User charles staats - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T12:51:45Z http://mathoverflow.net/feeds/user/5094 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130783/reasons-to-prefer-one-large-prime-over-another-to-approximate-characteristic-zero Reasons to prefer one large prime over another to approximate characteristic zero Charles Staats 2013-05-16T00:57:10Z 2013-05-16T00:57:10Z <p><strong>Background:</strong> In running algebraic geometry computations using software such as Macaulay2, it is often easier and faster to work over $\mathbb F_p = \mathbb Z / p\mathbb Z$ for a large prime $p$, rather than over $\mathbb Q$. (Note that working directly over $\mathbb C$ is not really possible for exact computations.) Some basic model theory implies (more or less) that if you have a question that is capable of being answered by an algorithm, and the question has the same answer for $k$ as for $\bar{k}$, then its answer over $\mathbb Q$ will be the same as its answer over $\mathbb F_p$ for all but finitely many primes $p$. The accepted wisdom is that with virtually no exceptions, if you want to answer an algebro-geometric question over $\mathbb Q$, you can get a reliable answer by picking a large prime such as $p=32003$ and doing your computations over $\mathbb F_p$.</p> <p>I think people generally pick a prime near the top of the range their software can handle, which is relatively easy to remember; $32003$ and $31667$, for instance, both fit the bill when using Macaulay2. However, I was wondering whether there are other mathematical characteristics of a prime that can affect how well it approximates characteristic zero. For instance, does some special pathology arise if $p$ is (or is not) a Mersenne prime, or if $(p-1)/2$ is (or is not) prime, or...?</p> <p>Note that I only am asking about behavior that affects <em>how well characteristic $p$ approximates characteristic $0$.</em> Questions that one would only ever ask in finite characteristic are not relevant. Also irrelevant are questions that cannot be answered by an algorithm; for instance, "is $n\cdot1=0$ for some positive integer $n$" cannot be asked unless you include a bound on $n$.</p> <p><strong>The Question:</strong> Are there valid mathematical reasons for preferring one large prime over another to approximate characteristic $0$, other than the assumption that larger primes are generally better?</p> http://mathoverflow.net/questions/92771/a-geometric-characterization-for-arithmetic-genus A geometric characterization for arithmetic genus Charles Staats 2012-03-31T20:21:46Z 2013-03-30T18:03:15Z <p>Let $X$ be a smooth projective variety over $\mathbb{C}$. The following information is all equivalent (any of these numbers can be computed by a linear equation from any of the others):</p> <ol> <li>the arithmetic genus of $X$</li> <li>the constant coefficient of the Hilbert polynomial of $X$</li> <li>$\chi(X, \mathscr{O}_X)$</li> <li>the "Todd genus" $\int_X \operatorname{td}(T_X)$, where $T_X$ is the tangent bundle of $X$ and $\operatorname{td}$ denotes the Todd class.</li> </ol> <blockquote> <p>Is there a geometric characterization for any of these numbers?</p> </blockquote> <p>If I understand correctly, characteristic classes (and in particular, Todd classes) can be defined entirely from the topology of $X$, or at least its structure as a smooth manifold. [Edit: <strong>This is not true</strong>--see the answer of "anonymous." If I understand correctly, the Todd class of a complex vector bundle is a smooth invariant. However, different complex structures on the same real manifold $X$ can give rise to non-isomorphic complex vector bundle structures on $T_X$; in fact, a complex vector bundle structure on $T_X$ is, by definition, an almost complex structure on a real manifold $X$.] Thus, in some sense, item 4 provides a "geometric characterization" for the arithmetic genus of $X$ (and the other items on the list). However, I personally find this description so far abstracted from actual geometric properties of $X$ as to be hardly geometric at all. (If anyone disagrees with me and can articulate a geometric intuition for the Todd genus, that would be a reasonable answer.) </p> <p>By comparison, I do consider the following characterizations of various properties "geometric":</p> <ul> <li>The self-intersection number of the diagonal embedding of $X$ into $X \times X$. (the Euler characteristic)</li> <li>The number of points in which a general linear space of complementary dimension meets $X \subset \mathbb P^n$. (the degree of $X \hookrightarrow \mathbb P^n$)</li> <li>The genus of the curve $X \cap L$, where $L$ is a general linear space of dimension one greater than $\operatorname{codim} X$. (I don't know of a standard name for this, but in a particular sense, it is one of the coefficients of the Hilbert polynomial of $X$.)</li> <li>The maximum number of copies of $S^1$ that can be removed from $X$ without disconnecting it. (the genus of $X$ if $X$ is a smooth curve, i.e., Riemann surface)</li> </ul> <p>Note that either the first or the last point gives a geometric characterization for (information equivalent to) the genus of a curve. Without one of these, I would not consider the third bullet a "geometric characterization" of anything. In a way, this provides part of my motivation for asking this question. Let $L_k$ be a general linear space of dimension $k$ in $\mathbb P^n$. Unless I am mistaken, knowing the Hilbert polynomial for $X$ is equivalent to knowing the arithmetic genus of $X \cap L_k$, for every $k \leq n$ such that this intersection is nonempty. Thus, a geometric characterization for arithmetic genus would automatically give a geometric characterization for the Hilbert polynomial. (Again, in some sense, this is already provided by the Hirzebruch-Riemann-Roch Theorem; but I find this formula so abstracted as to be hardly geometric at all.)</p> http://mathoverflow.net/questions/40632/what-is-the-inverse-image-sheaf-necessary-for-in-algebraic-geometry What is the inverse image sheaf necessary for in algebraic geometry? Charles Staats 2010-09-30T15:42:46Z 2013-03-14T18:13:15Z <p>Given a continuous map $f \colon X \to Y$ of topological spaces, and a sheaf $\mathcal{F}$ on $Y$, the inverse image sheaf $f^{-1}\mathcal{F}$ on $X$ is the sheafification of the presheaf $$U \mapsto \varinjlim_{V \supseteq f(U)} \Gamma(V, \mathcal{F}).$$ If $X$ and $Y$ happen to be ringed spaces, $f$ a morphism of ringed spaces, and $\mathcal{F}$ an $\mathcal{O}_X$-module, one then defines the pullback sheaf $f^* \mathcal{F}$ on $X$ as $$f^{-1}\mathcal{F} \otimes_{f^{-1} \mathcal{O}_Y} \mathcal{O}_X.$$ However, I cannot think of any other usage of the inverse image sheaf in algebraic geometry. Moreover, if $X$ and $Y$ are schemes and $\mathcal{F}$ is quasicoherent, there is an alternate way of defining $f^* \mathcal{F}$. Given $f \colon \mathrm{Spec} B \to \mathrm{Spec} A$, and $\mathcal{F} = \widetilde{M}$, where $M$ is an $A$-module, one defines $f^* \mathcal{F}$ to be the sheaf associated to the $B$-module $M \otimes_A B$. To extend this to arbitrary schemes, it is necessary to prove that it is well-defined; but I still think it is easier to work with than the other definition, which involves direct limits and two sheafifications of presheaves (the inverse image, and the tensor product). I have not checked, but I imagine that something similar can be done for formal schemes.