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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?

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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.


Once we have the result for prime ideals, it immediately follows for radical ideals, since radical ideals are just intersections of prime ideals.

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 $f\in (\sqrt{I})^c$.

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  • $\begingroup$ You've really gotta get back into math, you know? Your grasp of this stuff is phenomenal. $\endgroup$ – goblin Oct 12 '18 at 12:45

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