed fiber of $Z$, and the complement of $y$ in $Y$ is a disjoint union of open unit discs (assume for simplicity that $k$ is algebraically closed). Pick up one of them $D$, and consider the similar Banach algebra $\mathcal{H}'(D)$ which is the completion of the intersection of $k(Z)$ and $\mathcal{O}(D)$ with respect to the supremum norm on $D$. It coincides with the completion of the inductive limit of the $k$-affinoid algebras $A_X$ with respect to the supremum norm on $D$ taken over all affinoid domains $X$ in $Y$ that contain $D$. The spectrum of $\mathcal{H}'(D)$ coincides with the closure of $D$ in $Y$, <i>i.e.</i> the union of $D$ with the point $y$, and the non-Archimedean field $\mathscr{H}'(y)$ associated to $y$ with respect to the Banach algebra $\mathcal{H}'(D)$ coincides with the similar field $\mathscr{H}(y)$ of $y$ with respect to $Y$. It follows that the residue field of $\mathscr{H}'(y)$ is the field of rational functions on the closed fiber of $Z$. If the genus of $Z$ is positive, then so is the genus of its closed fiber. This implies that the functions on $X$ (in $Y$) do not necessarily restrict to analytic elements on $D$.</p> http://mathoverflow.net/questions/103047/minimal-degree-of-polynomial-vanishing-on-the-variety-of-small-degree/103547#103547 Answer by Jérôme Poineau for Minimal degree of polynomial vanishing on the variety of small degree. Jérôme Poineau 2012-07-30T20:38:09Z 2012-07-30T20:38:09Z <p>You can have a look at the article "Direct methods for primary decomposition" by Eisenbud, Huneke and Vasconselos (Inventiones 110, 1992), available on Eisenbud's webpage. In proposition 3.5, they prove (and say it is long known) that a homogeneous equidimensional ideal $I$ of degree $d$ is generated up to radical by forms of degree at most $d$. </p> <p>See also the remark following the proposition for an example showing it is sharp: $V$ is the union of $d$ skew lines that all meet a common line $L$. They also mention a conjecture in the case where $I$ is prime: up to a radical it should be generated by form of degree at most $d-$ codim $I + 1$. </p> http://mathoverflow.net/questions/92506/degree-of-generators-of-irreducible-components Degree of generators of irreducible components Jérôme Poineau 2012-03-28T20:53:54Z 2012-03-29T13:17:26Z <p>Let $V$ be a Zariski-closed subset of $\mathbb{A}^n_k$, where $k$ is an algebraically closed field. Assume that $V$ may be defined by polynomials of degree at most $d$ (or to put it otherwise $V$ is an intersection of hypersurfaces of degree at most $d$). My question is the following: is it also possible to define the irreducible components of $V$ by polynomials of degree at most $d$?</p> <p>This is true if $V$ is an hypersurface (the irreducible components are defined by the factors of a polynomial defining $V$), so the answer is positive when $n$ is at most 2. I have managed to prove it in a few other cases but not much and I would appreciate any advice.</p> <p>There exists algorithms to compute the irreducible components. I have checked a few of them but they could let the degree of generators grow. Any algorithm using Gröbner basis for instance will not fit. For the same reasons, trying to prove the results using projections is probably hopeless, since they may increase the degree of the generators.</p> <p>A few more remarks:</p> <ul> <li><p>I am not sure how relevant the fact that $k$ is algebraically closed is, but I suspect there could be very non-trivial arithmetic issues otherwise, even when $V$ is 0-dimensional.</p></li> <li><p>I do not mind to work in the projective space instead of the affine space (it implies the result anyway and makes it easier to deal with degrees).