Is $\mathbb Q[[x]]$ compact with the $(x)$-adic topology?

Question

I'd like to understand what the $(x)$-adic topology looks like here. I know that $\mathbb Q[[x]]$ is the completion of $\mathbb Q[x]$.

I feel like this should not be compact. To show its not compact we want to find some open cover which cannot be refined to a finite sub cover. The open sents in this topology are $0+(x^n)$ and translations. It seems a good candidate for an open cover would be include something like $0+(x^p)$ with $p$ running through the primes but this is not an cover.

score 2 · Accepted Answer · answered Dec 18 '20 at 14:01

2

I believe the answer is negative. The $(x)$-adic topology is induced by a norm, so compactness is equivalent to sequentual compactness. Taking an enumeration of $\mathbb{Q}$, $<q_i>$ gives us a sequence in which every 2 elements are at a fixed distance from each other, so it can not have a convergent subsequence.

answered Dec 18 '20 at 14:01

sss89

631

1

$\bigcup_{a\in \Bbb{Q}}a + (x)$ is an open cover with no finite subcover – reuns Dec 18 '20 at 14:04

Is $\mathbb Q[[x]]$ compact with the $(x)$-adic topology?

1 Answers1