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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