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I am a beginning graduate student. I have the following basic question I am very confused about:

Suppose $C$ is a smooth geometrically irreducible curve over a finite field $\mathbb{F}_q$, $q=p^m$, $p$ prime. Now take the ring $A$ of functions on $C$ regular away from a rational point, say $\infty$. Let $K$ be its function field and $K_\infty$ be the completed algebraic closure of $K$ wrt a valuation at $\infty$. Now define a rigid analytic function, say $f$, over $K_\infty$.

On the other hand suppose $B= \mathbb{F}_q[\theta]$, where $\theta$ is an indeterminate, and let $B_\infty$ be defined same way as $K_\infty$. Then define the Tate algebra to be $T_B = B_\infty[[T_1,\cdots, T_l]]$, where $T_i$ are indeterminates different from $\theta$. Again define the rigid analytic function, say $g$, over this Tate algebra $T_B$.

What is the difference (which one is more general) between theory developed around the functions $f$ and $g$?

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  • $\begingroup$ First, the Tate algebra $T_{B}$ is a particular proper subring of $B_{\infty}[[T_{1},\dots,T_{\ell}]]$, often denoted $B_{\infty}\langle T_{1},\dots,T_{\ell}\rangle$. Second, it's a little unclear what you mean by "a rigid analytic function over" $K_{\infty}$ or $T_{B}$. Do you mean elements $f$ of $K_{\infty}\langle X_{1},\dots,X_{m}\rangle$ and $g$ of $T_{B}\langle Y_{1},\dots,Y_{n}\rangle=B_{\infty}\langle T_{1},\dots,T_{\ell},Y_{1},\dots,Y_{n}\rangle$? If so, your first example is more general, since $B_{\infty}$ in your second example is just $K_{\infty}$ for $C=\mathbb{P}^{1}$. $\endgroup$
    – Tyler Foster
    Commented May 13, 2017 at 5:28
  • $\begingroup$ I believe your fields $K_{\infty}$ and $B_{\infty}$ are basically the same field. Let $\theta$ be any non-constant element of $A$, and so the map $\theta: C \to \mathbb{P}^1$ has a single pole only at $\infty$, which is given to be an $\mathbb{F}_q$-rational point. Therefore the field extension $K/\mathbb{F}_q(\theta)$ is only inert or ramified at $\infty$ (but not split). This implies the completion of $\mathbb{F}_q(\theta)$ at the infinite place is a subfield of the completion of $K$ at $\infty$ with the same norm. And so their algebraic closures and subsequent completions are the same. $\endgroup$
    – Matt Papanikolas
    Commented May 16, 2017 at 14:54

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