Showing a map, $R/Iotimes_R R/Jrightarrow R$, is well defined

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Let $R$ be a commutative ring, and $I,Jsubseteq R$ ideals. I'm trying to show that the map $$phi: R/Iotimes_R R/Jrightarrow R$$ given by $$(r+I)otimes (s+J)=rs(1+I)otimes (1+J)mapsto rs$$ is well defined.




This is what I've thought so far:



Let $r+I=r'+I$, and $s+J=s'+J$. Then I need to show that $(r+I)otimes (s+J)=(r'+I)otimes (s'+J)Rightarrow rs=r's'$. First I thought I could write $$(rs-r's')(1+I)otimes (1+J)=(r+I)otimes (s+J)-(r'+I)otimes (s'+J)=0tag1$$ and say that the lhs maps to $rs-r's'$ and the rhs maps to $0$, and that this implies $rs-r's'=0$. But this doesn't seem right, since I'm then assuming $phi$ is well defined.



I'm also not sure if I can say that $(1+I)otimes (1+J)$ generates $R/Iotimes_R R/J$, so if $(rs-r's')(1+I)otimes (1+J)$ then $rs-r's'=0$.



Am I on the right track? Any hint would be appreciated.




commutative-algebra modules tensor-products




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  • 1




    Try it with $R = Bbb Z$ and see. Say you have $Bbb Z/3oplus Bbb Z/4$, is your map well-defined?
    – Arthur
    Jul 19 at 15:34







  • 2




    There is a well-defined map with target $R/(I+J)$, but you can't do better than that.
    – Stephen
    Jul 19 at 15:37










  • @Arthur Oh, man I'm dumb. $bar 1otimes bar 1mapsto 1$ and $bar 4otimes bar 1mapsto 4$, and $1neq 4$, so my map is not well defined. Thank you! If you post your comment as an answer I'll gladly accept it, else I'll delete my question.
    – cansomeonehelpmeout
    Jul 19 at 15:41











  • @Stephen Thank you! I was trying to calculate the kernel and use the 1st. isomorphism theorem.
    – cansomeonehelpmeout
    Jul 19 at 15:44














up vote
1
down vote

favorite













Let $R$ be a commutative ring, and $I,Jsubseteq R$ ideals. I'm trying to show that the map $$phi: R/Iotimes_R R/Jrightarrow R$$ given by $$(r+I)otimes (s+J)=rs(1+I)otimes (1+J)mapsto rs$$ is well defined.




This is what I've thought so far:



Let $r+I=r'+I$, and $s+J=s'+J$. Then I need to show that $(r+I)otimes (s+J)=(r'+I)otimes (s'+J)Rightarrow rs=r's'$. First I thought I could write $$(rs-r's')(1+I)otimes (1+J)=(r+I)otimes (s+J)-(r'+I)otimes (s'+J)=0tag1$$ and say that the lhs maps to $rs-r's'$ and the rhs maps to $0$, and that this implies $rs-r's'=0$. But this doesn't seem right, since I'm then assuming $phi$ is well defined.



I'm also not sure if I can say that $(1+I)otimes (1+J)$ generates $R/Iotimes_R R/J$, so if $(rs-r's')(1+I)otimes (1+J)$ then $rs-r's'=0$.



Am I on the right track? Any hint would be appreciated.




commutative-algebra modules tensor-products




share|cite|improve this question

















  • 1




    Try it with $R = Bbb Z$ and see. Say you have $Bbb Z/3oplus Bbb Z/4$, is your map well-defined?
    – Arthur
    Jul 19 at 15:34







  • 2




    There is a well-defined map with target $R/(I+J)$, but you can't do better than that.
    – Stephen
    Jul 19 at 15:37










  • @Arthur Oh, man I'm dumb. $bar 1otimes bar 1mapsto 1$ and $bar 4otimes bar 1mapsto 4$, and $1neq 4$, so my map is not well defined. Thank you! If you post your comment as an answer I'll gladly accept it, else I'll delete my question.
    – cansomeonehelpmeout
    Jul 19 at 15:41











  • @Stephen Thank you! I was trying to calculate the kernel and use the 1st. isomorphism theorem.
    – cansomeonehelpmeout
    Jul 19 at 15:44












up vote
1
down vote

favorite









up vote
1
down vote

favorite












Let $R$ be a commutative ring, and $I,Jsubseteq R$ ideals. I'm trying to show that the map $$phi: R/Iotimes_R R/Jrightarrow R$$ given by $$(r+I)otimes (s+J)=rs(1+I)otimes (1+J)mapsto rs$$ is well defined.




This is what I've thought so far:



Let $r+I=r'+I$, and $s+J=s'+J$. Then I need to show that $(r+I)otimes (s+J)=(r'+I)otimes (s'+J)Rightarrow rs=r's'$. First I thought I could write $$(rs-r's')(1+I)otimes (1+J)=(r+I)otimes (s+J)-(r'+I)otimes (s'+J)=0tag1$$ and say that the lhs maps to $rs-r's'$ and the rhs maps to $0$, and that this implies $rs-r's'=0$. But this doesn't seem right, since I'm then assuming $phi$ is well defined.



