Variety of an ideal

Question

Let $F$ be a field, and $\underline{x}$ be the $n$-tuple $(x_1, x_2, ... , x_n) \in F^n$. Also, denote $F[x_1, x_2, ... , x_n]$ by $F[\underline{x}]$.

Let $J$ be an ideal of $F[\underline{x}]$. Define the variety of $J$, $V(J)= \lbrace \underline{a} \in F | f(\underline{a})=0, \forall f(\underline{x}) \in J \rbrace$.

Next, define $I(V(J))= \lbrace f(\underline{x}) \in F[\underline{x}] | f(\underline{a})=0, \forall \underline{a} \in V(J) \rbrace$. It's easy to show that $J \subseteq I(V(J))$, but the reverse may not be true. Give an explicit example to illustrate this.

I'm struggling to find an example, mostly because listing all of the elements in $V(J)$ seems challenging in general.

Answer 1

Take $p(x)=x^2+1$ in $\mathbb{R}[x]$ and let $J$ be the ideal generated by $p$. Then $V(J)=\emptyset$ since $x^2+1$ has no roots in $\mathbb{R}$. Thus $I(V(J))$ is all of $\mathbb{R}[x]$.

Answer 2

The key observation here is that the variety depends only on something being a root or not. The multiplicity of the root does not come into play.

Thus one needs to start with an ideal that would dictate even a multiple root, such as $(x^2)$. This information is then lost in the proecss.