Here is a fundamental notion in commutative algebra.
Definition 10.160.1. Let $(R, \mathfrak m)$ be a local ring. We say $R$ is a complete local ring if the canonical map to the completion of $R$ with respect to $\mathfrak m$ is an isomorphism1.
\[ R \longrightarrow \mathop{\mathrm{lim}}\nolimits _ n R/\mathfrak m^ n \]
Note that an Artinian local ring $R$ is a complete local ring because $\mathfrak m_ R^ n = 0$ for some $n > 0$. In this section we mostly focus on Noetherian complete local rings.
Lemma 10.160.2. Let $R$ be a Noetherian complete local ring. Any quotient of $R$ is also a Noetherian complete local ring. Given a finite ring map $R \to S$, then $S$ is a product of Noetherian complete local rings.
Proof. The ring $S$ is Noetherian by Lemma 10.31.1. As an $R$-module $S$ is complete by Lemma 10.97.1. Hence $S$ is the product of the completions at its maximal ideals by Lemma 10.97.8. $\square$
Lemma 10.160.3. Let $(R, \mathfrak m)$ be a complete local ring. If $\mathfrak m$ is a finitely generated ideal then $R$ is Noetherian.
Proof. See Lemma 10.97.5. $\square$
Definition 10.160.4. Let $(R, \mathfrak m)$ be a complete local ring. A subring $\Lambda \subset R$ is called a coefficient ring if the following conditions hold:
$\Lambda $ is a complete local ring with maximal ideal $\Lambda \cap \mathfrak m$,
the residue field of $\Lambda $ maps isomorphically to the residue field of $R$, and
$\Lambda \cap \mathfrak m = p\Lambda $, where $p$ is the characteristic of the residue field of $R$.
Let us make some remarks on this definition. We split the discussion into the following cases:
The local ring $R$ contains a field. This happens if either $\mathbf{Q} \subset R$, or $pR = 0$ where $p$ is the characteristic of $R/\mathfrak m$. In this case a coefficient ring $\Lambda $ is a field contained in $R$ which maps isomorphically to $R/\mathfrak m$.
The characteristic of $R/\mathfrak m$ is $p > 0$ but no power of $p$ is zero in $R$. In this case $\Lambda $ is a complete discrete valuation ring with uniformizer $p$ and residue field $R/\mathfrak m$.
The characteristic of $R/\mathfrak m$ is $p > 0$, and for some $n > 1$ we have $p^{n - 1} \not= 0$, $p^ n = 0$ in $R$. In this case $\Lambda $ is an Artinian local ring whose maximal ideal is generated by $p$ and which has residue field $R/\mathfrak m$.
The complete discrete valuation rings with uniformizer $p$ above play a special role and we baptize them as follows.
Definition 10.160.5. A Cohen ring is a complete discrete valuation ring with uniformizer $p$ a prime number.
Lemma 10.160.6. Let $p$ be a prime number. Let $k$ be a field of characteristic $p$. There exists a Cohen ring $\Lambda $ with $\Lambda /p\Lambda \cong k$.
Proof. First note that the $p$-adic integers $\mathbf{Z}_ p$ form a Cohen ring for $\mathbf{F}_ p$. Let $k$ be an arbitrary field of characteristic $p$. Let $\mathbf{Z}_ p \to R$ be a flat local ring map such that $\mathfrak m_ R = pR$ and $R/pR = k$, see Lemma 10.159.1. By Lemma 10.97.5 the completion $\Lambda = R^\wedge $ is Noetherian. It is a complete Noetherian local ring with maximal ideal $(p)$ as $\Lambda /p\Lambda = R/pR$ is a field (use Lemma 10.96.3). Since $\mathbf{Z}_ p \to R \to \Lambda $ is flat (by Lemma 10.97.2) we see that $p$ is a nonzerodivisor in $\Lambda $. Hence $\Lambda $ has dimension $\geq 1$ (Lemma 10.60.13) and we conclude that $\Lambda $ is regular of dimension $1$, i.e., a discrete valuation ring by Lemma 10.119.7. We conclude $\Lambda $ is a Cohen ring for $k$. $\square$
Lemma 10.160.7. Let $p > 0$ be a prime. Let $\Lambda $ be a Cohen ring with residue field of characteristic $p$. For every $n \geq 1$ the ring map is formally smooth.
\[ \mathbf{Z}/p^ n\mathbf{Z} \to \Lambda /p^ n\Lambda \]
Proof. If $n = 1$, this follows from Proposition 10.158.9. For general $n$ we argue by induction on $n$. Namely, if $\mathbf{Z}/p^ n\mathbf{Z} \to \Lambda /p^ n\Lambda $ is formally smooth, then we can apply Lemma 10.138.12 to the ring map $\mathbf{Z}/p^{n + 1}\mathbf{Z} \to \Lambda /p^{n + 1}\Lambda $ and the ideal $I = (p^ n) \subset \mathbf{Z}/p^{n + 1}\mathbf{Z}$. $\square$
Theorem 10.160.8 (Cohen structure theorem). Let $(R, \mathfrak m)$ be a complete local ring.
