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January 15, 2020, 9:55Pm 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) <p, g (x) >q , 1 i m, 1 j n , (2.1) { 2 || i | i | j | j } where m, n 0, p ,q R⇥, f ,g A,1 i m,1 j n. ≥ i j 2 + i j 2 Let us show already now that the Berkovich space M (A) has good topological properties. 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 suffices 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 suffices 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).
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