Transfer Between LCM, GCD for Rings?












3












$begingroup$


I am starting a chapter on divisibility in commutative rings, and I was wondering if there was a way to translate theorems about gcd to lcm and vice versa. I know the concepts are considered "dual" in some sense, so perhaps the theorems relating to them are also dual.






abstract-algebra ring-theory







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    I am starting a chapter on divisibility in commutative rings, and I was wondering if there was a way to translate theorems about gcd to lcm and vice versa. I know the concepts are considered "dual" in some sense, so perhaps the theorems relating to them are also dual.






    abstract-algebra ring-theory







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      2



      $begingroup$


      I am starting a chapter on divisibility in commutative rings, and I was wondering if there was a way to translate theorems about gcd to lcm and vice versa. I know the concepts are considered "dual" in some sense, so perhaps the theorems relating to them are also dual.






      abstract-algebra ring-theory







      share|cite|improve this question











      $endgroup$




      I am starting a chapter on divisibility in commutative rings, and I was wondering if there was a way to translate theorems about gcd to lcm and vice versa. I know the concepts are considered "dual" in some sense, so perhaps the theorems relating to them are also dual.






      abstract-algebra ring-theory




      abstract-algebra ring-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 18 '17 at 0:33







      Jacob Wakem

















      asked Mar 19 '14 at 3:44









      Jacob WakemJacob Wakem

      1,910621




      1,910621






















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          A prototypical example is proving $ rm gcd(a,b):lcm(a,b) = ab, $ using the $,overbrace{{rm involution}, x'! =, ab/x}^{rmlarge cofactor duality } $ on the divisors of $rm:ab. $ Notice that $rm xmid ycolor{#c00}iff y'mid x', $ by ${, rmdfrac{y}x = dfrac{x'}{y'} }$ by $rm, yy'! = ab = xx'., $ Thus



          $$begin{align}rm cmidgcd(a,b)!iff&rm cmid a,b\[2px]
          color{#c00}iff& rm a',b'mid c'\[2px]
          iff & rm lcm(a',b')mid c'\
          color{#c00}iff & rm cmid lcm(a',b')'\
          {rm Thus}rmquad gcd(a,b), = &rm , lcm(a',b')'= dfrac{ab}{lcm(b,a)}
          end{align}quad $$



          The black arrows above are the universal property (definition) of gcd and lcm, and the red arrows follow by cofactor duality.



          Noational abuse alert: the gcd, lcm "equalities" are up to unit factors (i.e. "equal" if associate)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Is your "involution" basically a formal division? I know there is no division per se without an identity, but what you wrote seems to make sense anyway as equivalent to x'x=ab.
            $endgroup$
            – Jacob Wakem
            Mar 19 '14 at 16:34










          • $begingroup$
            I don't know what you mean by "formal division" and "division per se without an identity". Please use standard math language. Do you know any ring theory, or only elementary number theory?
            $endgroup$
            – Bill Dubuque
            Mar 19 '14 at 16:48










          • $begingroup$
            I know some ring theory. I have never heard "involution" used in ring theory (up to the point I have studied). Are you assuming the ring is with identity here and x has an inverse?(If so, it isn't worthless but I would like to know of the limited scope). You say x'=ab/x which only makes sense as written with an identity, and x having an inverse. But we can alternatively define x'=ab/x to just mean x'x=ab to make it more general.
            $endgroup$
            – Jacob Wakem
            Mar 19 '14 at 17:03










          • $begingroup$
            @Jacob In any domain, if $,xmid ab,$ then the cofactor $,x',$ such that $,xx' = ab,$ is uniquely defined. You may also find of interest lattice-theoretic viewpoints, e.g. for starters see this Wikipedia article.
            $endgroup$
            – Bill Dubuque
            Mar 19 '14 at 18:25












          • $begingroup$
            See also this answer.
            $endgroup$
            – Bill Dubuque
            Sep 6 '14 at 5:22











          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%2f717760%2ftransfer-between-lcm-gcd-for-rings%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









          3












          $begingroup$

          A prototypical example is proving $ rm gcd(a,b):lcm(a,b) = ab, $ using the $,overbrace{{rm involution}, x'! =, ab/x}^{rmlarge cofactor duality } $ on the divisors of $rm:ab. $ Notice that $rm xmid ycolor{#c00}iff y'mid x', $ by ${, rmdfrac{y}x = dfrac{x'}{y'} }$ by $rm, yy'! = ab = xx'., $ Thus



          $$begin{align}rm cmidgcd(a,b)!iff&rm cmid a,b\[2px]
          color{#c00}iff& rm a',b'mid c'\[2px]
          iff & rm lcm(a',b')mid c'\
          color{#c00}iff & rm cmid lcm(a',b')'\
          {rm Thus}rmquad gcd(a,b), = &rm , lcm(a',b')'= dfrac{ab}{lcm(b,a)}
          end{align}quad $$



          The black arrows above are the universal property (definition) of gcd and lcm, and the red arrows follow by cofactor duality.



