Zero locus of polynomials in small field vanishing on a point in a larger field

Let $$L/K$$ be a Galois field extension. Fix $$n\geq 1$$ and a point $$\bf{x}$$ in $$L^n$$.

Let $$I$$ be the ideal of $$K[X_1,\dotsc ,X_n]$$ consisting of all polynomials vanishing on $$\bf{x}$$. Let $$X$$ be the subset of $$L^n$$ consisting of the common solutions for all polynomials in $$I$$. Let $$Y$$ be the $$\operatorname{Gal}(L/K)$$-orbit of $${\bf x}$$.

Clearly $$Y\subset X$$. Is $$X=Y$$?

I want to know this because I have a polynomial $$g\in K[X_1,\dotsc ,X_n]$$ that does not vanish on $$\bf{x}$$, but does vanish on some $${\bf y} \in L^n$$, and I want to prove that $${\bf y}$$ is not a common solution for the polynomials in $$I$$.

• Dec 4, 2022 at 14:58
• Certainly true when $n=1$. Dec 4, 2022 at 15:19

First, I claim that $$I$$ is a maximal ideal. Indeed, $$I$$ is the kernel of the homomorphism $$K[X_1,\dots,X_n]\to L$$ sending the $$X_i$$ to the coordinates of $$\mathbf{x}$$. The image of this homomorphism is a subring of $$L$$ which contains $$K$$ and is thus automatically a field since $$L$$ is algebraic over $$K$$.
Now an element of $$X$$ can be identified with a $$K$$-algebra homomorphism $$K[X_1,\dots,X_n]/I\to L$$ (by mapping the $$X_i$$ to the coordinates of your element of $$X$$). Thinking of $$K[X_1,\dots,X_n]/I$$ as an intermediate field between $$K$$ and $$L$$, any embedding of $$K[X_1,\dots,X_n]/I$$ into $$L$$ extends to an automorphism of $$L$$ since $$L$$ is normal over $$K$$. That is, $$Gal(L/K)$$ acts transitively on the set of homomorphisms $$K[X_1,\dots,X_n]/I\to L$$, so it acts transitively on $$X$$, so $$X=Y$$.