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Notes on condensed mathematics

Kiran S. Kedlaya

Section 11 Huber rings and solid rings

In this section, we introduce the topological rings that occur in Huber’s theory of adic spaces, and relate these to the constructions we have already introduced.

Reference.

This section is based on [8], Lectures 7 and 8. Remark 11.1.6 is taken from [8], Lecture 10 (see also [19], Remark 1.5.4).

Subsection 11.1 Huber rings

We first recall the definition of what Huber calls an \(f\)-adic ring in [16] and subsequent work.

Definition 11.1.1.

A (commutative) Huber ring is a topological commutative ring \(A\) admitting an open subring \(A_0\) (a ring of definition) which in turn contains a finitely generated ideal \(I\) (an ideal of definition) such that the topology on \(A_0\) is the \(I\)-adic topology. This implies in particular that \(A\) is linearly topologized (meaning that it admits a neighborhood basis of 0 consisting of subgroups).
An element \(x \in A\) is topologically nilpotent if \(1,x,x^2,\dots\) is a null sequence. We say that a Huber ring is Tate if it contains a topologically nilpotent unit.
A subset \(S\) is bounded if for every neighborhood \(U\) of \(0\) in \(A\text{,}\) there exists a neighborhood \(V\) of \(0\) in \(A\) such that \(V \cdot S \subseteq U\text{.}\)

Remark 11.1.2.

For \(A\) a Huber ring, the topological completion \(\hat{A}\) of \(A\) is again a Huber ring; if \(A_0\) is a ring of definition of \(A\) and \(I\) is an ideal of definition of \(A_0\text{,}\) then then the \(I\)-adic completion \(\hat{A}_0\) of \(A_0\) is a ring of definition of \(\hat{A}\) and \(I\hat{A}_0\) is an ideal of definition of \(\hat{A}_0\text{.}\)
We will mostly be interested in complete Huber rings in what follows; note that these are solid as condensed abelian groups. We will pick up this thread in Subsection 11.3.

Example 11.1.3.

Any discrete ring is Huber but not Tate. The ring \(\ZZ_p\) for the \(p\)-adic topology is Huber but not Tate. The ring \(\QQ_p\) for the \(p\)-adic topology is Huber and also Tate.

Example 11.1.4.

For any Huber ring \(A\text{,}\) with a ring of definition \(A_0\) and an ideal of definition \(I\text{,}\) we obtain a Huber pair \(A\langle T_1,\dots,T_n\rangle\) with a ring of definition \(A_0 \langle T_1,\dots,T_n \rangle\text{,}\) the \(I\)-adic completion of \(A[T_1,\dots,T_n]\text{,}\) and an ideal of definition \(I A_0 \langle T_1,\dots,T_n \rangle\text{.}\) This is called the Tate algebra in the variables \(T_1,\dots,T_n\) over \(A\text{;}\) however, it is a Tate Huber ring if and only if \(A\) is.
There is no analogue of the Hilbert basis theorem in this context: if \(A\) is noetherian (even a field) it can happen that \(A\langle T \rangle\) is not noetherian ([10], section 8.3). We say that \(A\) is strongly noetherian if \(A \langle T_1,\dots,T_n \rangle\) is noetherian for all \(n \in \NN\text{;}\) for instance this holds when \(A\) is a nonarchimedean field.
The concept of a Huber ring is closely related to the concept of a nonarchimedean Banach ring.

Definition 11.1.5.

For \(R\) a commutative ring, a submultiplicative seminorm on \(R\) is a function \(|\bullet|\colon R \to \RR_{\geq 0}\) satisfying the following conditions.
  1. We have \(|0| = 0\text{.}\) If \(R \neq 0\text{,}\) then also \(|1| = |-1| = 1\text{.}\)
  2. For all \(x,y \in R\text{,}\) \(|x+y| \leq |x| + |y|\text{.}\)
  3. For all \(x,y \in R\text{,}\) \(|xy| \leq |x| |y|\text{.}\)
Some variations:
  • If the first condition can be upgraded to say that \(|x| = 0\) if and only if \(x =0\text{,}\) we may say that \(|\bullet|\) is a norm rather than a seminorm.
  • If the second condition can be upgraded to say that \(|x+y| \leq \max\{|x|, |y|\}\text{,}\) we may say that \(|\bullet|\) is nonarchimedean.
  • If the third condition can be upgraded to say that \(|x^n| = |x|^n\) for all \(x \in R\) and all positive integers \(n\text{,}\) we may say that \(|\bullet|\) is power-multiplicative rather than submultiplicative.
  • If the third condition can be upgraded to say that \(|xy| = |x||y|\) for all \(x,y \in R\text{,}\) we may say that \(|\bullet|\) is multiplicative rather than submultiplicative.
Any submultiplicative seminorm defines a norm topology on \(R\) with respect to which the ring operations are continuous. A (commutative) Banach ring consists of a commutative ring \(R\) equipped with a submultiplicative norm such that \(R\) is complete for the norm topology. As for Huber rings, we say that a Banach ring is Tate if it contains a topologically nilpotent unit.