</p> <p>Hence, my question:</p> <blockquote> <p>What uses, if any, does the inverse image sheaf have in algebraic geometry, other than to define the pullback sheaf?</p> </blockquote> <p>A closely related question is</p> <blockquote> <p>In a course on schemes, is there a good reason to define the inverse image sheaf and the pullback sheaf for ringed spaces in general, rather than simply defining the pullback of a quasicoherent sheaf by a morphism of schemes?</p> </blockquote> <p>To go from the first question to the second question, I suppose one must also address whether there are $\mathcal{O}_X$-modules significant to algebraic geometers that are not quasicoherent.</p> <p><b>Edit:</b> I think the question deserves a certain amount of clarification. Several people have given interesting descriptions or explications of the inverse image sheaf. While I appreciate these, they are not the point of my question; I am, specifically, interested to know whether there are constructions or arguments in algebraic geometry that cannot reasonably be done without using the inverse image sheaf. So far, the answer seems to be that such things exist, but are not really within the scope of, say, a one-year first course on schemes. There are other constructions (such as the inverse image ideal sheaf) that do not, strictly speaking, require the inverse image sheaf, but for which it may be more appropriate to use the inverse image sheaf as a matter of taste.</p> http://mathoverflow.net/questions/48527/why-open-immersion-rather-than-open-embedding Why "open immersion" rather than "open embedding"? Charles Staats 2010-12-07T02:53:57Z 2013-02-09T04:23:27Z <p>When topologists speak of an "immersion", they are quite deliberately describing something that is not necessarily an "embedding." But I cannot think of any use of the word "embedding" in algebraic geometry, except sometimes as a word for an immersion of varieties. And the notion of an "immersion" of schemes, especially an "open immersion," seems much more similar to the topologists' "embedding" than their "immersion." [Closed immersions at least have the somewhat flimsy rationale that the scheme structure does not depend solely on the choice of subset.]</p> <blockquote> <p>Does anyone know of a good reason, other than cultural momentum, to use the word "immersion" rather than "embedding"?</p> </blockquote> <p>[Note: this has come up in Ravi Vakil's blog on his Algebraic Geometry notes.]</p> http://mathoverflow.net/questions/34673/kunneth-formula-for-sheaf-cohomology-of-varieties Kunneth formula for sheaf cohomology of varieties Charles Staats 2010-08-05T19:03:05Z 2013-01-25T03:51:18Z <p>What is a good reference for the following fact (the hypotheses may not be quite right):</p> <blockquote> <p>Let $X$ and $Y$ be projective varieties over a field $k$. Let $\mathcal{F}$ and $\mathcal{G}$ be coherent sheaves on $X$ and $Y$, respectively. Let $\mathcal{F} \boxtimes \mathcal{G}$ denote $p_1^*(\mathcal{F}) \otimes_{\mathcal{O}_{X \times Y}} p_2^* \mathcal{G}$. Then $$H^m(X \times Y, \mathcal{F} \boxtimes \mathcal{G}) \cong \bigoplus_{p+q=m} H^p(X,\mathcal{F}) \otimes_k H^q(Y, \mathcal{G}).$$</p> </blockquote> <p>Note: Wikipedia leads me to believe that this may be related to Theorem 6.7.3 in EGA III<sub>2</sub>, but I find this theorem quite intimidating. Although I would be willing to study this if there is no more basic reference, I would at least like some confirmation that I am studying the right thing.</p> http://mathoverflow.net/questions/115222/techniques-for-showing-that-a-curve-is-not-smoothable Techniques for showing that a curve is not smoothable Charles Staats 2012-12-03T01:14:11Z 2012-12-03T12:12:15Z <p>There are a number of techniques in algebraic geometry that can be used to show that a given reducible (often genus-zero) curve $C$ in a smooth variety $X$ becomes smooth and irreducible after a (generic) deformation. Such techniques are used, for instance, to show that (for smooth varieties in characteristic zero) rational chain-connectedness implies rational connectedness.</p> <p>However, suppose you have a genus-zero reducible curve that you suspect <em>cannot</em> be deformed (under a generic deformation) to a smooth rational curve--or even an irreducible rational curve. (More precisely, you suspect that for a given irreducible component of the Kontsevich moduli of stable maps, the general point corresponds to a reducible curve.) Are there any obvious criteria that would allow you to show this?</p> http://mathoverflow.net/questions/32968/slick-ways-to-make-annoying-verifications Slick ways to make annoying verifications Charles Staats 2010-07-22T16:07:16Z 2012-10-29T21:33:26Z <p>There are many times in mathematics that one needs to make verifications that are annoying and distract from the main point of the argument. Often, there are lemmas that can make this much easier, at least in many important cases. </p> <p>For instance, in topology, it can be quite annoying to verify directly from the definition that a particular quotient space is what you think it is, and not something else with the same underlying set. (In fact, I suspect that many topologists habitually skip this verification.) However, the following lemma can in many cases make this verification much simpler:</p> <blockquote> <p><strong>Lemma:</strong> If $X$ is compact, $Y$ is Hausdorff, and $f \colon X \to Y$ is surjective, then $f$ is a quotient map.</p> </blockquote> <p>This lemma can be made more powerful using the fact that it suffices to show a map is locally a quotient map.</p> <p>Another such difficulty is to verify that a category is abelian; if you go directly from the definition, there is an annoyingly long list of things to verify. However, unless I am mistaken, once you have an abelian category $\mathcal{A}$, there are a number of other categories that are guaranteed to give other abelian categories. These (I think) include the category of functors into $\mathcal{A}$ from a fixed other category, the category of sheaves in $\mathcal{A}$ on a (topological space? other category?) (assuming $\mathcal{A}$ is nice enough for this to make sense), and any full subcategory of $\mathcal{A}$ that is closed under 0, $\oplus$, kernels, and cokernels. Using these in combination, together with the fact that $R$-mod is an abelian category for every ring $R$, I believe <strike>one can get to every abelian category used in Hartshorne</strike>. (Note: I am not too confident in this example, so if someone wanted to elaborate this in an answer, it would be appreciated.)</p> <p><em>EDIT:</em> As is pointed out in the comments below, the category of $\mathcal{O}_X$-modules is not of this form. (I came up with this example while writing the question, and did not think it through too carefully.) Thus, I would doubly appreciate a good answer specifically addressing, "How do you show a category is abelian?"</p> <blockquote> <p><strong>Question:</strong> What are some more of these useful lemmas / collections of lemmas, and how are they used?</p> </blockquote> http://mathoverflow.net/questions/69096/why-is-the-degreerank-ratio-of-a-vector-bundle-called-its-slope Why is the degree:rank ratio of a vector bundle called its "slope"? Charles Staats 2011-06-29T06:00:00Z 2012-10-03T13:34:16Z <p>Whenever one studies moduli spaces of vector bundles on curves, one of the first things to be introduced is the "slope" of a vector bundle, i.e., its degree:rank ratio. Is there a nice (preferably geometric) intuition behind the use of the word "slope" for this?