Question on $(2, 1+\sqrt{-5})$ as a submodule of $\mathbb{Z}[\sqrt{-5}]$.

Solution 1:

The idea is to use that an ideal of a ring of integers is in fact projective (this is a general result).

Let $\varphi:(z,z')\in R^2\to 2z+z' (1+\sqrt{-5})\in I$. It is surjective, and then split since $I$ is projective. Few computations show that $2z+z' (1+\sqrt{-5})\in I\mapsto z (4,-1+\sqrt{-5 })+z'(2+2\sqrt{-5},-3)\in A^2$ is a section of $\varphi$.

Hence $I\simeq im(\varepsilon)$ and $R^2=im(\varepsilon)\oplus \ker(\varphi)\simeq I\times \ker(\varphi).$

What remains to do is to prove that $\ker(\varphi)\simeq I$, which is left as an exercise for you (Maybe it is worth noticing that $I=\{a+b\sqrt{-5}\mid a\equiv b \mod 2\}$ to achieve that).

Of course, if you know already that any ideal of $R$ is projective, you don't need to find an explicit section, you just need to prove the last part (indeed, since $im(\varphi)=I$, you have an exact sequence $0\to\ker(\varphi)\to R^2\to I\to 0$, which splits since $I$ is projective...)

Solution 2:

Here's another approach using results about modules over a Dedekind domain.

An intermediate result in the fundamental theorem of finitely generated modules over a Dedekind domain $R$ is that $I \oplus J \cong R \oplus IJ$ for any nonzero ideals $I, J \trianglelefteq R$. Cf., Dummit and Foote, $\S16.3$, Proposition $21$ (p. 769).

Thus $I \oplus I \cong R \oplus I^2$, so it suffices to show that $I^2$ is principal. Multiplying the generators in pairs, we find $$ I^2 = (2, 1 + \sqrt{-5})^2 = (4, 2 + 2\sqrt{-5}, -4 + 2 \sqrt{-5}) \subseteq (2) $$ and since $$ 2 = 2 + 2 \sqrt{-5} - (4 + -4 + 2 \sqrt{-5}) \in (4, 2 + 2\sqrt{-5}, -4 + 2 \sqrt{-5}), $$ then we have equality: $I^2 = (2)$.