In Chapter 11 of Atiyah-Macdonald, the authors prove that for a ring $A$, maximal ideal $\mathfrak{m}$, and $\mathfrak{m}$-primary ideal $\mathfrak{q}$, we have that if $M$ is a finitely-generated $A$-module, $x \in A$ is not a zero-divisor in $M$, and $M' = M / x M$, then $\deg \chi_\mathfrak{q}^{M'} \leq \deg \chi_\mathfrak{q}^M - 1$, where $\chi$ is the characteristic polynomial of $\mathfrak{q}$ (Proposition 11.8).
The authors then state the following corollary and proof:
Corollary 11.9: If $A$ is a Noetherian local ring, $x$ a non-zero-divisor in $A$, then $d(A / (x)) \leq d(A) - 1$
Proof: Put $M = A$ in (11.8).
It is clear to me that taking $M = A$, we have a finitely-generated $A$-module since $A$ is Noetherian, but in order to apply the result, we need to know that $(x)$ is $\mathfrak{m}$-primary. How do we know that any non-zero-divisor in a local Noetherian ring generates an $\mathfrak{m}$-primary ideal?
It seems to me that if $A$ has dimension greater than one then this should not be the case. For example, if we localize the coordinate ring of an irreducible (smooth) variety in $\mathbb{A}^3$ at a point on that variety that lies on some irreducible subvariety, shouldn't that give a local ring, where the ideal corresponding to the subvariety is a prime strictly contained in the maximal ideal? Then the ideal generated by any point inside that prime will have radical equal to the prime instead of equal to the maximal ideal, so will not be $\mathfrak{m}$-primary? For concreteness: Wouldn't any non-zero-divisor in the image of $\langle y - x^2, z \rangle$ in $(k[x,y,z]/\langle x^2 + y^2 - z \rangle)_{(x,y,z)}$ generate an ideal that is not $(x,y,z)$-primary?