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 $N$.
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2$\begingroup$ Do you mean for all finitely generated modules $N$? Otherwise this is clearly false. $\endgroup$– abxCommented Feb 2, 2019 at 8:00
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$\begingroup$ 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.). $\endgroup$– SashaCommented Feb 2, 2019 at 8:42
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$\begingroup$ which way are you comfortable with? $\endgroup$– Praphulla KoushikCommented Feb 2, 2019 at 10:12
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1 Answer
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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.
