Why is there only one way to make $M$ a $mathbb{Q}$-module, if possible?












3












$begingroup$


Suppose $M$ is a $mathbb{Q}$-module. Why is the given action of $mathbb{Q}$ on $M$ (whatever it may be) the only way to make $M$ a $mathbb{Q}$-module?



I know that an abelian group can only be made into a $mathbb{Z}$-module in a unique way, since $1cdot m=m$ for any $min M$. So for $p/qinmathbb{Q}$, $(p/q)cdot m=frac{1}{q}(pm)$. But $pcdot m$ is necessarily the sum of $m$ a total of $p$ times. Also, $frac{1}{q}(qm)=(frac{1}{q}q)cdot m=1cdot m=m$. With this, does it somehow follow that there is only a unique way to make an abelian group $M$ a $mathbb{Q}$-module when it is possible?










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The short answer is "yes."
    $endgroup$
    – Rasmus
    Oct 24 '11 at 7:28
















3












$begingroup$


Suppose $M$ is a $mathbb{Q}$-module. Why is the given action of $mathbb{Q}$ on $M$ (whatever it may be) the only way to make $M$ a $mathbb{Q}$-module?



I know that an abelian group can only be made into a $mathbb{Z}$-module in a unique way, since $1cdot m=m$ for any $min M$. So for $p/qinmathbb{Q}$, $(p/q)cdot m=frac{1}{q}(pm)$. But $pcdot m$ is necessarily the sum of $m$ a total of $p$ times. Also, $frac{1}{q}(qm)=(frac{1}{q}q)cdot m=1cdot m=m$. With this, does it somehow follow that there is only a unique way to make an abelian group $M$ a $mathbb{Q}$-module when it is possible?










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The short answer is "yes."
    $endgroup$
    – Rasmus
    Oct 24 '11 at 7:28














3












3








3


3



$begingroup$


Suppose $M$ is a $mathbb{Q}$-module. Why is the given action of $mathbb{Q}$ on $M$ (whatever it may be) the only way to make $M$ a $mathbb{Q}$-module?



I know that an abelian group can only be made into a $mathbb{Z}$-module in a unique way, since $1cdot m=m$ for any $min M$. So for $p/qinmathbb{Q}$, $(p/q)cdot m=frac{1}{q}(pm)$. But $pcdot m$ is necessarily the sum of $m$ a total of $p$ times. Also, $frac{1}{q}(qm)=(frac{1}{q}q)cdot m=1cdot m=m$. With this, does it somehow follow that there is only a unique way to make an abelian group $M$ a $mathbb{Q}$-module when it is possible?










share|cite|improve this question









$endgroup$




Suppose $M$ is a $mathbb{Q}$-module. Why is the given action of $mathbb{Q}$ on $M$ (whatever it may be) the only way to make $M$ a $mathbb{Q}$-module?



I know that an abelian group can only be made into a $mathbb{Z}$-module in a unique way, since $1cdot m=m$ for any $min M$. So for $p/qinmathbb{Q}$, $(p/q)cdot m=frac{1}{q}(pm)$. But $pcdot m$ is necessarily the sum of $m$ a total of $p$ times. Also, $frac{1}{q}(qm)=(frac{1}{q}q)cdot m=1cdot m=m$. With this, does it somehow follow that there is only a unique way to make an abelian group $M$ a $mathbb{Q}$-module when it is possible?







modules






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Oct 24 '11 at 7:22









CellariusCellarius

161




161








  • 1




    $begingroup$
    The short answer is "yes."
    $endgroup$
    – Rasmus
    Oct 24 '11 at 7:28














  • 1




    $begingroup$
    The short answer is "yes."
    $endgroup$
    – Rasmus
    Oct 24 '11 at 7:28








1




1




$begingroup$
The short answer is "yes."
$endgroup$
– Rasmus
Oct 24 '11 at 7:28




$begingroup$
The short answer is "yes."
$endgroup$
– Rasmus
Oct 24 '11 at 7:28










3 Answers
3






active

oldest

votes


















6












$begingroup$

Let $M$ be a $mathbb{Q}$-module. Then the additive group $(M,+)$ is torsion-free, because $nm=0$ for some $ninmathbb{N}$ implies $m=frac{1}{n}nm=0$.



In a torsion-free abelian group $G$ an equation of the form $nx=g$, $ninmathbb{N}$, $gin G$ has at most one solution.



In the $mathbb{Q}$-module $M$ the element $frac{1}{n}m$ thus is the unique solution of the equation $nx=m$, $min M$. Note that this holds whatever the operation of $mathbb{Q}$ on $M$ is like, while the equation $nx=m$ does only depend on the operation of $mathbb{Z}$ on $M$, which is unique as we already know.



