What is the field of definition of an invariant ideal?

Question

Let $K/k$ be a finitely generated field extension, such that $k=K^G$ for some (possibly infinite) set $G$ of automorphisms of $K$. Now, consider the extension of polynomial rings $$ k[X_1,\ldots,X_n]\subset K[X_1,\ldots,X_n] $$ and a $G$-invariant ideal $I\lhd K[X_1,\ldots,X_n]$ (with respect to the natural action of $G$ on coefficients). Is it true that $I$ can be defined by polynomials in $k$?

Answer 1

It seems that the answer is yes in general. I found the following lemma in good old André Weil's Foundations of algebraic geometry (I.7, p. 19)

Lemma 2 Let $K$ be a field, and let $\mathfrak{A}$ be an ideal in the ring $$ K[X]=K[X_1,\ldots,X_n]. $$ Then, of all the subfields $k$ of $K$ such that $\mathfrak{A}$ has a basis with coefficients in $k$, there is a smallest field $k_0$, contained in all the others. Morover, if $\sigma$ is any automorphism of $K$, then $\sigma$ transforms $\mathfrak{A}$ into itself if and only if it leaves invariant every element of the field $k_0$.

Answer 2

First, remark that a finitely generated field extension is actually a finite extension, by one of the versions of the Nullstellensatz. So, in fact, your $K/k$ is finite and Galois, and $G$ is the Galois group of $K$ over $k$.

Now I claim that if $K/k$ is Galois with Galois group $G$, and $I\subseteq K[X_1, \dots, X_n]$ is a $G$-invariant ideal, then $I$ is defined over $k$. To see this, consider the homogeneous situation, where $I$ is a homogeneous ideal in the graded ring $K[Y_0, \dots, Y_n]$. (By the process of homogenization and dehomogenization, we can always reduce to this case). By construction,

$$I= \bigoplus_{d=0}^\infty I_d,$$

where $I_d = I \cap K_d$. Then $G$ acts $k$-linearly on each $I_d$, and each $I_d$ is a finite-dimensional $K$-vector space, say of dimension $m_d$. According to the multi-variable Hilbert 90 theorem, the non-abelian cohomology $H^1(G, \text{GL}_{m_d}(K))$ vanishes, and this implies by the usual argument that the map $I_d^G \otimes_k K \to I_d$ is surjective, i.e. $I_d$ has a basis $B_d$ defined over $k$. The union of all $B_d$'s generates $I$, and is defined over $k$.