Lemma 10.160.11. Let $(R, \mathfrak m)$ be a Noetherian complete local domain. Then there exists a $R_0 \subset R$ with the following properties
$R_0$ is a regular complete local ring,
$R_0 \subset R$ is finite and induces an isomorphism on residue fields,
$R_0$ is either isomorphic to $k[[X_1, \ldots , X_ d]]$ where $k$ is a field or $\Lambda [[X_1, \ldots , X_ d]]$ where $\Lambda $ is a Cohen ring.
Proof. Let $\Lambda $ be a coefficient ring of $R$. Since $R$ is a domain we see that either $\Lambda $ is a field or $\Lambda $ is a Cohen ring.
Case I: $\Lambda = k$ is a field. Let $d = \dim (R)$. Choose $x_1, \ldots , x_ d \in \mathfrak m$ which generate an ideal of definition $I \subset R$. (See Section 10.60.) By Lemma 10.96.9 we see that $R$ is $I$-adically complete as well. Consider the map $R_0 = k[[X_1, \ldots , X_ d]] \to R$ which maps $X_ i$ to $x_ i$. Note that $R_0$ is complete with respect to the ideal $I_0 = (X_1, \ldots , X_ d)$, and that $R/I_0R \cong R/IR$ is finite over $k = R_0/I_0$ (because $\dim (R/I) = 0$, see Section 10.60.) Hence we conclude that $R_0 \to R$ is finite by Lemma 10.96.12. Since $\dim (R) = \dim (R_0)$ this implies that $R_0 \to R$ is injective (see Lemma 10.112.3), and the lemma is proved.
Case II: $\Lambda $ is a Cohen ring. Let $d + 1 = \dim (R)$. Let $p > 0$ be the characteristic of the residue field $k$. As $R$ is a domain we see that $p$ is a nonzerodivisor in $R$. Hence $\dim (R/pR) = d$, see Lemma 10.60.13. Choose $x_1, \ldots , x_ d \in R$ which generate an ideal of definition in $R/pR$. Then $I = (p, x_1, \ldots , x_ d)$ is an ideal of definition of $R$. By Lemma 10.96.9 we see that $R$ is $I$-adically complete as well. Consider the map $R_0 = \Lambda [[X_1, \ldots , X_ d]] \to R$ which maps $X_ i$ to $x_ i$. Note that $R_0$ is complete with respect to the ideal $I_0 = (p, X_1, \ldots , X_ d)$, and that $R/I_0R \cong R/IR$ is finite over $k = R_0/I_0$ (because $\dim (R/I) = 0$, see Section 10.60.) Hence we conclude that $R_0 \to R$ is finite by Lemma 10.96.12. Since $\dim (R) = \dim (R_0)$ this implies that $R_0 \to R$ is injective (see Lemma 10.112.3), and the lemma is proved. $\square$
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