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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 $I$.

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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 $X=Y$.

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