Equivalent definitions of $X$ being a generator for the category $Rtext-mathbfmod$

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I'm in trouble with the definition of generator of $Rtext-mathbfmod$ category of left $Rtext-mathbfmod$. I've found a lot of equivalent definitions. I'm using the one of Jacobson's Basic Algebra, so $X$ is a generator if every module $M$ is sum of submodules all of which are homomorphic images of $X$.



Does it mean that these submodules are images of $X$ via homomorphisms?



Then I've proved that the following are equivalent:



1) $X$ is a generator;



2) the functor $operatornamehom(X, -)$ is faithful;



3) $T(X)=R$ where $T(X) = sum_h in operatornamehom(X,R) h(X)$;



4) $R$ is a homomorphic image of $X^n$ for some $n$.



Now I need to use the fact that $X$ is a generator if and only if every left $R$-module $M$ is an epimorphic image of some direct sum $bigoplus_i X$.
Is this fact only a different way to write the definition I've used? I don't know what an epimorphic image is and I'm confused about this sentence.



If anybody can help me this will be very appreciated! Thanks!







share|cite|improve this question


























    up vote
    1
    down vote

    favorite












    I'm in trouble with the definition of generator of $Rtext-mathbfmod$ category of left $Rtext-mathbfmod$. I've found a lot of equivalent definitions. I'm using the one of Jacobson's Basic Algebra, so $X$ is a generator if every module $M$ is sum of submodules all of which are homomorphic images of $X$.



    Does it mean that these submodules are images of $X$ via homomorphisms?



    Then I've proved that the following are equivalent:



    1) $X$ is a generator;



    2) the functor $operatornamehom(X, -)$ is faithful;



    3) $T(X)=R$ where $T(X) = sum_h in operatornamehom(X,R) h(X)$;



    4) $R$ is a homomorphic image of $X^n$ for some $n$.



    Now I need to use the fact that $X$ is a generator if and only if every left $R$-module $M$ is an epimorphic image of some direct sum $bigoplus_i X$.
    Is this fact only a different way to write the definition I've used? I don't know what an epimorphic image is and I'm confused about this sentence.



    If anybody can help me this will be very appreciated! Thanks!







    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I'm in trouble with the definition of generator of $Rtext-mathbfmod$ category of left $Rtext-mathbfmod$. I've found a lot of equivalent definitions. I'm using the one of Jacobson's Basic Algebra, so $X$ is a generator if every module $M$ is sum of submodules all of which are homomorphic images of $X$.



      Does it mean that these submodules are images of $X$ via homomorphisms?



      Then I've proved that the following are equivalent:



      1) $X$ is a generator;



      2) the functor $operatornamehom(X, -)$ is faithful;



      3) $T(X)=R$ where $T(X) = sum_h in operatornamehom(X,R) h(X)$;



      4) $R$ is a homomorphic image of $X^n$ for some $n$.



      Now I need to use the fact that $X$ is a generator if and only if every left $R$-module $M$ is an epimorphic image of some direct sum $bigoplus_i X$.
      Is this fact only a different way to write the definition I've used? I don't know what an epimorphic image is and I'm confused about this sentence.



      If anybody can help me this will be very appreciated! Thanks!







      share|cite|improve this question














      I'm in trouble with the definition of generator of $Rtext-mathbfmod$ category of left $Rtext-mathbfmod$. I've found a lot of equivalent definitions. I'm using the one of Jacobson's Basic Algebra, so $X$ is a generator if every module $M$ is sum of submodules all of which are homomorphic images of $X$.



      Does it mean that these submodules are images of $X$ via homomorphisms?



      Then I've proved that the following are equivalent:



      1) $X$ is a generator;



      2) the functor $operatornamehom(X, -)$ is faithful;



      3) $T(X)=R$ where $T(X) = sum_h in operatornamehom(X,R) h(X)$;



      4) $R$ is a homomorphic image of $X^n$ for some $n$.



      Now I need to use the fact that $X$ is a generator if and only if every left $R$-module $M$ is an epimorphic image of some direct sum $bigoplus_i X$.
      Is this fact only a different way to write the definition I've used? I don't know what an epimorphic image is and I'm confused about this sentence.



      If anybody can help me this will be very appreciated! Thanks!