</p> http://mathoverflow.net/questions/87350/a-relative-algebraic-hartogs-lemma A Relative Algebraic Hartogs Lemma Charles Staats 2012-02-02T17:39:29Z 2012-09-09T01:43:13Z <p>The Algebraic Hartogs Lemma states that in a Noetherian normal scheme, a rational function that is regular outside a closed subset of codimension at least two, is in fact regular everywhere.</p> <p>In a research problem I was working on recently, I was (following suggestions by my advisor) using this to prove that a particular section of a line bundle existed on a space $\mathbb{A}^n_H$. The argument worked well when $H$ was a point. For more general $H$, I could show that the section was defined except on a codimension-two subset of every fiber; but it was not immediately obvious to me how to go from there, to showing that the section was defined "on all fibers simultaneously," i.e., over $\mathbb{A}^n_H$. This was especially problematic in that the base scheme $H$ in question was a component of a Hilbert scheme, and thus (as shown by Ravi Vakil) capable of exhibiting arbitrarily bad behavior. In particular, the fact that a composition of normal morphisms is normal (see EGA IV.2, section 6.8) would not come close to covering my situation.</p> <p>In order to deal with this, I have obtained what seems to be a proof of the following statement:</p> <blockquote> <p><strong>Lemma</strong> (Relative Algebraic Hartogs' Lemma) Let $X \to S$ be a flat, finite-type morphism of Noetherian schemes such that every associated fiber is normal. Let $\mathscr{L}$ be a line bundle on $X$. Suppose that $U \subset X$ is an open subscheme such that<br> (i) $U$ contains all the associated points of $X$,<br> (ii) for every $s \in S$, $U \cap X_s$ contains all the associated points of $X_s$, and<br> (iii) for every associated point $\eta$ of $S$, $U$ contains all but a codimension-two closed subset of $X_{\eta}$.<br> Then the restriction map $$\Gamma(X,\mathscr{L}) \to \Gamma(U,\mathscr{L})$$ is an isomorphism.</p> </blockquote> <p>(I'm using "associated fiber" to mean "fiber over an associated point," by analogy with the standard term "generic fiber.")</p> <p>I should note that suggestions of Will Sawin in <a href="http://mathoverflow.net/questions/86033/relative-integral-closure/86038#86038" rel="nofollow">this answer</a> were invaluable in coming up with the proof.</p> <p>This lemma is surprisingly strong, in that two key hypotheses--normality, and codimension-two-ness--only need to hold "generically" (i.e., in the fibers over the associated points). Another interesting feature is that, if you look at the proof, the fibers do not actually have to be normal, as long as they "satisfy the Hartogs property"--which I understand is equivalent to satisfying the S2 condition. I'm also not convinced that the finite-type hypothesis is necessary, but I have not yet verified the argument that I suspect would remove it.</p> <p>The proof is not incredibly long (4 pages from me; probably less from a more experienced mathematician who knows what details to leave out), uses only techniques that are extremely well known, and so far as I can tell, does not use them in an especially clever way. Thus, it seems likely that someone, at some point, has already written up a similar result. However, neither I, nor my advisor, nor anyone else I have spoken to was familiar with it (which suggests to me that, if nothing else, the result is less well-known than it should be). Hence, my question:</p> <blockquote> <p><strong>Question:</strong> Does anyone know of a similar statement in the literature? (To be safe, I'll also ask if anyone knows of any counterexamples, although I <em>think</em> the proof is solid.)</p> </blockquote> <p>One final note: From what I understand, the popularity of the name "algebraic hartogs lemma" is quite recent, possibly a result of its use in Ravi Vakil's notes, so a similar result in the older literature would probably not be called by the word "Hartogs."</p> <hr> <p><strong>Update:</strong> I have posted a <a href="http://math.uchicago.edu/~cstaats/Charles_Staats_III/Notes_and_papers_files/RelativeAlgebraicHartogs.pdf" rel="nofollow">proof of the Lemma above</a>.</p> http://mathoverflow.net/questions/106627/can-flatness-be-specified-by-a-natural-coherent-sheaf Can flatness be specified by a natural coherent sheaf? Charles Staats 2012-09-07T20:53:16Z 2012-09-08T07:06:57Z <p>More precisely:</p> <blockquote> <p>Given a finite-type morphism $f \colon X \to Y$ of nice schemes (say, both of finite type over a field), is there a "natural" coherent sheaf $\mathcal F_f$ on $X$ such that the support of $\mathcal F_f$ is precisely <code>$\{x \in X : f \text{ is not flat at }x\}$</code>?</p> </blockquote> <p>I'd rather not say what exactly "natural" should mean here, but if you held a gun to my head, I would guess it means that $\mathcal F_f$ is preserved under flat pullbacks; in other words, that if $g\colon Z \to Y$ is a flat finite-type morphism of nice schemes, then $\mathcal F_{f \circ g} = g^* \mathcal F_{f}$.</p> <p><strong>Example:</strong> If $X = \operatorname{Spec} B$, $Y = \operatorname{Spec} A$, and $A \to B$ is a local morphism of local noetherian rings, then $$\ker(B \otimes_A \mathfrak m \to B)$$ is a finitely generated $B$-module that is zero iff $B$ is flat over $A$ (by the local criterion for flatness). [Notation: $\mathfrak m$ is the maximal ideal of $A$.] This construction is respected under pullback by flat <em>local</em> morphisms $B \to C$ of local noetherian rings. Unfortunately, it runs into trouble if you look at flatness anywhere except the closed point of $\operatorname{Spec} B$.</p> <p><strong>Motivation:</strong> There are standard theorems that certain sets (including the one above) are "open," without really specifying a closed subscheme structure on the complement. In some cases, the complement can be described as the support of a coherent sheaf, which gives it a natural scheme structure. For instance, the "indeterminacy locus" of a rational function may be described by the cokernel of the ideal of denominators. I personally find such constructions more satisfying and memorable than more direct proofs that a particular subset is open.</p> http://mathoverflow.net/questions/100893/examples-of-nice-reduced-singularities-on-hilbert-schemes-edited Examples of nice reduced singularities on Hilbert schemes--Edited Charles Staats 2012-06-28T20:37:12Z 2012-08-12T20:55:53Z <p>In his "Murphy's Law" paper, Vakil showed that every "singularity type" (with a precise meaning) occurs on certain Hilbert schemes; for instance, the Hilbert scheme of <em>nonsingular</em> curves in projective space. He also gives a method for constructing such singularities; however, the process to construct even, say, a singularity of nodal type would be extremely involved. </p> <p>[As I understand it, one would have to blow up a plane at something like twenty points (conservative estimate), then take a certain eight-fold cover, then find an appropriate line bundle and take six "sufficiently general" sections,...]