# $P$ is projective if and only if $P\otimes N\cong Hom(Hom(P,R),N)$

Let $$R$$ be a commutative Noetherian ring and $$P$$ be finitely generated $$R$$-module.

How to prove the following.

$$P$$ is projective if and only if $$P\otimes N\cong Hom(Hom(P,R),N)$$ for all finitely generated $$R$$-modules $$N$$.

• Do you mean for all finitely generated modules $N$? Otherwise this is clearly false. – abx Feb 2 '19 at 8:00
• You should construct a natural transformation (in $N$) in one direction. Then consider the subclass of all objects on which the natural transformation is an isomorphism. Then check whether $R$ is in this subclass, and other things (if two objects from a short exact sequence are in the subclass, is the third as well?). Also, is the subclass closed under infinite direct sums? This will give you a feeling for the question, for which $N$ to expect an isomorphism (a finitely generated $N$ is the cokernel of a morphism between finite direct sums of $R$'s, etc.). – Sasha Feb 2 '19 at 8:42
• which way are you comfortable with? – Praphulla Koushik Feb 2 '19 at 10:12

## 1 Answer

That the canonical map $$P\otimes _{R}N\rightarrow \operatorname{Hom}_{R}(P^*,N)$$ is an isomorphism when $$P$$ is projective (for any $$R$$-module $$N$$) is classical, see e.g. Bourbaki Algebra II, §4, Proposition 2.

In the other direction, taking $$N=R$$ shows that $$P$$ is reflexive. Now take $$N=P^*$$. We get that the canonical map $$P\otimes _{R}P^*\rightarrow \operatorname{Hom}_{R}(P^*,P^*)$$ is an isomorphism. By the Remark 1 following the above Proposition, this implies that $$P^*$$ is projective, and therefore that its dual $$P$$ is projective.