15
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Completion of a local ring of a curve
My answer is most likely going to be rephrasing Francesco's answer above. Here's how I think about your question. IMHO, the Cohen Structure Theorem is too big a thermonuclear weapon to invoke, because ...
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14
votes
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Completion of a local ring of a curve
Let me expand and generalize my comments above. We can prove the following
Proposition. Let $X$ be a projective scheme of dimension $n$ which is defined over an algebraically closed field $k$. If $...
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12
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Can an intersection of ideals in a Noetherian ring be replaced by a countable intersection?
Yes it's true.
First, in a (commutative) noetherian ring $A$, every chain of ideals is well-ordered by reverse inclusion. The supremum of ordinal types of such chains is denoted $o(A)$.
A ...
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11
votes
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Is the completion of a CAT(0) open ball a closed ball?
The answer is "no".
Let $\Sigma$ be the suspension over Poincaré homology sphere.
It admits a polyhedral $\mathrm{CAT}[1]$-metric.
Let $B$ be the unit ball in the Euclidean cone $\mathrm{Cone}\,\...
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10
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Does Grothendieck's algebraization imply existence of colimits of schemes?
Here's one way to see what's going on. I will use the Tannakian duality theorem of Hall and Rydh (see Theorem 1.1 here). It is stated for algebraic stacks, but if you replace the word "algebraic ...
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9
votes
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Varieties with everywhere good reduction that are isomorphic over every completion have isomorphic generic fibers
Here's an explicit example. Let $R=\mathbb{Z}[\sqrt{2}]$, let $X=\mathbb{P}^1_R$, and let $Y$ be the smooth projective conic defined by the equation $$(2-\sqrt{2})x^2+y^2+(2-\sqrt{2})z^2+xy+yz+(3-2\...
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9
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How is a MacNeille completion "universal" like a beta-compactification is "universal"?
There are some differences on the categorical level.
The compact Hausdorff spaces are a reflective subcategory of topological spaces, and the Stone-Cech compactification is left adjoint to the ...
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8
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Completion of a local ring of a curve
Actually, if $x$ is a nonsingular point of an irreducible $n$-dimensional algebraic variety $X$ over an algebraically closed field $K$ then a choice of local parameters $u_1, \dots , u_n$ at $x$ gives ...
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8
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Rank of a finite group and its representations
The answer to your question is yes and is the main theorem of the paper Žmudʹ, È. M.
On isomorphic linear representations of finite groups.
Mat. Sb. N.S. 38(80) (1956), 417–430.
It can be found in ...
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7
votes
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Derived Nakayama for complete modules
Let $A$ be a commutative ring and $I\subset A$ be a finitely generated ideal. The basic facts are:
For any complex of derived $I$-complete $A$-modules $C^\bullet$, the cohomology modules $H^*(C^\...
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5
votes
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Vanishing tate of a $p$-complete spectra
That is how you prove this.
Recall that $(-)^\wedge_p \simeq L_{\mathbb{S}/p}(-)$, where $L_E(-)$ is the Bousfield localization with respect to $E$. Now, multiplication by $q$ mod $p$ is an ...
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5
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How is a MacNeille completion "universal" like a beta-compactification is "universal"?
A subset $A$ of a complete lattice $L$ is said to be join-dense if $L=\{\bigvee^{L} R\mid R\subseteq A\}$ and $A$ is said to be meet-dense in $L$ if $L=\{\bigwedge^{L}R\mid R\subseteq A\}$. It turns ...
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5
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Lie Algebras over DVRs and basechange to the completion
One can also give explicit high-rank counterexamples using special orthogonal groups of quadratic lattices over $R$ with non-degenerate reduction, and we can also arrange that $R$ has $K$ as its ...
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5
votes
Why is $K_{\upsilon}|K$ separable for a global field $K$?
Since there is already an answer, I want to give my slightly different answer in a special case: Say $K = k(t)$ where $k$ is a finite field and $K_v = k((t))$. We only need to check that
$$
K_v \...
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5
votes
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Why is $K_{\upsilon}|K$ separable for a global field $K$?
By definition an extension of fields $K'/K$ is separable when $K' \otimes_K F$ is reduced for all field extensions $F/K$, and by limit considerations it is the same to say that all finitely generated ...
