if $F\subseteq K$ real closed fields, $b\in K$ algebraic over F then $b\in F$

Question

I'm studying real closed field part in model theory. Because I'm not familiar with abstract algebra, I'm stuck with two (possibly) simple problems. Here are definitions and two problems I'm stuck with.

below $F,K$ are fields.

$\cdot$F is (formally) real if -1 is not sum of squares in F

$\cdot$F is real closed if F is real and has no proper real algebraic extension

$\cdot$K is real closure of F if K: real closed and K: algebraic extension of F

1. if $F\subseteq K$ real closed fields & $b\in K$ algebraic over $F$, then $b\in F$

2. $F\subseteq K$ $F$:real, $K$:real closed field $\Rightarrow$ $F^{alg}\cap K$ is a real closure of F ($F^{alg}$ is algebraic closure of F)

   I know definitions of each concept but I don't know how to prove these. For the first one, I can't proceed further from $K\models f(b)$ where $b\in K$ and $f(x)\in F[X]$. Since $K$ is real, $K$ is not proper algebraic extension of $F$. I think I need to use real closedness of $K$ but I don't know how to use this.

   For the second problem, I only need to prove $F^{alg}\cap K$ has no proper real algebraic extension which is hard to prove to me.

Answer 1

Suppose that $b$ is algebraic over $F$, $F(b)$ is an algebraic extension of $F$, suppose that $-1=\sum y_i^2, y_i\in F(b)\subset K$, it is impossible since $K$ is real. We deduce that $F(b)$ is a real extension of $F$, since $F$ is closed, we deduce that $F(b)=b$ and $b\in F$.