I'm also not sure if I can say that $(1+I)otimes (1+J)$ generates $R/Iotimes_R R/J$, so if $(rs-r's')(1+I)otimes (1+J)$ then $rs-r's'=0$.



Am I on the right track? Any hint would be appreciated.




commutative-algebra modules tensor-products




share|cite|improve this question














Let $R$ be a commutative ring, and $I,Jsubseteq R$ ideals. I'm trying to show that the map $$phi: R/Iotimes_R R/Jrightarrow R$$ given by $$(r+I)otimes (s+J)=rs(1+I)otimes (1+J)mapsto rs$$ is well defined.




This is what I've thought so far:



Let $r+I=r'+I$, and $s+J=s'+J$. Then I need to show that $(r+I)otimes (s+J)=(r'+I)otimes (s'+J)Rightarrow rs=r's'$. First I thought I could write $$(rs-r's')(1+I)otimes (1+J)=(r+I)otimes (s+J)-(r'+I)otimes (s'+J)=0tag1$$ and say that the lhs maps to $rs-r's'$ and the rhs maps to $0$, and that this implies $rs-r's'=0$. But this doesn't seem right, since I'm then assuming $phi$ is well defined.



I'm also not sure if I can say that $(1+I)otimes (1+J)$ generates $R/Iotimes_R R/J$, so if $(rs-r's')(1+I)otimes (1+J)$ then $rs-r's'=0$.



Am I on the right track? Any hint would be appreciated.





commutative-algebra modules tensor-products





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 19 at 15:35
























asked Jul 19 at 15:30









cansomeonehelpmeout

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  • 1




    Try it with $R = Bbb Z$ and see. Say you have $Bbb Z/3oplus Bbb Z/4$, is your map well-defined?
    – Arthur
    Jul 19 at 15:34







  • 2




    There is a well-defined map with target $R/(I+J)$, but you can't do better than that.
    – Stephen
    Jul 19 at 15:37










  • @Arthur Oh, man I'm dumb. $bar 1otimes bar 1mapsto 1$ and $bar 4otimes bar 1mapsto 4$, and $1neq 4$, so my map is not well defined. Thank you! If you post your comment as an answer I'll gladly accept it, else I'll delete my question.
    – cansomeonehelpmeout
    Jul 19 at 15:41











  • @Stephen Thank you! I was trying to calculate the kernel and use the 1st. isomorphism theorem.
    – cansomeonehelpmeout
    Jul 19 at 15:44












  • 1




    Try it with $R = Bbb Z$ and see. Say you have $Bbb Z/3oplus Bbb Z/4$, is your map well-defined?
    – Arthur
    Jul 19 at 15:34







  • 2




    There is a well-defined map with target $R/(I+J)$, but you can't do better than that.
    – Stephen
    Jul 19 at 15:37










  • @Arthur Oh, man I'm dumb. $bar 1otimes bar 1mapsto 1$ and $bar 4otimes bar 1mapsto 4$, and $1neq 4$, so my map is not well defined. Thank you! If you post your comment as an answer I'll gladly accept it, else I'll delete my question.
    – cansomeonehelpmeout
    Jul 19 at 15:41











  • @Stephen Thank you! I was trying to calculate the kernel and use the 1st. isomorphism theorem.
    – cansomeonehelpmeout
    Jul 19 at 15:44







1




1




Try it with $R = Bbb Z$ and see. Say you have $Bbb Z/3oplus Bbb Z/4$, is your map well-defined?
– Arthur
Jul 19 at 15:34





Try it with $R = Bbb Z$ and see. Say you have $Bbb Z/3oplus Bbb Z/4$, is your map well-defined?
– Arthur
Jul 19 at 15:34





2




2




There is a well-defined map with target $R/(I+J)$, but you can't do better than that.
– Stephen
Jul 19 at 15:37




There is a well-defined map with target $R/(I+J)$, but you can't do better than that.
– Stephen
Jul 19 at 15:37












@Arthur Oh, man I'm dumb. $bar 1otimes bar 1mapsto 1$ and $bar 4otimes bar 1mapsto 4$, and $1neq 4$, so my map is not well defined. Thank you! If you post your comment as an answer I'll gladly accept it, else I'll delete my question.
– cansomeonehelpmeout
Jul 19 at 15:41





@Arthur Oh, man I'm dumb. $bar 1otimes bar 1mapsto 1$ and $bar 4otimes bar 1mapsto 4$, and $1neq 4$, so my map is not well defined. Thank you! If you post your comment as an answer I'll gladly accept it, else I'll delete my question.
– cansomeonehelpmeout
Jul 19 at 15:41













@Stephen Thank you! I was trying to calculate the kernel and use the 1st. isomorphism theorem.
– cansomeonehelpmeout
Jul 19 at 15:44




@Stephen Thank you! I was trying to calculate the kernel and use the 1st. isomorphism theorem.
– cansomeonehelpmeout
Jul 19 at 15:44















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