          Noational abuse alert: the gcd, lcm "equalities" are up to unit factors (i.e. "equal" if associate)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Is your "involution" basically a formal division? I know there is no division per se without an identity, but what you wrote seems to make sense anyway as equivalent to x'x=ab.
            $endgroup$
            – Jacob Wakem
            Mar 19 '14 at 16:34










          • $begingroup$
            I don't know what you mean by "formal division" and "division per se without an identity". Please use standard math language. Do you know any ring theory, or only elementary number theory?
            $endgroup$
            – Bill Dubuque
            Mar 19 '14 at 16:48










          • $begingroup$
            I know some ring theory. I have never heard "involution" used in ring theory (up to the point I have studied). Are you assuming the ring is with identity here and x has an inverse?(If so, it isn't worthless but I would like to know of the limited scope). You say x'=ab/x which only makes sense as written with an identity, and x having an inverse. But we can alternatively define x'=ab/x to just mean x'x=ab to make it more general.
            $endgroup$
            – Jacob Wakem
            Mar 19 '14 at 17:03










          • $begingroup$
            @Jacob In any domain, if $,xmid ab,$ then the cofactor $,x',$ such that $,xx' = ab,$ is uniquely defined. You may also find of interest lattice-theoretic viewpoints, e.g. for starters see this Wikipedia article.
            $endgroup$
            – Bill Dubuque
            Mar 19 '14 at 18:25












          • $begingroup$
            See also this answer.
            $endgroup$
            – Bill Dubuque
            Sep 6 '14 at 5:22
















          3












          $begingroup$

          A prototypical example is proving $ rm gcd(a,b):lcm(a,b) = ab, $ using the $,overbrace{{rm involution}, x'! =, ab/x}^{rmlarge cofactor duality } $ on the divisors of $rm:ab. $ Notice that $rm xmid ycolor{#c00}iff y'mid x', $ by ${, rmdfrac{y}x = dfrac{x'}{y'} }$ by $rm, yy'! = ab = xx'., $ Thus



          $$begin{align}rm cmidgcd(a,b)!iff&rm cmid a,b\[2px]
          color{#c00}iff& rm a',b'mid c'\[2px]
          iff & rm lcm(a',b')mid c'\
          color{#c00}iff & rm cmid lcm(a',b')'\
          {rm Thus}rmquad gcd(a,b), = &rm , lcm(a',b')'= dfrac{ab}{lcm(b,a)}
          end{align}quad $$



          The black arrows above are the universal property (definition) of gcd and lcm, and the red arrows follow by cofactor duality.



          Noational abuse alert: the gcd, lcm "equalities" are up to unit factors (i.e. "equal" if associate)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Is your "involution" basically a formal division? I know there is no division per se without an identity, but what you wrote seems to make sense anyway as equivalent to x'x=ab.
            $endgroup$
            – Jacob Wakem
            Mar 19 '14 at 16:34










          • $begingroup$
            I don't know what you mean by "formal division" and "division per se without an identity". Please use standard math language. Do you know any ring theory, or only elementary number theory?
            $endgroup$
            – Bill Dubuque
            Mar 19 '14 at 16:48










          • $begingroup$
            I know some ring theory. I have never heard "involution" used in ring theory (up to the point I have studied). Are you assuming the ring is with identity here and x has an inverse?(If so, it isn't worthless but I would like to know of the limited scope). You say x'=ab/x which only makes sense as written with an identity, and x having an inverse. But we can alternatively define x'=ab/x to just mean x'x=ab to make it more general.
            $endgroup$
            – Jacob Wakem
            Mar 19 '14 at 17:03










          • $begingroup$
            @Jacob In any domain, if $,xmid ab,$ then the cofactor $,x',$ such that $,xx' = ab,$ is uniquely defined. You may also find of interest lattice-theoretic viewpoints, e.g. for starters see this Wikipedia article.
            $endgroup$
            – Bill Dubuque
            Mar 19 '14 at 18:25












