The A_2 discriminant, its Jacobian and the top coinvariant class in $\mathbb{C}[u,z]\mathrm{/}(uz,u^3+z^3)$

The Coxeter presentation maps onto $S_3$ by $s_1\mapsto (12)$ and $s_2\mapsto (23)$. Let $t=s_1{s}_2$; then $t^3=1$, $s_2=s_{1}t$, and $s_{1}t{s}_1=t^{-1}$. Every word is therefore one of $t^k$ or $s_1{t}^k$ for $k=0,1,2$, so $\mathrm{\mid }W\mathrm{\mid }\le 6$; the surjection onto $S_3$ gives $\mathrm{\mid }W\mathrm{\mid }\ge 6$ and hence $W\cong S_3$. The reflection in the $x$-axis has unit normal ${\alpha }_1=(0,1)$; the reflection in the line at angle $\pi \mathrm{/}3$ has unit normal ${\alpha }_2=(\sqrt{3}\mathrm{/}2,-1\mathrm{/}2)$, with $({\alpha }_1,{\alpha }_2)=-1\mathrm{/}2$. The Gram matrix of $({\alpha }_1,{\alpha }_2)$ is the $A_2$ Coxeter Gram matrix, so the map from the canonical simple-root basis to $({\alpha }_1,{\alpha }_2)$ is an isometry and conjugates the canonical representation to this model. The reflection actions are $s_1{\alpha }_1=-{\alpha }_1$, $s_1{\alpha }_2={\alpha }_1+{\alpha }_2$, $s_2{\alpha }_2=-{\alpha }_2$, and $s_2{\alpha }_1={\alpha }_1+{\alpha }_2$; hence $\{\pm {\alpha }_1,\pm {\alpha }_2,\pm ({\alpha }_1+{\alpha }_2)\}$ is exactly the root set and its positive elements are ${\alpha }_1,{\alpha }_2,{\alpha }_3$. The corresponding normalized forms are $y$, $\frac{\sqrt{3}}{2}x-\frac{1}{2}y$, and $\frac{\sqrt{3}}{2}x+\frac{1}{2}y$. Substitution $x=(u+z)\mathrm{/}2$, $y=(u-z)\mathrm{/}(2i)$ gives the three displayed scalar multiples in (1). Multiplying them gives $\mathrm{\Delta }=\frac{1}{8i}(u-z)(u-\zeta z)(u-{\zeta }^{2}z)=\frac{1}{8i}(u^3-z^3)$ because $1,\zeta ,{\zeta }^2$ are the distinct cube roots of unity; the product has degree $3=\mathrm{\mid }{\mathrm{\Phi }}_{+}\mathrm{\mid }$.

The reflection generator acts by $\sigma (u,z)=(z,u)$ and $R=s_2{s}_1$ acts by $R(u,z)=(\zeta u,{\zeta }^{-1}z)$. Since $s_2=R\sigma$, the pair $\sigma ,R$ generates $W$. Both $a=uz$ and $b=u^3+z^3$ are invariant. Let $f={\sum }_{p,q}{c}_{pq}{u}^p{z}^q$ be invariant. Rotation invariance and monomial independence imply $c_{pq}=0$ unless $3\mid (p-q)$; swap invariance gives $c_{pq}=c_{qp}$. Thus $f$ is a finite linear combination of $a^p$ and orbit sums $a^q(u^{3k}+z^{3k})$ with $k\ge 1$. Set $S_k=u^{3k}+z^{3k}$. Then $S_0=2$, $S_1=b$, and direct expansion gives $S_{k+1}=b{S}_k-a^3{S}_{k-1}$; induction shows each $S_k\in \mathbb{C}[a,b]$. Hence $\mathbb{C}[u,z{]}^W=\mathbb{C}[a,b]$.

The formal partial-derivative rules and the product rule give ${\mathrm{\partial }}_{u}a=z$, ${\mathrm{\partial }}_{z}a=u$, ${\mathrm{\partial }}_{u}b=3u^2$, and ${\mathrm{\partial }}_{z}b=3z^2$, so the Jacobian determinant is $\mathbf{J}=3(z^3-u^3)=-24i\mathrm{\Delta }$. It is not the zero polynomial. To prove $a,b$ algebraically independent, suppose a nonzero $H\in \mathbb{C}[Y_1,Y_2]$ of minimal total degree satisfies $H(a,b)=0$. Since the characteristic is zero, at least one partial derivative $H_j$ is nonzero. Expanding $H$ into monomials and applying the product rule gives the chain rule, so differentiating the relation with respect to $u,z$ gives $J^{\mathsf{T}}(H_1(a,b),H_2(a,b){)}^{\mathsf{T}}=0$. Multiply the displayed equation by the explicit adjugate $\left(\begin{array}{cc}3z^2& -3u^2\\ -u& z\end{array}\right)$ of $J^{\mathsf{T}}$. Its product with $J^{\mathsf{T}}$ is $3(z^3-u^3)I_2$, so the domain property and $3(z^3-u^3)\mathrm{\ne }0$ force both partial derivatives to vanish after substitution. A nonzero partial derivative is then a relation of strictly smaller total degree; if it were a nonzero constant it could not vanish, and otherwise this contradicts the minimal choice of $H$. Therefore $a,b$ are algebraically independent. They are homogeneous of degrees $2,3$ and generate $R$ by 2.1. Since $R_{+}=(a,b)R$, the coinvariant ideal is $(a,b)\mathbb{C}[u,z]$. Neither generator is in the ideal generated by the other: $a$ has degree $2<3$, while $b=u^3+z^3$ is not divisible by $uz$. Thus they form the basic family with degrees $2,3$ and exponents $1,2$. For anti-invariance, the swap sends $\mathrm{\Delta }$ to $-\mathrm{\Delta }$ and has determinant $-1$; the rotation sends $u^3-z^3$ to itself and has determinant $1$. Since these elements generate $W$, $w\cdot \mathrm{\Delta }=\det (w)\mathrm{\Delta }$ for every $w\in W$.

In $A=\mathbb{C}[u,z]\mathrm{/}(uz,u^3+z^3)$, all mixed monomials vanish, $z^3=-u^3$, and $u^4=z^4=0$, so $1,u,u^2,z,z^2,u^3$ span. The ideal is homogeneous: its degree-two part is spanned by $uz$, and its degree-three part by $u^{2}z,u{z}^2,u^3+z^3$. Thus $u^2,z^2$ are independent in degree two, and $u^3$ is nonzero in degree three because it is not in the span of those three degree-three relations; the degree-zero and degree-one classes are also independent because the ideal has no terms in those degrees. Hence the six displayed classes are a basis, $\mathrm{Hilb}(A,t)=1+2t+2t^2+t^3$, and $A_3$ is one-dimensional. Using $z^3=-u^3$ and the normalized scalar in step 1.1 gives $[\mathrm{\Delta }]=\frac{1}{8i}[u^3-z^3]=-\frac{i}{4}[u^3]\mathrm{\ne }0$, so it spans $A_3$ and carries the determinant character by step 3.1.

The degrees are $2,3$, so the exponents are $1,2$, their sum is $3=\mathrm{\mid }{\mathrm{\Phi }}_{+}\mathrm{\mid }$, and their product is $6=\mathrm{\mid }{S}_3\mathrm{\mid }$ by 1.1. The quotient basis in 4.1 has top degree $3$ and one-dimensional top component; the nonzero discriminant class is its generator. All constructions use explicit coordinates and finite polynomial identities, so no Choice is used. This proves the four clauses.