$M bigotimes_R (Pi_{iin I} B_i) ncong Pi_{i in I} (M bigotimes B_i)$ [closed]












2














let M and Bi be R-MODULE for all i in I
Show that $M bigotimes_R (Pi_{iin I} B_i) ncong Pi_{i in I} (M bigotimes B_i)$
Take $R=mathbb{Z}$, $M=mathbb{Q}$ and $B_n =frac{mathbb{Z}}{P^n mathbb{Z}}$
, n > 0










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closed as off-topic by Scientifica, user26857, Saad, Brahadeesh, Lord Shark the Unknown Nov 28 '18 at 5:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Scientifica, user26857, Saad, Brahadeesh, Lord Shark the Unknown

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 3




    Nice exercise! Where does it come from? I can see one side is zero, and the other isn't.
    – Lord Shark the Unknown
    Nov 27 '18 at 19:01
















2














let M and Bi be R-MODULE for all i in I
Show that $M bigotimes_R (Pi_{iin I} B_i) ncong Pi_{i in I} (M bigotimes B_i)$
Take $R=mathbb{Z}$, $M=mathbb{Q}$ and $B_n =frac{mathbb{Z}}{P^n mathbb{Z}}$
, n > 0










share|cite|improve this question















closed as off-topic by Scientifica, user26857, Saad, Brahadeesh, Lord Shark the Unknown Nov 28 '18 at 5:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Scientifica, user26857, Saad, Brahadeesh, Lord Shark the Unknown

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 3




    Nice exercise! Where does it come from? I can see one side is zero, and the other isn't.
    – Lord Shark the Unknown
    Nov 27 '18 at 19:01














2












2








2







let M and Bi be R-MODULE for all i in I
Show that $M bigotimes_R (Pi_{iin I} B_i) ncong Pi_{i in I} (M bigotimes B_i)$
Take $R=mathbb{Z}$, $M=mathbb{Q}$ and $B_n =frac{mathbb{Z}}{P^n mathbb{Z}}$
, n > 0










share|cite|improve this question















let M and Bi be R-MODULE for all i in I
Show that $M bigotimes_R (Pi_{iin I} B_i) ncong Pi_{i in I} (M bigotimes B_i)$
Take $R=mathbb{Z}$, $M=mathbb{Q}$ and $B_n =frac{mathbb{Z}}{P^n mathbb{Z}}$
, n > 0







commutative-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 14:44

























asked Nov 27 '18 at 19:00









Aaaaaa

253




253




closed as off-topic by Scientifica, user26857, Saad, Brahadeesh, Lord Shark the Unknown Nov 28 '18 at 5:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Scientifica, user26857, Saad, Brahadeesh, Lord Shark the Unknown

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Scientifica, user26857, Saad, Brahadeesh, Lord Shark the Unknown Nov 28 '18 at 5:54


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Scientifica, user26857, Saad, Brahadeesh, Lord Shark the Unknown

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 3




    Nice exercise! Where does it come from? I can see one side is zero, and the other isn't.
    – Lord Shark the Unknown
    Nov 27 '18 at 19:01














  • 3




    Nice exercise! Where does it come from? I can see one side is zero, and the other isn't.
    – Lord Shark the Unknown
    Nov 27 '18 at 19:01








3




3




Nice exercise! Where does it come from? I can see one side is zero, and the other isn't.
– Lord Shark the Unknown
Nov 27 '18 at 19:01




Nice exercise! Where does it come from? I can see one side is zero, and the other isn't.
– Lord Shark the Unknown
Nov 27 '18 at 19:01















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