Is $operatorname{Frac}(bigcap_{i in I}R_i)=K$ when each $R_i$ is integrally closed and...
$begingroup$
Let $K$ be a field. Let $left { R_iright }_{i in I} $ be a set of integrally closed domains
in $K$ whose field of fractions equals $K$.
It is easy to show that $R:=bigcap_{i in I}R_i$ is an integrally closed ring in its field of fractions. I was wondering if the field of fractions of $R$ is always $K$.
abstract-algebra algebraic-number-theory
edited Dec 2 '18 at 23:28
Bernard
119k740113
asked Dec 2 '18 at 23:25
CorneliusCornelius
1957
$endgroup$
add a comment |
$begingroup$
Let $K$ be a field. Let $left { R_iright }_{i in I} $ be a set of integrally closed domains
in $K$ whose field of fractions equals $K$.
It is easy to show that $R:=bigcap_{i in I}R_i$ is an integrally closed ring in its field of fractions. I was wondering if the field of fractions of $R$ is always $K$.
abstract-algebra algebraic-number-theory
edited Dec 2 '18 at 23:28
Bernard
119k740113
asked Dec 2 '18 at 23:25
CorneliusCornelius
1957
$endgroup$
add a comment |
$begingroup$
Let $K$ be a field. Let $left { R_iright }_{i in I} $ be a set of integrally closed domains
in $K$ whose field of fractions equals $K$.
It is easy to show that $R:=bigcap_{i in I}R_i$ is an integrally closed ring in its field of fractions. I was wondering if the field of fractions of $R$ is always $K$.
abstract-algebra algebraic-number-theory
edited Dec 2 '18 at 23:28
Bernard
119k740113
asked Dec 2 '18 at 23:25
CorneliusCornelius
1957
$endgroup$
Let $K$ be a field. Let $left { R_iright }_{i in I} $ be a set of integrally closed domains
in $K$ whose field of fractions equals $K$.
It is easy to show that $R:=bigcap_{i in I}R_i$ is an integrally closed ring in its field of fractions. I was wondering if the field of fractions of $R$ is always $K$.
abstract-algebra algebraic-number-theory
abstract-algebra algebraic-number-theory
edited Dec 2 '18 at 23:28
Bernard
119k740113
asked Dec 2 '18 at 23:25
CorneliusCornelius
1957
edited Dec 2 '18 at 23:28
Bernard
119k740113
asked Dec 2 '18 at 23:25
CorneliusCornelius
1957
edited Dec 2 '18 at 23:28
Bernard
119k740113
edited Dec 2 '18 at 23:28
Bernard
119k740113
edited Dec 2 '18 at 23:28
Bernard
119k740113
119k740113
asked Dec 2 '18 at 23:25
CorneliusCornelius
1957
asked Dec 2 '18 at 23:25
CorneliusCornelius
1957
asked Dec 2 '18 at 23:25
CorneliusCornelius
1957
1957
add a comment |
add a comment |
1 Answer
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$begingroup$
It is not true in general. Here is a counter-example.
Consider the domains $R_p:=mathbb Z[px]$, where $p$ is a prime number. Observe that the field of fractions of each $R_p$ is equal to $mathbb Q(x)$. Since each $R_p$ is isomorphic to $mathbb Z[x]$ then it is a UFD and hence integrally closed in its field of fractions. Then, observe that
begin{equation}
R:=bigcap_{p: prime} R_p=mathbb Z
end{equation}
where the field of fractions of $mathbb Z$ is $mathbb Q$.
answered Dec 2 '18 at 23:37
richarddedekindricharddedekind
711316
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add a comment |
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1 Answer
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$begingroup$
It is not true in general. Here is a counter-example.
Consider the domains $R_p:=mathbb Z[px]$, where $p$ is a prime number. Observe that the field of fractions of each $R_p$ is equal to $mathbb Q(x)$. Since each $R_p$ is isomorphic to $mathbb Z[x]$ then it is a UFD and hence integrally closed in its field of fractions. Then, observe that
begin{equation}
R:=bigcap_{p: prime} R_p=mathbb Z
end{equation}
where the field of fractions of $mathbb Z$ is $mathbb Q$.
answered Dec 2 '18 at 23:37
richarddedekindricharddedekind
711316
$endgroup$
add a comment |
$begingroup$
It is not true in general. Here is a counter-example.
Consider the domains $R_p:=mathbb Z[px]$, where $p$ is a prime number. Observe that the field of fractions of each $R_p$ is equal to $mathbb Q(x)$. Since each $R_p$ is isomorphic to $mathbb Z[x]$ then it is a UFD and hence integrally closed in its field of fractions. Then, observe that
begin{equation}
R:=bigcap_{p: prime} R_p=mathbb Z
end{equation}
where the field of fractions of $mathbb Z$ is $mathbb Q$.
answered Dec 2 '18 at 23:37
richarddedekindricharddedekind
711316
$endgroup$
add a comment |
$begingroup$
It is not true in general. Here is a counter-example.
Consider the domains $R_p:=mathbb Z[px]$, where $p$ is a prime number. Observe that the field of fractions of each $R_p$ is equal to $mathbb Q(x)$. Since each $R_p$ is isomorphic to $mathbb Z[x]$ then it is a UFD and hence integrally closed in its field of fractions. Then, observe that
begin{equation}
R:=bigcap_{p: prime} R_p=mathbb Z
end{equation}
where the field of fractions of $mathbb Z$ is $mathbb Q$.
answered Dec 2 '18 at 23:37
richarddedekindricharddedekind
711316
$endgroup$
It is not true in general. Here is a counter-example.
Consider the domains $R_p:=mathbb Z[px]$, where $p$ is a prime number. Observe that the field of fractions of each $R_p$ is equal to $mathbb Q(x)$. Since each $R_p$ is isomorphic to $mathbb Z[x]$ then it is a UFD and hence integrally closed in its field of fractions. Then, observe that
begin{equation}
R:=bigcap_{p: prime} R_p=mathbb Z
end{equation}
where the field of fractions of $mathbb Z$ is $mathbb Q$.
answered Dec 2 '18 at 23:37
richarddedekindricharddedekind
711316
answered Dec 2 '18 at 23:37
richarddedekindricharddedekind
711316
answered Dec 2 '18 at 23:37
richarddedekindricharddedekind
711316
answered Dec 2 '18 at 23:37
richarddedekindricharddedekind
711316
711316
add a comment |
add a comment |
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