          • $begingroup$
            See also this answer.
            $endgroup$
            – Bill Dubuque
            Sep 6 '14 at 5:22














          3












          3








          3





          $begingroup$

          A prototypical example is proving $ rm gcd(a,b):lcm(a,b) = ab, $ using the $,overbrace{{rm involution}, x'! =, ab/x}^{rmlarge cofactor duality } $ on the divisors of $rm:ab. $ Notice that $rm xmid ycolor{#c00}iff y'mid x', $ by ${, rmdfrac{y}x = dfrac{x'}{y'} }$ by $rm, yy'! = ab = xx'., $ Thus



          $$begin{align}rm cmidgcd(a,b)!iff&rm cmid a,b\[2px]
          color{#c00}iff& rm a',b'mid c'\[2px]
          iff & rm lcm(a',b')mid c'\
          color{#c00}iff & rm cmid lcm(a',b')'\
          {rm Thus}rmquad gcd(a,b), = &rm , lcm(a',b')'= dfrac{ab}{lcm(b,a)}
          end{align}quad $$



          The black arrows above are the universal property (definition) of gcd and lcm, and the red arrows follow by cofactor duality.



          Noational abuse alert: the gcd, lcm "equalities" are up to unit factors (i.e. "equal" if associate)






          share|cite|improve this answer











          $endgroup$



          A prototypical example is proving $ rm gcd(a,b):lcm(a,b) = ab, $ using the $,overbrace{{rm involution}, x'! =, ab/x}^{rmlarge cofactor duality } $ on the divisors of $rm:ab. $ Notice that $rm xmid ycolor{#c00}iff y'mid x', $ by ${, rmdfrac{y}x = dfrac{x'}{y'} }$ by $rm, yy'! = ab = xx'., $ Thus



          $$begin{align}rm cmidgcd(a,b)!iff&rm cmid a,b\[2px]
          color{#c00}iff& rm a',b'mid c'\[2px]
          iff & rm lcm(a',b')mid c'\
          color{#c00}iff & rm cmid lcm(a',b')'\
          {rm Thus}rmquad gcd(a,b), = &rm , lcm(a',b')'= dfrac{ab}{lcm(b,a)}
          end{align}quad $$



          The black arrows above are the universal property (definition) of gcd and lcm, and the red arrows follow by cofactor duality.



          Noational abuse alert: the gcd, lcm "equalities" are up to unit factors (i.e. "equal" if associate)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 6 '18 at 20:49

























          answered Mar 19 '14 at 4:06









          Bill DubuqueBill Dubuque

          212k29195650




          212k29195650












          • $begingroup$
            Is your "involution" basically a formal division? I know there is no division per se without an identity, but what you wrote seems to make sense anyway as equivalent to x'x=ab.
            $endgroup$
            – Jacob Wakem
            Mar 19 '14 at 16:34










          • $begingroup$
            I don't know what you mean by "formal division" and "division per se without an identity". Please use standard math language. Do you know any ring theory, or only elementary number theory?
            $endgroup$
            – Bill Dubuque
            Mar 19 '14 at 16:48










          • $begingroup$
            I know some ring theory. I have never heard "involution" used in ring theory (up to the point I have studied). Are you assuming the ring is with identity here and x has an inverse?(If so, it isn't worthless but I would like to know of the limited scope). You say x'=ab/x which only makes sense as written with an identity, and x having an inverse. But we can alternatively define x'=ab/x to just mean x'x=ab to make it more general.
            $endgroup$
            – Jacob Wakem
            Mar 19 '14 at 17:03










          • $begingroup$
            @Jacob In any domain, if $,xmid ab,$ then the cofactor $,x',$ such that $,xx' = ab,$ is uniquely defined. You may also find of interest lattice-theoretic viewpoints, e.g. for starters see this Wikipedia article.
            $endgroup$
            – Bill Dubuque
            Mar 19 '14 at 18:25












          • $begingroup$
            See also this answer.
            $endgroup$
            – Bill Dubuque
            Sep 6 '14 at 5:22


















          • $begingroup$
            Is your "involution" basically a formal division? I know there is no division per se without an identity, but what you wrote seems to make sense anyway as equivalent to x'x=ab.
            $endgroup$
            – Jacob Wakem
            Mar 19 '14 at 16:34










