Is Hartshorne's definition of associated points correct in the non-noetherian setting?

Question

Hartshorne defines (p. 257 of III.9) an associated point of a scheme $X$ as a point such that $\mathfrak{m}_x$ is an associated prime of the local ring which he says is equivalent to the maximal ideal consisting of only zero divisors.

I am wondering if these are actually equivalent in the non-noetherian setting? I am aware that associated points are badly behaved and weakly associated point is the better notion in this generality but I am wondering if these two definitions (maximal ideal consists of zero divisors and maximal ideal is associated) are actually still equivalent.

Answer 1

They aren't equivalent, unless Hartshorne's usage of "associated prime" is different from e.g. the definition in the Stacks project [Definition 00LA].

Indeed, [Example 05AI] gives the non-Noetherian local ring $A = k[x_1,x_2,\ldots]/(\{x_i^2\}_{i \geq 1})$ (for a field $k$) with no associated primes, but whose maximal ideal is locally nilpotent (i.e. every element in the maximal ideal is nilpotent).

Furthermore, the proposition afterwards (Hartshorne Algebraic Geometry Proposition III.9.7) is false without Noetherian hypotheses, again if "associated prime" has the definition as linked above, in the Stacks project. The proposition says that any scheme $X$ over an integral, regular, dimension $1$ scheme $Y$ is flat if and only if all associated points map to the generic point of $Y$.

Indeed, set $X = \mathrm{Spec} ~A$ where the ring $A$ is as above with $k$ of positive characteristic, and set $Y = \mathrm{Spec} ~\mathbb{Z}$. Then $X \rightarrow Y$ is not flat, but $X$ has no associated points so the hypotheses of the proposition are vacuously satisfied.

However, Hartshorne's claims here are ok if we interpret his "associated prime" as the notion called "weakly associated prime" in the Stacks project [Section 0546]. These notions are equivalent for Noetherian rings.