Remark 51.8.6. Let $X$ be a locally Noetherian scheme. Let $j : U \to X$ be the inclusion of an open subscheme with complement $Z$. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ U$-module. If there exists an $x \in \text{Ass}(\mathcal{F})$ and $z \in Z \cap \overline{\{ x\} }$ such that $\dim (\mathcal{O}_{\overline{\{ x\} }, z}) \leq 1$, then $j_*\mathcal{F}$ is not coherent. To prove this we can do a flat base change to the spectrum of $\mathcal{O}_{X, z}$. Let $X' = \overline{\{ x\} }$. The assumption implies $\mathcal{O}_{X' \cap U} \subset \mathcal{F}$. Thus it suffices to see that $j_*\mathcal{O}_{X' \cap U}$ is not coherent. This is clear because $X' = \{ x, z\}$, hence $j_*\mathcal{O}_{X' \cap U}$ corresponds to $\kappa (x)$ as an $\mathcal{O}_{X, z}$-module which cannot be finite as $x$ is not a closed point.

In fact, the converse of Lemma 51.8.4 holds true: given an open immersion $j : U \to X$ of integral Noetherian schemes and there exists a $z \in X \setminus U$ and an associated prime $\mathfrak p$ of the completion $\mathcal{O}_{X, z}^\wedge$ with $\dim (\mathcal{O}_{X, z}^\wedge /\mathfrak p) = 1$, then $j_*\mathcal{O}_ U$ is not coherent. Namely, you can pass to the local ring, you can enlarge $U$ to the punctured spectrum, you can pass to the completion, and then the argument above gives the nonfiniteness.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).