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.
Best Answer
user26857 linked a book where the result is proven: D. Bump's Algebraic Geometry, World Scientific. To get easier access, I will reproduce the result and proof from the book here (Proposition 5.24.3 below) The following excerpt has been OCR'd from the book using Mathpix. If anyone spots any scanning typo, please edit the answer and correct it.
Proposition 5.22. Let $A$ be an Noetherian integral domain, and let $x$ be an element of its field of fractions. Then $x$ is integral over $A$ if and only if there exists $c \in A$ such that $c x^r \in A$ for all $r>0$.
Proof. First suppose that $x$ is integral over $A$. Write $x=y / z$ where $y, z \in A$. We have a relation of integral dependence $x^n+a_{n-1} x^{n-1}+\ldots+a_0=0$ over $A$. This implies that $x^n \in A+A x+A x^2+\ldots+A x^{n-1}$. Now by induction we see that $x^r \in A+A x+A x^2+\ldots+A x^{n-1}$ for all $r$. Consequently, if $c=z^{n-1}$, we have $c x^r \in A$ for all $r>0$. $\square$
On the other hand, suppose that $c x^r \in A$ for all $r>0$. Then $A[x] \subseteq c^{-1} A$. Since $c^{-1} A$ is a finitely generated module over the Noetherian $\operatorname{ring} A$, so is $A[x]$ by the Hilbert Basis Theorem (Exercise 1.4). Now it follows that $x$ is integral over $A$ by Proposition 5.21 with $M=A[x]$.
Proposition 5.23. Let $A$ be a Noetherian local ring with maximal ideal $\mathfrak{m}$, and let $\mathfrak{a} \subseteq \mathfrak{m}$ be a proper ideal of $A$. Then $\bigcap_{n=0}^{\infty}\left(\mathfrak{a}+\mathfrak{m}^n\right)=\mathfrak{a}$.
Proof. If $\mathfrak{a}$ is the zero ideal, this is Exercise 5.7. The general case follows from this special case by the following argument. Let $\bar{A}=A / \mathfrak{a}$. This is also a Noetherian local ring. Let $\overline{\mathfrak{m}}$ be the image of $\mathfrak{m}$ in $\bar{A}$. Then by Exercise 5.7 , we have $\bigcap_{n=0}^{\infty} \overline{\mathfrak{m}}^n=0$ in $\bar{A}$. Now the preimage of $\bigcap_{n=0}^{\infty} \overline{\mathfrak{m}}^n$ in $A$ is $\bigcap_{n=0}^{\infty}\left(\mathfrak{a}+\mathfrak{m}^n\right)$ so this equals $\mathfrak{a}$. $\square$
We now return to the graded ring $G_{\mathfrak{m}}(A)$ associated with a Noetherian local ring $A$ with maximal ideal $\mathfrak{m}$. By Exercise $5.7, \bigcap_{n=0}^{\infty} \mathfrak{m}^n=0$. This implies that if $0 \neq a \in A$, there is an integer $k$ such that $a \in \mathfrak{m}^k$ but $a \notin \mathfrak{m}^{k+1}$. We will denote $k=v(a)$. If $a=0$, we will denote $v(a)=\infty$. It is our convention that $\infty+n=n+\infty=\infty$ for $n \in \mathbb{Z} \cup\{\infty\}$, and that $\infty>n$ for all $n \in \mathbb{Z}$. It is obvious that $v(x+y) \geq \min (v(x), v(y))$.
Also, if $a \in A$ let $G(a)$ be the element of $G_{\mathfrak{m}}(A)$ which is defined as follows: let $k=v(a)$, and define $G(a)$ to be the image of $a$ in $\mathrm{m}^k / \mathrm{m}^{k+1}$, which is homogeneous of degree $k$ in $G(a)$.
Proposition 5.24. Let $A$ be a Noetherian local ring with maximal ideal $\mathfrak{m}$.
The graded ring $G_{\mathfrak{m}}(A)$ is Noetherian.
If $G_{\mathfrak{m}}(A)$ is an integral domain, then so is $A$. In this case, if $x, y \in A$, then $v(x y)=v(x)+v(y)$, and $G(x y)=G(x) G(y)$.
If $G_{\mathfrak{m}}(A)$ is an integrally closed integral domain, then so is $A$.
