User b. cais - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T01:17:09Z http://mathoverflow.net/feeds/user/2215 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/7432/generalizing-miracle-flatness-matsumura-23-1-via-finite-tor-dimension Generalizing miracle flatness (Matsumura 23.1) via finite Tor-dimension B. Cais 2009-12-01T14:01:23Z 2013-01-13T03:25:46Z <p>Let $(A,m_A)$ and $(B,m_B)$ be noetherian local rings and $f:A\rightarrow B$ a local homomorphism. Let $F = B/m_AB$ be the fiber ring and assume that $$\mathrm{dim}(B) = \mathrm{dim}(A) + \mathrm{dim}(F).$$</p> <p>The following Theorem (23.1 in Matsumura's CRT) is really quite a miracle:</p> <p>Theorem: If $A$ is regular and $B$ is Cohen-Macaulay then $f$ is flat.</p> <p>I am wondering to what extent this theorem can be generalized. What I have in mind is a statement of the type:</p> <p>"Theorem": If $A$ is $X$ and $B$ is $Y$ then $f$ is of finite Tor-dimension (i.e. $\mathrm{Tor}^i_A(B,A)=0$ for all $i$ sufficiently large).</p> <p>Here, $X$ and $Y$ are ring-theoretic conditions which should be <em>strictly weaker</em> than "regular" and "CM" respectively. Is the "Theorem" above true just requiring $A$ and $B$ to be normal? How about both CM? Or maybe CM plus finitely many $(R_i)$? </p> <p>Any thoughts/ counterexamples? </p> http://mathoverflow.net/questions/84273/deformation-of-ordinary-p-divisible-groups-via-grothendieck-messing Deformation of ordinary p-divisible groups via Grothendieck-Messing B. Cais 2011-12-25T16:12:56Z 2011-12-25T16:12:56Z <p>I am hoping that someone can point out the error in the "proof" of the following "theorem":</p> <blockquote> Theorem: Let $k$ be a perfect field of characteristic $p>2$ and let $G$ be an ordinary $p$-divisible group over $W(k)$. Then the connected-etale sequence of G is split: $G\simeq G^m \times G^{et}$. </blockquote> <p>The "Theorem" is decidedly false. For example, extensions of $\mathbf{Q}_p/\mathbf{Z}_p$ by $\mathbf{G}_m[p^{\infty}]$ over $W(k)$ are classified by the abelian group $1+pW(k)$, so any nonzero element of this group gives an ordinary $p$-divisible group with non-split connected-etale sequence.</p> <blockquote> Proof: Let $G_0:=G\times_{W(k)} k$ be the special fiber of $G$. Since $k$ is perfect, $G_0= G_0^{m}\times G_0^{et}$. By Messing (LNM 264, Chap V. Theorem 1.6), there is an equivalence of categories between deformations of $G_0$ to $W(k)$ and free $W(k)$-submodules $L$ of $D(G_0)(W(k))$ lifting $\omega_{G_0}\hookrightarrow D(G_0)(k)$. This equivalence is induced by sending a lift $G'$ of $G_0$ to $\omega_{G'}$. Now $G' = G^{m}\times G^{et}$ and $G$ both lift $G_0$, and these lifts correspond to the submodules $\omega_{G'}$ and $\omega_G$ of $D(G_0)(W(k))$, respectively. But since $G$ is ordinary, so $G^0=G^{m}$, the pullback map $\omega_{G}\rightarrow \omega_{G^m}$ is an isomorphism. Via this isomorphism, the map $\omega_{G'}\rightarrow D(G_0)(W(k))$ coincides with the composite $$\omega_{G'}\simeq \omega_{G^m}\times \omega_{G^{et}} = \omega_{G^m}\simeq \omega_G\rightarrow D(G_0)(W(k))$$ and we conclude that $\omega_{G'}=\omega_G$ as submodules of $D(G_0)(W(k))$ lifting $\omega_{G_0}$. It then follows from Messing's Theorem above that $G\simeq G'$, as claimed. </blockquote> <p>I must be making some silly and obvious mistake...can you find it?</p> http://mathoverflow.net/questions/79087/subrings-of-invariants-in-divided-power-algebras Subrings of invariants in divided power algebras B. Cais 2011-10-25T14:49:13Z 2011-10-26T04:09:03Z <p>I am wondering to what extent the functors "ring of invariants under a group action $G$" and "divided power envelope with respect to a $G$-stable ideal" commute. </p> <p>To be precise, let $R$ be a commutative ring (with unit) and $G$ a group acting on $R$ by ring automorphisms. I am happy to assume that $R$ is a noetherian adic ring and that $G$ is profinite acting continuously on $R$, but I do not want to assume that $G$ is finite.</p> <p>Let $S:=R^G$ be the subring of invariants, and suppose given an ideal $I\subseteq R$ which is $G$-stable. Denote by $J:=I\cap S$ the contraction of $J$ to $S$, and let $D(R,I)$ and $D(S,J)$ be the divided power envelopes of $R$ with respect to $I$ and $S$ with respect to $J$. (I should probably also complete these algebras with respect to a $G$-stable ideal of $R$ that is contained in the ideal of definition, so the induced $G$-action on the completed PD algebras is continuous).</p> <p>Then $G$ acts on $D(R,I)$ and we have $D(S,J)\subseteq D(R,I)^G$. </p> <blockquote> Is this inclusion an equality? </blockquote> http://mathoverflow.net/questions/56289/orbits-of-sl-n-acting-on-matrices-of-determinant-p Orbits of SL_n acting on matrices of determinant p B. Cais 2011-02-22T15:07:39Z 2011-02-23T07:48:51Z <p>Fix a positive integer $n$ and let $S$ be the set of $n$ by $n$ matrices with entries in $\mathbf{Z}_p$ (the $p$-adic integers) whose determinant is $p$. The group $G:=\mathrm{SL}_n(\mathbf{Z}_p)$ acts freely on $S$ via left multiplication.