The two desired results follow from:
Claim. Let $I$ be an ideal of an integral domain $R$.
Assume that $I$ can be generated by $2$ elements but not less. Then the following are equivalent:
- The ideal $I$ is projective.
- $\bigwedge^2(I) = 0$.
- The second Fitting ideal $\operatorname{Fitt}_1(I)$ of $I$ is $R$.
Proof. See Proposition 20.7, Proposition 20.8 and Exercise 20.9.i of [1]. The latter exercise shows in particular that $\operatorname{Fitt}_1(I)$ is the annihilator of $\bigwedge^2(I)$, which is evidently a cyclic $R$-module.
If an ideal $I$ of a quadratic order $\mathfrak{O} \subseteq \mathbb{Q}(\sqrt{m})$ with $m \in \mathbb{Z}$, has standard basis ($a$, $d + ef \omega$) (see [2, Lemma 6.2 and 6.5] for definitions and proofs) where $$\omega = \left\{
\begin{array}{cc}
\sqrt{m} & \text{ if } m \not\equiv 1 \mod 4, \\
\frac{1 + \sqrt{m}}{2} & \text{ if } m \equiv 1 \mod 4, \\
\end{array}\right.$$
then $I$ is projective if and only if $$\frac{\gcd(a, d , ef)}{e} = 1.$$
Alternatively (this item contains self-promotion), the ideal $\operatorname{Fitt}_1(I)$ for such an ideal $I$ can be explicitly computed by means of [3, Lemma 6.5.$ii$].
Since $R_1$ is Dedekind, every ideal of $R_1$ is projective. Thus the above claim yields $\bigwedge^2(I) = 0$.
Since $(2, 2 \omega)$, with $\omega = \frac{1 + \sqrt{5}}{2}$, is a standard basis for $J$ in $R_2$, it follows from [2, Lemma 6.5] that $J$ is not projective. Therefore $\bigwedge^2(I) \neq 0$ by the above claim.
Note. The ideal $J$ of $R_2$ is the conductor of $R_2$ in the maximal order $\mathbb{Z}[\frac{1 + \sqrt{5}}{2}]$.
- D. Eisenbud, "Commutative Algebra", 1995.
- T. Ibukiyama and M. Kaneko, "Quadratic Forms and Ideal Theory of Quadratic", pages 75 - 93, 2014.
- L. Guyot, "Equivalent generating pairs of an ideal of a commutative ring", 2018.
