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Let $R$ be a commutative ring without zero divisors. Assume that ideal $a\subset R$ is a projective $R$-module. How to prove that $a$ is finitely generated ? I need only hints.

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    $\begingroup$ Do you know that a projective ideal is the same as an invertible ideal? Invertible ideals are finitely generated. $\endgroup$ – Crostul Jul 2 '15 at 22:00
  • $\begingroup$ No, I don't know this, but I'll try do prove it. Thanks ! $\endgroup$ – mikis Jul 3 '15 at 6:46
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Hint: Prove the statements below ( may assume $I \ne 0$ ):

  1. If $I$, $J$ are ideals of an integral domain $R$ with field of fractions $K$ then every morphism of $R$-modules $I \to J$ is given by the multiplication by an element in $K$.

  2. If $I$ is a projective module then there exists an imbedding of $I$ into a free $R$ module $i \colon I \hookrightarrow$ $R^{(\wedge)}$ and a projection $\pi \colon R^{(\wedge)} \to I$ so that $\pi \circ i = Id_I$.

  3. Assuming that $I$ is $\ne 0$, each component $i_{\alpha}$ of $i$ is given by multiplication by a unique $j_{\alpha}$ ( use 1.). Moreover, only finitely many of the $j_{\alpha}$'s are nonzero.

  4. The projection $\pi \colon R^{(\wedge)} \to I$ is given $\pi ( r_{\alpha} ) = \sum q_{\alpha} r_{\alpha}$.

  5. $j_{\alpha} \in I^{-1} \colon = \{ j \in K\ | \ j \cdot I \subset R\}$

  6. $q_{\alpha} \in I$ for all $\alpha$.

  7. $\sum q_{\alpha} j_{\alpha} = 1$ (a finite sum)

  8. $I \cdot I^{-1} = R$ ( that is, $I$ is invertible).

  9. $I$ is generated by the $q_{\alpha}$ in the above finite sum.

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