2. Binary cubics, their invariants, and covariants

Let ${\mathcal{BC}}(K)$ denote the space of binary cubic forms in $K[X,Y]$ with nonzero discriminant, and for each $\Delta \in K^*$ , let ${\mathcal{BC}}(K;\Delta )$ be the subset of those forms with discriminant $\Delta$ . Recall that for $g(X,Y) = aX^3+bX^2Y+cXY^2+dY^3\in {\mathcal{BC}}(K)$ , the discriminant $\mathrm{disc}(g)$ of $g$ is given by

\begin{equation*} \Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd; \end{equation*}

we also define the seminvariantsFootnote ¹

\begin{equation*} P = b^2-3ac, \qquad \hbox{and}\qquad U = 2b^3+27a^2d-9abc, \end{equation*}

which satisfy the syzygy

\begin{equation*} 4P^3 = U^2 + 27\Delta a^2. \end{equation*}

These are the leading coefficients of two covariant binary forms, the quadratic Hessian

\begin{equation*} H(X,Y) = (b^2-3ac)X^2 + (bc-9ad)XY + (c^2-3bd)Y^2, \end{equation*}

which has discriminant $-3\Delta$ , and the cubic

\begin{align*} G(X,Y) &= (2b^3+27a^2d-9abc)X^3 + 3(b^2c+9abd-6ac^2)X^2Y \\ &\quad -3(bc^2+9acd-6b^2d)XY^2 - (2c^3+27ad^2-9bcd)Y^3, \end{align*}

which has discriminant $729\Delta ^3$ ; these satisfy the syzygy

(1) \begin{equation} 4H(X,Y)^3 = G(X,Y)^2 + 27\Delta g(X,Y)^2, \end{equation}

extending the seminvariant syzygy, which is recovered on setting $(X,Y)=(1,0)$ .

We define the twisted action of $\mathrm{GL}(2,K)$ on ${\mathcal{BC}}(K)$ as follows: for $g\in {\mathcal{BC}}(K)$ and $M=\begin{pmatrix}r&&s\\t&&u\end{pmatrix}\in \mathrm{GL}(2,K)$ , set

\begin{equation*} g^M(X,Y) = \det (M)^{-1} g(r X + t Y, s X + u Y) = \det (M)^{-1} g(X',Y'), \end{equation*}

where $(X'\ Y')=(X\ Y)M$ . Both the untwisted action (without the determinant factor) and this action appear in the literature: for example, the untwisted action is used in [Reference Bhargava, Elkies and Shnidman2] and is also the subject of [Reference Kulkarni and Ure6], while the twisted action is used in [Reference Bhargava, Shankar and Tsimerman3]. Clearly, they are the same for $M\in \mathrm{SL}(2,K)$ , and the difference is only minor for our purposes since we will mostly be concerned with matrices with determinant $\pm 1$ and $-g = g^M$ with $M=-I$ .

The discriminant and other covariants of $g$ and $g^M$ are related byFootnote ²

\begin{align*} \Delta (g^M) &= \det (M)^2\Delta (g);\\ H(g^M) &= H(g)^M;\\ G(g^M) &= \det (M) G(g)^M; \end{align*}

indeed, the definition of covariants is precisely that such identities hold; the invariant $\Delta$ is just a covariant of degree $0$ .

Below we will also need that $g\in {\mathcal{BC}}(K)$ is coprime to its covariant $G$ , their resultant being $8\Delta ^3$ .

3. The resolvent algebra and the Cardano invariant

To each $\Delta \in K^*$ , we associate the resolvent algebra

\begin{equation*} L = K[\delta ] = K[T]/(T^2+3\Delta ), \end{equation*}

a quadratic étale algebra over $K$ , with a fixed generator $\delta$ (the image of $T$ in the quotient) such that $\delta ^2=-3\Delta$ ; it is a field unless $-3\Delta$ is a square in $K$ . We denote by a bar the nontrivial $K$ -automorphism of $L$ , mapping $u+v\delta \mapsto u-v\delta$ . Let $L^*$ denote the unit group of $L$ , which consists of the elements with nonzero norm; the norm map $\mathrm{N}_{L/K}: L \to K$ maps $z=u+v\delta \mapsto z\overline {z}=u^2+3\Delta v^2$ . We extend these to maps $L[X,Y]\to L[X,Y]$ and $L[X,Y]\to K[X,Y]$ , and also use the same notation for the induced group homomorphisms $L^*/L^*{}^3\to L^*/L^*{}^3$ and $L^*/L^*{}^3\to K^*/K^*{}^3$ .

For $M\in \mathrm{GL}(2,K)$ and $g\in {\mathcal{BC}}(K)$ , since $\Delta (g^M)=\det (M)^2\Delta (g)$ , the resolvent algebras of $g$ and $g^M$ are isomorphic via $\delta \mapsto \det (M)\delta$ . Transforms that preserve the discriminant, therefore, have determinant $\pm 1$ ; however, we regard $\delta$ as a fixed generator of $L$ , associated to $\Delta$ , which is the same for all cubics in ${\mathcal{BC}}(K;\Delta )$ .

Definition 1. The $\mathrm{Cardano\, covariant}$ $C(X,Y)$ of $g\in {\mathcal{BC}}(K;\Delta )$ is

\begin{equation*} C(X,Y) = \frac {1}{2}(G(X,Y)+3\delta g(X,Y)) \in L[X,Y]. \end{equation*}

Classically, this is called an “irrational” or algebraic covariant because its coefficients lie in the extension $L$ rather than in $K$ itself. It is a covariant for the action of $\mathrm{SL}(2,K)$ ; in general, we have

(2) \begin{equation} C(g^M) = \frac {1}{2}(\det (M)G + 3\delta g)^M; \end{equation}

this would be equal to $\det (M)C(g)^M$ if the value of $\delta$ for discriminant $\det (M)^2\Delta$ were fixed to be $\det (M)\delta$ , but it is not possible to do this consistently for transformations with both determinants $\pm \det (M)$ .

