This is probably just a stupid question about modules, but I'm having trouble proving the following, if it's true:
Let $R$ be a domain and $M$ a rank $1$ projective module over $R$ (i.e. $\dim_K(K \otimes_R M) = 1$, where $K = \mathrm{Frac}(R))$. Then $M \cong I$ for some ideal $I$ of $R$.
The reason I would even suspect this is the proof I'm following to show projective modules over a Dedekind domain are sums of ideals assumes for the base case the above fact, in the case $R$ is Dedekind. But they only proved the opposite direction in the article, namely an ideal in any domain has rank 1, which leads me to suspect this is also a general fact.
Yes, that's true. You can find a suitable map $M \to R$ giving $M$ as an ideal like this. Since $M$ is projective, it's torsion-free, so the map $M\to M\otimes_R K$ injects. When $M$ has rank one this gives $M$ as a submodule of $K$. If $M$ is finitely generated you can take a common denominator for the elements of a generating set (these elements thought of as lying in $K$). Then $M$ is contained in the module of fractions in $K$ with that denominator, and that module is isomorphic to $R$.
(You don't actually need to assume that $M$ is finitely generated, this follows from the hypothesis.)