4
votes
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link between completion of the universal enveloping algebra and an endomorphism of functor
It sounds like you want to prove that $\text{End}(F)$ is the profinite completion of the universal enveloping algebra $U(\mathfrak{g})$. I don't understand your strategy for proving this (in ...
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4
votes
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Completeness of Localizations of Completions of Commutative Rings
No, $\hat{R}_y$ need not be $\hat{x}$-adically complete. The polynomial ring $R=k[x,y]$ is a counterexample. The $x$-adic completion of $R$ is identified with $k[y][[x]]$, the ring of power series in $...
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4
votes
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Enriched Cauchy completions and underlying categories
It works the other way around --- you should have tried to answer your question by yourself before you posted it here :-)
Yes, there is an equivalence $[X^{op}, V] \simeq [\overline{X}^{op}, V]$. You ...
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4
votes
Lie Algebras over DVRs and basechange to the completion
That is false. Let $A_R$ be an Azumaya algebra with center $R$ and that is a free $R$-module of rank $n^2$. Consider $\mathfrak{g}_R = A_R$ with bracket the usual commutator. Assume that $A_R/\...
4
votes
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Is completion of isolated singularity isolated?
I believe that in your situation, $B$ indeed has an isolated singularity at the maximal ideal $\mathfrak{n} \subseteq B$. Let me first give two possible definitions for “isolated singularity”; please ...
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3
votes
Is Cauchy completion the largest extension with the same free cocompletion?
The answer is positive.
I found a published account with details to be chapter 6 and 7 of Handbook of Categorical Algebra 1 by Francis Borceux.
Thanks to the comments, useful links that summarize how ...
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3
votes
Completion of a local ring of a curve
One possible reference is Mumford, The red book of varieties and schemes, chap. III, § 6.
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3
votes
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Is there a complete local analogue of the Artin-Tate lemma?
Yes. In fact, if $B \subset C$ is a module-finite extension of noetherian rings with $C$ local and complete then $B$ is local and complete. Indeed, by standard prime-lifting stuff with module-finite ...
- 1,885
3
votes
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Does CZF prove there is a minimal cauchy completion of the rationals?
You can prove this from the regular extension axiom $\mathbf{REA}$ using the general theory of inductive definitions. See e.g. Theorem 5.11 in Aczel & Rathjen, Notes on Constructive Set Theory. I ...
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2
votes
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Doesn't completion of a representation ring preserve its indecomposables?
In fact $\text{dim }I(G)/I(G)^2=2$. Let $x=[V_{3L_1}]$, $y=[V_{2L_1+L_2}]$ and $z=[V_{3L_1+3L_2}]$. Then
\begin{eqnarray}R(PSU(3))\cong\mathbb{Z}[x, y, z]/(y^3-y^2-xz-2y(x+z)-x-y-z).\end{eqnarray}
...
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2
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Is a filtered colimit of complete module complete?
The answer should be negative. For example, just take $R=k[[t]]$ in your case. We can choose a series of indeterminates $x_i$, and let $k_i=k(x_1,x_2,...,x_i)$, $M_n=k_n[[t]]$, and take the filtered ...
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2
votes
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Completed stalks of the pushforward of the structure sheaf
This has little to do with morphisms, and follows immediately from the following commutative algebra lemma:
Lemma. Let $R$ be a Noetherian ring, and $f \colon M \to N$ a morphism of finite $R$-modules....
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2
votes
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Completion and extension by scalars
This follows by adapting the proof of Tag 00MA, even without assumption 5. We also don't need $S$ to be an algebra; a complete $R$-module suffices. Finally, we never use that $R$ is $I$-adically ...
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2
votes
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Mapping cone and derived tensor product
The comment by skd basically answers your question. I am writing to flesh it out with references, so your question doesn't stay open forever. The derived category of $I$-complete $A$-modules has ...
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1
vote
Derived Nakayama for complete modules
There is a version of Nakayama with finiteness of cohomology when the ring is noetherian. See Theorem 2.2 of the paper [PSY2].
For MGM Equivalance see [PSY1].
References:
[PSY1]: M. Porta, L. Shaul ...
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