User - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T06:52:34Z http://mathoverflow.net/feeds/user/5482 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/125164/functor-used-to-construct-sheafifications-property + functor (used to construct sheafification)'s property unknown (google) 2013-03-21T12:35:05Z 2013-03-23T06:34:09Z <p>Let $X$ be a topological space, $\mathcal{C}$ be a locally small category with "good properties" (such as having small inverse limit, small filtrant inductive limit...etc.) and $\mathcal{F}$ be a presheaf with values in $\mathcal{C}$.</p> <p>The + functor turns $\mathcal{F}$ to another presheaf $F^{+}$ such that: For any open subest $U$ of $X$, </p> <p>$$F^{+}(U) = \varinjlim_{\mathcal{U}} \ \mathcal{F}(\mathcal{U}),$$ here $\mathcal{U}$ is a open convering of $U$ and $\mathcal{F}(\mathcal{U})$ is the equalizer of</p> <p>$$\prod_{V \in \mathcal{U}} \mathcal{F}(V) \rightrightarrows \prod_{V_1, V_2 \in \mathcal{U}} \mathcal{F} (V_1 \ \cap \ V_2)$$</p> <p>One shows that $\mathcal{F}^{+}$ is a separated sheaf and if $\mathcal{F}$ is separated, $\mathcal{F}^{+}$ is a sheaf. So the sheafification is $\mathcal{F}^{++}$.</p> <p>If $\mathcal{C}$ is a concrete category (everything are sets and maps), then the proof for these are clear.</p> <p>My question is: For non-concrete categories, does one need some more conditions on $\mathcal{C}$ to prove these?</p> <p>For example, in order to prove that $\mathcal{F}^{+}$ is separated, which is equivalent to</p> <p>$$\mathcal{F}^{+} (U) \rightarrow \mathcal{F}^{+} (\mathcal{U}) \mathrm{\ is \ a \ monomorphism},$$</p> <p>I found I need some conditions such as:</p> <p>$$\varinjlim_{i} \ \mathrm{Hom}(A, X_i) \rightarrow \mathrm{Hom}(A, Y) \mathrm{ \ is \ injective}$$ $$<br> \varinjlim_{i} \ \mathrm{Hom}(A, X_i) \rightarrow \mathrm{Hom}(A, \varinjlim_{i} \ X_i) \mathrm{ \ is \ surjective}$$</p> <p>for some special filtrant inductive limits (those comes from open coverings of open subsets.)</p> <p>The above conditioins are just from trying to mimic the proof when $\mathcal{C}$ is concrete. For example, for an element $x$ in a filtrant inductive limit, one can find some index $i$ and an element $x_i \in X_i$ such that $x$ is the image of $x_i$.</p> <p>I would like to know if such conditions are really necessary. If not, how one proves the properties of $\mathcal{F}^{+}$?</p> <hr> <p>Edit: This question have been downvoted once, and I would like to know why. Is this question is not suitable for mathoverflow or it is just not the question of someone's favor?</p> http://mathoverflow.net/questions/69120/normal-crossing-divisor-v-s-strict-normal-crossing-divisor normal crossing divisor v.s. strict normal crossing divisor unknown (google) 2011-06-29T15:30:01Z 2013-01-22T01:27:34Z <p>On wikipedia, the normal crossing divisor is defined to be (by my understanding):</p> <p>(Assume $X/k$ be a smooth geometrically integral scheme of finite type over a field $k$).</p> <p>Let $D = \sum_{i=1}^n C_i$ be a Weil Divisor, here $C_i$ are irreducible closed subsets of codimension 1 of $X$. Endow $C_i$ with the reduced scheme structure (hence they are integral closed $k$-scheme of $X$ of codimension 1.) We call $D$ is a normal crossing divisor if each $D_i$ is smooth over $k$ and $D_i$'s intersect transversely.</p> <p>But in somewhere, I saw the notion "strict normal crossing divisor". For example,</p> <p>Definition 1.5.1, p.8, in <a href="http://math.arizona.edu/~swc/aws/07/KedlayaNotes10Mar.pdf" rel="nofollow">http://math.arizona.edu/~swc/aws/07/KedlayaNotes10Mar.pdf</a></p> <p>or</p> <p>5.1, p.16, in <a href="http://www.uni-due.de/~bm0032/publ/TubNbdDocumenta.pdf" rel="nofollow">http://www.uni-due.de/~bm0032/publ/TubNbdDocumenta.pdf</a></p> <p>The definition looks the same as "normal crossing divisor".</p> <p>I would like to know what's the difference between these two definitions. It would be great with examples. Also, if the base scheme $S$ is not a field, then is the definition the same as above with the modification "each $D_i$ is smooth over $S$"?</p> http://mathoverflow.net/questions/116010/complex-multiplication complex multiplication unknown (google) 2012-12-10T19:50:39Z 2012-12-10T21:00:12Z <p>For an abelian variety $A$, it is said to be have $complex \ multiplication$ if $\mathrm{End}(A) \otimes_{\mathbb{Z}} \mathbb{Q}$ contains a number filed $F$ of degree $2 \cdot \mathrm{dim} (A)$. (This is the definition I saw.)</p> <p>Now assume $A$ is simple. From sec. 1 in Ch. 1 in Lang's book "Complex Multiplication", if $A$ has complex multiplication, then $F = \mathrm{End}(A) \otimes_{\mathbb{Z}} \mathbb{Q}$. (If I understand Thm. 3.1, Lemma 3.2 and Thm. 3.3 correctly.)</p> <p>But I also saw a paragraph on another book:</p> <p><a href="http://postimage.org/image/tuoj3u709/" rel="nofollow">http://postimage.org/image/tuoj3u709/</a></p> <p>I don't understand this example well. Does it has an imaginary quadratic field? Or these above theorems in Lang's book require the abelian variety being over a characteristic 0 field?</p> http://mathoverflow.net/questions/99322/books-or-notes-on-complex-multiplication books (or notes) on complex multiplication unknown (google) 2012-06-11T23:32:21Z 2012-08-07T15:22:01Z <p>This would be a vague question, but I still want to ask here. Do you have any recommended book on complex multiplicaton. I know only 2 books: Shimura's book <code>Abelian Varieties with Complex Multiplication and Modular Functions</code> and Lang's book <code>Complex Multiplication</code>. Shimura's book used old-language (published in 1961), and I feel it would be nice to read this book when I have already learned complex multiplication. (But it would be great to know if there is something treated only or treated well in this book compared to other resource). There is also Milne's note.</p> <p>So other than these, do you have recommended books or notes?</p> http://mathoverflow.net/questions/100117/induced-maps-of-an-automorphism-of-a-curve-on-the-tangent-ot-its-jacobian-and-on Induced maps of an automorphism of a curve on the tangent ot its jacobian and on its differential forms unknown (google) 2012-06-20T11:34:51Z 2012-06-20T13:45:58Z <p>Let $C$ be a smooth projective curve of genus $g$ over a field $k$ and $J$ be its jacobian (defined over $k$). Let $\sigma: C \rightarrow C$ be a $k$-automorphsm of $C$. This automorphism $\sigma$ induces an automorphism (as abelian variety) on $J$, and hence on the tangent space $T$ at the origin $0_J$ of $J$. We know $\mathrm{dim}_k T = \mathrm{dim} \ J = g$. On the other hand, we can consider the sheaf of differential $\Omega_C$ on C. We know $\mathrm{dim}_k H^{0}(C, \Omega) = g$ and $\sigma$ induce an automorphism on $H^{0}(C, \Omega)$.</p> <p>My questions is:</p> <p>The induced maps of $\sigma$ on "the tangent $T$ of $J$" and on "the differential forms of $C$" are (canonical) related? (Over $\mathbb{C}$ only or true in general?)</p> <p>In a book, the author mentioned that these two maps on vector spaces are the same without giving the reason. I am OK with this fact, but I just don't know how to see this. I would like to know if the identification is canoncial also?