</p> <blockquote> Is it possible to write down an explicit list of representatives for the orbits of $G$ on $S$? If so, what is this list? If not, is there an algorithm for computing a complete list of orbit representatives? </blockquote> <p>For example, for $n=2$, I think that this is basically the "Hecke Operator at $p$" computation, and the $p+1$ orbits of $G$ on $S$ have representatives </p> <p>$$\left(\begin{matrix} p &amp; b \\ 0 &amp; 1 \end{matrix}\right),\ b=0,1,\ldots, p-1\ \text{and}\<br> \left(\begin{matrix} 0 &amp; -1 \\ p &amp; 0 \end{matrix}\right)$$ </p> <p>This is probably a question whose answer is well-known to many people, so I apologize in advance for my ignorance!</p> http://mathoverflow.net/questions/52625/tamely-ramified-p-adic-galois-representations Tamely ramified p-adic Galois representations B. Cais 2011-01-20T15:05:19Z 2011-01-20T18:34:29Z <p>The following question came up in a discussion with a colleague about local Galois representations:</p> <blockquote> To what extent is the classification of continuous $p$-adic representations of $G_{\mathbf{Q}_{\ell}}$ for $\ell\neq p$ similar to the classification of tamely ramified $p$-adic representations for $\ell=p$? </blockquote> <p>More precisely, let $\rho: G_{\mathbf{Q}_{\ell}}\rightarrow \mathrm{GL}_n(\mathbf{Q}_p)$ be a (continuous) $p$-adic representation. If $\ell\neq p$, then Grothendieck proved (using the observation that such a representation kills an open subgroup of wild inertia) that $\rho$ is determined by the associated Weil-Deligne representation (see, for example, the notes of Brinon and Conrad, pg. 111, or Taylor's 2002 ICM article). </p> <blockquote> When $\ell=p$ and $\rho$ is trivial on the wild inertia subgroup, is it the case that $\rho$ is necessarily de Rham? </blockquote> <p>What seems clear to me is that if one assumes that $\rho$ is Hodge-Tate, then the only Hodge-Tate weight is zero. If indeed $\rho$ were de Rham = pst, then the associated filtered $(\phi,N)$-module would have trivial filtration, and so one ought" to be able to recover it from the attached Weil-Deligne representation. In other words, the classification of $p$-adic representations of $G_{\mathbf{Q}_{\ell}}$ for $\ell\neq p$ is literally the same as the case $\ell=p$, provided one throws in the (rather drastic) condition that wild inertia is killed (or at least some open subgroup of it is killed).</p> <p>Does this sound correct? </p> http://mathoverflow.net/questions/7655/trace-map-attached-to-a-finite-homomorphism-of-noetherian-rings Trace map attached to a finite homomorphism of noetherian rings B. Cais 2009-12-03T10:59:50Z 2011-01-14T15:46:30Z <p>Let $f:A\rightarrow B$ be a homomorphism of noetherian rings which makes $B$ into a finite $A$-module. Under what conditions on $f$, $A$, $B$ can one associate to this map a canonical "trace map" $$\mathrm{Tr}_f:B\rightarrow A,$$ i.e. a homomorphism of $A$-modules $B\rightarrow A$ which is compatible with base change (perhaps with restrictions on the kind of allowable base changes), localization, and which recovers the "usual" thing when $B$ is free over $A$ (i.e. $\mathrm{Tr}_f(b) =$ the trace of the $A$-linear endomorphism given by multiplication by $b$ on the finite $A$-module $B$)?</p> <p>Here are my thoughts so far:</p> <p>1) If $f$ is flat, one always has $\mathrm{Tr}_f$. Just work locally, using that finite flat over a noetherian local ring is free.</p> <p>2) More generally, if $f$ is of finite Tor-dimension, I can construct $\mathrm{Tr}_f$ by taking a finite projective resolution</p> <p>$$0\rightarrow P_n\rightarrow \cdots\rightarrow P_0\rightarrow B\rightarrow 0$$</p> <p>of $B$ as an $A$-module: lift multiplication by $b$ on $B$ to a map of complexes $b:P_{\bullet}\rightarrow P_{\bullet}$ and define $$\mathrm{Tr}_f(b) := \sum_i (-1)^i \mathrm{Tr}_i(b)$$ where $\mathrm{Tr}_i(b)$ is the trace of the (lift of the) endomorphism $b$ of $B$ to $P_i$. One chekcs this is independent of the choice of projective resolution. It commutes with "tor-independent" base change (sometimes called "cohomologically transverse" base change).</p> <p>3) If $A$ and $B$ are normal, I can construct $\mathrm{Tr}_f$ as follows: the localization of $f$ at any height-1 prime ideal is automatically flat by Matsumura 23.1 as the corresponding localizations of $A$ and $B$ are regular and the dimensions work out (the fiber ring is 0-dimensional as $f$ is finite). Thus, one has a canonical trace map on each localization, and since $A$ and $B$ are normal, they are the intersections of these localizations so we win.</p> <p>4) If $A$ and $B$ are only assumed reduced, one can look at the injections $A\rightarrow A'$ and $B\rightarrow B'$ with $A'$ and $B'$ the normalizations of $A$ and $B$ in their total rings of fractions. Let $f':A'\rightarrow B'$ be the corresponding map. By 3), we get a trace map for $f'$ and the whole problem of constructing the trace map for $f$ comes down to showing that $\mathrm{Tr}_{f'}$ carries $B$ into $A$. Letting $C_A:=\mathrm{ann}_{A'}(A'/A)$ be the conductor ideal (with $C_B$ defined similarly), I think that a necessary condition for $\mathrm{Tr}_{f'}(B)$ to be contained in $A$ is $$f'(C_A) \supseteq C_B.