Let's start with the following general question. Let $k$ be the ground field. Let $K=k(x_1,\cdots, x_n)$ be a rational function field and let $L$ be a subfield of $K$. Is there a condition to guarantee that $L$ is algebraically closed in $K$? i.e., if $\alpha\in K$ is algebraic over $L$, then $\alpha\in L$.
To simplify the question, let me put some restrictions on $k$ and $L$:
$k=\mathbb{C}$ the complex number field.
$L$ is isomorphic to a rational function field. Namely, there is an embedding $i:k(y_1,\cdots, y_m)\to K$ whose image is $L$.
for $1\le k \le m$, $i(y_k)$ is a multilinear polynomial $f_k$ in $x_1,\cdots, x_n$, i.e., $f_k$ is a polynomial in $x_1,\cdots, x_n$ and $f_k$ is linear in each variable. (This condition is to avoid the following situation: $i:k(y)\to k(x)$ where $i(y)=x^2$)
My real question: For such $k,L$ and $K$, is it true that $L$ is algebraically closed in $K$? If it is not true, what conditions (non-trivial conditions) do we need to make it true?