</p> http://mathoverflow.net/questions/99015/kernel-of-an-isogeny-and-coker-of-its-induced-map-on-the-tate-module kernel of an isogeny and coker of its induced map on the Tate module unknown (google) 2012-06-07T06:48:36Z 2012-06-07T07:24:00Z <p>In a proof in Milne's note "Abelian Variety" (top on p.52), I saw an equality: $\mathrm{Ker}(\beta)(l) = \mathrm{Coker}(T_{l}(\beta))$, here $\beta$ is an (separable) isogeny of an abelian variety $A/k$, $l$ is a prime number different from the characteristic of the base field $k$, $(l)$ means the torsion points of order a power of $l$, $T_{l}(\beta)$ is the induced map of $\beta$ on the Tate module of $A$.</p> <p>I can't figure out why this equality holds. Do we have a natural map for these two groups?</p> http://mathoverflow.net/questions/86248/induced-map-on-algebraic-de-rham-cohomology Induced map on algebraic de Rham cohomology unknown (google) 2012-01-20T19:52:17Z 2012-01-28T03:02:48Z <p>Let $X/k$ and $Y/k$ be two smooth affine varieties over a field $k$ with $\mathrm{char}(k) = 0$ and $\varphi: X \rightarrow Y$ be a morphism. I would like to know under what conditions, the induced map $\varphi^{\ast}: H^i_{dR}(Y/k) \rightarrow H^i_{dR}(X/k)$ is injective. If $\varphi$ is dominant, then this is true for $i=0$. But for $i \geq 1$, I don't have idea.</p> http://mathoverflow.net/questions/85966/newton-method-in-p-adic-case Newton Method in $p$-adic case unknown (google) 2012-01-18T07:53:04Z 2012-01-18T11:08:04Z <p>The Newton Method over $\mathbb{R}$ has the property that the precision is doubled (under some continuous differentialbe assumption) in each iteration. For the ring $\mathbb{Z}_p$ of $p$-adic integers, we also have the Hensel's lemma, but the precision is increase by 1 in each iteration. The proof about the double precision in the case over $\mathbb{R}$ uses the Taylor expansion and its remainder, see <a href="http://www.answers.com/topic/newton-s-method#Analysis" rel="nofollow">http://www.answers.com/topic/newton-s-method#Analysis</a> . </p> <p>In a special case that is finding square root, one can use $x_{n+1} := \frac{1}{2} x_n - \frac{3}{2} x_n^3$ ( by considering $x^{-2}-c^{-1}$ instead of $x^2-c$) and one can show that the precision is doubled in the case over $\mathbb{Z}_p$.</p> <p>In general, is the precision is doubled in each iteration in the case over $\mathbb{Z}_p$, under some conditions?</p> <p>=======================================================================================================</p> <p>Sorry, I just fond that in fact the precision is doubled in $p$-adic case, without further assumption. I mentioned the precision increase by 1, since this is written in "A first course in $p$-adic Analysis" by Alain M. Robert.</p> <p><img src="http://i41.tinypic.com/f1d3qh.png" alt="alt text"></p> <p>But if one examines the proof ,we have $p(\hat{x}) \equiv 0 \ \mathrm{mod} \ p^{n-2k}$. And if $k = 0$, the precision is doubled.</p> http://mathoverflow.net/questions/27708/formal-completion formal completion unknown (google) 2010-06-10T15:35:14Z 2011-11-08T09:36:12Z <p>When I study formal completion and formal schemes, on p.194 of Hartshorne's "Algebraic Geometry", he said "One sees easily that the stalks of the sheaf $\mathcal{O}_{\hat{X}}$ are local rings." </p> <p>Notice that here $\mathcal{O}_{\hat{X}}$ is not the structure sheaf of X, there is a "hat" on the symbol $X$.</p> <p>But I can't see the reason for that the stalks of the sheaf $\mathcal{O}_{\hat{X}}$ are local rings.</p> <p>Could someone explains this for me, thanks.</p> http://mathoverflow.net/questions/78469/a-question-on-continuity-of-g-module-for-a-profinite-group-g a question on continuity of $G$-module for a profinite group $G$ unknown (google) 2011-10-18T16:47:15Z 2011-10-18T22:21:44Z <p>I have seen the following statment somewhere, for example in Appendix B2 on Silverman's book "The Arithmetic of Elliptic Curves" : Let $M$ be an abelian group with discrete topology and $G$ be a profinite group. Then an linear action ( which means that $\sigma(m_1+m_2)=\sigma(m_1)+\sigma(m_2)$, i.e it is a $G$-module) $\phi : G \times M \rightarrow M$ is continuous if and only if the stabilizer $\sigma \in G | \sigma(m)=m$ has finite index in $G$ for all $m \in M$. But what we need is that this stabilizer is open in $G$. I also saw that in a profinite group, not every subgroup of finite index is open. So is this statement correct? Or how to see that this stabilizer is open if it has finite rank?</p> http://mathoverflow.net/questions/76337/exhaustiveness-and-regularness-of-a-filtration-of-a-complex Exhaustiveness and regularness of a filtration of a complex unknown (google) 2011-09-25T16:00:54Z 2011-09-25T16:00:54Z <p>I am learning spectral sequence, but I didn't find a "clear" definition of exhaustiveness and regularness of a filtration $F^{\bullet} := \cdots \subset F^{p+1} \subset F^{p} \subset F^{p-1} \subset \cdots$ of a complex $K^{\bullet}$. On wikipedia, a filtration is exhaustive if </p> <blockquote> <p><em>the union of the set of all $F^{p}$ is the entire chain complex $K^{\bullet}$</em></p> </blockquote> <p>Naively, I think this means for any $q$, $K^q$ is the union of all $F^{p,q}$ (the degree $q$ component of $F^{p}$)</p> <p>And in one another book, it's defined by</p> <blockquote> <p>For each $q$, there is a number $p(q)$, suth that $K^q \subseteq F^{p(q)}$</p> </blockquote> <p>These two definitions are different. I am not sure about what "regularness" is for a filtration neither.</p> <p>And finally, since we may work on an abelian category, and in this case, how do one define these two properties? Intersection may be defined by fiber product, but for the union?</p> <p>p.s: I would like to know if there is a reference for spectral sequence with a good illustration of its power for computation. It would be great if it focus on "Leray spectral sequence" and using it to compute some sheaf cohomology (especially on schemes).</p> http://mathoverflow.net/questions/75164/relation-between-pull-back-of-cartier-divisors-invertible-sheaves-and-global-sec relation between pull-back of Cartier divisors, invertible sheaves and global sections unknown (google) 2011-09-11T21:56:12Z 2011-09-11T21:56:12Z <p>Let $\iota:X \hookrightarrow \mathbb{P}^n_k$ be a closed embedding of an irreducible non-singular projective $k$-scheme. For any hyperplane $H$ of $\mathbb{P}^n_k$ which doesn't contain $X$, since $X$ is reduced and $H$ doesn't contain $X$, one can pull-back the Cartier divisor defined by $H$ on $X$, see Lemma 1.29, p.260, Algebraic geometry and Arithmetic Curves by Qing Liu . This Cartier divisor on $X$ corresponds to a Weil divisor (via the isomorpihsm between Cartier divisors and Weil divisors on $X$), which we denote it by $H \cap X$. We have an isomorphism of invertible sheaf $\mathcal{O}_{\mathbb{P^n_k}} (H) \cong \mathcal{O}_{\mathbb{P}^n_k}(1)$ and $\mathcal{O}_X(H \cap X) \cong \mathcal{O}_X(1):= \iota^*(\mathcal{O}_{\mathbb{P}^n_k}(1))$, hence we have $\Gamma(\mathbb{P}^n_k, \mathcal{O}_{\mathbb{P}^n_k} (H) ) \rightarrow \Gamma(X, \mathcal{O}_X(H \cap X))$. But we know that $\Gamma(\mathbb{P}^n_k, \mathcal{O}_{\mathbb{P}^n_k} (H) ) = L(H)$ and $\Gamma(X, \mathcal{O}_X(H \cap X)) = L(H \cap X)$, so we have $L(H) \rightarrow L(H \cap X)$, here $L(D)$ is the usual Riemann-Roch space for a Weil divisor $D$. </p> <p>Now comes my question: If $f \in L(H) \subseteq K(\mathbb{P}^n_k)$ such that $\mathrm{Supp }(\mathrm{div} (f) )$ doesn't contain $X$, then one can pull-back $f$ on $X$, which is an element in the function field $K(X)$ of $X$ and turns out to be in $L(H \cap X)$. But if $\mathrm{Supp }(\mathrm{div} (f) )$ contains $X$, we can't pull-back $f$ on $X$ (as least, I don't know a way to do this). So it looks weired that we have a map $L(H) \rightarrow L(H \cap X)$ which is obtained from the corresponding invertible sheaf and on the other hand, we can't describe it naively by pull-back (only works for most $f$, but not all $f$ ). So could we describe $L(H) \rightarrow L(H \cap X)$ naturally?