Remark 11.1.6.

For \(A\) a nonarchimedean Banach ring, the set of \(A_0\) consisting of \(x \in A\) with \(|x| \leq 1\) is an open subring of \(A\text{.}\) If \(A\) is also Tate, then for any topologically nilpotent unit \(\varpi \in A\text{,}\) for \(n\) sufficiently large, \(\varpi^n \in A_0\) and the topology on \(A_0\) is the \(\varpi^n\)-adic topology. Consequently, in this case the underlying topological ring of \(A\) is a complete Huber ring; moreover, a subset \(S\) is bounded in the sense of Huber rings if and only if \(|S| \subseteq [0, \infty)\) is bounded away from \(\infty\text{.}\)
Conversely, for \(A\) a complete Huber ring which is Tate, we can promote \(A\) to a Banach ring as follows. Let \(A_0\) be a ring of definition, let \(I = (y_1,\dots,y_m)\) be an ideal of definition, and let \(\varpi \in A\) be a topologically nilpotent unit. By replacing \(\varpi\) with a power if needed, we can ensure that \(\varpi \in I\text{;}\) then \(\varpi = a_1 y_1 + \cdots + a_m y_m\) for some \(a_1,\dots,a_m \in A_0\text{,}\) so in particular \(I\) generates the unit ideal of \(A\text{.}\) Conversely, since multiplication by \(\varpi\) is a homeomorphism, \(\varpi A_0\) is open in \(A\) for every \(n\) so there exists some positive integer \(m\) such that \(y_1^n, \dots,y_m^n \in \varpi A_0\text{.}\) We deduce that \((\varpi)\) is also an ideal of definition of \(A_0\text{.}\)
By the previous paragraph, we have a well-defined function \(|\bullet|\colon A \to \RR\) taking \(x\) to \(0\) if \(x = 0\text{,}\) and otherwise to \(2^{-n}\) where \(n\) is the largest integer such that \(x \in \varpi^n A_0\text{.}\) It is easy to verify that this defines a submultiplicative norm on \(A\) giving the correct norm topology.

Subsection 11.2 Rings of integral elements

In Huber’s theory, to define a good notion of the adic spectrum (to be introduced in Subsection 12.3) one must work with not just a Huber ring, but a certain auxiliary ring attached to it. We will see later in this section that this extra structure admits a natural interpretation in the context of analytic rings (Subsection 11.3).

Definition 11.2.1.

Let \(A\) be a Huber ring. An element \(x \in A\) is power-bounded if the set \(\{1,x,x^2,\dots\}\) is bounded in \(A\text{.}\) The power-bounded elements form an open subring \(A^{\circ}\) of \(A\) containing any ring of definition. The topologically nilpotent elements form a radical ideal \(A^{\circ \circ}\) of \(A^{\circ}\text{.}\)
In general, \(A^\circ\) need not be bounded. When it is, we say that \(A\) is a uniform Huber ring.

Example 11.2.2.

The ring \(A = \QQ_p[T]/(T^2)\text{,}\) topologized as \(\QQ_p \times \QQ_p\text{,}\) is a complete Tate Huber ring: we may take \(A_0 = \ZZ_p \oplus \ZZ_p T\) as a ring of definition and \(I = p A_0\) as an ideal of definition. In this case, \(A^{\circ} = \ZZ_p \oplus \QQ_p T\) which is not bounded.
Using similar logic, one may see that \(A\) is a Tate Huber ring which is not reduced, then it is not uniform either.

Definition 11.2.3.

For \(A\) a Huber ring, a ring of integral elements of \(A\) is a subring \(A^+\) of \(A^{\circ}\) which is open and integrally closed in \(A\text{.}\) Like \(A^\circ\text{,}\) \(A^+\) need not be bounded.

Remark 11.2.4.

