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







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    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
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          $begingroup$

          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





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          $begingroup$

          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


















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