106 2. THE BERKOVICH SPECTRUM January 15, 2020, 9:55pm
2.1. The Berkovich spectrum of a seminormed ring A key construction in algebraic geometry is to associate a space, the prime ideal spectrum X = Spec(A) to any ring A. One can then view the elements of A as functions on X.Further,X is naturally equipped with a topology (the Zariski topology), and the assignment A Spec(A)defines a functor from the category of rings to the category of topological spaces. 7! Berkovich discovered a similar construction that to a seminormed ring A associates the Berkovich spectrum M (A), a compact1 topological space. This defines a functor from the category of semi- normed rings to the category of compact topological spaces. In algebraic geometry, it is not enough to consider prime ideal spectrum as a topological space: one needs to equip it with a structure sheaf. The analogous construction will be carried out (in a more restrictive setting) for Banach rings in later chapters. In this section, seminorms are not assumed to be non-Archimedean. 2.1.1. Definition and first properties. If (A, ) is a seminormed ring, then we say that a semivaluation on A is bounded if f f for all f kA·k. Note that since is power-multiplicative, this is equivalent|·| to the weaker condition| |k thatk there2 exists C 1 such that|·| f C f for all f. | | k k Indeed, the latter condition implies f = f n 1/n C1/n f n 1/n C1/n f for all n 1, and | | | | k k k k C1/n 1 as n . ! !1 Definition 2.1.1. Let (A, ) be a seminormed ring. The Berkovich spectrum M (A) of A is the set of bounded multiplicativek·k seminorms on A. It is equipped with the weakest topology in which the evaluation map M (A) f R 3|·|!| |2 + is continuous for every f A. 2 Example 2.1.2. When A is the zero ring, M (A) is the empty set since, by definition, a multi- plicative seminorm satisfies 1 = 1. More generally, M (A)isemptywhenA is equipped with the zero seminorm. | | We shall see shortly that if A is equipped with a seminorm di↵erent from the zero seminorm, then M (A) is a nonempty compact topological space. In particular, this happens when A is a nonzero Banach ring. Example 2.1.3. If k is a valued field, M (k) is a singleton, see Exercise 2.1.1. Complete valued fields play the same role for Berkovich spaces as fields do for schemes (recall that the prime ideal spectrum of a field is a singleton.) One should think of X = M (A) as a space, and of A as a ring of functions on X. It is therefore suggestive (and convenient) to denote the elements of X by symbols such as x, y,. . . , and to write f(x) for the value of the seminorm associated to x X on the element f A. At this point, this is merely| | a notational convention, but we will later make2 sense of the expression2 f(x) (without absolute value) as well. With this notation, the definition of the topology can be rephrased as follows: a general open set in X = M (A) is a union of sets of the form x X f (x)
1By our conventions, a compact space is Hausdor↵. 2.1.THEBERKOVICHSPECTRUMOFASEMINORMEDRING 107
Proposition 2.1.4. For any seminormed ring A, the Berkovich spectrum M (A) is compact.
Here we use the convention that the empty set is compact! We will see below that M (A)is empty i↵ A is equipped with the zero seminorm.
Proof. We proceed as in the proof of the Banach–Alaoglu Theorem in functional analysis. Since f(x) f for all f A and all x X, there is a natural injective map | |k k 2 2 ◆: X P := [0, f ]. ! k k f A Y2 Equip P with the product topology. Then P is compact (and nonempty) by Tychono↵’s Theorem. Note that ◆ is an embedding, that is, a subset of X is open i↵it is the preimage under ◆ of an open set in P ; this follows from the definition of the topologies on X and P . Finally note that ◆(X) P is ⇢ closed. Indeed, a point (tf )f A P lies in ◆(X)i↵t0 = 0, t1 = 1, t f = tf for f A, tf+g tf + tg, 2 2 2 and tfg = tf tg for f,g A; these conditions define closed subsets of P . Thus ◆: X P is a2 homeomorphism onto a closed subset of P .SinceP is compact, so is X. ! ⇤ 2.1.2. Functoriality. Next we note that the construction A M (A) is functorial. ! Definition 2.1.5. Let ': A B be a morphism of seminormed rings. Define '⇤ : M (B) ! ! M (A) as the map that sends y M (B) to the seminorm on A whose value on f A is '(f)(y) . 2 2 | | Proposition 2.1.6. The map '⇤ is well-defined and continuous. It is injective if the augmented image of A is dense in B.
