$R_p$ is a field iff for any $x in p$ there exists $y notin p$ such that $xy = 0$
$begingroup$
Show that the localization at p, prime ideal, $R_p$ is a field iff for any $x in p$ there exists $y notin p$ such that $xy = 0$
I know there is a similar question where R is an Noetherian ring, but this question does not have any conditions like that. I know there still is a one to one correspondence with the prime ideal in $R_p$ and the prime ideal in $R$ contained in $p$. So the only valid prime ideal is $pR_p$? And also the nilrad(R_p) is the intersection of all prime ideal, so nilrad(R_p) = $pR_p$. But I don't know how to put this all together and show the both if and only if direction.
abstract-algebra ring-theory maximal-and-prime-ideals localization
asked Dec 30 '18 at 8:32
ijk2438ijk2438
182
$endgroup$
add a comment |
$begingroup$
Show that the localization at p, prime ideal, $R_p$ is a field iff for any $x in p$ there exists $y notin p$ such that $xy = 0$
I know there is a similar question where R is an Noetherian ring, but this question does not have any conditions like that. I know there still is a one to one correspondence with the prime ideal in $R_p$ and the prime ideal in $R$ contained in $p$. So the only valid prime ideal is $pR_p$? And also the nilrad(R_p) is the intersection of all prime ideal, so nilrad(R_p) = $pR_p$. But I don't know how to put this all together and show the both if and only if direction.
abstract-algebra ring-theory maximal-and-prime-ideals localization
asked Dec 30 '18 at 8:32
ijk2438ijk2438
182
$endgroup$
$begingroup$
It's nothing as complicated as that: Elliot G's answer contains the seed for showing that $pR_p$ is the zero ideal of $R_p$, so that $R_p$ is a field, but it needs a little polishing.
$endgroup$
– rschwieb
Dec 30 '18 at 14:35
add a comment |
$begingroup$
Show that the localization at p, prime ideal, $R_p$ is a field iff for any $x in p$ there exists $y notin p$ such that $xy = 0$
I know there is a similar question where R is an Noetherian ring, but this question does not have any conditions like that. I know there still is a one to one correspondence with the prime ideal in $R_p$ and the prime ideal in $R$ contained in $p$. So the only valid prime ideal is $pR_p$? And also the nilrad(R_p) is the intersection of all prime ideal, so nilrad(R_p) = $pR_p$. But I don't know how to put this all together and show the both if and only if direction.
abstract-algebra ring-theory maximal-and-prime-ideals localization
asked Dec 30 '18 at 8:32
ijk2438ijk2438
182
$endgroup$
Show that the localization at p, prime ideal, $R_p$ is a field iff for any $x in p$ there exists $y notin p$ such that $xy = 0$
I know there is a similar question where R is an Noetherian ring, but this question does not have any conditions like that. I know there still is a one to one correspondence with the prime ideal in $R_p$ and the prime ideal in $R$ contained in $p$. So the only valid prime ideal is $pR_p$? And also the nilrad(R_p) is the intersection of all prime ideal, so nilrad(R_p) = $pR_p$. But I don't know how to put this all together and show the both if and only if direction.
abstract-algebra ring-theory maximal-and-prime-ideals localization
abstract-algebra ring-theory maximal-and-prime-ideals localization
asked Dec 30 '18 at 8:32
ijk2438ijk2438
182
asked Dec 30 '18 at 8:32
ijk2438ijk2438
182
asked Dec 30 '18 at 8:32
ijk2438ijk2438
182
asked Dec 30 '18 at 8:32
ijk2438ijk2438
182
asked Dec 30 '18 at 8:32
ijk2438ijk2438
182
182
$begingroup$
It's nothing as complicated as that: Elliot G's answer contains the seed for showing that $pR_p$ is the zero ideal of $R_p$, so that $R_p$ is a field, but it needs a little polishing.
$endgroup$
– rschwieb
Dec 30 '18 at 14:35
add a comment |
$begingroup$
It's nothing as complicated as that: Elliot G's answer contains the seed for showing that $pR_p$ is the zero ideal of $R_p$, so that $R_p$ is a field, but it needs a little polishing.
$endgroup$
– rschwieb
Dec 30 '18 at 14:35
$begingroup$
It's nothing as complicated as that: Elliot G's answer contains the seed for showing that $pR_p$ is the zero ideal of $R_p$, so that $R_p$ is a field, but it needs a little polishing.
$endgroup$
– rschwieb
Dec 30 '18 at 14:35
$begingroup$
It's nothing as complicated as that: Elliot G's answer contains the seed for showing that $pR_p$ is the zero ideal of $R_p$, so that $R_p$ is a field, but it needs a little polishing.
$endgroup$
– rschwieb
Dec 30 '18 at 14:35
add a comment |
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$begingroup$
It's nothing as complicated as that: Elliot G's answer contains the seed for showing that $pR_p$ is the zero ideal of $R_p$, so that $R_p$ is a field, but it needs a little polishing.
$endgroup$
– rschwieb
Dec 30 '18 at 14:35