Prove that $X_1+2X_2+\cdots+nX_n$ is a primitive element in a field extension of rational functions

Question

I want to prove that $X_1+2X_2+\cdots+nX_n$ is a primitive element for the field extension $K \subset L$ with $L=k(X_1,...,X_n)$ and $K=k(\sigma_1,...,\sigma_n)$ and $\mathrm{char}(k)=0$.

Answer 1

Recall $L/K$ is Galois with group naturally isomorphic to $S_n$. Let $w=X_1+2X_2+\cdots+nX_n$.

It is enough to show that the Galois group of $L/K(w)$ is the trivial subgroup of $S_n$, by the Galois correspondence theorem.

Now, if $\sigma$ is in the Galois group of $L/K(w)$, then for some permutation $\alpha$ of $\{1,\ldots,n\}$, $\sigma(X_i)=X_{\alpha^{-1}(i)}$.

(This is because the $X_i$ are the only roots of $\sum_{k=0}^n{T^k\sigma_{n-k}(-1)^{n-k}} \in K[T]$ with $\sigma_0=1$).

Then $\sum_{i=1}^n{\alpha(i)X_i}=\sigma(w)=w=\sum_{i=1}^n{iX_i}$, so $\alpha=\mathrm{Id}$.

Answer 2

The extension $L/K$ is Galois. If $w = X_1 + \cdots + n X_n$ does not generate $L$, then some nontrivial $\sigma \in Gal(L/K) \cong S_n$ fixes $w$. But when $\sigma \in S_n$ is nontrivial, we cannot have $$X_1 + \cdots + n X_n = X_{\sigma(1)} + \cdots + n X_{\sigma(n)}$$ unless $0 < \mathrm{char}(K) < n$.