A reflexive module which is not free Here is an exercise in Christian Peskine's book An Algebraic Introduction to Complex Projective Geometry, pg. 25:   

Show that $(X_0,X_1)/(X_0X_3-X_1X_2)$ is a reflexive but not free ideal of the ring $$\mathbb{C}[X_0,X_1,X_2,X_3]/(X_0X_3-X_1X_2).$$

I have no idea how to prove it.
 A: Let $k$ be a commutative ring, consider $R=k[a,b,c,d]/(ad-bc)$ and $I=(a,b)$. We claim that $I$ is a reflexive $R$-module.
Sketch: There is an exact sequence of $R$-modules $R^2 \xrightarrow{\phi} R^2 \xrightarrow{\psi} I \to 0$ with $\phi(e_1)=(d,-c)$, $\phi(e_2)=(b,-a)$ and $\psi(e_1)=a$, $\psi(e_2)=b$. It induces an exact sequence $0 \to I^* \to \hom(R^2,R) \to \hom(R^2,R)$. This shows $I^* \cong \ker(\alpha)$, where $\alpha  : R^2 \to R^2$ is defined by $\alpha(e_1)=(d,b)$, $\alpha(e_2)=-(c,a)$. We have $(a,b) \in \ker(\alpha)$ since $a(d,b)-b(c,a)=0$, and similarly $(c,d) \in \ker(\alpha)$. In fact, they generate the kernel, and we have an exact sequence $R^2 \xrightarrow{\gamma} R^2 \xrightarrow{\beta} R^2 \xrightarrow{\alpha} R$, where $\beta(e_1)=(a,b)$, $\beta(e_2)=(c,d)$, $\gamma(e_1)=(d,-b)$, $\gamma(e_2)=(c,-a)$. Hence, there is an exact sequence $R^2 \xrightarrow{\gamma} R^2 \to I^* \to 0$, which in turn induces the exact sequence $0 \to I^{**} \to R^2 \xrightarrow{\delta} R^2$, where $\delta(e_1)=(d,c)$, $\delta(e_2)=-(b,a)$. The sequence $R^2 \xrightarrow{\phi} R^2 \xrightarrow{\epsilon} R^2 \xrightarrow{\delta} R^2$ is exact, where $\epsilon(e_1)=(b,d)$, $\epsilon(e_2)=(a,c)$. Hence, $I^{**} \cong \ker(\delta) \cong \mathrm{coker}(\phi) \cong I$. One checks that this is, in fact, the canonical homomorphism $I \to I^{**}$.
