# On calculating the restriction of the ideals in polynomial rings.

Let $K$ be a field and $F$ a Galois extension of $K$ and $G$ the Galois group of extension $F/K$. For any $\sigma\in G$, define an $K$-algebra automorphism $\sigma^*$ on $F[X_1,\dots,X_n]$ by $X_i\mapsto X_i, a\mapsto\sigma a\ (\forall a\in F$), then $\sigma\mapsto\sigma^*$ defines an action of $G$ on $F[X_1,\dots,X_n]$.

Now consider the embedding $i:K[X_1,\dots,X_n]\to F[X_1,\dots,X_n]$. My question is if for any ideal $I$ of $F[X_1,\dots,X_n]$, we have

$$F[X_1,\dots,X_n]i(i^{-1}I)=\bigcap_{\sigma\in G}\sigma^*I.$$

• @user26857 It's natural to define a action of Galois group $G_{F/E}$ on $F[X_1,\cdots,X_n]$. And $i^{-1}(I)$ is not an ideal of $F[X_1,\cdots,X_n]$. – Censi LI Feb 18 '16 at 22:43
• Could you explain. The intersection of all Galois conjugates of the ideal is still an ideal of $F[X_1,\ldots,X_n]$. Are you asking whether $I\cap K[X_1,\ldots,X_n]$ is the same thing as $\displaystyle \left(\bigcap_{\sigma}\sigma(I)\right)\cap K[X_1,\ldots,X_n]$? – Alex Youcis Feb 18 '16 at 22:44
• @AlexYoucis No. I'm asking if $F[X_1,\cdots,X_n](I\cap K[X_1,\cdots,X_n])=\cap_{\sigma}\sigma(I)$ – Censi LI Feb 18 '16 at 22:54
• @user26857: Censi means (the ideal generated by) $i(i^{-1}(I)) \subseteq F[X_1, \dots X_n]$. Censi, I think you should rewrite your question to be clearer about this. – Qiaochu Yuan Feb 18 '16 at 22:54
• @QiaochuYuan OK, I've expanded it. – Censi LI Feb 19 '16 at 15:14

Let $I = (X-i,Y-i)(X+i,Y+i) \subset \mathbb{C}[X,Y]$.
Then $f=(X-i)(Y+i)$ and its complex conjugate are both in $I$, but $f$ is not generated by $I\cap \mathbb{R}[X,Y] = (X^2+1,Y^2+1,XY+1)$.