At first, we will restrict our attention to the action of $\mathrm{SL}(2,K)$ .

In terms of $C$ , the covariant syzygy may be simply written

\begin{equation*} \mathrm{N}_{L/K}(C) = C\overline {C} = H^3. \end{equation*}

Hence, for all $x,y\in K$ , not both $0$ , except for those satisfying $H(x,y)=0$ (which only exist when $-3\Delta$ is a square in $K$ ), the value $C(x,y)$ lies in $L^*$ and satisfies $\mathrm{N}_{L/K}(C(x,y))=H(x,y)^3$ with $H(x,y)\in K^*$ .

We will define the Cardano invariant in terms of values of the Cardano covariant. Before we give the general definition, first consider cubics with $P=H(1,0)\not =0$ . Set $z = C(1,0) = \frac {1}{2}(U+3a\delta )$ ; then $\mathrm{N}_{L/K}(z)=P^3 \in K^*{}^3$ , so $z\in L^*$ , and a provisional definition of the Cardano invariant of $g$ with $P\not =0$ is simply this quantity $z$ . One can show (see Proposition 5 below) that $z\in L^*{}^3$ if and only if $g$ has a linear factor in $K[X,Y]$ and that for two cubics $g_1,g_2$ with the same discriminant $\Delta$ and hence the same resolvent algebra, they are $\mathrm{SL}(2,K)$ -equivalent if and only if their $z$ -invariants are equal modulo $L^*{}^3$ . Instead, however, we proceed as follows, leading to a general definition of the Cardano invariant $z(g)$ as a well-defined element of $L^*/L^*{}^3$ for all $g\in {\mathcal{BC}}(K;\Delta )$ .

Observe that $C$ , whose norm $H^3$ is a cube in $K[X,Y]$ , is itself the cube of a linear form in $L[X,Y]$ , up to a constant factor. Explicitly, we have

\begin{equation*} 27 c_0^2 C(X,Y) = (3 c_0 X + c_1 Y)^3, \end{equation*}

where $C(X,Y) = c_0X^3+c_1X^2Y+c_2XY^2+c_3Y^3$ , so $c_0=z$ (as defined above) and $c_1=\frac {3}{2}(b^2 c - 6 a c^2 + 9 a b d + b \delta )$ . (This algebraic identity may be readily checked directly or derived by writing $C$ as a constant times the cube of a linear form and differentiating twice.) This already shows that if $P\not =0$ , so that $c_0\in L^*$ , then for every $x,y\in K$ such that $H(x,y)\not =0$ , we have $C(x,y)\in L^*$ and $C(x,y)\equiv c_0\mod {L^*{}^3}$ . For the general case, we use the following identity:

Proposition 3. The identity

\begin{equation*} C(X_1,Y_1)^2 C(X_2,Y_2) = F(X_1,Y_1,X_2,Y_2)^3 \end{equation*}

holds, where $F\in L[X_1,Y_1,X_2,Y_2]$ is given by

\begin{equation*} 3F(X_1,Y_1,X_2,Y_2) = (3c_0X_1^2 + 2c_1X_1Y_1 + c_2Y_1^2) X_2 + (c_1X_1^2 + 2c_2X_1Y_1 + 3c_3Y_1^2) Y_2. \end{equation*}

Also, $\mathrm{N}_{L/K} F(X_1,Y_1,X_2,Y_2) = H(X_1,Y_1)^2 H(X_2,Y_2)$ and $C(X,Y) = F(X,Y,X,Y)$ .

Proof. By the proposition, for $x_1,y_1,x_2,y_2$ such that $H(x_1,y_1),H(x_2,y_2)\not =0$ , we have $C(x_1,y_1),$ $C(x_2,y_2) \in L^*$ and, modulo $L^*{}^3$ , we have

\begin{equation*} C(x_1,y_1)^{-1}C(x_2,y_2) \equiv C(x_1,y_1)^2C(x_2,y_2) \equiv F(x_1,y_1,x_2,y_2)^3 \equiv 1. \end{equation*}

Hence, we may define the Cardano invariant $z(g)$ of $g\in {\mathcal{BC}}(K;\Delta )$ as an element of $L^*/L^*{}^3$ , by setting

\begin{equation*} z(g) = C(x,y) \in L^*/L^*{}^3, \end{equation*}

for any choice of $x,y\in K$ such that $C(x,y)$ is a unit. This is well-defined by the corollary, and if $P=H(1,0)\not =0$ , then we may take $z(g)=C(1,0)=z$ (modulo cubes), as above. In all cases we have, modulo  $K^*{}^3$ ,

\begin{equation*} \mathrm{N}_{L/K}(z(g)) \equiv \mathrm{N}_{L/K}(C(x,y)) \equiv H(x,y)^3 \equiv 1, \end{equation*}

so that $z(g) \in (L^*/L^*{}^3)_{N=1} := \ker (\mathrm{N}_{L/K}\,:\, L^*/L^*{}^3 \to K^*/K^*{}^3)$ , the Cardano group. This establishes the first part of Theorem1. The second part is the following:

Proof. Writing $g=aX^3+bX^2Y+cXY^2+dY^3$ , we first observe that when $a=0$ we have $P=b^2\not =0$ (as otherwise $\Delta =0$ ), so $z(g)=z=U/2=b^3\in K^*{}^3$ , and $g$ has the factor $Y$ .

