Lemma 62.3.1. Let $K/k$ be a field extension. Let $Z$ be an integral locally algebraic scheme over $k$. The multiplicity $m_{Z', Z_ K}$ of an irreducible component $Z' \subset Z_ K$ is $1$ or a power of the characteristic of $k$.

**Proof.**
If the characteristic of $k$ is zero, then $k$ is perfect and the multiplicity is always $1$ since $X_ K$ is reduced by Varieties, Lemma 33.6.4. Assume the characteristic of $k$ is $p > 0$. Let $L$ be the function field of $Z$. Since $Z$ is locally algebraic over $k$, the field extension $L/k$ is finitely generated. The ring $K \otimes _ k L$ is Noetherian (Algebra, Lemma 10.31.8). Translated into algebra, we have to show that the length of the artinian local ring $(K \otimes _ k L)_\mathfrak q$ is a power of $p$ for every minimal prime ideal $\mathfrak q$.

Let $L'/L$ be a finite purely inseparable extension, say of degree $p^ n$. Then $K \otimes _ k L \subset K \otimes _ k L'$ is a finite free ring map of degree $p^ n$ which induces a homeomorphism on spectra and purely inseparable residue field extensions. Hence for every minimal prime $\mathfrak q$ as above there is a unique minimal prime $\mathfrak q' \subset K \otimes _ k L'$ lying over it and

by Algebra, Lemma 10.52.12 applied to $M = (K \otimes _ k L')_{\mathfrak q'} \cong (K \otimes _ k L)_{\mathfrak q}^{\oplus p^ n}$. Since $[\kappa (\mathfrak q') : \kappa (\mathfrak q)]$ is a power of $p$ we conclude that it suffices to prove the statement for $L'$ and $\mathfrak q'$.

By the previous paragraph and Algebra, Lemma 10.45.3 we may assume that we have a subfield $L/k'/k$ such that $L/k'$ is separable and $k'/k$ is finite purely inseparable. Then $K \otimes _ k k'$ is an Artinian local ring. The argument of the preceding paragraph (applied to $L = k$ and $L' = k'$) shows that $\text{length}(K \otimes _ k k')$ is a power of $p$. Since $L/k'$ is the localization of a smooth $k'$-algebra (Algebra, Lemma 10.158.10). Hence $S = (K \otimes _ k L)_\mathfrak q$ is the localization of a smooth $R = K \otimes _ k k'$-algebra at a minimal prime. Thus $R \to S$ is a flat local homomorphism of Artinian local rings and $\mathfrak m_ R S = \mathfrak m_ S$. It follows from Algebra, Lemma 10.52.13 that $\text{length}(K \otimes _ k k') = \text{length}(R) = \text{length}(S) = \text{length}((K \otimes _ k L)_\mathfrak q)$ and the proof is finished. $\square$

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