## 86.14 Rig-closed points

We develop just enough theory to be able to use this for testing rig-flatness in a later section. The reader can find more theory in [BL-I] who discuss (among other things) the case of locally Noetherian formal schemes.

Lemma 86.14.1. Let $A$ be a Noetherian adic topological ring. Let $\mathfrak q \subset A$ be a prime ideal. The following are equivalent

1. for some ideal of definition $I \subset A$ we have $I \not\subset \mathfrak q$ and $\mathfrak q$ is maximal with respect to this property,

2. for some ideal of definition $I \subset A$ the prime $\mathfrak q$ defines a closed point of $\mathop{\mathrm{Spec}}(A) \setminus V(I)$,

3. for any ideal of definition $I \subset A$ we have $I \not\subset \mathfrak q$ and $\mathfrak q$ is maximal with respect to this property,

4. for any ideal of definition $I \subset A$ the prime $\mathfrak q$ defines a closed point of $\mathop{\mathrm{Spec}}(A) \setminus V(I)$,

5. $\dim (A/\mathfrak q) = 1$ and for some ideal of definition $I \subset A$ we have $I \not\subset \mathfrak q$,

6. $\dim (A/\mathfrak q) = 1$ and for any ideal of definition $I \subset A$ we have $I \not\subset \mathfrak q$,

7. $\dim (A/\mathfrak q) = 1$ and the induced topology on $A/\mathfrak q$ is nontrivial,

8. $A/\mathfrak q$ is a $1$-dimensional Noetherian complete local domain whose maximal ideal is the radical of the image of any ideal of definition of $A$, and

Proof. It is clear that (1) and (2) are equivalent and for the same reason that (3) and (4) are equivalent. Since $V(I)$ is independent of the choice of the ideal of definition $I$ of $A$, we see that (2) and (4) are equivalent.

Assume the equivalent conditions (1) – (4) hold. If $\dim (A/\mathfrak q) > 1$ we can choose a maximal ideal $\mathfrak q \subset \mathfrak m \subset A$ such that $\dim ((A/\mathfrak q)_\mathfrak m) > 1$. Then $\mathop{\mathrm{Spec}}((A/\mathfrak q)_\mathfrak m) - V(I(A/\mathfrak q)_\mathfrak m)$ would be infinite by Algebra, Lemma 10.61.1. This contradicts the fact that $\mathfrak q$ is closed in $\mathop{\mathrm{Spec}}(A) \setminus V(I)$. Hence we see that (6) holds. Trivially (6) implies (5).

Conversely, assume (5) holds. Let $I \subset A$ be an ideal of definition. Since $A/\mathfrak q$ is complete with respect to $I(A/\mathfrak q)$ (for example by Algebra, Lemma 10.97.1) we see that all closed points of $\mathop{\mathrm{Spec}}(A/\mathfrak q)$ are contained in $V(IA/\mathfrak q)$ by Algebra, Lemma 10.96.6. Since $\dim (A/\mathfrak q) = 1$ and since $I \not\subset \mathfrak q$ we conclude two things: (a) $V(IA/\mathfrak q)$ must contain all points distinct from the generic point of $\mathop{\mathrm{Spec}}(A/\mathfrak q)$, and (b) $V(IA/\mathfrak q)$ must be a (finite) discrete set. From (a) we see that $\mathfrak q$ is a closed point of $\mathop{\mathrm{Spec}}(A) \setminus V(I)$ and we conclude that (2) holds.

Continuing to assume (5) we see that the finite discrete space $V(IA/\mathfrak q)$ must be a singleton by More on Algebra, Lemma 15.11.16 for example (and the fact that complete pairs are henselian pairs, see More on Algebra, Lemma 15.11.4). Hence we see that (8) is true. Conversely, it is clear that (8) implies (5).

At this point we know that (1) – (6) and (8) are equivalent. We omit the verification that these are also equivalent to (7). $\square$

In order to comfortably talk about such primes we introduce the following nonstandard notation.

Definition 86.14.2. Let $A$ be a Noetherian adic topological ring. Let $\mathfrak q \subset A$ be a prime ideal. We say $\mathfrak q$ is rig-closed if the equivalent conditions of Lemma 86.14.1 are satisfied.

We will need a few lemmas which essentially tell us there are plenty of rig-closed primes even in a relative settting.

Lemma 86.14.3. Let $\varphi : A \to B$ in $\textit{WAdm}^{Noeth}$. Denote $\mathfrak a \subset A$ and $\mathfrak b \subset B$ the ideals of topologically nilpotent elements. Assume $A/\mathfrak a \to B/\mathfrak b$ is of finite type. Let $\mathfrak q \subset B$ be rig-closed. The residue field $\kappa$ of the local ring $B/\mathfrak q$ is a finite type $A/\mathfrak a$-algebra.