Recall that the augmented image of ' is the set of elements in B if the form '(f)/'(g), with f,g A and '(g)invertibleinB. 2 As an obvious consequence, '⇤ is injective if '(A)isdenseinB, but the more general condition will be very useful.
Proof. Write X = M (A) and Y = M (B). Pick C>0 such that '(f) C f for f A. Consider y Y . We claim that the assignment f '(f)(y) definesk a boundedk k multiplicativek 2 seminorm on2A. Indeed, it is clear that '(0)(y) = 0,7!' |(1)(y) =| 1, '( f)(y) = '(f)(y) , '(f + g)(y) '(f)(y) + '(g)(y) , and '(fg)(| y) = |'(f)(y|) '(g)(| y) , for| f,g A|,so| f '(|f)(| y) is a multiplicative|| seminorm| | on| A. It| is furthermore| | bounded,|·| since| '(f)(y) 2 '(f) !|C f (and| | |k k k k hence '(f)(y) f ) for all f A. Thus we get a well-defined map '⇤ : Y X. | |k k 2 ! To prove continuity of '⇤, it su ces to show that the function y '(f)(y) is continuous on Y for any f A.Butsince'(f) B, this is clear by the definition of the7! topology | on| Y . (Alternatively, 2 2 simply observe that the preimage under '⇤ of an open set in Y of the type (2.1) is a set of the same type.) Finally we prove the injectivity criterion. Let B0 B be the subset of elements of the form ⇢ '(f)/'(g)withf,g A and '(g)invertibleinB and assume that B0 is dense in B. Consider two 2 points y ,y Y such that '⇤(y )='⇤(y )=:x.Ifh = '(f)/'(g) B0,then 1 2 2 1 2 2 h(y ) = '(f)(y ) / '(g)(y ) = f(x) / g(x) | i | | i | | i | | | | | for i =1, 2. The density of B0 in B now implies that y1 = y2, see Exercise 2.1.6. ⇤
In this way can view M as a contravariant functor from the category of seminormed rings with bounded maps to the category of topological spaces. Some properties of this functor are explored in the exercises. 108 2. THE BERKOVICH SPECTRUM
2.1.3. Completion and uniformization. While the Berkovich spectrum is defined for general seminormed rings, we shall mainly consider the case of Banach rings. The general case is reduced to the case of Banach rings in view of the following result. Proposition 2.1.7. Let A be a seminormed ring, and let Aˆ be the separated completion. Then Aˆ is a Banach ring, and the canonical morphism A Aˆ induces a homeomorphism M (Aˆ) ⇠ M (A). ! ! The same result also holds when Aˆ is replaced by the separation or completion. Proof. Write ⌧ : A Aˆ for the canonical map. Then ⌧ is bounded, and induces a continuous ! map M (Aˆ) M (A). The latter is injective since ⌧ has dense image; see Proposition 2.1.6. ! Let us prove that it is also surjective. Pick x M (A) and define a multiplicative seminorm 0 2 |·| on ⌧(A) Aˆ by ⌧(f) 0 := f(x) .Thisiswelldefinedsinceif⌧(f )=⌧(f ), then (f f )(x) ⇢ | | | | 1 2 | 1 2 | f f = 0, and hence f (x) = f (x) . It is straightforward to see that 0 is a bounded k 1 2k | 1 | | 2 | |·| multiplicative seminorm on ⌧(A). Since ⌧(A)isdenseinAˆ, 0 extends to a bounded multiplicative |·| seminorm on Aˆ, see Exercise 2.1.7, and hence a point in M (Aˆ). This proves that the continuous map M (Aˆ) M (A) is bijective, and hence a homeomorphism ! since M (Aˆ) is quasicompact and M (A) Hausdor↵. ⇤ To some extent, it also su ces to consider uniform Banach rings, that is, Banach rings for which the norm is power multiplicative. Let A be an arbitrary seminormed ring and Au be its uniformization, that is, the separated completion of A with respect to the spectral radius seminorm ⇢: A R . ! + Proposition 2.1.8. The canonical morphism A Au induces a homeomorphism M (Au) ⇠ ! ! M (A). Proof. The continuous map M (Au) M (A) is injective since the canonical morphism ⇡ : A ! ! Au has dense image. To prove surjectivity, pick x M (A). For f A and n 1wehave 2 2 f(x) = f n(x) 1/n f 1/n. | | | | k k Letting n we get f(x) ⇢(f). In particular, f(x) =0whenever⇢(f) = 0, so the assignment !1 | | | | ⇡(f) f(x) defines a bounded multiplicative seminorm on Au. This defines a point y M (Au) !