</p> <p>For a smooth variety $X$, let $H_X$ denote its Hilbert scheme. A point of $H_X$ corresponds to a subscheme $V$ of $X$. I am interested in cases in which $V$ is also smooth. <strong>[Edit: I am also requiring that $H^1(T_X) = 0$, i.e., that the ambient variety $X$ admit no infinitesimal deformations. To put it another way, the complex structure on the smooth manifold $X$ cannot be deformed. This holds in particular if $X=\mathbb P^n$.]</strong> Certainly, explicit examples of such pairs $(V,X)$ corresponding to singular points of $H_X$ have been described; however, in the very few examples I have seen, the technique is to show that $V$ is contained in an irreducible component of $H_X$ that is <em>generically non-reduced.</em></p> <blockquote> <p>Can anyone give explicit examples of a smooth projective variety $X$ <strong>[such that $H^1(X,T_X) = 0$]</strong>, together with a smooth subvariety $V$, such that the point $[V]\in H_X$ is both singular and reduced? [A method of constructing explicit examples will not suffice unless you can show, by example, that this method is actually practical to carry out.]</p> </blockquote> http://mathoverflow.net/questions/53567/why-is-line-bundle-appropriate-rational-section-not-a-standard-kind-of-divisor Why is (line bundle, appropriate rational section) not a standard kind of divisor? Charles Staats 2011-01-28T04:22:40Z 2012-07-30T16:12:56Z <p>In algebraic geometry, there are two standard "kinds" of divisors: Weil divisors and Cartier divisors. Weil divisors provide better geometric intuition, while Cartier divisors are more general (if not precisely a generalization). In both of these kinds of divisors, the "key" results seem to be their relation to the Picard group (of isomorphism classes of line bundles).</p> <p>However, one could also define a "divisor" to be an equivalence class of pairs $(s, \mathcal{L})$, where $s$ is an invertible rational section of $\mathcal{L}$. (By "invertible" I mean that there exists a rational section $s'$ of $\mathcal{L}^{\vee}$ such that $s' s = 1$.) We say $(s, \mathcal{L}) \sim (s', \mathcal{L}')$ if there is an isomorphism $\mathcal{L} \to \mathcal{L}'$ taking $s \mapsto s'$.</p> <p>This definition works well at least for all Noetherian schemes (on which associated points behave nicely), and possibly more generally. It also seems less confusing than the definition of a Cartier divisor, and the relationship between divisors and line bundles is already embodied in the definition.</p> <p>So why have I never seen this definition given as a kind of divisor?</p> <p>[Note: I am aware of what "data" defines a Cartier divisor (the $(U_i, f_i)$, etc.) and how this provides a reasonably natural way to think of Cartier divisors geometrically (the "subscheme" defined locally by the $f_i$). So while I appreciate the thought, please don't waste your time writing a note for the sole purpose of explaining this.]</p> <p><b>Edit:</b> Since this issue has come up in an answer, I thought I would explain that by "rational section," I mean a (maximally extended) section over an open subset containing all the associated points; or equivalently, a section of $\mathcal{K} \otimes \mathcal{L}$, where $\mathcal{K}$ is the sheaf of total quotient rings of $\mathcal{O}_X$. (Actually, $\mathcal{K} \otimes \mathcal{L}$ is isomorphic to $\mathcal{K}$, but not naturally so.) A rational section can fail to be invertible if it vanishes on one or more associated points.</p> http://mathoverflow.net/questions/96378/can-the-holomorphic-image-of-mathbbcn-be-open-but-not-dense Can the holomorphic image of $(\mathbb{C}^*)^n$ be open but not dense Charles Staats 2012-05-08T21:36:54Z 2012-07-15T17:55:24Z <blockquote> <p>Let $M$ be a compact complex connected [but <em>not necessarily kähler</em>] $n$-manifold, and suppose we have a holomorphic map $$(\mathbb{C}^*)^n \to M$$ such that the image is open. Is the image necessarily dense in $M$?</p> </blockquote> <p>Motivation: My intuition (which comes from the algebraic world) says that the answer ought to be "yes." On the other hand, I know that many properties of smooth algebraic varieties do not hold for complex manifolds in general. Knowing whether this statement has a counterexample would improve my intuition about the complex world.</p> http://mathoverflow.net/questions/97352/reasonable-implementation-of-finding-grobner-bases-over-non-field-coefficient-rin Reasonable implementation of finding Gröbner bases over non-field coefficient rings Charles Staats 2012-05-18T21:13:23Z 2012-05-20T10:55:01Z <p>Gröbner bases are usually considered in the ring of polynomials over a field. However, there are useful definitions and algorithms for Gröbner bases over other coefficient rings; see, for instance, Chapter 4 of Adams and Loustaunau, <em>An introduction to Gröbner bases</em>. Unfortunately, I have had little success tracking down actual implementations that would allow me to compute these Gröbner bases; this may be either because they do not exist, or because I am not familiar enough with programs like Singular and Macaulay2 to identify such algorithms in the documentation.</p> <blockquote> <p>Are there existing and practical implementations of algorithms to compute Gröbner bases in polynomial rings with coefficients in a general ring? I am, in particular, interested in computing Gröbner bases in the polynomial ring $A[x_1, \dotsc, x_n]$ where the coefficient ring $A$ is a domain of finite type over a field.</p> </blockquote> <p><strong>Motivation:</strong> The generic freeness lemma tells us that, given a finite-type morphism $f \colon X \to Y$ of affine noetherian integral schemes, there is a dense open subset $U \subset Y$ such that $f^{-1}(U) \to U$ is flat and surjective. In particular, all the fibers over $U$ necessarily have the same dimension and are equivalent up to flat deformation." So, in some sense, a computable version of generic freeness would allow us to classify all the fibers of a morphism (and not just, say, the general fiber).</p> <p>Such a computable version is described in Vasconcelos' 1997 paper "Flatness testing and torsionfree morphisms," Theorem 2.1, although Vasconcelos gives the impression that this "computable version" was already well-known in certain circles. If we are looking at a ring homomorphism $$A \longrightarrow B = A[T_1,\dotsc,T_n]/I,$$ the only computationally nontrivial part of the algorithm is to compute a Gröbner basis for $I$, in the sense described in this question.</p> <hr> <p>[Qualification 1: If you actually want to compute a stratification, you also need to be able to find the irreducible components of $A/(f)$ so that the next step will have an integral base. This is not computationally trivial, but if $A$ is of finite type over a field, it does have standard implementations, for instance, in Macaulay2.]</p> <p>[Qualification 2: Judging by the next remark in Vasconcelos' paper, it <em>may</em> be possible to get by if you can compute a Gröbner basis for $I$ over the fraction field of $A$. Macaulay2 can do this--I think--but according to the documentation, it is not remotely efficient.]</p> http://mathoverflow.net/questions/20551/sources-for-bibtex-entries Sources for Bibtex entries Charles Staats 2010-04-06T21:42:10Z 2012-05-07T18:55:31Z <p>Does anyone know of a good place to find already-done bibtex entries for standard books in advanced math? Or is this impossible because the citation should include items specific to your copy? (I am seeing the latter as potentially problematic because the only date I can find in my copy of Hartshorne is 2006, whereas the citations I can find all put the publication date at 1977.)