How do you prove a valuation ring is a subring? Let's say I have a field $\mathbb{F}$. 
Now suppose I take the set $R = \{x \in \mathbb F^{\times}: \ y(x) \ge 0\} \cup \{0\}$ where $y$ is a function $y:\mathbb F^{\times} \rightarrow \mathbb{Z}$ satisfying $y(ab) = y(a) + y(b)$, $y$ is surjective, and $y(a + b) \ge \min\{y(a), y(b)\}$ for all $a,b \in\mathbb F^{\times}$ with $a + b \neq 0$. $\mathbb F^{\times}$ is of course the units of $\mathbb{F}$.
How could I show that $R$ is a subring of $\mathbb F$ which contains the identity?
Note: A subring is a nonempty subset which is closed under subtraction and multiplication.  But how do I show that this set is closed like that?

Let's show that $R$ contains the identity. Indeed, $y(1) = y(1 \cdot 1) = y(1) + y(1) \Rightarrow y(1) = 0$.
Let now $\alpha, \beta \in R$ be arbitrary. First of all, $y(\alpha - \beta) \ge \min\{y(\alpha), y(-\beta)\}$. Furthermore, $y(-\beta) = y(-1) + y(\beta)$. Further, $y(1) = y((-1)(-1)) = y(-1) + y(-1)$ and hence $y(-1) = 0$. It follows $\alpha - \beta \in R$.
From $y(\alpha \beta) = y(\alpha) + y(\beta)$ directly follows $\alpha \beta \in R$.
Hence, $R$ is indeed a subring.
