We know that $\mathcal{I(V(}I)) = \sqrt{I}$ when the underlying field is algebraically closed. So, algebraic sets and radical idels correspond.
The question that I wonder about is how much relation of this holds for polynomials over the field $\mathbb{R}$?
Certainly we need $\mathcal{V}(I) \neq \emptyset$. But certainly this is not sufficient. Take $x^2+y^2$(Easy to check).
So, First lets look at one variable polynomials. Here $\mathcal{I(V(}I)) = \sqrt{I}$ only holds for ideals generated by a polynomial with only real roots. This is a complete and clean description.
So, my question is: Do we have such descriptions, or at least sufficient conditions for $\mathcal{I(V(}I)) = \sqrt{I}$ in more variables case?. Is there something in terms of like if $\mathcal{V}(I)$ is sufficiently big? I say this because I considered the following example in one of my previous questions:
Are there any non trivial polynomial identities in $\cos(\theta)$ and $\sin(\theta)$? and so $\mathcal{I(V}(x^2+y^2-1)) = (x^2+y^2-1)$ over $\mathbb{R}$
Is this a conincidence? Can we derive any sufficient conditions at least in two variable case?
I believe that Theorem 5.1 in https://arxiv.org/abs/1606.03127 will give you a sufficient condition. If $I$ is prime and $\mathcal V(I)$ contains a smooth real point, then the Theorem tells you that $\overline{\mathcal V_{\mathbb R}(I)}=\mathcal V(I)$. By definition of the Zariski closure, this is saying that $$ \mathcal V(\mathcal I(\mathcal V_{\mathbb R}(I)))=\mathcal V(I)$$ We take $\mathcal I$ on both sides and we use the Nullstellensatz. Recall that $\mathcal I$ applied to anything results in a radical ideal: $$ \mathcal I(\mathcal V_{\mathbb R}(I))=\sqrt{\mathcal I(\mathcal V_{\mathbb R}(I))}=\mathcal I(\mathcal V(\mathcal I(\mathcal V_{\mathbb R}(I))))=\mathcal I(\mathcal V(I))=\sqrt I$$ This explains why you see nice behavior in $(x^2+y^2-1)$, which cuts out a smoooth variety, as opposed to $(x^2+y^2)$, whose real point is singular.