# An extension field induces new variety

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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