Automorphism-stable Laurent ideals descend to rational coefficients

Finite stable linear spaces have rational bases. Let $W\subseteq {\mathbb{C}}^N$ be a complex linear subspace invariant under coordinatewise application of all the automorphisms in [F1]. Choose any finite basis and put its row matrix in reduced row-echelon form, deleting zero rows. Finite elimination terminates by choosing successive pivot columns in the fixed order. The resulting nonzero rows are independent of the initial basis: the pivot columns are characterized successively by the dimensions of the projections of $W$ to initial coordinate blocks; once these columns are fixed, each row is the unique vector of $W$ whose pivot coordinates are its specified standard basis vector. Indeed the pivot-coordinate projection is an isomorphism $W\to {\mathbb{C}}^{\dim W}$ by row reduction. Applying a field automorphism preserves row-echelon equations, zero entries, leading ones and the increasing pivot order. It takes the row space to $W$, so uniqueness fixes every entry of this matrix. By [F1] all entries are rational. Thus its rows belong to $W\cap {\mathbb{Q}}^N$ and their complex span is $W$. If $W=0$ the row list is empty and the same conclusion holds.

Apply finite descent to each support box. For $d\ge 0$ let $V_d$ be the finite-dimensional complex span of monomials $z^{\alpha }$ with $-d\le {\alpha }_i\le d$, in a fixed lexicographic order. For $n=0$ take the single monomial $1$. Coefficientwise automorphisms preserve $V_d$ and preserve $W_d=I\cap V_d$. Step 1.1 expresses each element of $W_d$ as a finite complex linear combination of elements of $I\cap R_{\mathbb{Q}}$. Every Laurent polynomial has finite support and belongs to some $V_d$. Hence every element of $I$ belongs to $(I\cap R_{\mathbb{Q}})R_{\mathbb{C}}$. The reverse inclusion holds because $I$ is an ideal containing that rational intersection. This proves the asserted scalar-extension equality without choosing an infinite family of generators or invoking an elimination theorem.

Radical contraction and stable point sets. If $f\in R_{\mathbb{Q}}$ and $f^k\in I\cap R_{\mathbb{Q}}$ for an integer $k\ge 1$, radicality of $I$ gives $f\in I\cap R_{\mathbb{Q}}$. Thus the contracted ideal is radical. For the point-set assertion let $I(S)$ be the Laurent polynomials vanishing on every point of $S$. It is radical, since a complex number whose positive power vanishes is zero. For an automorphism $\sigma$ and $s\in S$, one has $(\sigma f)(s)=\sigma (f({\sigma }^{-1}s))=0$ whenever $f\in I(S)$, so $\sigma I(S)=I(S)$ using the inverse automorphism too. The Zariski closure has exactly the same vanishing ideal: each zero set of a polynomial vanishing on $S$ is closed and contains the closure, and the reverse inclusion follows from $S\subseteq \bar{S}$. Apply step 2.1 to $I(S)$, then the radical contraction just proved. Empty $S$ gives the unit ideal and the full torus gives zero; step 2.1 also handles $n=0$ directly. These are finite row computations plus the precise full-AC fixed-field theorem, with no extra choice assumption.