My question refers to a step in @KReiser's answer in following thread: regularity of finite type $k$-schemes by base changing to $\bar{k}$ and an equality of dimensions of tangent spaces
Here the complete answer:
Here we consider the preimage $\{p_1, ..., p_d\}= \pi^{-1}(p)$ of a closed point $p \in X$ (so it corresponds to a maximal ideal $m$) under the morphism $\pi: X \otimes {\bar{k}} =X_{\bar{k}} \to X$.We know that it has the shape $\pi^{-1}(p) = \sqcup_d \mathrm{Spec}(\bar{k})$.
Let the residual field $k' := Frac(\mathcal{O}_{X,p})$ at $p$ be a separable (!) extension of degree $d$ of $k$. Denote futhermore by $m_i$ the maximal ideals corresponding to preimages $p_i$ of $p$.
My question is how to see that $$(\mathfrak{m}\otimes_{k'}\overline{k})=\mathfrak{m}_1\mathfrak{m}_2\cdots\mathfrak{m}_d$$ holds?
The cruical point was according to @KReiser's explanation that $k'$ is separable but here I don't see how to use it.
Attempts: I know that one can characterize separableness $k'$ by criterion $rad(k' \otimes _k \overline{k}) =0$. Can I somehow make usage of this fact?
$\mathfrak{m}\otimes_k\overline{k}$ is the ideal of the fiber over $p$. The fiber is Spec of $A'/(\mathfrak{m}\otimes\overline{k})$, and this ring is isomorphic to $(A/\mathfrak{m})\otimes_{k} \overline{k}$, which is further isomorphic to $k'\otimes_k \overline{k}$, so this ring is isomorphic to $\overline{k}^d$ (this last isomorphism is where we used separability). From here, we know that $\mathfrak{m}\otimes_k\overline{k}$ factors as a product of maximal ideals by considering it's primary decomposition (every primary ideal in the decomposition must be a power of a maximal ideal, but since the quotient is reduced, all of these powers must be 1, and since they're all comaximal, the intersection is exactly the product).