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2.4. EXAMPLE: , DISCRETE RINGS, AND DEDEKIND DOMAINS 125 January 26, 2020, 11:59am 2.4. Example: integers, discrete valuation rings, and Dedekind domains Our next class of examples appear in number theory. First we consider the Z equipped with the usual Archimedean absolute value. We then generalize this to the in a number field. Finally, we consider a Dedekind domain equipped with the trivial absolute value. 2.4.1. The integers. Let denote the usual absolute value on Z. In other words: n = n if n 0 and n = n if n<0.|·| We1 want to describe the Berkovich spectrum | |1 | |1 X := (Z), M i.e. the set of bounded semivaluations on Z. Now, the chosen norm on Z is canonical in the sense that it is dominates any other ring seminorm on Z, (see Exercise 1.2.3. This means that (Z)issimply the set of semivaluations on Z; the boundedness assumption is irrelevant. To describeM the Berkovich spectrum, we now use Ostrowski’s Theorem, see 1.3.2. Pick any point x X, corresponding to a semivaluation on Z. § 2 |·|x If x is a valuation, it extends to a valuation on Q, and Ostrowski’s Theorem applies. Thus there are|·| three cases:

(a) x = 0 is the trivial valuation on Q; |·| |·|⇢ (b) x = for a unique ⇢ (0, 1]; (c) |·| = |·|1 is a p-adic valuation2 normalized by p = ",where" (0, 1). |·|x |·|p," | | 2 Now suppose has nontrivial kernel. The kernel is a prime of Z, and hence of the form pZ |·|x for some prime p. It follows that x induces a valuation on the finite field Fp = Z/pZ.Sincethe only valuation on a finite field is the|·| trivial valuation, it follows that = , where the latter |·|x |·|p,0 semivaluation is defined by n p,0 =0ifp n, and n p,0 =1ifp - n. Let us draw X schematically| | as a tree| with a| single| branch point at the trivial norm ,a |·|0 branch connecting 0 to p,0 for each p, and a branch connecting 0 and . See Figure 2.1. The point of describing|·| X |·|this way, is that the Berkovich topology can|·| be described|·|1 in terms of the tree structure on this tree. We will discuss this more generally in 3.1, but let describe the topology § in terms on Figure 2.1. By definition, the topology on X is the weakest one for which x n x is continuous for all n Z. One can then show (Exercise 2.4.1) that 7! | | 2 (a) the map (0, 1] ⇢ ⇢ X is a homeomorphism onto its image; (b) the map [0, 1) 3 " 7! | · |1 2 X is a homeomorphism onto its image for every prime p; 3 7! | · |p," 2 (c) a basis of open neighborhoods of 0 in X is given by complements in X of finite unions ⇢ |·| of segments of the form ⇢ ⇢0 , ⇢0 (0, 1], or p," " "0 , "0 [0, 1). {| · |1 | } 2 {| · | |  } 2 Note that conditions (a)–(c) determine a basis of open neighborhoods of any point in X, and therefore completely describe the topology. 2.4.2. The ring of integers in a number field. We can generalize the discussion above as follows. Let K be a number field, i.e. a finite extension of Q. An element ↵ K is an algebraic if it satisfies a monic equation 2 n n 1 ↵ + a ↵ + + a =0, 1 ··· n where n 1 and ai Z for all i,ThesubsetoK K of algebraic integers is a subring of K. It is an with2 fraction field K. ⇢ The set of absolute values on oK is described by a generalization of Ostrowski’s theorem. We shall leave the proof as an exercise (Exercise 2.4.3) but let us explain the statement. First, since oK is an integral domain, the trivial norm 0 is a valuation on oK . Second, for any p of o , define a function v : o N |·|by K p K ! [{1} v (↵) := max n N ↵ pn . p { 2 | 2 } 126 2. THE BERKOVICH SPECTRUM

|·|1

⇢ |·|1

|·|0

|·|p," |·|q,"

|·|p,0 |·|q,0 Figure 2.1. The Berkovich spectrum (Z) of the integers, see 2.4.1. The branches at the bottom correspond to primeM numbers. §