</p> http://mathoverflow.net/questions/95521/what-is-known-about-singularity-types-in-the-murphys-law-sense What is known about "singularity types" in the Murphy's Law sense? Charles Staats 2012-04-29T21:31:30Z 2012-04-29T21:31:30Z <p>In his "Murphy's Law" paper, Vakil gives a definition equivalent to the following:</p> <blockquote> <p>The <em>singularity type</em> of a pointed scheme $(X,p)$ its equivalence class, under the following equivalence relation: $(X,p) \sim (Y,q)$ if there exists a pointed scheme $(Z,r)$ admitting smooth morphisms $(Z,r) \to (X,p)$ and $(Z,r) \to (Y,q)$.</p> </blockquote> <p>This relation is obviously reflexive and symmetric; to check transitivity, use the fact that pullbacks of smooth morphisms are smooth.</p> <p>On its surface, this appears, at least to me, to have a different flavor from other ways in which singularities are studied. (I'm thinking in particular of the study of deformations of singularities, but I know very little about this subject, so my comment should not be taken too seriously.) Thus, I'd like to know what is known about the "singularity types" of this definition, other than the fact that all of them show up in each of a large number of moduli spaces.</p> <p>I'm really interested in getting a general "flavor" of the theory, but since this is a question-and-answer site, here's a specific question:</p> <blockquote> <p>Can infinitely many singularity types occur on a single scheme $X$ that is of finite type over a field?</p> </blockquote> <p>Obviously, infinitely many singularity types show up in the moduli spaces for which Vakil proves Murphy's Law. But if I understand correctly, these spaces are only locally noetherian, so it is still possible that only finitely many singularity types show up on each connected component.</p> http://mathoverflow.net/questions/18271/what-out-of-print-books-would-you-like-to-see-re-printed/39031#39031 Answer by Charles Staats for What out-of-print books would you like to see re-printed? Charles Staats 2010-09-16T22:39:08Z 2012-02-20T07:25:51Z <p>"Topologie Algébrique et Théorie des Faisceaux," by Roger Godement. The classic reference on sheaf theory. The edition I'm reading right now (checked out from the library) is beginning to fall apart, and it's really making my eyes water.</p> http://mathoverflow.net/questions/45347/why-does-the-s2-property-of-a-ring-correspond-to-the-hartogs-phenomenon/88681#88681 Answer by Charles Staats for Why does the (S2) property of a ring correspond to the Hartogs phenomenon? Charles Staats 2012-02-17T03:02:52Z 2012-02-17T03:02:52Z <p>Here's a somewhat more elementary argument that (S2) implies the Hartogs condition. More precisely, I will show that if $X$ is an (S2) noetherian scheme, then any rational function defined outside a closed subset of codimension two can be extended to the whole domain. (This extension is unique by definition of a rational function.)</p> <p>Assume, by way of contradiction, that $X$ is an (S2) noetherian scheme and $f$ is a rational function on $X$ that is defined outside a closed set of codimension at least two, but cannot be extended to the whole domain. Let $\mathscr{I}$ be the ideal of denominators of $f$; in other words, over an open affine $\operatorname{Spec} A$, <code>$$I = \{g \in A \mid g f \in A\}.$$</code> This is well-defined as a sheaf since the ideal of denominators is preserved under flat pullback (and in particular, localization); see <a href="http://mathoverflow.net/questions/74529/is-the-ideal-of-denominators-preserved-under-flat-pullback" rel="nofollow">this mathoverflow question</a>.</p> <p>If $g \in A$ is a nonzerodivisor, then $g \in I$ if and only if $f = a / g$ for some $a \in A$, hence the name "ideal of denominators." One can check that the closed subscheme $Z \subset X$ corresponding to $\mathscr{I}$ is, set-theoretically, the "indeterminacy locus of $f$": the smallest closed subset such that $f$ is defined over $X \smallsetminus Z$. By hypothesis, $f$ can be defined outside a closed subset of codimension two, so $\operatorname{codim} Z \geq 2$. Equivalently, whenever $W$ is an irreducible component of $Z$, then the local ring <code>$\mathscr{O}_{X,W}$</code> has dimension at least two. Since $X$ is assumed to be (S2), every maximal regular sequence in $\mathscr{O}_{X,W}$ has length at least two.</p> <p>Since $W$ is an irreducible component of the subscheme corresponding to $\mathscr{I}$, it follows that the radical of <code>$\mathscr{I}_W \subset \mathscr{O}_{X,W}$</code> is precisely the maximal ideal $\mathfrak{p}$. (Algebraically, $\mathfrak{p}$ is a minimal prime over $I$, and corresponds to the generic point of $W$.) Let $g,h \in \mathfrak{p}$ form a regular sequence (which exists since $X$ is (S2)). Replacing $g$ and $h$ by appropriate powers, we may assume that they are both contained in $\mathscr{I}_W$. By definition of regular sequence, $g$ is a nonzerodivisor. Since $h,g$ is a also a regular sequence, $h$ is a nonzerodivisor. Thus, $g$ and $h$, being nonzerodivisors that lie in the ideal of denominators, are in fact denominators of $f$: there exist $a, b \in \mathscr{O}_X,W$ such that $$\frac{a}{g} = \frac{b}{h} = f$$ $$ah = bg.$$ Since $g,h$ is a regular sequence, $h$ is a nonzerodivisor mod $g$. When we mod out by $g$, the equation above becomes $ah \equiv 0$, which would imply $a \equiv 0 \pmod{g}$. In other words, $a \in (g)$. But since $f = a/g$, this would imply that <code>$f \in \mathscr{O}_{X,W}$</code>, a contradiction since $f$ cannot be extended over $W$. </p> http://mathoverflow.net/questions/88138/is-the-support-of-a-flat-sheaf-flat Is the support of a flat sheaf flat? Charles Staats 2012-02-10T20:30:22Z 2012-02-11T15:54:58Z <p>Note: in the following, all scheme/algebra morphisms should be assumed essentially of finite type.</p> <blockquote> <p><strong>Geometric version:</strong> Let $X$ be a scheme flat over $S$ (both noetherian), and let $\mathscr{F}$ be a coherent sheaf on $X$, also flat over $S$. The scheme-theoretic support $\mathfrak{X}$ for $\mathscr{F}$ is a closed subscheme of $X$. Is it necessarily true that $\mathfrak{X}$ is flat over $S$?</p> <p><strong>Algebraic version:</strong> Let $B$ be a flat $A$-algebra (both noetherian), and let $M$ be a finitely generated $B$-module, also flat over $A$. Is it necessarily true that $B/\operatorname{Ann}(M)$ is flat over $A$?</p> </blockquote> <p>Motivation: the only way I know how to visualize a coherent sheaf is to visualize its support, which is a closed subscheme. I justify this by the fact that many of the properties of a coherent sheaf are shared by (the structure sheaf of) its scheme-theoretic support. <strike>For instance, they have the same associated points. In case $A$ is a DVR, this even provides a proof for the algebraic version above, since a module is flat over a DVR iff all its associated points map to the generic point.</strike> (see Angelo's comment below)</p> <p>This general "visual intuition" tells me that the two (equivalent) statements above should be true. However, I cannot think of a good argument for this. Although it is not really essential to anything I am doing, it is bothering the heck out of me not to know whether this actually works, and distracting me from my other, more "essential" work. Thus, I would appreciate some help here. A positive answer will help me sleep at night (figuratively speaking); a negative answer will, hopefully, give me a useful counterexample against which to test my intuition in the future.