The problem is to show that if $\mathfrak{p}\ne 0$ is a prime ideal of
$\mathbb{Z}[\sqrt{-3}]=R$, and we denote $S=\mathbb{Z}[\frac{1+\sqrt{-3}}{2}]$ then we have $R\subset S$ and for the localizations $R_{\mathfrak{p}}\subset S_{\mathfrak{p}}$, and $R_{\mathfrak{p}}=S_{\mathfrak{p}}$ if $\mathfrak{p}$ does not contain $2$.
The inclusion is certainly still true since injectivity is a local property. For equality, I believe the intuition is clear but I am having trouble with the details. If $2\not \in \mathfrak{p}$, then it gets inverted, and there is no difference between $R$ and $S$ and whatever else gets inverted at this point. I imagine there is probably more to say. How do I fill in the details?
For added context on the problem: the first part was to show that $S$ is finitely generated as an $R$ module, which is clear as $S$ is finitely generated as $\mathbb{Z}$ module, $\frac{1+\sqrt{-3}}{2}$ being integral over $\mathbb{Z}$. The final part is to show that $S$ is not a flat $R$ module, which I have not yet tackled.
Thanks in advance for any help.
As you said, if $2\not \in \mathfrak{p}$, then $1/2\in R_{\mathfrak{p}}$, hence also $\frac{1+\sqrt{-3}}{2}\in R_{\mathfrak{p}}$, so $R_{\mathfrak{p}}=S_{\mathfrak{p}}$.