# Associated graded is integrally closed domain implies original ring was integrally closed

Let $$A$$ be a local ring (I am also happy to assume it is Noetherian) with maximal ideal $$\mathfrak{m}$$.

Claim: If the associated graded ring $$G_{\mathfrak{m}}(A)=\bigoplus_{n\geq 0} \mathfrak{m}^n/\mathfrak{m}^{n+1}$$ is an integrally closed domain, then $$A$$ is an integrally closed domain.

This claim appears without proof in Chapter 11 of Atiyah-MacDonald (page 123). I see why $$A$$ must be a domain, but have been unsuccessful in verifying the rest of the claim.

References for a proof or suggestions for how to proceed would be much appreciated.

• I am not sure the claim holds: take $A’=A\oplus Aq$ with $q^2=q$, $\mathfrak{m}’=\mathfrak{m}\oplus Aq$, then $G_{\mathfrak{m}}(A)=G_{\mathfrak{m}’}(A’)$, and $(A’,\mathfrak{m}’)$ is a local ring (I believe it is Noetherian) but not a domain. – Mindlack Jun 28 at 22:08
• @Mindlack Thanks - though I'm not sure I follow your comment. What exactly is $Aq$? If it is a nonzero ring (and $A$ is nonzero), then $A'$ is not local. In any event, it is true that if $G_\mathfrak{m}(A)$ is an integral domain, then $A$ is also an integral domain. This is Lemma 11.23 in Atiyah-MacDonald. I am interested in the corresponding claim for the property of being integrally closed. – 351910953 Jun 29 at 0:48
• See here. – user26857 Jun 29 at 21:19