Lemma 15.39.4. Let $S \to R$ and $S' \to R$ be surjective maps of complete Noetherian local rings. Then $S \times _ R S'$ is a complete Noetherian local ring.
Proof. Let $k$ be the residue field of $R$. If the characteristic of $k$ is $p > 0$, then we denote $\Lambda $ a Cohen ring (Algebra, Definition 10.160.5) with residue field $k$ (Algebra, Lemma 10.160.6). If the characteristic of $k$ is $0$ we set $\Lambda = k$. Choose a surjection $\Lambda [[x_1, \ldots , x_ n]] \to R$ (as in the Cohen structure theorem, see Algebra, Theorem 10.160.8) and lift this to maps $\Lambda [[x_1, \ldots , x_ n]] \to S$ and $\varphi : \Lambda [[x_1, \ldots , x_ n]] \to S$ and $\varphi ' : \Lambda [[x_1, \ldots , x_ n]] \to S'$ using Lemmas 15.39.1 and 15.37.5. Next, choose $f_1, \ldots , f_ m \in S$ generating the kernel of $S \to R$ and $f'_1, \ldots , f'_{m'} \in S'$ generating the kernel of $S' \to R$. Then the map
\[ \Lambda [[x_1, \ldots , x_ n, y_1, \ldots , y_ m, z_1, \ldots , z_{m'}]] \longrightarrow S \times _ R S, \]
which sends $x_ i$ to $(\varphi (x_ i), \varphi '(x_ i))$ and $y_ j$ to $(f_ j, 0)$ and $z_{j'}$ to $(0, f'_ j)$ is surjective. Thus $S \times _ R S'$ is a quotient of a complete local ring, whence complete. $\square$
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