Is it true that if there exists a rational place then $K$ is algebraically closed in $F$? I am working through "Algebraic function fields and codes-Henning Stichtenoth" (2nd Edition) and I got a little confused about the following:
First of all an algebraic function field $F/K$ of one variable over $K$ is an extension field $F$ of $K$ such that $F$ is a finite algebraic extension of $K(x)$ for some element $x \in F$ which is transcendental over $K$.
Remark 1.11.17 says that if $K$ is an algebraic closed field, then all places are rational ( A Place of $F/K$ is defined as the unique maximal ideal of a valuation ring $\mathcal{O}$ of $F/K$). So far so good, but isn't the following also true:
If there exists at least one rational place, then $K$ is algebraically closed in $F$(!), therefore all places are rational.
Because: The field of constants $\tilde K :=\{z \in F:z \text{ is algebraic over }K\}$ of $ F/K$ is embedded in $F_p:=\mathcal{O}_p/P$ (The residue class field of a Place $P$). So it follows that $[\tilde K:K]\leq[F_p:K]$. This holds for any Place $P$ of $F/K$. (part of proof of Corollary 1.1.16)
Therefore if there exists a Place $P$, which is rational (defined by deg$P=[F_p:K]=1$). Then $[\tilde K:K]\leq[F_p:K]=1$, therefore $K$ is already algebraically closed in $F$. 
Then for any other Place $\hat P$, since deg$\hat P=[F_{\hat p}:K]$ has to be finite (Proposition 1.1.15), $F_{\hat p}:K$ is an algebraic extensions of $K$, which uses only elements of $F$, so it can't get any bigger than $K$. Therefore $\hat P$ is rational.
This strikes me odd, since there seems to be only two possibilties:
1) Either no Place $P$ of $F/K$ is rational,
2) or every Place $P$ of $F/K$ is rational(which is the case if $K$ is algebraic closed or algebraically closed in $F$). 
Am I missing something? If yes could you please give me a hint where I went wrong.
 A: I think I found my mistake. It is in the statement "$F_p$ is an algebraic extensions of $K$, which uses only elements of $F$, so it can't get any bigger than $K$." Because $K$ is only embedded in $\mathcal{O}_P/P=F_p$, it is not an actual subset. So the statement actually makes no real sense. Therefore I am right about the question I asked, but not on my further statement, that if there is one rational place, there can only be rational places.
For example, for the rational function field $K(x)$ a valuation ring is given by:
\begin{align}
\mathcal{O}_{p(x)}= \Big\{ \frac{f(x)}{g(x)}| f(x),g(x)\in K[x], p(x) \text{ does not divide }g(x) \Big\}
\end{align}
for an irreducible monic polynomial $p(x)\in K[x]$, the place associated to that valuation ring is:
\begin{align}
P_{p(x)}= \Big\{ \frac{f(x)}{g(x)}| f(x),g(x)\in K[x],p(x) \text{ does divide }f(x), p(x) \text{ does not divide }g(x) \Big\}
\end{align}
The degree of a Place $P$ is defined by deg$P=[\mathcal{O}_{p(x)}/P:K]$, and $\mathcal{O}_{p(x)}/P$ is isomorphic to $K[x]/(p(x))$, with an isomorphism given by: $f(x)+(p(x)) \mapsto f(x)+P$. Therefore if $K$ is algebraically closed, every irreducible polynomial has degree one, therefore every place has degree one. 
If $K=\mathbb{R}$ on the other hand take $p(x)=x^2+1$ and you get a place with degree $2$. Note that $\mathbb{R}$ is algebraically closed in $\mathbb{R}(x)$, this means $\tilde K=\mathbb{R}$.
If you look at $\mathcal{O}_{x^2+1}/P$, you see it contains more algebraic Elements over $\mathbb{R}$ than $\mathbb{R}$ itself does ($x^2+P$), but this is not a problem because $\tilde K$ is defined as $\{z \in F: z \text{ is algebraic over } K\}$, so in this case:
\begin{align}
\tilde K=\{z \in \mathbb{R}(x): z \text{ is algebraic over } \mathbb{R}\}=\mathbb{R}
\end{align}
and $x^2+P \in \mathcal{O}_{x^2+1}/P$  but on the other hand $x^2+P \not \in \mathbb{R}(x)$.
To sum things up; the conclusion I drew is partly wrong, if there is a rational place there still can be other places with a higher degree than $1$, although $K$ has to be algebraically closed in $F$.
