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.

  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.