Linear independence of $sqrt{2}, sqrt[3]{2}, sqrt[5]{2}$ over $mathbb{Q}$












2












$begingroup$


In other words, show that for $a,b,cin mathbb{Q}$, $asqrt{2}+b sqrt[3]{2}+c sqrt[5]{2}=0$ implies that $a=b=c=0$.



There might be some questions on this forum that are similar to mine (e.g. Linear independence of $sqrt{2}$, $sqrt[3]{2}$, $sqrt[4]{2}$, . . .). However, I was given the following hint:



"Note: If organized smartly, this comes out as first the third order radical, then the second order is added to form an extension of degree 6 (since it can't belong to the first), and then, the radical of order 5 cannot be in the extension since it would violate the degree."



So I would love to see a solution to this using the field extensions, and how the hint makes sense. Huge thanks in advance!






abstract-algebra field-theory







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    In other words, show that for $a,b,cin mathbb{Q}$, $asqrt{2}+b sqrt[3]{2}+c sqrt[5]{2}=0$ implies that $a=b=c=0$.



    There might be some questions on this forum that are similar to mine (e.g. Linear independence of $sqrt{2}$, $sqrt[3]{2}$, $sqrt[4]{2}$, . . .). However, I was given the following hint:



    "Note: If organized smartly, this comes out as first the third order radical, then the second order is added to form an extension of degree 6 (since it can't belong to the first), and then, the radical of order 5 cannot be in the extension since it would violate the degree."



    So I would love to see a solution to this using the field extensions, and how the hint makes sense. Huge thanks in advance!






    abstract-algebra field-theory







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      In other words, show that for $a,b,cin mathbb{Q}$, $asqrt{2}+b sqrt[3]{2}+c sqrt[5]{2}=0$ implies that $a=b=c=0$.



      There might be some questions on this forum that are similar to mine (e.g. Linear independence of $sqrt{2}$, $sqrt[3]{2}$, $sqrt[4]{2}$, . . .). However, I was given the following hint:



      "Note: If organized smartly, this comes out as first the third order radical, then the second order is added to form an extension of degree 6 (since it can't belong to the first), and then, the radical of order 5 cannot be in the extension since it would violate the degree."



      So I would love to see a solution to this using the field extensions, and how the hint makes sense. Huge thanks in advance!






      abstract-algebra field-theory







      share|cite|improve this question









      $endgroup$




      In other words, show that for $a,b,cin mathbb{Q}$, $asqrt{2}+b sqrt[3]{2}+c sqrt[5]{2}=0$ implies that $a=b=c=0$.



      There might be some questions on this forum that are similar to mine (e.g. Linear independence of $sqrt{2}$, $sqrt[3]{2}$, $sqrt[4]{2}$, . . .). However, I was given the following hint:



      "Note: If organized smartly, this comes out as first the third order radical, then the second order is added to form an extension of degree 6 (since it can't belong to the first), and then, the radical of order 5 cannot be in the extension since it would violate the degree."



      So I would love to see a solution to this using the field extensions, and how the hint makes sense. Huge thanks in advance!






      abstract-algebra field-theory




      abstract-algebra field-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 29 '18 at 2:08









      AlexAlex

      757




      757






















          1 Answer
          1






          active

          oldest

          votes


















          5












          $begingroup$


          • $x^6-2$ and $x^5-2$ both are irreducible over $mathbb{Q}$ thus $mathbb{Q}(sqrt[6]{2}) cong mathbb{Q}[x]/(x^6-2)$ and $mathbb{Q}(sqrt[5]{2})cong mathbb{Q}[x]/(x^5-2)$ are $mathbb{Q}$ vector spaces of dimension $6$ and $5$.



          • Let $K = mathbb{Q}(sqrt[6]{2},sqrt[5]{2})$ a $mathbb{Q}$ vector space of dimension $n$.



            By the multiplicativity of degree of extensions we know that $5|n$ and $6|n$ thus $30| n$.




          • Assume that $asqrt{2}+b sqrt[3]{2}= sqrt[5]{2}$ with $a,b in mathbb{Q}$.



            Then $sqrt[5]{2} in mathbb{Q}(sqrt[6]{2})$ and $n= 6$ which is a contradiction.



          • The same idea applies to show that $sqrt[3]{2} not in mathbb{Q}(sqrt{2})$ so $asqrt{2}+bsqrt[3]{2} =0$ implies $a=b=0$.


          • An alternative is to develop a similar contradiction over some finite field $mathbb{F}_p$ instead of $mathbb{Q}$ and then to lift back the contradiction to $mathbb{Q}$.