Hence in your case the equation $ qx=pm $ has at most one solution $ frac{1}{q}(pm) $ -- and it has one.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Let $M$ be an abelian group, then



    $M$ is a $mathbb{Q}-$module iff $M$ is torsion free and divisible.



    the $mathbb{Q}-$module structure is $(p/q).m=y$ where $p.m = q.(y)$ such a $y$ exists by divisbility requirement and can be checked to be well defined by torsion freeness.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      This (old) question already has a nice, explicit answer, but I thought I might offer an alternative perspective for future visitors.



      Let $R$ be a commutative ring. To give an $R$-module structure on an abelian group $M$ is equivalent to giving a ring homomorphism $R to mathrm{End}_{mathbb{Z}}(M)$. This gives another explanation why an abelian group $M$ can be given exactly one $mathbb{Z}$-module structure: $mathbb{Z}$ is initial in the category of rings, so there is a unique homomorphism $mathbb{Z} to mathrm{End}_{mathbb{Z}}(M)$.



      Now, let $R$ be a ring, and let $S subset R$ be a multiplicative subset, and let $i colon R to S^{-1}R$ be the canonical localization map. Recall that the universal property of localization says that given a ring morphism $alpha colon R to T$ with $alpha(S) subset T^{times}$, there is a unique map $beta colon S^{-1}R to T$ such that $beta circ i = alpha$.



      Note that $mathbb{Q}$ is the localization of $mathbb{Z}$ at the multiplicative subset $S = mathbb{Z} setminus {0}$. Again, since $mathbb{Z}$ is initial in the category of rings, there is exactly one ring morphism $mathbb{Z} to T$ for any ring $T$, so by the universal property of localization, there is at most one ring homomorphism $mathbb{Q} to T$, which exists if and only if the image of $mathbb{Z}$ in $T$ lies in $T^{times}$. Since $mathbb{Q}$-module structures on an abelian group $M$ are in bijection with ring morphisms $mathbb{Q} to mathrm{End}_{mathbb{Z}}(M)$, this shows the desired result.






      share|cite|improve this answer











      $endgroup$














        Your Answer








        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%2f75313%2fwhy-is-there-only-one-way-to-make-m-a-mathbbq-module-if-possible%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        6












        $begingroup$

        Let $M$ be a $mathbb{Q}$-module. Then the additive group $(M,+)$ is torsion-free, because $nm=0$ for some $ninmathbb{N}$ implies $m=frac{1}{n}nm=0$.



        In a torsion-free abelian group $G$ an equation of the form $nx=g$, $ninmathbb{N}$, $gin G$ has at most one solution.



        In the $mathbb{Q}$-module $M$ the element $frac{1}{n}m$ thus is the unique solution of the equation $nx=m$, $min M$. Note that this holds whatever the operation of $mathbb{Q}$ on $M$ is like, while the equation $nx=m$ does only depend on the operation of $mathbb{Z}$ on $M$, which is unique as we already know.



        Hence in your case the equation $ qx=pm $ has at most one solution $ frac{1}{q}(pm) $ -- and it has one.






        share|cite|improve this answer











        $endgroup$


















          6












          $begingroup$

          Let $M$ be a $mathbb{Q}$-module. Then the additive group $(M,+)$ is torsion-free, because $nm=0$ for some $ninmathbb{N}$ implies $m=frac{1}{n}nm=0$.



          In a torsion-free abelian group $G$ an equation of the form $nx=g$, $ninmathbb{N}$, $gin G$ has at most one solution.



          In the $mathbb{Q}$-module $M$ the element $frac{1}{n}m$ thus is the unique solution of the equation $nx=m$, $min M$. Note that this holds whatever the operation of $mathbb{Q}$ on $M$ is like, while the equation $nx=m$ does only depend on the operation of $mathbb{Z}$ on $M$, which is unique as we already know.



          Hence in your case the equation $ qx=pm $ has at most one solution $ frac{1}{q}(pm) $ -- and it has one.






          share|cite|improve this answer











          $endgroup$
















            6












            6








            6





            $begingroup$

            Let $M$ be a $mathbb{Q}$-module. Then the additive group $(M,+)$ is torsion-free, because $nm=0$ for some $ninmathbb{N}$ implies $m=frac{1}{n}nm=0$.



            In a torsion-free abelian group $G$ an equation of the form $nx=g$, $ninmathbb{N}$, $gin G$ has at most one solution.



            In the $mathbb{Q}$-module $M$ the element $frac{1}{n}m$ thus is the unique solution of the equation $nx=m$, $min M$. Note that this holds whatever the operation of $mathbb{Q}$ on $M$ is like, while the equation $nx=m$ does only depend on the operation of $mathbb{Z}$ on $M$, which is unique as we already know.



