Does locally DVR implies Dedekind Domain when it is 1-dimensional, semi-local domain but Noetherian not given

Question

Let R be a semi-local integral domain of dimension 1 such that $\forall P \in Spec{R} $ such that $P \ne 0$ we have, $R_P$ to be a Discrete Valuation Ring. Then prove that $R$ is a Dedekind Domain?

If it was given that $R$ is Noetherian then I know that it would follow that, a Noetherian domain of dimension 1 which is locally DVR $\implies$ it is a Dedekind Domain. But I am not sure how to get the Noetherian property from the localisations to the ring $R$.

Thanks for help.

Answer 1

I expand my comment. So assume $R$ has just two maximal ideals, $P,Q$. Let us show that $P$ is finitely generated. One knows $PR_P, PR_Q$ are both finitely generated (in fact, in your case one generated). So, you can find (by clearing denominators), $x,y\in P$ such that $xR_P=PR_P, yR_Q=PR_Q$. Then, obviously $(x,y)R_P=PR_P$ and $(x,y)R_Q=PR_Q$. Now, can you show that $(x,y)R=P$?