$$ Is this condition sufficient? As an example of how things can go wrong if this condition is violated, consider the finite map between reduced $k$-algebras ($k$ a field) $$k[x,y]/(xy) \rightarrow k[x]$$ given by sending $y$ to 0. The normalization of $k[x,y]/(xy)$ is the product $k[x]\times k[y]$ and the trace map attached to $f':k[x]\times k[y]\rightarrow k[x]$ sends $b\in k[x]$ to $(b,0)$. But $(b,0)\in k[x]\times k[y]$ lies in the image of $k[x,y]/(xy)\rightarrow k[x]\times k[y]$ if and only if $b(0)=0$. It follows that the trace map on normalizations doesn't restrict to a trace map on the original rings.</p> <p>I'd be happy to assume that $A$ and $B$ are flat $R$-algebras, for a regular local ring $R$, and that $f:A\rightarrow B$ is an $R$-algebra homomorphism. I'd also be happy to assume that $A$, $B$, and $f$ are local, with $A$ and $B$ reduced complete intersections over $R$. I'm wondering if there is a framework for trace maps in this context that is general enough to handle the different constructions given in 1) -- 4) above.</p> http://mathoverflow.net/questions/7655/trace-map-attached-to-a-finite-homomorphism-of-noetherian-rings/52083#52083 Answer by B. Cais for Trace map attached to a finite homomorphism of noetherian rings B. Cais 2011-01-14T15:46:30Z 2011-01-14T15:46:30Z <p>This is really a response to Karl's beautiful example; I'm posting it as an "answer" only because there isn't enough room to leave it as a comment.</p> <p>The condition on conductor ideals is one that I had come across by thinking about the dual picture. Namely, let $f:Y\rightarrow X$ be a finite map of 1-dimensional proper and reduced schemes over an algebraically closed field $k$. Then $Y$ and $X$ are Cohen-Macaulay by Serre's criterion, so the machinery of Grothendieck duality applies. In particular, the sheaves $f_*O_Y$ and $f_*\omega_Y$ are dual via the duality functor $\mathcal{H}om(\cdot,\omega_X)$, as are $O_X$ and $\omega_X$. Here, $\omega_X$ and $\omega_Y$ are the ralative dualizing sheaves of $X$ and $Y$, respectively. Thus, the existence of a trace morphism $f_*O_Y\rightarrow O_X$ is equivalent by duality to the existence of a pullback map on dualizing sheaves $\omega_X\rightarrow f_*\omega_Y$. </p> <p>In the reduced case which we are in, one has Rosenlicht's explicit description of the dualizing sheaf: for any open $V$ in $X$, the $O_X(V)$-module $\omega_X(V)$ is exactly the set of meromorphic differentials $\eta$ on the normalization $\pi:X'\rightarrow X$ with the property that $$\sum_{x'\in \pi^{-1}(x)} res_{x'}(s\eta)=0$$ for all $x\in V(k)$ and all $s\in O_{X,x}$. </p> <p>It is not difficult to prove that if $C$ is the conductor ideal of $X'\rightarrow X$ (which is a coherent ideal sheaf on $X'$ supported at preimages of non-smooth points in $X$), then one has inclusions $$\pi_*\Omega^1_{X'} \subseteq \omega_X \subseteq \pi_*\Omega^1_{X'}(C).$$ Since $X'$ and $Y'$ are smooth, so one has a pullback map on $\Omega^1$'s, our question about a pullback map on dualizing sheaves boils down the following concrete question:</p> <p><Blockquote> When does the pullback map on meromorphic differentials $\Omega^1_{k(X')}\rightarrow \pi_*\Omega^1_{k(Y')}$ carry the subsheaf $\omega_X$ into $\pi_*\omega_Y$? </Blockquote></p> <p>By looking at the above inclusions, I was led to conjecture the necessity of conductor ideal containment as in my original post. As Karl's example shows, this containment is not sufficient.<br> Here is Karl's example re-worked on the dual side:</p> <p>Set $B:=k[x,y]/(xy)$ and $A:=k[u,v]/(uv)$ and let $f:A\rightarrow B$ be the $k$-algebra map taking $u$ to $x^2$ and $v$ to $y$. Writing $B'$ and $A'$ for the normalizations, we have $B'$ and $A'$ as in Karl's example, and the conductor ideals are $(x,y)$ and $(u,v)$. Now the pullback map on meromorphic differentials on $A'$ is just $$(f(u)du,g(v)dv)\mapsto (2xf(x^2)dx,g(y)dy).$$ The condition of being a section of $\omega_A$ is exactly $$res_0(f(u)du)+res_0(g(v)dv)=0,$$ and similarly for being a section of $\omega_B$. But now we notice that $$res_0(2xf(x^2)dx)+res_0(g(y)dy) = 2 res_0(f(u)du) + res_0(g(v)dv) = res_0(f(u)du)$$ if $(f(u)du,g(v)dv)$ is a section of $\omega_A$. Thus, as soon as $f$ is not holomorphic (i.e. has nonzero residue) the pullback of the section $(f(u)du,g(v)dv)$, as a meromorphic differential on $B'$, will NOT lie in the subsheaf $\omega_B$.