So I want to prove that if I have $L/K/F$ then given that the transcendence degrees are finite, then $tr_F(L)=tr_K(L)+tr_F(K)$
$\bf{Definition:}$ We say that a set $X=\{x_i\}_{i\in I}$ is algebraically independent over $F$ if $f\in F[\{t_i\}_{i\in I}]$ such that $f((x_i)_{i\in I})=0$ implies that $f=0$.
Now I basically started like this: Let $B_1=\{x_1,...,x_m\}$ be a transcendence basis for $L$ over $K$, and $B_2=\{y_1,...,y_n\}$ a transcendence basis for $K$ over $F$. Let $B_3=\{x_1,...,y_n\}$.
We shall prove that $B_3$ is a transcendental basis for $L$ over $F$. Say $f(t_1,...,t_{m+n})$ is a polynomial with coefficients in $F$, such that $f(x_1,...,y_n)=0$. Then define:$h(t_1,...,t_m)=f(t_1,...,t_m,y_1,...,y_n)$Then $h$ is a polynomial with coefficients in $K$ since $B_2\subset K$, and $h(x_1,...,x_m)=0$. This means that $h=0$ by the algebraic independence of $B_1$. Then, we can let $g(t_{m+1},...,t_{m+n})=f(x_1,....,x_m,t_{m+1},...,t_{m+n})$However, I cannot use the fact that $g$ will be zero because $g$ might have coefficients in $F$. Any hints?
Thanks