In the valuation ring $R$, I expected that
the sum of a finite number of element of R with different values cannot be zero.
Is this correct? If it is correct, please give me a proof, and if it is incorrect, please give me a counterexample. This is a question that came to me when I was studying elliptic curves.
What do you mean by different value ? If you mean that the values of the valuation, say $\nu$, of these elements are different then it is correct. We can reduce easily to the case of two elements.
Let $x_1, \ldots, x_n \in R$ such that $\nu(x_i) \not = \nu(x_j)$ for all distinct $i, j \in \{1, \ldots, n\}$. Without loss of generality, we can assume that $i < j$ implies $\nu(x_i) < \nu(x_j)$. But $$\nu(x_2+\ldots+x_{n}) \geqslant \min_{j=2..n} \nu(x_j) \geqslant \nu(x_2) > \nu(x_1).$$ If $x_1 + \cdots + x_n = 0$, then $\nu(x_1) = \nu(-x_2-\cdots-x_n) = \nu(x_2 + \cdots + x_n)$.