Lagrange resolvent

Remarks

Since $\sigma$ fixes $\zeta$ and $\sigma^n = \id$, one has $\sigma((\zeta, \alpha)) = \sum_{i} \zeta^i \sigma^{i+1}(\alpha) = \zeta^{-1} (\zeta, \alpha)$. Hence $(\zeta, \alpha)^n$ is fixed by $\sigma$ and lies in $K$. By the independence of characters applied to the automorphisms $\id, \sigma, \ldots, \sigma^{n-1}$, some $\alpha$ has $(\zeta, \alpha) \ne 0$. For such $\alpha$, the automorphism $\sigma^j$ multiplies $(\zeta, \alpha)$ by $\zeta^{-j}$, which equals $1$ only when $n$ divides $j$. So only the identity of the Galois group fixes $(\zeta, \alpha)$, hence $L = K((\zeta, \alpha))$ is generated by an $n$-th root of an element of $K$.

Example: let $K = \Q$, $n = 2$, $\zeta = -1$, $L = \Q(\sqrt{2})$ and $\sigma(\sqrt{2}) = -\sqrt{2}$. For $\alpha = \sqrt{2}$, the resolvent is $\sqrt{2} - \sigma(\sqrt{2}) = 2\sqrt{2}$, and its square $8$ lies in $\Q$.

Non-example: with $\zeta = 1$ the same sum is $\alpha + \sigma(\alpha) + \cdots + \sigma^{n-1}(\alpha)$, the trace of $\alpha$. This is not a Lagrange resolvent for $n \ge 2$, because $1$ is not a primitive $n$-th root of unity, and it lies in $K$ instead of generating $L$.

Classically, for a polynomial with roots $x_1, \ldots, x_n$, the Lagrange resolvent is the expression $x_1 + \zeta x_2 + \cdots + \zeta^{n-1} x_n$.