If one have an irreducible affine algebraic varieties $X$, one can define the function field of $X$ as the field of fraction of its coordinate ring $K[X]$. This definition it makes sense by the irreducibility of $X$, that permit us to say $K[X]$ is a domain.
Conversely:
-we consider the finite extension field $L/K$, where $L=K(a_1,\cdots , a_n)$, $a_i\in L$. Then it is possible to define a new variety on the field $K$ in the following way:
we coinsider the natural evaluation map $\psi: K[x_1,\cdots, x_n]\to K$. In this way $\ker(\psi)=\{f\in K[x_1,\cdots, x_n]: f(a_1,\cdots, a_n)=0\}$ result to be a prime ideal.
The (irreducible) variety associated to the finite extension $L$ is $Z(\ker(\psi))\subseteq K^n$. Is it correct?
Is it possible the association is a bijection map?
I have also another problem, with the following example:
We consider the finite extension field of $\mathbb{C}(s, t)$, $\mathbb{C}(s, t)(s^{\frac{1}{2}},t^{\frac{1}{2}})$. Is the following variety the associated variety of the extension?
$Z(x^2-s,y^2-t)\subseteq \mathbb{C}(s, t)^2$
I have in mind there is always a natural map between this variety and the variety associated to $C(s, t)$, that is $\mathbb{C}^2$. How is it possible to define this map $\pi: Z(x^2-s,y^2-t)\to \mathbb{C}^2$?