      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 28 at 16:56









      Jendrik Stelzner

      7,63121037




      7,63121037










      asked Aug 28 at 16:30









      robbis

      528




      528




















          1 Answer
          1






          active

          oldest

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          up vote
          1
          down vote



          accepted










          If $R$ is a unital ring, you know that every module $M$ can be described as a homomorphic image of some free module $F$; since
          $$
          F cong bigoplus_i in IR
          $$
          for some index set $I$, you are pretty much done. Explicitly, let $M$ be a left $R$-module and let $F$ be a free module chosen such that you can realize $M$ as the homomorphic image of some map $phi:F to M$, i.e., $operatornameImage(phi) cong M$. Then since there exists an $n in mathbbN$ such that $R$ is the homomorphic image of $X^n$, you get that
          $$
          F cong bigoplus_i in I R cong bigoplus_i in I X^n cong bigoplus_i in Ileft(bigoplus_k=1^nXright) cong bigoplus_j in J X
          $$
          where $I$ is some index set and $J$ is the evident new index set given by reindexing the double direct sum. Call the isomorphism from $F$ to this direct sum above $alpha$, i.e.,
          $$
          alpha:bigoplus_j in J X to F.
          $$
          Then pre-composing $phi$ with $alpha$ gives a surjection
          $$
          (phicirc alpha):bigoplus_j in J X to M
          $$
          where
          $$
          operatornameImage(phi circ alpha) = operatornameImage(phi) = M
          $$
          because $alpha$ is an isomorphism and $phi$ is a surjection.






          share|cite|improve this answer




















          • Your answer is very clear, thanks! So epimorphic image only means that there is a surjection (an epimorphism) such that $M$ is image via this epimorphism?
            – robbis
            Aug 29 at 7:30










          • sorry but I still have some questions if you can answer. Is being a generator equivalent to say that there is a split epimorphism $G^n to R to 0$? and from this how can I see that $G$ is a generator if and only if $G^n$ is the direct sum of $R$ and $R'$ for some module $R'$? Thanks again
            – robbis
            Aug 29 at 9:17











          • @robbis Well a morphism in the category of left $R$-modules is epic if and only if it is surjective, so in this case it means that $M$ is the image of an epimorphism out of some domain you've chosen prior. Be careful in general, as there are categories in which epimorphisms are not surjective on underlying sets, but in module categories you're okay. Also if you like my answer, please consider upvoting it and accepting it! I'll answer your second comment in a second from here.
            – Geoff
            Aug 29 at 14:57






          • 1




            I used the fact that $R$ is free so is projective and I can find that map $s$. Now I consider the exact sequence $0 to ker(e) to G^n to R to 0$ and I apply the splitting lemma (if I'm not wrong, this is related to Rank Nullity Theorem) and I can say that $G^n= R bigoplus ker(e)$
            – robbis
            Aug 29 at 15:31






          • 1




            Thanks, now it's all clear to me! I'll upvote and accept surely! Thanks :)
            – robbis
            Aug 29 at 16:13










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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          If $R$ is a unital ring, you know that every module $M$ can be described as a homomorphic image of some free module $F$; since
          $$
          F cong bigoplus_i in IR
          $$
          for some index set $I$, you are pretty much done. Explicitly, let $M$ be a left $R$-module and let $F$ be a free module chosen such that you can realize $M$ as the homomorphic image of some map $phi:F to M$, i.e., $operatornameImage(phi) cong M$. Then since there exists an $n in mathbbN$ such that $R$ is the homomorphic image of $X^n$, you get that
          $$
          F cong bigoplus_i in I R cong bigoplus_i in I X^n cong bigoplus_i in Ileft(bigoplus_k=1^nXright) cong bigoplus_j in J X
          $$
          where $I$ is some index set and $J$ is the evident new index set given by reindexing the double direct sum. Call the isomorphism from $F$ to this direct sum above $alpha$, i.e.,
          $$
          alpha:bigoplus_j in J X to F.
          $$
          Then pre-composing $phi$ with $alpha$ gives a surjection
          $$
          (phicirc alpha):bigoplus_j in J X to M
          $$
          where
          $$
          operatornameImage(phi circ alpha) = operatornameImage(phi) = M
          $$
          because $alpha$ is an isomorphism and $phi$ is a surjection.