</p> <p>Clearly what goes wrong is that the ramification indices of the map $f:A'\rightarrow B'$ over the two preimages of the nonsmooth point are NOT equal. With this in mind, I propose the following addendum to my original number 4):</p> <blockquote> In the notation of 4) above and of Karl's post, assume that $f'(C_A)=C_B^e$ for some positive integer $e$. Then the trace map $B'\rightarrow A'$ carries $B$ into $A$. </blockquote> <p>Certainly this rules out Karl's example. I think another way of stating the condition is that the map $f':Spec(B')\rightarrow Spec(A')$ should be "equi-ramified" over the nonsmooth locus of $Spec(A)$, i.e. that the ramification indices of $f'$ over all $x'\in Spec(A')$ which map to the same nonsmooth point in $Spec(A)$ are all equal.</p> <p>Is this the right condition?</p> http://mathoverflow.net/questions/42284/geometric-inertia-action Geometric Inertia Action B. Cais 2010-10-15T13:33:53Z 2010-10-15T13:33:53Z <p>Let $K$ be a finite extension of $Q_p$ and $K'/K$ a totally ramified Galois extension with Galois group $G$. For $g\in G$ and any scheme $X$ over $O_{K'}$, write $X_g$ for the base change of $X$ along the automorphism of $Spec(O_{K'})$ induced by $g$.</p> <p>Suppose that $X$ is a smooth and proper scheme over $Spec(O_{K'})$ whose generic fiber descends to $K$. By Galois descent, there is a "semi-linear" action of $G$ on $X_{K'}$, i.e. for each $g\in G$ a $K'$-morphism</p> <p>$$g: X_{K'}\rightarrow (X_{K'})_g$$</p> <p>which is compatible with change in $g$ in the obvious way.</p> <p>My question is:</p> <blockquote> <p>When does this semi-linear action of $G$ on $X_{K'}$ extend to $X$ over $Spec(O_{K'})$? </p> </blockquote> <p>Actually, I am really only interested in the following weakened version of the above question:</p> <blockquote> <p>When does the semi linear action of $G$ on $X_{K'}$ induce a (linear) action of $G$ on the special fiber of $X$?</p> </blockquote> <p>Certainly a positive answer to the first question implies a positive answer to the second one, since the projection map $X_g\rightarrow X$ is an isomorphism on special fibers over the residue field, since $K'/K$ is totally ramified.</p> <p>Note that $O_{K'}/O_K$ is NOT etale, so etale descent does not apply here (i.e. the existence of such a semi-linear $G$ action on $X$ does not imply that $X$ admits a descent to $Spec(O_K)$.)</p> <p>Here is an example of when one does get the kind of "geometric inertia" action I am seeking: Let $X$ be an abelian scheme. Then the $K'$-maps $X_{K'}\rightarrow (X_{K'})_g$ extend to maps $X\rightarrow X_g$ by the Neron mapping property. </p> http://mathoverflow.net/questions/37536/quotient-of-abelian-variety-by-an-abelian-subvariety Quotient of abelian variety by an abelian subvariety B. Cais 2010-09-02T20:26:19Z 2010-09-02T21:48:48Z <p>Let $k$ be a field and $A$ an abelian variety over $k$. Suppose that $B$ is an abelian subvariety of $A$. Consider the following fact:</p> <p>There exists an abelian variety $C$ over $k$ and a surjective morphism $A\twoheadrightarrow C$ with kernel exactly $B$. </p> <p>This is proved in section 9.5 of the book "Abelian varieties, theta functions and the Fourier transform" By Alexander Polishchuk on the way to proving Poincare reducibility. The proof there seems (to me) to be a bit complicated, so I'm wondering if anyone knows of a "simple" proof. I think I could probably devise a proof of the above fact using Poincare reducibility (employing the proof of the latter result in Milne's chapter of Cornell-Silverman, Proposition 12.1 to avoid circular logic), but somehow I'm not so satisfied by this as it seems like it ought to be an "easy" fact.</p> http://mathoverflow.net/questions/18765/lifting-abelian-varieties-in-the-closed-fiber-of-a-fixed-neron-model Lifting abelian varieties in (the closed fiber of) a fixed Neron model B. Cais 2010-03-19T16:00:02Z 2010-03-20T21:52:10Z <p>Suppose that $R$ is a dvr with field of fractions $K$ and residue field $k$ and that $A_K$ is an abelian variety over $K$ with Neron model $A$ over $R$. Then the closed fiber $A_k$ is a smooth commutative group scheme over $k$. Suppose now that $B_k$ is an abelian variety over $k$ that is a quotient of $A_k$. </p> <blockquote> <p>Does there exist an abelian scheme $B$ over $R$ and a morphism $A\rightarrow B$ of smooth commutative $R$-groups whose base change to $k$ is the quotient mapping $A_k \rightarrow B_k$?</p> </blockquote> <p>The answer is NO, I'm fairly sure, as the generic fiber of $A$ could be simple with $A_k$ an extension of an abelian variety by a torus. Let us therefore assume that $B_k$ can be lifted up to isogeny, i.e. that there exists $C$ an abelian scheme over $R$ such that 1) $C_k$ is isogenous to $B_k$ and 2) There exists a map of smooth groups $A\rightarrow C$ over $R$ whose base change to $k$ is the quotient map $A_k\rightarrow B_k$ followed by the isogeny $B_k\rightarrow C_k$. With this added assumption, is the answer to the question above still NO? </p> <p>I'm inclined to think that this is the case, but can't immediately convince myself of this.