If $k$ is an algebraically closed field, Hilbert's Nullstellensatz says that functions I,V form a bijection between the radical ideals of $k[X_1,\dots,X_n]$ and the algebraic sets of $k^n$. Function $V:\mathcal{P}(k[X_1,\dots,X_n])\to \mathcal{P}(k^n)$ gives the set of common zeros, and function $I:\mathcal{P}(k^n)\to\mathcal{P}(k[X_1,\dots,X_n])$ gives the radical ideal of polynomials that have the input zeros.
When the field $k$ is no longer algebraically closed, we still have an injection from algebraic sets of $k^n$ into radical ideals $J$ of $k[X_1,\dots,X_n]$ such that if $J$ contains a polynomial without zeros, then $J=k[X_1,\dots,X_n]$. Are those ideals the image of the closure operator $I\circ V$? If not, can we further characterize this image?
Best Answer
No, "radical ideals $J$ such that if there exists $f\in J$ with $V(f)=\emptyset$, then $1\in J$" is not the image of $I(V(-))$. For instance, if $k=\Bbb R$, then $(x^2+y^2)\subset \Bbb R[x,y]$ is a radical ideal which satisfies those criteria, but is not the image of $I(V(-))$. (Why? $I(V(-))$ is an idempotent operator, but $I(V(x^2+y^2))=(x,y)$.)
I'm not sure about general results concerning the image of this operator. There are some circumstances where we do know more, though. For instance, when $k=\Bbb R$ (or equivalently, any real-closed field) the ideals that are the image of the $I(V(-))$ operator are exactly those that are real radical, since the operation is designed to return the largest ideal with the same vanishing locus as the given ideal. (In general, I think this is a tough problem which needs to take in to account the "shape" of all the polynomials over $k$ which don't have roots - the case of a real-closed field is in some sense the simplest one, as the only fields where the extension $k\subset \overline{k}$ is finite are the real-closed fields, and the algebraic closure is obtained by adjoining the square root of $-1$.)
Further, while this is an interesting problem to think about, the main thrust of algebraic geometry has figured out a way around it which suffices for developing theory. Instead of thinking about a variety as a zero locus in $k^n$ (or related concepts), upgrading to schemes lets you keep the nullstellensatz with no alterations to the conclusions no matter what your base field looks like. If you're interested in pursuing algebraic geometry over fields that are not algebraically closed, this is material you'll want to cover at some point.