Let $F/K$ be a field extension of transcendence degree $r$ and let $\{x_1,\ldots,x_n\}$ be a set of elements which are algebraically independent over $K$. Then is it true that $tr.deg. F(x_1,\ldots,x_n)/K(x_1,\ldots,x_n)=r$?
If $\{t_1,\ldots,t_r\}$ is a transcendence base of the extension $F/K$ then we know $F/K(t_1,\ldots,t_r)$ is algebraic and $K(t_1,\ldots,t_r)/K$ is purely transcendental. Thus $F(x_1,\ldots,x_n)/K(x_1,\ldots,x_n)(t_1,\ldots,t_r)$ is also algebraic. Now it will be enough to show that $\{t_1,\ldots,t_r\}$ is algebraically independent over $K(x_1,\ldots,x_n)$. How should I approach?
My above queries are from the book of Lang's Algebra book in the context of being free from one field to another. I have added a photo from Lang's Algebra book: I want to understand why $tr. deg. F(y)/K(y)=r$ as shown in the diagram.
If I understood the question correctly, it is about why the extension $F(y)/k(y)$ described in the excerpt has transcendence degree $r$.
This follows from the assumptions made in that claim. Let $(t):=\{t_1,t_2,\ldots,t_r\}$ be a transcendence basis of $F/k$. Then $(t)$ is a subset of $K$ that is algebraically independent over the base field $k$. Because $K$ was assumed to be free from $L$ over $k$, it follows that $(t)$ remains algebraically independent over $L$. Therefore $(t)$ is also algebraically independent over the subfield $k(y)\subseteq L$. Therefore the transcendence degree of $F(y)/k(y)$ is at least $r$. Because $F(y)$ is algebraic over $k(y)(t)$ the transcendence degree cannot be higher. Hence the transcendence degree of $F(y)/k(y)$ is exactly $r$ as the diagram claims.
The assumption of freeness of $K$ from $L$ is absolutely essential in this argument, and the claim may be false otherwise.
In the prescribed context $F$ is a subfield of $K$, so the title question is a bit strange.