How do we show that the set of module homomorphisms is a commutative group?












0












$begingroup$


Assume $A$ is a commutative ring with 1, and further let $M, N$ be its modules. Then, the set, S, of all homomorphisms from $M$ to $N$ is an $A-$module as well where operations are: For $f_1, f_2 in S$, $(f_1+f_2)(x) = f_1(x)+f_2(x), (af_1)(x)=af_1(x)$ for any $a in A$.



From the definition of a module, $S$ has to be an abelian group. I can show that the said set is an $A-$module, only if I start from the assumption that the set is a group in the first place. Showing existence of unity, associativity, and commutativity is not an issue. I just do not see how I can prove the existence of inverses since the elements are homomorphisms.






abstract-algebra abelian-groups







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$endgroup$








  • 2




    $begingroup$
    Suppose you want $g$ to be an inverse of $f$. Then what does $g(x)$ need to be?
    $endgroup$
    – Eric Wofsey
    Jan 15 at 7:31
















0












$begingroup$


Assume $A$ is a commutative ring with 1, and further let $M, N$ be its modules. Then, the set, S, of all homomorphisms from $M$ to $N$ is an $A-$module as well where operations are: For $f_1, f_2 in S$, $(f_1+f_2)(x) = f_1(x)+f_2(x), (af_1)(x)=af_1(x)$ for any $a in A$.



From the definition of a module, $S$ has to be an abelian group. I can show that the said set is an $A-$module, only if I start from the assumption that the set is a group in the first place. Showing existence of unity, associativity, and commutativity is not an issue. I just do not see how I can prove the existence of inverses since the elements are homomorphisms.






abstract-algebra abelian-groups







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Suppose you want $g$ to be an inverse of $f$. Then what does $g(x)$ need to be?
    $endgroup$
    – Eric Wofsey
    Jan 15 at 7:31














0












0








0





$begingroup$


Assume $A$ is a commutative ring with 1, and further let $M, N$ be its modules. Then, the set, S, of all homomorphisms from $M$ to $N$ is an $A-$module as well where operations are: For $f_1, f_2 in S$, $(f_1+f_2)(x) = f_1(x)+f_2(x), (af_1)(x)=af_1(x)$ for any $a in A$.



From the definition of a module, $S$ has to be an abelian group. I can show that the said set is an $A-$module, only if I start from the assumption that the set is a group in the first place. Showing existence of unity, associativity, and commutativity is not an issue. I just do not see how I can prove the existence of inverses since the elements are homomorphisms.






abstract-algebra abelian-groups







share|cite|improve this question











$endgroup$




Assume $A$ is a commutative ring with 1, and further let $M, N$ be its modules. Then, the set, S, of all homomorphisms from $M$ to $N$ is an $A-$module as well where operations are: For $f_1, f_2 in S$, $(f_1+f_2)(x) = f_1(x)+f_2(x), (af_1)(x)=af_1(x)$ for any $a in A$.



From the definition of a module, $S$ has to be an abelian group. I can show that the said set is an $A-$module, only if I start from the assumption that the set is a group in the first place. Showing existence of unity, associativity, and commutativity is not an issue. I just do not see how I can prove the existence of inverses since the elements are homomorphisms.






abstract-algebra abelian-groups




abstract-algebra abelian-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 20:51









user26857

39.4k124183




39.4k124183










asked Jan 15 at 7:24









NawajNawaj

283




283








  • 2




    $begingroup$
    Suppose you want $g$ to be an inverse of $f$. Then what does $g(x)$ need to be?
    $endgroup$
    – Eric Wofsey
    Jan 15 at 7:31














  • 2




    $begingroup$
    Suppose you want $g$ to be an inverse of $f$. Then what does $g(x)$ need to be?
    $endgroup$
    – Eric Wofsey
    Jan 15 at 7:31








2




2




$begingroup$
Suppose you want $g$ to be an inverse of $f$. Then what does $g(x)$ need to be?
$endgroup$
– Eric Wofsey
Jan 15 at 7:31




$begingroup$
Suppose you want $g$ to be an inverse of $f$. Then what does $g(x)$ need to be?
$endgroup$
– Eric Wofsey
Jan 15 at 7:31