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks very much for your answer! It made a lot more sense to me. So the idea is first to show that $sqrt[3]{2}notin mathbb{Q}(sqrt{2})$, showing that $sqrt{2}$ and $sqrt[3]{2}$ are linearly independent. Then by the same idea, show that $sqrt[5]{2}notin mathbb{Q}(sqrt[6]{2})$, so that $sqrt[5]{2}notin mathbb{Q}(sqrt[3]{2}), mathbb{Q}(sqrt[3]{2})$, since $mathbb{Q}(sqrt[3]{2}),mathbb{Q}(sqrt{2})subseteq mathbb{Q}(sqrt[6]{2})$.
            $endgroup$
            – Alex
            Dec 29 '18 at 2:55










          • $begingroup$
            @Alex And to show that $x^6-2,x^5-2$ are irreducible ! Factorize them, let $f_6,f_5$ be the minimal polynomials of $sqrt[6]{2},sqrt[5]{2}$ then look at $f_6(0),f_5(0)$
            $endgroup$
            – reuns
            Dec 29 '18 at 5:08








          • 1




            $begingroup$
            @renus Aren't they irreducible by Eisenstein's Criterion?
            $endgroup$
            – Alex
            Dec 29 '18 at 5:10











          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%2f3055470%2flinear-independence-of-sqrt2-sqrt32-sqrt52-over-mathbbq%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









          5












          $begingroup$


          • $x^6-2$ and $x^5-2$ both are irreducible over $mathbb{Q}$ thus $mathbb{Q}(sqrt[6]{2}) cong mathbb{Q}[x]/(x^6-2)$ and $mathbb{Q}(sqrt[5]{2})cong mathbb{Q}[x]/(x^5-2)$ are $mathbb{Q}$ vector spaces of dimension $6$ and $5$.



          • Let $K = mathbb{Q}(sqrt[6]{2},sqrt[5]{2})$ a $mathbb{Q}$ vector space of dimension $n$.



            By the multiplicativity of degree of extensions we know that $5|n$ and $6|n$ thus $30| n$.




          • Assume that $asqrt{2}+b sqrt[3]{2}= sqrt[5]{2}$ with $a,b in mathbb{Q}$.



            Then $sqrt[5]{2} in mathbb{Q}(sqrt[6]{2})$ and $n= 6$ which is a contradiction.



          • The same idea applies to show that $sqrt[3]{2} not in mathbb{Q}(sqrt{2})$ so $asqrt{2}+bsqrt[3]{2} =0$ implies $a=b=0$.


          • An alternative is to develop a similar contradiction over some finite field $mathbb{F}_p$ instead of $mathbb{Q}$ and then to lift back the contradiction to $mathbb{Q}$.







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks very much for your answer! It made a lot more sense to me. So the idea is first to show that $sqrt[3]{2}notin mathbb{Q}(sqrt{2})$, showing that $sqrt{2}$ and $sqrt[3]{2}$ are linearly independent. Then by the same idea, show that $sqrt[5]{2}notin mathbb{Q}(sqrt[6]{2})$, so that $sqrt[5]{2}notin mathbb{Q}(sqrt[3]{2}), mathbb{Q}(sqrt[3]{2})$, since $mathbb{Q}(sqrt[3]{2}),mathbb{Q}(sqrt{2})subseteq mathbb{Q}(sqrt[6]{2})$.
            $endgroup$
            – Alex
            Dec 29 '18 at 2:55










          • $begingroup$
            @Alex And to show that $x^6-2,x^5-2$ are irreducible ! Factorize them, let $f_6,f_5$ be the minimal polynomials of $sqrt[6]{2},sqrt[5]{2}$ then look at $f_6(0),f_5(0)$
            $endgroup$
            – reuns
            Dec 29 '18 at 5:08








          • 1




            $begingroup$
            @renus Aren't they irreducible by Eisenstein's Criterion?
            $endgroup$
            – Alex
            Dec 29 '18 at 5:10
















          5












          $begingroup$


          • $x^6-2$ and $x^5-2$ both are irreducible over $mathbb{Q}$ thus $mathbb{Q}(sqrt[6]{2}) cong mathbb{Q}[x]/(x^6-2)$ and $mathbb{Q}(sqrt[5]{2})cong mathbb{Q}[x]/(x^5-2)$ are $mathbb{Q}$ vector spaces of dimension $6$ and $5$.



          • Let $K = mathbb{Q}(sqrt[6]{2},sqrt[5]{2})$ a $mathbb{Q}$ vector space of dimension $n$.



            By the multiplicativity of degree of extensions we know that $5|n$ and $6|n$ thus $30| n$.




          • Assume that $asqrt{2}+b sqrt[3]{2}= sqrt[5]{2}$ with $a,b in mathbb{Q}$.



            Then $sqrt[5]{2} in mathbb{Q}(sqrt[6]{2})$ and $n= 6$ which is a contradiction.



