Let $K$ be a field and $\alpha\_i$ ($i=1,\dots,n$) be Galois conjugates. Let $L=K(\alpha\_1,\dots,\alpha\_n)$ and $G=Gal(L/K)$. We embed $G$ in $S\_n$ by its action on $\alpha\_i$. Let $H$ be another subgroup of $S\_n$, acting on the variables $X\_i$ ($i=1,\dots,n$). Let $R=K[X\_1,\dots,X\_n]^H$ be the ring of invariant polynomials under $H$. Suppose for all $f\in R$, $f(\alpha\_1,\dots,\alpha\_n)\in K$. Does it follow that $G$ is contained in $H$?


Note that when $H=A\_n$ we can take $f$ to be the discriminant, and this question reduces to the equivalence between the discriminant being in $K$ and the Galois group being in $A\_n$.



Galois group and invariant polynomials
**Lemma 1:** Let $H$ be a subgroup of $\mathfrak{S}\_n$, and $K$ a field. If $(\xi\_1,\ldots,\xi\_n)$ and $(\xi'\_1,\ldots,\xi'\_n)$ are two $n$-tuples of elements of $K$, then t.f.a.e.:


1. there exists $\sigma\in H$ such that $\xi'\_i = \xi\_{\sigma(i)}$ for all $i$,
2. for every $f \in K[X\_1,\ldots,X\_n]^H$ we have $f(\xi'\_1,\ldots,\xi'\_n) = f(\xi\_1,\ldots,\xi\_n)$,
3. every $f \in K[X\_1,\ldots,X\_n]^H$ which vanishes at $(\xi\_1,\ldots,\xi\_n)$ also vanishes at $(\xi'\_1,\ldots,\xi'\_n)$.


**Proof:** (1)⇒(2) and (2)⇒(3) are trivial. Assume (1) does not hold: then one can construct a polynomial $h \in K[X\_1,\ldots,X\_n]$ which vanishes at $(\xi\_1,\ldots,\xi\_n)$ but at none of the $(\xi'\_{\sigma(1)},\ldots,\xi'\_{\sigma(n)})$; then $f = \prod\_{\sigma\in H} h(X\_{\sigma(1)},\ldots,X\_{\sigma(n)})$ is in $K[X\_1,\ldots,X\_n]^H$, vanishes at $(\xi\_1,\ldots,\xi\_n)$ but not at $(\xi'\_1,\ldots,\xi'\_n)$, so (3) does not hold.


**Lemma 2:** Let $H$ be a subgroup of $\mathfrak{S}\_n$, and $K \subseteq L$ a field extension. If $(\xi\_1,\ldots,\xi\_n)$ and $(\xi'\_1,\ldots,\xi'\_n)$ are two $n$-tuples of elements of $L$, then t.f.a.e.:


1. there exists $\sigma\in H$ such that $\xi'\_i = \xi\_{\sigma(i)}$ for all $i$,
2. for every $f \in K[X\_1,\ldots,X\_n]^H$ we have $f(\xi'\_1,\ldots,\xi'\_n) = f(\xi\_1,\ldots,\xi\_n)$.


**Proof:** (1)⇒(2) is again trivial. Assume (1) does not hold: then by the previous lemma (applied to $L$ instead of $K$), there exists $f \in L[X\_1,\ldots,X\_n]^H$ such that $f(\xi'\_1,\ldots,\xi'\_n) \neq f(\xi\_1,\ldots,\xi\_n)$. Write $f$ as a $L$-linear combination of polynomials which are each the sum of the orbit of a monomial under the action of $H$: such polynomials are in $K[X\_1,\ldots,X\_n]^H$, and at least one of them, call it $g$, must satisfy $g(\xi'\_1,\ldots,\xi'\_n) \neq g(\xi\_1,\ldots,\xi\_n)$. So (2) does not hold.




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Now in the case of your question, assume there exists $\sigma$ in $G$ which is not in $H$. Define $(\alpha'\_1,\ldots,\alpha'\_n) = (\sigma(\alpha\_1),\ldots,\sigma(\alpha\_n))$, which is also $(\alpha\_{\sigma(1)},\ldots,\alpha\_{\sigma(n)})$ by definition of the inclusion of $G$ in $\mathfrak{S}\_n$; note that the $\alpha\_i$ being implicitly assumed distinct, $\sigma$ is the only $\tau \in \mathfrak{S}\_n$ such that $(\alpha'\_1,\ldots,\alpha'\_n) = (\alpha\_{\tau(1)},\ldots,\alpha\_{\tau(n)})$, and it is not in $H$. So by lemma 2, there exists $f \in K[X\_1,\ldots,X\_n]^H$ such that $f(\alpha'\_1,\ldots,\alpha'\_n) \neq f(\alpha\_1,\ldots,\alpha\_n)$, which means exactly $\sigma(f(\alpha\_1,\ldots,\alpha\_n)) \neq f(\alpha\_1,\ldots,\alpha\_n)$. This contradicts $f(\alpha\_1,\ldots,\alpha\_n) \in K$.


