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 at 18:27









      KarlPeter

      4871313




      4871313






















          1 Answer
          1






          active

          oldest

          votes


















          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 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 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 at 18:46













          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2715107%2fapplication-chinese-remainder-theorem-to-dedekind-domains%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          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 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 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 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 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 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 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 at 18:39









          jgon

          12.2k21840




          12.2k21840












          • 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 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 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 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 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 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 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 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 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 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 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 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 at 18:46




















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2715107%2fapplication-chinese-remainder-theorem-to-dedekind-domains%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Wiesbaden

          Image
          Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
          Read more

          Marschland

          Image
          Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
          Read more

          Dieringhausen

          Image
          Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
          Read more