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. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #8740 by pkas on
Comment #9337 by Stacks Project on