Proof. Let $\mathfrak q \subset \mathfrak m \subset B$ be the unique maximal ideal containing $\mathfrak q$. Then $\mathfrak b \subset \mathfrak m$. Hence $A/\mathfrak a \to B/\mathfrak b \to B/\mathfrak m = \kappa$ is of finite type. $\square$

Lemma 86.14.4. Let $\varphi : A \to B$ be an arrow of $\textit{WAdm}^{Noeth}$ which is adic and topologically of finite type. Let $\mathfrak q \subset B$ be rig-closed. Let $\mathfrak p = \varphi ^{-1}(\mathfrak q) \subset A$. Let $\mathfrak a \subset A$ be the ideal of topologically nilpotent elements. The following are equivalent

1. the residue field $\kappa$ of $B/\mathfrak q$ is finite over $A/\mathfrak a$,

2. $\mathfrak p \subset A$ is rig-closed,

3. $A/\mathfrak p \subset B/\mathfrak q$ is a finite extension of rings.

Proof. Assume (1). Recall that $B/\mathfrak q$ is a Noetherian local ring of dimension $1$ whose topology is the adic topology coming from the maximal ideal. Since $\varphi$ is adic, we see that $A \to B/\mathfrak q$ is adic. Hence $\varphi (\mathfrak a)$ is a nonzero ideal in $B/\mathfrak q$. Hence $B/\mathfrak q + \varphi (\mathfrak a)$ has finite length. Hence $B/\mathfrak q + \varphi (\mathfrak a)$ is finite as an $A/\mathfrak a$-module by our assumption. Thus $B/\mathfrak q$ is finite over $A$ by Algebra, Lemma 10.96.12. Thus (3) holds.

Assume (3). Then $\mathop{\mathrm{Spec}}(B/\mathfrak q) \to \mathop{\mathrm{Spec}}(A/\mathfrak p)$ is surjective by Algebra, Lemma 10.36.17. This implies (2).

Assume (2). Denote $\kappa '$ the residue field of $A/\mathfrak p$. By Lemma 86.14.3 (and Lemma 86.12.4) the extension $\kappa /\kappa '$ is finitely generated as an algebra. By the Hilbert Nullstellensatz (Algebra, Lemma 10.34.2) we see that $\kappa /\kappa '$ is a finite extension. Hence we see that (1) holds. $\square$

Lemma 86.14.5. Let $\varphi : A \to B$ be an arrow of $\textit{WAdm}^{Noeth}$ which is adic and topologically of finite type. Let $\mathfrak q \subset B$ be rig-closed. If $A/I$ is Jacobson for some ideal of definition $I \subset A$, then $\mathfrak p = \varphi ^{-1}(\mathfrak q) \subset A$ is rig-closed.

Proof. By Lemma 86.14.3 (combined with Lemma 86.12.4) the residue field $\kappa$ of $B/\mathfrak q$ is of finite type over $A/\mathfrak a$. Since $A/\mathfrak a$ is Jacobson, we see that $\kappa$ is finite over $A/\mathfrak a$ by Algebra, Lemma 10.35.18. We conclude by Lemma 86.14.4. $\square$

Lemma 86.14.6. Let $\varphi : A \to B$ be an arrow of $\textit{WAdm}^{Noeth}$ which is adic and topologically of finite type. Let $\mathfrak p \subset A$ be rig-closed. Let $\mathfrak a \subset A$ and $\mathfrak b \subset B$ be the ideals of topologically nilpotent elements. If $\varphi$ is flat, then the following are equivalent

1. the maximal ideal of $A/\mathfrak p$ is in the image of $\mathop{\mathrm{Spec}}(B/\mathfrak b) \to \mathop{\mathrm{Spec}}(A/\mathfrak a)$,

2. there exists a rig-closed prime ideal $\mathfrak q \subset B$ such that $\mathfrak p = \varphi ^{-1}(\mathfrak q)$.

and if so then $\varphi$, $\mathfrak p$, and $\mathfrak q$ satisfy the conclusions of Lemma 86.14.4.

