# Projective and reflexive modules

Is there an easy way to prove that if $M$ is finitely generated projective $R$-module, then $M$ is a reflexive module using just the definition of the natural map?

The natural map is $\theta: M \rightarrow M^{**}$ is defined by $\theta(m)=\widehat{m},$ where given $f \in M^{*},$ we have $\widehat{m}(f)=f(m).$

Using just the natural map and the dual basis for projective modules, it is easy to see that $M$ is torsionless (or that $\theta$ is injective), but I'm struggling to see that $\theta$ is surjective.

Hint:

$M$ is a direct summand of a finitely generated free $R$-module $F\simeq R^n$ for some $n$. As the $\operatorname{Hom}$ functor commutes with direct sums, you can suppose $M$ is free, and ultimately that it is (isomorphic to) $R$.

Now, any linear form on $R$ is multiplication $m_\lambda$ by some $\lambda\in R$. A linear form $u\colon \operatorname{Hom}(R,R)\to R$ satisfies $$u(m_\lambda)=u(\lambda\operatorname{Id})=\lambda u(\operatorname{Id})=m_\lambda(u(\operatorname{Id}))$$ so that $u=\theta( u(\operatorname{Id}))$.

• Hi, sorry to bother, why can I suppose $M$ is free? – ZekeJay Apr 11 '17 at 22:07
• You don't bother at all:. The reason is that if $M\oplus P=F$, the natural morphism $F\to F^{**}$ is the direct sum of the natural morphisms $M\to M^{**}$ and $P\to P^{**}$, and if the first one is surjective, the last two are surjective too. – Bernard Apr 11 '17 at 22:19
• Wow! Thanks. Do you mind if i ask something else? Is there a explicit way of showing its surjective? – ZekeJay Apr 12 '17 at 0:46

Let $$M\oplus N=R^k$$; then applying $$\operatorname{Hom}_R(-,R)$$ to the split exact sequence $$0\to M\to R^k\to N\to0$$ preserves split exactness. Applying it again you get the commutative diagram with split exact rows: $$\require{AMScd} \begin{CD} 0 @>>> M @>>> R^k @>>> N @>>> 0 \\ @. @VVV @VVV @VVV @. \\ 0 @>>> M^{**} @>>> (R^k)^{**} @>>> N^{**} @>>> 0 \end{CD}$$ The middle vertical arrow is easily seen to be an isomorphism. By diagram chasing, $$M\to M^{**}$$ is injective and by symmetry also $$N\to N^{**}$$ is injective. Hence, by diagram chasing, $$M\to M^{**}$$ is surjective.

• One can use the snake lemma to show the left vertical map is bijective (having trivial kernel and cokernel). – Lao-tzu Feb 14 at 10:32
• @Lao-tzu Certainly so. In these cases I find diagram chasing easier. – egreg Feb 14 at 10:52