$K$ is a finite Galois extension of $k$, $I=(f_1,\dots,f_m)$ an prime ideal in $K[X_1,\dots,X_n]$, then what is $I\cap k[X_1,\dots,X_n]$? Is it $\sqrt{(\Pi_{\sigma}{\sigma(f_1)},\dots,\Pi_{\sigma}{\sigma(f_m)})}$?

Conversely, suppose $J=(g_1,\dots,g_m)$ is an ideal in $k[X_1,\dots,X_n]$, can we write out all its extensions explicitly with the $\sigma$s?

  • 2
    $\begingroup$ A simple counterexample to your first guess is the case $k=\Bbb{R}$, $K=\Bbb{C}$, $n=1$, $f_1(x)=x+i$, $f_2(x)=x-i$. Then $\prod_\sigma \sigma(f_i)=x^2+1$, for $i=1,2,$ so the radical ideal is $\langle x^2+1\rangle$. But $I$ is all of $\Bbb{C}[x]$, and hence the intersection is all of $\Bbb{R}[x]$. Under some reasonable assumptions it may be possible to say more. Awaiting comments from experts. $\endgroup$ – Jyrki Lahtonen Nov 23 '13 at 9:03
  • $\begingroup$ @JyrkiLahtonen Your example is nice. The background (I didn't mention above) requires I to be a proper prime ideal $\endgroup$ – Qixiao Nov 23 '13 at 13:50
  • $\begingroup$ Ok. So use $n=2$, $f_1(x_1,x_2)=x_2(x_1+i)$, $f_2(x_1,x_2)=x_2(x_1-i)$. Now $I=\langle x_2\rangle$, $\Bbb{R}[x_1,x_2]\cap I =\langle x_2\rangle$. But the radical ideal is generated by $x_2(x_1^2+1)$. $\endgroup$ – Jyrki Lahtonen Nov 23 '13 at 14:40
  • $\begingroup$ Nice example! Is there an explicit display of the contracted ideal? $\endgroup$ – Qixiao Nov 24 '13 at 1:00

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