---
tag: F7FYTD7Z56
kind: definition
name: Lagrange resolvent
msc: [12F10, "12E12"]
defines: 2GDFFH7Y7T
uses: [VFE5B71E7Y, 0J817RTVG6, 0XVR72M51M, N43G50YHG2]
sources:
  - url: https://en.wikipedia.org/wiki/Resolvent_(Galois_theory)
---

## Definition

Let $n \ge 1$ be an integer and let $K$ be a field that contains a [[primitive $n$-th root of unity|0J817RTVG6]] $\zeta$.
Let $L/K$ be a [[cyclic extension|VFE5B71E7Y]] of degree $n$, and let $\sigma$ be a generator of its [[Galois group|0XVR72M51M]].
For $\alpha \in L$, the *Lagrange resolvent* of $\alpha$ with respect to $\zeta$ is
$$(\zeta, \alpha) = \sum_{i=0}^{n-1} \zeta^i \sigma^i(\alpha) = \alpha + \zeta \sigma(\alpha) + \cdots + \zeta^{n-1} \sigma^{n-1}(\alpha).$$

## 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|N43G50YHG2]] 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 $x_1 + \zeta x_2 + \cdots + \zeta^{n-1} x_n$.