          • The same idea applies to show that $sqrt[3]{2} not in mathbb{Q}(sqrt{2})$ so $asqrt{2}+bsqrt[3]{2} =0$ implies $a=b=0$.


          • An alternative is to develop a similar contradiction over some finite field $mathbb{F}_p$ instead of $mathbb{Q}$ and then to lift back the contradiction to $mathbb{Q}$.







          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks very much for your answer! It made a lot more sense to me. So the idea is first to show that $sqrt[3]{2}notin mathbb{Q}(sqrt{2})$, showing that $sqrt{2}$ and $sqrt[3]{2}$ are linearly independent. Then by the same idea, show that $sqrt[5]{2}notin mathbb{Q}(sqrt[6]{2})$, so that $sqrt[5]{2}notin mathbb{Q}(sqrt[3]{2}), mathbb{Q}(sqrt[3]{2})$, since $mathbb{Q}(sqrt[3]{2}),mathbb{Q}(sqrt{2})subseteq mathbb{Q}(sqrt[6]{2})$.
            $endgroup$
            – Alex
            Dec 29 '18 at 2:55










          • $begingroup$
            @Alex And to show that $x^6-2,x^5-2$ are irreducible ! Factorize them, let $f_6,f_5$ be the minimal polynomials of $sqrt[6]{2},sqrt[5]{2}$ then look at $f_6(0),f_5(0)$
            $endgroup$
            – reuns
            Dec 29 '18 at 5:08








          • 1




            $begingroup$
            @renus Aren't they irreducible by Eisenstein's Criterion?
            $endgroup$
            – Alex
            Dec 29 '18 at 5:10














          5












          5








          5





          $begingroup$


          • $x^6-2$ and $x^5-2$ both are irreducible over $mathbb{Q}$ thus $mathbb{Q}(sqrt[6]{2}) cong mathbb{Q}[x]/(x^6-2)$ and $mathbb{Q}(sqrt[5]{2})cong mathbb{Q}[x]/(x^5-2)$ are $mathbb{Q}$ vector spaces of dimension $6$ and $5$.



          • Let $K = mathbb{Q}(sqrt[6]{2},sqrt[5]{2})$ a $mathbb{Q}$ vector space of dimension $n$.



            By the multiplicativity of degree of extensions we know that $5|n$ and $6|n$ thus $30| n$.




          • Assume that $asqrt{2}+b sqrt[3]{2}= sqrt[5]{2}$ with $a,b in mathbb{Q}$.



            Then $sqrt[5]{2} in mathbb{Q}(sqrt[6]{2})$ and $n= 6$ which is a contradiction.



          • The same idea applies to show that $sqrt[3]{2} not in mathbb{Q}(sqrt{2})$ so $asqrt{2}+bsqrt[3]{2} =0$ implies $a=b=0$.


          • An alternative is to develop a similar contradiction over some finite field $mathbb{F}_p$ instead of $mathbb{Q}$ and then to lift back the contradiction to $mathbb{Q}$.







          share|cite|improve this answer











          $endgroup$




          • $x^6-2$ and $x^5-2$ both are irreducible over $mathbb{Q}$ thus $mathbb{Q}(sqrt[6]{2}) cong mathbb{Q}[x]/(x^6-2)$ and $mathbb{Q}(sqrt[5]{2})cong mathbb{Q}[x]/(x^5-2)$ are $mathbb{Q}$ vector spaces of dimension $6$ and $5$.



          • Let $K = mathbb{Q}(sqrt[6]{2},sqrt[5]{2})$ a $mathbb{Q}$ vector space of dimension $n$.



            By the multiplicativity of degree of extensions we know that $5|n$ and $6|n$ thus $30| n$.




          • Assume that $asqrt{2}+b sqrt[3]{2}= sqrt[5]{2}$ with $a,b in mathbb{Q}$.



            Then $sqrt[5]{2} in mathbb{Q}(sqrt[6]{2})$ and $n= 6$ which is a contradiction.



          • The same idea applies to show that $sqrt[3]{2} not in mathbb{Q}(sqrt{2})$ so $asqrt{2}+bsqrt[3]{2} =0$ implies $a=b=0$.


          • An alternative is to develop a similar contradiction over some finite field $mathbb{F}_p$ instead of $mathbb{Q}$ and then to lift back the contradiction to $mathbb{Q}$.