          share|cite|improve this answer




















          • Your answer is very clear, thanks! So epimorphic image only means that there is a surjection (an epimorphism) such that $M$ is image via this epimorphism?
            – robbis
            Aug 29 at 7:30










          • sorry but I still have some questions if you can answer. Is being a generator equivalent to say that there is a split epimorphism $G^n to R to 0$? and from this how can I see that $G$ is a generator if and only if $G^n$ is the direct sum of $R$ and $R'$ for some module $R'$? Thanks again
            – robbis
            Aug 29 at 9:17











          • @robbis Well a morphism in the category of left $R$-modules is epic if and only if it is surjective, so in this case it means that $M$ is the image of an epimorphism out of some domain you've chosen prior. Be careful in general, as there are categories in which epimorphisms are not surjective on underlying sets, but in module categories you're okay. Also if you like my answer, please consider upvoting it and accepting it! I'll answer your second comment in a second from here.
            – Geoff
            Aug 29 at 14:57






          • 1




            I used the fact that $R$ is free so is projective and I can find that map $s$. Now I consider the exact sequence $0 to ker(e) to G^n to R to 0$ and I apply the splitting lemma (if I'm not wrong, this is related to Rank Nullity Theorem) and I can say that $G^n= R bigoplus ker(e)$
            – robbis
            Aug 29 at 15:31






          • 1




            Thanks, now it's all clear to me! I'll upvote and accept surely! Thanks :)
            – robbis
            Aug 29 at 16:13














          up vote
          1
          down vote



          accepted










          If $R$ is a unital ring, you know that every module $M$ can be described as a homomorphic image of some free module $F$; since
          $$
          F cong bigoplus_i in IR
          $$
          for some index set $I$, you are pretty much done. Explicitly, let $M$ be a left $R$-module and let $F$ be a free module chosen such that you can realize $M$ as the homomorphic image of some map $phi:F to M$, i.e., $operatornameImage(phi) cong M$. Then since there exists an $n in mathbbN$ such that $R$ is the homomorphic image of $X^n$, you get that
          $$
          F cong bigoplus_i in I R cong bigoplus_i in I X^n cong bigoplus_i in Ileft(bigoplus_k=1^nXright) cong bigoplus_j in J X
          $$
          where $I$ is some index set and $J$ is the evident new index set given by reindexing the double direct sum. Call the isomorphism from $F$ to this direct sum above $alpha$, i.e.,
          $$
          alpha:bigoplus_j in J X to F.
          $$
          Then pre-composing $phi$ with $alpha$ gives a surjection
          $$
          (phicirc alpha):bigoplus_j in J X to M
          $$
          where
          $$
          operatornameImage(phi circ alpha) = operatornameImage(phi) = M
          $$
          because $alpha$ is an isomorphism and $phi$ is a surjection.






          share|cite|improve this answer




















          • Your answer is very clear, thanks! So epimorphic image only means that there is a surjection (an epimorphism) such that $M$ is image via this epimorphism?
            – robbis
            Aug 29 at 7:30










          • sorry but I still have some questions if you can answer. Is being a generator equivalent to say that there is a split epimorphism $G^n to R to 0$? and from this how can I see that $G$ is a generator if and only if $G^n$ is the direct sum of $R$ and $R'$ for some module $R'$? Thanks again
            – robbis
            Aug 29 at 9:17











          • @robbis Well a morphism in the category of left $R$-modules is epic if and only if it is surjective, so in this case it means that $M$ is the image of an epimorphism out of some domain you've chosen prior. Be careful in general, as there are categories in which epimorphisms are not surjective on underlying sets, but in module categories you're okay. Also if you like my answer, please consider upvoting it and accepting it! I'll answer your second comment in a second from here.
            – Geoff
            Aug 29 at 14:57






          • 1




            I used the fact that $R$ is free so is projective and I can find that map $s$. Now I consider the exact sequence $0 to ker(e) to G^n to R to 0$ and I apply the splitting lemma (if I'm not wrong, this is related to Rank Nullity Theorem) and I can say that $G^n= R bigoplus ker(e)$
            – robbis
            Aug 29 at 15:31






