Terminology: Let $I\subset\mathbb{R}[X_1,...,X_n]=:A$ be an ideal. We call $\sqrt[\mathbb{R}]{I}:=\{f\in A\ |\ \exists k\in\mathbb{N},\ g_1,...,g_m\in A:\ f^{2k}+\sum_{i=1}^mg_i^2\in I\}$ the real radical of $I$. We call $I$ real if $\sqrt[\mathbb{R}]{I}=I$.
It is clear that every real ideal is a radical ideal (i.e. $f^k\in I$ implies $f\in I$) and that every prime ideal is a radical ideal. Also, the principal ideal generated by an irreducible polynomial need not be real.
This naturally leads to the question whether every proper real ideal is prime. I'm pretty sure this is not the case but I haven't been able to come up with a counterexample. Does anyone know of one? (Preferably one which also works in the case $n\leq 3$.)
What can be said about $A/I$ for a real ideal $I$? (I.e. is there some algebraic property $P$ such that $I$ is real if and only if $A/I$ satisfies $P$?) I know that if $I$ is real and prime, then the field of fractions of $A/I$ is a real field.