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May 17 |
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Lattices as invertible modules. No. The case of quadratic fields is misleading, or rather has a special property that fails in higher degree: the ring of integers is monogenic over $\mathbf{Z}$. Once you drop that property, the invertibility can fail (and all orders in rings of integers of number fields are Cohen-Macaulay). I think there are counterexamples given in Shimura's introductory book on modular forms, around where he discusses the invertibility in the quadratic case. |
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May 17 |
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extending truncated Barsotti-Tate group Yes; see Illusie's paper that surveys Grothendieck's work on the deformation theory of p-divisible groups. Early in there he gives Dieudonne-module arguments of Gabber and Ekedahl to handle the extension problem over any perfect field. |
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May 17 |
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Algebraic variety geometrically reduced Another proof: $\Omega^1_{K(X)/k}$ has $K)X)$-dimension $d$ by computing after extension to the perfect closure of $k$ (preserves topology -- irreducibility -- and reducedness by hypothesis), using existence of separating transcendence bases over perfect fields. Since $\Omega^1_{K(X)/k}$ is $K(X)$-spanned by elements ${\rm{d}}f$ for $f \in K(X)$, there is $\{f_j\}$ so ${\rm{d}}f_j$'s are a $K(X)$-basis. The $f_j$'s are algebraically independent over the perfect $k$ (differentiate a potential irreducible relation) and $L=k(t_1,\dots,t_d)\rightarrow K(X)$ has $\Omega^1_{K(X)/L}=0$. QED |
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May 7 |
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Shafarevich’s theorem for elliptic curves defined over function field of algebraic curve over algebraically closed field Yes. Let $U$ be the open affine curve away from $S$ over the constant field $k$. For any elliptic curve $E \rightarrow U$ there's a finite etale cover $U' \rightarrow U$ of universally bounded degree over which $E$ acquires a point of order 1728. Since $\pi_1(U)$ has only finitely many quotients of any given size, there are only finitely many possibilities for $U'$ up to $U$-isomorphism. Each connected component $U'_i$ of $U'$ has a specified map to $Y_1(1728)$ and it is dominant if $j(E) \not\in k$. There are only finitely many such maps since $X_1(1728)$ has genus $> 1$. QED |
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May 7 |
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examples of “exotic” moduli problems for elliptic curves? Aside from the fact that the phrase "doesn't have to do with torsion data" is vague, if you consider a $\mathbf{Z}[1/N]$-schemes $S$ that is sufficiently disconnected then the set of level-$N$ structures on a fixed elliptic curve $E$ over $S$ can be arbitrarily large and in particular not finite (akin to global sections of a constant sheaf on a disconnected space). So do you mean just that for elliptic curves over algebraically closed fields the associated set should be finite? If so, then how about assigning to any $E \rightarrow S$ the automorphism group? |
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May 2 |
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Reason for studying coherent sheaves on complex manifolds. It's the same reason that one develops a general theory of finitely generated modules over noetherian rings rather than focus exclusively on locally free modules (akin to vector bundles): we get more robust notions of kernel and cokernel, and a wider framework in which to prove results about ideals of holomorphic functions by viewing them as instance of coherent sheaves (for which many operations are available which cannot be expressed purely within the setting of ideal sheaves: tensor product, Hom-sheaves, etc.). And one can perform normalization of coherent sheaves of algebras, etc., etc. |
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May 1 |
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Example of codim 1 regular embedding that is not an effective Cartier divisor? Stripping away the geometric terminology, it sounds like you seek a commutative ring $A$ and ideal $I$ such $I_{\mathfrak{p}}$ is invertible as an $A_{\mathfrak{p}}$-module for all prime ideals $\mathfrak{p}$ of $A$ but $I$ is not invertible as an $A$-module (equivalently, $I$ is not finitely presented as an $A$-module). Is that correct? |
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May 1 |
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are moduli stacks deligne-mumford stacks in general @Jason: For another example, $X_0(N)$ (appropriately defined as a proper flat Artin stack over $\mathbf{Z}$) is not Deligne-Mumford in characteristic $p$ when $p^2|N$. |
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Apr 30 |
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are moduli stacks deligne-mumford stacks in general It would be a bit more accurate to say "finite etale automorphism schemes" (though since offered just as an "expectation", perhaps one cannot insist on too much precision). |
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Apr 30 |
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Weyl group of the restriction of scalars of split reductive group [I assume when you wrote "maximal (split) torus" you mean that $T$ is a split maximal $E$-torus of $G$ that is also maximal as an $E$-torus.] |
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Apr 30 |
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Weyl group of the restriction of scalars of split reductive group They're equal. This can be seen in multiple ways. For example, the evident isomorphism $S'_E \rightarrow T$ induced by $G_{E'} \twoheadrightarrow G$ gives an identification ${\rm{X}}_F(S') = {\rm{X}}_F(T)$ under which $\Phi(G',S')$ is carried isomorphically onto $\Phi(G,T)$ (using the equality of $\mathfrak{g}'$ with the underlying $F$-vector space of $\mathfrak{g}$ to match root spaces), and the Weyl group of this common root system is naturally identified with the finite constant groups $W$ and $W'$. This respects the induced map $W'_E \rightarrow W$, so the latter is an isomorphism. |
