2 The orbit parametrization
A \(C_3\)-algebra is either a cyclic cubic field with a specified \(C_3\)-action, or \(\mathbb {Q}^3\) with cyclic permutation of its factors. An element is normal when its three conjugates form a \(\mathbb {Q}\)-basis. We identify pairs \((K,x)\) up to \(C_3\)-equivariant \(\mathbb {Q}\)-algebra isomorphism taking one distinguished element to the other.
There is a bijection between rational points of \(T\) and equivalence classes of \(C_3\)-algebras \(K/\mathbb {Q}\) equipped with a trace-one normal element \(x\in K\).
Construct the algebra and normal element from a rational quotient point, and recover the quotient point from the conjugates of the normal element. These constructions are inverse on isomorphism classes. This node states the rational-point part of the paper’s theorem.
Let \(K\) be a cubic abelian number field. Then \(K\), equipped with its canonical Galois action, is a \(C_3\)-algebra. The twist \(K'\) of the \(C_3\)-algebra \(K\) by the outer automorphism \(g \mapsto g^{-1}\) of \(C_3\) (with twisted action \(g \ast x = g^{-1}x\)) is not isomorphic to \(K\) as a \(C_3\)-algebra. 1
The Galois group of a cyclic cubic field is abelian. An isomorphism to the inverse-action twist would therefore force a generator to equal its inverse, a contradiction. The cocycles describe the two rational points.
Let \(K_{\text{spl}} = \mathbb {Q}^3\), the split cubic algebra. Then \(C_3\subset S_3 = \mathrm{Aut}_{\mathbb {Q}\text{-alg}}(K_{\text{spl}})\) and \(K_{\text{spl}}\), equipped with its canonical \(C_3\)-action, is a \(C_3\)-algebra. Any transposition gives an isomorphism of \(C_3\)-algebras from \(K_{\text{spl}}\) to its twist \(K_{\text{spl}}'\).
Every algebra automorphism of \(\mathbb {Q}^3\) permutes its three coordinate idempotents. A transposition conjugates cyclic translation to its inverse, giving the asserted isomorphism with the twist.
An element \(x\) of trace one in the split \(C_3\)-algebra \(K_{\text{spl}}\) is normal if and only if \(x\) either has distinct coordinates or exactly two identical coordinates. For such a normal element \(x\), the pairs \((K_{\text{spl}},x)\) and \((K_{\text{spl}}',x)\) are equivalent if and only if \(x\) has exactly two identical coordinates (swapping the identical coordinates gives the required isomorphism); in particular, if \(x\) has distinct coordinates then \((K_{\text{spl}},x)\) and \((K_{\text{spl}}',x)\) determine different rational points of \(\mathcal G_1/C_3\), even though \(K_{\text{spl}}\) and \(K_{\text{spl}}'\) are isomorphic as \(C_3\)-algebras.
Evaluate the explicit algebra and cyclic action, then compute the characteristic polynomial and the normality condition.