This is a question on algebraic geometry/commutative algebra. Let $K,L$ be fields of characteristics zero and let $K\subset L$ be a field extension (I am interested in the case when this is transcendental). Assume that there are elements $\sigma_1,\dots,\sigma_k\in{\rm Gal}(L/K)$ so that $K=\{ a\in L\mid \forall i,\ \sigma_i(a)=a\}$.
We consider a polynomial ring $L[x_1,\dots,x_n]$ and an ideal $I$ of it. Assume $I$ is invariant by $\sigma_i$, i.e., the ideal $I_i:=\{ p^{\sigma_i}\mid p\in I\}$ is equal to $I$ for any $i$. Here, $p^{\sigma_i}$ is an element of $L[x_1,\dots,x_n]$ obtained from $p$ by replacement $ax^\alpha\mapsto \sigma_i(a)x^\alpha$ $(a\in L)$. Is it true that there exists an ideal $I_0\subset K[x_1,\dots,x_n]$ such that its extension $L\otimes_KI_0\subset L[x_1,\dots,x_n]$ is equal to $I$?
I do not know what this type of problem is called. I checked Hartshorne and found exercise 8 in Chap 2.Section 4 which is close to our setup. Unfortunately, it is when $L=\mathbb{C}$, $K=\mathbb{R}$ and $\sigma$ is the complex conjugate. I think the solution to this exercise relies on the fact that $\sigma$ is an involution so I do not know how to generalize it.
Background: I am interested in an affine variety defined over $L=\mathbb{C}(s)\supset \mathbb{C}=K$, where $s$ is a new variable. $\sigma:L\to L$ is the shift operator $s\mapsto s+1$. I guess that the statement above is true at least in this setup.
Thank you very much in advance.
