Lemma 58.12.1. Let $A$ be a normal domain whose fraction field $K$ is separably algebraically closed. Let $\mathfrak p \subset A$ be a nonzero prime ideal. Then the residue field $\kappa (\mathfrak p)$ is algebraically closed.

**Proof.**
Assume the lemma is not true to get a contradiction. Then there exists a monic irreducible polynomial $P(T) \in \kappa (\mathfrak p)[T]$ of degree $d > 1$. After replacing $P$ by $a^ d P(a^{-1}T)$ for suitable $a \in A$ (to clear denominators) we may assume that $P$ is the image of a monic polynomial $Q$ in $A[T]$. Observe that $Q$ is irreducible in $K[T]$. Namely a factorization over $K$ leads to a factorization over $A$ by Algebra, Lemma 10.38.5 which we could reduce modulo $\mathfrak p$ to get a factorization of $P$. As $K$ is separably closed, $Q$ is not a separable polynomial (Fields, Definition 9.12.2). Then the characteristic of $K$ is $p > 0$ and $Q$ has vanishing linear term (Fields, Definition 9.12.2). However, then we can replace $Q$ by $Q + a T$ where $a \in \mathfrak p$ is nonzero to get a contradiction.
$\square$

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