I am stuck at the following exercise:
Let $v$ be a valuation on $K$ and $f, g, h$ monic polynomials in $K[x]$ with $f(x) = g(x)h(x)$. If $f \in R_v[x]$ then $g$ and $h$ are in $R_v[x]$.
My attempt thus far is: Since $f \in R_v$ we have $v(f) \ge 0$, and since $f$ is monic we have $v(f) = v(1) = 0$. Now we should probably use that both $g$ and $h$ are monic too, to probably conclude that $v(g) = v(h) = 0$, but I do not see how.
Could you help me?
Below you can find all the relevant definitions and information I know about valuations thus far:
In lecture I learned the following definition of valuation: Let $(K,+,\cdot)$ be a field and let $(G,+)$ be a totally ordered group. A map $v: K \longrightarrow G\cup\{\infty\}$ is a valuation if the following properties hold:
- $v(ab) = v(a)+v(b)$
- $v(a+b) \ge \min\{v(a),v(b)\}$
- $v(a) = \infty \iff a = 0$
Then we proved the basic properties:
- $v(1) = 0$
- $v(a^{-1}) = -v(a)$
- $v(-a) = v(a)$
- $v(a - b) \ge \min\{v(a), v(b)\}$
- $ \text{If }v(a) \ne v(b), \text{ then } v(a+b) = \min\{v(a), v(b)\}$
Let $v$ be a valuation on a field $K$. For $f = \sum_{i=0}^n a_ix^i \in K[X]$ we define $v(f) = \min \{v(a_i) \mid 0 \le i \le n\}$. We see that for $f,g \in K[x]$ holds
$$v(fg) = v(f) + v(g).$$ And finally we defined $R_v := \{k \in K \mid v(k) \ge 0\}$.
Since $g,h$ are monic we have $v(g)\le 0$ and $v(h)\le 0$, so $v(gh)=v(g)+v(h)\le 0$. But we know that $v(gh)=v(f)=0$, and thus $v(g)=v(h)=0$.