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Let $K$ be a complete valued division ring and $S$ be a compact subset of $D$. It is easy to see that if $K$ is commuative, the $K$-algebra $\mathscr C(S,K)$ of continuous functions on $S$ in $K$ is $K[x]$ if and only if $S$ is finite. Does one have such a characterization on $S$ when $K$ is not commutative?

Thanks in advance.

joaopa
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    What do you mean by "is $K[x]$"? Maybe you mean that every continuous function is given by a polynomial? Note that this does not mean that $\mathscr{C}(S,K)$ is isomorphic to $K[x]$ though since the map from $K[x]$ may not be injective. – Eric Wofsey Dec 29 '18 at 01:18
  • Yes, I meant that every continuous function is given by a polynomial. – joaopa Dec 29 '18 at 01:26
  • Interesting question (when fixed as suggested above). Indeed, for either implication, I run into the problems discussed here: https://math.stackexchange.com/q/122898/96384. – Torsten Schoeneberg Jan 01 '19 at 18:13
  • May I ask what is the motivation for this? Precisely because of the problems discussed in the link above, standard polynomials (where the variable commutes with coefficients) are an unnatural object and a not very helpful tool over non-commutative rings, even division rings. Instead, something like skew polynomials might make a good replacement. – Torsten Schoeneberg Jan 02 '19 at 21:20
  • standard polynomials are not skew polynomials (for the identity endomorphism and the null derivation)? – joaopa Jan 03 '19 at 18:00
  • They are a special case which, as explained in that link, do not work well here. The basic problem is that evaluation does not commute with multiplication. But now I think that even more non-trivial skew polynomials will not repair that, because there the variable $x$ would have a fixed commutation relation, which is not true after evaluating $x$ at arbitrary elements ... rather, one would need an object like $D\langle x\rangle$ explained in Qiaochu Yuan's answer and Arturo Magidin's comment in there. But then it's totally unclear to me how to perform calculations in that algebra. – Torsten Schoeneberg Jan 03 '19 at 18:25

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