$R$ has a coefficient ring (see Definition 10.160.4),
if $\mathfrak m$ is a finitely generated ideal, then $R$ is isomorphic to a quotient
where $\Lambda $ is either a field or a Cohen ring.
\[ \Lambda [[x_1, \ldots , x_ n]]/I \]
Proof. Let us prove a coefficient ring exists. First we prove this in case the characteristic of the residue field $\kappa $ is zero. Namely, in this case we will prove by induction on $n > 0$ that there exists a section
to the canonical map $R/\mathfrak m^ n \to \kappa = R/\mathfrak m$. This is trivial for $n = 1$. If $n > 1$, let $\varphi _{n - 1}$ be given. The field extension $\kappa /\mathbf{Q}$ is formally smooth by Proposition 10.158.9. Hence we can find the dotted arrow in the following diagram
This proves the induction step. Putting these maps together
gives a map whose image is the desired coefficient ring.
Next, we prove the existence of a coefficient ring in the case where the characteristic of the residue field $\kappa $ is $p > 0$. Namely, choose a Cohen ring $\Lambda $ with $\kappa = \Lambda /p\Lambda $, see Lemma 10.160.6. In this case we will prove by induction on $n > 0$ that there exists a map
whose composition with the reduction map $R/\mathfrak m^ n \to \kappa $ produces the given isomorphism $\Lambda /p\Lambda = \kappa $. This is trivial for $n = 1$. If $n > 1$, let $\varphi _{n - 1}$ be given. The ring map $\mathbf{Z}/p^ n\mathbf{Z} \to \Lambda /p^ n\Lambda $ is formally smooth by Lemma 10.160.7. Hence we can find the dotted arrow in the following diagram
This proves the induction step. Putting these maps together
gives a map whose image is the desired coefficient ring.
The final statement of the theorem follows readily. Namely, if $y_1, \ldots , y_ n$ are generators of the ideal $\mathfrak m$, then we can use the map $\Lambda \to R$ just constructed to get a map
Since both sides are $(x_1, \ldots , x_ n)$-adically complete this map is surjective by Lemma 10.96.1 as it is surjective modulo $(x_1, \ldots , x_ n)$ by construction. $\square$
Remark 10.160.9. If $k$ is a field then the power series ring $k[[X_1, \ldots , X_ d]]$ is a Noetherian complete local regular ring of dimension $d$. If $\Lambda $ is a Cohen ring then $\Lambda [[X_1, \ldots , X_ d]]$ is a complete local Noetherian regular ring of dimension $d + 1$. Hence the Cohen structure theorem implies that any Noetherian complete local ring is a quotient of a regular local ring. In particular we see that a Noetherian complete local ring is universally catenary, see Lemma 10.105.9 and Lemma 10.106.3.
Lemma 10.160.10. Let $(R, \mathfrak m)$ be a Noetherian complete local ring. Assume $R$ is regular.
If $R$ contains either $\mathbf{F}_ p$ or $\mathbf{Q}$, then $R$ is isomorphic to a power series ring over its residue field.
If $k$ is a field and $k \to R$ is a ring map inducing an isomorphism $k \to R/\mathfrak m$, then $R$ is isomorphic as a $k$-algebra to a power series ring over $k$.
Proof. In case (1), by the Cohen structure theorem (Theorem 10.160.8) there exists a coefficient ring which must be a field mapping isomorphically to the residue field. Thus it suffices to prove (2). In case (2) we pick $f_1, \ldots , f_ d \in \mathfrak m$ which map to a basis of $\mathfrak m/\mathfrak m^2$ and we consider the continuous $k$-algebra map $k[[x_1, \ldots , x_ d]] \to R$ sending $x_ i$ to $f_ i$. As both source and target are $(x_1, \ldots , x_ d)$-adically complete, this map is surjective by Lemma 10.96.1. On the other hand, it has to be injective because otherwise the dimension of $R$ would be $< d$ by Lemma 10.60.13. $\square$
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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