Proof. To see that $G_{\mathfrak{m}}(A)$ is Noetherian, let $x_1, \cdots, x_d$ generate $\mathfrak{m}$ as an ideal. Let $\bar{x}_i$ be the image of $x_i$ in $\mathfrak{m} / \mathfrak{m}^2$. Then it is clear that the $\bar{x}_i$ generate $G_{\mathfrak{m}}(A)$ as an algebra over the field $A / \mathfrak{m}$, which is the homogeneous part of degree zero. Hence $G_{\mathfrak{m}}(A)$ is Noetherian by the Hilbert Basis Theorem (Exercise 1.4). This proves part 1.
Suppose that $G_{\mathfrak{m}}(A)$ is an integral domain. Let $x, y \in A$, and suppose that neither $x$ nor $y$ is 0. As we have pointed out, since $\bigcap_{n=1}^{\infty} \mathfrak{m}^n=0$ by Exercise 5.7, there exist integers $n=v(x)$ and $m=v(y)$ such that $x \in \mathfrak{m}^n$ but $x \notin \mathfrak{m}^{n+1}$, and $y \in \mathfrak{m}^m$ but $y \notin \mathfrak{m}^{m+1}$. Thus the images $G(x)$ and $G(y)$ of $x$ and $y$ in $\mathfrak{m}^n / \mathfrak{m}^{n+1}$ and $\mathfrak{m}^m / \mathfrak{m}^{m+1}$ are nonzero. Since $G_{\mathfrak{m}}(A)$ is an integral domain, this implies that the image $G(x y)$ of $x y$ in $\mathfrak{m}^{n+m} / \mathfrak{m}^{n+m+1}$ is nonzero, so that $x y \notin \mathfrak{m}^{n+m+1}$. In particular, $x y \neq 0$, and $v(x y)=k+l$. This proves that $A$ is an integral domain, and that $v(x y)=v(x)+v(y)$. It is obvious that this implies $G(x y)=G(x) G(y)$. This proves part 2.
Now suppose that $G_{\mathfrak{m}}(A)$ is an integrally closed integral domain. By part 2 which we have already established, $A$ is an integral domain. Since $A$ and $G_{\mathfrak{m}}(A)$ are both Noetherian integral domains, we may use the criterion of Proposition 5.22 to test integral closure. Thus let $x$ lie in the field of fractions of $A$. Suppose that $x$ is integral over $A$. Then there exists $c \in A$ such that $c x^n \in A$ for all $n$. We must prove that $x \in A$.
Let $x=y / z$ where $y, z \in A$. We will prove that if there exists $\mu \in A$ such that $v(z(x-\mu))=k$, then $y \in(z)+\mathfrak{m}^k$, and moreover, there exists a $\mu^{\prime} \in A$ such that $v\left(z\left(x-\mu^{\prime}\right)\right)>k$. Of course if we can prove this, we may start the induction with $k=v(y)$ and $\mu=0$. We will then have $y \in(z)+\mathfrak{m}^k$ for all $k$, which by Proposition 5.23 implies that $y \in(z)$, so that $x \in A$.
Suppose that $v(z(x-\mu))=k$. since $z(x-\mu)=y-z \mu$, this implies that $y \in(z)+\mathfrak{m}^k$. We must show that we can find $\mu^{\prime}$ so that $v\left(z\left(x-\mu^{\prime}\right)\right)>k$.
Since $c x^n \in A$ for all $n$, expanding $c(x-\mu)^n$ by the Binomial Theorem shows that $c(x-\mu)^n \in A$ for all $n$, and so $z^n \mid c(y-\mu z)^n$ for all $n$. Since $G(a b)=G(a) G(b)$, we have $G(z)^n \mid G(c) G(y-\mu z)^n$ in $G_{\mathfrak{m}}(A)$ for all $n$. Since $G_{\mathfrak{m}}(A)$ is an integrally closed integral domain, Proposition 5.22 implies that $G(y-\mu z) / G(z) \in G_{\mathfrak{m}}(A)$, so $G(z) \mid G(y-\mu z)$. Thus there exists $w \in A$ such that $G(y-\mu z)=G(w) G(z)=G(w z)$. Since $v(y-\mu z)=k$, this means that $w z \in \mathfrak{m}^k$, and so $y-\mu z-w z \in \mathfrak{m}^{k+1}$. Thus we may take $\mu^{\prime}=\mu+w$. $\square$