Is there an isomorphism between $ text{Spec}(R_{mathfrak{p}}) $ and the prime ideals of $ R $ which are...












1












$begingroup$


Suppose $ R $ is a ring, and $ mathfrak{p} in text{Spec}(R). $



I have been told that $ text{Spec}(R_{mathfrak{p}}) cong lbrace mathfrak{q} in text{Spec}(R);| mathfrak{q} subset mathfrak{p} rbrace, $ and I am trying to figure out whether this is true or not, but I am stuck trying to show that there is an injective map $ alpha : text{Spec}(R_{mathfrak{p}}) longrightarrow lbrace mathfrak{q} in text{Spec}(R);| mathfrak{q} subset mathfrak{p} rbrace.$



I suspect that there is a way to show this by exploiting the fact that any (prime) ideal of $ text{Spec}(R_{p}) $ cannot contain units of $ R_{mathfrak{p}} $, but I can't find a way through.



Any assistance would be much appreciated.










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    1












    $begingroup$


    Suppose $ R $ is a ring, and $ mathfrak{p} in text{Spec}(R). $



    I have been told that $ text{Spec}(R_{mathfrak{p}}) cong lbrace mathfrak{q} in text{Spec}(R);| mathfrak{q} subset mathfrak{p} rbrace, $ and I am trying to figure out whether this is true or not, but I am stuck trying to show that there is an injective map $ alpha : text{Spec}(R_{mathfrak{p}}) longrightarrow lbrace mathfrak{q} in text{Spec}(R);| mathfrak{q} subset mathfrak{p} rbrace.$



    I suspect that there is a way to show this by exploiting the fact that any (prime) ideal of $ text{Spec}(R_{p}) $ cannot contain units of $ R_{mathfrak{p}} $, but I can't find a way through.



    Any assistance would be much appreciated.










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Suppose $ R $ is a ring, and $ mathfrak{p} in text{Spec}(R). $



      I have been told that $ text{Spec}(R_{mathfrak{p}}) cong lbrace mathfrak{q} in text{Spec}(R);| mathfrak{q} subset mathfrak{p} rbrace, $ and I am trying to figure out whether this is true or not, but I am stuck trying to show that there is an injective map $ alpha : text{Spec}(R_{mathfrak{p}}) longrightarrow lbrace mathfrak{q} in text{Spec}(R);| mathfrak{q} subset mathfrak{p} rbrace.$



      I suspect that there is a way to show this by exploiting the fact that any (prime) ideal of $ text{Spec}(R_{p}) $ cannot contain units of $ R_{mathfrak{p}} $, but I can't find a way through.



      Any assistance would be much appreciated.










      share|cite|improve this question









      $endgroup$




      Suppose $ R $ is a ring, and $ mathfrak{p} in text{Spec}(R). $



      I have been told that $ text{Spec}(R_{mathfrak{p}}) cong lbrace mathfrak{q} in text{Spec}(R);| mathfrak{q} subset mathfrak{p} rbrace, $ and I am trying to figure out whether this is true or not, but I am stuck trying to show that there is an injective map $ alpha : text{Spec}(R_{mathfrak{p}}) longrightarrow lbrace mathfrak{q} in text{Spec}(R);| mathfrak{q} subset mathfrak{p} rbrace.$



      I suspect that there is a way to show this by exploiting the fact that any (prime) ideal of $ text{Spec}(R_{p}) $ cannot contain units of $ R_{mathfrak{p}} $, but I can't find a way through.



      Any assistance would be much appreciated.







      commutative-algebra






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      asked Jan 11 at 2:53









      Overwhelmed AG ApprenticeOverwhelmed AG Apprentice

      458




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          $begingroup$

          In a word, yes.



          In general, for a multiplicatively closed subset $S$ of $R$,
          there is a natural correspondence between the prime ideals of
          $S^{-1}R$ and the prime ideals of $R$ disjoint from $S$. Your
          example is the case where $S=R-newcommand{fp}{mathfrak{p}}
          newcommand{fq}{mathfrak{q}}fp$
          .



          The prime ideals of $S^{-1}R$ all have the form $S^{-1}fq$
          where $fqsubseteq fp$ is a prime ideal of $S$. It is straightforward
          to check that these $S^{-1}fq$ are prime in $S^{-1}R$. Conversely,
          let $Q$ be a prime ideal of $S^{-1}R$. Then
          $fq={ain R:a/1in Q}$ is a prime ideal of $R$, being the inverse image
          of $Q$ under the map $phi:amapsto a/1$ from $R$ to $S^{-1}R$.
          If $b/sin Q$, then $b/1=(b/s)(s/1)in Q$ so $binfq$ and so $Q=S^{-1}fq$
          etc.



          Texts on commutative algebra will have more details.



          Your $alpha$ is $Qmapstophi^{-1}(Q)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for this. I think I see your argument. It's quite nice. Arguments like this make the effort of trying to understand these things worth it in the end.
            $endgroup$
            – Overwhelmed AG Apprentice
            Jan 20 at 4:29













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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          In a word, yes.



          In general, for a multiplicatively closed subset $S$ of $R$,
          there is a natural correspondence between the prime ideals of
          $S^{-1}R$ and the prime ideals of $R$ disjoint from $S$. Your
          example is the case where $S=R-newcommand{fp}{mathfrak{p}}
          newcommand{fq}{mathfrak{q}}fp$
          .



