2
$\begingroup$

Let $k$ be a finite field. Let $K$ a finite extension of $k(t)$. Let $S$ be a finite set of places of $K$. Let $$K_S = \prod_{v\in S} K_v$$ where $K_v$ is the completion of $K$ at $v$. For $v\in S$, let $\mathfrak{m}_v\subset\mathcal{O}_{K_v}$ be the maximal ideal of the valuation ring $\mathcal{O}_{K_v}$ of $K_v$. Let $$\mathcal{O}_{K,S} = \{\alpha\in K : v(\alpha)\geq 0, \forall v\not\in S\}\subset K$$ be the $S$-integers and embed $\mathcal{O}_{K,S}$ into $K_S$ diagonally. Let $\mathfrak{m}_S = \prod_{v\in S}\mathfrak{m}_v$. By the product formula, $\mathcal{O}_{K,S}\cap \mathfrak{m}_S = \{0\}$, so we get an embedding $$\mathfrak{m}_S\hookrightarrow \frac{K_S}{\mathcal{O}_{K,S}}.$$ My question is, is this embedding surjective?

The reason why I stated weak approximation in the title is because this question can be restated as follows: Given $\alpha = (\alpha_v)_{v\in S}\in K_S$, does there exists $\beta\in \mathcal{O}_{K,S}$ such that for all $v\in S$, $\alpha_v - \beta\in \mathfrak{m}_v$? If $\mathcal{O}_{K,S}$ is replaced by $K$ in this question, then the answer is yes by weak approximation (and we can replace the $\mathfrak{m}_v$'s by any power of the $\mathfrak{m}_v$'s).

My intuition says the answer to the question is yes, but after trying for a while I cannot prove it or find a reference. The most trivial example is the fact that $$k((t^{-1})) = k[t] \oplus t^{-1}k[[t^{-1}]]$$ where the direct sum is as abelian groups. In this example, $K = k(t)$ and $S = \{v\}$ where $v$ is the valuation with uniformizer $t^{-1}$. Then $\mathcal{O}_{K,S} = k[t]$ and $K_S = k((t^{-1}))$.

Any hint, answer, or reference would be appreciated!

$\textbf{Edit}$: By Felipe's comment, the answer to the original question is no, and a necessary condition for $\beta$ to exist is that $$\sum_{v\in S}res_v(\alpha_v\omega) = 0$$ for all differentials $\omega$ of $K$. My two new questions are, is there a sufficient condition for such a $\beta$ to exist? And is there a description or meaning/significance to the quotient $\displaystyle \frac{K_S}{\mathcal{O}_{K,S} + \mathfrak{m}_S}$?

$\endgroup$
4
  • 3
    $\begingroup$ This the Riemann-Roch problem. Given a bunch of power series at a finite set of places (the $\alpha_v$) is there a global element with these polar parts and regular elsewhere? The answer is no, in general and there is an additional condition for $\beta$ to exist, namely $\sum res_v(\alpha_v \omega) = 0$ for every differential $\omega$. $\endgroup$ Commented Apr 5, 2019 at 6:36
  • $\begingroup$ Hi Felipe, thank you that is very useful. I'm familiar with Riemann-Roch telling one the dimension of the space of global functions with prescribed zeros and poles, but not when one specifies the whole polar parts of the function (the $\alpha_v$'s) as in the question. Do you know a reference where what you've mentioned is explained? Also, is there a description or meaning/significance of the quotient $K_S/(\mathcal{O}_{K,S} + \mathfrak{m}_S)$? $\endgroup$ Commented Apr 5, 2019 at 20:56
  • 1
    $\begingroup$ Weil's adelic proof of RR goes this route. A reference I like is Lang's "introduction to algebraic and abelian functions" but watch out for the usual typos. $\endgroup$ Commented Apr 5, 2019 at 21:32
  • $\begingroup$ Thanks for the reference! I took a look at ch 1 in Lang. In the language of Lang, by differential, you mean differential form of the first kind? Then the sum in your first comment would follow from theorem 5.2 in Lang applied to differential forms of the first kind. This gives a necessary condition for $\beta$ to exist. I couldn't find a sufficient condition in Lang. I'm going to edit my question to include this. Thank you! $\endgroup$ Commented Apr 6, 2019 at 0:10

0

You must log in to answer this question.