I'm working on a problem where I need to find all the prime ideals of $\mathbb{Z}[X]/(2X, X^2)$ and then computing the localization at those primes, but I'm having trouble computing the localization at $(2, X)$.
I'm aware of the identity $(R/I)_{\mathfrak{p}} \cong R_{\mathfrak{p}}/IR_{\mathfrak{p}}$. That doesn't seem to help in this case, though, as $(2X, X^2) \mathbb{Z}[X]_{(2, X)}$ doesn't obviously simplify, since $2$ isn't invertible (unlike the $(p, X)$ case for odd primes $p$, or the $(X)$ case).
Is there any other way to simplify $(\mathbb{Z}[X]/(2X, X^2))_{(2, X)}$?
Finally got around to working up a full solution to this question. The quickest way to show it is to construct $\pi \colon \mathbb{Z}[X]/(2X, X^2) \to \mathbb{Z}_{(2)}[X]/(2X, X^2)$ satisfying the universal property of localization at $(2, X)$, i.e. that $\pi$ sends elements not in $(2, X)$ to units, and is initial among such functions. Then uniqueness up to unique isomorphism automatically shows that $\mathbb{Z}_{(2)}[X]/(2X, X^2) \cong (\mathbb{Z}[X]/(2X, X^2))_{(2, X)}$.