Decomposing Noetherian hereditary rings of Krull dimension $1$ into product of hereditary domains (i.e. Dedekind domains) Let $R$ be a commutative Noetherian  hereditary ring (https://en.wikipedia.org/wiki/Hereditary_ring) of Krull dimension $1$. Then is it true that $R$ is a finite direct product of Dedekind domains ? 
 A: Certainly finite products of Dedekind rings with fields work as well. These are the only ones, even without the assumption on the Krull dimension:


Lemma. Let $R$ be a Noetherian (commutative) ring. Then $R$ is hereditary if and only if $R$ is a finite product $\prod_{i = 1}^r R_i$ where each $R_i$ is a Dedekind domain or a field.


Proof. If $R$ is of said form, then $R$ is hereditary since each $R_i$ is, and we have
$$\operatorname{\underline{Mod}}_R = \prod_{i=1}^r \operatorname{\underline{Mod}}_{R_i},$$
where a module $M = \prod M_i$ is projective iff each $M_i$ is.
Conversely, suppose $R$ is hereditary. We claim that for any prime $\mathfrak p \subseteq R$, the local ring $R_{\mathfrak p}$ is hereditary. Indeed, if $M$ is a projective module over $R_{\mathfrak p}$, then $M$ is free [Tag 0593], so in particular $M \cong M_0 \otimes_R R_{\mathfrak p}$ for some projective $R$-module $M_0$. Let $N \subseteq M$ be a submodule, and consider the $R$-module
$$N_0 = \left\{n \in M_0\ \bigg|\ n \otimes 1 \in N \subseteq M_0 \otimes_R R_{\mathfrak p}\right\} \subseteq M_0.$$
Then $N_0$ is projective by the hereditary assumption on $R$, and $N_0 \otimes_R R_{\mathfrak p} = N$ as submodules of $M = M_0 \otimes_R R_{\mathfrak p}$. Indeed, clearly $N_0 \otimes_R R_{\mathfrak p} \subseteq N$, and it spans since we can clear denominators. Thus, $N$ is free as well, showing that $R_{\mathfrak p}$ is hereditary.
This immediately forces $R_{\mathfrak p}$ to be a principal ideal ring, for a submodule or $R_{\mathfrak p}$ cannot be free of rank bigger than $1$. Then a theorem of Zariski and Samuel implies $R_{\mathfrak p}$ is either Artinian or a DVR (since we are in the local case). In the Artinian case, the maximal ideal $\mathfrak m = \mathfrak pR_{\mathfrak p}$ has to be free as well. This is only possible if $\mathfrak m = 0$, for $\operatorname{length}(\mathfrak m) = \operatorname{length}(R_{\mathfrak p}) - 1$ should be a multiple of $\operatorname{length}(R_{\mathfrak p})$.
Thus, every localisation $R_{\mathfrak p}$ is either a DVR or a field. In particular, $R$ is normal, hence a product of normal domains [Tag 030C], which each have to be either a Dedekind domain or a field. $\square$