</p></li> <li><p>I do not mind to get the irreducible components only as a set (i.e. the ideal up to a radical).</p></li> <li><p>I tried to make a few computations but found it rather hard. If you know a way to compute the least integer $d$ such that a given Zariski-closed subset may be defined by polynomials of degree $d$, I would also be glad to know. </p></li> </ul> http://mathoverflow.net/questions/90100/for-which-fields-is-the-inverse-galois-problem-known/90110#90110 Answer by Jérôme Poineau for For which fields is the inverse Galois problem known? Jérôme Poineau 2012-03-03T08:15:55Z 2012-03-03T08:15:55Z <p>You should find what you want in the following article by Jochen Koenigsmann: The regular inverse Galois problem over non-large fields. J. Europ. Math. Soc., 6(4):425–434, 2004.</p> http://mathoverflow.net/questions/88643/projectivity-of-one-tate-algebra-over-another/88822#88822 Answer by Jérôme Poineau for Projectivity of one Tate algebra over another Jérôme Poineau 2012-02-18T10:13:47Z 2012-02-18T10:13:47Z <p>Let me try to adapt Konstantin Ardakov's answer to the case of Tate algebras over a valued field $k$.</p> <p>If $k$ is trivially valued, then $k\langle T_1,T_2\rangle = k[T_1,T_2]$ is free over $k\langle T_1\rangle = k[T_1]$.</p> <p>If $k$ is not trivially valued, there exists an element $\alpha \in k$ with $0 &lt; |\alpha|&lt;1$. If $k\langle T_1,T_2\rangle$ were projective over $k\langle T_1\rangle$, it would be free by Bass' theorem on big projective modules. So there would be a basis $(x_1,\dots)$. By mutiplying by powers of $\alpha$, we may assume that the $x_i$'s have norm at most 1. The series $\sum_{n\ge 1} \alpha^n T_1^n x_n$ is Cauchy, hence convergent in $k\langle T_1,T_2\rangle$. Now we can reduce modulo $T_1^m$ for $m$ big enough to get a contradiction just as in Konstantin Ardakov's answer.</p> http://mathoverflow.net/questions/88072/how-to-compute-the-henselization-of-some-simple-rings/88147#88147 Answer by Jérôme Poineau for how to compute the henselization of some simple rings? Jérôme Poineau 2012-02-10T21:36:20Z 2012-02-10T21:36:20Z <p>In both cases, the henselization $A^h$ of your ring $A$ is its algebraic closure in its completion $\hat{A}$.This follows for instance from Artin approximation (Algebraic approximation of structures over complete local rings, Publications Mathématiques de l'IHÉS, 36, 1969, p. 23-58): any system of polynomial equations that has a solution in $\hat{A}$ has a solution in $A^h$ (and you may find a solution as close as you want to the original one).</p> http://mathoverflow.net/questions/69620/k-th-powers-in-the-field-of-p-adics/69626#69626 Answer by Jérôme Poineau for k-th powers in the field of p-adics Jérôme Poineau 2011-07-06T11:27:49Z 2011-07-06T11:27:49Z <p>The basic idea is that it is possible to extract $k^\mathrm{th}$-roots of elements of $\mathbb{Z}_p$ that are close enough to 1. This is because you can compute explicitely the radius of convergence of the series $(1+X)^{1/k}$ (because you know the $p$-adic valuation of the factorials, hence of the binomial coefficients). Let us say that the radius of convergence is greater that $p^{-r}$.</p> <p>Now you can write $x= n + yp^r$, with $y\in \mathbb{Z}_p$ and $n$ invertible in $\mathbb{Z}_p$, since $v(x)=0$. Then $x/n = 1+ n^{-1}yp^r$ has a $k^\mathrm{th}$-root.