Application Chinese Remainder Theorem to Dedekind Domains












1














I have a question about the application of CRT in a proof of following thread:
Dedekind domain with a finite number of prime ideals is principal



The claim is that a Dedekind domain with a finite number of prime ideals is already principal:



In his answer @pki uses following argument:



Let $R$ be a Dedekind ring and assume that the prime ideals are $mathfrak{p}_1,ldots,mathfrak{p}_n$. Then $mathfrak{p}_1^2,mathfrak{p}_2,ldots,mathfrak{p}_n$ are coprime. Pick an element $pi in mathfrak{p}_1setminus mathfrak{p}_1^2$ and by CRT you can find an $xin R$ s.t.



$$ xequiv pi,(textrm{mod } mathfrak{p}_1^2),;; xequiv 1,(textrm{mod } mathfrak{p}_k),; k=2,ldots,n $$



Factoring we must have $(x)=mathfrak{p}_1$ (???)



Indeed the CRT provides a $x$ such that $ xequiv pi,(textrm{mod } mathfrak{p}_1^2),;; xequiv 1,(textrm{mod } mathfrak{p}_k),; k=2,ldots,n $ holds.



But why we get $(x)=mathfrak{p}_1$?



Factoring just implies $(bar{x}) = mathfrak{p}_1/ mathfrak{p}_1^2$. How to deduce $(x)=mathfrak{p}_1$?






commutative-algebra dedekind-domain







share|cite|improve this question



























    1














    I have a question about the application of CRT in a proof of following thread:
    Dedekind domain with a finite number of prime ideals is principal



    The claim is that a Dedekind domain with a finite number of prime ideals is already principal:



    In his answer @pki uses following argument:



    Let $R$ be a Dedekind ring and assume that the prime ideals are $mathfrak{p}_1,ldots,mathfrak{p}_n$. Then $mathfrak{p}_1^2,mathfrak{p}_2,ldots,mathfrak{p}_n$ are coprime. Pick an element $pi in mathfrak{p}_1setminus mathfrak{p}_1^2$ and by CRT you can find an $xin R$ s.t.



    $$ xequiv pi,(textrm{mod } mathfrak{p}_1^2),;; xequiv 1,(textrm{mod } mathfrak{p}_k),; k=2,ldots,n $$



    Factoring we must have $(x)=mathfrak{p}_1$ (???)



    Indeed the CRT provides a $x$ such that $ xequiv pi,(textrm{mod } mathfrak{p}_1^2),;; xequiv 1,(textrm{mod } mathfrak{p}_k),; k=2,ldots,n $ holds.



    But why we get $(x)=mathfrak{p}_1$?



    Factoring just implies $(bar{x}) = mathfrak{p}_1/ mathfrak{p}_1^2$. How to deduce $(x)=mathfrak{p}_1$?






    commutative-algebra dedekind-domain







    share|cite|improve this question

























      1












      1








      1







      I have a question about the application of CRT in a proof of following thread:
      Dedekind domain with a finite number of prime ideals is principal



      The claim is that a Dedekind domain with a finite number of prime ideals is already principal:



      In his answer @pki uses following argument:



      Let $R$ be a Dedekind ring and assume that the prime ideals are $mathfrak{p}_1,ldots,mathfrak{p}_n$. Then $mathfrak{p}_1^2,mathfrak{p}_2,ldots,mathfrak{p}_n$ are coprime. Pick an element $pi in mathfrak{p}_1setminus mathfrak{p}_1^2$ and by CRT you can find an $xin R$ s.t.



      $$ xequiv pi,(textrm{mod } mathfrak{p}_1^2),;; xequiv 1,(textrm{mod } mathfrak{p}_k),; k=2,ldots,n $$



      Factoring we must have $(x)=mathfrak{p}_1$ (???)



      Indeed the CRT provides a $x$ such that $ xequiv pi,(textrm{mod } mathfrak{p}_1^2),;; xequiv 1,(textrm{mod } mathfrak{p}_k),; k=2,ldots,n $ holds.



      But why we get $(x)=mathfrak{p}_1$?



