Lemma 110.38.1. There exists a local homomorphism $A \to B$ of local domains which is essentially of finite type and such that $A/\mathfrak m_ A \to B/\mathfrak m_ B$ is finite such that for every prime $\mathfrak q \not= \mathfrak m_ B$ of $B$ the ring map $A \to B/\mathfrak q$ is not the localization of a quasi-finite ring map.

## 110.38 Topology of a finite type ring map

Let $A \to B$ be a local map of local domains. If $A$ is Noetherian, $A \to B$ is essentially of finite type, and $A/\mathfrak m_ A \subset B/\mathfrak m_ B$ is finite then there exists a prime $\mathfrak q \subset B$, $\mathfrak q \not= \mathfrak m_ B$ such that $A \to B/\mathfrak q$ is the localization of a quasi-finite ring map. See More on Morphisms, Lemma 37.52.6.

In this section we give an example that shows this result is false $A$ is no longer Noetherian. Namely, let $k$ be a field and set

and

The inclusion $A \to C$ is of finite type as $C$ is generated by $y$ over $A$. We claim that $A$ is a local ring with maximal ideal $\mathfrak m = \{ a_1 x + a_2 x^2 + \ldots \in A\} $ and no prime ideals besides $(0)$ and $\mathfrak m$. Namely, an element $f = a_0 + a_1 x + a_2 x^2 + \ldots $ of $A$ is invertible as soon as $a_0 \not= 0$. If $\mathfrak q \subset A$ is a nonzero prime ideal, and $f = a_ i x^ i + \ldots \in \mathfrak q$, then using properties of power series one sees that for any $g \in k((y))$ the element $g^{i + 1} x^{i + 1} \in \mathfrak q$, i.e., $gx \in \mathfrak q$. This proves that $\mathfrak q = \mathfrak m$.

As to the spectrum of the ring $C$, arguing in the same way as above we see that any nonzero prime ideal contains the prime $\mathfrak p = \{ a_1 x + a_2 x^2 + \ldots \in C\} $ which lies over $\mathfrak m$. Thus the only prime of $C$ which lies over $(0)$ is $(0)$. Set $\mathfrak m_ C = yC + \mathfrak p$ and $B = C_{\mathfrak m_ C}$. Then $A \to B$ is the desired example.

**Proof.**
See the discussion above.
$\square$

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