Any ring of integral elements contains \(A^{\circ \circ}\text{:}\) if \(x \in A\) is topologically nilpotent, then the openness of \(A^+\) ensures that for some positive integer \(n\) we have \(x^n \in A^+\text{,}\) and the integral closure in \(A\) then forces \(x \in A^+\text{.}\) Since \(A^{\circ \circ}\) is itself an open subset of \(A\text{,}\) we see from this that rings of integral elements of \(A\) are in bijection with integrally closed subrings of \(A^\circ/A^{\circ \circ}\text{.}\)
Conversely, \(A^{\circ \circ}\) is itself an open neighborhood of \(0\) in \(A^{\triangleright}\text{.}\) We may thus redefine a ring of integral elements of \(A\) to be simply an integrally closed subring of \(A^{\circ}\) containing \(A^{\circ \circ}\text{.}\) For example, \(A^\circ\) is such a subring; this includes the assertion that it is integrally closed, which we will recover later by proving an analogue for solid rings (Lemma 11.3.5).

Definition 11.2.5.

A Huber pair is a pair \(A = (A^{\triangleright}, A^+)\) in which \(A^{\triangleright}\) is a Huber ring and \(A^+\) is a ring of integral elements of \(A^{\triangleright}\text{.}\) We will apply adjectives like complete and Tate to a Huber pair by passing through to the underlying Huber ring.

Example 11.2.6.

We can construct multiple complete Huber pairs \(A\) with \(A^{\triangleright} = \QQ_p \langle T \rangle\text{:}\) in one case we take \(A^+ = \ZZ_p \langle T \rangle\text{,}\) and in the other we take the smaller ring \(A^+ = \ZZ_p + p \ZZ_p \langle T \rangle\text{.}\)

Remark 11.2.7.

In Huber’s theory, the difference between a Huber ring and a Huber pair manifests in the definition of the adic spectrum; see Subsection 12.3. the elements of the adic spectrum are (equivalence classes of) continuous (semi)valuations on the underlying Huber ring whose valuation rings contains the ring of integral elements. This has some effect when glueing adic spectra together to make adic spaces.
However, this difference can also be detected from the point of view of solid modules, which means that different Huber pairs with the same underlying Huber ring will give rise to distinct solid analytic rings. This is one indication that analytic rings will provide a satisfactory basis for constructing analytic stacks in a manner that can absorb adic spaces.

Subsection 11.3 Solid modules and Huber rings

We have already observed that complete Huber rings give rise to solid condensed rings. This prompts us to upgrade our definitions to apply more directly to solid rings.

Definition 11.3.1.

Write \(\ZZ[T]_\solid\) for the analytic ring \((\underline{\ZZ}[T], \Mod_{\underline{\ZZ}[T] \liquid \{1, T\}})\text{.}\) Beware that this is not consistent with the notation in Section 1.

Definition 11.3.2.

Let \(A^{\triangleright}\) be a solid condensed ring. For any \(f \in A^{\triangleright}(*)\text{,}\) we have a unique homomorphism \(\underline{\ZZ}[T] \to A^{\triangleright}\) mapping \(T\) to \(f\text{.}\) We say \(f\) is power-bounded if \(A^{\triangleright} \in \Mod_{\ZZ[T]_\solid}\text{.}\) Let \(A^{\circ} \subseteq A^{\triangleright}(*)\) be the set of power-bounded elements.
We say \(f\) is topologically nilpotent if \(\underline{\ZZ}[T] \to A^{\triangleright}\) factors through \(P_\solid \cong \underline{\ZZ}\llbracket T \rrbracket\text{.}\) Since we are assuming that \(A^{\triangleright}\) is solid as a condensed abelian group, this is equivalent to the condition given in Definition 10.4.5. Let \(A^{\circ \circ} \subseteq A^{\circ}\) be the set of topologically nilpotent elements.
When \(A^{\triangleright}\) is a Huber ring, these definitions agree with Definition 11.2.1.

Proof.

Form the map \(\underline{\ZZ}[x_1,\dots,x_n][T]/(F) \to A^{\triangleright}\) via \(x_i \mapsto f_i, T \mapsto f\text{.}\) The hypothesis that \(A^{\triangleright}\) is solid and \(f_1,\dots,f_n \in A^\circ\) means that \(A^{\triangleright} \in \Mod_{\underline{\ZZ}[x_1,\dots,x_n]\liquid \{1,f_1,\dots,f_n\}}\text{.}\) Since \(\ZZ[x_1,\dots,x_n][T]/(F)\) is finite free as a module over \(\ZZ[x_1,\dots,x_n]\text{,}\) we may formally upgrade the previous conclusion to \(A^{\triangleright} \in \Mod_{\underline{\ZZ}[x_1,\dots,x_n][T]/(F)\liquid \{1,f_1,\dots,f_n,T\}}\) and hence \(f \in A^\circ\text{.}\)

Proof.

