Over an algebraically closed field, affine geometry sees exactly radicals: $I(V(J))=\sqrt J$.Suggest a slogan
(Hilbert's Nullstellensatz)
Let \(k\) be algebraically closed and let \(J \subseteq k[x_1, \ldots, x_n]\) be an ideal. Then \[ I(V(J)) = \sqrt{J} . \] [@Har10a, Theorem I.1.3A]
The weak form is the case \(J \neq k[x_1,\ldots,x_n] \implies V(J) \neq \emptyset\), and it is equivalent: the strong form follows from the weak one by the Rabinowitsch trick, adjoining a variable \(y\) and the polynomial \(1 - yf\).
Algebraic closure is not a convenience here. Over \(\RR\) the ideal \((x^2+1)\) is prime, hence radical, but \(V(x^2+1) = \emptyset\) in \(\AA^1_\RR\), so \(I(V(J)) = (1) \neq \sqrt{J}\).
State the Nullstellensatz.
What is $I(V(J))$?
Where does the Nullstellensatz use that $k$ is algebraically closed?
How is the Nullstellensatz proved?
If $L/k$ is a field extension and $L$ is a finitely generated $k$-algebra, show that $L/k$ is algebraic.
State the weak Nullstellensatz in terms of maximal ideals and in terms of systems of polynomial equations, and deduce the strong form by the Rabinowitsch trick.
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