This series of posts is aimed to be a rudiment of the Berkovich line, a locally compact space that “fills up” a non-archimedean field $K$. The materials here are more or less my summary of the lecture note by (M. Baker 2007), with some extra materials that I found meaningful to append.

Non-archimedean Fields

A non-archimedean field $(K,\vert\cdot\vert)$ is a field $K$ with an absolute value $\vert\cdot\vert$, that satisfies the followings.

1. $\vert a\vert=0$ if and only if $a=0$.
2. $\vert ab\vert=\vert a\vert\cdot\vert b\vert$,
3. $\vert a+b\vert\leq\max\left(\vert a\vert,\vert b\vert\right)$,
4. …with equality if $\vert a\vert\neq\vert b\vert$.¹

The first one is simply the nondegeneracy of the absolute value, and the second is quite natural among absolute values on fields (unlike algebras, where we demand submultiplicative $\vert ab\vert\leq\vert a\vert\cdot\vert b\vert$).

The third is the oddest line. It resembles the triangle inequality $\vert a+b\vert\leq\vert a\vert+\vert b\vert$, but the right-hand side is replaced with the maximum of two norms, instead of their sum.

The fourth line is the equality condition: if one quantity is strictly smaller than the other, then we have equality. Compare this with the equality condition of the triangle inequality: when “two quantities have equal sign/direction.”

An equivalent way of saying this is to declare a valuation of the field $K$. For that, choose a number $q>1$, and set $v(a):=-\log_q\vert a\vert$ (note the negated logarithm). This declares a function $v\colon K\to\mathbb{R}\cup{\infty}$ such that

1. $v(a)=\infty$ if and only if $a=0$.
2. $v(ab)=v(a)+v(b)$.
3. $v(a+b)\geq\min\left(v(a),v(b)\right)$,
4. …with equality if $v(a)\neq v(b)$.

I will later be interested in a non-archimedean field $K$ which is algebraically closed and Cauchy complete. But before I continue, let me list some first examples of the notion.

Example 1: formal Laurant series, Puiseux Series

The field $k((t))$ of formal Laurent series on any (algebraically closed) field $k$ is one of the first examples of a non-archimedean field. It is defined by the valuation (assume $a_{-N}\neq 0$)

so we have $v(1)=v(t^0)=0$, $v(t^2+t^3+t^5+\ldots)=2$, $v(\exp(t)-1)=1$, etc. Function $v$ measures the order of zero $t=0$, while a pole is interpreted as a negative zero.

The first two axioms of valuations are thus quite natural if we recall how to do multiplication of (formal) Laurent series. The third axiom is meant to cover the following case: if

are two Laurent series with same valuation $v(a)=v(b)=v$, their sum

will have (1) valuation $v$ if $a_v+b_v\neq 0$, but (2) valuation $>v$ if $a_v+b_v=0$ and one needs to go up the exponents to find a nonzero term. This is how we understand $v(a+b)\geq\min(v(a),v(b))$.

On the other hand, if $v(a)=v<v(b)=v+w$ say, then writing

we have

so by $a_v\neq 0$, $v(a+b)=v=v(a)=\min(v(a),v(b))$ follows.

The algebraic closure of the field of formal Laurent series is the field of Puiseux series,

where the valuation on $k((t))$ extends to Puiseux series naturally, so that $v(t^{1/i})=\frac1i$.

Note that this field is not (Cauchy) complete. To verify so, think of the series

This series does not fit into $k((t^{1/i}))$ for any $i$, but the partial sums are Cauchy (as the exponents increase to infinity). Because of that, one needs to Cauchy complete the Puiseux field $k\{\{t\}\}$ if one requires a non-archimedean algebraically-closed Cauchy-complete field.

Example 2: p-adic fields

Any introduction of $p$-adic fields $\mathbb{Q}_p$ involves some notes that goes as follows. On the rational field $\mathbb{Q}$, one may make a metric completion not just with the standard absolute value (and get $\mathbb{R}$), but also with $p$-adic absolute values $\vert\cdot\vert_p$. This absolute value is defined with the $p$-adic valuation $v_p$² that satisfies the following axioms.