      Factoring just implies $(bar{x}) = mathfrak{p}_1/ mathfrak{p}_1^2$. How to deduce $(x)=mathfrak{p}_1$?






      commutative-algebra dedekind-domain







      share|cite|improve this question













      I have a question about the application of CRT in a proof of following thread:
      Dedekind domain with a finite number of prime ideals is principal



      The claim is that a Dedekind domain with a finite number of prime ideals is already principal:



      In his answer @pki uses following argument:



      Let $R$ be a Dedekind ring and assume that the prime ideals are $mathfrak{p}_1,ldots,mathfrak{p}_n$. Then $mathfrak{p}_1^2,mathfrak{p}_2,ldots,mathfrak{p}_n$ are coprime. Pick an element $pi in mathfrak{p}_1setminus mathfrak{p}_1^2$ and by CRT you can find an $xin R$ s.t.



      $$ xequiv pi,(textrm{mod } mathfrak{p}_1^2),;; xequiv 1,(textrm{mod } mathfrak{p}_k),; k=2,ldots,n $$



      Factoring we must have $(x)=mathfrak{p}_1$ (???)



      Indeed the CRT provides a $x$ such that $ xequiv pi,(textrm{mod } mathfrak{p}_1^2),;; xequiv 1,(textrm{mod } mathfrak{p}_k),; k=2,ldots,n $ holds.



      But why we get $(x)=mathfrak{p}_1$?



      Factoring just implies $(bar{x}) = mathfrak{p}_1/ mathfrak{p}_1^2$. How to deduce $(x)=mathfrak{p}_1$?






      commutative-algebra dedekind-domain




      commutative-algebra dedekind-domain






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 30 '18 at 18:27









      KarlPeterKarlPeter

      5971315




      5971315






















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          First $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$ for some $e_iinnewcommand{NN}{mathbb{N}}NN$. If $e_i ge 1$ for $ige 2$, then $xin(x)subset pp_i$, however, $xequiv 1 pmod{pp_i}$, so this is impossible. Hence $(x)=pp_1^{e_1}$, for some $e_1ge 1$ (since $(x)subset pp_1$ by assumption, so $(x)ne (1)$). However, $xnotin pp_1^2$ either, so $xin(x)=pp_1^{e_1}$ forces $e_1=1$. Thus $(x)=pp_1$ as desired.






          share|cite|improve this answer





















          • Hi, thank you for the answer. One step stays unclear: How do you get $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$?
            – KarlPeter
            Mar 30 '18 at 18:43










          • @KarlPeter Ah, that's a definition of Dedekind domain, every nonzero ideal factors into primes. See Wikipedia: en.wikipedia.org/wiki/Dedekind_domain#Alternative_definitions
            – jgon
            Mar 30 '18 at 18:45










          • @KarlPeter For a proof you'll want to look at a resource on algebraic number theory, e.g. Milne's course notes here: jmilne.org/math/CourseNotes/ant.html It's a little too involved for an answer here.
            – jgon
            Mar 30 '18 at 18:46













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          First $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$ for some $e_iinnewcommand{NN}{mathbb{N}}NN$. If $e_i ge 1$ for $ige 2$, then $xin(x)subset pp_i$, however, $xequiv 1 pmod{pp_i}$, so this is impossible. Hence $(x)=pp_1^{e_1}$, for some $e_1ge 1$ (since $(x)subset pp_1$ by assumption, so $(x)ne (1)$). However, $xnotin pp_1^2$ either, so $xin(x)=pp_1^{e_1}$ forces $e_1=1$. Thus $(x)=pp_1$ as desired.






          share|cite|improve this answer





















          • Hi, thank you for the answer. One step stays unclear: How do you get $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$?
            – KarlPeter
            Mar 30 '18 at 18:43










          • @KarlPeter Ah, that's a definition of Dedekind domain, every nonzero ideal factors into primes. See Wikipedia: en.wikipedia.org/wiki/Dedekind_domain#Alternative_definitions
            – jgon
            Mar 30 '18 at 18:45










          • @KarlPeter For a proof you'll want to look at a resource on algebraic number theory, e.g. Milne's course notes here: jmilne.org/math/CourseNotes/ant.html It's a little too involved for an answer here.
            – jgon
            Mar 30 '18 at 18:46


















          2














          First $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$ for some $e_iinnewcommand{NN}{mathbb{N}}NN$. If $e_i ge 1$ for $ige 2$, then $xin(x)subset pp_i$, however, $xequiv 1 pmod{pp_i}$, so this is impossible. Hence $(x)=pp_1^{e_1}$, for some $e_1ge 1$ (since $(x)subset pp_1$ by assumption, so $(x)ne (1)$). However, $xnotin pp_1^2$ either, so $xin(x)=pp_1^{e_1}$ forces $e_1=1$. Thus $(x)=pp_1$ as desired.






          share|cite|improve this answer





















          • Hi, thank you for the answer. One step stays unclear: How do you get $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$?
            – KarlPeter
            Mar 30 '18 at 18:43