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Apr 23 |
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Weyl group an maximal tori An abstract group of points over an algebraically closed field is not an algebraic group. One has to specify the algebro-geometric structure over the ground field. Do you mean for $G$ to be the linear algebraic group ${\rm{GL}}_n$ over $k = \mathbf{F}_p$, or in other words is your implicit Galois action of Frobenius given by the usual $p$-power on matrix entries? If so, then since $T$ as a $k$-group is a split torus, the answer to your question is negative since every element of $W(\overline{k})$ arises from a point in $N_G(T)(k)$ modulo right translation by a point in $T(\overline{k})$. |
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Apr 22 |
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About isomorphism of $PGL(2)$ and $SO(3)$ Dear Robert: For a non-degenerate quadratic space $(V,q)$ over a field $k$, usually ${\rm{SO}}(q)$ denotes the algebraic $k$-group classifying automorphisms of $(V,q)$ (over extensions of $k$). If $q$ is the standard split quadratic form $q_n$ on $k^n$ ($x_1 x_2 + x_3 x_4 + \dots + x_{n-1}x_n$ for even $n$, $x_0^2 + q_{n-1}$ for odd $n > 1$), it is common for algebraists to write ${\rm{SO}}_n$ to denote ${\rm{SO}}(q_n)$. So for $k = \mathbf{R}$, the Lie group ${\rm{SO}}_n(\mathbf{R})$ is not the same as ${\rm{SO}}(n)$. And ${\rm{PGL}}_2 = {\rm{SO}}_3$ as algebraic groups (over any $k$)! |
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Apr 18 |
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reduction types of elliptic curves The formation of the Neron model over henselian discrete valuation rings commutes with scalar extension to the maximal unramified extension and its completion, so for any "table" of reduction types it is often sufficient to consider only separably closed residue fields. Hence, to the extent the residue field is perfect, it usually may as well be algebraically closed. (It isn't clear if Tate's algorithm works for imperfect residue field $k$ of char. 2 or 3, due to the existence of non-smooth regular Weierstrass cubics over such $k$.) For much more, read 10.2 in Qing Liu's textbook. |
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Apr 14 |
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Weil restriction of abelian schemes along finite étale (resp. finite locally free) morphisms Do you seek a proof that avoids algebraic spaces entirely? That seems unlikely to be possible, and if one is going to be using algebraic spaces then the Faltings-Chai proof is an extremely natural one (building off of the normal case in a clever way). For your purposes with Deligne 1-motives, is it a problem to work throughout with algebraic spaces (which are well-suited to questions related to quotients in algebraic geometry)? |
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Mar 31 |
awarded | ● Supporter |
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Mar 20 |
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How many proofs of the Weil conjectures are there? My recollection is that Kedlaya's proof is modeled on Deligne's 2nd proof, using a $p$-adic cohomology theory. Laumon's proof via $\ell$-adic Fourier transform is rather different from Deligne's proofs (or at least I remember it seeming so when I read it). |
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Mar 20 |
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Reductive Lie Groups and Complexification @PDC: There are commutative compact complex Lie groups which have nothing to do with linear algebraic groups, namely the "complex tori" in the sense of $V/L$ for a finite-dimensional complex vector space $V$ and full rank lattice $L$ in $V$. So the structure of the center needs to be brought out in the analytic theory to "rule out" problematic cases. The description I gave with the Lie algebra and the center (and a bit for the component group) seems as "good" as one can hope to say over $\mathbf{C}$ (life is harder over $\mathbf{R}$), or maybe someone else has a better idea... |
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Mar 20 |
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Reductive Lie Groups and Complexification Correction to my previous comment: I should have written $Z^0_{H^0}$ rather than $Z_{H^0}$. |
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Mar 20 |
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Reductive Lie Groups and Complexification The functor $G \rightarrow G(\mathbf{C})$ is an equivalence from linear algebraic $\mathbf{C}$-groups $G$ with reductive $G^0$ to complex Lie groups $H$ with reductive Lie algebra and finite $\pi_0(H)$ such that $Z_{H^0}$ is a power of $\mathbf{C}^{\times}$. For any such $H$ and maximal compact subgroup $K$, denote by $K'$ the unique linear algebraic $\mathbf{R}$-group with $K'(\mathbf{R})=K$ meeting every component of $K'$. Then $K'(\mathbf{C})=H$ and $H$ is the complexification of $K'$ as defined in Bourbaki. See D.3.2 and D.3.3 in the Luminy notes on reductive group schemes (use Google). |
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Mar 20 |
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equivariant Serre Duality. @Angelo: Perhaps the OP seeks a reference to justify the invariance you mention? Many references on Serre duality make the construction of the trace in a manner that is not sufficiently intrinsic to render the triviality apparent. It is equivalent to show that the natural composite map $H^n(X,\Omega^n_{X/k}) \rightarrow H^n(X,g^{\ast}(\Omega^n_{X/k})) \rightarrow H^n(X,\Omega^n_{X/k})$ is the identity (1st step pullback, 2nd step canonical at sheaf level); settling projective spaces "by bare hands" is a bit unpleasant (though easy by using the structure of the automorphism group). |