1. $v_p(a)<\infty$ if $a\neq 0$, and $v_p(0)=\infty$.
2. $v_p(ab)=v_p(a)+v_p(b)$.
3. $v_p(a+b)\geq\min\left(v_p(a),v_p(b)\right)$.
4. $v_p(a+b)=\min\left(v_p(a),v_p(b)\right)$ holds if $v_p(a)>v_p(b)$ holds.

Adding $v_p(p)=1$ and $v_p(a)=0$ if $(a,p)=1$, the axioms completely determines $v_p$. Morally, if $a\in\mathbb{Z}$, the valuation $v_p(a)$ find the number $v$ in which $p^v \vert\vert a$ (i.e., $p^v\mid a$ but $p^{v+1}\nmid a$) holds.

The corresponding $p$-adic absolute value is declared as follows:

By this, we Cauchy complete the metric space $(\mathbb{Q},\vert\cdot\vert_p)$ to obtain the $p$-adic field $\mathbb{Q}_p$.

This field is not algebraically closed, and we denote the algebraic closure as $\overline{\mathbb{Q}}_p$. Still the metric $\vert\cdot\vert_p$ extends to there; say by $\vert p^{1/4}\vert_p=p^{-1/4}$, etc. However, one can also find a series in $\overline{\mathbb{Q}}_p$ that is Cauchy but not convergent in $\overline{\mathbb{Q}}_p$ (see (Koblitz 1996), Sec. III.4, Theorem 12). Hence one Cauchy-completes the metric spcae $(\overline{\mathbb{Q}}_p,\vert\cdot\vert_p)$, and obtain a non-archimedean, algebraically-closed, Cauchy-complete field $(\mathbb{C}_p,\vert\cdot\vert_p)$.

<sup>1. This axiom is in fact redundant, by the following argument (proof from Junekey Jeon). Suppose $\vert a\vert<\vert b\vert$. Then if $\vert a+b\vert\neq\vert b\vert$, we have $\vert a+b\vert<\vert b\vert$. But then as $\vert b\vert=\max(\vert b\vert,\vert-a\vert)\leq\max(\vert a+b\vert,\vert -a\vert)$, this contradicts unless $\max(\vert b\vert,\vert -a\vert)=\vert-a\vert$, which is another contradiction to $\vert a\vert<\vert b\vert$.↩
2. I believe that $\nu_p$ (nu, not vee) is more common notation for this. Nonetheless, I also believe that the ‘vee’ notation is making more sense, as this is compatible with the word ‘Valuations.’↩</sup>

Gelfand theory?

An algebraically-closed Cauchy-complete normed field that we first see, is the field of complex numbers $\mathbb{C}$. This field is perfect to build a theory of Banach algebras (algebra with a complete, submultiplicative norm) over $\mathbb{C}$. (See Gelfand duality and Gelfand–Mazur Theorem.)

Now as another algebraically-closed Cauchy-complete field $K$, but non-archimedean, we would expect something similar. That is, for a commutative Banach algebra $A$ over $K$, we expect that the space of $K$-algebra maps $A\to K$ forms the ‘spectrum’ of $A$ and continuous maps on the ‘spectrum’ of $A$ recovers $A$.

No, not the case for non-archimedean $K$. This is because there is a field extension $L/K$ (yes, must be transcendental) and an extension norm $\vert\cdot\vert_L$ of $(K,\vert\cdot\vert)$. This makes $L$ a normed $K$-algebra, but the ‘spectrum’ of $L$ is empty (there is no $K$-algebra map $L\to K$), so the “usual” Gelfand duality is not the case here.

Example. Let $L$ be the field of rational functions $K(T)$ in 1 variable. Since any $f\in K(T)$ may be written

$$f(T) = a\cdot\frac{(T-x_1)(T-x_2)\ldots(T-x_n)}{(T-y_1)(T-y_2)\ldots(T-y_m)},$$

if we declare

$$\vert f(T)\vert_L = \vert a\vert\cdot\frac{\max(1,\vert x_1\vert)\cdot \max(1,\vert x_2\vert)\cdots\max(1,\vert x_n\vert)}{\max(1,\vert y_1\vert)\cdot\max(1,\vert y_2\vert)\cdots\max(1,\vert y_m\vert)},$$

this norm $\vert\cdot\vert_L$ declares a multiplicative nondegenerate norm on the field $L$, that extends the norm $\vert\cdot\vert$ on $K$. (Here, that $\vert f+g\vert_L\leq\vert f\vert_L+\vert g\vert_L$ is left as a nontrivial task for later.)