            Hence in your case the equation $ qx=pm $ has at most one solution $ frac{1}{q}(pm) $ -- and it has one.






            share|cite|improve this answer











            $endgroup$



            Let $M$ be a $mathbb{Q}$-module. Then the additive group $(M,+)$ is torsion-free, because $nm=0$ for some $ninmathbb{N}$ implies $m=frac{1}{n}nm=0$.



            In a torsion-free abelian group $G$ an equation of the form $nx=g$, $ninmathbb{N}$, $gin G$ has at most one solution.



            In the $mathbb{Q}$-module $M$ the element $frac{1}{n}m$ thus is the unique solution of the equation $nx=m$, $min M$. Note that this holds whatever the operation of $mathbb{Q}$ on $M$ is like, while the equation $nx=m$ does only depend on the operation of $mathbb{Z}$ on $M$, which is unique as we already know.



            Hence in your case the equation $ qx=pm $ has at most one solution $ frac{1}{q}(pm) $ -- and it has one.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 19 at 2:31









            user549397

            1,7141618




            1,7141618










            answered Oct 24 '11 at 8:10









            Hagen KnafHagen Knaf

            6,9521318




            6,9521318























                0












                $begingroup$

                Let $M$ be an abelian group, then



                $M$ is a $mathbb{Q}-$module iff $M$ is torsion free and divisible.



                the $mathbb{Q}-$module structure is $(p/q).m=y$ where $p.m = q.(y)$ such a $y$ exists by divisbility requirement and can be checked to be well defined by torsion freeness.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  Let $M$ be an abelian group, then



                  $M$ is a $mathbb{Q}-$module iff $M$ is torsion free and divisible.



                  the $mathbb{Q}-$module structure is $(p/q).m=y$ where $p.m = q.(y)$ such a $y$ exists by divisbility requirement and can be checked to be well defined by torsion freeness.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    Let $M$ be an abelian group, then



                    $M$ is a $mathbb{Q}-$module iff $M$ is torsion free and divisible.



                    the $mathbb{Q}-$module structure is $(p/q).m=y$ where $p.m = q.(y)$ such a $y$ exists by divisbility requirement and can be checked to be well defined by torsion freeness.






                    share|cite|improve this answer









                    $endgroup$



                    Let $M$ be an abelian group, then



                    $M$ is a $mathbb{Q}-$module iff $M$ is torsion free and divisible.



                    the $mathbb{Q}-$module structure is $(p/q).m=y$ where $p.m = q.(y)$ such a $y$ exists by divisbility requirement and can be checked to be well defined by torsion freeness.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jul 7 '14 at 11:59









                    user2902293user2902293

                    1,119820




                    1,119820























                        0












                        $begingroup$

                        This (old) question already has a nice, explicit answer, but I thought I might offer an alternative perspective for future visitors.



                        Let $R$ be a commutative ring. To give an $R$-module structure on an abelian group $M$ is equivalent to giving a ring homomorphism $R to mathrm{End}_{mathbb{Z}}(M)$. This gives another explanation why an abelian group $M$ can be given exactly one $mathbb{Z}$-module structure: $mathbb{Z}$ is initial in the category of rings, so there is a unique homomorphism $mathbb{Z} to mathrm{End}_{mathbb{Z}}(M)$.



                        Now, let $R$ be a ring, and let $S subset R$ be a multiplicative subset, and let $i colon R to S^{-1}R$ be the canonical localization map. Recall that the universal property of localization says that given a ring morphism $alpha colon R to T$ with $alpha(S) subset T^{times}$, there is a unique map $beta colon S^{-1}R to T$ such that $beta circ i = alpha$.



                        Note that $mathbb{Q}$ is the localization of $mathbb{Z}$ at the multiplicative subset $S = mathbb{Z} setminus {0}$. Again, since $mathbb{Z}$ is initial in the category of rings, there is exactly one ring morphism $mathbb{Z} to T$ for any ring $T$, so by the universal property of localization, there is at most one ring homomorphism $mathbb{Q} to T$, which exists if and only if the image of $mathbb{Z}$ in $T$ lies in $T^{times}$. Since $mathbb{Q}$-module structures on an abelian group $M$ are in bijection with ring morphisms $mathbb{Q} to mathrm{End}_{mathbb{Z}}(M)$, this shows the desired result.






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          This (old) question already has a nice, explicit answer, but I thought I might offer an alternative perspective for future visitors.



                          Let $R$ be a commutative ring. To give an $R$-module structure on an abelian group $M$ is equivalent to giving a ring homomorphism $R to mathrm{End}_{mathbb{Z}}(M)$. This gives another explanation why an abelian group $M$ can be given exactly one $mathbb{Z}$-module structure: $mathbb{Z}$ is initial in the category of rings, so there is a unique homomorphism $mathbb{Z} to mathrm{End}_{mathbb{Z}}(M)$.



