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If $Kcong K(X)$ then must $K$ be a field of rational functions in infinitely many variables?
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If $k$ is any field, then the field $K=k(X_0,X_1,dots)$ of rational functions in infinitely many variables satisfies $K(X)cong K$ (by mapping $X$ to $X_0$ and $X_n$ to $X_n+1$). My question is, does the converse hold? That is:
Suppose $K$ is a field such that $Kcong K(X)$. Must there exist a field $k$ such that $Kcong k(X_0,X_1,dots)$?
(If such a $k$ exists, then we can in fact take $k=K$, by splitting the variables into two infinite sets.)
Note that if we were talking about polynomial rings instead of fields of rational functions, the answer would be no: there exists an integral domain $R$ such that $Rcong R[X]$ but $Rnotcong S[X_0,X_1,dots]$ for any ring $S$. A counterexample is given at A ring isomorphic to its finite polynomial rings but not to its infinite one. However, for that example the field of fractions of $R$ actually is a field of rational functions in infinitely many variables, so it does not give a counterexample to this question.
In any case, I suspect the answer is no and it may be possible to find a counterexample using some idea similar to the example there (some sort of "all but finitely many..." construction), but don't have any concrete idea of how to make it work.
Some other (unanswered) questions that may be related: Is there a field $F$ which is isomorphic to $F(X,Y)$ but not to $F(X)$? (also on MO), The field of fractions of the rational group algebra of a torsion free abelian group. (In particular, an answer to the former question would give a field $F$ such that $Fcong F(X,Y)$ but $F$ is not isomorphic to a field of rational functions in infinitely many variables, which is very close to this question.)
abstract-algebra field-theory
asked Jul 18 at 22:05
Eric Wofsey
162k12189300
add a comment |Â
up vote
12
down vote
favorite
If $k$ is any field, then the field $K=k(X_0,X_1,dots)$ of rational functions in infinitely many variables satisfies $K(X)cong K$ (by mapping $X$ to $X_0$ and $X_n$ to $X_n+1$). My question is, does the converse hold? That is:
Suppose $K$ is a field such that $Kcong K(X)$. Must there exist a field $k$ such that $Kcong k(X_0,X_1,dots)$?
(If such a $k$ exists, then we can in fact take $k=K$, by splitting the variables into two infinite sets.)
Note that if we were talking about polynomial rings instead of fields of rational functions, the answer would be no: there exists an integral domain $R$ such that $Rcong R[X]$ but $Rnotcong S[X_0,X_1,dots]$ for any ring $S$. A counterexample is given at A ring isomorphic to its finite polynomial rings but not to its infinite one. However, for that example the field of fractions of $R$ actually is a field of rational functions in infinitely many variables, so it does not give a counterexample to this question.
In any case, I suspect the answer is no and it may be possible to find a counterexample using some idea similar to the example there (some sort of "all but finitely many..." construction), but don't have any concrete idea of how to make it work.
Some other (unanswered) questions that may be related: Is there a field $F$ which is isomorphic to $F(X,Y)$ but not to $F(X)$? (also on MO), The field of fractions of the rational group algebra of a torsion free abelian group. (In particular, an answer to the former question would give a field $F$ such that $Fcong F(X,Y)$ but $F$ is not isomorphic to a field of rational functions in infinitely many variables, which is very close to this question.)
abstract-algebra field-theory
asked Jul 18 at 22:05
Eric Wofsey
162k12189300
Might it be worth asking this interesting question on MathOverflow?
– Watson
Jul 23 at 17:56
Sure. Given that your related question hasn't gotten an answer there I would be surprised if this one got an answer there quickly, though. Feel free to crosspost it if you want.
– Eric Wofsey
Jul 23 at 18:10
add a comment |Â
up vote
12
down vote
favorite
up vote
12
down vote
favorite
If $k$ is any field, then the field $K=k(X_0,X_1,dots)$ of rational functions in infinitely many variables satisfies $K(X)cong K$ (by mapping $X$ to $X_0$ and $X_n$ to $X_n+1$). My question is, does the converse hold? That is:
Suppose $K$ is a field such that $Kcong K(X)$. Must there exist a field $k$ such that $Kcong k(X_0,X_1,dots)$?
(If such a $k$ exists, then we can in fact take $k=K$, by splitting the variables into two infinite sets.)
