Quadratic extensions of function fields

Question

Let $F = k(t_1,\dots,t_r)$ be the function field in $r$ variables of a field $k$, and let $F'$ be a quadratic extension of $F$.

Does there exist a quadratic extension $k'$ of $k$ such that $F' = k'(t_1,\dots,t_r)$?

I have tried the following: we can write an element of $F'$ as $\alpha + u \beta$ with $\alpha,\beta \in F$ and $u\in F'\setminus F$ such that $f(u) = 0$ with $f$ polynomial of degree two with coefficients in $F$.

Now, since $\alpha,\beta$ are rational function in $t_1,\dots,t_r$ we can write $\alpha + u \beta$ as a rational function in $t_1,\dots,t_r$ whose coefficients are of the form $\alpha_i + u\beta_i$ with $\alpha_i,\beta_i\in k$. Furthermore, evaluating $f$ in some fixed $t_1,\dots,t_r$ we se that $u$ is a root of a degree two polynomial with coefficients in $k$. So $\alpha + u \beta \in k'(t_1,\dots,t_r)$ with $k' = k(u)$.

Is this correct? Thank you.

Answer 1

It's not true in general. The field $k$ can be algebraically closed in $F'$, and so there is no such an extension. For example: for $char(k)\neq 2$, take $F=k(x)$ and $F':=F(y)=k(x,y)$ with $y^2=p(x)$, where $p(x)$ is an irreducible monic polynomial over $k$. We have $|F': F|=2$ and $F'=F(y)$, but $k$ is algebraically closed in $F'$.

Furthermore, evaluating $f$ in some fixed $t_1,…,t_r$ we se that $u$ is a root of a degree two polynomial with coefficients in $k$

The element $u \in F'$ is also a function that depends on $t_1,...,t_r$, so you can't just calculate the function $f$ on $t_1,...,t_r$ in $f(u)=0$.