12
votes
Accepted
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 ...
- 60.4k
11
votes
Accepted
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 ...
- 2,524
11
votes
Accepted
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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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 $$(2-\sqrt{2})x^2+y^2+(2-\sqrt{2})z^2+xy+yz+(3-2\...
- 22.2k
8
votes
Rank of a finite group and its representations
The answer to your question is yes and is the main theorem of the paper Žmudʹ, È. M.
On isomorphic linear representations of finite groups.
Mat. Sb. N.S. 38(80) (1956), 417–430.
It can be found in ...
- 37.4k
7
votes
Accepted
Malcev completion of free groups
Here is a proof that the statement is, as you expected, false (for $n\ge 2$), in case the field $K$ is $\mathbf{R}$ or $\mathbf{C}$. Unfortunately it does not give anything explicit.
Let $B$ be its ...
- 60.4k
7
votes
Accepted
Derived Nakayama for complete modules
Let $A$ be a commutative ring and $I\subset A$ be a finitely generated ideal. The basic facts are:
For any complex of derived $I$-complete $A$-modules $C^\bullet$, the cohomology modules $H^*(C^\...
- 15.2k
5
votes
Why is $K_{\upsilon}|K$ separable for a global field $K$?
Since there is already an answer, I want to give my slightly different answer in a special case: Say $K = k(t)$ where $k$ is a finite field and $K_v = k((t))$. We only need to check that
$$
K_v \...
- 1,376
5
votes
Accepted
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 ...
5
votes
Accepted
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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4
votes
Accepted
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 ...
- 1,663
4
votes
Accepted
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
Accepted
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 $...
- 8,941
4
votes
“Geometric” vs Homotopical completion
In the affine case, this is more-or-less proved in [Bhargav Bhatt, Completions and derived de Rham cohomology]. More precisely, Kathryn Hess' completion is more akin to Carlsson's Adams completion, ...
- 2,003
3
votes
Accepted
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 ...
- 3,846
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 ...
- 5,038
2
votes
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 ...
- 806
2
votes
Accepted
“Geometric” vs Homotopical completion
Yes, there's a way to relate the two. First, it's helpful to think of both in terms of universal properties. Since geometric completion is a fiber product, it's a pullback. Meanwhile, the homotopical ...
- 29.8k
2
votes
Accepted
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
votes
Accepted
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....
- 27.3k
2
votes
Accepted
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 ...
- 27.3k
2
votes
Accepted
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 ...
- 29.8k
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 ...
- 580
1
vote
Accepted
Is a filtered colimit of complete module complete?
The answer is no. Here is another example. Let the ring $R=k[[h]]$ and the modules be $M_n=k[h]/h^n$ with the bounding maps $M_n\to M_{n+1}$ sending 1 to $h$.
Then all $M_n$ are complete (they are $h$...
- 16.2k
1
vote
How is a MacNeille completion "universal" like a beta-compactification is "universal"?
For a more recent perspective on the universal property of the Dedekind–MacNeille (extended from posets to categories), see the paper Tight limits and completions from Dedekind-MacNeille to Lambek-...
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