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7 votes
1 answer
347 views

$\mathbb{Z}$-homomorphism and $\mathbb{Z}_p$-homomorphism

$\newcommand{\cts}{\mathrm{cts}}$Thanks for your reading. Let $A,B$ be two $\mathbb{Z}_p$-modules, where $\mathbb{Z}_p$ is the $p$-adic integer ring. I have two questions. Is $\mathrm{Hom}_{\mathbb{Z}...
Rellw's user avatar
  • 319
4 votes
1 answer
259 views

On inverse limits of $\pi$-adically complete algebras

Consider the following situation, let $\mathcal{R}$ be a discrete valuation ring with uniformizer $\pi$ (say the valuation ring of a finite extension $K$ of $\mathbb{Q}_{p}$. Let $\{ A_{n}\}_{n\in\...
FPV's user avatar
  • 541
5 votes
0 answers
204 views

What are all of the topological (commutative) monoid structures on a closed interval?

Consider a closed real interval $[a,b]$ as a toplogical space. Up to homeomeorphism it doesn't matter, but I like to take $[a,b] = [0,\infty]$. Question 1: What are all of the topological commutative ...
Tim Campion's user avatar
  • 63.9k
5 votes
0 answers
260 views

When is the profinite completion of a Noetherian group ring also Noetherian?

Let $G$ be a group, and let $\mathbb{Z}[G]$ denote its group ring. Its profinite completion is the inverse limit over all ideals of finite index. By Benjamin Steinberg's answer here, this profinite ...
stupid_question_bot's user avatar
1 vote
0 answers
201 views

Is there a natural topology on $\mathbb{C}(t)[x_1,\ldots, x_n]$ with this property?

Is there a good topology on $A=\mathbb{C}(t)[x_1,\ldots, x_n]$ so that $A$ is a topological algebra with the following property: For any $N>0$ and a polynomial $F\in\mathbb{C}[x_1,\ldots, x_n]$ ...
solver6's user avatar
  • 291
3 votes
0 answers
78 views

Is $X$ closed in $Aut_{\mathbb{C}(t)}(\mathbb{C}(t)[x_1,\ldots, x_n])$?

Consider $\mathbb{C}$-algebras $$A = \mathbb{C}[t][x_1,\ldots, x_n]\subset\mathbb{C}(t)[x_1,\ldots, x_n] = B$$ Group $\operatorname{Aut}_{k(t)}(k(t)[x_1,\ldots, x_n])$ carry a power series topology (...
A.Skutin's user avatar
  • 329
3 votes
0 answers
110 views

Is this a lattice?

Let $R$ be a locally compact ring (commutative with unit) and let $D\subset R$ be a discrete cocompact subring (cocompact means the additive group $R/D$ is compact). Let $G$ be a semisimple linear ...
user avatar
19 votes
2 answers
566 views

Ostrowski's Theorem for topological rings?

Ostrowski's theorem classifies all absolute values on a number field $K$. Questions: More generally, can one classify all Hausdorff topologies on $K$ making $K$ into a topological field? In ...
Tim Campion's user avatar
  • 63.9k
13 votes
1 answer
442 views

Is the identity function a unique multiplicative homeomorphism of $\mathbb N$?

Endow the set $\mathbb N$ of positive integers with the topology $\tau$ generated by the base consisting of arithmetic progressions $a+b\mathbb N_0$ where $\mathbb N_0=\{0\}\cup\mathbb N$, where $a,b\...
Taras Banakh's user avatar
  • 41.8k
3 votes
0 answers
420 views

Discrete vs. finitely generated subgroups of the adèles

If $U\subseteq\mathbf{R}^n$ is an additive subgroup, discrete with respect to the induced topology, then $U$ is a finitely generated abelian group. Question. Given a discrete additive subgroup $U\...
user avatar
10 votes
1 answer
818 views

Is $k(\!(x,y)\!)$ a topological field?

More generally, let $(R,m)$ be a Noetherian local domain with fraction field $K$. The $m$-adic topology turns $R$ into a topological ring. When $R$ is a discrete valuation ring, this topology extends ...
Laurent Moret-Bailly's user avatar
2 votes
1 answer
304 views

Can we Characterise Rings of Continuous Functions?

Suppose $K$ is some nice space, for example $\mathbb R$ or $\mathbb C$. Let $X$ be a set and $C$ a ring of functions $X \to K$. Is there any way to determine, from the algebraic structure of $C$, ...
Daron's user avatar
  • 1,955
8 votes
1 answer
351 views

When is a valued field second-countable?

Let $R$ be a valuation ring, with fraction field $K$ and residue field $k$. Denote by $\Gamma=K^{\times}/R^{\times}$ the valuation group (assumed nontrivial). The valuation $v:K^{\times}\to\Gamma$ ...
Laurent Moret-Bailly's user avatar
5 votes
3 answers
676 views

Does every compact Hausdorff ring admit a decomposition into primitive idempotents?

Let $\mathbf{R} = (R,\mathcal{T},+,\cdot,0,1)$ be a compact Hausdorff topological unitary ring, and consider the set $I(\mathbf{R}) := \{ e \in R \mid e \cdot e = e \}$ (of idempotents in $\mathbf{R}$)...
Niemi's user avatar
  • 1,498