</p> <p>I am thinking this question when I consider the following question :</p> <p>Let $D$ be an effective Weil divisor on $X$ such that $\mathcal{O}_X(D) \cong O_X(1)$, and assume that $\Gamma(\mathbb{P}^n_k, \mathcal{O}_{\mathbb{P}^n_k} (1) ) \rightarrow \Gamma(X, \mathcal{O}_X(1))$ is surjective, then could we find a hyperplane $H$ on $\mathbb{P}^n_k$ such that $H \cap X = \mathrm{Supp} (D)$. If the above map $L(H) \rightarrow L(H \cap X)$ can be described by pull-back for all $f \in L(H)$, then this could be done. </p> http://mathoverflow.net/questions/74684/push-forward-of-the-structure-sheaf-stein-factorization-birational-and-connecte push-forward of the structure sheaf, stein factorization, birational and connected fibers unknown (google) 2011-09-06T20:46:47Z 2011-09-07T10:28:50Z <p>I am confused with the following observation: Let $f : X:=\mathrm{Spec}(K) \rightarrow \mathrm{Spec}(k) =: Y$ be a scheme morphism corresponding to a non-trivial finite field extension ( hence $f$ is proper and $X$, $Y$ are noetherian). Its Stein factorization is $X \overset { \mathrm{id}_X} \longrightarrow X \overset {f} \longrightarrow Y$. But it's clear that $f$ is not a birational, even if $char=0$. We don't have</p> <p>$f_{*} \mathcal{O}_X = \mathcal{O}_Y$</p> <p>in this example ( $char = 0$ or not) either, although we have $f$ has connected fiber and $Y$ is normal.</p> <hr> <p>When I read the following discussion, I got a question.</p> <blockquote> <p><a href="http://mathoverflow.net/questions/63301/when-will-the-pushforward-of-a-structure-sheaf-still-be-a-structure-sheaf" rel="nofollow">When will the pushforward of a structure sheaf still be a structure sheaf?</a> </p> </blockquote> <p>Let $f: X \rightarrow Y$ be a proper morphism of noetherian shcemes and $X \overset {f'} \rightarrow Z \overset {g} \rightarrow Y$ be its Stein factorization. In J.C. Ottem response, he mentioned that if the fibers of $f$ are connected, then $g$ must be birational ( I think that we assume both $X$ and $Y$ are integral schemes), and from this, one gets $g_{*} \mathcal{O}_Z = \mathcal{O}_Y$, hence we have </p> <p>$f_{*} \mathcal{O}_X = \mathcal{O}_Y$</p> <p>He also give the reference, which is Hartshorne III.10.3, for which I think the right one is Hartshorne III.11.3, but 11.3 is telling that the isomorphism gives connected fibers.</p> <p>So I would like to know :</p> <p>(1) How to see that $g$ is birational? The correct reference?</p> <p>(2) In which part we need the characteristic 0 condition?</p> <p>(3) Could we replace the characteristic 0 condition to another condition, e.g $f$ has integral fibers?</p> http://mathoverflow.net/questions/72631/automorphism-groups-of-elliptic-curves-as-galois-module Automorphism groups of Elliptic curves as Galois module unknown (google) 2011-08-10T22:29:50Z 2011-08-11T12:18:57Z <p>Let $E/k$ be an elliptic curve over a field of characteristic $\neq$ 2, 3. Then we have an isomorphism $[ \ \ ] :\mu_n \rightarrow\mathrm{Aut}_{\overline{k}}(E)$, $[ \zeta ] : (x,y) \rightarrow (\zeta^2x, \zeta^3y)$, here $n=2, 4,6$, depending on the $j$-invariant $j(E)$. See Corollary 10.2 on Ch3 in "The arithmetic of Elliptic Curves" by Silverman. There it mentioned that this isomorphism commutes with the Galois action, but I am confused. For example, let $\sigma \in G=\mathrm{Gal}(\overline{k}/k)$, then $[\zeta^\sigma] : (x,y) \rightarrow ( (\zeta^\sigma)^2x, (\zeta^\sigma)^3y)$, but $\sigma( [\zeta])$ is $(x,y) \rightarrow (\zeta^2x, \zeta^3y) \rightarrow ( (\zeta^2x)^\sigma, (\zeta^3y)^\sigma)$, hence they are different. Am I thinking something in the wrong way? ( Sorry about such level of question....)</p> http://mathoverflow.net/questions/72579/geometrical-reducedness-of-the-identity-connected-component-reference-request geometrical reducedness of the identity connected component (reference request) unknown (google) 2011-08-10T11:38:43Z 2011-08-10T13:28:04Z <p>I think there are references for this question, but I didn't find it. We know that for a simple abelian variety $A/k$, the rign $\mathrm{End}^0 (A)$ is a division algebra. One use the fact that every homomorphism $\phi : A \rightarrow A$ is either an isogeny or $0$. To see this, one consider the identity connected component $G^0$ of $G := \mathrm{ker}(\phi)$ and one shows that $G^0_{\mathrm{red}}$ is an abelian subvariety of $A$. But I can't find a reference for the geometrical reducedness of $G^0_{\mathrm{red}}$ in the general case ($k$ not necessary being perfect). It would be appreciated for a reference or an explanation.</p> http://mathoverflow.net/questions/71878/two-different-properties-for-the-quotient two different properties for the quotient unknown (google) 2011-08-02T13:32:40Z 2011-08-03T17:12:54Z <p>(Updated)</p> <p>I have looked the draft of Ch4 of the book "Abelian Varieties" by Gerard van der Geer and Ben Moonen. It looks like in order to see the group scheme structure on G/H, one should consider the fppf quotient. It is eaiser to see the group scheme structure on the fppf quotient. And one can prove that the fppf quotient is equal to the category quotient. Is this the standard way? ( In fact, I am curious if we need the notion of grothendieck topology to see the group scheme structure on G/H)</p> <p>===</p> <p>Let me just mention the original question for this topic which is about the quotient of a scheme by a finite group scheme action. In SGA 3, the (general) definition is as following: Consider a diagram</p> <p>$$X_1 { \xrightarrow[]{d_0} \atop \xrightarrow[d_1]{} } X_0 \xrightarrow{ \ p \ } Y$$</p> <p>We call $(Y,p)$ is a quotient if $p \circ d_0 = p \circ d_1$ and for any $q: X_0 \rightarrow Z$ such that $q \circ d_0 = q \circ d_1$, there exists a unique $r: Y \rightarrow Z$ such that $q = r \circ p$. The existence of the quotient $Y$ is equivalent to the representability of the functor $K: T \rightarrow K(T)$, i.e $K=\mathrm{Hom}(Y,-)$, here $K(T)$ is the kernel of </p> <p>$$\mathrm{Hom}(X_0, T) { \xrightarrow[]{T(d_0)} \atop \xrightarrow[T(d_1)]{} } \mathrm{Hom}(X_1, T)$$ </p> <p>In SGA 3, it's proved that the quotient exists in some case.</p> <p>On the other hand, on wikipedia(group scheme), it's written that:</p> <p><em>"For a subgroup scheme H of a group scheme G, the functor that takes an S-scheme T to G(T)/H(T) is in general not a sheaf, and even its sheafification is in general not representable as a scheme. However, if H is finite, flat, and closed in G, then the quotient is representable, and admits a canonical left G-action by translation. If the restriction of this action to H is trivial, then H is said to be normal, and the quotient scheme admits a natural group law. Representability holds in many other cases, such as when H is closed in G and both are affine.[1]"</em></p> <p>It looks like that the two definitons of quotient are different. The first one considers morphisms to an object $T$ and the second definition considers morphisms from $T$. The first definition seems more natural to me for the quotient.</p> <p>My question is : are these two definitions equivalent, under the following assumptions:</p> <p>$X_0 = G$ is a group scheme, $X_1 = H \times G$ for a finite closed subgroup scheme $H$, $d_0 = m$ being the induced morphism from the multiplication and $d_1$ is the second projection.