Now assume that $a\not =0$ . The equations $z=\frac {1}{2}(U+3a\delta )=(x+y\delta )^3$ , together withFootnote ³ $\mathrm{N}_{L/K}(x+y\delta )=P$ , have a solution $x,y\in K$ if and only if

\begin{equation*} f(x) = 0 \quad \text{and}\quad y = 9a / f'(x), \end{equation*}

where $f(X) = 8X^{\,3} - 6PX - U = a^{-1}g(2X+b,-3a)$ . Hence, a solution $x,y\in K$ exists if and only if $g$ has a linear factor (with $(2x+b)/(-3a)$ a root of $g(X,1)$ ), noting that $x$ cannot be a double root of $f$ since $\Delta \not =0$ , so $f'(x)\not =0$ .

A special case of Theorem8 below, combined with Proposition 5, is that all cubics with the same discriminant that have linear factors over $K$ are $\mathrm{SL}(2,K)$ -equivalent; this is also implied by the following.

Remark 3. Cardano’s formula for the roots of the cubic expresses them in terms of the cube root of a quantity in $L=K(\sqrt {-3\Delta })$ . From the previous proof we see that, when $P\not =0$ , the roots of $g(X,1)$ are given by

\begin{equation*} x = - (b + \sqrt [3]{z} + P/\sqrt [3]{z}) / 3a, \end{equation*}

where $z = (U + 3a\sqrt {-3\Delta })/2$ . This is essentially Cardano’s formula. If $w=\sqrt [3]{z}\in L$ with $w\overline {w}=\sqrt [3]{\mathrm{N}_{L/K}(z)}=P$ then we have $x=-(b+w+\overline {w})/(3a)\in K$ . All three roots are in $K$ when $\sqrt {-3}\notin K$ but $\sqrt {-3}\in L$ , so $L=K(\sqrt {-3})$ and $\Delta \in (K^*)^2$ , for then $L$ contains a primitive roots of unity $\zeta _3$ and $\overline {\zeta _3}=\zeta _3^2$ , so that replacing $w$ by $\zeta _3w$ or $\zeta _3^2w$ in the formula again gives an element of $K$ .

Remark 5. The formulas given here may be used to parametrize all cubic extensions of $K$ as follows. (This construction may also be obtained from the Delone-Faddeev parametrization of cubic orders by binary cubic forms in [ Reference Delone and Faddeev5 ], replacing their base ring $\mathbb Z$ by a field.) Each such extension $M=K(\alpha )$ has a discriminant $\Delta \in K^*$ , well-defined modulo squares as the discriminant of the irreducible cubic minimal polynomial of $\alpha$ in $K[X]$ , and $\Delta =1$ (modulo squares) if and only if the extension $M/K$ is Galois. To each $\Delta \in K^*$ , we form the cubic algebra $L=K[T]/(T^2+3\Delta )$ as above; then the cubics with discriminant $\Delta$ (modulo squares) are parametrized by the subgroups of the Cardano group $(L^*/L^*{}^3)_{N=1}$ . Explicitly, for $z\in (L^*/L^*{}^3)_{N=1}$ with $\mathrm{N}_{L/K}(z)=P^3$ and $\mathrm{Tr}_{L/K}(z)=U$ , the associated cubic is $f_z(X)=X^3-3PX-U$ . (Note that $\overline {z}$ has the same trace and norm as $z$ , and hence $f_{\overline {z}}=f_z$ , in accordance with the fact that $z\overline {z}=P^3$ , so that $\overline {z}$ generates the same subgroup of the Cardano group as $z$ .) The root(s) of $f_z(X)$ are $\alpha =w+P/w$ where $w^3=z$ .

A refinement of this construction may be used when $K$ is a number field and $S$ a finite set of primes of $K$ , to determine all cubic extensions of $K$ which are unramified outside $S$ . For simplicity, we assume that $6$ is an $S$ -unit; otherwise, the cubic extensions constructed may be ramified at primes dividing $6$ , which are not in $S$ . Recall that for an integer $m\ge 2$ the subgroup

\begin{equation*} K(S,m) = \{x\in K^*/(K^*)^m\mid \mathrm{ord}_{{\mathfrak p}}(x)\equiv 0\pmod {m}\;\forall {\mathfrak p}\notin S\} \end{equation*}

of $K^*/(K^*)^m$ is finite and may be computed in terms of the class group and unit group of $K$ . First, we determine the possible quadratic subfields $K(\sqrt {\Delta })$ of the normal closure of the cubic, which must also be unramified outside $S$ , by restricting $\Delta$ to lie in $K(S,2)$ . For each such $\Delta$ , we also have that $K(\sqrt {-3\Delta })/K$ is unramified outside $S$ ; we then restrict $z$ to the subgroup $\ker (\mathrm{N}_{L/K}\,:\, L(S,3) \to K(S,3))$ in the construction of the preceding paragraph.

4. Equivalence of binary cubics

In this section, we consider necessary and sufficient conditions for two binary cubic forms over $K$ with the same discriminant $\Delta$ to be $\mathrm{GL}(2,K)$ -equivalent. We first observe that we need only consider transformations with determinant $\pm 1$ and reduce the problem to testing $\mathrm{SL}(2,K)$ -equivalence.

For $i=1,2$ , denote the coefficients of $g_i\in {\mathcal{BC}}(K;\Delta )$ by $a_i,b_i,c_i,d_i$ , their seminvariants by $P_i,U_i$ , and their Cardano covariants and invariants by $C_i, z_i$ . Since $\mathrm{disc}(g_1)=\mathrm{disc}(g_2)=\Delta$ , they have the same covariant algebra $L=K[\delta ]$ where $\delta ^2=-3\Delta$ .

