Let $R$ be a noetherian ring. There is a natural map $R\rightarrow \prod_p R_p$, where $R_p$ is the localizations of $R$ at prime ideals $p$ of $R$.
Must this map be injective?
(I'm happy to consider the case where $R$ is moreover finitely generated over a Dedekind domain)
(I can show that the kernel must be contained in the nilradical)
The answer is yes - the map must be injective.
Indeed, let $K$ be the kernel of the map, and let $x\in K$. This means, that for every prime $p$, there is a $y_p\in R - p$ satisfying $xy_p = 0$ in $R$.
Viewing $K$ as an $R$-module, this implies that the localization $K\otimes_R R_p = 0$ for every prime $p$ of $R$, but we know from the theory of quasicoherent sheaves that the modules $K\otimes_R R_p$ are precisely the stalks of the sheaf $\tilde{K}$, and a sheaf is 0 if and only if its stalks are zero, and hence $K = 0$.
Alternatively, the fact that $K = 0$ follows from the results of https://stacks.math.columbia.edu/tag/0546
Moreover, when $R$ is noetherian, one gets an injective map to the product of the localizations at the associated primes of $R$.