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If $F$ algebraically closed and $L/F$ is an extension, is there $x_1,...,x_n$ s.t. $L=F(x_1,...,x_n)$ ? To me $L=F$. Indeed, $L$ contain $F$ as a subfield. Therefore, if $L=F(x_1,...,x_n)$ since $F$ is algebraically closed, $F(x_1,...,x_n)=F$, and thus $L=F$.

Edit This is the problem : Let $F$ algebraically closed. I want to show that every maximal ideal of $F[x_1,...,x_n]$ is of the form $(x_1-c_1,...,x_n-c_n)$. The proof goes like : Let $m\subset F[x_1,...,x_n]=:R$ maximal. Let $\pi: R\to R/m$. We set $k=R/m$ which is a field. Why $$k=F[\bar x_1,...,\bar x_n]\ \ ?$$

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    $\pi$ is surjective as a ring morphism. Then, $k\cong F[x_1,..., x_n]/\ker \pi$ as ring. But, $\pi|_F$ is a ring morphism, then $k$ is the extension of $F$ by $\bar x_1,...,\bar x_n$, i.e. $k=F[\bar x_1,...,\bar x_n]$. – 2017-01-22

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$F$ being algebraically closed only means that $L=F$ if $L$ is algebraic. $F$ also has proper transcendental extensions ; for example if $\overline{\Bbb Q}$ is the algebraic closure of $\Bbb Q$, then $\overline{\Bbb Q}(\pi)$ is an extension of $\overline{\Bbb Q}$. But such extensions need not be of the form $\overline{\Bbb Q}(x_1,\dots,x_n)$ ; for example $\Bbb C$ cannot be generated by a finite number of elements since it is uncountable.

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Note that you can have trancendental extensions. For instance, $\Bbb C(x)$ is the field of rational functions in one variable with complex coefficients. You can even do the same thing with infinitely many variables, $\Bbb C(x_1, x_2, \ldots)$, which contains elements such as $$ \frac1{x_1} \quad\text{and}\quad \frac{5x_{1592} - (3+\sqrt2i)x_{3}}{x_{1\,507\,392\,588}^2} $$ But you're right. If $F$ is algebraically closed, and the $x_i$ are algebraic over $F$, then $F(x_1, \ldots,x_n) = F$.