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 equationis of course an elliptic curve over . Therefore, this more difficult question is equivalent to asking for the structure of the Mordell-Weil group of .
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 onThis gives an injection Now we run into the following problem: The Galois cohomology group is often infinite-dimensional. To illustrate this, let us suppose for the moment that the -torsion of is defined over . Then , and therefore by the Kummer sequence which is very infinite-dimensional.
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,and therefore it suffices to show that Furthermore, by considering the exact sequence arising from multiplication by on , we obtain (as in the case for ) an injection
Computing with Galois cohomology
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 thereforewhere the last isomorphism is non-canonical, i.e. depends on a choice of basis.
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, whereFurthermore, the Galois group acts on the ring , and we claim that in fact is equal to the full automorphism group of over , i.e. is a Galois cover with Galois group . Indeed, this follows from the fact that any automorphism of the generic fiber preserves , since .
We leave it as an exercise for the reader to verify that splits over .
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 sequenceThe low degree terms of this spectral sequence give rise to an exact sequence
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 injection10).
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 sequenceBut because it receives a surjection from which is also zero since has class number . In summary, we have shown that as abstract abelian groups,
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 bysatisfying the relations , . For any , we want to give an explicit description of and via the isomorphism (11). To write down such an action explicitly, it makes sense intuitively that we must also compute the (usual) Galois action on the finite-dimensional -vector space , where by “usual” we mean the one coming from the Galois action on .
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 byHere satisfies and is the fundamental unit of . Furthermore, the restriction of the Galois action on to is given in terms of this basis as follows: For , we have
Proposition 2: For , we havewhere by , etc we mean the -action on given by Proposition 1.
The reader may refer to the expanded version of this article here for proofs of these propositions.
Proof of Theorem 1
Now suppose that is invariant under and . Then the following relations must hold:The first relation says which implies that In other words,14), this reads Comparing coefficients, we obtain and therefore as well. Now we summarize what we have deduced about and so far:
We’re nearly there. Consider the third relation . The product corresponds to adding the vectors and . But and thereforeThis must equal , which in terms of , says In other words, . The final relation does not yield any extra information since . In summary, we have proven that , i.e. is spanned by . This completes the proof of Theorem 1.