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In his solution of the equation $x^3 + dy^3 = 1$, Nagell comes across the equation $$ u^3 + 6u^2v + 3uv^2 - v^3 = w^3. $$ He then observes that $$ (u^3 + 6u^2v + 3uv^2 - v^3) U^3 = V^3 + W^3 $$ for $$ U = u^2+uv+v^2, \quad V = u^3+3u^2v-v^3, \quad W = 3u^2v+3uv^2, $$ and then appeals to Fermat's Last Theorem for the exponent $3$.

My question: what is known about representing a binary cubic form $$ au^3 + bu^2v + cuv^2 + dv^3 $$ as a sum of two rational cubes in the polynomial ring ${\mathbb Z}[u,v]$? Where is Nagell's solution coming from?

P.S. Dehomogenizing by setting $v = 1$ gives us an elliptic surface. Still I wonder how Nagell dreamed up his solution.

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up vote 13 down vote accepted

This seems to be two separate questions, one for writing a given cubic form $C(u,v)$ as the sum of two polynomial cubes, the other for a sum of two rational cubes of given form.

In the polynomial case we want $C=x^3+y^3$ with $x,y$ linear. Assume we're not in characteristic $3$ (maybe I also have to exclude characteristic $2$). Then over an algebraically closed field it's not hard to see what's going on because ${\rm GL}_2$ acts simply transitively on ordered pairs $(x,y)$ of independent linear forms (a.k.a. bases for a 2-dimensional space). The stabilizer of $x^3+y^3$ is just the obvious $18$-element group: permute $\{x,y\}$ and multiply each by a cube root of unity. So, given $C$, we can recover $x,y$ as follows: locate the roots of $C$ on the projective $(u:v)$ line, then apply a projective linear transformation to map them to the cube roots of unity; this yields linear forms $x,y$ such that $x^3-y^3$ is a multiple of $C$, and then scaling them lets you match $C$ exactly. Over a field like ${\bf Q}$ you must then check for rationality. If $C$ has one double root and one simple root then it cannot be written as $x^3+y^3$ (not even in ${\bf C}(u,v)$, because over that field $x^3+y^3=C$ is equivalent with the curve $x^3+y^3=t$ over ${\bf C}(t)$ which has rank zero); while if $C$ has a triple root it is equivalent to $cu^3$ and then it's "just" a matter of writing the scalar $c$ as a sum of two cubes.

But this does not immediately explain Nagell's formulas, in which $x=V/U$ and $y=W/U$. One way to describe the situation is as follows. Let $C$ be a homogeneous cubic form with distinct roots, and thus of nonzero discriminant $\Delta$; and denote by $S_C$ the plane curve $w^3=C(u,v)$. Then $S_C$ is smooth, and thus of genus $1$. Let $J_C$ be its Jacobian. Since $S_C$ has a $3$-cycle with fixed points (multiply $w$ by a cube root of unity), so does $J_C$. Hence $J_C$ has $j$-invariant zero and can be written as $y^2 = x^3 + a_6$ for some $a_6$, determined up to multiplication by a sixth power; if I did this right we can take $a_6 = -27\Delta/4$. Now in general $y^2 = x^3 + a_6$ is $3$-isogenous with $y^2 = x^3 - 27 a_6$. Here the degree-9 map $S_C \rightarrow J_C$ factors through the $3$-isogeny between $J_C$ and the curve $y^2 = x^3 + \Delta/4$, call it $J'_C$. So we get a degree-$3$ map $S_C \rightarrow J'_C$. When $\Delta$ is a square, say $\Delta = c^2/4$, the target curve $J'_C$ can also be written as $x^3 + y^3 = c$. In Nagell's case, $\Delta = 729$, so $c=27$, and $x^3 + y^3 = 27$ is isomorphic with $x^3 + y^3 = 1$; Nagell gives in effect the map $S_C \rightarrow J'_C$ composed with this isomorphism. Another nice feature of this case $\Delta=c^2$ is that the $3$-cycles in ${\rm PGL}_2$ that cyclically permute the roots of $C$ are defined over the ground field; in Nagell's case these are the cyclic permutations of $u$, $v$, and $-(u+v)$. Note that $\{U\}$ and $\{V,W\}$ are bases for the spaces of homogeneous quadrics and cubics invariant under this $3$-cycle. Perhaps Nagell found them first and, looking for identities among these invariants, noticed that $U^3$ divides $V^3+W^3$, "etc.". BTW the recipe in the previous paragraph doesn't write this $C$ as the sum of two linear polynomials over ${\bf Q}$, because while the $3$-cycle is defined over $\bf Q$ the roots of $C$ are defined only over a cyclic cubic field, namely ${\bf Q}(2 \cos 2\pi/9)$ [a.k.a. the real subfield of the ninth cyclotomic field].

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Noam, is there any way I could talk you into writing a book on Diophantine Analysis? – Franz Lemmermeyer Aug 9 '11 at 17:35

A paper by Bruce Reznick and Jeremy Rouse, preprint available at completely solves the equation $x^3+y^3=f^3+g^3$, where
$f$ and $g$ are homogeneous rational functions in $x$ and $y$.

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