Lemma 10.113.2. Let $A \to B$ be a ring map. Assume

$A \subset B$ is an extension of domains,

the induced extension of fraction fields is finite,

$A$ is Noetherian, and

$A \to B$ is of finite type.

Let $\mathfrak p \subset A$ be a prime of height $1$. Then there are at most finitely many primes of $B$ lying over $\mathfrak p$ and they all have height $1$.

**Proof.**
By the dimension formula (Lemma 10.113.1) for any prime $\mathfrak q$ lying over $\mathfrak p$ we have

\[ \dim (B_{\mathfrak q}) \leq \dim (A_{\mathfrak p}) - \text{trdeg}_{\kappa (\mathfrak p)} \kappa (\mathfrak q). \]

As the domain $B_\mathfrak q$ has at least $2$ prime ideals we see that $\dim (B_{\mathfrak q}) \geq 1$. We conclude that $\dim (B_{\mathfrak q}) = 1$ and that the extension $\kappa (\mathfrak p) \subset \kappa (\mathfrak q)$ is algebraic. Hence $\mathfrak q$ defines a closed point of its fibre $\mathop{\mathrm{Spec}}(B \otimes _ A \kappa (\mathfrak p))$, see Lemma 10.35.9. Since $B \otimes _ A \kappa (\mathfrak p)$ is a Noetherian ring the fibre $\mathop{\mathrm{Spec}}(B \otimes _ A \kappa (\mathfrak p))$ is a Noetherian topological space, see Lemma 10.31.5. A Noetherian topological space consisting of closed points is finite, see for example Topology, Lemma 5.9.2.
$\square$

## Comments (0)