For any " (0, 1], the function 2 := "vp |·|p," is then a valuation on o . Note that = , the trivial norm. One can check that if p = p , K |·|p,1 |·|0 1 6 2 then p1,"1 = p2,"2 for all "1,"2 (0, 1), see Exercise 2.4.4. Third,|·| any6 |·| embedding : K 2C defines an absolute value ( ) .If(K) R,then is called a real embedding. When !is not real, its conjugate is another| · |1 embedding⇢ which of course gives rise to the same absolute value. Let r1 be the number of real embeddings of K, and r2 the number of pairs of complex-conjugate embeddings of K. The following result is well known *** REFERENCE ***

Lemma 2.4.1. We have r1 +2r2 =[K : Q]. Further, two di↵erent embeddings of K into C give rise to the same absolute value i↵they are complex conjugate. Ostrowski’s Theorem now reads: Theorem 2.4.2. Any absolue value on on o is of one of the following three types: |·| K (a) is the trivial valuation; (b) |·|= ( ) ⇢ for some ⇢ (0, 1] and some embedding : K C; (c) |·|= | · |1for some maximal2 ideal p o and some " (0!, 1). |·| |·|p," ⇢ K 2 We now want to compute the Berkovich spectrum X = (o ). M K of oK . For this, we need a norm on oK . We pick this as ↵ = max (↵) , k k | |1 where the maximum is taken over all embeddings K, C. One can check that is the largest ! k·k possible power-multiplicative norm on oK (Exercise 2.4.6). In particular, any semivaluation on oK is automatically bounded, so X is simply the set of all semivaluations on oK . The set of valuations on o is described by Ostrowski’s theorem, so consider a semivaluation on o that is not a valuation. K |·| K Then its kernel p := ker( ) is a maximal idel of oK . The induced valuation on oK /p must be the trivial valuation, since o |·|/p is a finite field, so = , where the latter semivaluation is defined K |·| |·|p,0 by ↵ p,0 =0if↵ p, and ↵ p,0 = 1 otherwise. | The| topological2 structure| | of X is now very similar to that of (Z). The only di↵erence is that there may now be several Archimedean branches, see Figure 2.2. M 2.4. EXAMPLE: INTEGERS, DISCRETE VALUATION RINGS, AND DEDEKIND DOMAINS 127

1 2 | |1 | |1

⇢ ⇢ 1 1 | |1 | |1

|·|0

|·|p," |·|q,"

|·|p,0 |·|q,0

Figure 2.2. The Berkovich spectrum of the ring of integers oK in a number field K,see 2.4.2. The infinitely many branches at the bottom correspond to maximal § ideals of oK . The finitely many branches at the top correspond to real embeddings and conjugacy classes of complex embeddings of K.

2.4.3. Discrete valuation rings. Next, let A be a discrete .ThusA is a with (nonzero) maximal ideal m, and m is principal, say generated by some element g.Despite the term “valuation ring”, we equip A with the trivial norm 0, which is a valuation since A is an integral domain. |·| Consider a point x (A). We have f(x) f 0 1 for all f A. This implies that f(x) =1whenf A 2m,sinceM f is then invertible.| | It| now|  follows that x2is uniquely determined by| the| number " :=2 g(\x) [0, 1]. Indeed, any f A 0 can be written uniquely as f = gnh, where n 0 and h |A m|,2 and it follows that f(x2) = \{"n. } Denote by 2 the\ semivaluation on A defined| | by f = "n in the notation above. Then |·|" | | " " gives a map from the interval [0, 1] onto (A), and this bijection is easily seen to be a homeomorphism,7! | · | see Exercise 2.4.8. M Note that in multiplicative notation, the discrete valuation on A is exactly of the form for some |·|" " (0, 1). If we equip A with this valuation, then the Berkovich spectrum becomes (A, ")=[0,"] using2 the identification above. Again see Exercise 2.4.8. M |·|

2.4.4. Dedekind domains. Next we consider the case when A is a trivially valued Dedekind domain. One may keep in mind the examples when A is the ring of integers in a number field (but equipped with the trivial norm) or the ring in one variable. The case of a that we just treated is another example. Consider a point x (A), with associated semivaluation x on A. Note that a x a 0 =1 for all a = 0. We now have2M two natural prime ideals associated|·| to x, namely p = | | =0| | and 6 x {| · |x } qx := < 1 . Note that qx is well-defined since a x 1 for all a A. Since{| · |A is} a Dedekind ring, there are now three| cases:|  2

(i) px = qx = 0; (ii) px = qx = m for some maximal ideal m; (iii) px = 0 and qx = m for some maximal ideal m.