</p> <p>Second motivation: If the statement is true, then it provides evidence for a morphism from the Quot scheme to the Hilbert scheme, that--loosely speaking--takes a coherent sheaf to its support. (Thinking about it in these terms may also suggest solutions to mathematicians who--unlike me--have a great deal of experience with Quot and Hilbert schemes.)</p> http://mathoverflow.net/questions/60442/an-effective-way-to-tell-if-the-saturation-of-a-homogeneous-ideal-is-the-irreleva An effective way to tell if the saturation of a homogeneous ideal is the irrelevant ideal Charles Staats 2011-04-03T15:54:53Z 2012-02-08T06:04:34Z <p>Let $\Bbbk$ be an algebraically closed field, let $R$ denote the graded ring $\Bbbk[x_0, \dotsc, x_N]$, and let $f_1, \dotsc, f_n \in R_m$ be nonconstant homogeneous polynomials. Then the common vanishing locus $V(f_1, \dotsc, f_n) \subset \mathbb{P}^N$ is empty if and only if the saturation of the homogeneous ideal $I = (f_1, \dotsc, f_n)$ is the entire irrelevant ideal $R_+$. This is true iff for some $d > 0$, $I_d = R_d$ (the degree-$d$ parts are equal), in which case $I_e = R_e$ for all $e \geq d$.</p> <p>It is not hard to compute the vector subspace $I_d \subset R_d$ for successive values of $d$: if $I_d$ is generated as a $\Bbbk$-module by $a_1, \dotsc, a_k \in R_d$, then $I_{d+1}$ is generated by the $x_i a_j$, plus any of the $f_i$ that are of degree $d+1$. </p> <p>If you want to show that $I$ does, in fact, have saturation equal to the entire ideal $R_+$, you can start computing the vector subspaces $I_d$; if you're right, then sooner or later you'll get $I_d = R_d$ and have your answer. But suppose you go on and on, and $I_d$ remains stubbornly a proper subspace of $R_d$. Is there some point--when $d$ is a thousand, a million, $10^{100}$&mdash;at which you can say, "If $I_d$ does not contain all $R_d$ by now, it never will"?</p> <blockquote> <p>Does there exist $D$, depending only on the degrees of the $f_i$, sufficiently large that if $I_d = R_d$ for any $d$, then $I_D = R_D$?</p> </blockquote> <p>I'm reasonably confident that the answer to that question is yes, based on the following sketch: Look at the space $V$ of all $n$-tuples of homogeneous polynomials $(f_1, \dotsc, f_n)$ with fixed degrees $d_1, \dotsc, d_n$. Let $S_d \subset V$ be the subset of those for which $I_d = R_d$. Since the condition on $S_d$ comes down to the condition that some linear map of vector spaces is surjective, $S_d$ is Zariski-open. Thus, $S_d \subset S_{d+1} \subset S_{d+2} \subset \dotsb$ is an increasing union of Zariski-open sets, and consequently must stabilize at some $D$.</p> <p>Unfortunately, this argument is entirely non-effective. We have no idea what the value of $D$ is, and so if we actually want to show that $I^{sat} \neq R_+$, we're out of luck. This motivates the following question:</p> <blockquote> <p>Assuming an affirmative answer to the previous question, what is a (preferably computable) function $$D = D(d_1, \dotsc, d_n)$$ that works?</p> </blockquote> http://mathoverflow.net/questions/87847/explaining-the-concept-of-projective-space-notes-for-students/87852#87852 Answer by Charles Staats for Explaining the concept of projective space: notes for students Charles Staats 2012-02-07T23:17:43Z 2012-02-07T23:17:43Z <p>Appendix A of <em>Rational Points on Elliptic Curves</em>, by Silverman and Tate, may be helpful. This unfortunately deals only with the projective plane, not projective spaces in general, but a reasonably well-motivated definition is given in pages 220-224. Later sections of the appendix include an elementary proof of Bezout's Theorem.</p> http://mathoverflow.net/questions/87480/explanation-on-a-scheme-which-is-not-affine-scheme/87498#87498 Answer by Charles Staats for explanation on a scheme which is not affine scheme Charles Staats 2012-02-04T00:56:23Z 2012-02-04T00:56:23Z <p>Compute the ring $R$ of globally defined regular functions. If the scheme were affine, then there would be a bijective correspondence between closed points of the scheme and maximal ideals of $R$, given by taking a closed point to the ideal of functions that vanish at that point. But the two points represented by the colon in your diagram give the same ideal of $R$, so this "correspondence" is not injective. Therefore, the scheme is not affine.</p> http://mathoverflow.net/questions/86033/relative-integral-closure Relative integral closure Charles Staats 2012-01-18T20:45:04Z 2012-01-28T20:47:17Z <p><strong>Definition:</strong> Let $R$ be an $A$-algebra, where $R$ and $A$ are both commutative rings with unit. Let $S \subset R$ be the (multiplicatively closed) subset consisting of those nonzerodivisors $s \in R$ such that $R/sR$ is flat over $A$. I will call $R$ <em>integrally closed over $A$</em> if it satisfies</p> <ol> <li>$R$ is flat over $A$, and</li> <li>$R$ is integrally closed in $S^{-1}R$.</li> </ol> <blockquote> <p>Is $A[x_1, \dotsc, x_n]$ necessarily integrally closed over $A$? </p> </blockquote> <p>If it is helpful, you may assume $A$ is noetherian, or even of finite type over a field. But do not assume it is integral or reduced or anything like that. (In the case I care about, $A$ is likely to be an open affine on a Hilbert or Quot scheme.)</p> <p>Some thoughts: assuming things are noetherian, and I have not made an error, the condition for $s \in S$ is equivalent to the condition that $s$ pulls back to a nonzerodivisor on every fiber of $\operatorname{Spec} R$ over $\operatorname{Spec} A$. This, in turn, is equivalent to the condition that $V(s)$ contains no associated component of any fiber.</p> <p>If this "relatively integrally closed" condition is preserved under pullbacks, then the question has a positive answer. However, the only reason I might even imagine that this is true is my hope that the condition I have given is a reasonably "natural" condition for a morphism to have. I can't think of a way to begin arguing for preservation under pullbacks, and I don't even find it all that plausible.</p> <p>Another, more plausible (to me) statement that would imply a positive answer is if normal morphisms (in the sense of Grothendieck) are necessarily "relatively integrally closed." A morphism is normal if it is flat and its (geometric) fibers are normal. I have a flawed argument for this: let $\ell \in S^{-1}R$ be integral over $R$. Then over each (geometric) fiber, $\ell$ is still integral over $R$, so $R \to R[\ell]$ pulls back to an isomorphism over each (geometric) fiber, and consequently must have been an isomorphism to begin with. The flaw (that I've found and cannot seem to repair) is that, at least in principle, $R[\ell] \to S^{-1}R$ need not pull back to an injection.</p> <hr> <p><strong>Note:</strong> Will Sawin's answer below was incredibly helpful to me. Please upvote it.</p> http://mathoverflow.net/questions/85685/subtle-examples-of-morphisms-that-are-finite-but-not-flat Subtle examples of morphisms that are finite but not flat Charles Staats 2012-01-14T20:13:44Z 2012-01-18T04:41:21Z <p>Let $R$ be a ring (commutative noetherian with unit), and let $K(R)$ be its total ring of fractions (obtained by inverting all nonzerodivisors). Thus, $R \hookrightarrow K(R)$. Let $a \in K(R)$ be integral over $R$; in other words, $R[a]$ is a subring of $K(R)$ that is finite over $R$. </p> <p>Is it necessarily true that $R[a]$ is flat over $R$? More to the point, since I'm expecting a negative answer: what is a concrete counterexample?