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  "body": "\n## Definition\n\nLet $n \\ge 1$ be an integer and let $K$ be a field that contains a [[primitive $n$-th root of unity|0J817RTVG6]] $\\zeta$.\nLet $L/K$ be a [[cyclic extension|VFE5B71E7Y]] of degree $n$, and let $\\sigma$ be a generator of its [[Galois group|0XVR72M51M]].\nFor $\\alpha \\in L$, the *Lagrange resolvent* of $\\alpha$ with respect to $\\zeta$ is\n$$(\\zeta, \\alpha) = \\sum_{i=0}^{n-1} \\zeta^i \\sigma^i(\\alpha) = \\alpha + \\zeta \\sigma(\\alpha) + \\cdots + \\zeta^{n-1} \\sigma^{n-1}(\\alpha).$$\n\n## Remarks\n\nSince $\\sigma$ fixes $\\zeta$ and $\\sigma^n = \\id$, one has $\\sigma((\\zeta, \\alpha)) = \\sum_{i} \\zeta^i \\sigma^{i+1}(\\alpha) = \\zeta^{-1} (\\zeta, \\alpha)$.\nHence $(\\zeta, \\alpha)^n$ is fixed by $\\sigma$ and lies in $K$.\nBy the [[independence of characters|N43G50YHG2]] applied to the automorphisms $\\id, \\sigma, \\ldots, \\sigma^{n-1}$, some $\\alpha$ has $(\\zeta, \\alpha) \\ne 0$.\nFor such $\\alpha$, the automorphism $\\sigma^j$ multiplies $(\\zeta, \\alpha)$ by $\\zeta^{-j}$, which equals $1$ only when $n$ divides $j$.\nSo 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 $K$.\n\nExample: let $K = \\Q$, $n = 2$, $\\zeta = -1$, $L = \\Q(\\sqrt{2})$ and $\\sigma(\\sqrt{2}) = -\\sqrt{2}$.\nFor $\\alpha = \\sqrt{2}$, the resolvent is $\\sqrt{2} - \\sigma(\\sqrt{2}) = 2\\sqrt{2}$, and its square $8$ lies in $\\Q$.\n\nNon-example: with $\\zeta = 1$ the same sum is $\\alpha + \\sigma(\\alpha) + \\cdots + \\sigma^{n-1}(\\alpha)$, the trace of $\\alpha$.\nThis 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 $L$.\n\nClassically, for a polynomial with roots $x_1, \\ldots, x_n$, the Lagrange resolvent is the expression $x_1 + \\zeta x_2 + \\cdots + \\zeta^{n-1} x_n$.\n",
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