**Proof.**
By construction $R^ h$ is a colimit of étale $R$-algebras, see Algebra, Lemma 10.155.1. Since étale ring maps are flat (Algebra, Lemma 10.143.3) we see that $R^ h$ is flat over $R$ by Algebra, Lemma 10.39.3. As a flat local ring homomorphism is faithfully flat (Algebra, Lemma 10.39.17) we see that $R \to R^ h$ is faithfully flat. The ring map $R^ h \to R^{sh}$ is a colimit of finite étale ring maps, see proof of Algebra, Lemma 10.155.2. Hence the same arguments as above show that $R^ h \to R^{sh}$ is faithfully flat.

Part (2) follows from Algebra, Lemmas 10.155.1 and 10.155.2. Part (3) follows from Algebra, Lemma 10.101.1 because $R/\mathfrak m \to R^ h/\mathfrak mR^ h$ is an isomorphism and $R/\mathfrak m^ n \to R^ h/\mathfrak m^ nR^ h$ is flat as a base change of the flat ring map $R \to R^ h$ (Algebra, Lemma 10.39.7). Let $\kappa ^{sep}$ be the residue field of $R^{sh}$ (it is a separable algebraic closure of $\kappa $). Choose $x_ i \in R^{sh}$ mapping to a basis of $\kappa ^{sep}$ as a $\kappa $-vector space. Then (4) follows from Algebra, Lemma 10.101.1 in exactly the same way as above.
$\square$

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