</p> http://mathoverflow.net/questions/85685/subtle-examples-of-morphisms-that-are-finite-but-not-flat/85692#85692 Answer by Charles Staats for Subtle examples of morphisms that are finite but not flat Charles Staats 2012-01-14T21:25:50Z 2012-01-14T21:25:50Z <p>Combining insight from Mahdi's and Yusuf's answers, it looks as thought the finite map in question is virtually never flat.</p> <p>Specifically, assume $R$ is an integral domain, and $K = K(R)$ is its fraction field. Let $a \in K$ be integral over $R$. Consider the finite morphism $$\operatorname{Spec} R[a] \to \operatorname{Spec} R.$$ Since $R[a]$ and $R$ have the same fraction field $K$, this morphism has degree one (is birational). If it is flat, then every fiber must have the same degree (one); in other words, this morphism is an isomorphism in every fiber. Finiteness, together with Nakayama's Lemma, then implies it is an isomorphism above every stalk. Hence, it is an isomorphism, i.e., $a \in R$.</p> <p>Contrapositively, if $a \not\in R$, then the morphism is not flat.</p> <p>For a specific example suggested by Yusuf's answer, consider $R = \Bbbk[x^2, x^3]$, and $a = x = x^3 / x^2 \in K(R)$.</p> http://mathoverflow.net/questions/82403/relationship-between-the-notions-of-excellent-ring-and-universally-catenary-na Relationship between the notions of "excellent ring" and "universally catenary Nagata ring" Charles Staats 2011-12-01T19:37:07Z 2011-12-01T19:37:07Z <p>Every excellent ring is both universally catenary and Nagata. How "close" is a universally catenary Nagata ring to being excellent?</p> <p><b>Context:</b> I have not worked very much with the notions described above. While I certainly would not turn down a counterexample, I am more interested in getting an idea of how these notions are used in practice, and when it is appropriate to use one versus the other.</p> http://mathoverflow.net/questions/82023/what-does-compose-direct-mean-in-mathematical-french What does "composé direct" mean in mathematical French? Charles Staats 2011-11-27T19:47:33Z 2011-11-29T17:49:06Z <p>In EGA IV, Lemma 6.14.1.1, the first sentence is </p> <blockquote> <p>Soit $R$ un anneau composé direct d'un nombre fini de corps.</p> </blockquote> <p>I'm guessing this means</p> <blockquote> <p>Let $R$ be a ring that is the direct product of a finite number of fields,</p> </blockquote> <p>but I'm not positive. For one thing, "produit direct" seems to be the more usual term for "direct product." I've tried looking up "composé direct" on Google and on Wikipedia.fr, and I have not been able to find a definitive reference. I'm also confused by the fact that "direct" appears to be an adjective in the French, but I can't see what noun it might be modifying if my translation is correct. Google translate gives</p> <blockquote> <p>Let R be a ring composed of a direct finite number of bodies.</p> </blockquote> <p>Obviously, "bodies" should be "fields," but it still does not seem to be helpful. Thus, my question:</p> <p><b>Question:</b> What does "composé direct" mean in this context? In particular, why is "direct" being used as an adjective?</p> <p><b>Secondary Question:</b> Is there a good reason that "composé direct" is being used here rather than another, more searchable term such as "produit direct"?</p> http://mathoverflow.net/questions/23268/geometric-intuition-for-limits Geometric intuition for limits Charles Staats 2010-05-02T17:54:55Z 2011-11-12T18:28:09Z <p>I'm the sort of mathematician who works really well with elements. I really enjoy point-set topology, and category theory tends to drive me crazy. When I was given a bunch of exercises on subjects like limits, colimits, and adjoint functors, I was able to do them, although I am sure my proofs were far longer and more laborious than they should have been. However, I felt like most of the understanding I gained from these exercises was gone within a week. I have a copy of MacLane's "Categories for the Working Mathematician," but whenever I pick it up, I can never seem to get through more than two or three pages (except in the introduction on foundations).</p> <p>A couple months ago, I was trying to use the statements found in Hartshorne about glueing schemes and morphisms and realized that these statements were inadequate for my purposes. Looking more closely, I realized that Hartshorne's hypotheses are "wrong," in roughly the same way that it is "wrong" to require, in the definition of a basis for a topology that it be closed under finite intersections. (This would, for instance, exclude the set of open balls from being a basis for $\mathbb{R}^n$.) Working through it a bit more, I realized that the "right" statement was most easily expressed by saying that a certain kind of diagram in the category of schemes has a colimit. At this point, the notion of "colimit" began to seem much more manageable: a colimit is a way of gluing objects (and morphisms).</p> <p>However, I cannot think of any similar intuition for the notion of "limit." Even in the case of a fibre product, a limit can be anything from an intersection to a product, and I find it intimidating to try to think of these two very different things as a special cases of the same construction. I understand how to show that they are; it just does not make intuitive sense, somehow.</p> <p>For another example, I think (and correct me if I am wrong) that <strike>the sheaf condition on a presheaf can be expressed as stating that the contravariant functor takes colimits to limits</strike>. [This is not correct as stated. See Martin Brandenburg's answer below for an explanation of why not, as well as what the correct statement is.] It seems like a statement this simple should make everything clearer, but I find it much easier to understand the definition in terms of compatible local sections gluing together. I can (I think) prove that they are the same, but by the time I get to one end of the proof, I've lost track of the other end intuitively.</p> <p>Thus, my question is this: Is there a nice, preferably geometric intuition for the notion of limit? If anyone can recommend a book on category theory that they think would appeal to someone like me, that would also be appreciated.</p> http://mathoverflow.net/questions/80151/is-a-union-of-local-complete-intersections-a-local-complete-intersection Is a union of local complete intersections, a local complete intersection? Charles Staats 2011-11-05T20:55:03Z 2011-11-05T22:03:33Z <p>Let $X$ be a smooth variety over a field $\Bbbk$, and let $Y, Z \subset X$ be closed reduced subschemes of the same dimension, both of which are local complete intersections. Is $Y \cup Z$ necessarily a local complete intersection?