</p> <p>ps: When I tried to figure out how to give a multiplication on the quotient $G/H$ (of course, one needs the condition "normal"), I have the first definition of quotient in mind, and can't see why "<em>...and admits a canonical left G-action by translation"</em>. Using the second definition, it is easy to see it. </p> http://mathoverflow.net/questions/69453/dual-isogeny-for-abelian-varieties-over-a-general-field dual isogeny for abelian varieties over a general field unknown (google) 2011-07-04T09:40:06Z 2011-07-26T22:36:06Z <p>Let $\phi: A \rightarrow B$ be a separable isogeny between two abelian varieties over a field $k$. One knows that there is a dual isogeny $\hat {\phi} : B \rightarrow A$ such that $\hat{\phi} \circ \phi =$ multiplication by $\mathrm{deg}(\phi)$.</p> <p>When I studied elliptic curves and abelian varieties, most of the references deal with a base field which is perfect. In this case, the proof for the existence of the dual isogeny is as following:</p> <p>One makes a base change to work on the algebraic closure $\overline{k}$ of $k$. One considers $\mathrm{ker} (\phi)$, the closed points of the fiber of $\phi$ at the origin of $B$. It's a finite group which acts on $A$ and we have $\mathrm{ ker } (\phi) \subset \mathrm{ker} ( \times \mathrm{deg} (\phi))$. Then using the results about the quotien of a scheme by a finite group, one get the dual isogeny $\hat{\phi}$. Finally, one uses the action of $\mathrm{Gal}(\overline{k}/k)$ to get the result on the original base filed $k$.</p> <p>Now if $k$ is not perfect, I didn't figure out how to do this. I have a feeling that the reason we work on the algebraic closure $\overline{k}$ is to get a finite group action. Of course, one can think about the action of $\mathrm{ker} (\phi)(k)$, but I think it doesn't work, because it's too small (We need all its closed points).</p> <p>I would like to know if the dual isogeny exists in a general base field. If yes, what's the idea to see it, (and some reference) </p> <p>======</p> <p>For elliptic curves, one can treat the inseparable case (with the base field $k$ being perfect or not.) The idea is as following: </p> <p>(1) For every morphism $\psi : C_1 \rightarrow C_2$ of smooth curves (geometrically integral, complete) over a filed of char($k$) $=p > 0$, it factors as $C_1 \ \xrightarrow{\phi} \ C_1^{(q)} \ \xrightarrow{\lambda} \ C_2$, here $\phi$ is the $q^{\mathrm{th}}$-power Frobenius morphism, $q=$ inseparable degree of $\psi$, and $\lambda$ is separable.</p> <p>(2) Using (1), one only need to treat with separable case and Frobenius morphism case. Furthermore, one only need to treat with the $p^{\mathrm{th}}$-power Frobenius morphism $\phi$ only. In elliptic curves case, one shows that the morphism $\times p$ is not separable, hence using (1), its factroization must contains $\phi$, i.e $\times p = \lambda \circ \phi^r$ and we are done.</p> <p>Does this method can be used to higher dimensional abelian varieties? I have some difficulities to di this.</p> <p>(a) The factorization in (1): In elliptic curves case, the degree of the $p^{\mathrm{th}}$-power Frobenius morphism $\phi$ is $p$, (not a power of $p$), which make it possible to construct this factorization. (Silverman, The Arithmetic of Elliptic Curves, p.30, Corollary 2.12). But for an abelian varieties of dimension $g$, the degree is $p^g$. In order to construct this factorization, one should have a result "the inseparable degree of an isogeny is a power of $p^g$", which I don't know if this is correct.</p> <p>(b) One need to show that $\times p$ is not separable, which is ok.</p> <p>So for (a), is it possible to do this for higher dimensional abelian varieties?</p> <p>=========</p> <p>I have learned the quotient of a scheme by a finite group scheme action, but still can't give a proof for this for purely inseparable case. Let $K$ be the kernel of $\phi$, and consider the action $m: K \times_k A \rightarrow A$ which is induced by the multiplication on $A$ and let $\pi : K \times_k A \rightarrow A$ be the natural projection. Let $\psi : A \rightarrow A$ be the multiplication by $\mathrm{deg}(\phi)$ map. Then one needs to show that $\psi \circ m = \psi \circ \pi$ to get the desired induced morphism, but I can't get a proof for this. If $k$ is perfect, then by working on the function fields of $A$ and $B$, one only need to deal with $\phi =$ Frobenius morphism. (But still need to show that the multiplication by $p$ factors through by the Frobeinus morphism.) I need help to give a proof for this case ( $k$ not necessary being perfect), i.e the equality $\psi \circ m = \psi \circ \pi$. I think one just need to show that the induced morphism of $\psi$ on $K$ factors throguth the structure morphism $K \rightarrow k$ followed by the constant morphism $e: k \rightarrow K$. </p> <p>=========</p> <p>Thank Qing Liu for the reference. Today I thought I had a way to prove this. Suppose $\phi : A \rightarrow B$ has degree $p^r$. Let $\psi := \times p^r : A \rightarrow A$. In order to show that the existence of $\hat{ \phi}$, one only need to show that $\psi^* (K(A))$ is contained in $\phi^* (K(B))$, here $K(A)$ and $K(B)$ are the function fields of $A$ and $B$. (Then we have a rational map from $B$ to $A$ and extends to a morphism $\hat{ \phi }$.) But notice that for each element $g \in K(A)$, $g^{p^r} \in \phi^* (K(B))$. If one can show that for each $f \in K(A)$, there exists $g \in K(A)$ such that $\psi^* (f) = g^{p^r}$, then we are done.</p> <p>Since the multiplication by $p$ morphism $[p] : A \rightarrow A$ is the composition of the Frobenius morphism $F: A \rightarrow A^{(p)}$ with the Verschiebung morphism $V : A^{(p)} \rightarrow A$, one sees that for any $f \in K(A)$, $[p]^* (f) = g^p$ for some $g \in K(A)$, if $k$ is perfect. In particular, this holds for an algebraically closed field $k$. So in this case, the statement in the end of the previous paragraph is true.</p> <p>Finally, for $k$ not necessary being algebraically closed, we take a base change and work on $\overline{k}$ first. We then get $\overline{\psi} = \Phi \circ \overline{\phi}$, here $\Phi : \overline{B} \rightarrow \overline{A}$ and the $\ \overline{ \ }$ means the schemes and the morphisms obtained from the base change. For any $f \in K(A)$, if one can show that $\Phi^* (f) \in \phi^* (K(B))$, then one completes the proof. But we have $\Phi^* (f) \in K(A) \cap \overline{\phi}^* (\overline{K}(B)) = K(A) \cap \phi^* (K(B)) \otimes_k \overline{k}$. Consider the subfield extension $\phi^* (K(B)) (\Phi^* (f))$ of $K(A)/\phi^* (K(B))$, then $\Phi^* (f) \in K(A) \cap \overline{\phi}^* (\overline{K}(B))$ gives $\phi^* (K(B)) (\Phi^* (f)) = \phi^* (K(B))$, and we are done.</p> <p>Just a remark about the factorization of the morphism $[p] = V \circ F$. This is somehow different with our $\phi$ and $\hat{ \phi }$, since $\mathrm{deg}(F) = p^g$, not $p$, so it's a stronger result. Also one can show that the kernel of $F$ is of this form (via a $k$-isomorphism) $k[X_1, \cdots, X_n]/(X_1^p, \cdots, X_n^p)$. For a commutative group scheme of this form, I think one can show that the morphism $[p]$ on it is the zero morphism $0$ by dealing with symmetric functions. (I am not sure about this, but I saw somewhere that the Verschiebung morphism $V$ also uses the symmetric functions.) I like this way of approach since it uses the description of $[p]$ by the Frobenius morphism $F$, and the kernel $K$ of $F$ is explicit (if one chooses a coordinate), also the factorization $[p] = V \circ F$ seems can be proved without further knowledge. However, in Mumford's book, there is a proof that $[p] = 0$ for height one commutative group scheme by using the Lie-algebra. It would be appreciated for any comment on this.