          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 29 '18 at 2:34

























          answered Dec 29 '18 at 2:26









          reunsreuns

          21.5k21352




          21.5k21352












          • $begingroup$
            Thanks very much for your answer! It made a lot more sense to me. So the idea is first to show that $sqrt[3]{2}notin mathbb{Q}(sqrt{2})$, showing that $sqrt{2}$ and $sqrt[3]{2}$ are linearly independent. Then by the same idea, show that $sqrt[5]{2}notin mathbb{Q}(sqrt[6]{2})$, so that $sqrt[5]{2}notin mathbb{Q}(sqrt[3]{2}), mathbb{Q}(sqrt[3]{2})$, since $mathbb{Q}(sqrt[3]{2}),mathbb{Q}(sqrt{2})subseteq mathbb{Q}(sqrt[6]{2})$.
            $endgroup$
            – Alex
            Dec 29 '18 at 2:55










          • $begingroup$
            @Alex And to show that $x^6-2,x^5-2$ are irreducible ! Factorize them, let $f_6,f_5$ be the minimal polynomials of $sqrt[6]{2},sqrt[5]{2}$ then look at $f_6(0),f_5(0)$
            $endgroup$
            – reuns
            Dec 29 '18 at 5:08








          • 1




            $begingroup$
            @renus Aren't they irreducible by Eisenstein's Criterion?
            $endgroup$
            – Alex
            Dec 29 '18 at 5:10


















          • $begingroup$
            Thanks very much for your answer! It made a lot more sense to me. So the idea is first to show that $sqrt[3]{2}notin mathbb{Q}(sqrt{2})$, showing that $sqrt{2}$ and $sqrt[3]{2}$ are linearly independent. Then by the same idea, show that $sqrt[5]{2}notin mathbb{Q}(sqrt[6]{2})$, so that $sqrt[5]{2}notin mathbb{Q}(sqrt[3]{2}), mathbb{Q}(sqrt[3]{2})$, since $mathbb{Q}(sqrt[3]{2}),mathbb{Q}(sqrt{2})subseteq mathbb{Q}(sqrt[6]{2})$.
            $endgroup$
            – Alex
            Dec 29 '18 at 2:55










          • $begingroup$
            @Alex And to show that $x^6-2,x^5-2$ are irreducible ! Factorize them, let $f_6,f_5$ be the minimal polynomials of $sqrt[6]{2},sqrt[5]{2}$ then look at $f_6(0),f_5(0)$
            $endgroup$
            – reuns
            Dec 29 '18 at 5:08








          • 1




            $begingroup$
            @renus Aren't they irreducible by Eisenstein's Criterion?
            $endgroup$
            – Alex
            Dec 29 '18 at 5:10
















          $begingroup$
          Thanks very much for your answer! It made a lot more sense to me. So the idea is first to show that $sqrt[3]{2}notin mathbb{Q}(sqrt{2})$, showing that $sqrt{2}$ and $sqrt[3]{2}$ are linearly independent. Then by the same idea, show that $sqrt[5]{2}notin mathbb{Q}(sqrt[6]{2})$, so that $sqrt[5]{2}notin mathbb{Q}(sqrt[3]{2}), mathbb{Q}(sqrt[3]{2})$, since $mathbb{Q}(sqrt[3]{2}),mathbb{Q}(sqrt{2})subseteq mathbb{Q}(sqrt[6]{2})$.
          $endgroup$
          – Alex
          Dec 29 '18 at 2:55




          $begingroup$
          Thanks very much for your answer! It made a lot more sense to me. So the idea is first to show that $sqrt[3]{2}notin mathbb{Q}(sqrt{2})$, showing that $sqrt{2}$ and $sqrt[3]{2}$ are linearly independent. Then by the same idea, show that $sqrt[5]{2}notin mathbb{Q}(sqrt[6]{2})$, so that $sqrt[5]{2}notin mathbb{Q}(sqrt[3]{2}), mathbb{Q}(sqrt[3]{2})$, since $mathbb{Q}(sqrt[3]{2}),mathbb{Q}(sqrt{2})subseteq mathbb{Q}(sqrt[6]{2})$.
          $endgroup$
          – Alex
          Dec 29 '18 at 2:55












          $begingroup$
          @Alex And to show that $x^6-2,x^5-2$ are irreducible ! Factorize them, let $f_6,f_5$ be the minimal polynomials of $sqrt[6]{2},sqrt[5]{2}$ then look at $f_6(0),f_5(0)$
          $endgroup$
          – reuns
          Dec 29 '18 at 5:08






          $begingroup$
          @Alex And to show that $x^6-2,x^5-2$ are irreducible ! Factorize them, let $f_6,f_5$ be the minimal polynomials of $sqrt[6]{2},sqrt[5]{2}$ then look at $f_6(0),f_5(0)$
          $endgroup$
          – reuns
          Dec 29 '18 at 5:08






          1




          1




          $begingroup$
          @renus Aren't they irreducible by Eisenstein's Criterion?
          $endgroup$
          – Alex
          Dec 29 '18 at 5:10




          $begingroup$
          @renus Aren't they irreducible by Eisenstein's Criterion?
          $endgroup$
          – Alex
          Dec 29 '18 at 5:10


















          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.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3055470%2flinear-independence-of-sqrt2-sqrt32-sqrt52-over-mathbbq%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