We will give two criteria for the equivalence of a pair of binary cubic forms with the same discriminant: one in terms of their Cardano invariants and a second one in terms of a cubic bicovariant; the latter will also directly give a matrix (or matrices) $M\in \mathrm{SL}(2,K)$ transforming one to the other. Both criteria are similar to criteria (established in [Reference Cremona and Fisher4]) for the equivalence of a pair of binary quartic forms with the same invariants.

4.1. Cubic equivalence in terms of equality of Cardano invariants

We restate parts (3) and (4) of Theorem1.

Theorem 8. Let $g_1,g_2\in {\mathcal{BC}}(K;\Delta )$ , with common resolvent algebra $L=K[\delta ]$ . Then

1. (1) $g_2=g_1^M$ with $\det (M)=+1$ if and only if $z(g_1)=z(g_2)$ in $L^*/L^*{}^3$ ;

2. (2) $g_2=g_1^M$ with $\det (M)=-1$ if and only if $z(g_1)=z(g_2)^{-1}$ in $L^*/L^*{}^3$ .

Hence $g_1$ and $g_2$ are $\mathrm{GL}(2,K)$ -equivalent if and only if $z(g_1)$ and $z(g_2)$ generate the same subgroup of $L^*/L^*{}^3$ .

Proof. (1) Suppose that $g_2=g_1^M$ with $M\in \mathrm{SL}(2,K)$ . Then $g_2(X_2,Y_2) = g_1(X_1,Y_1)$ , where $(X_1\ Y_1) = (X_2\ Y_2)M$ , and by the covariant property of $C_1$ we have that

\begin{equation*} C_2(X_2,Y_2) = C_1^M(X_2,Y_2) = C_1(X_1,Y_1). \end{equation*}

It is now clear that the unit values taken by $C_1$ and $C_2$ are the same. Explicitly, we similarly have $H_2(X_2,Y_2)=H_1^M(X_2,Y_2)=H_1(X_1,Y_1)$ ; let $x_2,y_2\in K$ be such that $H_2(x_2,y_2)\not =0$ , and set $(x_1\ y_1)=(x_2\ y_2)M$ , so then also $H_1(x_1,y_1)=H_2(x_2,y_2)\not =0$ . (This will be true for all choices of $(x_2,y_2)\not =(0,0)$ when $-3\Delta$ is not a square, and for all but at most two choices, up to scaling, when $-3\Delta$ is a square in $K.)$ Then $C_1(x_1,y_1) = C_2(x_2,y_2) \in L^*$ , and so $z(g_1)=z(g_2)$ .

For the converse, suppose that $z(g_1)=z(g_2)$ . If $z(g_1)=z(g_2)=1$ , then by Proposition 5, both cubics have linear factors, and then Proposition 6 shows that both are $\mathrm{SL}(2,K)$ -equivalent to $Y(X^2 - \frac {1}{4}\Delta Y^2)$ , and hence to each other.

Now we may assume that neither cubic has a linear factor; in particular, $a_1,a_2\not =0$ . Then also $U_1,{U_2\not =0}$ , since otherwise $g_1(-b_1,3a_1)=a_1U_1=0$ and $g_1(X,Y)$ would be divisible by $3a_1X+b_1Y$ ; similarly for $g_2$ . By taking suitable $\mathrm{SL}(2,K)$ -transforms of $g_1$ and $g_2$ if necessary, we can assume that $P_1,P_2\not =0$ also. Then we may take as representatives of the Cardano covariants the elements $z_i=(U_i+3\delta a_i)/2 \in L^*$ for $i=1,2$ , with $\mathrm{N}_{L/K}(z_i)=P_i^3$ .

Since $z_i\overline {z_i} = \mathrm{N}_{L/K}(z_i) = P_i^3$ , the condition that $z_2/z_1 \in L^*{}^3$ is equivalent to $\overline {z_1}z_2 \in L^*{}^3$ . A computation similar to that in the proof of Proposition 5 shows that $\overline {z_1}z_2 =(x+y\delta )^3$ with $\mathrm{N}_{L/K}(x+y\delta )=P_1P_2$ implies that $x$ is a root of $f(X)$ , where

\begin{equation*} f(X) = 16X^3 - 12P_1P_2X - (U_1U_2+27a_1a_2\Delta ). \end{equation*}

We now reduce to the case $a_2U_1-a_1U_2=0$ . If this does not hold, set $B(X,Y) = U_1g_2(X,Y)-a_1G_2(X,Y) = aX^3+bX^2Y+cXY^2+dY^3$ , where $a=a_2U_1-a_1U_2\not =0$ . One may check that $B(4P_1X+b,-3a) = 4aP_1^3 \,f(X)$ . Hence, since $z(g_1)=z(g_2)$ implies that $f(X)$ has a root in $K$ , it follows that $B$ has a linear factor over $K$ . After an $\mathrm{SL}(2,K)$ -transformation of $g_2$ , we can take this linear factor to $Y$ , so that $a_2U_1-a_1U_2=0$ . This transform does not change $P_1$ , so we still have $P_1\not =0$ .

Assuming that $a_2U_1-a_1U_2=0$ , let $\lambda = a_2/a_1 = U_2/U_1 \in K^*$ . The seminvariant syzygy then gives $P_2^3=\lambda ^2 P_1^3$ . Since $P_1\not =0$ , then also $P_2\not =0$ , and we may set $\mu =\lambda P_1/P_2$ , so that $\lambda =\mu ^3$ and $P_2/P_1=\mu ^2$ . Let $g_3(X,Y)=g_1(\mu X,\mu ^{-1}Y)$ , the transform of $g_1$ by $\mathrm{diag}(\mu, \mu ^{-1})\in \mathrm{SL}(2,K)$ ; then the seminvariants $a_3,P_3,U_3$ of $g_3$ are the same as those of $g_2$ , and we may check that $g_2=g_3^M$ with $M=\begin{pmatrix}1&&0\\(b_2-b_3)/3a_3&&1\end{pmatrix}\in \mathrm{SL}(2,K)$ . Hence $g_1$ and $g_2$ are $\mathrm{SL}_2(K)$ -equivalent.

