Projective ideal in $\mathbb{Z}[\sqrt{-5}]$ This is a question from my past Qual
"Set $R=\mathbb{Z}[\sqrt{-5}]$ and $I=(2,1+\sqrt{-5})$. Is $I$ prime? Is $I$ projective as an $R$-module?"
Clearly $R/I \cong \mathbb{Z}/2\mathbb{Z}$, hence $I$ is prime.
About the other part, I do not how to approach since I only know 2 facts that may be helpful: the definition and the fact that a module is projective iff it is a direct summand of a free module. Both of them do not seem to solve this problem. Can we do it directly using the above two facts?
I googled and it turns out $R$ here is a Dedekind domain and hence all ideals are projective modules. But I did not learn about Dedekind domains in class and hence this would trivialize the problem.
 A: In general an $R$-module $M$ is projective as soon as it is finitely presented and punctually free: Matsumura, Theorem 7.12.
Punctually free means that for all maximal ideals $\mathfrak m\subset R$, the localized module $M_\mathfrak m$ is free over the localized ring $R_\mathfrak m$.
In your case $R=\mathbb{Z}[\sqrt{-5}]$ is noetherian so that finitely presented is equivalent to  finitely generated, which your $I$ certainly is.
It remains for us to prove that $I$ is punctually free. 
So let $\mathfrak m\subset R$ be maximal:
a) If $I=\langle 2, 1+\sqrt -5\rangle\not\subset \mathfrak m$, then $I_\mathfrak m=R_\mathfrak m$,  which is trivially free of dimension one over $R_\mathfrak m$.
b)  If $I=\langle 2, 1+\sqrt -5\rangle\subset \mathfrak m$, then $2\in \mathfrak m$ so that $3\not\in \mathfrak m$.
But then $$2=\frac {(1+\sqrt -5)\cdot (1-\sqrt -5)}{3}\in I_\mathfrak m$$
Thus $I_\mathfrak m=\langle 2, 1+\sqrt -5\rangle=\langle1+\sqrt -5\rangle\subset R_\mathfrak m$, a free $R_\mathfrak m$-module of dimension one.
Hence $I$ is indeed punctually free and thus projective.
Remark
Some authors call "locally free" what I have called punctually free.
Their terminology is incompatible with that in algebraic geometry where locally free is the stronger condition  that $\operatorname {Spec}R$ can be covered by open subsets $D(f)$ such that $M_f$ is free over $R_f$.
Edit
Here is a completely elementary and explicit decomposition of the free module $R^2$ as a direct sum of two $R$ modules, the first of them, $s(I)$,  isomorphic to $I$: $$R^2=s(I)\oplus \operatorname {ker}(p)\cong I\oplus \operatorname {ker}(p)$$ where $p$ is the $R$-linear morphism $R^2\to I$ sending $(r_1,r_2)$ to $2r_1+(1+\sqrt -5)r_2$
and $s$ is the $R$-linear morphism $I\to R^2$ sending $i$ to $(-i, i\frac {1-\sqrt{-5}}{2})$.
Note carefully that
a) $ (\frac {1-\sqrt{-5}}{2})I\subset R$ even though $\frac {1-\sqrt{-5}}{2}\notin R$
b)  $2\cdot (-1)+(1+\sqrt -5)\cdot (\frac {1-\sqrt{-5}}{2})=1$
I learned the fine trick in this Edit from @Jyrki Lahtonen: see here and here.
