Isomorphism of quotient by prime ideal and quotient of localization The Next CEO of Stack OverflowCorrespondence between valuations and prime idealsFind primitive element such that conductor is relatively prime to an ideal (exercise from Neukirch)Atiyah - Macdonald Exericse 9.7 via LocalizationIdeals in a Dedekind domain localized at a prime idealfractional ideals in the localization of a DedekindA proof in Janusz Algebraic Number Fieldideal and ideal classes in the ring of integers.Non maximal prime ideals and localizationIntegral and prime ideal in Dedekind domainThere exist an integral ideal prime to a given nonzero integral ideal

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Isomorphism of quotient by prime ideal and quotient of localization



The Next CEO of Stack OverflowCorrespondence between valuations and prime idealsFind primitive element such that conductor is relatively prime to an ideal (exercise from Neukirch)Atiyah - Macdonald Exericse 9.7 via LocalizationIdeals in a Dedekind domain localized at a prime idealfractional ideals in the localization of a DedekindA proof in Janusz Algebraic Number Fieldideal and ideal classes in the ring of integers.Non maximal prime ideals and localizationIntegral and prime ideal in Dedekind domainThere exist an integral ideal prime to a given nonzero integral ideal










0












$begingroup$


Let $R$ be an integral domain.



$mathfrakp$ be a prime ideal of $R$.



Let $R_mathfrakp$ be localization of $R$ at $mathfrakp$.



Then an exercise in Algebraic Number Fields-Janusz asks




There is isomorphism between the fields $R/mathfrakp$ and $R_mathfrakp/mathfrakpR_mathfrakp$.




I think, $mathfrakp$ should be maximal, otherwise $R/mathfrakp$ is an integral domain and it always embeds in $R_mathfrakp/mathfrakpR_mathfrakp$; but isomorphism is not always possible. Am I right?



So question is simply that $mathfrakp$ should be maximal ideal to prove above assertion, is this right?










share|cite|improve this question









$endgroup$











  • $begingroup$
    Here you can find a generalization and a proof of the statement.
    $endgroup$
    – Fabio Lucchini
    yesterday















0












$begingroup$


Let $R$ be an integral domain.



$mathfrakp$ be a prime ideal of $R$.



Let $R_mathfrakp$ be localization of $R$ at $mathfrakp$.



Then an exercise in Algebraic Number Fields-Janusz asks




There is isomorphism between the fields $R/mathfrakp$ and $R_mathfrakp/mathfrakpR_mathfrakp$.




I think, $mathfrakp$ should be maximal, otherwise $R/mathfrakp$ is an integral domain and it always embeds in $R_mathfrakp/mathfrakpR_mathfrakp$; but isomorphism is not always possible. Am I right?



So question is simply that $mathfrakp$ should be maximal ideal to prove above assertion, is this right?










share|cite|improve this question









$endgroup$











  • $begingroup$
    Here you can find a generalization and a proof of the statement.
    $endgroup$
    – Fabio Lucchini
    yesterday













0












0








0





$begingroup$


Let $R$ be an integral domain.



$mathfrakp$ be a prime ideal of $R$.



Let $R_mathfrakp$ be localization of $R$ at $mathfrakp$.



Then an exercise in Algebraic Number Fields-Janusz asks




There is isomorphism between the fields $R/mathfrakp$ and $R_mathfrakp/mathfrakpR_mathfrakp$.




I think, $mathfrakp$ should be maximal, otherwise $R/mathfrakp$ is an integral domain and it always embeds in $R_mathfrakp/mathfrakpR_mathfrakp$; but isomorphism is not always possible. Am I right?



So question is simply that $mathfrakp$ should be maximal ideal to prove above assertion, is this right?










share|cite|improve this question









$endgroup$




Let $R$ be an integral domain.



$mathfrakp$ be a prime ideal of $R$.



Let $R_mathfrakp$ be localization of $R$ at $mathfrakp$.



Then an exercise in Algebraic Number Fields-Janusz asks




There is isomorphism between the fields $R/mathfrakp$ and $R_mathfrakp/mathfrakpR_mathfrakp$.




I think, $mathfrakp$ should be maximal, otherwise $R/mathfrakp$ is an integral domain and it always embeds in $R_mathfrakp/mathfrakpR_mathfrakp$; but isomorphism is not always possible. Am I right?



So question is simply that $mathfrakp$ should be maximal ideal to prove above assertion, is this right?







algebraic-number-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked yesterday









BeginnerBeginner

4,00611226




4,00611226











  • $begingroup$
    Here you can find a generalization and a proof of the statement.
    $endgroup$
    – Fabio Lucchini
    yesterday
















  • $begingroup$
    Here you can find a generalization and a proof of the statement.
    $endgroup$
    – Fabio Lucchini
    yesterday















$begingroup$
Here you can find a generalization and a proof of the statement.
$endgroup$
– Fabio Lucchini
yesterday




$begingroup$
Here you can find a generalization and a proof of the statement.
$endgroup$
– Fabio Lucchini
yesterday










0






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