Analog of finite set in valued complete division ring
1
$begingroup$
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
asked Dec 29 '18 at 1:11
joaopajoaopa
35618
$endgroup$
|
show 1 more comment
1
$begingroup$
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
asked Dec 29 '18 at 1:11
joaopajoaopa
35618
$endgroup$
1
$begingroup$
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.
$endgroup$
– Eric Wofsey
Dec 29 '18 at 1:18
$begingroup$
Yes, I meant that every continuous function is given by a polynomial.
$endgroup$
– joaopa
Dec 29 '18 at 1:26
$begingroup$
Interesting question (when fixed as suggested above). Indeed, for either implication, I run into the problems discussed here: math.stackexchange.com/q/122898/96384.
$endgroup$
– Torsten Schoeneberg
Jan 1 at 18:13
$begingroup$
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.
$endgroup$
– Torsten Schoeneberg
Jan 2 at 21:20
$begingroup$
standard polynomials are not skew polynomials (for the identity endomorphism and the null derivation)?
$endgroup$
– joaopa
Jan 3 at 18:00
|
show 1 more comment
1
1
1
1
$begingroup$
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
asked Dec 29 '18 at 1:11
joaopajoaopa
35618
$endgroup$
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
analysis p-adic-number-theory noncommutative-geometry
asked Dec 29 '18 at 1:11
joaopajoaopa
35618
asked Dec 29 '18 at 1:11
joaopajoaopa
35618
asked Dec 29 '18 at 1:11
joaopajoaopa
35618
asked Dec 29 '18 at 1:11
joaopajoaopa
35618
asked Dec 29 '18 at 1:11
joaopajoaopa
35618
35618
1
$begingroup$
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.
$endgroup$
– Eric Wofsey
Dec 29 '18 at 1:18
$begingroup$
Yes, I meant that every continuous function is given by a polynomial.
$endgroup$
– joaopa
Dec 29 '18 at 1:26
$begingroup$
Interesting question (when fixed as suggested above). Indeed, for either implication, I run into the problems discussed here: math.stackexchange.com/q/122898/96384.
$endgroup$
– Torsten Schoeneberg
Jan 1 at 18:13
$begingroup$
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.
$endgroup$
– Torsten Schoeneberg
Jan 2 at 21:20
$begingroup$
standard polynomials are not skew polynomials (for the identity endomorphism and the null derivation)?
$endgroup$
– joaopa
Jan 3 at 18:00
|
show 1 more comment
1
$begingroup$
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.
$endgroup$
– Eric Wofsey
Dec 29 '18 at 1:18
$begingroup$
Yes, I meant that every continuous function is given by a polynomial.
$endgroup$
– joaopa
Dec 29 '18 at 1:26
$begingroup$
Interesting question (when fixed as suggested above). Indeed, for either implication, I run into the problems discussed here: math.stackexchange.com/q/122898/96384.
$endgroup$
– Torsten Schoeneberg
Jan 1 at 18:13
$begingroup$
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.
$endgroup$
– Torsten Schoeneberg
Jan 2 at 21:20
$begingroup$
standard polynomials are not skew polynomials (for the identity endomorphism and the null derivation)?
$endgroup$
– joaopa
Jan 3 at 18:00
1
1
$begingroup$
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.
$endgroup$
– Eric Wofsey
Dec 29 '18 at 1:18
$begingroup$
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.
$endgroup$
– Eric Wofsey
Dec 29 '18 at 1:18
$begingroup$
Yes, I meant that every continuous function is given by a polynomial.
$endgroup$
– joaopa
Dec 29 '18 at 1:26
$begingroup$
Yes, I meant that every continuous function is given by a polynomial.
$endgroup$
– joaopa
Dec 29 '18 at 1:26
$begingroup$
Interesting question (when fixed as suggested above). Indeed, for either implication, I run into the problems discussed here: math.stackexchange.com/q/122898/96384.
$endgroup$
– Torsten Schoeneberg
Jan 1 at 18:13
$begingroup$
Interesting question (when fixed as suggested above). Indeed, for either implication, I run into the problems discussed here: math.stackexchange.com/q/122898/96384.
$endgroup$
– Torsten Schoeneberg
Jan 1 at 18:13
$begingroup$
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.
$endgroup$
– Torsten Schoeneberg
Jan 2 at 21:20
$begingroup$
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.
$endgroup$
– Torsten Schoeneberg
Jan 2 at 21:20
$begingroup$
standard polynomials are not skew polynomials (for the identity endomorphism and the null derivation)?
$endgroup$
– joaopa
Jan 3 at 18:00
$begingroup$
standard polynomials are not skew polynomials (for the identity endomorphism and the null derivation)?
$endgroup$
– joaopa
Jan 3 at 18:00
|
show 1 more comment
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1
$begingroup$
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.
$endgroup$
– Eric Wofsey
Dec 29 '18 at 1:18
$begingroup$
Yes, I meant that every continuous function is given by a polynomial.
$endgroup$
– joaopa
Dec 29 '18 at 1:26
$begingroup$
Interesting question (when fixed as suggested above). Indeed, for either implication, I run into the problems discussed here: math.stackexchange.com/q/122898/96384.
$endgroup$
– Torsten Schoeneberg
Jan 1 at 18:13
$begingroup$
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.
$endgroup$
– Torsten Schoeneberg
Jan 2 at 21:20
$begingroup$
standard polynomials are not skew polynomials (for the identity endomorphism and the null derivation)?
$endgroup$
– joaopa
Jan 3 at 18:00