Exercise 11.1

Proof. Here $f(u)=1+a_{1}u+a_2{u}^2+\cdots \in ℂ[[u]]$ is a formal series in the variable $u$.

We suppose that $f(u)=P(u)∕Q(u)$, where we may assume, after simplification, that the two polynomials are relatively prime. Then $P(0)∕Q(0)=1$. Write $c=P(0)=Q(0)\in F$.

If $c=0$, then $u\mid P(u)$ and $u\mid Q(u)$. This is impossible since $P\wedge Q=1$. So $c\ne 0$.

Define $P_1(u)=(1∕c)P(u),Q_1(u)=(1∕c)Q(u)$. Then $f(u)=P_1(u)∕Q_1(u)$ and $P_1(0)=Q_1(0)=1$. If we replace $P,Q$ by $P_1,Q_1$, then the pair $(P_1,Q_1)$ has the required properties. □