</p> http://mathoverflow.net/questions/70489/whats-a-finite-group-scheme-action-on-a-variety what´s a finite group scheme action on a variety? unknown (google) 2011-07-16T09:31:01Z 2011-07-18T08:17:15Z <p>Let $G/k$ be a finite group scheme over a field $k$ and $X$ be $k$-scheme of finite type. An action of $G$ on $X$ is a $k$-morphism $\mu : G \times_k X \rightarrow X$ satisfying the usual conditions. In SGA3-V-4 and 5, it states that the quotient $X/G$ exists if $\mu$ is a finite flat morphism with other conditions. But in somewhere, I saw that the quotient $X/G$ exists if $G$ is a finite group scheme over $k$, but it doesn't require that the action morphism $\mu$ to be finite or flat. My question is that if we need $\mu$ to be finite flat to ensure the existence of the quotient $X/G$.</p> <p>Now I will show that under the assumption that $X$ is geometrically reduced, an action of a finite group scheme over $k$ on $X$ is always finite flat. So the above question reduces to if we really need the geometrically reduced assumption. This assumption enables us to get a scheme morphism from a "morphism" on its closed points. Also, even under this assumption, is there another way to show that the finiteness of $\mu$ without working on the base change to the algebraically closure of $k$ first.</p> <p>Notice that a finite $k$-scheme $G$ is a finite disjoint union of $U_g := \mathrm{Spec}(A_g)$ with each $A_g$ being a finite $k$-algebra ( $\mathrm{dim}(A_g) = 0$ ) such that $U_g$ is a one-point set and each $U_g$ is both open and closed in $G$. For each point $g \in G$, we denote the one point set $g$ by $U_g$, or simply by $g$, and $(U_g)_\mathrm{red}$ by $\overline{g}$.</p> <p>We also denote the point in $G_{\mathrm{red}}$ which corresponds to the point $g$ in $G$ by $\overline{g}$.</p> <p>Let first assume that $k$ is algebraically closed. In this case, each point $g$ of $G$ is a $k$-rational point and that $\overline{g}$ is an open affine subset of $G_{\mathrm{red}}$, which is $k$-isomorphic to $\mathrm{Spec}(k)$. For each $\overline{g}$, we consider the natural morphism $\overline{g} \times_k X = (U_g)_\mathrm{red} \times_k X \rightarrow X$. From the conditions of a group scheme acting on a scheme, we know that each point $\overline{g}$ gives an isomorphism from $X(k)$ to $X(k)$. The assumption on geometrically reduced tells us that it in fact gives an $k$-isomorphism of $X$. Hence the composition $\mu_{\overline{g}} : {\overline{g}} \times_k X \rightarrow g \times_k X \rightarrow X$ is a $k$-isomorphism. Using the fact that the natural morphism $i : Y_{\mathrm{red}} \rightarrow Y$ has a property that $i(U)$ is an open affine subset of $Y$ for any open affine subset $U$ of $Y_{\mathrm{red}}$ ( for any noetherian scheme $Y$ ), one sees that $\mu_{\overline{g}}$ is an affine morphism. For any open affine subset $\mathrm{Spec}(A)$ of $X$, let $\mathrm{Spec}(B)$ be its inverse image under $\mu_g : g \times_k X \rightarrow X$. We have that the composition of $A \leftarrow B \leftarrow A$ is $\mathrm{id}_A$. Also notice that $A \leftarrow B$ is the quotient of $B$ by a nilpotent ideal. This implies that for each $b \in b$, there exists $a \in A$ such that $b-a$ is nilpotent. Hence $B \leftarrow A$ is an integral homomorphism. It's easy to see that $\mu_g$ is of finite type. Hence $B \leftarrow A$ is in fact finite. This proves that $\mu_g : g \times_k X \rightarrow X$ is finite flat. Since $\mu : G \times_k X \rightarrow X$ is the finite disjoint union of $\mu_g$, we know that $\mu$ is finite flat.</p> <p>Now for $k$ not necessary being algebraically close, we take the base change to $\overline{k}$ and apply the above result. So the question is that, if the base change $\overline{\mu}$ of $\mu : G \times_k X \rightarrow X$ to $\overline{k}$ is finite flat, then could we conclude that $\mu$ is finite flat. Notice that $\mu$ is of finite type. One can show that $\mu$ is separated and universally closed from the separateness of $\overline{\mu}$ which is at the same time universally closed ( since $\overline{\mu}$ is finite). So one concludes that $\mu$ is proper. It's easy to show that $\mu$ is quasi-finite hence $\mu$ is finite. Finally, it's easy to show that $\mu$ is flat.</p> http://mathoverflow.net/questions/70335/a-question-about-affiness a question about affiness unknown (google) 2011-07-14T15:34:31Z 2011-07-15T22:47:14Z <blockquote> <p><strong>Possible Duplicate:</strong><br> <a href="http://mathoverflow.net/questions/95/is-there-an-example-ofa-scheme-x-whose-reduction-x-red-is-affine-but-x-is-not-af" rel="nofollow">Is there an example of&#160;a scheme X whose reduction X_red is affine but X is not affine?</a> </p> </blockquote> <p>I got a question, which may be very easy, but I didn't figure out it. Let $X$ be a scheme such that $X_{\mathrm{red}}$ is an affine scheme. Could we conclude that $X$ itself must be affine also? This question came into my mind because I am thinking if the natural morpihsm $i: X_{\mathrm{red}} \rightarrow X$ has such a property that for any open affine subset $U$ of $X_{\mathrm{red}}$, $i(U)$ is an open affine subset of $X$. Notice that $i$ is an homeomorphism, hence $i(U)$ is an open subset of $X$.</p> http://mathoverflow.net/questions/70018/quotient-of-a-variety-by-a-finite-group-scheme quotient of a variety by a finite group scheme unknown (google) 2011-07-11T14:51:18Z 2011-07-11T18:39:31Z <p>Let $X/k$ be a scheme of finite type over a field $k$, $G/k$ be a finite group scheme and suppose $G$ acts on $X$, i.e we have a $k$-morphism $\mu : G \times_k X \rightarrow X$ satisfies some conditions, see Mumford's book "Abelian Varieties", p. 108.</p> <p>Suppose that $k$ is algebraically closed and consider the map $\mu$ induced on closed points $G(k) \times X(k) \rightarrow X(k)$, which gives $G(k) \rightarrow \mathrm{Aut}(X(k))$. Under the assumption that each $x \in X$ has an open affine neighborhood $U$ such that $U$ is invariant under $G$, one can form the quotient $X/G$. In particular, quasi-projective varieties have this property. This assumption reduces the construction from general variety to affine case.</p> <p>For a general field $k$, if one can cover $X$ by open affine subset $U_i$ such that the image of $G \times_k U_i$ under $\mu$ is contained in $U$, then one reduces the construction to the affine case. But I couldn't figure out if a quasi-projective variety $X$ over $k$ always has such a covering. By working with base change to the algebraic closure of $k$, one gets a $G$-invariant open affine covering of $X_{\overline{k}}$. The images of these open affine subsets of $X_{\overline{k}}$ under the projection to $X$ is still $G$-invariant, but not necessary affine. So for quasi-projective varieties over $k$, do we have the existence of the quotient $X/G$?</p> <p>Another question is that if it exists, then if it is also quasi-projective? What I know is that, in the classical case ( over algebracially closed field and giving finite group action $H \rightarrow \mathrm{Aut}_k (X))$, if the quotient exists and $X$ is complete, then $X/H$ is also complete.