Is there an integral domain $R$ with ideal $I$ such that $I^2 = I$ and $I$ is nontrivial?

Question

Where $I$ being nontrivial means $I \neq 0$ and $I \neq R$. I have found that such a ring must not be Noetherian (since otherwise, roughly, you could take some nonzero nonunit $x_1 \in I$ and recursively take $x_i \in I$ such that $x_i^2 = x_{i-1}$ to create a strictly increasing chain of ideals), and if we don't require our ring to be an integral domain, the ring $\mathbb{Z}/6\mathbb{Z}$ with ideal $I = (2)$ works.

Answer 1

$R=\Bbb{Q}[X, X^{1/2} , X^{1/4}, X^{1/8}, \dots , X^{1/2^n}, \dots]$ does the job with $I=(X, X^{1/2} , X^{1/4}, X^{1/8}, \dots , X^{1/2^n}, \dots)$.