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

Question

Let $R$ be the ring $\mathbb{Z}[\sqrt{-5}]$ and $I$ the ideal generated by $2$ and $1+\sqrt{-5}$, $I=(2, 1+\sqrt{-5})$.

Show that $I$ is not R-module isomorphic to $R$ but $I\bigoplus I$ is R-module isomorphic to $R^2$.

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My attempt thus far:

It can be shown that $I$ is not principal, and thus not generated by a single element. Since $R$ is generated by $1$, if there were an isomorphism $\phi: R \rightarrow I$, $\phi(1)$ must generate $I$. By contradiction, $I$ cannot be isomorphic to $R$.

However, I am unable to prove the second part of the question. I see that $R^2$ has rank 2, so if we can find a basis for $I\bigoplus I$, say $(x,y)$, then there is a direct isomorphism $\phi: R^2 \rightarrow I \bigoplus I$, $\phi(1,0)=x$, $\phi(0,1)=y$. However, it's not immediately obvious to me what I should take as the basis for $I\bigoplus I$; is there a nonconstructive way of showing the two modules are isomorphic?

Answer 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...)

Answer 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: $I^2 = (2)$.