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Let $(k,|\cdot|)$ be an algebraically closed field, complete wrt a (multiplicative) norm as in the framework of the Berkovich's analytic geometry. Given a commutative Banach $k$-algebra $\mathcal{A}\neq 0$, let $X=\mathcal{M}(\mathcal{A})$ its spectrum and let $0\neq f\in \mathcal{A}$.

For every point $P=||\cdot||\in X$, the kernel $\mathfrak{B}_P$ of $P$ is a closed prime ideal of $\mathcal{A}$. Then $P$ induces a seminorm on the integral domain $\mathcal{A}/\mathfrak{B}_P$ and therefre in its fraction field $F$. The completion of $F$ wrt the norm induced by $P$ is denoted $\mathcal{H}(P)$. It is a normed field extension of $k$. The image of $f$ under the above sequence of transformations is an element of $\mathcal{H}(P)$ which is denoted $f(P)$.

My point is: the function induced by $f$ on $X$ take values at different field extensions of $k$ (unless, for example, if $k=\mathbb{C}$ with the euclidean norm). $$f:X\rightarrow\bigcup_{P\in X}\mathcal{H}(P),$$ via $P\mapsto f(P).$

Does anyone know if there is a nice way to give a topology to the above disjoint union (which can be identified with $\mathbb{C}$ in the "classical" case...) so that we have properties like path-connectedness, Hausdorff, etc as we do have in the aforementioned particular case?

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  • $\begingroup$ Sorry, but I am not sure that I understand. You want the nice properties on the disjoint union, right? But if your start with a nonarchimedean $k$, then every $\mathcal{H}(P)$ is nonarchimedean, hence certainly not path-connected. So the topology induced on $\mathcal{H}(P)$ would not be the usual one? $\endgroup$ Commented May 27, 2014 at 19:53

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If you want to think of $f$ in a way not too far from complex intuition, you should rather consider the induced morphism $\varphi$ from $X$ to the Berkovich affine line. If $x\in X$ then $\varphi(x)$ is the semi-norm $P\mapsto |P(f)(x)|$.

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  • $\begingroup$ The notation seems confusing. I guess here $P$ is meant to be a point(seminorm) in the affine line, unlike its role in the above question (which is played by $x$ now). $\endgroup$
    – Uri Bader
    Commented Apr 21, 2016 at 18:18
  • $\begingroup$ No. $P$ is meant to be a polynomial. And my formula defines a semi-norm, i.e., something that takes a polynomial and provides a non-negative real number. $\endgroup$ Commented Apr 21, 2016 at 18:22
  • $\begingroup$ Got it, thanks. Still, it is confusing because in the quetion $P$ was a point. $\endgroup$
    – Uri Bader
    Commented Apr 21, 2016 at 22:02
  • $\begingroup$ Thanks. I'm sorry for being so out of this conversation. $\endgroup$
    – amateur
    Commented May 1, 2017 at 17:46

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