# Galois group and invariant polynomials

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 $A_n$.

• Just a note to say that these invariant polynomials $R$ form the basis of the "resolvent method" or the "Stauduhar method" to compute the Galois group of a given polynomial, precisely by using the above result to prove that $G$ is contained in some $H$. The paper "Computation of Galois groups of rational polynomials" by Fieker and Kluners is a good place to read about this. Feb 10 '17 at 16:00

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.

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 $f(\alpha_1,\ldots,\alpha_n) \in K$.