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Feb 25 |
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Power of ideals and exact sequences Typo: meant "right to left" on the first line above. |
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Feb 25 |
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Power of ideals and exact sequences Both are false in general. For (1), the map from left to right is surjective but has huge kernel in general: consider $\mathfrak{a} = (x,y) \subset k[x,y] = R$ for a field $k$ (in which case the left side has dimension $n+1$ over $k$ whereas the right side has dimension $2^n$ over $k$. For (2), the map from 2nd to 3rd term doesn't make sense, and using the same $R$ and $\mathfrak{a}$ and $M = \mathfrak{a}$ makes the quotient of the middle term by the left have $k$-dimension $n+2$ whereas the third term has $k$-dimension $3(n+1)$. |
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Feb 25 |
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a question about Beauville-Laszlo In the noetherian case, BL is a special case of usual faithfully flat descent. For any noetherian ring $R$ (such as $V[[u,v]]/(uv-\pi)$) and any $\pi \in R$, an element of $R$ divisible by $\pi$ in the $\pi$-adic completion $R'$ of $R$ is divisible by $\pi$ in $R$ because the natural map $R/\pi R \rightarrow R'/\pi R'$ is injective (even an isomorphism), and the diagonal map $R \rightarrow R[1/\pi] \times R'$ is injective (by faithful flatness considerations locally along the zeros of $\pi$ in Spec($R$)), so the $F$ you ask about is always the original noetherian ring $R$. |
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Feb 25 |
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Is strong multiplicity one (obviously) stronger than multiplicity one? The $q$-expansion principle has nothing to do with multiplicity one (at least if you mean "$q$-expansion principle" in the sense that arises in arithmetic geometry, controlling the field or ring generated by the $q$-expansion coefficients in terms of the first few of them, which is a rather different thing from saying exactly what specific such coefficients are equal to). Put another way, multiplicity one is an analytic or representation-theoretic fact, whereas the $q$-expansion principle is an algebraic (or algebro-geometric) fact. Quite different beasts. |
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Feb 24 |
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Restricting a Soft Sheaf to an Open is again Soft? Every separable locally compact Hausdorff topological space is paracompact, so under your hypotheses (including separability) every subspace of $X$ is paracompact. |
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Feb 18 |
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normal subgroup scheme of a group scheme @unknown: You are still making a mistake. Your modified version is generally not a subfunctor since normality is not functorial in group homomorphisms. So I ask again: please explain why you do not want to work with the notion of scheme-theoretic normalizer that is the one which "works" in SGA3. |
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Feb 17 |
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Classification of Tori of GL2, up to conjugation The $H^1(k,X)$ in this answer should be $H^1(k,{\rm{Aut}}_{X/k})$ (and one should note the relevance of projectivity hypotheses to ensure effectivity of descent, unless "object" is understood in some sense like algebraic spaces). |
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Feb 13 |
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Weierstrass Models and Canonical Models Let $R$ be a strictly henselian dvr with fraction field $K$ and residue field $k$, and $X$ a proper flat $R$-scheme with generic fiber $X_K$ geometrically connected and smooth of dimension 1 and positive genus. Assume $X$ is the minimal regular proper model of $X_K$. Choose a prime $\ell \in R^{\times}$. For $G_K := {\rm{Gal}}(K_s/K)$, you want $H^1(X_k, \mathbf{Q}_{\ell}) \rightarrow H^1(X_{K_s},\mathbf{Q}_{\ell})^{G_K}$ to be an isomorphism. This problem can be "localized" via vanishing cycles, and in general is subtle (but is tractable when $X_k$ is semistable). Let's see what Q. Liu says. |
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Feb 12 |
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Why the Abel-Jacoby map is algebraic morphism? If $C$ were initially given over a subfield $k \subset \mathbf{C}$ do you also want to know that this analytically defined map arises from an algebraic map over $k$ (using the algebraic theory of the Jacobian)? Do you need to know something similar for "algebraic families" of such curves? The argument given by wccanard is the "best" one for such purposes (except that it masks the necessity to prove that the analytic map you have written down is actually computing the same thing as a map built using the functorial perspective in the algebraic theory, up to a sign). |
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Feb 9 |
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on trivialisation of T-torsors @prochet: Oops, I should have remembered the vanishing result over fields of cohomological dimension 1. Strictly speaking, I think Steinberg's paper is restricted to perfect fields. The vanishing over any field of cohomological dimension 1 (such as the function field of a curve over an algebraically closed field, which is imperfect in positive characteristic) is not true for general smooth connected affine groups but is true for connected reductive groups (such as tori); see Remark 1 in section 2.3 of Chapter III of Serre's "Galois cohomology" book. |
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Nov 27 |
awarded | ● Commentator |
