On some basic property of a valuation ring

Question

Let $R$ be a valuation ring. Suppose its field of fractions $k$ is algebraically closed. Let $(a_0, \ldots, a_n)$ be a tuple of elements in $k$. Then I would like to deduce that there exists some $b \in k$ such that $b a_j$ are all in $R$, but not all of them in the unique maximal ideal of $R$. How can I prove this? Thank you.

Answer 1

Fist of all, since $a_i \in k$, they are of the form, $\alpha_i/\beta_i$, where $\alpha_i, \beta_i \in R$. Let $b = \beta_0 \cdots \beta_n$. Then $ba_i$ are all in $R$.

If $(ba_0,\dots,ba_n) = R$, then you are done. If not, then since $R$ is a valuation domain, every finitely generated ideal is principal. Thus, there exists $a \in R$ such that $(a) = (ba_0,\dots,ba_n)$. Now divide this ideal by $a$ to obtain the unit ideal.

Answer 2

We can assume that $\nu (a_0) \leq \nu (a_1) \leq \cdots \leq \nu (a_n)$. In particular this gives $a_0 \neq 0$. Choose $b = a_0^{-1}$. Observe that for $i > 0$ we have $\nu (b a_i) = \nu (a_i) - \nu (a_0) \geq 0$. And $b a_0 = 1 \in R \setminus \mathfrak{m}$.

I don't think you need the algebraically closed hypothesis.