Is an ideal prime when its complex extension is prime?

Question

Let $I = \langle f_1,\dots,f_k\rangle$ be an ideal in $\mathbb R[x_1,\dots,x_n]$. The same $f_i$ generate an ideal $\widetilde I$ in $\mathbb C[x_1,\dots,x_n]$. When $\widetilde I$ is prime in $\mathbb C[x_1,\dots,x_n]$, do we have $I$ prime in $\mathbb R[x_1,\dots,x_n]$?

For $k=1$ this is just the statement that a real polynomial that's irreducible over $\mathbb C$ is also irreducible over $\mathbb R$. For $k>1$ I'm not sure if this is true.

Answer 1

Yes. In fact for any ring homomorphism $\phi:R\to S$, and any prime $P\subset S$, the preimage $\phi^{-1}(P)$ is also prime. In your situation, $\phi$ is the inclusion of $\mathbb{R}[x_1,\ldots,x_n]$ into $\mathbb{C}[x_1,\ldots,x_n]$.