(One cannot make such a field extension $L/\mathbb{C}$, that extends the modulus on $\mathbb{C}$. This is what Gelfand–Mazur theorem asserts.)

Thus instead, we demand a novel concept for the ‘spectrum’ of a normed $K$-algebra. Following Berkovich, one defines the notion of the Berkovich spectrum as below. (See (Baker–Rumely 2010) or (Berkovich 1990).)

Berkovich Spectrum

From now on, we will deal with a variety of norms defined on an algebra. Thus I will denote $(K,\vert\cdot\vert_K)$ ($K$ emphasized in the subscript) for my non-archimedean field of interest.

Let $A$ be a normed $K$-algebra. A multiplicative seminorm on $A$ is a map $\vert\cdot\vert\colon A\to[0,\infty)$ that has the following properties.

1. $\vert 0\vert = 0$.
2. $\vert f\cdot g\vert = \vert f\vert\cdot\vert g\vert$, for $f,g\in A$.
3. $\vert f+g\vert\leq\vert f\vert + \vert g\vert$.³
4. $\vert a\vert = \vert a\vert_K$ if $a\in K$.

(Note that the first property does not assert that $\vert f\vert = 0$ only if $f=0$. This is why the term is called seminorm, not a norm.)

The Berkovich spectrum of $A$, denoted $\widehat{A}$, is the set of all multiplicative seminorms of $A$. Here, we denote $x\in\widehat{A}$ for points of $\widehat{A}$, while actually it is a seminorm $\vert\cdot\vert_x$ on $A$.

The spectrum $\widehat{A}$ is a topological space endowed with the following subbasic open sets.

Remark. By the evaluation map

$$\widehat{A}\hookrightarrow\prod_{f\in A}[0,\infty)\cong\prod_{f\in A}(-\infty,\infty], \\ x\mapsto\left(\vert f\vert_x\right)_{f\in A}\mapsto (-\log_q\vert f\vert_x)_{f\in A},$$

we can (topologically) embed $\widehat{A}$ into a product of $(-\infty,\infty]=\mathbb{R}\cup{\infty}$’s. This is how the subbasic open sets described above are designed.

^{3. In fact, one can show that $\vert f+g\vert\leq\max(\vert f\vert,\vert g\vert)$ if the base field $K$ is non-archimedean.↩}

Examples

In my experience, the notion of multiplicative seminorms was new when I happened to learn Berkovich spectra. There were a couple of examples that remedied this unfamiliarity.

• Some preliminary theory

The following proposition is found to be useful in understanding seminorms.

Proposition. For a multiplicative seminorm $\vert\cdot\vert$ on $A$, define

$$\mathrm{ker}(\vert\cdot\vert) = \{f\in A : \vert f\vert=0\}.$$

Then $\mathrm{ker}(\vert\cdot\vert)$ is a prime ideal of $A$.

(Proof) Let $I=\mathrm{ker}(\vert\cdot\vert)$. Then the followings verify that $I$ is a prime ideal.

• If $f,g\in I$, we have $\vert f+g\vert\leq\vert f\vert+\vert g\vert=0$, so $f+g\in I$ follows.
• If $f\in I$ and $g\in A$, then $\vert fg\vert=\vert f\vert\cdot\vert g\vert = 0\cdot\vert g\vert = 0$, so $fg\in I$ follows.
• If $fg\in I$, then $\vert f\cdot g\vert=\vert f\vert\cdot\vert g\vert=0$, so either $\vert f\vert=0$ or $\vert g\vert=0$; hence $f\in I$ or $g\in I$.

So for any multiplicative seminorm $\vert\cdot\vert$, there is a prime ideal $I$ that factors the norm as $\vert\cdot\vert\colon A\to A/I\to[0,\infty)$. Because $A/I$ is an integral domain and $\vert\cdot\vert$ is a multiplicative norm on it, the norm extends to the field of fractions $\mathsf{Frac}(A/I)\to[0,\infty)$. This gives the following

Corollary. For a multiplicative seminorm $\vert\cdot\vert_x$ on $A$, it has a residue field $\mathcal{H}(x)=\mathsf{Frac}(A/\mathrm{ker}(\vert\cdot\vert_x))$ in which

1. the seminorm $\vert\cdot\vert_x$ factors $A\to\mathcal{H}(x)\to[0,\infty)$,
2. and the induced map $\mathcal{H}(x)\to[0,\infty)$ is a multiplicative norm on $\mathcal{H}(x)$.