| | 2 such that f(x) = ⇡(f)(y) for every f A. Thus the continuous map M (Au) M (A)isbijective. | u | | | 2 ! Since M (A ) is quasicompact and M (A) Hausdor↵, the map is a homeomorphism. ⇤ 2.1.4. Fundamental theorem. — The following result is crucial to the development of Berkovich spaces. Theorem 2.1.9. Let A be a ring equipped with a nonzero seminorm. Then the Berkovich spec- trum M (A) is a nonempty compact topological space. We have already seen in Proposition 2.1.4 that the spectrum is compact, so what remains is to prove that the spectrum is nonempty. This is obvious in many concrete situations, such as when the norm on A is multiplicative. To prove it in general, we need Zorn’s Lemma. The following terminology will be convenient. Definition 2.1.10. A nonzero seminorm on a ring A is minimal if the only nonzero seminorm k·k 0 on A satisfying f 0 f for all f A is 0 = . k·k k k k k 2 k·k k·k Note that any minimal nonzero seminorm is power-multiplicative, that is, f n = f n for f A n 1/n k k k k 2 and n 1. Indeed, the spectral radius seminorm ⇢(f):=limn f is a nonzero seminorm on A satisfying ⇢ ,sowemusthave⇢ = . !1 k k k·k k·k Lemma 2.1.11. Let (A, ) be a normed field. Then the following conditions are equivalent: k·k 2.1.THEBERKOVICHSPECTRUMOFASEMINORMEDRING 109
(i) the norm on A is minimal; k·1k 1 (ii) we have f = f for every f A⇥ ; (iii) the normk on Ak isk multiplicative,k that2 is, a} valuation. Note that on a field, a nonzero seminorm is the same thing as a norm. Proof. The equivalence between (ii) and (iii) is straightforward, see Exercise 1.2.11, as is the implication (ii) = (i): if 0 is a norm on A, then for every f = 0, we have ) k·k k·k 6 1 1 1 1 1= 1 0 = ff 0 f 0 f 0 f f = ff = 1 =1. k k k k k k ·k k k k·k k k k k k 1 1 Since f 0 f and f 0 f ,thisimplies 0 = . Itk onlyk k remainsk tok provek k that (i)k implies (ii). Ask·k notedk· above,k the norm is automatically power multiplicative. We first consider the case when it is also complete, thatk·k is, A is a Banach 1 1 field. Arguing by contradiction, suppose that there exists f A⇥ with f > f .Weneedto construct a nonzero seminorm on A strictly smaller than the2 given norm.k k k k 1 1 To this end, set r := f and consider the relative disc algebra k k 1 1 i 1 i B := A r T = g = a T g := a r < + h i { i |k k k ik 1} i=0 i=0 X X 1 introduced in 1.8. We claim that the element f T is noninvertible in A r T .Indeed,since § h i 1 i i 1 1 i i 1 f r = f r = 1=+ , k k k k 1 i=0 i=0 i=0 X X X 1 1 the element 1 f T is not invertible in A r T , see Lemma 1.8.1; hence neither is f T . It follows h i 1 1 that the closure a of the principal ideal (f T )inA r T is a proper ideal. Endow A0 := A r T /a h i h i with the quotient norm 0.ThenA0 is a nonzero Banach ring, and we have a contractive morphism k·k 1 1 ': A A0, defined as the composition of A, A r T and the canonical map A r T A0. ! ! h i h i! Since A is a field, ' is injective. Thus a '(a) 0 defines a norm on A bounded by .Further, 7! k k k·k '(f) 0 = T 0 = r< f . This contradicts the minimality assumption on . k k k k k k k·k Finally we treat the case when (A, ) is not necessarily complete. Let (A,ˆ ) be the (separated) k·k k·k completion; this is a nonzero Banach ring. We claim that Aˆ is a field. (In general, the completion of a normed field is not necessarily a field.) Indeed, pick any maximal ideal n Aˆ and equip A0 := A/ˆ n ⇢ with the quotient norm 0.ThenA0 is a Banach field. The composition ': A A0 of the isometry k·k ! A Aˆ and the contractive morphism Aˆ A0 is contractive. It is also injective, as A is a field. ! ! Hence f '(f) 0 defines a norm on A dominated by . By minimality, 0 = ,which 7! k k k·k k·k k·k means that ': A A0 is an isometry. Since the image of A is dense in A0,wemusthaveA0 = Aˆ,so ! n = 0 and Aˆ is a Banach field. We further claim that the norm on Aˆ is minimal. Indeed, suppose 0 is a norm on Aˆ with k·k 0 .Since is minimal on A Aˆ,wehave f 0 = f on A. Now consider f Aˆ and k·k k·k k·k ⇢ k k k k 2 pick a sequence (fn)n in A such that limn fn f 0. Then limn fn f 0 0, and hence !1 k k! !1 k k ! f 0 =limn fn 0 =limn fn = f . k k !1 k k !1 k k k k Since the norm on Aˆ is minimal, it follows from what precedes that (ii) holds for f Aˆ⇥, and 2 hence afortiorialso for f A⇥. 