</p> http://mathoverflow.net/questions/67843/algebraic-extensions-of-p-adic-closed-fields/67848#67848 Answer by Jérôme Poineau for Algebraic extensions of p-adic closed fields Jérôme Poineau 2011-06-15T08:17:55Z 2011-06-15T16:30:08Z <p>For $\mathbb{Q}_p$, you can have a look at Laurent Berger's notes <a href="http://perso.ens-lyon.fr/laurent.berger/coursM2/Poly2010.pdf" rel="nofollow">http://perso.ens-lyon.fr/laurent.berger/coursM2/Poly2010.pdf</a>, especially theorem 15.3. They are written in French, but guessing from your name, I would not be surprised if you could read French.</p> <p>Basically, the argument is as follows. Pick an extension $K$ of $\mathbb{Q}_p$ of finite degree d. Let me denote by e the ramification index and by f the degree of the residual extension. We have d=ef. The maximal unramified subextension $L$ is obtained by adjoining a root of unity (of order $p^f-1$). Then you go from $L$ to $K$ by adjoining a root of an Eisenstein polynomial of degree e. Now, the coefficients of such a polynomial lives in the ring of integers of $L$, which is compact. By Krasner's lemma, the extension will not change if you change the Eisenstein polynomial by another one which is close enough. Hence you only have a finite number of extensions and they are generated by elements which are algebraic over $\mathbb{Q}$.</p> <p>Unfortunately, I am not really familiar with general p-adically closed fields and cannot tell you if the previous argument still works. By the way, do you have any good reference for them?</p> http://mathoverflow.net/questions/62703/is-analytic-quillen-suslin-simple/63335#63335 Answer by Jérôme Poineau for Is analytic Quillen-Suslin simple? Jérôme Poineau 2011-04-28T20:52:05Z 2011-04-28T20:52:05Z <p>I would be rather surprised to see a proof that is significantly simpler than a proof of Cartan's lemma.</p> <p>Indeed, suppose you know that every vector bundle on $\mathbb{C}^n$ is trivial. Consider two open subsets U and V that cover $\mathbb{C}^n$ and pick a matrix A in $GL_n(\mathscr{O}(U\cap V))$. Now define a vector bundle $\mathscr{E}$ on $\mathbb{C}^n$ by glueing $\mathscr{O}^n$ on U with $\mathscr{O}^n$ on V using A. Take bases of global sections $(e_1,\ldots,e_n)$ and $(f_1,\ldots,f_n)$ of $\mathscr{O}^n$ over U and V. The vector bundle $\mathscr{E}$ is isomorphic to $\mathscr{O}^n$, by hypothesis. Now express a basis of global sections of $\mathscr{O}^n$ over $\mathbb{C}^n$ in terms of the $e_i$'s and $f_i$'s using the isomorphism. You end up with two matrices $A_U \in GL_n(\mathscr{O}(U))$ and $A_V \in GL_n(\mathscr{O}(V))$ such that $A_U A_V^{-1} = A$. This is quite close to Cartan's lemma.</p> <p>As regards the answers by Georges Elencwajg and SGP, if I am not mistaken, both use the fact that $\mathbb{C}^n$ is Stein and I do not know how to prove this without using Cartan's lemma.</p> http://mathoverflow.net/questions/25630/major-mathematical-advances-past-age-fifty/27320#27320 Answer by Jérôme Poineau for Major mathematical advances past age fifty Jérôme Poineau 2010-06-07T07:05:00Z 2010-06-07T07:05:00Z <p>When Khare and Wintenberger proved Serre's conjecture, Wintenberger was older than fifty.</p> http://mathoverflow.net/questions/15366/which-journals-publish-expository-work/15681#15681 Answer by Jérôme Poineau for Which journals publish expository work? Jérôme Poineau 2010-02-18T08:01:53Z 2010-02-18T08:01:53Z <p>I just wanted to comment on Pete's answer to Felipe Voloch but it seems I can't so I'll write an answer instead. I think that L'Enseignement Mathématique does publish purely expository papers. I have two examples in mind (hope I'm not mistaken):</p> <ul> <li>L. Illusie "Catégories dérivées et dualité: travaux de J.