Note that if we were talking about polynomial rings instead of fields of rational functions, the answer would be no: there exists an integral domain $R$ such that $Rcong R[X]$ but $Rnotcong S[X_0,X_1,dots]$ for any ring $S$. A counterexample is given at A ring isomorphic to its finite polynomial rings but not to its infinite one. However, for that example the field of fractions of $R$ actually is a field of rational functions in infinitely many variables, so it does not give a counterexample to this question.
In any case, I suspect the answer is no and it may be possible to find a counterexample using some idea similar to the example there (some sort of "all but finitely many..." construction), but don't have any concrete idea of how to make it work.
Some other (unanswered) questions that may be related: Is there a field $F$ which is isomorphic to $F(X,Y)$ but not to $F(X)$? (also on MO), The field of fractions of the rational group algebra of a torsion free abelian group. (In particular, an answer to the former question would give a field $F$ such that $Fcong F(X,Y)$ but $F$ is not isomorphic to a field of rational functions in infinitely many variables, which is very close to this question.)
abstract-algebra field-theory
asked Jul 18 at 22:05
Eric Wofsey
162k12189300
If $k$ is any field, then the field $K=k(X_0,X_1,dots)$ of rational functions in infinitely many variables satisfies $K(X)cong K$ (by mapping $X$ to $X_0$ and $X_n$ to $X_n+1$). My question is, does the converse hold? That is:
Suppose $K$ is a field such that $Kcong K(X)$. Must there exist a field $k$ such that $Kcong k(X_0,X_1,dots)$?
(If such a $k$ exists, then we can in fact take $k=K$, by splitting the variables into two infinite sets.)
Note that if we were talking about polynomial rings instead of fields of rational functions, the answer would be no: there exists an integral domain $R$ such that $Rcong R[X]$ but $Rnotcong S[X_0,X_1,dots]$ for any ring $S$. A counterexample is given at A ring isomorphic to its finite polynomial rings but not to its infinite one. However, for that example the field of fractions of $R$ actually is a field of rational functions in infinitely many variables, so it does not give a counterexample to this question.
In any case, I suspect the answer is no and it may be possible to find a counterexample using some idea similar to the example there (some sort of "all but finitely many..." construction), but don't have any concrete idea of how to make it work.
Some other (unanswered) questions that may be related: Is there a field $F$ which is isomorphic to $F(X,Y)$ but not to $F(X)$? (also on MO), The field of fractions of the rational group algebra of a torsion free abelian group. (In particular, an answer to the former question would give a field $F$ such that $Fcong F(X,Y)$ but $F$ is not isomorphic to a field of rational functions in infinitely many variables, which is very close to this question.)
abstract-algebra field-theory
asked Jul 18 at 22:05
Eric Wofsey
162k12189300
asked Jul 18 at 22:05
Eric Wofsey
162k12189300
asked Jul 18 at 22:05
Eric Wofsey
162k12189300
asked Jul 18 at 22:05
Eric Wofsey
162k12189300
162k12189300
Might it be worth asking this interesting question on MathOverflow?
– Watson
Jul 23 at 17:56
Sure. Given that your related question hasn't gotten an answer there I would be surprised if this one got an answer there quickly, though. Feel free to crosspost it if you want.
– Eric Wofsey
Jul 23 at 18:10
add a comment |Â
Might it be worth asking this interesting question on MathOverflow?
– Watson
Jul 23 at 17:56
Sure. Given that your related question hasn't gotten an answer there I would be surprised if this one got an answer there quickly, though. Feel free to crosspost it if you want.
– Eric Wofsey
Jul 23 at 18:10
Might it be worth asking this interesting question on MathOverflow?
– Watson
Jul 23 at 17:56
Might it be worth asking this interesting question on MathOverflow?
– Watson
Jul 23 at 17:56
Sure. Given that your related question hasn't gotten an answer there I would be surprised if this one got an answer there quickly, though. Feel free to crosspost it if you want.
– Eric Wofsey
Jul 23 at 18:10
Sure. Given that your related question hasn't gotten an answer there I would be surprised if this one got an answer there quickly, though. Feel free to crosspost it if you want.
– Eric Wofsey
Jul 23 at 18:10
add a comment |Â
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Might it be worth asking this interesting question on MathOverflow?
– Watson
Jul 23 at 17:56
Sure. Given that your related question hasn't gotten an answer there I would be surprised if this one got an answer there quickly, though. Feel free to crosspost it if you want.
– Eric Wofsey
Jul 23 at 18:10