Since
\begin{equation*} \underline{\ZZ}\llbracket x_1 \rrbracket \otimes^L_\solid \cdots \otimes^L_\solid \underline{\ZZ}\llbracket x_n \rrbracket \cong \underline{\ZZ}\llbracket x_1,\dots,x_n \rrbracket, \end{equation*}
the conditions in question imply that we have a homomorphism \(\underline{\ZZ}\llbracket x_1,\dots,x_n,T\rrbracket/(F) \to A^{\triangleright}\) acting via \(x_i \mapsto f_i, T \mapsto f\text{.}\) We thus deduce that \(f \in A^{\circ \circ}\text{.}\)

Proof.

We check the first claim by repeated application of Lemma 11.3.3: we check that \(1 \in A^\circ\) by taking \(F = T-1\text{;}\) we check closure under ring operations by taking \(F = T - (x_1+x_2)\) and \(F = T - (x_1x_2)\text{;}\) we check integral closure by taking \(F\) arbitrary.
We similarly check the second claim by repeated application of Lemma 11.3.4.

Remark 11.3.6.

Both aspects of Lemma 11.3.5 are peculiar to solid analytic rings: for instance they fail for \(\RR_{\liquid}\) because the interval \((0,1)\) is not closed under addition.

Remark 11.3.7.

For \(A^{\triangleright}\) a condensed ring, an element \(f \in A^{\triangleright}(*)\) which is universally power-bounded in the sense of Definition 10.4.5 is also power-bounded, but not conversely even if \(A^{\triangleright}\) is represented by a Tate Huber ring. The point is that the conclusion of Lemma 10.4.6 does not hold.
Mirroring the transition from Huber rings to Huber pairs, we now transition from solid condensed rings to solid analytic rings.

Definition 11.3.8.

Let \(A = (A^{\triangleright}, \Mod_A)\) be a solid analytic ring. Let \(A^+\subseteq A^{\triangleright}(*)\) be the set of \(f\) for which the unique homomorphism \(\underline{\ZZ}[T] \to A^{\triangleright}\) taking \(T\) to \(f\) factors through a morphism \(\ZZ[T]_\solid \to A\) of analytic rings. We have an evident inclusion \(A^+ \subseteq A^\circ\text{;}\) we can also identify \(A^+\) with the core of \(A\) in the sense of Definition 10.4.8.

Proof.

Argue as in Lemma 11.3.3 but with \(\underline{\ZZ}[x_1,\dots,x_n]\) replaced by the analytic completion of \(A^{\triangleright}[x_1,\dots,x_n]\text{.}\)

Proof.

The containments \(A^{\circ \circ} \subseteq A^+ \subseteq A^{\circ}\) are straightforward. To check that \(A^+\) is an integrally closed subring of \(A^\circ\text{,}\) argue as in Lemma 11.3.5 but with Lemma 11.3.3 replaced by Lemma 11.3.9.

Proof.

The fact that the functor does yield a ring of integral elements is the content of Proposition 11.3.10. The left adjoint takes a Huber pair \((A^{\triangleright}, A^+)\) to \((\underline{A^{\triangleright}}, \Mod_{\underline{A^{\triangleright}} \liquid A^+})\text{;}\) the inclusion \(A^+ \subseteq A^\circ\) ensures that \(A^{\triangleright} \in \Mod_{\underline{A^{\triangleright}} \liquid A^+}\text{.}\)

Example 11.3.12.

We have the following (for the discrete topologies on \(\ZZ\) and \(\ZZ[T]\)):
\begin{align*} (\ZZ, \ZZ)_\solid &= (\underline{\ZZ}, \Mod_{\underline{\ZZ} \solid})\\ (\ZZ[T], \ZZ)_\solid &= (\underline{\ZZ[T]}, \Mod_{\underline{\ZZ}[T] \liquid 1})\\ (\ZZ[T], \ZZ[T])_\solid &= (\underline{\ZZ[T]}, \Mod_{\underline{\ZZ}[T]\liquid\{1,T\}})\text{.} \end{align*}

Proof.

Straightforward.

Remark 11.3.14.

If \(A\) is a solid analytic ring, then we cannot necessarily equip \(A^{\triangleright}(*)\) with the structure of a Huber ring in such a way that \(A^\circ\) and \(A^{\circ \circ}\) are the sets of power-bounded and topologically nilpotent elements, respectively. (One issue is finding a finitely generated ideal of definition.) However, we can still treat \(A(*) := (A^{\triangleright}(*), A^+)\) as a complete Huber pair for the discrete topology on \(A^{\triangleright}(*)\text{;}\) this will be useful for transferring information about Huber pairs to solid analytic rings.
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