-L. Verdier" Enseign. Math. (2) 36 (1990), no. 3-4, 369--391</li> <li>J. Nicaise "Formal and rigid geometry: an intuitive introduction and some applications." Enseign. Math. (2) 54 (2008), no. 3-4, 213--249</li> </ul> http://mathoverflow.net/questions/15673/an-unfamiliar-to-me-form-of-hensels-lemma/15679#15679 Answer by Jérôme Poineau for An unfamiliar (to me) form of Hensel's Lemma Jérôme Poineau 2010-02-18T07:48:55Z 2010-02-18T07:48:55Z <p>What you say at the beginning of your post is right: Hensel-Kurschak's lemma may be deduced from some refined version of Hensel's lemma. Actually, it's what Neukirch does in Algebraic Number Theory (see chapter II, corollary 4.7). His proof relies on the following (see 4.6)</p> <p>Hensel's lemma: Let $(K,|.|)$ be a complete valued field with valuation ring $R$, maximal ideal $\mathfrak{m}$. Let $f(x) \in R[x]$ be a primitive polynomial (ie $f\ne 0$ mod $\mathfrak{m}$). Suppose $f=\bar{g}\bar{h}$ mod $\mathfrak{m}$, with $\bar{g}$ and $\bar{h}$ relatively prime. Then you can lift $\bar{g}$ and $\bar{h}$ to polynomials $g$ and $h$ in $R[x]$ such that $\textrm{deg}(g)=\textrm{deg}(\bar{g})$ and $f=gh$. </p> <p>Neukirch goes on with proving that the valuation on $K$ extends uniquely to any algebraic extension (see corollary 4.7), as you say Roquette does.</p> <p>As regards your last question, you may want to have a look at chapter II, paragraph 6 (appropriately called Henselian Fields) in the book of Neukirch. His definition of Henselian field is that it should satisfy Hensel-Kurschak's lemma. In theorem 6.6, he shows that this property is equivalent to the unique extension of the valuation to algebraic extensions.</p> http://mathoverflow.net/questions/131126/why-do-rigid-spaces-have-not-enough-points/131274#131274 Comment by Jérôme Poineau Jérôme Poineau 2013-05-22T20:57:01Z 2013-05-22T20:57:01Z Let me also add a warning: your example is probably the simplest, but it will not work if $\sqrt{|k^\times|}=\mathbb{R}_{&gt;0}$. http://mathoverflow.net/questions/131126/why-do-rigid-spaces-have-not-enough-points/131274#131274 Comment by Jérôme Poineau Jérôme Poineau 2013-05-22T20:54:03Z 2013-05-22T20:54:03Z @ampit6: Yes, there is a notion of point of a topos: it is a morphism from the punctual topos to the topos (see SGA4, IV.6). And if the topos is the topos associated to a topological space, yes you can recover such a space from the topos (see SGA4, IV.7.1). Moreover, if the topological space is sober (every irreducible subspace has exactly one generic point), then it is unique (see SGA4, IV.4.2). And Huber's construction provides such a sober space associated to the rigid topos (see his paper &quot;Continuous valuations&quot; Math. Z. 212 (1993)). http://mathoverflow.net/questions/131126/why-do-rigid-spaces-have-not-enough-points/131274#131274 Comment by Jérôme Poineau Jérôme Poineau 2013-05-21T05:25:16Z 2013-05-21T05:25:16Z You are right, thanks for the comment Andrew. And since we know we get every point of the topos from the points of the Huber's space, we will not get any other example for the disc. http://mathoverflow.net/questions/129061/morphism-from-adic-spaces-to-schemes Comment by Jérôme Poineau Jérôme Poineau 2013-04-29T15:39:09Z 2013-04-29T15:39:09Z Then I am not sure that it is possible to find such a morphism. How would you find $\hat{A} \to A$ that induces $\mathrm{Spa}A \simeq \mathrm{Spa}\hat{A} \to \mathrm{Spec}\hat{A}$? http://mathoverflow.net/questions/129061/morphism-from-adic-spaces-to-schemes Comment by Jérôme Poineau Jérôme Poineau 2013-04-29T08:35:23Z 2013-04-29T08:35:23Z What's wrong with the obvious morphism between rings of global sections? http://mathoverflow.net/questions/127622/tame-morphism-from-a-curve-to-mathbbp1 Comment by Jérôme Poineau