</p> http://mathoverflow.net/questions/66858/resolution-of-singular-points-on-plane-curves-and-base-change resolution of singular points on plane curves and base change unknown (google) 2011-06-03T23:24:33Z 2011-06-03T23:24:33Z <p>Let $k$ be a field and $C/k$ be an affine plane curve over $k$, namely $C = \mathrm{Spec}(A)$ for some $A = k[x,y]/(f(x,y))$, here $f(x,y) \in k[x,y]$ is an irreducible polynomial. Let $B$ be the integral closure of $A$ in the quotient field $\mathrm{Frac}(A)$ of $A$. Then $C' := \mathrm{Spec}(B)$ is the normalization of $C$, which is an non-singular affine curve over $k$. Consider the base change $C_{\overline{k}} := C \times_{k} \overline{k}$. Suppose $f(x,y)$ is geometrically irreducible, then $C_{\overline{k}}$ is an affine plane curve over $\overline{k}$ defined by an equation $f(x,y) \in \overline{k}[x,y]$. Suppose $C_{\overline{k}}$ has only ordinary singular points, (namely, the tangent lines at each singular point of $C_{\overline{k}}$ are all distinct.) One can blow up the curve $C_{\overline{k}}$ and get a non-singular affine plane curve $C'_{\overline{k}}$. (Each time, we take a linear change such that the singular point at which we want to blow up becomes origin, then use the well-known coordinate change $(x, y) \rightarrow (x, xy)$. On the other hand, if $k$ is perfect, then $C' \times_k \overline{k}$ is also a non-singular affine curve over $\overline{k}$ and we have a natural dominant morphism $C' \times_k \overline{k} \rightarrow C \times_k \overline{k} = C_{\overline{k}}$. Hence we have an $\overline{k}$-isomorphism $C'_{\overline{k}} \rightarrow C' \times_k \overline{k}$. This means that $C'$ is $\overline{k}$-isomorphic to a plane curve which is defined by a single equation if we work with $\overline{k}$. I heard from other people that the non-singular model of a plane curve is not always able to be defined by a single equation. (Although some of them said this with thinking the projective case.) </p> <p>My question is that: With the above assumptions, does $C'$ is $k$-isomorphism to a plane curve? If yes, how to get the equation? If not, the counter-example is appreciated. (If all the singular points are $k$-rational, then the answer is positive, and the result of the blow up process for $C_{\overline{k}}$ gives the require equation.)</p> http://mathoverflow.net/questions/66548/resolution-of-singular-points-on-curve resolution of singular points on curve unknown (google) 2011-05-31T13:45:45Z 2011-05-31T15:05:25Z <p>After reading Fulton's book "Algebraic Curves", I know how to do resolution of singular points on curves. Given an affine equation, I can get it's non-singular affine model, i.e the normalization of its affine coordinate ring. The problem is that in Fulton's book, he worked with algebraically closed field and he worked with the origin (0,0) also. If one do blowing-up (p.165 in Fulton's book) of a curve defined a field $k$ which is not necessary algebraically colsed and with a singular point which is not necessary $k$-rational, one get an eqaution which is defined over a finite extension of the field $k$. But for any affine curve (i.e integral scheme of dimension 1), one can form the normalization of its coordinate ring and hence get a non-singular affine curve which is birational to the original curve. My question is that how one can get the equation for this normalized affine curve? I know there are algorithms to compute the integral closure of function fields, but the result is an integral basis rather that an equation. Moreover, I would like to know if there is a systematic method to get this normalized affine equation with the blowing-up method.</p> http://mathoverflow.net/questions/64391/computation-for-composition-of-polynomials Computation for composition of polynomials unknown (google) 2011-05-09T13:49:48Z 2011-05-09T14:37:40Z <p>Let $R$ be a ring, $f(X)$ be a polynomial with coefficients in $R$ of degree $n$. It's known that for any $\alpha \in R$, one can evaluate $f$ at $\alpha$, i.e compute $f( \alpha)$ in $O(n)$ operations in $R$ for some $k >0$ . But could we compute $f( X + \alpha )$, or more generally compute $f( \alpha X + \beta)$ in $O(n)$ or $O(n * ( \mathrm{log} \; n)^k )$ operations in $R$. Since the result is a polynomial in $X$ of degree at most $n$, I would guess there is a such way. But from the naive way, I can't just figure out a $O(n^2)$ algorithm to do this as follows: If $f(X) = \sum_{i=1}^n a_i X^i$, then one compute $g \leftarrow a_0, h \leftarrow \alpha X + \beta, g \leftarrow g+ a_{i+1} * h, h \leftarrow h* (\alpha X + \beta)$ for $0 \leq i \leq n-1$. In the $i$-th step, $g$ is a polynomial of degree $i$ and $h$ is a polynomial of degree $i+1$, so we need $O(i)$ operations in $R$ to compute $g+ a_{i+1} * h$ and $h* (\alpha X + \beta)$, so the total operations needed is $O(n^2)$.</p> <p>Is there any way to do so in $O(n)$ or $O(n * ( \mathrm{log} \; n)^k )$ operations in $R$? I don't really need to know how the algorithm works, I just need an explicit reference which says this is possible.</p> http://mathoverflow.net/questions/60847/best-deterministic-complexity-for-factoring-polynomials-over-finite-field best deterministic complexity for factoring polynomials over finite field unknown (google) 2011-04-06T19:42:40Z 2011-04-07T05:49:52Z <p>I would like to know currently what's the best deterministic complexity for factoring polynomials over finite field (without the assumption of GRH)? I have searched on google, there are many source, but I can't find an explicit result. Some of them are only for some special cases. Some of them are used the assumption of GRH. </p> http://mathoverflow.net/questions/56403/universal-finite-differential-module-of-affinoid-algebra universal finite differential module of affinoid algebra unknown (google) 2011-02-23T15:19:29Z 2011-03-23T18:22:14Z <p>Let $k$ be a value field (archimedean), for example $k = \mathbb{Q}_p$, the p-adic field. The free Tate algebra is $$T_n := \left\{ \ \sum a_I X^I, \ a_I \in k, \ a_I \rightarrow 0 \text{ as } |I| \rightarrow \infty \ \right\}$$.</p> <p>I want to compute the universal finite differential module $\Omega^f_{T_n/k}$ of $T_n$ over $k$. For any $A$-algebra $B$, the "<em>universal finite differential module</em>" $\Omega^f_{B/A}$ is a finitely generated $B$-module with an $A$-derivation $d : B \rightarrow \Omega^f_{B/A}$ such that for any $A$-derivation $d^{'} : B \rightarrow M$ with $M$ being a finite generated $B$-module, there exists a $B$-module homomorphism $\phi : \Omega^f_{B/A} \rightarrow M$ such that $d^{'} = \phi \circ d$. It doesn't always exist.</p> <p>I would like to show that $\Omega^f_{T_n/k}$ is the free $T_n$ module of rank $n$. Let me explan the case $n=1$. Similar to the case of formal power series ring $k[[ X ]]$, given a $k$-derivation $d_1 : T_1 \rightarrow M$ to a finitely genterated $T_1$-module $M$, we definte $\phi : T_1 * dX \rightarrow M$ by sending $dX$ to $d_1 X$ and extend it $T_1$-linearly. We want to show that $d_2 := d_1 - \phi \circ d = 0$. $d_2$ is still a $k$-derivation and one shows that $d_2 (f) = 0$ if $f \in k[X]$. In the case of $k[[X]]$, we know for any $f \in k[[X]]$, $d_2 (f) \in (X^r)M$ for any $r > 0$. By Krull Intersection Theorem, we know there is a $g \in (X)$ such that $(1-g) N = 0$, here $N := \cap_{ r > 0 } (X^r)M$. Since for $g \in (X)$, $1-g$ is invertible in $k[[X]]$, we know $N = 0$ and hence $d_2 = 0$.