(2) Every $M\in \mathrm{GL}(2,K)$ with $\det (M)=-1$ can be written as $M=DM_1$ , where $M_1\in \mathrm{SL}(2,K)$ and $D=\mathrm{diag}(-1,1)$ . By part (1), it suffices (for both implications) to show that $z(\tilde {g})=z(g_1)^{-1}$ , where $\tilde {g} = g_1^D$ . We have $\tilde {g}(X,Y) = g_1(X,-Y)$ , which has covariants $\tilde {H}(X,Y)=H_1(X,-Y)$ and $\tilde {G}(X,Y)=-G_1(X,-Y)$ , so its Cardano covariant $\tilde {C}$ satisfies

\begin{equation*} \tilde {C}(X,-Y) = -\overline {C_1}(X,Y), \end{equation*}

from which the syzygy implies

\begin{equation*} C_1(X,Y) \tilde {C}(X,-Y) = (-H_1(X,Y))^3. \end{equation*}

Hence $z(\tilde {g}) = z(g_1)^{-1}$ as required.

The last part is now clear, using Proposition 7(1).

One consequence of this result, combined with Proposition 5, is that two irreducible cubics cannot be equivalent via matrices of both determinants, $+1$ and $-1$ ; equivalently, only reducible cubics can have automorphisms with determinant $-1$ . We will return to automorphisms in Section 5.

From the preceding proof, we may extract the following criterion for $\mathrm{SL}(2,K)$ -equivalence for cubic forms with no linear factor, complementing Proposition 6 in the reducible case.

Corollary 9. Let $g_1,g_2\in {\mathcal{BC}}(K;\Delta )$ be irreducible. Then $g_2=g_1^M$ with $M\in \mathrm{SL}(2,K)$ if and only if the polynomial $f(X) = 16X^3 - 12P_1P_2X - (U_1U_2+27a_1a_2\Delta )$ has a root in $K$ .

To complete the proof of Theorem1, we only need to show that the Cardano invariant map is surjective.

Proposition 10. Every $z \in (L^*/L^*{}^3)_{N=1}$ arises as $z(g)$ for some $g\in {\mathcal{BC}}(K;\Delta )$ .

Proof. Let $z=x+y\delta \in L^*$ be such that $\mathrm{N}_{L/K}(z)=x^2+3\Delta y^2=P^3$ with $P\in K^*$ . Let $g\in {\mathcal{BC}}(K)$ have coefficients $(2y/3,0,-P/2y,x/6y^2)$ if $y\not =0$ or $(0,x/P,0,-\Delta /4x)$ if $y=0$ . In each case, one may check that $\mathrm{disc}(g)=\Delta$ and that $U(g)=2x$ and $P(g)=P$ , so that $z(g)=z$ as required.

4.2. Cubic equivalence in terms of factors of a cubic bicovariant

In the proof of Theorem8, we saw two conditions equivalent to the $\mathrm{SL}(2,K)$ -equivalence of the cubics $g_1$ and $g_2$ : that a monic cubic (denoted $f(X)$ in the proof) had a root in $K$ , or that a third cubic form $B(X,Y)$ had a linear factor over $K$ . However, we only established these conditions under certain extra conditions, such as the irreducibility of the $g_i$ . The cubic form $B(X,Y)$ used in the proof is the specialization at $(X_1,Y_1,X_2,Y_2) = (1,0,X,Y)$ of $g_2(X_2,Y_2)G_1(X_1,Y_1)-G_2(X_2,Y_2)g_1(X_1,Y_1)$ . We now show how linear factors of this “bicovariant” directly give matrices $M\in \mathrm{SL}(2,K)$ such that $g_1=g_2^M$ (of which there are at most three), in all cases. A twist of this bicovariant similarly produces matrices (again, at most three) of determinant $-1$ transforming $g_1$ into $g_2$ .

We continue to use the earlier notation, where $g_1,g_2\in {\mathcal{BC}}(K; \Delta )$ have covariants $H_i$ and $G_i$ . To the pair $(g_1,g_2)$ we associate a bicovariant in $K[X_1,Y_1,X_2,Y_2]$ :

\begin{equation*} B_{g_1,g_2}(X_1,Y_1,X_2,Y_2) = g_2(X_2,Y_2)G_1(X_1,Y_1) - G_2(X_2,Y_2)g_1(X_1,Y_1). \end{equation*}

This is a bihomogeneous polynomial of bidegree $(3,3)$ ; that is, it is homogeneous of degree $3$ in each pair of variables separately. It is also a bicovariant with respect to $\mathrm{SL}(2,K)\times \mathrm{SL}(2,K)$ , meaning that, for $M_1,M_2\in \mathrm{SL}(2,K)$ , we have

\begin{align*} B_{g_1^{M_1},g_2^{M_2}}(X_1,Y_1,X_2,Y_2) &= B_{g_1,g_2}(X_1,Y_1,X_2,Y_2)^{(M_1,M_2)} \\&= B_{g_1,g_2}(X_1',Y_1',X_2',Y_2') \end{align*}

where $(X_i'\ Y_i') = (X_i\ Y_i)M_i$ for $i=1,2$ .

Lemma 11. $B_{g_1,g_2}$ is divisible by $X_1Y_2-X_2Y_1$ if and only if $g_2=\pm g_1$ .

Proof. Clearly, $B_{g,g}$ is divisible by $X_1Y_2-X_2Y_1$ , and so is $B_{g,-g}=-B_{g,g}$ .

