# why is this ideal projective but not free?

Let $R=\mathbb{Z}[\sqrt{-5}]$ and $I=(2,1+\sqrt{-5})$ how can I prove that $I$ is projective but not free?

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If it were free, it would have to be principal, right? What is its norm? –  Hurkyl May 12 '12 at 13:16
Compute the fractional ideal that is its inverse. Using the ideal product then allows you to write it as a split summand of a free module of rank two. This implies that it is projective. –  Jyrki Lahtonen May 12 '12 at 17:51
Please see $\S 3.5.4$ -- "Projective verus free" -- in my commutative algebra notes. In particular, Proposition 27 and the exercise follow it step you through showing that the ideal $\langle 3, 1+ \sqrt{-5} \rangle$ is projective but not free.
The same techniques apply to $I = \langle 2, 1+ \sqrt{-5} \rangle$. (In fact, I view it as a happy accident that your question is similar but not identical to what is treated in my notes. You'll learn more by figuring out what slight modifications you need to make.)