2 ⇤ Proof of Theorem 2.1.9. We already know from Proposition 2.1.4 that X := M (A) is com- pact. It remains to show that X is nonempty. First, we claim that it su ces to consider the case when A is a Banach field, i.e. a field equipped with a complete but possibly non-multiplicative norm. Indeed, since the seminorm A is not the zero seminorm, the separated completion Aˆ of A is a nonzero Banach ring. Further, the canonical map M (Aˆ) M (A) is homeomorphism by Proposition 2.1.7, so it su ces to consider the case when A ! 110 2. THE BERKOVICH SPECTRUM is a nonzero Banach ring. Now let m A be any maximal ideal, By Proposition 1.2.18, m is closed in A, so the quotient seminorm on the⇢ field A/m is a complete norm, i.e. A/m is a Banach field. By Proposition 2.1.6, the bounded map A A/m induces a continuous map M (A/m) M (A), so if ! ! M (A/m) is nonempty, so is M (A). From now on, assume that A is a Banach field. Let S be the set of all norms on A satisfying |·| .ThenS is nonempty since it contains the given norm on A. Partially order S by 0 |·| k·k |·| |·| i↵ f f 0 for all f A. | The|| poset| (S, )2 satisfies the conditions of Zorn’s Lemma, that is, any chain contains a minimal element. Indeed, a chain is a nonempty collection S0 S such that for any seminorms , in ⇢ |·|1 |·|2 S0,wehave f f for all f A or f f for all f A. Define a function 0 : A R⇥ by | |1 | |2 2 | |1 | |2 2 |·| ! + f 0 := inf f S0 . Clearly 0 0 = 0, 1 0 = 1 and f 0 f for all f A. It therefore su ces | | {| |||·|2 } | | | | | | k k 2 to prove f f 0 f 0 + f 0 and f f 0 f 0 f 0 for all f A, i =1, 2. Given ">0pick | 1 2| | 1| | 2| | 1 2| | 1| ·| 2| i 2 S0 such that f f 0 + ", i =1, 2. Since S0 is a chain, we may assume .Then |·|i 2 | i|i | i| |·|1 |·|2 f f 0 f f f + f f + f f 0 + f 0 +2", | 1 2| | 1 2|1 | 1|1 | 2|1 | 1|1 | 2|2 | 1| | 2| and
f f 0 f f f f f f ( f 0 + ")( f 0 + "), | 1 2| | 1 2|1 | 1|1 ·| 2|1 | 1|1 ·| 2|2 | 1| | 2| so letting " 0 we get f1 f2 0 f1 0 + f2 0 and f1f2 0 f1 0 f2 0, proving that 0 is a seminorm in S. ! | | | | | | | | | | ·| | |·| By Zorn’s Lemma, there exists a minimal element S. Applying Lemma 2.1.11 to the (not |·|2 necessarily complete) normed field (A, ) we see that the norm is a valuation, so M (A) = . |·| |·| 6 ; This completes the proof. ⇤ Remark 2.1.12. Unless A is a Banach field, the proof of Theorem 2.1.9 only produces a bounded multiplicative seminorm on A, and not necessarily a multiplicative norm. One may ask whether A admits a bounded multiplicative norm. This is clearly not the case when A has zero divisors. As a consequence of the Maximum Modulus Principle (Theorem 2.2.1), it is also not the case if A has quasinilpotent elements, that is, elements f such that ⇢(f) = 0. Finally, as shown in 2.3, it is not true when A is a complex Banach algebra. For a non-Archimedean Banach ring A,itdoesn’tseem§ to be known whether the absence of zero divisors and quasinilpotent elements is enough to guarantee that M (A) contains any norms, see Exercise 2.1.15. However, we will show in 5.2.2 that this is true when A is a k-a noid algebra, for a non-Archimedean field k. §
