Nullstellensatz non-valid for non-algebraically closed fields

Question

I want an example (with details, please) showing that Nullstellensatz may be false over non-algebraically closed fields. Thanks in advance!

Answer 1

Consider the ideal $J = (x^2 + 1)$ in $\mathbb{R}[x]$. We have $V(J) = \emptyset$ and $I(V(J)) = I(\emptyset) = \mathbb{R}[x]$ but $\sqrt{J} = J \neq \mathbb{R}[x]$.

Answer 2

Not exactly the answer but related. One can get Nullstellensatz over any field $F$ provided you take the affine space over the closure $\overline{F}$. In the above example if we took the variety of $\mathbb{C}$, then $V(J)=\lbrace i,-i\rbrace$ and then $I(V(J))=\lbrace f\in \mathbb{R}[x] \mid f(\pm i)=0\rbrace$. But $f(i)=0\Rightarrow f(x)=(x-i)g(x)$ and similarly $f(x)=(x+i)h(x)$. Thus, $f(x)=(x+i)(x-i)k(x) \Rightarrow f(x)=(x^2+1)k(x)$ and since $f$ and $x^2+1$ are in $\mathbb{R}[x]$, it is forced that $k(x)\in \mathbb{R}[x]$. We get $I(V(J))=J$.

In general, $\lbrace F-\text{closed subsets of }\mathbb{A}^n({\overline{F}})\rbrace \leftrightarrow \lbrace \text{radical ideals of } F[x_1,\ldots,x_n]\rbrace$.