                          Now, let $R$ be a ring, and let $S subset R$ be a multiplicative subset, and let $i colon R to S^{-1}R$ be the canonical localization map. Recall that the universal property of localization says that given a ring morphism $alpha colon R to T$ with $alpha(S) subset T^{times}$, there is a unique map $beta colon S^{-1}R to T$ such that $beta circ i = alpha$.



                          Note that $mathbb{Q}$ is the localization of $mathbb{Z}$ at the multiplicative subset $S = mathbb{Z} setminus {0}$. Again, since $mathbb{Z}$ is initial in the category of rings, there is exactly one ring morphism $mathbb{Z} to T$ for any ring $T$, so by the universal property of localization, there is at most one ring homomorphism $mathbb{Q} to T$, which exists if and only if the image of $mathbb{Z}$ in $T$ lies in $T^{times}$. Since $mathbb{Q}$-module structures on an abelian group $M$ are in bijection with ring morphisms $mathbb{Q} to mathrm{End}_{mathbb{Z}}(M)$, this shows the desired result.






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            This (old) question already has a nice, explicit answer, but I thought I might offer an alternative perspective for future visitors.



                            Let $R$ be a commutative ring. To give an $R$-module structure on an abelian group $M$ is equivalent to giving a ring homomorphism $R to mathrm{End}_{mathbb{Z}}(M)$. This gives another explanation why an abelian group $M$ can be given exactly one $mathbb{Z}$-module structure: $mathbb{Z}$ is initial in the category of rings, so there is a unique homomorphism $mathbb{Z} to mathrm{End}_{mathbb{Z}}(M)$.



                            Now, let $R$ be a ring, and let $S subset R$ be a multiplicative subset, and let $i colon R to S^{-1}R$ be the canonical localization map. Recall that the universal property of localization says that given a ring morphism $alpha colon R to T$ with $alpha(S) subset T^{times}$, there is a unique map $beta colon S^{-1}R to T$ such that $beta circ i = alpha$.



                            Note that $mathbb{Q}$ is the localization of $mathbb{Z}$ at the multiplicative subset $S = mathbb{Z} setminus {0}$. Again, since $mathbb{Z}$ is initial in the category of rings, there is exactly one ring morphism $mathbb{Z} to T$ for any ring $T$, so by the universal property of localization, there is at most one ring homomorphism $mathbb{Q} to T$, which exists if and only if the image of $mathbb{Z}$ in $T$ lies in $T^{times}$. Since $mathbb{Q}$-module structures on an abelian group $M$ are in bijection with ring morphisms $mathbb{Q} to mathrm{End}_{mathbb{Z}}(M)$, this shows the desired result.






                            share|cite|improve this answer











                            $endgroup$



                            This (old) question already has a nice, explicit answer, but I thought I might offer an alternative perspective for future visitors.



                            Let $R$ be a commutative ring. To give an $R$-module structure on an abelian group $M$ is equivalent to giving a ring homomorphism $R to mathrm{End}_{mathbb{Z}}(M)$. This gives another explanation why an abelian group $M$ can be given exactly one $mathbb{Z}$-module structure: $mathbb{Z}$ is initial in the category of rings, so there is a unique homomorphism $mathbb{Z} to mathrm{End}_{mathbb{Z}}(M)$.



                            Now, let $R$ be a ring, and let $S subset R$ be a multiplicative subset, and let $i colon R to S^{-1}R$ be the canonical localization map. Recall that the universal property of localization says that given a ring morphism $alpha colon R to T$ with $alpha(S) subset T^{times}$, there is a unique map $beta colon S^{-1}R to T$ such that $beta circ i = alpha$.



                            Note that $mathbb{Q}$ is the localization of $mathbb{Z}$ at the multiplicative subset $S = mathbb{Z} setminus {0}$. Again, since $mathbb{Z}$ is initial in the category of rings, there is exactly one ring morphism $mathbb{Z} to T$ for any ring $T$, so by the universal property of localization, there is at most one ring homomorphism $mathbb{Q} to T$, which exists if and only if the image of $mathbb{Z}$ in $T$ lies in $T^{times}$. Since $mathbb{Q}$-module structures on an abelian group $M$ are in bijection with ring morphisms $mathbb{Q} to mathrm{End}_{mathbb{Z}}(M)$, this shows the desired result.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jan 19 at 7:39

























                            answered Jan 19 at 3:22









                            Alex WertheimAlex Wertheim

                            16.3k22848




                            16.3k22848






























                                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%2f75313%2fwhy-is-there-only-one-way-to-make-m-a-mathbbq-module-if-possible%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

                                Human spaceflight

                                Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

                                張江高科駅