Examples over $\mathbb{C}$

The notion of multiplicative seminorms pretty much applies to any normed field. So one can think of multiplicative seminorms on an arbitrary normed $\mathbb{C}$-algebras.

Given a normed $\mathbb{C}$-algebra $A$, if $\vert\cdot\vert_x$ is a multiplicative seminorm, then its residue field $\mathcal{H}(x)$ can only be $\mathbb{C}$ (by Gelfand–Mazur), and thus the induced norm on $\mathcal{H}(x)$ is the complex modulus.

Because of that, each multiplicative seminorm uniquely corresponds to a ring map $A\to\mathcal{H}(x)=\mathbb{C}$. This is precisely what the Gelfand spectrum collects. That is,

Over $\mathbb{C}$, Gelfand spectrum and Berkovich spectrum coincide.

To elaborate, below is how the correspondence works.

• (Gelfand to Berkovich) For a ring map $x\colon A\to\mathbb{C}$, we define a multiplicative seminorm $\vert f\vert_x=\vert x(f)\vert_\mathbb{C}$.
• (Berkovich to Gelfand) For a multiplicative seminorm $\vert\cdot\vert_x$ on $A$, as $A/\mathrm{ker}(\vert\cdot\vert_x)$ is an integral domain, it is naturally isomorphic to $\mathbb{C}$.⁴ Thus the natural ring map $x\colon A\to A/\mathrm{ker}(\vert\cdot\vert_x)\cong\mathbb{C}$ is induced.

^{4. A version of Gelfand–Mazur (also called Kaplansky’s theorem) claims that, for a (multiplicatively) normed integral domain $B$ over $\mathbb{C}$, the map $\mathbb{C}\to B$, $c\mapsto c\cdot 1_B$, is an isomorphism. See Cabrera Garcia and Rodriguez-Palacios 1995, as mentioned in Bhatt and Kulkarni 2018.↩}

0-dimensional case

Next, we work on a case where we can explicitly compute the spectrum.

Say put $A=K\times K$. For any $x\in\widehat{A}$, there are two choices for the $\mathrm{ker}(\vert\cdot\vert_x)$:

• $\mathrm{ker}(\vert\cdot\vert_x)=0\times K$: this corresponds to $\vert (f_1,f_2)\vert_1=\vert f_1\vert_K$.
• $\mathrm{ker}(\vert\cdot\vert_x)=K\times 0$: this corresponds to $\vert(f_1,f_2)\vert_2=\vert f_2\vert_K$.

Thus $\widehat{A}$ consists of two points. Likewise, for $A=K^m$ ($m$-fold ring product), its spectrum $\widehat{A}$ consists of $m$ points. Thus this classifies all 0-dimensional (finitely generated) Berkovich spectrum.

1-dimensional case

The next complicated example will rise with $A=K[T]$. The Berkovich spectrum of $A$ gains special attention; it is called the Berkovich line and denoted $\mathbb{A}^1_{\mathrm{Berk}}$ or $(\mathbb{A}^1)^{an}$.

Recall that $K$ is algebraically closed. A well-known fact from algebraic geometry classifies all prime ideals of $K[T]$ as follows.

• Maximal ideals all have the form $(T-a)$ for some $a\in K$.
• The only non-maximal prime ideal is $(0)$.

For the former, the corresponding norm is then described as

in which I will call as evaluation seminorms. Such norms uniquely correspond to each $a\in K$, thus we see that the Berkovich line $\mathbb{A}^1_{\mathrm{Berk}}$ has a (natural) copy of $K$ inside it. (One can even show that the inclusion $K\hookrightarrow\mathbb{A}^1_{\mathrm{Berk}}$ is a topological embedding.)

For the latter, the residue field is then $\mathcal{H}(x)=K(T)$ (field of rational functions in $T$), and we ask what are multiplicative norms on there. It turns out that there are a lot of norms in this class, which will be introduced in later posts.

Next

This post is continued at part 2.

Update Log

• 220528: Created
• 220704: Some remarked added according to Junekey Jeon
• 220727: Recoined ‘evaluation norm’ → ‘evaluation seminorm’