Analog of finite set in valued complete division ring












1














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.










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  • 1




    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 1:18










  • Yes, I meant that every continuous function is given by a polynomial.
    – joaopa
    Dec 29 '18 at 1:26












  • Interesting question (when fixed as suggested above). Indeed, for either implication, I run into the problems discussed here: math.stackexchange.com/q/122898/96384.
    – Torsten Schoeneberg
    Jan 1 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 2 at 21:20










  • standard polynomials are not skew polynomials (for the identity endomorphism and the null derivation)?
    – joaopa
    Jan 3 at 18:00
















1














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.










share|cite|improve this question


















  • 1




    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 1:18










  • Yes, I meant that every continuous function is given by a polynomial.
    – joaopa
    Dec 29 '18 at 1:26












  • Interesting question (when fixed as suggested above). Indeed, for either implication, I run into the problems discussed here: math.stackexchange.com/q/122898/96384.
    – Torsten Schoeneberg
    Jan 1 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 2 at 21:20










  • standard polynomials are not skew polynomials (for the identity endomorphism and the null derivation)?
    – joaopa
    Jan 3 at 18:00














1












1








1


1





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.










share|cite|improve this question













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.







analysis p-adic-number-theory noncommutative-geometry






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 29 '18 at 1:11









joaopajoaopa

34418




34418








  • 1




    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 1:18










  • Yes, I meant that every continuous function is given by a polynomial.
    – joaopa
    Dec 29 '18 at 1:26












  • Interesting question (when fixed as suggested above). Indeed, for either implication, I run into the problems discussed here: math.stackexchange.com/q/122898/96384.
    – Torsten Schoeneberg
    Jan 1 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 2 at 21:20










  • standard polynomials are not skew polynomials (for the identity endomorphism and the null derivation)?
    – joaopa
    Jan 3 at 18:00














  • 1




    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 1:18










  • Yes, I meant that every continuous function is given by a polynomial.
    – joaopa
    Dec 29 '18 at 1:26












  • Interesting question (when fixed as suggested above). Indeed, for either implication, I run into the problems discussed here: math.stackexchange.com/q/122898/96384.
    – Torsten Schoeneberg
    Jan 1 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 2 at 21:20










  • standard polynomials are not skew polynomials (for the identity endomorphism and the null derivation)?
    – joaopa
    Jan 3 at 18:00








1




1




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 1:18




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 1:18












Yes, I meant that every continuous function is given by a polynomial.
– joaopa
Dec 29 '18 at 1:26






Yes, I meant that every continuous function is given by a polynomial.
– joaopa
Dec 29 '18 at 1:26














Interesting question (when fixed as suggested above). Indeed, for either implication, I run into the problems discussed here: math.stackexchange.com/q/122898/96384.
– Torsten Schoeneberg
Jan 1 at 18:13




Interesting question (when fixed as suggested above). Indeed, for either implication, I run into the problems discussed here: math.stackexchange.com/q/122898/96384.
– Torsten Schoeneberg
Jan 1 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 2 at 21:20




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 2 at 21:20












standard polynomials are not skew polynomials (for the identity endomorphism and the null derivation)?
– joaopa
Jan 3 at 18:00




standard polynomials are not skew polynomials (for the identity endomorphism and the null derivation)?
– joaopa
Jan 3 at 18:00










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