</p> <hr> <h2>Reformulation</h2> <p>Consider the following theorem of Chevalley (see 9.2/1 of the book "Neron Models" by Bosch, Lutkebohmert and Raynaud):</p> <p><strong>Theorem:</strong> Let $k$ be a perfect field and $G$ a smooth and connected algebraic $k$-group. Then there exists a smallest connected linear subgroup $L$ of $G$ such that the quotient $G/L$ is an abelian variety. Furthermore, $L$ is smooth and of formation compatible with extension of $k$.</p> <p><strong>Definition:</strong> We write $av(G)$ for $G/L$ as in the Theorem.</p> <p>Now fix a dvr $R$ of mixed characteristic $(0,p)$ with fraction field $K$ and residue field $k$. Let $A_K$ be an abelian variety over $K$. There exists an abelian variety quotient $B_K$ of $A_K$, unique up to isogeny, with the following properties:</p> <ol> <li>$B_K$ has good reduction</li> <li>Any abelian variety quotient $A_K\rightarrow C_K$ of $A_K$ having good reduction factors through $A_K\rightarrow B_K$.</li> </ol> <p>If we impose the additional assumption that the kernel of $A_K\rightarrow B_K$ is connected (i.e. an abelian sub-variety of $A_K$), then $B_K$ is uniquely determined. We call this $B_K$ the maximal good reduction quotient of $A_K$. </p> <p>The surjection $A_K\rightarrow B_K$ induces a mapping $A\rightarrow B$ on Neron models over $R$ and hence a mapping on identity components of closed fibers $A^0_k \rightarrow B_k$ which yields a homomorphism of abelian varieties $$\varphi:av(A^0_k)\rightarrow B_k.$$</p> <blockquote> <p><strong>Question:</strong> Is the kernel of $\varphi$ an abelian sub-variety of $av(A^0_k)$?</p> </blockquote> http://mathoverflow.net/questions/15407/relationship-between-line-bundles-with-isomorphic-ring-of-sections/15451#15451 Answer by B. Cais for Relationship between Line Bundles with isomorphic ring of sections B. Cais 2010-02-16T14:19:11Z 2010-02-16T14:19:11Z <p>If $X$ is a smooth projective algebraic variety of dimension $d$ over a field and $L$ is an ample line bundle on $X$, then $R=\bigoplus_{m=0}^{\infty} H^0(X,mL)$ is a graded $k$-algebra of dimension $d+1$ and one has $X\simeq \mathrm{Proj}(R)$.</p> http://mathoverflow.net/questions/15336/is-symn-v-cong-symn-v-ast-naturally-in-positive-characteristic/15339#15339 Answer by B. Cais for Is $Sym^n (V^*) \cong Sym^n (V)^\ast$ naturally in positive characteristic? B. Cais 2010-02-15T15:32:52Z 2010-02-15T17:14:14Z <p>As several MO'ers have noted, the answer is in general no, contrary to what I falsely claimed in my previous post. To make amends for writing nonsense before, let me point you to Eisenbud, Commutative Algebra with a View, where this issue is nicely discussed in A.2.4. What turns out to be true, as Marty had hinted at, is that the graded dual of the symmetric algebra is naturally isomorphic to the divided power algebra of the dual. When you are working over $\mathbf{Q}$, this divided power algebra is isomorphic to the symmetric algebra, but in general it need not even be noetherian. </p> http://mathoverflow.net/questions/15141/why-is-2-so-odd/15145#15145 Answer by B. Cais for Why is 2 so odd? B. Cais 2010-02-12T22:23:53Z 2010-02-12T22:23:53Z <p>Another reason that 2 is a strange prime is that for a prime $p$, the divided powers $\frac{p^n}{n!}$ tend to zero $p$-adically <em>unless</em> $p=2$. This makes many things in the theory of crystalline cohomology, $p$-divisible groups and integral $p$-adic Hodge theory more subtle (and in some cases just false) when $p=2$. </p> http://mathoverflow.net/questions/15133/rosenlicht-differentials-for-possibly-non-reduced-curves Rosenlicht differentials for possibly non-reduced curves B. Cais 2010-02-12T18:03:55Z 2010-02-12T18:03:55Z <p>Let $X$ be proper Cohen-Macaulay scheme of pure dimension 1 over an algebraically closed field $k$. When $X$ is moreover reduced, Rosenlicht's theory of regular differential forms gives a beautiful <em>explicit</em> description of Grothendieck duality for $X$. I am wondering if there is an analogue of Rosenlicht's theory in the general context of proper CM curves.</p> <p>More precisely, in the reduced case Rosenlicht defines a sheaf of $\cal{O}_X$-modules $\omega$</p> <p>whose sections over an open $U$ in $X$ are those meromorphic differentials $\eta$ on the inverse image of $U$ in the normalization $X'$ of $X$ with the property that for each closed point $x\in U$, the sum of the residues of $f\eta$ over all points $y$ of $X'$ mapping to $x$ is zero for all $f\in \mathcal{O}_{X,x}$. He also defines a trace morphism $Tr:H^1(X,\omega)\rightarrow k$ in terms of "sums of residues" with the property that the pair $(\omega,Tr)$ is canonically isomorphic to the relative dualizing sheaf with its (Grothendieck) trace mapping (one proves this by showing that Roesnlicht's construction satisfies the right universal property).</p> <p>Is there a similarly explicit (i.e. in terms of certain kinds of differential forms and residues) description of the sections of the dualizing sheaf and of the trace map in the general proper CM setting?