          The prime ideals of $S^{-1}R$ all have the form $S^{-1}fq$
          where $fqsubseteq fp$ is a prime ideal of $S$. It is straightforward
          to check that these $S^{-1}fq$ are prime in $S^{-1}R$. Conversely,
          let $Q$ be a prime ideal of $S^{-1}R$. Then
          $fq={ain R:a/1in Q}$ is a prime ideal of $R$, being the inverse image
          of $Q$ under the map $phi:amapsto a/1$ from $R$ to $S^{-1}R$.
          If $b/sin Q$, then $b/1=(b/s)(s/1)in Q$ so $binfq$ and so $Q=S^{-1}fq$
          etc.



          Texts on commutative algebra will have more details.



          Your $alpha$ is $Qmapstophi^{-1}(Q)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for this. I think I see your argument. It's quite nice. Arguments like this make the effort of trying to understand these things worth it in the end.
            $endgroup$
            – Overwhelmed AG Apprentice
            Jan 20 at 4:29


















          2












          $begingroup$

          In a word, yes.



          In general, for a multiplicatively closed subset $S$ of $R$,
          there is a natural correspondence between the prime ideals of
          $S^{-1}R$ and the prime ideals of $R$ disjoint from $S$. Your
          example is the case where $S=R-newcommand{fp}{mathfrak{p}}
          newcommand{fq}{mathfrak{q}}fp$
          .



          The prime ideals of $S^{-1}R$ all have the form $S^{-1}fq$
          where $fqsubseteq fp$ is a prime ideal of $S$. It is straightforward
          to check that these $S^{-1}fq$ are prime in $S^{-1}R$. Conversely,
          let $Q$ be a prime ideal of $S^{-1}R$. Then
          $fq={ain R:a/1in Q}$ is a prime ideal of $R$, being the inverse image
          of $Q$ under the map $phi:amapsto a/1$ from $R$ to $S^{-1}R$.
          If $b/sin Q$, then $b/1=(b/s)(s/1)in Q$ so $binfq$ and so $Q=S^{-1}fq$
          etc.



          Texts on commutative algebra will have more details.



          Your $alpha$ is $Qmapstophi^{-1}(Q)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for this. I think I see your argument. It's quite nice. Arguments like this make the effort of trying to understand these things worth it in the end.
            $endgroup$
            – Overwhelmed AG Apprentice
            Jan 20 at 4:29
















          2












          2








          2





          $begingroup$

          In a word, yes.



          In general, for a multiplicatively closed subset $S$ of $R$,
          there is a natural correspondence between the prime ideals of
          $S^{-1}R$ and the prime ideals of $R$ disjoint from $S$. Your
          example is the case where $S=R-newcommand{fp}{mathfrak{p}}
          newcommand{fq}{mathfrak{q}}fp$
          .



          The prime ideals of $S^{-1}R$ all have the form $S^{-1}fq$
          where $fqsubseteq fp$ is a prime ideal of $S$. It is straightforward
          to check that these $S^{-1}fq$ are prime in $S^{-1}R$. Conversely,
          let $Q$ be a prime ideal of $S^{-1}R$. Then
          $fq={ain R:a/1in Q}$ is a prime ideal of $R$, being the inverse image
          of $Q$ under the map $phi:amapsto a/1$ from $R$ to $S^{-1}R$.
          If $b/sin Q$, then $b/1=(b/s)(s/1)in Q$ so $binfq$ and so $Q=S^{-1}fq$
          etc.



          Texts on commutative algebra will have more details.



          Your $alpha$ is $Qmapstophi^{-1}(Q)$.






          share|cite|improve this answer









          $endgroup$



          In a word, yes.



          In general, for a multiplicatively closed subset $S$ of $R$,
          there is a natural correspondence between the prime ideals of
          $S^{-1}R$ and the prime ideals of $R$ disjoint from $S$. Your
          example is the case where $S=R-newcommand{fp}{mathfrak{p}}
          newcommand{fq}{mathfrak{q}}fp$
          .



          The prime ideals of $S^{-1}R$ all have the form $S^{-1}fq$
          where $fqsubseteq fp$ is a prime ideal of $S$. It is straightforward
          to check that these $S^{-1}fq$ are prime in $S^{-1}R$. Conversely,
          let $Q$ be a prime ideal of $S^{-1}R$. Then
          $fq={ain R:a/1in Q}$ is a prime ideal of $R$, being the inverse image
          of $Q$ under the map $phi:amapsto a/1$ from $R$ to $S^{-1}R$.
          If $b/sin Q$, then $b/1=(b/s)(s/1)in Q$ so $binfq$ and so $Q=S^{-1}fq$
          etc.



          Texts on commutative algebra will have more details.



          Your $alpha$ is $Qmapstophi^{-1}(Q)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 11 at 4:20









          Lord Shark the UnknownLord Shark the Unknown

          104k1160132




          104k1160132












          • $begingroup$
            Thanks for this. I think I see your argument. It's quite nice. Arguments like this make the effort of trying to understand these things worth it in the end.
            $endgroup$
            – Overwhelmed AG Apprentice
            Jan 20 at 4:29




















          • $begingroup$
            Thanks for this. I think I see your argument. It's quite nice. Arguments like this make the effort of trying to understand these things worth it in the end.
            $endgroup$
            – Overwhelmed AG Apprentice
            Jan 20 at 4:29


















          $begingroup$
          Thanks for this. I think I see your argument. It's quite nice. Arguments like this make the effort of trying to understand these things worth it in the end.
          $endgroup$
          – Overwhelmed AG Apprentice
          Jan 20 at 4:29






          $begingroup$
          Thanks for this. I think I see your argument. It's quite nice. Arguments like this make the effort of trying to understand these things worth it in the end.
          $endgroup$
          – Overwhelmed AG Apprentice
          Jan 20 at 4:29




















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