          • 1




            Thanks, now it's all clear to me! I'll upvote and accept surely! Thanks :)
            – robbis
            Aug 29 at 16:13












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          If $R$ is a unital ring, you know that every module $M$ can be described as a homomorphic image of some free module $F$; since
          $$
          F cong bigoplus_i in IR
          $$
          for some index set $I$, you are pretty much done. Explicitly, let $M$ be a left $R$-module and let $F$ be a free module chosen such that you can realize $M$ as the homomorphic image of some map $phi:F to M$, i.e., $operatornameImage(phi) cong M$. Then since there exists an $n in mathbbN$ such that $R$ is the homomorphic image of $X^n$, you get that
          $$
          F cong bigoplus_i in I R cong bigoplus_i in I X^n cong bigoplus_i in Ileft(bigoplus_k=1^nXright) cong bigoplus_j in J X
          $$
          where $I$ is some index set and $J$ is the evident new index set given by reindexing the double direct sum. Call the isomorphism from $F$ to this direct sum above $alpha$, i.e.,
          $$
          alpha:bigoplus_j in J X to F.
          $$
          Then pre-composing $phi$ with $alpha$ gives a surjection
          $$
          (phicirc alpha):bigoplus_j in J X to M
          $$
          where
          $$
          operatornameImage(phi circ alpha) = operatornameImage(phi) = M
          $$
          because $alpha$ is an isomorphism and $phi$ is a surjection.






          share|cite|improve this answer












          If $R$ is a unital ring, you know that every module $M$ can be described as a homomorphic image of some free module $F$; since
          $$
          F cong bigoplus_i in IR
          $$
          for some index set $I$, you are pretty much done. Explicitly, let $M$ be a left $R$-module and let $F$ be a free module chosen such that you can realize $M$ as the homomorphic image of some map $phi:F to M$, i.e., $operatornameImage(phi) cong M$. Then since there exists an $n in mathbbN$ such that $R$ is the homomorphic image of $X^n$, you get that
          $$
          F cong bigoplus_i in I R cong bigoplus_i in I X^n cong bigoplus_i in Ileft(bigoplus_k=1^nXright) cong bigoplus_j in J X
          $$
          where $I$ is some index set and $J$ is the evident new index set given by reindexing the double direct sum. Call the isomorphism from $F$ to this direct sum above $alpha$, i.e.,
          $$
          alpha:bigoplus_j in J X to F.
          $$
          Then pre-composing $phi$ with $alpha$ gives a surjection
          $$
          (phicirc alpha):bigoplus_j in J X to M
          $$
          where
          $$
          operatornameImage(phi circ alpha) = operatornameImage(phi) = M
          $$
          because $alpha$ is an isomorphism and $phi$ is a surjection.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 28 at 17:24









          Geoff

          549513




          549513











          • Your answer is very clear, thanks! So epimorphic image only means that there is a surjection (an epimorphism) such that $M$ is image via this epimorphism?
            – robbis
            Aug 29 at 7:30










          • sorry but I still have some questions if you can answer. Is being a generator equivalent to say that there is a split epimorphism $G^n to R to 0$? and from this how can I see that $G$ is a generator if and only if $G^n$ is the direct sum of $R$ and $R'$ for some module $R'$? Thanks again
            – robbis
            Aug 29 at 9:17











          • @robbis Well a morphism in the category of left $R$-modules is epic if and only if it is surjective, so in this case it means that $M$ is the image of an epimorphism out of some domain you've chosen prior. Be careful in general, as there are categories in which epimorphisms are not surjective on underlying sets, but in module categories you're okay. Also if you like my answer, please consider upvoting it and accepting it! I'll answer your second comment in a second from here.
            – Geoff
            Aug 29 at 14:57






          • 1




            I used the fact that $R$ is free so is projective and I can find that map $s$. Now I consider the exact sequence $0 to ker(e) to G^n to R to 0$ and I apply the splitting lemma (if I'm not wrong, this is related to Rank Nullity Theorem) and I can say that $G^n= R bigoplus ker(e)$
            – robbis
            Aug 29 at 15:31






          • 1




            Thanks, now it's all clear to me! I'll upvote and accept surely! Thanks :)
            – robbis
            Aug 29 at 16:13
















