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 ? 



Decomposing Noetherian hereditary rings of Krull dimension $1$ into product of hereditary domains (i.e. Dedekind domains)
Certainly finite products of Dedekind rings with fields work as well. These are the only ones, even without the assumption on the Krull dimension:



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> > **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.*
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*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](https://stacks.math.columbia.edu/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](https://stacks.math.columbia.edu/tag/030C)], which each have to be either a Dedekind domain or a field. $\square$


