Consider the function field $F/\mathbb{R}$, where $F:=\mathbb{R}(x,y)$ with $x^2+y^2+1=0$. Show that $F/\mathbb{R}$ has genus $0$ but is not a rational function field. Furthermore, all places of $F/\mathbb{R}$ have degree $2$.
I've already proven that $F/\mathbb{R}$ is not a rational function field and has genus $0$. Now I'm trying to prove the degree $2$ part.
It's obvious that $\deg(P)\geq 2$ for all $P$ because if there were $P$ with degree $1$, the fact that $g=0$ would imply $F/\mathbb{R}$ is rational, which I've already proven is not true.
Now since $[F:K(x)]=[F:K(y)]=2$ and $x^2+y^2+1=0$, we have $(x)_\infty=(y)_\infty=P_0$ for some $P_0$ with degree $2$. So for $Q\neq P_0$, we must prove $\deg(Q)= 2$.
Since $g=0$, there is $z\in F$ with $(z)_\infty=Q$. We may write $z=\frac{a+by}{c}$, where $a,b,c\in\mathbb{R}[x]$. Since $Q$ is the only pole of $z$, we must have that $\deg(a)\leq \deg(c)$ and $\deg(b)\leq \deg(c)-1$ and that $Q$ is the only zero of $c$. In particular, $(c)=Q-\deg(c)P_0$, so $\deg(Q)=2\deg(c)$, i.e, $\deg(Q)$ is even.
I still wasn't able to prove that $\deg(c)=1$. Any suggestions?