          • Your answer is very clear, thanks! So epimorphic image only means that there is a surjection (an epimorphism) such that $M$ is image via this epimorphism?
            – robbis
            Aug 29 at 7:30










          • sorry but I still have some questions if you can answer. Is being a generator equivalent to say that there is a split epimorphism $G^n to R to 0$? and from this how can I see that $G$ is a generator if and only if $G^n$ is the direct sum of $R$ and $R'$ for some module $R'$? Thanks again
            – robbis
            Aug 29 at 9:17











          • @robbis Well a morphism in the category of left $R$-modules is epic if and only if it is surjective, so in this case it means that $M$ is the image of an epimorphism out of some domain you've chosen prior. Be careful in general, as there are categories in which epimorphisms are not surjective on underlying sets, but in module categories you're okay. Also if you like my answer, please consider upvoting it and accepting it! I'll answer your second comment in a second from here.
            – Geoff
            Aug 29 at 14:57






          • 1




            I used the fact that $R$ is free so is projective and I can find that map $s$. Now I consider the exact sequence $0 to ker(e) to G^n to R to 0$ and I apply the splitting lemma (if I'm not wrong, this is related to Rank Nullity Theorem) and I can say that $G^n= R bigoplus ker(e)$
            – robbis
            Aug 29 at 15:31






          • 1




            Thanks, now it's all clear to me! I'll upvote and accept surely! Thanks :)
            – robbis
            Aug 29 at 16:13















          Your answer is very clear, thanks! So epimorphic image only means that there is a surjection (an epimorphism) such that $M$ is image via this epimorphism?
          – robbis
          Aug 29 at 7:30




          Your answer is very clear, thanks! So epimorphic image only means that there is a surjection (an epimorphism) such that $M$ is image via this epimorphism?
          – robbis
          Aug 29 at 7:30












          sorry but I still have some questions if you can answer. Is being a generator equivalent to say that there is a split epimorphism $G^n to R to 0$? and from this how can I see that $G$ is a generator if and only if $G^n$ is the direct sum of $R$ and $R'$ for some module $R'$? Thanks again
          – robbis
          Aug 29 at 9:17





          sorry but I still have some questions if you can answer. Is being a generator equivalent to say that there is a split epimorphism $G^n to R to 0$? and from this how can I see that $G$ is a generator if and only if $G^n$ is the direct sum of $R$ and $R'$ for some module $R'$? Thanks again
          – robbis
          Aug 29 at 9:17













          @robbis Well a morphism in the category of left $R$-modules is epic if and only if it is surjective, so in this case it means that $M$ is the image of an epimorphism out of some domain you've chosen prior. Be careful in general, as there are categories in which epimorphisms are not surjective on underlying sets, but in module categories you're okay. Also if you like my answer, please consider upvoting it and accepting it! I'll answer your second comment in a second from here.
          – Geoff
          Aug 29 at 14:57




          @robbis Well a morphism in the category of left $R$-modules is epic if and only if it is surjective, so in this case it means that $M$ is the image of an epimorphism out of some domain you've chosen prior. Be careful in general, as there are categories in which epimorphisms are not surjective on underlying sets, but in module categories you're okay. Also if you like my answer, please consider upvoting it and accepting it! I'll answer your second comment in a second from here.
          – Geoff
          Aug 29 at 14:57




          1




          1




          I used the fact that $R$ is free so is projective and I can find that map $s$. Now I consider the exact sequence $0 to ker(e) to G^n to R to 0$ and I apply the splitting lemma (if I'm not wrong, this is related to Rank Nullity Theorem) and I can say that $G^n= R bigoplus ker(e)$
          – robbis
          Aug 29 at 15:31




          I used the fact that $R$ is free so is projective and I can find that map $s$. Now I consider the exact sequence $0 to ker(e) to G^n to R to 0$ and I apply the splitting lemma (if I'm not wrong, this is related to Rank Nullity Theorem) and I can say that $G^n= R bigoplus ker(e)$
          – robbis
          Aug 29 at 15:31




          1




          1




          Thanks, now it's all clear to me! I'll upvote and accept surely! Thanks :)
          – robbis
          Aug 29 at 16:13




          Thanks, now it's all clear to me! I'll upvote and accept surely! Thanks :)
          – robbis
          Aug 29 at 16:13

















           

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