Proof. The implication (2) $\Rightarrow$ (1) is immediate. Assume (1). To prove the existence of $\mathfrak q$ we may replace $A$ by $A/\mathfrak p$ and $B$ by $B/\mathfrak p B$ (some details omitted). Thus we may assume $(A, \mathfrak m, \kappa )$ is a local complete $1$-dimensional Noetherian ring, $\mathfrak m = \mathfrak a$, and $\mathfrak p = (0)$. Condition (1) just says that $B_0 = B \otimes _ A \kappa = B/\mathfrak m B = B/\mathfrak a B$ is nonzero. Note that $B_0$ is of finite type over $\kappa$. Hence we can use induction on $\dim (B_0)$. If $\dim (B_0) = 0$, then any minimal prime $\mathfrak q \subset B$ will do (flatness of $A \to B$ insures that $\mathfrak q$ will lie over $\mathfrak p = (0)$). If $\dim (B_0) > 0$ then we can find an element $b \in B$ which maps to an element $b_0 \in B_0$ which is a nonzerodivisor and a nonunit, see Algebra, Lemma 10.130.4. By Algebra, Lemma 10.99.2 the ring $B' = B/bB$ is flat over $A$. Since $B'_0 = B' \otimes _ A \kappa = B_0/(b_0)$ is not zero, we may apply the induction hypothesis to $B'$ and conclude. The final statement of the lemma is clear from Lemma 86.14.4. $\square$

We introduce some notation.

Definition 86.14.7. Let $A$ be an adic topological ring which has a finitely generated ideal of definition. Let $f \in A$. The completed principal localization $A_{\{ f\} }$ of $A$ is the completion of $A_ f = A[1/f]$ of the principal localization of $A$ at $f$ with respect to any ideal of definition of $A$.

To be sure, if $f$ is topologically nilpotent, then $A_{\{ f\} }$ is the zero ring.

Lemma 86.14.8. Let $A$ be an adic Noetherian topological ring. Let $\mathfrak p \subset A$ be a prime ideal. Let $f \in A$ be an element mapping to a unit in $A/\mathfrak p$. Then

$\mathfrak p A_{\{ f\} } = \mathfrak p(A_ f)^\wedge = \mathfrak p \otimes _ A (A_ f)^\wedge = (\mathfrak p_ f)^\wedge$

is a prime ideal with quotient

$A/\mathfrak p = (A/\mathfrak p) \otimes _ A (A_ f)^\wedge = (A_ f)^\wedge / \mathfrak p (A_ f)^\wedge = A_{\{ f\} }/\mathfrak p A_{\{ f\} }$

Proof. Since $A_ f$ is Noetherian the ring map $A \to A_ f \to (A_ f)^\wedge$ is flat. For any finite $A$-module $M$ we see that $M \otimes _ A (A_ f)^\wedge$ is the completion of $M_ f$. If $f$ is a unit on $M$, then $M_ f = M$ is already complete. See discussion in Algebra, Section 10.97. From these observations the results follow easily. $\square$

Lemma 86.14.9. Let $\varphi : A \to B$ be an arrow of $\textit{WAdm}^{Noeth}$ which is adic and topologically of finite type. Let $\mathfrak q \subset B$ be rig-closed. There exists an $f \in A$ which maps to a unit in $B/\mathfrak q$ such that we obtain a diagram

$\vcenter { \xymatrix{ B \ar[r] & B_{\{ f\} } \\ A \ar[r] \ar[u]_\varphi & A_{\{ f\} } \ar[u]_{\varphi _{\{ f\} }} } } \quad \text{with primes}\quad \vcenter { \xymatrix{ \mathfrak q \ar@{-}[r] \ar@{-}[d] & \mathfrak q' \ar@{-}[d] \ar@{=}[r] & \mathfrak q B_{\{ f\} } \\ \mathfrak p \ar@{-}[r] & \mathfrak p' } }$

such that $\mathfrak p'$ is rig-closed, i.e., the map $A_{\{ f\} } \to B_{\{ f\} }$ and the prime ideals $\mathfrak q'$ and $\mathfrak p'$ satisfy the equivalent conditions of Lemma 86.14.4.

Proof. Please see Lemma 86.14.8 for the description of $\mathfrak q'$. The only assertion the lemma makes is that for a suitable choice of $f$ the prime ideal $\mathfrak p'$ has the property $\dim ((A_ f)^\wedge /\mathfrak p') = 1$. By Lemma 86.14.4 this in turn just means that the residue field $\kappa$ of $B/\mathfrak q = (B_ f)^\wedge /\mathfrak q'$ is finite over $(A_ f)^\wedge /\mathfrak a' = (A/\mathfrak a)_ f$. By Lemma 86.14.3 we know that $A/\mathfrak a \to \kappa$ is a finite type algebra homomorphism. By the Hilbert Nullstellensatz in the form of Algebra, Lemma 10.34.2 we can find an $f \in A$ which maps to a unit in $\kappa$ such that $\kappa$ is finite over $A_ f$. This finishes the proof. $\square$

Lemma 86.14.10. Let $A$ be a Noetherian adic topological ring. Denote $A\{ x_1, \ldots , x_ n\}$ the restricted power series over $A$. Let $\mathfrak q \subset A\{ x_1, \ldots , x_ n\}$ be a prime ideal. Set $\mathfrak q' = A[x_1, \ldots , x_ n] \cap \mathfrak q$ and $\mathfrak p = A \cap \mathfrak q$. If $\mathfrak q$ and $\mathfrak p$ are rig-closed, then the map

$A[x_1, \ldots , x_ n]_{\mathfrak q'} \to A\{ x_1, \ldots , x_ n\} _\mathfrak q$

defines an isomorphism on completions with respect to their maximal ideals.