1 Answer
1






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oldest

votes


















0












$begingroup$

So you only have problems with inverses? Then here you go:



first of all: $mathrm{Mod}_A(M,N)subset mathrm{Set}(M,N)$, where $mathrm{Mod}_A(M,N)$ denotes the morphisms in the category of $A$-modules and $mathrm{Set}(M,N)$ denotes only maps. Now since $N$ is an $A$ module, the set $mathrm{Set}(M,N)$ carries the structure of an $A$-module, and we will show that $mathrm{Mod}_A(M,N)$ is a submodule. To prove this we need to show that it contains the neutral element, additive inverses, is closed under addition and scalar multiplication:



neutral element:



$0:M to N$ is clearly an $A$-homomorphism, hence $mathrm{Mod}_A(M,N)$ contains the neutral element.



additive inverses: let $f:M to N$ be an $A$-homomorphism, then we have that, since $f$ is a groupmorphism that $-f$ is a groupmorphism, so we only need to check that $-f$ is $A$-linear. but this is clear since: $-f(ax)=-(f(ax))=-(af(x))=a(-f(x))$.



additive closed: let $f,g: Mto N$ then we know again since both are morphisms of commutative groups that $f+g$ is a group homomorphism. Now check again: $$f+g(ax)=f(ax)+g(ax)=af(x)+ag(x)=a(f(x)+g(x))= a(f+g)(x)$$



closed under scalar multiplication: let $f:M to N$ then compute again $$af(x+a'y)=a(f(x) + f(a'y))=af(x)+aa'f(y)=af(x) + a'af(x)$$
which precisely means that $af$ is $A$ linear.



Please forgive that I skipped all the additive calcultions, but they are completely analogous. And also observe that oneactually would not need additive inverses, since they come for free from scalar multiplication as $-1 in A$.






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    $begingroup$

    So you only have problems with inverses? Then here you go:



    first of all: $mathrm{Mod}_A(M,N)subset mathrm{Set}(M,N)$, where $mathrm{Mod}_A(M,N)$ denotes the morphisms in the category of $A$-modules and $mathrm{Set}(M,N)$ denotes only maps. Now since $N$ is an $A$ module, the set $mathrm{Set}(M,N)$ carries the structure of an $A$-module, and we will show that $mathrm{Mod}_A(M,N)$ is a submodule. To prove this we need to show that it contains the neutral element, additive inverses, is closed under addition and scalar multiplication:



    neutral element:



    $0:M to N$ is clearly an $A$-homomorphism, hence $mathrm{Mod}_A(M,N)$ contains the neutral element.



    additive inverses: let $f:M to N$ be an $A$-homomorphism, then we have that, since $f$ is a groupmorphism that $-f$ is a groupmorphism, so we only need to check that $-f$ is $A$-linear. but this is clear since: $-f(ax)=-(f(ax))=-(af(x))=a(-f(x))$.



    additive closed: let $f,g: Mto N$ then we know again since both are morphisms of commutative groups that $f+g$ is a group homomorphism. Now check again: $$f+g(ax)=f(ax)+g(ax)=af(x)+ag(x)=a(f(x)+g(x))= a(f+g)(x)$$



    closed under scalar multiplication: let $f:M to N$ then compute again $$af(x+a'y)=a(f(x) + f(a'y))=af(x)+aa'f(y)=af(x) + a'af(x)$$
    which precisely means that $af$ is $A$ linear.



    Please forgive that I skipped all the additive calcultions, but they are completely analogous. And also observe that oneactually would not need additive inverses, since they come for free from scalar multiplication as $-1 in A$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      So you only have problems with inverses? Then here you go:



      first of all: $mathrm{Mod}_A(M,N)subset mathrm{Set}(M,N)$, where $mathrm{Mod}_A(M,N)$ denotes the morphisms in the category of $A$-modules and $mathrm{Set}(M,N)$ denotes only maps. Now since $N$ is an $A$ module, the set $mathrm{Set}(M,N)$ carries the structure of an $A$-module, and we will show that $mathrm{Mod}_A(M,N)$ is a submodule. To prove this we need to show that it contains the neutral element, additive inverses, is closed under addition and scalar multiplication:



      neutral element:



      $0:M to N$ is clearly an $A$-homomorphism, hence $mathrm{Mod}_A(M,N)$ contains the neutral element.



      additive inverses: let $f:M to N$ be an $A$-homomorphism, then we have that, since $f$ is a groupmorphism that $-f$ is a groupmorphism, so we only need to check that $-f$ is $A$-linear. but this is clear since: $-f(ax)=-(f(ax))=-(af(x))=a(-f(x))$.



      additive closed: let $f,g: Mto N$ then we know again since both are morphisms of commutative groups that $f+g$ is a group homomorphism. Now check again: $$f+g(ax)=f(ax)+g(ax)=af(x)+ag(x)=a(f(x)+g(x))= a(f+g)(x)$$



      closed under scalar multiplication: let $f:M to N$ then compute again $$af(x+a'y)=a(f(x) + f(a'y))=af(x)+aa'f(y)=af(x) + a'af(x)$$
      which precisely means that $af$ is $A$ linear.



