# Is contraction of radical ideals and/or prime ideals surjective in this case?

Let $$K$$ denote a field and $$\overline{K}$$ denote it's algebraic closure. Then we have an inclusion of polynomial rings $$K[x_1,\ldots,x_n] \subseteq \overline{K}[x_1,\ldots,x_n].$$ Since the preimage of a prime (resp. radical) ideal is prime (resp. radical), the contraction of an ideal $$I \subseteq \overline{K}[x_1,\ldots,x_n]$$ to an ideal $$I^c \subseteq K[x_1,\ldots,x_n]$$ defined by $$I^c := I \cap K[x_1,\ldots,x_n]$$ preserves both radicalness and primeness.

This proceed needn't be injective. For example, let $$K = \mathbb{R}$$ and $$\overline{K} = \mathbb{C}$$. Then $$(x+i)^c = (x^2+1)$$ and $$(x-i)^c = (x^2+1)$$.

However, I have a hunch it might be surjective on prime ideals and/or radical ideals. If so, it should follow that contraction commutes with radicals $$(\sqrt{I})^c = \sqrt{I^c}.$$

Question. Is any or all of this true?

Yes, this is true. For prime ideals, this is a special case of the "lying over theorem" for integral extensions. If $$B$$ is any commutative ring with a subring $$A$$ such that $$B$$ is integral over $$A$$, then every prime ideal of $$A$$ is the contraction of some prime ideal of $$B$$. In this case, $$B=\overline{K}[x_1,\ldots,x_n]$$ is integral over $$A=K[x_1,\ldots,x_n]$$ since it is generated by $$\overline{K}$$ as an $$A$$-algebra and every element of $$\overline{K}$$ is integral over $$K$$.
As a sketch of a proof, let $$P\subset A$$ be a prime ideal. Localizing at $$P$$, we may assume $$P$$ is the unique maximal ideal of $$A$$. It then suffices to show that $$PB$$ is a proper ideal of $$B$$, so it is contained in some maximal ideal of $$B$$ (which can then only lie over $$P$$ since $$P$$ is maximal). If $$PB=B$$, then this equation remains true when we replace $$B$$ with some finitely generated $$A$$-subalgebra $$B_0\subseteq B$$. But then $$B_0$$ is a finitely generated $$A$$-module since $$B$$ is integral over $$A$$, so $$PB_0=B_0$$ implies $$B_0=0$$ by Nakayama. This is a contradiction, and hence $$PB$$ must be a proper ideal of $$B$$, as desired.
Note that the statement $$(\sqrt{I})^c = \sqrt{I^c}$$ is comparatively trivial and holds for any extension of commutative rings $$A\subseteq B$$. Indeed, if $$f\in (\sqrt{I})^c$$, then $$f\in \sqrt{I}$$ and so $$f^d\in I$$ for some $$d$$. But then $$f^d\in A$$ as well and so $$f^d\in I^c$$ and $$f\in \sqrt{I^c}$$. Conversely, if $$f\in\sqrt{I^c}$$, then $$f^d\in I^c$$ for some $$d$$ and so $$f^d\in I$$ and hence $$f\in (\sqrt{I})^c$$.