bio | website | |
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location | ||
age | ||
visits | member for | 2 years, 8 months |
seen | Jun 22 '13 at 6:39 | |
stats | profile views | 488 |
Mar 20 |
awarded | Yearling |
Aug 20 |
awarded | Yearling |
Jun 15 |
comment |
Schemes over a noetherian local ring and its completion
@ayanta: Doesn't the descent datum, at least a priori, entail specifying additional data over $\hat{R} \otimes_R \hat{R}$ (satisfying a cocycle condition) in addition to the isomorphism over $K \otimes_R \hat{R}$? |
Jun 15 |
comment |
Schemes over a noetherian local ring and its completion
@ayanta: Thanks; of course, I meant (and should have said) to keep track of the ample line bundle. How does one get around having to worry about $\hat{R} \otimes_R \hat{R}$ while directly applying fpqc descent? |
Jun 15 |
comment |
Schemes over a noetherian local ring and its completion
stacks.math.columbia.edu/tag/05E5 |
Jun 15 |
comment |
Schemes over a noetherian local ring and its completion
For any noetherian pair $(R,\mathfrak{m})$, one knows: giving a quasicoherent (qc) sheaf $F$ on $S := \mathrm{Spec}(R)$ is the same as giving a qc sheaf $\hat{F}$ on $\hat{S} := \mathrm{Spec}(\hat{R})$, a qc sheaf $F_U$ on $U = S - V(\mathfrak{m})$, and an isomorphism between the two over $U \times_S \hat{S}$. The same comment applies to "quasicoherent sheaves" replaced by "affine/quasiprojective morphisms." |
Jun 2 |
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are these functors exact?
$j_!$ does not preserve quasicoherent sheaves... |
May 25 |
comment |
Orders in number fields
The $r=1$ case can be handled as follows. Each order in $R$ of index $p$ is uniquely determined by its image in $R/pR$. Now $R/pR$ is an \'etale $k := \mathbf{F_p}$-algebra of rank $n$ as $p$ is unramified. Now the set of such rings injects into the set of index $p$ $k$-subalgebras of $R/pR \otimes_k \overline{k} \simeq \overline{k}^n$. The latter set is a combinatorial object (it's the set of surjective maps from an $n$-element set to an $(n-1)$-element set), and can be bounded above by the set of index $p$ subrings of $\mathbf{Z}^n$. |
Apr 21 |
awarded | Enlightened |
Apr 21 |
awarded | Nice Answer |
Apr 20 |
revised |
Is the derived double dual of a quasi-coherent sheaf a sheaf?
Complete rewrite with clarification on which derived category we work in |
Apr 20 |
comment |
pushforward of injective sheaf acyclic for cohomology with supports
A variation on Angelo's question: where does the argument in the last paragraph use the Zariski topology? All you need is that $\pi_* F(Y) \to \pi_* F(Y - Z)$ is surjective, which is the same as asking that $F(X) \to F(X - f^{-1} Z)$ be surjective, which is what the last paragraph shows. |
Apr 19 |
revised |
Is the derived double dual of a quasi-coherent sheaf a sheaf?
more clarification |
Apr 19 |
revised |
Is the derived double dual of a quasi-coherent sheaf a sheaf?
Added clarification about missing details |
Apr 19 |
answered | Is the derived double dual of a quasi-coherent sheaf a sheaf? |
Apr 11 |
comment |
$k[[x]]$ as a $(k[[x]])^p$ module for ugly fields
@Filippo: The $R$-submodule $x^p S \subset S$ is exactly the ideal $(x^p) \subset S$, so I don't understand the question. |
Apr 10 |
answered | $k[[x]]$ as a $(k[[x]])^p$ module for ugly fields |
Apr 10 |
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$k[[x]]$ as a $(k[[x]])^p$ module for ugly fields
@Karl: By passing to direct summands (which preserves completeness), it suffices to show that $M :== \oplus_{i=1}^\infty R$ is not $x$-adically complete for any ring $R$ with a regular element $x$. Consider the elements $m_n:=\sum_{i=1}^n x^i \cdot e_i$, where $e_i$ is the $i$-th basis vector. Then $\{m_n\}$ is a Cauchy sequence in $M$ with no limit (as any $m := \sum_{i=1}^N a_i \cdot e_i \in M$ admits an open neighbourhood $m + x^{N+2} M$ that does not contain a single $m_n$). |
Apr 10 |
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$k[[x]]$ as a $(k[[x]])^p$ module for ugly fields
It seems to me that the larger ring $S$ is the $x$-adic completion of the extension by scalars along $k^p \to k$ of the power series ring $R$ over $k^p$. Since $k$ is a free $k^p$-module of infinite rank, the question becomes: is the $x$-adic completion of a free $R$-module of infinite rank also free? The answer to this should be no, as an infinite rank free $R$-module is never $x$-adically complete. |
Apr 6 |
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Algebraic $p$-adic integers mod $p$
The Fontaine-Wintenberger theorem identifies $\overline{\mathbf{Z}_p}/p$ with its characteristic $p$ counterpart, i.e., with $R/t$, where $R$ is the ring of integers in an algebraic closure of $\mathbf{F}_p ((t))$. |