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Could you help me solve the following exercise (which I saw in the book: An Algebraic Introduction to Complex Projective Geometry by Peskine, page 210).

Let $R$ be a Noetherian normal domain. Let $I$ be a reflexive ideal. Assume that $I = (a, b)$. Show that the kernel of the surjective map $R^2 \overset{(a, b)}{\to } I$ is isomorphic to $I^* = \mathrm{Hom}(I, R)$.

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    $\begingroup$ In fact, $I^*\simeq I^{-1}$, where $I^{-1}=\{x\in K:xI\subseteq R\}$. ($K$ is the field of fractions of $R$.) Now define a map $I^{-1}\stackrel{\psi}\to R^2$ by $x\mapsto(xb,-xa)$ and show that this gives you a short exact sequence $0\to I^{-1}\stackrel{\psi}\to R^2\overset{(a, b)}{\to } I\to 0$. $\endgroup$ – user26857 Sep 5 '16 at 9:23
  • $\begingroup$ Oh, thanks, could you put it as an answer, I will accept it. There is another exercise in page 200, exercise 3 which I can not solve. May you give me some helps. $\endgroup$ – Pham Hung Quy Sep 5 '16 at 10:01

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