          • @KarlPeter Ah, that's a definition of Dedekind domain, every nonzero ideal factors into primes. See Wikipedia: en.wikipedia.org/wiki/Dedekind_domain#Alternative_definitions
            – jgon
            Mar 30 '18 at 18:45










          • @KarlPeter For a proof you'll want to look at a resource on algebraic number theory, e.g. Milne's course notes here: jmilne.org/math/CourseNotes/ant.html It's a little too involved for an answer here.
            – jgon
            Mar 30 '18 at 18:46
















          2












          2








          2






          First $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$ for some $e_iinnewcommand{NN}{mathbb{N}}NN$. If $e_i ge 1$ for $ige 2$, then $xin(x)subset pp_i$, however, $xequiv 1 pmod{pp_i}$, so this is impossible. Hence $(x)=pp_1^{e_1}$, for some $e_1ge 1$ (since $(x)subset pp_1$ by assumption, so $(x)ne (1)$). However, $xnotin pp_1^2$ either, so $xin(x)=pp_1^{e_1}$ forces $e_1=1$. Thus $(x)=pp_1$ as desired.






          share|cite|improve this answer












          First $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$ for some $e_iinnewcommand{NN}{mathbb{N}}NN$. If $e_i ge 1$ for $ige 2$, then $xin(x)subset pp_i$, however, $xequiv 1 pmod{pp_i}$, so this is impossible. Hence $(x)=pp_1^{e_1}$, for some $e_1ge 1$ (since $(x)subset pp_1$ by assumption, so $(x)ne (1)$). However, $xnotin pp_1^2$ either, so $xin(x)=pp_1^{e_1}$ forces $e_1=1$. Thus $(x)=pp_1$ as desired.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 30 '18 at 18:39









          jgonjgon

          13.4k21941




          13.4k21941












          • Hi, thank you for the answer. One step stays unclear: How do you get $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$?
            – KarlPeter
            Mar 30 '18 at 18:43










          • @KarlPeter Ah, that's a definition of Dedekind domain, every nonzero ideal factors into primes. See Wikipedia: en.wikipedia.org/wiki/Dedekind_domain#Alternative_definitions
            – jgon
            Mar 30 '18 at 18:45










          • @KarlPeter For a proof you'll want to look at a resource on algebraic number theory, e.g. Milne's course notes here: jmilne.org/math/CourseNotes/ant.html It's a little too involved for an answer here.
            – jgon
            Mar 30 '18 at 18:46




















          • Hi, thank you for the answer. One step stays unclear: How do you get $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$?
            – KarlPeter
            Mar 30 '18 at 18:43










          • @KarlPeter Ah, that's a definition of Dedekind domain, every nonzero ideal factors into primes. See Wikipedia: en.wikipedia.org/wiki/Dedekind_domain#Alternative_definitions
            – jgon
            Mar 30 '18 at 18:45










          • @KarlPeter For a proof you'll want to look at a resource on algebraic number theory, e.g. Milne's course notes here: jmilne.org/math/CourseNotes/ant.html It's a little too involved for an answer here.
            – jgon
            Mar 30 '18 at 18:46


















          Hi, thank you for the answer. One step stays unclear: How do you get $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$?
          – KarlPeter
          Mar 30 '18 at 18:43




          Hi, thank you for the answer. One step stays unclear: How do you get $newcommand{pp}{mathfrak{p}}newcommand{mm}{mathfrak{m}}(x)=pp_1^{e_1}cdots pp_n^{e_n}$?
          – KarlPeter
          Mar 30 '18 at 18:43












          @KarlPeter Ah, that's a definition of Dedekind domain, every nonzero ideal factors into primes. See Wikipedia: en.wikipedia.org/wiki/Dedekind_domain#Alternative_definitions
          – jgon
          Mar 30 '18 at 18:45




          @KarlPeter Ah, that's a definition of Dedekind domain, every nonzero ideal factors into primes. See Wikipedia: en.wikipedia.org/wiki/Dedekind_domain#Alternative_definitions
          – jgon
          Mar 30 '18 at 18:45












          @KarlPeter For a proof you'll want to look at a resource on algebraic number theory, e.g. Milne's course notes here: jmilne.org/math/CourseNotes/ant.html It's a little too involved for an answer here.
          – jgon
          Mar 30 '18 at 18:46






          @KarlPeter For a proof you'll want to look at a resource on algebraic number theory, e.g. Milne's course notes here: jmilne.org/math/CourseNotes/ant.html It's a little too involved for an answer here.
          – jgon
          Mar 30 '18 at 18:46




















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