Lemma 10.153.12. Let $\varphi : R \to S$ be a local homomorphism of strictly henselian local rings. Let $P_1, \ldots , P_ n \in R[x_1, \ldots , x_ n]$ be polynomials such that $R[x_1, \ldots , x_ n]/(P_1, \ldots , P_ n)$ is étale over $R$. Then the map

\[ R^ n \longrightarrow S^ n, \quad (h_1, \ldots , h_ n) \longmapsto (\varphi (h_1), \ldots , \varphi (h_ n)) \]

induces a bijection between

\[ \{ (r_1, \ldots , r_ n) \in R^ n \mid P_ i(r_1, \ldots , r_ n) = 0, \ i = 1, \ldots , n \} \]

and

\[ \{ (s_1, \ldots , s_ n) \in S^ n \mid P^\varphi _ i(s_2, \ldots , s_ n) = 0, \ i = 1, \ldots , n \} \]

where $P^\varphi _ i \in S[x_1, \ldots , x_ n]$ are the images of the $P_ i$ under $\varphi $.

**Proof.**
The first solution set is canonically isomorphic to the set

\[ \mathop{\mathrm{Hom}}\nolimits _ R(R[x_1, \ldots , x_ n]/(P_1, \ldots , P_ n), R). \]

As $R$ is henselian the map $R \to R/\mathfrak m_ R$ induces a bijection between this set and the set of solutions in the residue field $R/\mathfrak m_ R$, see Lemma 10.153.3. The same is true for $S$. Now since $R[x_1, \ldots , x_ n]/(P_1, \ldots , P_ n)$ is étale over $R$ and $R/\mathfrak m_ R$ is separably algebraically closed we see that $R/\mathfrak m_ R[x_1, \ldots , x_ n]/(\overline{P}_1, \ldots , \overline{P}_ n)$ is a finite product of copies of $R/\mathfrak m_ R$ where $\overline{P}_ i$ is the image of $P_ i$ in $R/\mathfrak m_ R[x_1, \ldots , x_ n]$. Hence the tensor product

\[ R/\mathfrak m_ R[x_1, \ldots , x_ n]/(\overline{P}_1, \ldots , \overline{P}_ n) \otimes _{R/\mathfrak m_ R} S/\mathfrak m_ S = S/\mathfrak m_ S[x_1, \ldots , x_ n]/ (\overline{P}^\varphi _1, \ldots , \overline{P}^\varphi _ n) \]

is also a finite product of copies of $S/\mathfrak m_ S$ with the same index set. This proves the lemma.
$\square$

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