Discriminants of cyclic cubic orders (Q301411)

Let \(\alpha\) be a cubic algebraic integer and \(\Pi_\alpha(X)=X^3-aX^2+bX-c\) be its minimal polynomial. Let \(\alpha_1\), \(\alpha_2\) and \(\alpha_3\) be the complex conjugates of \(\alpha\) and set \(\Omega_3=\{1, \alpha_1, \alpha_1^2, \alpha_2, \alpha_2\alpha_1, \alpha_2\alpha_1^2\}\), then \(\Omega_3\) is a \(\mathbb Z\)-generating system of the order \(\mathbb M_3=\mathbb Z[\alpha_1, \alpha_2, \alpha_3]\). If the cubic number field \(\mathbb K=\mathbb Q(\alpha)\) is not Galois, then its normal closure \(\mathbb N=\mathbb Q(\alpha_1, \alpha_2, \alpha_3)\) is a sextic number field with Galois group isomorphic to the symmetric group \(\mathfrak{G}_3\), \(\Omega_3\) is a \(\mathbb Z\)-basis of the \(\mathbf{Gal}(\mathbb N/\mathbb Q)\)-invariant sextic order \(\mathbb M_3\) and its discriminant \(d_{\mathbb M_3}\) is given by \(d_{\mathbb M_3}=d_{\alpha}^3\), where \(d_\alpha\in\mathbb Z-\{0\}\) is the discriminant of \(\Pi_\alpha(X)\), this result was proved by Jun Ho Lee and Louboutin.NEWLINENEWLINEIn this paper, the authors give us an explicit \(\mathbb Z\)-basis and the discriminant of the \(\mathbf{Gal}(\mathbb K/\mathbb Q)\)-invariant totally real cubic order \(\mathbb M_3\), if \(\mathbb K\) is Galois. The new result is as follow: Since the cubic number field \(\mathbb K=\mathbb Q(\alpha)\) is supposed Galois, then the discriminant \(d_{\alpha} =-4a^3c-4b^3+a^2b^2+18abc-27c^2\) of \(\Pi_\alpha(X)\) is a perfect square, i.e. \(d_\alpha= D^2\) where \(D\in\mathbb Z\). Let \(\alpha_1\), \(\alpha_2\) and \(\alpha_3\) be the three complex conjugates of \(\alpha\). Set NEWLINE\[NEWLINE\Delta=\gcd(D, 3b-a^2, 3ac-b^2).NEWLINE\]NEWLINE Let \(x, y, z\in \mathbb Z\) be such that NEWLINE\[NEWLINE\Delta=xD+y(3b-a^2)+z(3ac-b^2)NEWLINE\]NEWLINE and set NEWLINE\[NEWLINE\eta=x\alpha_1^2+y\alpha_2+z\alpha_2\alpha_1^2.NEWLINE\]NEWLINE Then \(\{1, \alpha_1, \eta\}\) is a \(\mathbb Z\)-basis of the \(\mathbf{Gal}(\mathbb K/\mathbb Q)\)-invariant order \(\mathbb M_3\) and \(d_{\mathbb M_3}=\Delta^2\) divides \(d_\alpha\).