2.1.5. Direct limits and products. Let (Ai,'ij)i I be a directed system of seminormed rings, where ' : A A is a contractive map. Let A := lim2 A be the direct limit (in the category of ij i ! j i i rings), and write : A A for the natural map. As ! in 1.1.13, we equip A with the seminorm i i ! § a := inf a a = (a ),i I,a A (2.2) k k {k iki | i i 2 i 2 i} so that i is contractive. The seminormed ring (A, ) is is the direct limit of (Ai,'ij)inthe category of seminormed rings, see Exercise ??.??. k·k Proposition 2.1.13. There is a natural homeomorphism
M (lim Ai) ' lim M (Ai), i I ! i I !2 2 where the inverse limit is taken in the category of topological spaces. The proof is left as an exercise (see Exercise 2.1.3). We have an analogous result for Banach rings. If (A ,' ) is a direct system of Banach rings, then the complete direct limit lim A is i ij i I i I i 2 2 the direct limit of (Ai,'ij)i I in the category of Banach rings. ! 2 c 2.1.THEBERKOVICHSPECTRUMOFASEMINORMEDRING 111
Corollary 2.1.14. There is a natural homeomorphism
(lim A ) ' lim (A ), M i I i M i 2 ! i I ! 2 where the inverse limit is taken in the categoryc of topological spaces. Proof. The complete direct limit is the separated completion of the direct limit as seminormed group. The result therefore follows from Proposition 2.1.13 and Proposition 2.1.7. ⇤
Proposition 2.1.15. Let Ai i I be a finite set of seminormed rings. There is a natural home- omorphism { } 2
M ( Ai)= M (Ai), i I i I Y2 a2 where the coproduct is taken in the category of topological spaces. The proof of Proposition 2.1.15 is left as an exercise; see Exercise 2.1.4. See also Exercise 2.1.5 for the case of an infinite product in which each factor is a complete valued field. 2.1.6. Points, complete residue fields and characters. We now give various interpretations of the points in the Berkovich spectrum of a seminormed ring, emphasizing the analogy to algebraic geometry. First recall how an element f of a ring A defines a function on the prime ideal spectrum Spec(A). A point ⇠ Spec(A) is given by a prime ideal p of A.Theresiduefield(⇠) of ⇠ is defined as the 2 ⇠ fraction field of the integral domain A/p⇠.Wethenwritef(⇠) for the image of f in (⇠). In this way, f induces a function f :Spec(A) (⇠) ! ⇠ Spec(A) 2 a such that f(⇠) (⇠) for all ⇠ Spec(A). Note that the element f A is only uniquely determined by this function2 when A is reduced.2 In general, two elements f,g 2 A define the same function i↵ f g belongs to all prime ideals of A, that is, f g is nilpotent. 2 Now let A be a seminormed ring, and x M (A) a point in the Berkovich spectrum. Recall that this means that x defines a bounded multiplicative2 seminorm on A. It follows immediately that |·|x p := f A f(x) =0 x { 2 || | } is a prime ideal of A.Theseminorm x defines a valuation on the integral domain A/px, and this valuation extends uniquely to a valuation|·| on the fraction field Frac(A/p ), still denoted . x |·|x Definition 2.1.16. The complete residue field H (x) at x is the completion of field Frac(A/px) with respect to the valuation .Wewritef(x) H (x) for the image of f under the composition |·|x 2 A Frac(A/p ) H (x). ! x ! Note that the value f(x) of the induced valuation on H (x) on the element f(x) H (x)is equal to f , so the notation| | is consistent with our earlier usage f(x) for the value2 on f of a | |x | | seminorm in M (A). We see that f A defines a function 2 f : M (A) H (x) ! x M (A) 2a such that f(x) H (x) for every x M (x). Following the analogy with complex Banach algebras, see 2.3, Berkovich2 calls this induced2 function the Gelfand transform of f. We shall describe the kernel§ of the Gelfand transform in . § Going back to the analogy with algebraic geometry, recall that if A is a ring, then the points of Spec(A) can be understood as equivalence classes of characters of A, that is ring homomorphisms : A K,whereK is a field, and where two characters : A K , i =1, 2, are equivalent i↵there ! i ! i 112 2. THE BERKOVICH SPECTRUM is a third character : A K and characters ⌧i : K Ki such that i = ⌧i .Indeed,thekernel of any character : A K! defines a prime ideal and! hence a point ⇠ Spec( A), and the character ! 