Mixing
On integral closedness of formal power series ring over an integrally closed domain satisfying Krull...
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Let $R$ be a normal domain (i.e. an integral domain integrally closed in its fraction field) such that for every non-unit $tin R$, $cap_{nge 1} (t^n)=(0)$ ; then is it true that $R[[X]]$ is normal (i.e. integrally closed in its fraction field) ?
If this is not true in general, what if we strengthen the hypothesis to say $cap_{nge 1} J^n=(0)$ for every proper ideal $J$ of $R$ ?
ring-theory commutative-algebra ideals formal-power-series integral-extensions
asked Jan 24 at 22:27
user521337user521337
1,1981417
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add a comment |
$begingroup$
Let $R$ be a normal domain (i.e. an integral domain integrally closed in its fraction field) such that for every non-unit $tin R$, $cap_{nge 1} (t^n)=(0)$ ; then is it true that $R[[X]]$ is normal (i.e. integrally closed in its fraction field) ?
If this is not true in general, what if we strengthen the hypothesis to say $cap_{nge 1} J^n=(0)$ for every proper ideal $J$ of $R$ ?
ring-theory commutative-algebra ideals formal-power-series integral-extensions
asked Jan 24 at 22:27
user521337user521337
1,1981417
$endgroup$
$begingroup$
If $R[[X]]$ is integrally closed, then $R$ is integrally closed and $cap_{nge 1} (t^n)=(0)$ for every non-unit $tin R$, but the converse doesn't hold. See here.
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– user26857
Jan 25 at 10:33
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@user26857: thanks, I will take a look into that ... and what if we assume the stronger condition that $cap_{nge 1} J^n=(0)$ for every proper ideal $J$ ?
$endgroup$
– user521337
Jan 25 at 21:47
add a comment |
$begingroup$
Let $R$ be a normal domain (i.e. an integral domain integrally closed in its fraction field) such that for every non-unit $tin R$, $cap_{nge 1} (t^n)=(0)$ ; then is it true that $R[[X]]$ is normal (i.e. integrally closed in its fraction field) ?
If this is not true in general, what if we strengthen the hypothesis to say $cap_{nge 1} J^n=(0)$ for every proper ideal $J$ of $R$ ?
ring-theory commutative-algebra ideals formal-power-series integral-extensions
asked Jan 24 at 22:27
user521337user521337
1,1981417
$endgroup$
Let $R$ be a normal domain (i.e. an integral domain integrally closed in its fraction field) such that for every non-unit $tin R$, $cap_{nge 1} (t^n)=(0)$ ; then is it true that $R[[X]]$ is normal (i.e. integrally closed in its fraction field) ?
If this is not true in general, what if we strengthen the hypothesis to say $cap_{nge 1} J^n=(0)$ for every proper ideal $J$ of $R$ ?
ring-theory commutative-algebra ideals formal-power-series integral-extensions
ring-theory commutative-algebra ideals formal-power-series integral-extensions
asked Jan 24 at 22:27
user521337user521337
1,1981417
asked Jan 24 at 22:27
user521337user521337
1,1981417
asked Jan 24 at 22:27
user521337user521337
1,1981417
asked Jan 24 at 22:27
user521337user521337
1,1981417
asked Jan 24 at 22:27
user521337user521337
1,1981417
1,1981417
$begingroup$
If $R[[X]]$ is integrally closed, then $R$ is integrally closed and $cap_{nge 1} (t^n)=(0)$ for every non-unit $tin R$, but the converse doesn't hold. See here.
$endgroup$
– user26857
Jan 25 at 10:33
$begingroup$
@user26857: thanks, I will take a look into that ... and what if we assume the stronger condition that $cap_{nge 1} J^n=(0)$ for every proper ideal $J$ ?
$endgroup$
– user521337
Jan 25 at 21:47
add a comment |
$begingroup$
If $R[[X]]$ is integrally closed, then $R$ is integrally closed and $cap_{nge 1} (t^n)=(0)$ for every non-unit $tin R$, but the converse doesn't hold. See here.
$endgroup$
– user26857
Jan 25 at 10:33
$begingroup$
@user26857: thanks, I will take a look into that ... and what if we assume the stronger condition that $cap_{nge 1} J^n=(0)$ for every proper ideal $J$ ?
$endgroup$
– user521337
Jan 25 at 21:47
$begingroup$
If $R[[X]]$ is integrally closed, then $R$ is integrally closed and $cap_{nge 1} (t^n)=(0)$ for every non-unit $tin R$, but the converse doesn't hold. See here.
$endgroup$
– user26857
Jan 25 at 10:33
$begingroup$
If $R[[X]]$ is integrally closed, then $R$ is integrally closed and $cap_{nge 1} (t^n)=(0)$ for every non-unit $tin R$, but the converse doesn't hold. See here.
$endgroup$
– user26857
Jan 25 at 10:33
$begingroup$
@user26857: thanks, I will take a look into that ... and what if we assume the stronger condition that $cap_{nge 1} J^n=(0)$ for every proper ideal $J$ ?
$endgroup$
– user521337
Jan 25 at 21:47
$begingroup$
@user26857: thanks, I will take a look into that ... and what if we assume the stronger condition that $cap_{nge 1} J^n=(0)$ for every proper ideal $J$ ?
$endgroup$
– user521337
Jan 25 at 21:47
add a comment |
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$begingroup$
If $R[[X]]$ is integrally closed, then $R$ is integrally closed and $cap_{nge 1} (t^n)=(0)$ for every non-unit $tin R$, but the converse doesn't hold. See here.
$endgroup$
– user26857
Jan 25 at 10:33
$begingroup$
@user26857: thanks, I will take a look into that ... and what if we assume the stronger condition that $cap_{nge 1} J^n=(0)$ for every proper ideal $J$ ?
$endgroup$
– user521337
Jan 25 at 21:47