Lemma 10.47.7. Let $k$ be a field. Let $S$ be a geometrically irreducible $k$-algebra. Let $R$ be any $k$-algebra. The map

induces a bijection on irreducible components.

Lemma 10.47.7. Let $k$ be a field. Let $S$ be a geometrically irreducible $k$-algebra. Let $R$ be any $k$-algebra. The map

\[ \mathop{\mathrm{Spec}}(R \otimes _ k S) \longrightarrow \mathop{\mathrm{Spec}}(R) \]

induces a bijection on irreducible components.

**Proof.**
Recall that irreducible components correspond to minimal primes (Lemma 10.26.1). As $R \to R \otimes _ k S$ is flat we see by going down (Lemma 10.39.19) that any minimal prime of $R \otimes _ k S$ lies over a minimal prime of $R$. Conversely, if $\mathfrak p \subset R$ is a (minimal) prime then

\[ R \otimes _ k S/\mathfrak p(R \otimes _ k S) = (R/\mathfrak p) \otimes _ k S \subset \kappa (\mathfrak p) \otimes _ k S \]

by flatness of $R \to R \otimes _ k S$. The ring $\kappa (\mathfrak p) \otimes _ k S$ has irreducible spectrum by assumption. It follows that $R \otimes _ k S/\mathfrak p(R \otimes _ k S)$ has a single minimal prime (Lemma 10.30.5). In other words, the inverse image of the irreducible set $V(\mathfrak p)$ is irreducible. Hence the lemma follows. $\square$

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