Conversely, if $B_{g_1,g_2}$ is divisible by $X_1Y_2-X_2Y_1$ then $B_{g_1,g_2}(X,Y,X,Y)=0$ , so $g_2(X,Y)G_1(X,Y)=g_1(X,Y)G_2(X,Y)$ . By the coprimality of $g_i$ and $G_i$ for $i=1,2$ , this implies that $g_2=\lambda g_1$ and $G_2=\lambda G_1$ for some $\lambda \in K^*$ ; but the first of these equations implies (using covariance of the $G_i$ ) that $G_2=\lambda ^3 G_1$ ; so $\lambda ^2=1$ .

We now consider bilinear factors of $B_{g_1,g_2}$ : factors of degree $(1,1)$ , of the form

\begin{equation*} L_M = (X_1\ Y_1)M\begin{pmatrix}0&&1\\-1&&0\end{pmatrix} \begin{pmatrix}X_2\\Y_2\end{pmatrix} = -s X_1X_2 + r X_1Y_2 - u Y_1X_2 + t Y_1Y_2, \end{equation*}

where $M=\begin{pmatrix}r&&s\\t&&u\end{pmatrix}$ is a nonzero matrix. The previous lemma concerned the factor $L_I$ associated with the identity matrix $I$ .

Lemma 12. $L_M$ is irreducible if and only if $\det (M)\not =0$ .

Proof. $(aX_1+bY_1)(cX_2+dY_2) = L_M$ with $M=\begin{pmatrix}a\\b\end{pmatrix} \begin{pmatrix}d&-c \end{pmatrix}$ , so $M$ has rank $1$ if and only if $L_M$ is reducible.

Lemma 13. If $L_M$ divides $B_{g_1,g_2}$ for $g_1,g_2\in {\mathcal{BC}}(K;\Delta )$ , then $\det (M)\in K^*{}^2$ and $g_2^M=\pm \det (M)^{1/2}g_1$ .

Proof. Suppose that $L_M$ divides $B_{g_1,g_2}$ . If $\det (M)=0$ , then by the previous lemma, $B_{g_1,g_2}$ is divisible by $(aX_1+bY_1)$ for some $a,b\in K$ not both zero. Then we have $B_{g_1,g_2}(b,-a,X_2,Y_2) = 0$ identically, so $g_2(X_2,Y_2)G_1(b,-a) = G_2(X_2,Y_2)g_1(b,-a)$ . Since $g_1$ and $G_1$ are coprime, $g_1(b,-a)$ and $G_1(b,-a)$ are not both zero, so this equation contradicts the coprimality of $g_2$ and $G_2$ .

Let $\mu =\det (M)$ . Working over $\overline {K}$ , we have $M = \mu ^{1/2}M_1$ with $M_1\in \mathrm{SL}(2,\overline {K})$ . Then $L_M=\mu ^{1/2}L_{M_1}$ , so $B_{g_1,g_2}$ is also divisible (over $\overline {K}$ ) by $L_{M_1}$ . Hence, $B_{g_1^{M_1^{-1}},g_2} = B_{g_1,g_2}^{M_1^{-1},I}$ is divisible by $L_{M_1}^{M_1^{-1},I} = L_I = X_1Y_2-X_2Y_1$ (over $\overline {K}$ ), so $g_1=\pm g_2^{M_1}$ by Lemma 11. Hence, $g_2^M=\pm \mu ^{1/2}g_1$ , which implies that $\mu ^{1/2}\in K^*$ .

Proposition 14. For $g_1,g_2\in {\mathcal{BC}}(K;\Delta )$ :

1. (1) Every bilinear factor of $B_{g_1,g_2}(X_1,Y_1,X_2,Y_2)$ in $K[X_1,Y_1,X_2,Y_2]$ is (up to scaling) of the form $L_M$ with $M\in \mathrm{SL}(2,K)$ satisfying $g_1=g_2^M$ .

2. (2) Every bilinear factor of

   \begin{equation*} g_2(X_2,Y_2)G_1(X_1,Y_1) + G_2(X_2,Y_2)g_1(X_1,Y_1). \end{equation*}
   in $K[X_1,Y_1,X_2,Y_2]$ is (up to scaling) of the form $s X_1X_2 + r X_1Y_2 - u Y_1X_2 - t Y_1Y_2$ , where $M=\begin{pmatrix}r&&s\\t&&u\end{pmatrix}$ satisfies $g_1=g_2^M$ and $\det (M)=-1$ .

3. (3) There are at most six $M\in \mathrm{GL}(2,K)$ such that $g_1=g_2^M$ , one for each bilinear factor in $K[X_1,Y_1,X_2,Y_2]$ of

   \begin{equation*} g_2(X_2,Y_2)^2G_1(X_1,Y_1)^2 - G_2(X_2,Y_2)^2g_1(X_1,Y_1)^2 \end{equation*}
   (which is equal to $g_2(X_2,Y_2)^2H_1(X_1,Y_1)^3 - H_2(X_2,Y_2)^3g_1(X_1,Y_1)^2$ ).

Proof. (1) By the lemma, we may scale each linear factor $L_M$ so that $\det (M)=1$ .

(2) Replace $g_1(X_1,Y_1)$ by $g_1(X_1,-Y_1)$ and $G_1(X_1,Y_1)$ by $G_1(X_1,-Y_1)$ , which is minus the cubic covariant of $g_1(X_1,-Y_1)$ , and apply (1).

(3) Immediate from (1) and (2); for the second expression, apply the covariant syzygy.

All that remains to establish Theorem2 is to check that every transform from $g_2$ to $g_1$ arises from bilinear factors in this way.