</p> <p>But in the case of $T_1$, for $g \in (X)$, $1-g$ may not be invertible in $T_1$. For example, $g=X$, its inverse is $1+X+X^2+ \cdot \cdot \cdot$ which is not in $T_1$.</p> <p>So the question is how to prove $d_2 = 0$. In p.64 of the book <em>Rigid analytic geometry and its applications</em> by Jean Fresnel and Marius van der Put, there is a proof. Instead of considering only the ideal (X), they consider any maximal ideal $m$ of $T_1$. They said that any maximal ideal $m$ is generated by polynomials ( It's ok.), hence $d_2 ( T_1)$ is contained in $m^r M$ for any $r > 0$ for which I don't understand why.</p> http://mathoverflow.net/questions/58927/galois-fixed-divisor-class-group Galois-fixed divisor class group unknown (google) 2011-03-19T16:29:12Z 2011-03-19T17:27:20Z <p>Let $k$ be a perfect field. For a geometrically integral smooth $k$-variety $X$, let $Z^{1}(X)$ be the group of Weil divisors ( free $\mathbb{Z}$-module generated by irreducible closed subsets of codimension 1 of $X$ ), $P^{1}(X)$ be its subgroup of principal divisors and $Cl(X) := Z^{1}(X) / P^{1}(X)$ be the divisor class gruop.</p> <p>Let $\pi : X_{\overline{k}} \rightarrow X$ be the projection from the base change $k \rightarrow \overline{k}$. We have that the pullback $\pi^* : Z^{1}(X) \rightarrow Z^{1}(X_{\overline{k}})$ is injective and via this identification, $Z^{1}(X) = Z^{1}(X_{ \overline{k} })^G$, here $G := Gal( \overline{k} / k)$, and if we assume that $X/k$ is complete, $Z^{1}(X) \cap P^{1}(X_{\overline{k}}) = P^1(X)$ and hence $\pi^*$ induces an injective homomorphsim $\pi^{*} : Cl(X) \rightarrow Cl(X_{\overline{k}})$. The Galois group still acts on $Cl(X_{\overline{k}})$ and the image of $\pi^*$ is in $Cl(X_{\overline{k}})^{G}$.</p> <p>In general, we don't have $Cl(X) = Cl(X_{\overline{k}})^{G}$. In Milne's note ( abelian variety ), he mentioned that this is true if $X$ is a curve and we have a $k$-rational point on it. But I don't know the reason, and do we have this equality for any complete geometrically integral smooth $k$-variety which has an irreducible closed subset $Z$ of codimension 1 of $X$ such that $Z_{red} \times_k \overline{k}$ is still an integral closed subvariety of $X_\overline{k}$ ?</p> http://mathoverflow.net/questions/56403/universal-finite-differential-module-of-affinoid-algebra/56420#56420 Answer by unknown (google) for universal finite differential module of affinoid algebra unknown (google) 2011-02-23T17:45:17Z 2011-02-23T17:45:17Z <p>I got an idea. Let $\overline{k}$ be the algebraic closure of $k$ and extend everything to be over $\overline{k}$, i.e extend the $k$-derivations $d_2$ to $D_2: T_n \otimes_{k}\overline{k} \rightarrow M \otimes_{k} \overline{k}$. Since the natural morphism $M \rightarrow M \otimes_{k} \overline{k}$ is injective, in order to show that $d_2 = 0$, we only need to show that $D_2 = 0$. So we can assume that $k$ is algebraically closed at the beginning.</p> <p>One can show that the residue field $T_n/m$ is a finite extension of $k$ for any maximal $m$. In our case, since $k$ is algebraically closed, we have $T_n/m = k$. It follows that any maximal ideal $m$ is of the form $m = ( X - \lambda )$ for some $\lambda \in k$ ( with $| \lambda | &lt; 1$) . For any $f \in T_1$ and any $r > 0$, we can write $f$ as $f = \sum_{i=0}^{r} \ a_i (X - \lambda)^i + (X - \lambda )^{r+1} g(X)$ with $a_i \in k$ and $g(X) \in T_1$. Using the rules of derivations and the fact that $d_2$ is zero on polynomials, we get $d_2 (f) \in m^{r+1}M$. It follows that $d_2 (f) \in \cap_{r>0} I^rM$, here $I = \cap_{m \in \text{Max}(T_1)} m$ is the Jacobson radical. By Krull Intersection Theorem, we get the result.</p> <p>I apology for answering my own question. It will be appreciated for other solutions. </p> http://mathoverflow.net/questions/55891/module-of-differentials-of-formal-power-series-ring-and-of-its-field-of-quotiens module of differentials of formal power series ring and of its field of quotiens unknown (google) 2011-02-18T18:45:38Z 2011-02-18T22:46:06Z <p>For any $A$-algebra $B$ ( commutative ring with 1 ), we have the existence of $\Omega_{B/A}$, the module of relative differentials of $B$ over $A$, which can be defined by an universal property. In the case $A = k$ being a field and $B = k[[X]]$ being the formal power series ring over $k$, $\Omega_{B/A}$ is not always a finite $B$-module. ( And I don't know if it is a free $B$-module. ) For example, if char$(k) = 0$, since $k[[X]]$ has a infinite subset whose elements are algebraic independent over $k$, one can show that $\Omega_{B/A}$ is not a finite $B$-module. I have seen another notion in a book as following: the "<em>universal finite differential module</em>" $\Omega^f_{B/A}$ is a $B$-module with an $A$-derivation $d : B \rightarrow \Omega^f_{B/A}$ such that for any $A$-derivation $d^{'} : B \rightarrow M$ with $M$ being a finite generated $B$-module, there exists a $B$-module homomorphism $\phi : \Omega^f_{B/A} \rightarrow M$ such that $d^{'} = \phi \circ d$. With this definition, one can show that, in the case of formal power series ring, $\Omega^f_{B/A}$ is a free $B$-module of rank $1$ with $dX$ as a basis, and for any $f \in k[[X]], df = f' dX$, here $f'$ is defined by the natural way, which is a result I can't deduce for $\Omega_{B/A}$. One need to use Krull Intersection Theorem or the structure theorem for finitely generated modules over a $PID$.</p> <p>Now I try to compute $\Omega^f_{B/A}$ for $B = k ((X))$, the field of quotients of $k[[X]]$, but I can't get the result. The problem is that the $A$-derivation $d'$ is not necessary continuous with respect to the $(X)$-adic topology.</p> <p>My question is that: do we have $\Omega^f_{B/A}$, for $B = k((X))$, is a free $k((X))$-module of rank $1$ with $dX$ as a basis?</p> http://mathoverflow.net/questions/48082/does-the-fundamental-equality-control-finiteness Does the fundamental equality control finiteness ? unknown (google) 2010-12-02T19:20:52Z 2010-12-02T22:03:36Z <p>There is a fundamental equality in algebraic number theory:</p> <p>Let $(A,\mathfrak{p})$ be a DVR (Discrete Valuation Ring), $K$ be its field of quotient, $L/K$ be a finite field extension of degree $[L:K] = n$ and $B$ is a subing of $K$ of field of quotient $L$ containing $A$. Assume $B$ is a finite $A$-module, then we know $B$ is a Dedekind domain. Let $\mathfrak{p}B =\mathfrak{P}_1^{e_1} \mathfrak{P}_2^{e_2} \cdot \cdot \cdot \mathfrak{P}_r^{e_r}$ and $f_i = [ B / \mathfrak{P}_i : A / \mathfrak{p} ]$, then we have <code>$n = \sum_{i=1}^r e_i f_i$</code>. (I have to assume that $B$ is integrally closed. If not, we still have such a formula but with $e_i$ being the length of some modules as below.)</p> <p>Notice that we don't have to assume $L/K$ is separable if we assume $B$ is finite over $A$.</p> <p>In general, if we only assume $B$ is integral over $A$ (not necessary finite), then we have <code>$l_{A} (B / \mathfrak{p}B) = \sum_{i=1}^r e_i f_i$</code>, here <code>$l_A (B / \mathfrak{p}B)$</code> means the lenght of the $A$-module $B / \mathfrak{p}B$ and $e_i$ is the length of the <code>$B_{\mathfrak{P}_i}$</code>-module <code>$B_{\mathfrak{P}_i} / \mathfrak{p} B_{\mathfrak{P}_i}$</code>. We also have $n = [ L:K ]$ $\geq l_A (B / \mathfrak{p}B)$.</p> <p>My question is that: Under the above assumption ( $B$ is integral over $A$, not necessary finite over $A$), if we have <code>$n = e_i f_i$</code>, or equivalently, $[ L:K ]$ $= l_A (B / \mathfrak{p}B)$, if we can conclude that $B$ is in fact finite over $A$?