Let $I$ be an ideal of $\mathbb{R}[x_1,\cdots,x_n]$. The real radical $^{\mathbb{R}}\sqrt{I}$ of $I$ is defined as the set of all polynomials of the form $p^{2m}+\sum_{j \in J} q_j^2$, with $p \in I$, $m$ positive integer, $q_j \in \mathbb{R}[x_1,\cdots,x_n]$, $J$ finite set. $I$ is called real radical if it is equal to its real radical.
Question: Suppose that $V_{\mathbb{R}}$ is finite and that $I$ is real radical. How can we show that $V_{\mathbb{C}}(I)=V_{\mathbb{R}}(I)$?