2 is equivalent to the one defined by the canonical map A Frac(A/p⇠). Note also that the point in Spec(A) defined by a character is the image of the unique! point in Spec(K) under the induced map Spec(K) Spec(A). The analogue! for seminormed rings is as follows. Let A be a seminormed ring. A character on A is a bounded ring homomorphism : A K,whereK is a complete valued field. Two characters ! i : A Ki, i =1, 2, are equivalent if there is a third character : A K and characters ⌧i : K Ki such that! = ⌧ . ! ! i i Proposition 2.1.17. The relation on characters on A is an equivalence relation, and the set of equivalence classes coincides with the Berkovich spectrum M (A). Remark 2.1.18. The proof shows the following more precise statement. Any character : A ! K defined a point x M (A), and factors through the canonical character : A H (x). 2 x ! Proof. First, suppose x M (A) and let H (x) be the complete residue field. Then the 2 canonical map x : A H (x)definedbyf f(x) is a character on A. Conversely, if : A K is a character, then f ! (f) defines a bounded7! multiplicative seminorm on A, and hence a point! 7! | | x M (A); it is the image of the induced map ⇤ : M (K) M (A). We claim that the character 2 ! factors through x. To see this, note that both and x factor through the canonical map ⇡ : A Frac(A/p ), say = ◆ ⇡ and =˜ ⇡ . x ! x x x x x ⇡x ˜ A Frac(A/px) K ◆ x ⌧
H (x)
From the universal property of the (separated) completion it now follows that ˜ factors through ◆x as in the diagram, which proves the claim. Now consider two equivalent characters i : A Ki, i =1, 2 factoring through a common character : A K. The claim above shows that ! , and hence , i =1, 2, factors through the ! i canonical character x : A H (x), where x M (A) is the point defined by . This implies that the relation on characters is! indeed an equivalence2 relation. Namely, reflexivity and symmetry are obvious, so we only need to verify transitivity. But if : A K , i =1, 2, 3 are characters with i ! i , and , being equivalent, then all the define the same point x M (A), and all factor 1 2 2 3 i 2 through A H (x). This completes the proof. ! ⇤ Finally we consider morphisms and complete residue fields. First recall the situation in alge- braic geometry, so let ': A B be a ring homomorphism. Then ' induces a continuous map ! '⇤ :Spec(B) Spec(A). Consider a point ⌘ Spec(B) and let ⇠ = '⇤(⌘) Spec(A) be its image. The composition! A B (⌘) factors through2 A (⇠) and induces an2 embedding (⇠) , (⌘) of residue fields. ! ! ! ! Now consider a bounded morphism ': A B of seminormed rings. Let y M (B) be a point and ! 2 x M (A) its image under the induced map M (B) M (A). The composition A B H (y)is 2 ! ! ! a character on A which by Proposition 2.1.17 must factor through the character A H (x), defining ! a character H (x) H (y). This leads to a commutative diagram ! A / B
✏ ✏ H (x) / H (y) 2.1.THEBERKOVICHSPECTRUMOFASEMINORMEDRING 113
Note that the morphism H (x) H (y) must be an isometric embedding, see Exercise 1.2.4. ! 2.1.7. Invertibility criterion. Recall that if A is a ring, then an element f A is invertible i↵it does not lie in any prime ideal, that is, for any point ⇠ Spec A, the image 2f(⇠) of f in the residue field (⇠) is nonzero. The following result gives an alternative2 criterion for Banach rings. Proposition 2.1.19. If A is a Banach ring, then an element f A is invertible i↵ f(x) =0for 2 6 all x M (A). 