      Please forgive that I skipped all the additive calcultions, but they are completely analogous. And also observe that oneactually would not need additive inverses, since they come for free from scalar multiplication as $-1 in A$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        So you only have problems with inverses? Then here you go:



        first of all: $mathrm{Mod}_A(M,N)subset mathrm{Set}(M,N)$, where $mathrm{Mod}_A(M,N)$ denotes the morphisms in the category of $A$-modules and $mathrm{Set}(M,N)$ denotes only maps. Now since $N$ is an $A$ module, the set $mathrm{Set}(M,N)$ carries the structure of an $A$-module, and we will show that $mathrm{Mod}_A(M,N)$ is a submodule. To prove this we need to show that it contains the neutral element, additive inverses, is closed under addition and scalar multiplication:



        neutral element:



        $0:M to N$ is clearly an $A$-homomorphism, hence $mathrm{Mod}_A(M,N)$ contains the neutral element.



        additive inverses: let $f:M to N$ be an $A$-homomorphism, then we have that, since $f$ is a groupmorphism that $-f$ is a groupmorphism, so we only need to check that $-f$ is $A$-linear. but this is clear since: $-f(ax)=-(f(ax))=-(af(x))=a(-f(x))$.



        additive closed: let $f,g: Mto N$ then we know again since both are morphisms of commutative groups that $f+g$ is a group homomorphism. Now check again: $$f+g(ax)=f(ax)+g(ax)=af(x)+ag(x)=a(f(x)+g(x))= a(f+g)(x)$$



        closed under scalar multiplication: let $f:M to N$ then compute again $$af(x+a'y)=a(f(x) + f(a'y))=af(x)+aa'f(y)=af(x) + a'af(x)$$
        which precisely means that $af$ is $A$ linear.



        Please forgive that I skipped all the additive calcultions, but they are completely analogous. And also observe that oneactually would not need additive inverses, since they come for free from scalar multiplication as $-1 in A$.






        share|cite|improve this answer









        $endgroup$



        So you only have problems with inverses? Then here you go:



        first of all: $mathrm{Mod}_A(M,N)subset mathrm{Set}(M,N)$, where $mathrm{Mod}_A(M,N)$ denotes the morphisms in the category of $A$-modules and $mathrm{Set}(M,N)$ denotes only maps. Now since $N$ is an $A$ module, the set $mathrm{Set}(M,N)$ carries the structure of an $A$-module, and we will show that $mathrm{Mod}_A(M,N)$ is a submodule. To prove this we need to show that it contains the neutral element, additive inverses, is closed under addition and scalar multiplication:



        neutral element:



        $0:M to N$ is clearly an $A$-homomorphism, hence $mathrm{Mod}_A(M,N)$ contains the neutral element.



        additive inverses: let $f:M to N$ be an $A$-homomorphism, then we have that, since $f$ is a groupmorphism that $-f$ is a groupmorphism, so we only need to check that $-f$ is $A$-linear. but this is clear since: $-f(ax)=-(f(ax))=-(af(x))=a(-f(x))$.



        additive closed: let $f,g: Mto N$ then we know again since both are morphisms of commutative groups that $f+g$ is a group homomorphism. Now check again: $$f+g(ax)=f(ax)+g(ax)=af(x)+ag(x)=a(f(x)+g(x))= a(f+g)(x)$$



        closed under scalar multiplication: let $f:M to N$ then compute again $$af(x+a'y)=a(f(x) + f(a'y))=af(x)+aa'f(y)=af(x) + a'af(x)$$
        which precisely means that $af$ is $A$ linear.



        Please forgive that I skipped all the additive calcultions, but they are completely analogous. And also observe that oneactually would not need additive inverses, since they come for free from scalar multiplication as $-1 in A$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 15 at 8:22









        EnkiduEnkidu

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