The rank of
via Mazur-Tate methods
When I was a young kid, I heard the mathematical fact that the only (positive) integer that is one more than a square and one less than a cube is . Said differently, the only integer solutions
to
are given by
. There are elementary methods to prove this, using the fact that the ring of integers of
is a unique factorization domain. However, what if we now ask for rational solutions? The equation
Let us first analyze the structure of the torsion subgroup of , namely
. The discriminant of
is
and so
has good reduction outside of
and
. One checks that
and
, and thus
. It now remains to compute the rank of
. In Silverman’s Arithmetic of Elliptic Curves, one finds an algorithm to compute the rank of an elliptic curve over
– in the case that the
-torsion is defined over
. However, the
-torsion of
is not defined over
, so the method in Silverman cannot be applied. In this article, we prove the following theorem:
Theorem 1: The rank of is equal to
.
We remark that since is non-torsion, it is enough to show that the rank of
is bounded by
. Certainly, it is true that there are computer programs such as MAGMA that can calculate the rank of
. Nonetheless, we believe that there is value in bounding the rank of
“by hand.” Reason being, the latter method requires non-trivial input about the arithmetic of a certain
-extension of
. In addition, the calculations turns out to be very explicit (computing invariants in terms of generators and relations) which is always fun!
The method of 2-descent
We will use the method of -descent to compute the rank of
. As in the proof of the weak Mordell-Weil theorem, we want to compute the dimension of the
-vector space
, and in fact for the purposes of Theorem 1, we want to show that
To this end, consider the short exact sequence of étale sheaves on
The get-out-of-jail-free card, as introduced in the Mazur-Tate article Points of order on elliptic curves, is to work integrally. The reason this is great is because for a number field
, the group of non-squares
is infinite-dimensional, but
is not, thanks to Dirichlet’s unit theorem. In view of this, we will now modify our approach above as follows. First, observe that the elliptic curve
has bad reduction at
and
and nowhere else. Therefore, we may extend
to an elliptic scheme
over
. In simple terms,
is simply the vanishing locus of the same equation for
in
, since
is already defined integrally. The more advanced reader may note that
is also the Néron model of
over
, since any abelian scheme over a Dedekind base is the Néron model of its generic fiber.
Now by the valuative criteria for properness,
Computing with Galois cohomology
Preliminaries
Let denote the splitting field of
. It is a basic exercise in Galois theory that
, where
is a primitive third root of unity. By definition of the group law on an elliptic curve, the
-torsion on
is precisely the zero locus of
, and therefore
In addition, we make the following important observation. Let be the set of primes in
lying over
and
. It is known (e.g. by Keith Conrad’s article here) that
where
. Consequently
is finite étale, where
We leave it as an exercise for the reader to verify that splits over
.
Kummer Theory
Since is isomorphic to
over
, it seems only natural to pass to this extension. Even better, we now have the Kummer sequence at our disposal, which as we will see makes everything very explicit. To this end, recall that
is Galois with Galois group
. Therefore, we have a Hochschild-Serre spectral sequence
Lemma 1:
Proof: Do inflation-restriction with the normal subgroup .
Now recall that our goal is to show that the dimension of (as an
-vector space) is bounded by
. By (9) and Lemma 1, we have an injection
Let us spell out the right side of (10) without the Galois invariants. We know that the -torsion
is abstractly isomorphic to
. Therefore, the calculation of
reduces to one using the Kummer sequence
Note this is exact on the étale site of , precisely because
is an
-unit.
Now pass to the long exact sequence in cohomology. We get a short exact sequence
Great, so we now want to take the Galois invariants of . But how do we do this precisely ? It is tempting to think that the Galois action on
is coordinate-wise given by the usual Galois action on
. However, this is false because the Galois action on
is “twisted,” in the sense that it comes from the Galois action on
.
The Galois action on units mod squares
Recall that where
is a primitive third root of unity. The Galois group of
(always denoted
) is abstractly isomorphic to
, with explicit generators given by
This is summarized in the two propositions below:
Proposition 1: The group of units mod squares is a finite-dimensional
-vector space with basis given by
Proposition 2: For , we have
The reader may refer to the expanded version of this article here for proofs of these propositions.
Proof of Theorem 1
By (8) and (10), we have injections
Now suppose that is invariant under
and
. Then the following relations must hold:
We’re nearly there. Consider the third relation . The product
corresponds to adding the vectors
and
. But
and therefore