Theorem 15. Let $g_1,g_2\in {\mathcal{BC}}(K;\Delta )$ . Then $g_1$ and $g_2$ are $\mathrm{SL}(2,K)$ -equivalent if and only if $B_{g_1,g_2}$ has a bilinear factor in $K[X_1,Y_1,X_2,Y_2]$ , and every bilinear factor of $B_{g_1,g_2}$ has the form $L_M$ with $M\in \mathrm{SL}(2,K)$ , where $g_1=g_2^M$ .

Proof. If $g_1=g_2^M$ with $M\in \mathrm{SL}(2,K)$ then $B_{g_1,g_2}=B_{g_2,g_2}^{M,I}$ , which is divisible by $(X_1Y_1-X_2Y_1)^{M,I}=L_M$ . The converse is Corollary 13.

Remark 6. It follows from Theorem 15 and Proposition 14 that the number of matrices $M\in \mathrm{GL}(2,K)$ with $g_1=g_2^M$ is at most six, with at most three with each determinant $\pm 1$ . This number, if nonzero, is also the number of automorphisms in $\mathrm{GL}(2,K)$ of $g_1$ (or of $g_2$ ); in the next section, we study automorphisms in more detail and will see that the number of automorphisms in $\mathrm{SL}(2,K)$ is either $1$ or $3$ , the latter occurring if and only if $\Delta \in K^*{}^2$ ; hence $B_{g_1,g_2}$ splits completely into bilinear factors if and only if $g_1,g_2$ are $\mathrm{SL}(2,K)$ -equivalent and $\Delta \in K^*{}^2$ .

5. Automorphisms of cubic forms

For $g\in {\mathcal{BC}}(K)$ , the automorphism group of $g$ is the subgroup ${\mathcal{A}}(g)$ of matrices $M\in \mathrm{GL}(2,K)$ such that $g^M=g$ . Since for $M=\lambda I$ , we have $g^M=\lambda g$ , and the onlyFootnote ⁴ scalar matrix in ${\mathcal{A}}(g)$ is the identity. By Proposition 7, every $M\in {\mathcal{A}}(g)$ satisfies $\det (M)=\pm 1$ .

Each $M\in {\mathcal{A}}(g)$ acts on the roots of $g$ , viewed as lying in ${\mathbb P}^1(\overline {K})$ , by linear fractional transformations. Since $\mathrm{PGL}(2,\overline {K})$ acts faithfully and transitively on ordered triples of distinct points on the projective line, the action on the roots induces an injective homomorphism ${\mathcal{A}}(g)\longrightarrow S_3$ . Hence, ${\mathcal{A}}(g)$ has order at most $6$ . Set ${\mathcal{A}}_1(g) = {\mathcal{A}}(g)\cap \mathrm{SL}(2,K)$ . Proposition 16 below implies that the above homomorphism restricts to an injection of ${\mathcal{A}}_1(g)$ into the alternating group $A_3$ , so ${\mathcal{A}}_1(g)$ is either trivial or has order $3$ .

Also, since an automorphism $M$ has finite order, it has distinct eigenvalues either both in $K$ or in a quadratic extension, so $M$ has exactly two fixed points in ${\mathbb P}^1(\overline {K})$ .

We now determine when $g$ has automorphisms of order $2$ and $3$ , respectively.

Putting these parts together, we see that the automorphism group of a binary cubic form $g$ depends only on the Galois group of the splitting field of $g$ .

Part (2) of this result is the same as Theorem 3.1 in the work [Reference Xiao8] of Xiao, which gives the result only for $K={\mathbb Q}$ , though the proof Xiao gives is general and similar to ours. He also states (in Proposition 2.1 of [Reference Xiao8]) that ${\mathcal{A}}(g)\cong S_3$ when $K={\mathbb C}$ . Automorphisms of binary cubic forms over $\mathbb Z$ are also the subject of [Reference Neusel7], which also considers the singular case; there, the author’s motivation is to find all subgroups of $\mathrm{GL}(2,{\mathbb Z})$ whose invariant subring in ${\mathbb Z}[X,Y]$ contains a cubic form; they embed ${\mathcal{A}}(g)$ into $\mathrm{GL}(2,{\mathbb F}_3)$ by reduction modulo $3$ , and then consider all possible subgroups of that group, eventually reaching the same conclusion as here.

Part (2) also follows from the Delone-Faddeev parametrization of cubic orders by binary cubic forms from [Reference Delone and Faddeev5], whose proof goes through with their base ring $\mathbb Z$ replaced by any field: see Proposition 12 of [Reference Bhargava, Shankar and Tsimerman3].

6. Integral equivalence

Our results so far concern binary cubic forms over a field $K$ (with $\mathrm{char}(K)\not =2,3$ ) and their equivalence under the actions of $\mathrm{SL}(2,K)$ and $\mathrm{GL}(2,K)$ . In applications, one may be interested in forms with coefficients in some subring $R$ of $K$ and their orbits under $\mathrm{SL}(2,R)$ and $\mathrm{GL}(2,R)$ . One classical example is when $R={\mathbb Z}$ and $K={\mathbb Q}$ , or more generally when $K$ is a number field and $R$ its ring of integers.

Using Theorem18 of [Reference Bhargava, Elkies and Shnidman2], it would be possible to write down an equivalence test in terms of a generalized Cardano invariant, at least when $R$ is a Dedekind domain. However, it is simpler to use our second criterion. Since, $\mathrm{GL}(2,R)$ -equivalence implies $\mathrm{GL}(2,K)$ -equivalence, and similarly for $\mathrm{SL}(2)$ , it is easy to adapt our results to give a test for $\mathrm{GL}(2,R)$ - or $\mathrm{SL}(2,R)$ -equivalence. First, let $g_1,g_2$ be two cubic forms with coefficients in $R$ and the same nonzero discriminant $\Delta$ . Using Theorem15, we can find all $M\in \mathrm{SL}(2,K)$ such that $g_1=g_2^M$ , if any. If there are none, then certainly the forms are not $\mathrm{SL}(2,R)$ -equivalent; if any such $M$ exist, their number will be either $1$ or $3$ , the latter if and only if $\Delta$ is a square, and the forms are $\mathrm{SL}(2,R)$ -equivalent if and only if at least one of the matrices has entries in  $R$ . For $\mathrm{GL}(2,R)$ -equivalence, we repeat using $g_1(X,Y)$ and $g_2(X,-Y)$ .

