Is $D\otimes_{R}A\cong \mathrm{Hom}_{R}(D,A)$ ever true if $D$ is a projective $R$-module?

I am trying to prove that a short exact sequence $0\rightarrow A\xrightarrow{f} B\xrightarrow{g}C\rightarrow 0$ of left $R$-modules induces a short exact sequence $0\rightarrow D\otimes_{R} A\xrightarrow{1_{D}\otimes f} D\otimes_{R} B\xrightarrow{1_{D}\otimes g}D\otimes_{R} C\rightarrow 0$ of abelian groups, given that $D$ is a projective unitary right $R$-module and $R$ has identity.

There is a theorem that states that if $D$ is projective, then the given short exact sequence induces a short exact sequence $0\rightarrow \mathrm{Hom}_{R}(D,A)\xrightarrow{\overline{f}} \mathrm{Hom}_{R}(D,B)\xrightarrow{\overline{g}}\mathrm{Hom}_{R}(D,C)\rightarrow 0$.

If the $D\otimes_{R}A\cong \mathrm{Hom}_{R}(D,A)$ is ever true, then I believe the result should follow. Should I try to construct an isomorphism between these two groups, or is there no such isomorphism?

I am not sure how to use the standard definition of projective in this context, since the factor $D$ is stuck inside a tensor product. Any suggestions with how to approach this problem are appreciated!

• I don't see any reason why $D\otimes_R A$ should be isomorphic to $\mbox{Hom}_R(D,A)$, so I wouldn't personally recommend going down that path...
– John
Commented Feb 14, 2017 at 23:40

The isomorphism you ask for doesn't even make sense in general: $\operatorname{Hom}_R(D,A)$ is not defined because $D$ is a right $R$-module and $A$ is a left $R$-module.
Instead, I would suggest first proving the result in the case that $D$ is a free module, and then using the fact that any projective module is a direct summand of a free module.
• Thank you, that is very helpful since I already proved true for a free module. But I was confused how to start with that fact... should I consider $F\cong K\oplus D$ for $F$ free and $K$ a module, and consider $0\rightarrow (K\oplus D)\otimes_{R} A\rightarrow (K\oplus D)\otimes_{R} B\rightarrow(K\oplus D)\otimes_{R} C\rightarrow 0$ Commented Feb 15, 2017 at 0:25
• Yes, consider that sequence, and note that it is just the termwise direct sum of the two sequences $0\to K\otimes A\to K\otimes B\to K\otimes C\to 0$ and $0\to D\otimes A\to D\otimes B\to D\otimes C\to 0$. Now use the fact that your sequence is exact to prove that both of these summand sequences must be exact. Commented Feb 15, 2017 at 0:32