</p> <p>Things to note: A random normal proper and flat curve (=scheme of pure relative dimension 1) $\cal{X}$ over $W(k)$ will always have CM special fiber. However, this special fiber is "very often" not reduced, so there are many examples of non-reduced CM curves.</p> <p>I asked a previous question on MO hinting at this one: <a href="http://mathoverflow.net/questions/14012/" rel="nofollow">http://mathoverflow.net/questions/14012/</a></p> <p>I'm happy to assume that $X$ is Gorenstein, if that is at all useful. </p> http://mathoverflow.net/questions/14508/galois-representations-attached-to-newforms Galois representations attached to newforms B. Cais 2010-02-07T16:40:46Z 2010-02-07T17:43:55Z <p>Suppose that $f$ is a weight $k$ newform for $\Gamma_1(N)$ with attached $p$-adic Galois representation $\rho_f$. Denote by $\rho_{f,p}$ the restriction of $\rho_f$ to a decomposition group at $p$. When is $\rho_{f,p}$ semistable (as a representation of $\mathrm{Gal}(\overline{\mathbf{Q}}_p/\mathbf{Q}_p)$?</p> <p>To make things really concrete, I'm happy to assume that $k=2$ and that the $q$-expansion of $f$ lies in $\mathbf{Z}[[q]]$. </p> <p>Certainly if $N$ is prime to $p$ then $\rho_{f,p}$ is in fact crystalline, while if $p$ divides $N$ exactly once then $\rho_{f,p}$ is semistable (just thinking about the Shimura construction in weight 2 here, and the corresponding reduction properties of $X_1(N)$ over $\mathbf{Q}$ at $p$). For $N$ divisible by higher powers of $p$, we know that these representations are de Rham, hence potentially semistable. Can we say more? For example, are there conditions on "numerical data" attached to $f$ (e.g. slope, $p$-adic valuation of $N$, etc.) which guarantee semistability or crystallinity over a specific extension? Can we bound the degree and ramification of the minimal extension over which $\rho_{f,p}$ becomes semistable in terms of numerical data attached to $f$? Can it happen that $N$ is highly divisible by $p$ and yet $\rho_{f,p}$ is semistable over $\mathbf{Q}_p$?</p> <p>I feel like there is probably a local-Langlands way of thinking about/ rephrasing this question, which may be of use... </p> <p>As a possible example of the sort of thing I have in mind: if $N$ is divisible by $p$ and $f$ is ordinary at $p$ then $\rho_{f,p}$ becomes semistable over an abelian extension of $\mathbf{Q}p$ and even becomes crystalline over such an extension provided that the Hecke eigenvalues of $f$ for the action of $\mu_{p-1}\subseteq (\mathbf{Z}/N\mathbf{Z})^{\times}$ via the diamond operators are not all 1.</p> http://mathoverflow.net/questions/13977/is-there-something-like-ech-cohomology-for-p-adic-varieties/14146#14146 Answer by B. Cais for Is there something like Čech cohomology for p-adic varieties? B. Cais 2010-02-04T13:59:47Z 2010-02-04T13:59:47Z <p>The first comment to make is that Cech theory is really extremely general, and can be set up to compute the cohomology of any complex of abelian sheaves on any site (provided you have coverings that are cohomologically trivial). This is explained at least somewhat in SGA4, Expose 5 and EGA III, Chap 0, section 12. </p> <p>I think you should be working with the rigid analytic space attached to $X$, and not with the $\mathbf{Q}_p$-points of $X$, and the latter really has no good topology on it besides the totally disconnected one induced from the topology on $\mathbf{Q}_p$. </p> <p>Let's assume that $X$ has a model $\mathcal{X}$ over $\mathbf{Z}_p$ that is smooth and proper and write $\widehat{\mathcal{X}}$ for the formal completion of $\mathcal{X}$ along its closed fiber. Then the (Berthelot) generic fiber $\widehat{\mathcal{X}}^{rig}$ of $\widehat{\mathcal{X}}$ is a rigid analytic space that is canonically identified with the rigid analytification of $X$ (using properness here). Moreover, one has a "specialization morphism" of ringed sites $$sp:X^{an}\simeq \widehat{\mathcal{X}}^{rig}\rightarrow \widehat{\mathcal{X}}$$ with the property that for any (Zariski) locally closed subset $W$ of the target, the inverse image $sp^{-1}(W)$ is an admissible open of the rigid space $X^{an}$ (called the open tube over W). In this way, coverings of the special fiber by locally closed subsets give coverings of the rigid generic fiber by admissible opens, and you can use Cech theory with these coverings and or your favorite spectral sequence to compute sheaf cohomology in the rigid analytic world. Again using properness, by rigid GAGA this cohomology agrees with usual (Zariski) cohomology on the scheme $X$ (provided your sheaf is a coherent sheaf of $\mathcal{O}_X$-modules, say). </p> <p>This idea of computing cohomology using admissible coverings of the associated rigid space is a really important one as it allows you to use the geometry of the special fiber. It occurs (allowing $\mathcal{X}$ to have semistable reduction) in the work of Gross on companion forms, of Coleman on $\mathcal{L}$-invariants and most prominently in Iovita-Coleman (see their article on "Frobenius and Monodromy operators"). This latter article might be a good place to start.