It is known that if a commutative Noetherian ring $R$ is hereditary then for any maximal ideal $M$ the localization $R_M$ is also hereditary. Is the Noetherian assumption necessary?


You do not need to suppose your ring to be noetherian. In fact, the following stronger statement is true:

If $R$ is a hereditary commutative ring and $S\subseteq R$ is a subset, then the ring $S^{-1}R$ is hereditary.

In order to prove this we use the following two facts:

1) A commutative ring $R$ is hereditary if and only if any epimorphism of $R$-modules with injective source has injective target (i.e., quotients of injectives are injective). (See e.g. T.Y. Lam, Lectures on modules and rings, Theorem 3.22.)

2) If $R$ is a commutative ring, $S\subseteq R$ is a subset an $M$ is an $S^{-1}R$-module, then $M$ is injective if and only if it is so considered as an $R$-module by means of scalar restriction. (See e.g. M.P. Brodmann, R.Y. Sharp, Local cohomology, Lemma 10.1.12 (where the noetherian hypothesis is not used), or E.C. Dade, Localization of injective modules, J. Algebra 69 (1981), 416--425.)

Now, suppose that $R$ is a hereditary commutative ring and consider a subset $S\subseteq R$. Let $M\rightarrow N$ be an epimorphism of $S^{-1}R$-modules with injective source. By means of scalar restriction to $R$ we can consider this as an epimorphism $M\rightarrow N$ of $R$-modules with injective source by 2). Hence, its target is injective by 1), and therefore the $S^{-1}R$-module $N$ is injective by 2). It follows that $S^{-1}R$ is hereditary.

  • $\begingroup$ Is commutativity of $R$ necessary or is it sufficient to suppose that $S$ satisfies the Ore conditions ? $\endgroup$ – tj_ Nov 14 '13 at 8:56
  • $\begingroup$ @tj_: Statement 1) is proven in the non-commutative case in Lam's book. I think statement 2) also generalises (and hence so does the whole thing), but since I usually do not work with non-commutative rings it might be safer for you to check this again. $\endgroup$ – Fred Rohrer Nov 14 '13 at 9:24
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    $\begingroup$ Can't you proceed directly? Assume $M$ is a submodule of a free $S^{-1}R$-module $F$. Then $F=S^{-1}F_0$ with $F_0$ $R$-free with same basis. Let $j:F_0\to F$ be the canonical map. Put $M_0=j^{-1}(M)$. Then $M=S^{-1}M_0$ (easy) and $M_0$ is $R$-projective by assumption on $R$, so $M$ is $S^{-1}R$-projective. Did I miss something? $\endgroup$ – Laurent Moret-Bailly Nov 14 '13 at 16:11
  • $\begingroup$ @Laurent: Your proof is definitely the easier one. (I just happen to like injectives so much...) $\endgroup$ – Fred Rohrer Nov 14 '13 at 20:32

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