Lemma 110.39.1. 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.

## 110.39 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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)