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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 $L$:

  1. $k=\mathbb{C}$ the complex number field.

  2. $L$ is isomorphic to a rational function field. Namely, there is an embedding $i:k(y_1,\cdots, y_m)\to K$ whose image is $L$.

  3. 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?

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  • $\begingroup$ If $m=n$ clearly the only option is $L=K$. If $m<n$, if $f_k$ is just linear, you are alright. So, what kind of condition are you looking for? $\endgroup$
    – Mohan
    Sep 23, 2015 at 19:38
  • $\begingroup$ Hi Mohan, $f_k$ is not linear but multilinear. For example, $f(x_1,x_2)$ could be $x_1x_2$. $\endgroup$
    – kennyyeke
    Sep 24, 2015 at 17:47
  • $\begingroup$ I did not mean what your functions look like, but want to know what kind of answers do you expect? $\endgroup$
    – Mohan
    Sep 24, 2015 at 19:54
  • $\begingroup$ Hi Mohan, I have changed my question. Thanks! $\endgroup$
    – kennyyeke
    Sep 24, 2015 at 21:35

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