Lemma 53.14.1. In the situation above, let $Z = \mathop{\mathrm{Spec}}(k')$ where $k'$ is a field and $Z' = \mathop{\mathrm{Spec}}(k'_1 \times \ldots \times k'_ n)$ with $k'_ i/k'$ finite extensions of fields. Let $x \in X$ be the image of $Z \to X$ and $x'_ i \in X'$ the image of $\mathop{\mathrm{Spec}}(k'_ i) \to X'$. Then we have a fibre product diagram

$\xymatrix{ \prod \nolimits _{i = 1, \ldots , n} k'_ i & \prod \nolimits _{i = 1, \ldots , n} \mathcal{O}_{X', x'_ i}^\wedge \ar[l] \\ k' \ar[u] & \mathcal{O}_{X, x}^\wedge \ar[u] \ar[l] }$

where the horizontal arrows are given by the maps to the residue fields.

Proof. Choose an affine open neighbourhood $\mathop{\mathrm{Spec}}(A)$ of $x$ in $X$. Let $\mathop{\mathrm{Spec}}(A') \subset X'$ be the inverse image. By construction we have a fibre product diagram

$\xymatrix{ \prod \nolimits _{i = 1, \ldots , n} k'_ i & A' \ar[l] \\ k' \ar[u] & A \ar[u] \ar[l] }$

Since everything is finite over $A$ we see that the diagram remains a fibre product diagram after completion with respect to the maximal ideal $\mathfrak m \subset A$ corresponding to $x$ (Algebra, Lemma 10.97.2). Finally, apply Algebra, Lemma 10.97.8 to identify the completion of $A'$. $\square$

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