</p> <p>I would also highly recommend the articles of Berthelot:</p> <p><a href="http://perso.univ-rennes1.fr/pierre.berthelot/publis/Cohomologie_Rigide_I.pdf" rel="nofollow">http://perso.univ-rennes1.fr/pierre.berthelot/publis/Cohomologie_Rigide_I.pdf</a></p> <p><a href="http://perso.univ-rennes1.fr/pierre.berthelot/publis/Finitude.pdf" rel="nofollow">http://perso.univ-rennes1.fr/pierre.berthelot/publis/Finitude.pdf</a></p> <p>I'd also suggest the AWS 2007 notes by Brian Conrad for learning about rigid geometry, which seems generally quite pertinent to your situation. For etale cohomology of rigid spaces, you might want to look at the article of Berkovich, though this would require learning about his analytic spaces. </p> <p>In any case, I hope this is a good start.</p> http://mathoverflow.net/questions/14012/adjunction-for-underlying-reduced-subschemes Adjunction for underlying reduced subschemes B. Cais 2010-02-03T18:03:34Z 2010-02-03T19:49:10Z <p>Let $k$ be a perfect field (so reduced = geometrically reduced) and $f:X\rightarrow \mathrm{Spec}(k)$ a Cohen-Macaulay morphism. Denote by $i:X_{red}\rightarrow X$ the underlying reduced subscheme and by $\omega_{X}$ and $\omega_{X_{red}}$ the relative dualizing sheaves of $X$ and $X_{red}$ over $k$. What can one say about the relationship between these two sheaves? </p> <p>One might hope that there is an "adjunction formula" relating them, but I only know the adjunction formula in the context of a pair of maps $g:Y\rightarrow X$ and $f:X\rightarrow Z$ that are flat, of finite type, and CM, so this doesn't apply to the closed immersion $i:X_{red}\rightarrow X$ unless $X$ is already reduced (failure of flatness).</p> <p>Certainly one has a trace morphism $i_*: i_*\omega_{X_{red}}\rightarrow \omega_X$. Can one describe the image and kernel of $i_*$, say in terms of the ideal sheaf defining $i$?</p> http://mathoverflow.net/questions/7567/when-can-one-localize-ext/7651#7651 Answer by B. Cais for When can one localize Ext? B. Cais 2009-12-03T09:38:38Z 2009-12-03T09:38:38Z <p>I would suggest having a look at the article <a href="http://www.sciencedirect.com/science?%5Fob=ArticleURL&amp;%5Fudi=B6W9F-4CRY60R-1H3&amp;%5Fuser=2459438&amp;%5Frdoc=1&amp;%5Ffmt=&amp;%5Forig=search&amp;%5Fsort=d&amp;%5Fdocanchor=&amp;view=c&amp;%5Facct=C000057302&amp;%5Fversion=1&amp;%5FurlVersion=0&amp;%5Fuserid=2459438&amp;md5=deca64ad5b5501727ed918a10bc1ffd9" rel="nofollow">"Compactifying the Picard Scheme"</a> by Altman-Kleiman. They discuss base change issues for Ext. I'm not sure if this will be applicable in your particular context, but it may be a start.</p> http://mathoverflow.net/questions/84273/deformation-of-ordinary-p-divisible-groups-via-grothendieck-messing Comment by B. Cais B. Cais 2011-12-25T18:26:44Z 2011-12-25T18:26:44Z Or explicitly, as the case may be :), and this must be where the error lies. The formation of $$0\rightarrow \omega_G\rightarrow D(G)(W)\rightarrow Lie(G^t)\rightarrow 0$$ is functorial in $G$; applying this to the connected-etale sequence of $G$ shows that the map $\omega_{G^m}\rightarrow D(G^m)(W)$ coincides with the composition of $\omega_{G^m}\simeq \omega_G\rightarrow D(G)(W)$ with the projection to $D(G^m)(W)$, so I guess the problem is that the isomorphism $D(G)(W)\simeq D(G^m)(W)\times D(G^{et})(W)$ does not coincide with projection to $D(G^m)$ followed by inclusion (clearly). http://mathoverflow.net/questions/79087/subrings-of-invariants-in-divided-power-algebras Comment by B. Cais B. Cais 2011-10-26T18:01:13Z 2011-10-26T18:01:13Z The following example, which shows the answer to my question is &quot;no&quot;, was explained to me by BCnrd: Let $R:=\mathbf{Z}[\zeta_n][X]$ for $n&gt;1$ and $\zeta_n$ a primitive $n$-th root of unity, with $G:=\mu_n$ acting on $R$ via $\zeta\cdot X:=\zeta X$. The PD envelope of $R$ relative to $I:=(x)$ contains the $G$-invariant element $X^n/n!$, whereas the PD envelope of $R^G=\mathbf{Z}[\zeta_n][X^n]$ relative to $I^G=(x^n)$ does not (as can be easily checked by grading everything by degree). http://mathoverflow.net/questions/56289/orbits-of-sl-n-acting-on-matrices-of-determinant-p/56369#56369 Comment by B. Cais B. Cais 2011-02-23T13:47:31Z 2011-02-23T13:47:31Z Thank you for the reference! http://mathoverflow.net/questions/56289/orbits-of-sl-n-acting-on-matrices-of-determinant-p/56296#56296 Comment by B. Cais B. Cais 2011-02-23T13:47:04Z 2011-02-23T13:47:04Z Thanks, Qiaochu, your answer looks good to me! http://mathoverflow.net/questions/52625/tamely-ramified-p-adic-galois-representations/52645#52645 Comment by B. Cais B. Cais 2011-01-20T20:15:17Z 2011-01-20T20:15:17Z Dear Florian, Thank you for the very nice answer! http://mathoverflow.net/questions/51002/cyclotomic-periods-in-cohen-ring-of-norm-field Comment by B. Cais B. Cais 2011-01-03T18:20:15Z 2011-01-03T18:20:15Z Hi Keerthi, If $R=\varprojlim O_{\mathbf{C}_p}/p$ is Fontaine's ring, then you are probably viewing the norm field $k((u))$ as a subfield of the algebraically closed field $Frac(R)$ via $u\mapsto (\pi^{1/p^n})_n$. Doesn't the field of norms machinery imply that $Frac(R)$ is the algebraic closure of $k((u))$? I think what you are asking is tantamount to an explicit description of the (algebraic!) extension of $k((u))$ obtained by adjoining $(\varphi^{-n}(\epsilon))_n$ for $\epsilon=(1,\zeta_p,\ldots)$. Already &quot;writing down&quot; the min poly of $\epsilon$ over $k((u))$ seems difficult... http://mathoverflow.net/questions/42284/geometric-inertia-action Comment by B. Cais B. Cais 2010-10-15T14:45:39Z 2010-10-15T14:45:39Z Can I use the old trick of &quot;closure of graph of generic fiber morphism in $X\times X_g$&quot; and then Gruson-Raynaud to flatten, if needed? http://mathoverflow.net/questions/42284/geometric-inertia-action Comment by B. Cais B. Cais 2010-10-15T14:39:44Z 2010-10-15T14:39:44Z Thnks, Brian. What I had in mind for q. 2 is the existence of a smooth and proper $O_{K'}$-scheme $Y$ with finite flat maps $Y\rightarrow X$ and $Y\rightarrow X_{g}$ such that &quot;pullback and trace&quot; on crystalline cohomology of special fibers coincides with the action on de Rham cohomology of generic fibers after extending scalars to $K'$ and making the usual identification of dR and crys cohomology (after extending scalars). So yes, the motivation is to see a $G$-action on crystalline cohomology of smooth and proper schemes given the &quot;pot crys&quot; condition on etale coh of generic fiber descent http://mathoverflow.net/questions/37536/quotient-of-abelian-variety-by-an-abelian-subvariety/37544#37544 Comment by B. Cais B. Cais 2010-09-03T16:01:41Z 2010-09-03T16:01:41Z Brian--Ah, I see...thanks! http://mathoverflow.net/questions/37536/quotient-of-abelian-variety-by-an-abelian-subvariety/37544#37544 Comment by B. Cais B. Cais 2010-09-03T14:18:06Z 2010-09-03T14:18:06Z I see. My primary interest is this question is in positive characteristic, so I'd very much like to find a &quot;simple&quot; argument that does not rely on the the complex-analytic picture (i.e. a purely algebraic argument). I agree that exactness of duality as used above is easy over $\mathbb{C}$, but I think it is more subtle in general. http://mathoverflow.net/questions/37536/quotient-of-abelian-variety-by-an-abelian-subvariety Comment by B. Cais B. Cais 2010-09-03T13:08:19Z 2010-09-03T13:08:19Z Hey Brian, and thanks for your thoughts. I now agree that one has to use some things which are not &quot;trivial&quot;, whether it be representability of alg. space quotients or Poincare reducibility or properties of $\mathcal{E}xt(\cdot,\mathbb{G}_m)$, as in my comment to Francesco's answer below. http://mathoverflow.net/questions/37536/quotient-of-abelian-variety-by-an-abelian-subvariety/37544#37544 Comment by B. Cais B. Cais 2010-09-03T12:29:13Z 2010-09-03T12:29:13Z Thanks, Francesco. Comments: 1. Are you working over $\mathbb{C}$ to ensure $\mathrm{ker} \hat{u}$ is reduced? For general {\em perfect} $k$, \{($\mathrm{ker}\hat{u}$)_0\}_{red} is an abelian variety, but then one loses exactness. 2. The fact that &quot;dualizing&quot; is exact isn't obvious to me. I had to apply $\mathcal{H}om(\cdot,\mathbb{G}_m)$ and use that $\mathcal{E}xt^i(G,\mathbb{G}_m)$ vanishes for $i\ge 2$ (Oort, Comm gp sch, II 12.3) and is isomorphic to $G^{\vee}$ when $G$ is an ab. var. and $i=1$. 3. Can you explain why the kernels of $u$ and its dual have same no. of conn comps? http://mathoverflow.net/questions/15979/motivation-for-the-etale-topology-over-other-possibilities Comment by B. Cais B. Cais 2010-02-21T19:05:51Z 2010-02-21T19:05:51Z I'm a bit confused by the statement that &quot;there can't be a Weil cohomology with coefficients in those fields.&quot; Perhaps I am mis-remembering what the axiomatic definition of a Weil cohomology is, but isn't rigid cohomology supposed to be a Weil cohomology? http://mathoverflow.net/questions/15336/is-symn-v-cong-symn-v-ast-naturally-in-positive-characteristic/15339#15339 Comment by B. Cais B. Cais 2010-02-15T17:16:36Z 2010-02-15T17:16:36Z I edited my previous post (which was nonsense) to be more useful. My apologies if this makes the resulting comments seem a little out of context. http://mathoverflow.net/questions/14508/galois-representations-attached-to-newforms/14522#14522 Comment by B. Cais B. Cais 2010-02-08T00:49:39Z 2010-02-08T00:49:39Z Rob: What Wiles shows is that $p$ not dividing $a_p$ implies potentially ordinary, as he only works with his Galois representations up to $\overline{\mathbf{Q}}_p$-equivalence. The point is that the ordinary filtration on the Galois side may not be defined over $\mathbf{Q}_p$. Any weight 2 neworm of level $p^r$ primitive nebentypus has associated Galois representation that is potentially crystalline but non-crystalline over $\mathbf{Q}_p$ if $r&gt;0$. Wiles + Perrin-Riou as you indicate only gives $p$ not dividing $a_p$ implies potentially semistable, which one has anyway...