          • $begingroup$
            I don't know what you mean by "formal division" and "division per se without an identity". Please use standard math language. Do you know any ring theory, or only elementary number theory?
            $endgroup$
            – Bill Dubuque
            Mar 19 '14 at 16:48










          • $begingroup$
            I know some ring theory. I have never heard "involution" used in ring theory (up to the point I have studied). Are you assuming the ring is with identity here and x has an inverse?(If so, it isn't worthless but I would like to know of the limited scope). You say x'=ab/x which only makes sense as written with an identity, and x having an inverse. But we can alternatively define x'=ab/x to just mean x'x=ab to make it more general.
            $endgroup$
            – Jacob Wakem
            Mar 19 '14 at 17:03










          • $begingroup$
            @Jacob In any domain, if $,xmid ab,$ then the cofactor $,x',$ such that $,xx' = ab,$ is uniquely defined. You may also find of interest lattice-theoretic viewpoints, e.g. for starters see this Wikipedia article.
            $endgroup$
            – Bill Dubuque
            Mar 19 '14 at 18:25












          • $begingroup$
            See also this answer.
            $endgroup$
            – Bill Dubuque
            Sep 6 '14 at 5:22
















          $begingroup$
          Is your "involution" basically a formal division? I know there is no division per se without an identity, but what you wrote seems to make sense anyway as equivalent to x'x=ab.
          $endgroup$
          – Jacob Wakem
          Mar 19 '14 at 16:34




          $begingroup$
          Is your "involution" basically a formal division? I know there is no division per se without an identity, but what you wrote seems to make sense anyway as equivalent to x'x=ab.
          $endgroup$
          – Jacob Wakem
          Mar 19 '14 at 16:34












          $begingroup$
          I don't know what you mean by "formal division" and "division per se without an identity". Please use standard math language. Do you know any ring theory, or only elementary number theory?
          $endgroup$
          – Bill Dubuque
          Mar 19 '14 at 16:48




          $begingroup$
          I don't know what you mean by "formal division" and "division per se without an identity". Please use standard math language. Do you know any ring theory, or only elementary number theory?
          $endgroup$
          – Bill Dubuque
          Mar 19 '14 at 16:48












          $begingroup$
          I know some ring theory. I have never heard "involution" used in ring theory (up to the point I have studied). Are you assuming the ring is with identity here and x has an inverse?(If so, it isn't worthless but I would like to know of the limited scope). You say x'=ab/x which only makes sense as written with an identity, and x having an inverse. But we can alternatively define x'=ab/x to just mean x'x=ab to make it more general.
          $endgroup$
          – Jacob Wakem
          Mar 19 '14 at 17:03




          $begingroup$
          I know some ring theory. I have never heard "involution" used in ring theory (up to the point I have studied). Are you assuming the ring is with identity here and x has an inverse?(If so, it isn't worthless but I would like to know of the limited scope). You say x'=ab/x which only makes sense as written with an identity, and x having an inverse. But we can alternatively define x'=ab/x to just mean x'x=ab to make it more general.
          $endgroup$
          – Jacob Wakem
          Mar 19 '14 at 17:03












          $begingroup$
          @Jacob In any domain, if $,xmid ab,$ then the cofactor $,x',$ such that $,xx' = ab,$ is uniquely defined. You may also find of interest lattice-theoretic viewpoints, e.g. for starters see this Wikipedia article.
          $endgroup$
          – Bill Dubuque
          Mar 19 '14 at 18:25






          $begingroup$
          @Jacob In any domain, if $,xmid ab,$ then the cofactor $,x',$ such that $,xx' = ab,$ is uniquely defined. You may also find of interest lattice-theoretic viewpoints, e.g. for starters see this Wikipedia article.
          $endgroup$
          – Bill Dubuque
          Mar 19 '14 at 18:25














          $begingroup$
          See also this answer.
          $endgroup$
          – Bill Dubuque
          Sep 6 '14 at 5:22




          $begingroup$
          See also this answer.
          $endgroup$
          – Bill Dubuque
          Sep 6 '14 at 5:22


















          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%2f717760%2ftransfer-between-lcm-gcd-for-rings%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

          How to change which sound is reproduced for terminal bell?

          Image
          3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
          Read more

          Can I use Tabulator js library in my java Spring + Thymeleaf project?

          Image
          0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
          Read more

          Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

          Image
          up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
          Read more