First, I would like to offer a more rigorous statement of Hilbert's Nullstellensatz from Dummit and Foote's *Abstract Algebra*:

Let $E$ be an algebraically closed field. Then $\mathcal{I}(\mathcal{Z}(I)) = \mathop{\mathrm{rad}} I$ for every ideal $I$ of $E[x_1, \ldots, x_n].$ Moreover the maps $\mathcal{Z}$ and $\mathcal{I}$ in the correspondence $$\{ \mbox{affine algebraic sets} \} \xleftarrow[\mathcal{Z}]{\xrightarrow{\mathcal{I}}} \{\mbox{radical ideals} \}$$ are bijections of each other.

Now, it should be absolutely obvious why the bijection breaks down when the field considered is $\mathbb{R}$. $\mathbb{R}$ is not algebraically closed (consider $x^2+1 \in \mathbb{R}[x]$, which has a variety that includes $i\not\in \mathbb{R}$) so the Nullstellensatz does not apply and the bijection does not happen.

As for whether or not this is a failure to be injective or surjective, that depends on whether you are talking about $\mathcal{Z}$ or $\mathcal{I}.$

notthe same. Their sets of $\mathbb{R}$-valued points are. – Robert Kucharczyk Dec 17 '11 at 21:34