Adele space of rational function field

Question

Let $F=K(x)$ and $K(x)/K$ be the rational function field. I am then trying to prove \begin{align*} A_F = A_F(0) + F \end{align*} where $A_F$ is the adele space, $A_F(0)$ the set of adeles $\alpha$ where $v_P(\alpha_P)\geq 0$ for each place $P$, and $F$ in the equation is the embedding of $F$ into $A_F$ as principal adeles. The $\supseteq$ inclusion is trivial, but I am stuck trying to prove the converse.

Answer 1

Found an answer myself. As $\ell(0)=1$ we then have for any positive divisor $B\geq 0$ that $\ell(B)=\deg(B)+1$. Now consider $\alpha\in A_F$ and $B_1\geq 0$ such that $\alpha\in A_F(B_1)$. Then \begin{align} \dim_K\left( A_F(B_1)+F)/(A_F(0)+F) \right) &= (\deg(B_1)-\ell(B_1))- (\deg(0)-\ell(0)) \\ &= \deg(B_1) -\deg(B_1)-1 +1 \\ &= 0 \end{align} so $A_F(B_1)+F=A_F(0)+F$, and as $\alpha\in A_F(B_1)$ implies $\alpha\in A_F(B_1)+F$ then $\alpha\in A_F(0)+F$.