For example, let $g_1(X,Y)=X^{\,3}-16Y^3$ and $g_2(X,Y)=g_1(2X,Y/2)=8X^3-2Y^3$ . Both have integral coefficients, and they are $\mathrm{SL}(2,{\mathbb Q})$ -equivalent via $\mathrm{diag}(2,1/2)$ . Since the common discriminant is $-2^83^3$ which is not a square, this is the unique $\mathrm{SL}(2,{\mathbb Q})$ -equivalence, so they are not $\mathrm{SL}(2,{\mathbb Z})$ -equivalent. Since the cubics are irreducible, neither has an automorphism of determinant $-1$ , hence there are no other $\mathrm{GL}(2,{\mathbb Q})$ -equivalences between them and they are therefore not $\mathrm{GL}(2,{\mathbb Z})$ -equivalent.

As a second example, let $g(X,Y)=(X-Y)(X-2Y)(X-3Y)$ . This has cubic covariant $G(X,Y)=-2Y(3X-7Y)(3X-5Y)$ . Its self-bicovariant $B_{g,g}$ has three linear factors:

\begin{align*} &Y_1X_2-X_1Y_2,\\ -3&X_1X_2+7X_1Y_2+5Y_1X_2-13Y_1Y_2,\\ -3&X_1X_2+5X_1Y_2+7Y_1X_2-13Y_1Y_2. \end{align*}

From the coefficients of the three factors we find that ${\mathcal{A}}_1(g) = \{I, M, M^2\}$ , where $M=\frac {1}{2}\begin{pmatrix}-5&&-3\\13&&7 \end{pmatrix}$ and $M^2=\frac {1}{2}\begin{pmatrix}7&&3\\-13&&-5 \end{pmatrix}$ . This shows that it is possible for two integral forms to be $\mathrm{SL}(2,{\mathbb Q})$ -equivalent via a non-integral matrix $M$ , while also being $\mathrm{SL}(2,{\mathbb Z})$ -equivalent. Hence when testing for $\mathrm{SL}(2,{\mathbb Z})$ -equivalence (and similarly for $\mathrm{GL}(2,{\mathbb Z})$ -equivalence) it is important to consider all possible $\mathrm{GL}(2,{\mathbb Q})$ -equivalences, to determine whether one is integral.

7. Applications to elliptic curves

Our work [Reference Cremona and Fisher4] with Fisher, on binary quartic forms and their equivalence, was motivated by the application to $2$ -descent on elliptic curves. Specifically, there is a bijection between orbits of binary quartic forms with classical invariants $I,J$ under a suitably twisted group action, and $2$ -covers of the elliptic curve $Y^2=X^3-27IX-27J$ , with the $2$ -covering maps being given by the syzygy between covariants of the quartic.

There is a similar application for binary cubic forms, this time to $3$ -isogeny descent on elliptic curves with $j$ -invariant $0$ . This connection is treated in detail by Bhargava et al. in [Reference Bhargava, Elkies and Shnidman2]. We briefly describe the connection here, summarizing Section 3 of [Reference Bhargava, Elkies and Shnidman2]

Let $E_k$ denote the elliptic curve with Weierstrass equation $Y^2=X^3+k$ , where $k\in K^*$ . There is a $3$ -isogeny $\phi :E_k\to E_{-27k}$ with dual $\hat {\phi }$ . Elements of the Galois cohomology group $H^1(G_K,E_{-27k}[\hat {\phi }])$ , which are locally trivial and so represent $\phi$ -Selmer elements, can be represented by cubic curves ${\mathcal{C}}_g\,:\,Z^3=g(X,Y)$ , where $g\in {\mathcal{BC}}(K;-108k)$ ; these are $\phi$ -coverings of $E_k$ . They are trivial (coming from $K$ -rational points on $E_k$ ) if and only if ${\mathcal{C}}(K)\not =\emptyset$ . The covering map ${\mathcal{C}}_g\to E_k$ is given by the covariant syzygy (1), which implies that

\begin{equation*} (x,y,z) \in {\mathcal{C}}_g(K) \quad \implies \quad \left (\frac {H(x,y)}{(3z)^2},\frac {G(x,y)}{2(3z)^3}\right )\in E_k(K). \end{equation*}

This is stated (with slightly different notation) in Remark 30 in [Reference Bhargava, Elkies and Shnidman2], the syzygy being equation (16) there. In [Reference Bhargava, Elkies and Shnidman2] the $\mathrm{GL}(2,K)$ -action is untwisted, but as the results there concern orbits under $\mathrm{SL}(2,K)$ this is immaterial: for example, [Reference Bhargava, Elkies and Shnidman2, Theorem 27] states that there is a bijection between $H^1(G_K,E_{-27k}[\hat {\phi }])$ and the set of $\mathrm{SL}(2,K)$ -orbits on ${\mathcal{BC}}(K;-108k)$ .

The untwisted $\mathrm{GL}(2)$ -action is also studied by Kulkarni and Ure in [Reference Kulkarni and Ure6], who give (under the assumption that $K$ contains the cube roots of unity) a different relation between the (untwisted) $\mathrm{GL}(2,K)$ -orbits on ${\mathcal{BC}}(K)$ and elliptic curves over $K$ with $j$ -invariant $0$ .