</p> http://mathoverflow.net/questions/28326/a-nontrivial-surface-on-which-any-two-curves-intersect A nontrivial surface on which any two curves intersect Charles Staats 2010-06-16T00:21:13Z 2011-10-27T18:18:22Z <p>One interesting property of the projective plane is that any two plane curves intersect. (More generally, if $V$ and $W$ are subvarieties of any projective space, and codim $V$ + codim $W \geq 0$, then $V$ and $W$ intersect.) However, the same does not seem to hold for most other easy examples of surfaces. For instance, any ruled surface $S \to C$ has non-intersecting curves (take the fibers over any two distinct points of $C$). Furthermore, any surface $S$ obtained from $\mathbb{P}^2_k$ by blowing up points $p_1, \ldots, p_n$ has two non-intersecting curves: take two lines that intersect transversely at $p_1$ and avoid $p_2, \ldots, p_n$, and lift them to curves on $S$.</p> <p>Thus, my question:</p> <blockquote> <p>Is there any nonsingular algebraic surface other than the projective plane such that any two curves on the surface have nontrivial intersection?</p> </blockquote> <p>(Note: assume the base field is algebraically closed.)</p> http://mathoverflow.net/questions/126021/a-technical-question-on-injective-hull Comment by Charles Staats Charles Staats 2013-03-30T18:28:18Z 2013-03-30T18:28:18Z Alternative notation: If I understand correctly, the module $M$ might be described as a $\mathbb C[x^{\pm 1}]$-module as $$\bigoplus_{i=1}^{\infty} \mathbb C[x^{\pm 1}] y^{-i};$$ to make it a $\mathbb C[x^{\pm 1},y]$-module, the $y$-action is the obvious one, except that $y^{-1}\cdot y = 0$. The proposed injective hull might similarly be described as $$\bigoplus_{i=1}^{\infty} \mathbb C(x) y^{-i}.$$ http://mathoverflow.net/questions/92771/a-geometric-characterization-for-arithmetic-genus/125934#125934 Comment by Charles Staats Charles Staats 2013-03-30T17:52:22Z 2013-03-30T17:52:22Z A worthwhile point. I assume what is happening here is that Chern classes are an invariant of the complex vector bundle $T_X$--but different complex structures on the same smooth manifold $X$ can give rise to non-isomorphic descriptions $T_X$ as a complex vector bundle over $X$. http://mathoverflow.net/questions/125381/why-kx-y-z-x-cap-y-z-is-complete Comment by Charles Staats Charles Staats 2013-03-23T16:06:38Z 2013-03-23T16:06:38Z To elaborate Ralph's comment: The ring you have specified is not a local ring. Thus, it is not at all clear what you mean by stating that this ring &quot;is complete.&quot; Are you sure you are correctly understanding the statement in the paper? http://mathoverflow.net/questions/116123/how-to-find-define-eigenvectors-as-a-continuous-function-of-matrix/116131#116131 Comment by Charles Staats Charles Staats 2012-12-13T14:24:51Z 2012-12-13T14:24:51Z Are you sure about &quot;nice properties of the derivative&quot;? http://mathoverflow.net/questions/115900/how-should-a-number-theorist-learn-a-modest-amount-of-algebraic-geometry Comment by Charles Staats Charles Staats 2012-12-11T00:49:23Z 2012-12-11T00:49:23Z A note on Mumford's Red Book: The very first section is much more technical than most of the book, and can be skipped (and returned to at a later time, if desired) without significant loss. http://mathoverflow.net/questions/115900/how-should-a-number-theorist-learn-a-modest-amount-of-algebraic-geometry/115903#115903 Comment by Charles Staats Charles Staats 2012-12-11T00:46:08Z 2012-12-11T00:46:08Z Note that the &quot;functor point of view&quot; is also fairly well explained in Mumford's Red Book. (And the same explanation, word-for-word, appears also in Lecture 3 of Mumford's <i>Lectures on Curves on an Algebraic Surface</i>.) http://mathoverflow.net/questions/116022/which-local-ringed-spaces-are-schemes Comment by Charles Staats Charles Staats 2012-12-11T00:37:42Z 2012-12-11T00:37:42Z It's actually not true that &quot;every restriction map of the [structure] sheaf is a localization of ring&quot;; this is only guaranteed for restrictions to distinguished open sets. http://mathoverflow.net/questions/115222/techniques-for-showing-that-a-curve-is-not-smoothable/115280#115280 Comment by Charles Staats Charles Staats 2012-12-03T18:59:42Z 2012-12-03T18:59:42Z Additional note: I've just looked up the definition of &quot;Hirzebruch surface,&quot; and actually my &quot;more basic example&quot; is a special case of your example. http://mathoverflow.net/questions/115222/techniques-for-showing-that-a-curve-is-not-smoothable Comment by Charles Staats Charles Staats 2012-12-03T17:05:40Z 2012-12-03T17:05:40Z Roy: Thanks for the excellent suggestion. Unless I misunderstand, the tangent space dimension will typically be harder to compute than the &quot;expected dimension&quot; (tangent space dimension minus obstruction dimension), since the latter is often accessible via Riemann-Roch.  One advantage of this technique is that it is (in principle) capable of working even if we are looking at an irreducible component of the moduli space, some of whose points lie in the closure of the moduli of irreducible curves. http://mathoverflow.net/questions/115222/techniques-for-showing-that-a-curve-is-not-smoothable/115280#115280 Comment by Charles Staats Charles Staats 2012-12-03T17:04:09Z 2012-12-03T17:04:09Z Edit: Probably, I mean to say that the general curve cannot be infinitesimally deformed to be irreducible. (But some points of the moduli component might lie in the closure of the space of irreducible curves.) http://mathoverflow.net/questions/115222/techniques-for-showing-that-a-curve-is-not-smoothable/115280#115280 Comment by Charles Staats Charles Staats 2012-12-03T16:58:37Z 2012-12-03T16:58:37Z I'm familiar with this technique in the case of surfaces--a more basic example is to take $X$ to be the blowup of $\mathbb P^2$ at a point, and consider reducible curves that have negative intersection number with the exceptional divisor $E$. But I have two questions: 1. Can this sort of argument be applied if the ambient $X$ is not a surface? 2. Is there any hope for this sort of argument if we want to show that the general curve is reducible on an irreducible component of the moduli space (but not the whole connected component)? http://mathoverflow.net/questions/83097/are-there-any-related-rates-calculus-problems-that-dont-feel-contrived Comment by Charles Staats Charles Staats 2012-11-30T16:54:51Z 2012-11-30T16:54:51Z In some circumstances, the ratio of two related rates is precisely the mechanical advantage. http://mathoverflow.net/questions/111464/self-intersection-and-the-normal-bundle Comment by Charles Staats Charles Staats 2012-11-04T19:09:34Z 2012-11-04T19:09:34Z In more senses than one. http://mathoverflow.net/questions/111464/self-intersection-and-the-normal-bundle Comment by Charles Staats Charles Staats 2012-11-04T18:58:22Z 2012-11-04T18:58:22Z ...the points that are &quot;infinitesimally fixed&quot; by the deformation, i.e., not moved off of $C$. This is not an unusual situation: what is true locally in topology, is true infinitesimally in algebraic geometry. http://mathoverflow.net/questions/111464/self-intersection-and-the-normal-bundle Comment by Charles Staats Charles Staats 2012-11-04T18:54:02Z 2012-11-04T18:54:02Z There are a couple of things that may be helpful here. One is the notion of &quot;deformation to the normal cone.&quot; Another is the following: a section of the normal bundle $\mathcal N_{X/C}$ is precisely a first-order deformation of $C$, i.e., a tangent vector to the point $[C]$ in the space [Hilbert scheme] $H$ of closed subschemes of $X$. Whenever you can move $C$ via a smooth, one-dimensional family of curves in $X$, this corresponds to a curve in $H$ passing through $[C]$. The tangent vector at $[C]$ corresponds to a section of $\mathcal N_{X/C}$, and the zeros of this section are precisely...