Ideal of an affine variety generated by a single polynomial

Question

This is perhaps a trivial question, but I would like to still verify. Is it not true that given a variety $X=V(f)$ where $f$ is a polynomial, we always have $I(X)=\sqrt{(f)}$?

Answer 1

No this is not true in general. It depends on the field you are working over. Consider the ideal $(x^2+1) \subset \mathbb{R}[x]$, then $I(V(x^2 +1)) = I(\emptyset) = \mathbb{R}[x]$. But if the field is algebraically closed, then your claim is indeed true. It's called Hilbert's Nullstellensatz.

Answer 2

Yes, if $k$ is algebraically closed. That also holds more generally when $X = V(\mathfrak{a})$ is given by some arbitrary ideal $\mathfrak{a} \subset k[T_1,...T_n]$. Some people call that Hilbert's Nullstellensatz (for me that is only an equivalent formulation and not the Nullstellensatz itself). There I was assuming the ground field to be algebraically closed though. The latter is necessary - Just try it out over the reals for example to find some counterexamples.