# Adding transcendental elements to an algebraic extensions

Suppose $\mathbb{K}\subseteq\mathbb{L}$ algebraic and $\mathbb{L}\subseteq\mathbb{F}$, and take $S\subseteq\mathbb{F}$ a set of transcendental elements over $\mathbb{L}$: then is $\mathbb{K}(S)\subseteq\mathbb{L}(S)$ algebraic?

I'm not sure if this line of reasoning is correct: the following holds $$\mathbb{L}(S)=\mathbb{K}(S)(\mathbb{L})\label{*}\tag{*}$$ Because the smallest subfield of $\mathbb{F}$ containing $\mathbb{K}$, $\mathbb{L}$ and $S$ is exactly the one containing $\mathbb{L}$ and $S$.

The field $\mathbb{L}(S)$ is being generated over $\mathbb{K}(S)$ by elements which are algebraic over $\mathbb{K}$ and hence over $\mathbb{K}(S)$ (for it contains $\mathbb{K}$), so the extension is in fact algebraic. Is this sufficient, or are there further considerations to be made?

• I find that sufficient. If you want, you can say any element of $L(S)$ is a a rational function in the elements of $S$ with coefficients from $L$; if we simply adjoin the coefficients to $K(S)$ (which will be an algebraic extension) we get an extension field containing that arbitrary element of $L(S)$. – arctic tern Dec 1 '16 at 15:47

You need to show that an element of $L(S)$ is algebraic over $K(S)$.
Since quotients of algebraic elements are algebraic, it is not restrictive to assume the element you pick in $L(S)$ actually belongs to $L[S]$. But then you just need to show that a monomial in $L[S]$ is algebraic over $K(S)$, because sums of algebraic elements are algebraic. Now use the fact that products of algebraic elements are algebraic.