$\newcommand{\aaa}{\mathfrak{a}} \newcommand{\mmm}{\mathfrak{m}} \newcommand{\qqq}{\mathfrak{q}} $

- Volume 2, Chapter VIII, Section 10 (Theory of Multiplicities), Proof of Theorem 22: in the case of zero-divisors in dimension 1, the set up is this:
- $A$ is a local ring with maximal ideal $\mmm$ such that $A/\mmm$ is an infinite field,
- $\qqq$ is an ideal of $A$ with $\sqrt{\qqq} = \mmm$,
- $x \in A$ is outside of all isolated prime ideals of $0$,
- $x$ is also superficial of order $1$ for $\qqq$, i.e. there is a nonnegative integer $c$ such that $(\qqq^n: Ax) \cap \qqq^c = \qqq^{n-1}$ for $n \gg 1$,
- $\dim(A) = 1$,
- $\mmm$ is an embedded prime ideal of $0$, i.e. all elements of $\mmm$ are zero divisors.

With this, they define $\aaa := (0: Ax)$ and at some point claim that**$x$ is a non zero-divisor in $A/\aaa$**. This is**false**, as seen from the following example: $A := k[[x, y]]/\langle x^2y, y^2 \rangle$, where $k$ is an infinite field, and $\qqq = \mmm = \langle x, y \rangle$. Then every element of $A$ can be represented as $f(x) + ay + bxy$ with $f \in k[[x]]$, $a, b \in k$. It is straightforward to check that $\langle x^2y, y^2 \rangle = \langle y \rangle \cap \langle x^2, y^2 \rangle$ induces a primary decomposition of the zero ideal in $A$, and $x$ satisfies all the above properties. However, in this case $\aaa = Axy$, and $x$ is a zero-divisor in $A/\aaa$, since $xy = 0 \in A/\aaa$. This gives the required counterexample. (The proof can be salvaged by defining $\aaa$ as $\bigcup_n (0: Ax^n)$.)