# Is an ideal prime when its complex extension is prime?

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.

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 $\mathbb{C}[x_1,\ldots,x_n]$.
• But why is the preimage of $\widetilde I$ under the inclusion equal to $I$? Dec 5, 2014 at 12:03
• Take an arbitrary $f=g_1f_1+\cdots+g_kf_k\in \tilde{I}$ for some $g_i\in\mathbb{C}[x_1,\ldots,x_n]$. If $f$ is a real polynomial, then we have $$f=\overline{f}=\overline{g_1f_1+ \cdots+g_kf_k}=\overline{g_1}f_1+\cdots+ \overline{g_k}f_k$$ so that $$f=\frac{f+\overline{f}}{2}= \Bigl(\frac{g_1+\overline{g_1}}{2}\Bigr)f_1+\cdots+\Bigl(\frac{g_k+ \overline{g_k}}{2}\Bigr)f_k$$ exhibits $f$ as an element of $I$.
• For a general ring morphism $\phi:R\to S$ a sufficient condition for having $\phi^{-1} (\phi (I).B)=I$ for all ideals $I\subset R$ is that $S$ be faithfully flat over $R$ , which is the case here. Of course in the case at hand Sal's very concrete argument in his comment is much, much nicer than an invocation of this abstract result . Dec 5, 2014 at 13:29
• @GeorgesElencwajg Actually the general result which I find appropriate here is the following: if $R\subset S$ is such that $R$ is a direct summand in $S$ (viewed as $R$-module), then $IS\cap R=I$ for any ideal $I$ of $R$. Dec 5, 2014 at 14:46