Timeline for Comparing homomorphisms over different base rings
Current License: CC BY-SA 2.5
3 events
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| Oct 23, 2010 at 10:06 | comment | added | TonyS | Yes, i tried the Hom-Tensor adjunction. This should also give $Hom_R(R\otimes_A M,N)\cong Hom_A(M,N)$.But then i couldn't see why $R\otimes_A M$ is isomorphic to $M^r$ as an $R$-module. The same problem you encountered. | |
| Oct 22, 2010 at 19:57 | comment | added | Hailong Dao | Hold on, perhaps there is a problem: $Hom_A(R,N)$ as an $R$-module, might not be $N^r$?. I have to run now, so I will come back to this... | |
| Oct 22, 2010 at 19:46 | history | answered | Hailong Dao | CC BY-SA 2.5 |