Proof. By Lemma 86.14.4 the ring map $A/\mathfrak p \to A\{ x_1, \ldots , x_ n\} /\mathfrak q$ is finite. For every $m \geq 1$ the module $\mathfrak q^ m/\mathfrak q^{m + 1}$ is finite over $A$ as it is a finite $A\{ x_1, \ldots , x_ n\} /\mathfrak q$-module. Hence $A\{ x_1, \ldots x_ n\} /\mathfrak q^ m$ is a finite $A$-module. Hence $A[x_1, \ldots , x_ n] \to A\{ x_1, \ldots , x_ n\} /\mathfrak q^ m$ is surjective (as the image is dense and an $A$-submodule). It follows in a straightforward manner that $A[x_1, \ldots , x_ n]/(\mathfrak q')^ m \to A\{ x_1, \ldots , x_ n\} /\mathfrak q^ m$ is an isomorphism for all $m$. From this the lemma easily follows. Hint: Pick a topologically nilpotent $g \in A$ which is not contained in $\mathfrak p$. Then the map of completions is the map

$\mathop{\mathrm{lim}}\nolimits _ m \left(A[x_1, \ldots , x_ n]/(\mathfrak q')^ m\right)_ g \longrightarrow \left(A\{ x_1, \ldots , x_ n\} /\mathfrak q^ m\right)_ g$

Some details omitted. $\square$

Lemma 86.14.11. Let $\varphi : A \to B$ be an arrow of $\textit{WAdm}^{Noeth}$. Assume $\varphi$ is adic, topologically of finite type, flat, and $A/I \to B/IB$ is étale for some (resp. any) ideal of definition $I \subset A$. Let $\mathfrak q \subset B$ be rig-closed such that $\mathfrak p = A \cap \mathfrak q$ is rig-closed as well. Then $\mathfrak p B_\mathfrak q = \mathfrak q B_\mathfrak q$.

Proof. Let $\kappa$ be the residue field of the $1$-dimensional complete local ring $A/\mathfrak p$. Since $A/I \to B/IB$ is étale, we see that $B \otimes _ A \kappa$ is a finite product of finite separable extensions of $\kappa$, see Algebra, Lemma 10.143.4. One of these is the residue field of $B/\mathfrak q$. By Algebra, Lemma 10.96.12 we see that $B/\mathfrak p B$ is a finite $A/\mathfrak p$-algebra. It is also flat. Combining the above we see that $A/\mathfrak p \to B /\mathfrak p B$ is finite étale, see Algebra, Lemma 10.143.7. Hence $B/\mathfrak p B$ is reduced, which implies the statement of the lemma (details omitted). $\square$

Lemma 86.14.12. Let $A$ be an adic Noetherian topological ring. Let $\mathfrak p \subset A$ be a rig-closed prime. For any $n \geq 1$ the ring map

$A/\mathfrak p \longrightarrow A\{ x_1, \ldots , x_ n\} \otimes _ A A/\mathfrak p = A/\mathfrak p\{ x_1, \ldots , x_ n\}$

is regular. In particular, the algebra $A\{ x_1, \ldots , x_ n\} \otimes _ A \kappa (\mathfrak p)$ is geometrically regular over $\kappa (\mathfrak p)$.

Proof. We will use some fact on regular ring maps the reader can find in More on Algebra, Section 15.41. Since $A/\mathfrak p$ is a complete local Noetherian ring it is excellent (More on Algebra, Proposition 15.52.3). Hence $A/\mathfrak p[x_1, \ldots , x_ n]$ is excellent (by the same reference). Hence $A/\mathfrak p[x_1, \ldots , x_ n] \to A/\mathfrak p\{ x_1, \ldots , x_ n\}$ is a regular ring homomorphism by More on Algebra, Lemma 15.50.14. Of course $A/\mathfrak p \to A/\mathfrak p[x_1, \ldots , x_ n]$ is smooth and hence regular. Since the composition of regular ring maps is regular the proof is complete. $\square$

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