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
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