2 Proof. If A is the zero ring, then there is nothing to prove. Now assume A = 0. If f A is invertible, say fg =1whereg A,then1= (fg)(x) = f(x) g(x) , and hence6 f(x) = 02 for 2 | | | |·| | 6 every x M (A). Conversely, if f A is not invertible, then there exists a maximal ideal m of A containing2 f.Sincem is automatically2 closed, see Proposition 1.2.14, B := A/m is a Banach field under the quotient norm, so its Berkovich spectrum M (B) is nonempty. If x M (A)liesinthe 2 image of the map M (B) M (A) induced by the canonical morphism A B,thenf(x) = 0. ! ! 6 ⇤ 2.1.8. Fibers. Recall that if A B is a ring homomorphism, then the fiber of a point ⇠ Spec A under the induced map Spec B! Spec A is homeomorphic to Spec((⇠) B). In particular,2 ! ⌦A the fiber is empty i↵ (⇠) A B = 0. In this section we establish analogues for seminormed rings and Banach rings. Let ⌦ ': A B ! be a morphism of seminormed rings, and let f : Y X ! be the induced continuous map between the Berkovich spectra Y = M (B) and X = M (A).
Proposition 2.1.20. For any x X, the canonical contractive morphism B H (x) A B 2 1 !1 ⌦ induces a homeomorphism of M (H (x) B) onto f (x) Y . In particular, f (x)= i↵the ⌦A ⇢ ; seminorm on H (x) B is the zero seminorm. ⌦A Usually we will work with Banach rings, in which case the following reformulation is perhaps more natural.
Corollary 2.1.21. For any x X, the canonical contractive morphism B H (x) ˆ AB in- 2 1 1 ! ⌦ duces a homeomorphism of M (H (x) ˆ B) onto f (x). In particular, f (x)= i↵the seminorm ⌦A ; on H (x) ˆ B =0. ⌦A 6 This follows from Proposition 2.1.20 since the separated completion morphism H (x) B ⌦A ! H (x) ˆ B induces a homeomorphism on Berkovich spectra, see Proposition 2.1.7. ⌦A Remark 2.1.22. The seminormed rings A and B are not assumed non-Archimedean, so the tensor product above above is the general seminormed tensor product H (x) sum B, i.e. the seminorm on ⌦A H (x) B is the one in (1.9). When A and B are non-Archimedean, we can also use the non- ⌦A Archimedean seminorm in (1.11), leading to the seminormed ring H (x) max B. The two seminorms ⌦A on H (x) A B are in general not equivalent, but they have the same uniformization, and hence the same Berkovich⌦ spectrum, see Exercise 2.1.7.
Proof of Proposition 2.1.20. Set Bx := H (x) A B and Yx := M (Bx). We have the following two commutative diagrams. ⌦
Bx Yx
H (x) B M (H (x)) Y
' f A X 114 2. THE BERKOVICH SPECTRUM
Here M (H (x)) is a singleton, whose image in X is x. It follows that the image of Yx in Y is 1 1 contained in f (x). On the other hand, if y f (x) Y , then the composition A B H (y) 2 ⇢ ! ! is a character equivalent to A H (x), and hence must factor through a character H (x) H (y). The universal property of the! seminormed tensor product now shows that the canonical morphism! B H (y) factors through B . Let z Y = M (B ) be the point determined by the character ! x 2 x x B H (y). Then the image of z under the map Y Y is equal to y. x ! z ! It only remains to prove that the map Yx Y is injective. Indeed, by what proceeds, Yx Y !1 1 ! then defines a continuous bijection of Yx onto f (x). Since Yx is quasicompact and f (x) Hausdor↵, 1 it will follow that Y f (x) is a homeomorphism. x ! The injectivity of Y Y follows from Lemma 3.3.3. Indeed, by construction of H (x), the x ! augmented image of the canonical map A H (x)isdenseinH (x), see Exercise 2.1.11. Hence the augmented image of the morphism B B! is also dense. By Proposition 2.1.6, the map Y Y is ! x x ! injective. ⇤
2.1.9. Finite and nonempty fibers. Consider a ring homomorphism ': A B and let f :Spec(B) Spec(A) be the induced map between tprime ideal spectra. The! following re- sults are well-known,! see [Sta, Tag 00GH]: (1) if ' is finite then f has finite fibers, and in fact 1 sup⇠ Spec(A) #f (x) < ; (2) if ' is injective and integral, then f is surjective. 2 1 In this section we establish analogues for morphisms of seminormed rings. Thus let ': A B be a morphism of seminormed rings. *** Not yet finished. Commented out for now *** !