Jérôme Poineau 2013-04-16T09:40:52Z 2013-04-16T09:40:52Z @Peter: Thanks for your comment. This $S^4$ makes more sense but I don't know how to make the argument work in characteristic 2 either. On the other hand, I may be able to do it using patching techniques from inverse Galois theory (maybe if $k$ is algebraically closed but I don't mind too much). http://mathoverflow.net/questions/127622/tame-morphism-from-a-curve-to-mathbbp1 Comment by Jérôme Poineau Jérôme Poineau 2013-04-16T09:17:27Z 2013-04-16T09:17:27Z Thanks Will. There are a few things I don't understand. What is the link with $S^4$? How do you know the dimension of the moduli space of covers? (Since $M_{0,4}$ has dimension 1, I would have thought it would be bigger, but this will not cause any trouble anyway.) I don't have time to check the details right now but will come back to it next week. http://mathoverflow.net/questions/124466/what-about-schemes-built-up-out-of-graded-rings Comment by Jérôme Poineau Jérôme Poineau 2013-03-14T07:09:24Z 2013-03-14T07:09:24Z You could have a look at Michael Temkin's paper &quot;On local properties of non-Archimedean spaces II&quot; Isr. J. of Math. 140 (2004), 1-27 (see <a href="http://www.math.huji.ac.il/~temkin/papers/Local_Properties_II.pdf" rel="nofollow">math.huji.ac.il/~temkin/papers/&hellip;</a>), especially the first section. http://mathoverflow.net/questions/122915/bornological-vector-spaces-over-a-non-archimedean-field Comment by Jérôme Poineau Jérôme Poineau 2013-02-25T20:01:33Z 2013-02-25T20:01:33Z I think that Francesco Baldassarri was interested by this kind of things recently. Maybe you should ask him. http://mathoverflow.net/questions/121187/reference-for-rigid-analytic-gaga/121215#121215 Comment by Jérôme Poineau Jérôme Poineau 2013-02-21T12:37:22Z 2013-02-21T12:37:22Z @ Simon Wadsley: I did not see K&#246;pf's paper myself, so I cannnot tell. But I am not sure why you would expect the results you need to be there. http://mathoverflow.net/questions/121187/reference-for-rigid-analytic-gaga/121215#121215 Comment by Jérôme Poineau Jérôme Poineau 2013-02-10T20:52:30Z 2013-02-10T20:52:30Z Thanks for the clarifications, nosr! http://mathoverflow.net/questions/109213/uniqueness-of-analytic-continuation-in-rigid-analytic-geometry Comment by Jérôme Poineau Jérôme Poineau 2012-10-09T16:57:52Z 2012-10-09T16:57:52Z Have a look at Berkovich's book &quot;Spectral theory and analytic geometry over non-archimedean fields&quot;, especially the end of section 3.3. http://mathoverflow.net/questions/98002/how-to-solve-a-fifth-degree-polynomial/98005#98005 Comment by Jérôme Poineau Jérôme Poineau 2012-05-26T12:37:03Z 2012-05-26T12:37:03Z Par ailleurs, en fran&#231;ais, on &#233;crit &#171; un Fran&#231;ais &#187;. http://mathoverflow.net/questions/94503/domain-of-holomorphy Comment by Jérôme Poineau Jérôme Poineau 2012-04-19T09:14:49Z 2012-04-19T09:14:49Z You can use the same arguments as for a Hartogs' figure to show that any holomophic function on $D$ extends to $\mathbb{C}^2$. Also beware that in general a smallest domain of holomorphy $S\subset \mathbb{C}^2$ containing $D$ need not exist. http://mathoverflow.net/questions/93276/a-game-on-noetherian-rings Comment by Jérôme Poineau Jérôme Poineau 2012-04-06T10:37:32Z 2012-04-06T10:37:32Z @FAE: I don't believe that Krull dimension is the right thing to consider. In a geometric situation like that of the OP, the minimal dimension of an irreducible component should play a role (you can win in one move if you have the union of a line and a plane for instance). And there are reduction issues too (start with a double line).