In (i), x = 0 is the trivial valuation. In (ii), x defines a valuation on the field A/m.But this valuation|·| |·| is bounded by the trivial valuation,|·| and must hence be equal to it. It follows that x is the semivaluation m,0 defined by a m,0 =0ifa m, and a m,0 =1ifa m. Finally, we consider|·| the situation in (iii).|·| The maximal| ideal| m is principal,2 say| generated| by 62b m.Ifweset 2 n " := b x (0, 1), then x is the valuation m," defined as follows: given a A 0 ,writea = b c, where| |n 2 0 and c m|·|;then a = "n. |·| 2 \{ } 62 | |m," 128 2. THE BERKOVICH SPECTRUM

It follows that the Berkovich spectrum (A) of A is a tree with a single branch point at the M trivial valuation 0, and one branch for each maximal ideal m. See Figure 2.3. Note that when A = Z, we get the|·| same picture as in Figure 2.1 but with the Archimedean branch pruned.

|·|0

|·|m," |·|n,"

|·|m,0 |·|n,0 Figure 2.3. The Berkovich spectrum of a trivially valued Dedekind ring, see 2.4.4. There is one branch for every maximal ideal. §

It is possible to generalize this example to the case when A is a one-dimensional integral domain whose integral closure A (in the fraction field of A) is a Dedekind domain. See Exercise 2.4.9.

Exercises for Section 2.4 EX70242 (1) Prove the description of the topology on (Z)in 2.4.1. M § EX70248 (2) Let n 1 be an integer, and " (0, 1). Let denote the n-adic norm on Q defined in 2 k·kn," Exercise 1.2.25. Describe the Berkovich spectrum M(Q, n,"). Also describe M(Z(n), n,"), where Z Q is the set of rational numbers with denominatork·k relatively prime to n. k·k (n) ⇢ EX70243 (3) Prove Ostrowski’s Theorem for number fields.

EX70244 (4) Let K be a number field and let p1 = p2 be maximal ideals in oK . Prove that p1,"1 = p2,"2 for all " ," (0, 1). Hint; use the Chinese6 remainder theorem. |·| 6 |·| 1 2 2 EX70251 (5) Let K be a number field. Prove that any Archimedean valuation on K is of the form ⇢ ,where : K C is a complex (or real) embedding, and ⇢ (0, 1]. | |1 ! 2 EX70245 (6) Let o be the ring of integers of a number field K. Let : K C, i I,bethe(finite)set K i ! 2 of embeddings : K C, and define a norm on oK by a = maxi I i(a) . Prove that is the largest possible! power-multiplicativek· normk on o .k k 2 | |1 k·k K EX70246 (7) Let K be a number field, and let S be a finite set of places of K, including all non-Archimedean places, and let o K be the set of S-integers in K. In practice, this means that we are given K,S ⇢ finitely many prime ideals p1,...,pn of oK , and that oK,S := a K a p," 1 for all prime ideals p of o not in p ,...,p . Here " (0, 1) is any number.{ 2 Prove|| that| o } o , so that K { 1 n} 2 K ⇢ K,S inclusion oK , oK,S induces a continuous map (oK,S) (oK ). Prove that this map is a homeomorphism! onto its image, and describe theM image. !M EX70247 (8) Let A be a discrete valuation ring, equipped with the trivial valuation. Prove that the map [0, 1] " " (A)in 2.4.3 is a homeomorphism. Also prove that (A, ")=[0,"] under3 this7! identification. | · | 2M § M |·| EX70249 (9) Analyze the Berkovich spectrum (A) in the case when A is a trivially valued integral domain of dimension one, whose integralM closure (in the fraction field of A) is a Dedekind domain.