Exercises for Section 2.1 (1) Prove that if k is a valued field, then M (k) is a singleton. (2) Give an example of a Banach ring A that is not a valued field, but whose Berkovich spectrum is a singleton. (3) Prove Proposition 2.1.13. (4) Prove Proposition 2.1.15. (5) * Let ki i I be an arbitrary set of complete valued fields. Prove that M ( ki) is the Stone- { } 2 i I Cechˇ compactification of the discrete set I. 2 Q (6) Let and 0 be bounded multiplicative seminorms on a seminormed ring (A, ). Assume |·| |·| k·k that f = f 0 for all f A0,whereA0 A is a dense subset. Prove that = 0. | | | | 2 ⇢ |·| |·| (7) Let (A, ) be a seminormed ring and A0 A a dense subring. Prove that any bounded k·k ⇢ multiplicative seminorm on A0 extends uniquely to a bounded multiplicative seminorm on A. (8) Give an example showing that Proposition 2.1.19 does not hold for general normed rings. (9) Let A be a non-Archimedean seminormed ring and B, C non-Archimedean seminormed A- algebras. Consider the following seminorms on the tensor product B C: ⌦A
f sum =inf bi ci and f max = inf max bi ci , k k k k·k k k k i k k·k k i X where, in both cases, the infimum is taken over all representations f = b c .WriteD and i i ⌦ i sum Dmax for B A C equipped with these two seminorms. Prove that the identity map Dsum Dmax ⌦ P ! induces a homeomorphism M (D ) ⇠ M (D ). Hint: see Exercise 1.6.6. max ! sum (10) Let A be a valued ring. Prove that if a A is a unit, then a(x) = 1 for all x M (A). 2 | | 2 (11) Let A be a seminormed ring and x M (A) a point. Prove that the augmented image of the 2 character A H (x)isdense. (12) Let A B be! a morphism of seminormed rings, Y X the induced continuous map between ! ! Berkovich spectra, and y Y a point. Is it always true that canonical morphism H (f(y)) 2 ! H (y) is an isomoetric isomorphism? EXERCISES FOR SECTION 2.1 115
(13) Let A be a seminormed ring, Aˆ its separated completion, and Au its uniformization. Consider a point x M (A) and writex ˆ M (Aˆ), xu M (Au) for its unique preimages under the canonical 2 2 2 maps M (Aˆ) M (A) and M (Au) M (A), respectively, see 2.1.3. Prove that the induced ! ! § isometries H (x) H (ˆx) and H (x) H (xu) are isomorphisms. ! ! (14) Let (A, ) be a non-Archimedean Banach ring. Given t R⇥, show that M (A, ) and k·k 2 + k·k M (A, t) are homeomorphic. (15) * Let kA·bek a uniform non-Archimedean nonzero Banach ring A without zero divisors. Does M (A) contain any element that is a norm (rather than seminorm) on A? (16) Let A B be a morphism of seminormed rings, f : Y = M (B) M (A)=X the induced ! ! continuous map between Berkovich spectral. Fix a point x X, and set Z = M (C), where 2 C = H (x) B. Let y Y be any point with f(y)=x. It induces a character C H (z). ⌦A 2 ! Prove that the induced morphism H (z) H (y) is an isometric isomorphism. ! Categorical exercises. (17) Consider the functor A M (A) from the category of seminormed rings to the category of compact topological spaces.! (a) Does M commute with products? (b) Is M essentially surjective? (c) Is M full? (d) Is M faithful? Later. (18)