</p> <p>In Kaplansky's book "Commutative rings" (Theorem 100), he gave an example of $L/K$ being a purely inseparable extension of degree 2 and $B$ and $A$ are DVRs such that $B$ is not finite over $A$. But in this example, $2 = [ L:K ]$ $> l_A (B / \mathfrak{p}B) = 1$. In order to find a counter-example for my guess, we have to consider another example than that one in Kaplansky's book, for which I can't figure out one. On the other hand, I can't give a proof for it neither.</p> http://mathoverflow.net/questions/42775/morphism-which-is-open-but-not-universally-open morphism which is open but not universally open unknown (google) 2010-10-19T12:20:59Z 2010-10-19T12:20:59Z <p>In someone's note, I have seen such an example, but I can't show that it is not universally open. Here is the example:</p> <p>Let $k$ be a field and $A = k[T]_{(T)}$, the discrete valuation ring obtained from the polynomial ring $k[T]$ localized at the prime ideal $(T)$. Let $\hat{A}$ be the completion with respect to $(T)$, which is just the power series ring $k[[ \ T \ ]]$.</p> <p>Now the natural map $A \rightarrow \hat{A}$ gives a open morphism $i : Spec(\hat{A}) \rightarrow Spec(A)$. Consider the base change $j : Spec(\hat{A}) \rightarrow Spec(A)$, we obtain a morphism $i^{'} : Spec( \hat{A} \otimes_{A} \hat{A} ) \rightarrow Spec( \hat{A} )$. Then the author said this is not an open morphism.</p> <p>There is a unique maximal ideal, called $m$ in $\hat{A} \otimes_{A} \hat{A}$, whose pullback in $\hat{A}$ under $i^{'}$ is the maximal ideal in $\hat{A}$. In order to show $i^{'}$ is not open, I need to show $m$ is also a minimal prime ideal. But I don't even know if $\hat{A} \otimes_{A} \hat{A}$ is an integral domain or not? </p> <p>There is another example in EGA, but I still want to know if the above example is right and how to see it.</p> http://mathoverflow.net/questions/125164/functor-used-to-construct-sheafifications-property Comment by 2013-03-23T06:31:07Z 2013-03-23T06:31:07Z @Zhen Lin, thank you for the comment, but if you could provide me some reference to these in order that I can get more precise ideas about these in the future. http://mathoverflow.net/questions/125164/functor-used-to-construct-sheafifications-property Comment by 2013-03-22T08:26:50Z 2013-03-22T08:26:50Z @Zhen Lin, I don't understand what this menas &quot;lifting constructions on Set to constructions on Sh(X)&quot;. Also could you explain or give a reference that illustrate what is the relation between this with local ring, thank you. http://mathoverflow.net/questions/116010/complex-multiplication Comment by 2012-12-11T06:53:06Z 2012-12-11T06:53:06Z @kreck, I felt strange that it is written that &quot;...be a quaternion algebra over a quadratic field...&quot;. It is good to know that it contains an imaginary quadratic field and the theorem Tate proved. Could you give the reference for it. Thank you very much. http://mathoverflow.net/questions/116010/complex-multiplication/116018#116018 Comment by 2012-12-10T21:14:37Z 2012-12-10T21:14:37Z Thanks. It seems that those theorems I listed above require characteristic zero. http://mathoverflow.net/questions/99015/kernel-of-an-isogeny-and-coker-of-its-induced-map-on-the-tate-module/99016#99016 Comment by 2012-06-07T08:34:55Z 2012-06-07T08:34:55Z Thank you, that's clear. http://mathoverflow.net/questions/96214/divisor-criterion-for-a-curve-being-rational Comment by 2012-05-07T14:16:21Z 2012-05-07T14:16:21Z Sorry for this trivial mistake...I will delete this question my self http://mathoverflow.net/questions/86248/induced-map-on-algebraic-de-rham-cohomology Comment by 2012-01-20T21:50:43Z 2012-01-20T21:50:43Z I need this property since I am considering morphisms between curves and the relationship between their cohomology groups. I computed some examples and found that this is injective. So I am wondering if this is true. Since in the case of curves, the morphisms are finite surjective, so you mentioned this is always true. This simplifies the computation. But do you know a reference of this, or the argument is easy? http://mathoverflow.net/questions/78469/a-question-on-continuity-of-g-module-for-a-profinite-group-g Comment by 2011-10-18T22:18:45Z 2011-10-18T22:18:45Z @Kevin Ventullo: Thanks. But I made a mistake in my question. In fact, the statement is for $G$-modules, i.e it requires the action satisfies $\sigma (m_1 + m_2) = \sigma(m_1) + \sigma(m_2)$. The example that $G$ acts on $G/H$ for a subgroup $H$ of $G$ is not a $G$-module. I should add this condition in my question, sorry. http://mathoverflow.net/questions/78469/a-question-on-continuity-of-g-module-for-a-profinite-group-g/78470#78470 Comment by 2011-10-18T19:05:51Z 2011-10-18T19:05:51Z Without continuity, I can't see why the stabilizer is closed....:( Could you give me a hint? Thanks. http://mathoverflow.net/questions/75164/relation-between-pull-back-of-cartier-divisors-invertible-sheaves-and-global-sec Comment by 2011-09-12T10:23:59Z 2011-09-12T10:23:59Z @ Fei YE, could you explain why surjectivity implies that $D$ can be extended to a hyperplane in $\mathbb{P}^n$, this is just what I would like to know. http://mathoverflow.net/questions/74684/push-forward-of-the-structure-sheaf-stein-factorization-birational-and-connecte Comment by 2011-09-07T18:09:28Z 2011-09-07T18:09:28Z @ Karl, yes and it should be finite as a $k$-vector space here to get the conclusion also. And I think that finite algebra is the common terminology for algebras which are finite as modules (over base ring), see p.30 of Atiyah &amp; Macdonald's book &quot;Introduction to Commutative Algebra&quot; http://mathoverflow.net/questions/70489/whats-a-finite-group-scheme-action-on-a-variety/70495#70495 Comment by 2011-07-16T21:44:36Z 2011-07-16T21:44:36Z That&#180;s so easy, why I didn&#180;t figure out this http://mathoverflow.net/questions/70335/a-question-about-affiness/70373#70373 Comment by 2011-07-15T11:52:22Z 2011-07-15T11:52:22Z I totally agree with brunoh. This moring, I have thought this question. It looks like that we get an open affine covering $U_i$ of $X$, each $U_i$ is a principal open affine subset of an (not necessary principal ) open affine subset $V$ of $X$, which may not be of a form $X_f$ for some global section $f$ of $O_X$ http://mathoverflow.net/questions/69453/dual-isogeny-for-abelian-varieties-over-a-general-field Comment by 2011-07-04T10:37:02Z 2011-07-04T10:37:02Z @Fran&#231;ois: You are right. Since I read this on a book on Elliptic Curves, so $E$ is isomorphic to it's dual ( I guess so ). @Jason: Thanks. For an elliptic curve over a perfect field $k$, the existence of $\hat{\phi}$ is for all isogeny. One needs to decompose $\phi$ as a separably isogeny followed by a Frobenius morphism. I have to think about why this doesn't work when $k$ is not perfect (Hence in this case, one need to focus on separably isogeny only) http://mathoverflow.net/questions/66858/resolution-of-singular-points-on-plane-curves-and-base-change Comment by 2011-06-04T09:10:36Z 2011-06-04T09:10:36Z Yes, you are right. Before I have done the computations for some curves, like $y^2 - x^3$ or $y^2 - x^2(x+1)$ and what I got is a finite morphism. For example, $z := y/x$ in the quotient field of $k[x,y]/( y^2 - x^2(x+1) )$ or $k[x,y]/(y^2-x^3)$ is an integral element. But in general, we don't get an integral element. Thank you.