Mathematics Asked on January 3, 2022

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 $L=K(a_1,cdots , a_n)$, $a_iin L$. 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 $ker(psi)={fin K[x_1,cdots, x_n]: f(a_1,cdots, a_n)=0}$

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 $pi: Z(x^2-s,y^2-t)to mathbb{C}^2$?

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