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For an elliptic curve $E$ over $\Bbb{Q}$, we know from the proof of the Mordell-Weil theorem that the weak Mordell-Weil group of $E$ is $E(\Bbb{Q})/2E(\Bbb{Q})$. It is well known that $$ 0 \rightarrow E(\Bbb{Q})/2E(\Bbb{Q}) \rightarrow S^{(2)}(E/\Bbb{Q}) \rightarrow Ш(E/\Bbb{Q})[2] \rightarrow 0 $$ is an exact sequence which gives us a procedure to compute the generators for $E(\Bbb{Q})/2E(\Bbb{Q})$.

(Relatively) recently I found out that there is another way to compute the rank of $E$ using $3$-descent. I was wondering, since the natural structure of the weak Mordell-Weil group is $E(\Bbb{Q})/2E(\Bbb{Q})$, what is the motivation behind using $3$-descent? Also does $3$-descent similarly produce the generators of $E(\Bbb{Q})/2E(\Bbb{Q})$ or does it simply tell us the structure of $E(\Bbb{Q})$ via the Mordell-Weil theorem by giving us only the rank of $E$? Finally does it help us get around the issue of $Ш(E/\Bbb{Q})$ containing an element that is infinitely $2$-divisible?

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2 Answers 2

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An $n$-descent will compute the $n$-Selmer group, which sits in a s.e.s. $$0 \to E(\mathbb Q)/n E(\mathbb Q) \to S^{(n)}(E/\mathbb Q) \to Ш(E/\mathbb Q)[n] \to 0.$$

If you do a $2$-descent, it will give an upper bound on the size of $E(\mathbb Q)/2E(\mathbb Q).$ If you do a $3$-descent, it will give you an upper bound on the size of $E(\mathbb Q)/3 E(\mathbb Q).$

The advantage of one over the other will depend on the structure as a Galois module of $E(\mathbb Q)[n]$, and the structure of the $n$-torsion in Sha.

If $E$ contains a rational $2$-torsion point, then the $2$-Selmer group may be easier to compute than the $n$-Selmer groups for other $n$.

As an example of a descent at a different choice of $n$, note that the elliptic curve $X_0(11)$ contains a rational $5$-torsion point, and Mazur does a $5$-descent to prove that it has no other rational points besides the five points generated by this $5$-torsion point. (See this answer for more details on this.) The elliptic curve $X_0(17)$ has a rational $3$-torsion point, and for this Mazur does a $3$-descent.

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There is no "natural" structure. The point is that if $E(\mathbb{Q})\cong \Delta\oplus \mathbb{Z}^r$, where $\Delta$ is a finite group, then for any positive integer $n$, $$ E(\mathbb{Q})/nE(\mathbb{Q}) \cong \Delta/n\Delta\oplus (\mathbb{Z}/n\mathbb{Z})^r.$$ Note that this is a finite group. Since we can basically assume that we know everything about $\Delta$, computing $|E(\mathbb{Q})/nE(\mathbb{Q})|$ is equivalent to computing the rank $r$.

Moreover, for any $n$, we have the short exact sequence$^1$ $$ 0 \rightarrow E(\Bbb{Q})/nE(\Bbb{Q}) \rightarrow S^{(n)}(E/\Bbb{Q}) \rightarrow Ш(E/\Bbb{Q})[n] \rightarrow 0.$$

The point is now that if you can compute the middle term, you get an upper bound on the rank. Depending on the precise shape of your elliptic curve, this term can be easier of harder to compute for various $n$. Moreover, depending on the size of the Tate-Shafarevich group, the upper bound you get might be tighter or less tight for different $n$. That's the reason people try to do $n$-descent for as many different $n$ as possible. The number 2 plays no special role here. You should read up on all this in Silverman.

To recapitulate, 3-descent produces (under ideal circumstances!) the generators of $E(\Bbb{Q})/3E(\Bbb{Q})$. But it might fail to do so, if the Tate-Shafarevich group has non-trivial 3-torsion. Then you might need to try a different $n$.

$^1$ In fact, such a sequence exists for any isogeny between two elliptic curves, but let's not go there for now. This is all nicely explained in Silverman.

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I'm confused. Doesn't the natural weak Mordell-Weil group of $E/\Bbb{Q}$ come from lemma 4.1 and theorem 3.1 (descent procedure) in section VIII of Silverman? Do we have that $h([3]P) \geq 9h(P) - C_2$ for some constant $C_2$ depending on the Weierstrass equation for $E$? –  Eugene Jun 19 '12 at 4:48
    
@Eugene: Dear Eugene, There is no single "weak Mordell--Weil group"; the weak Mordell--Weil theorem is about $E(\mathbb Q)/nE(\mathbb Q)$ for some, or any, $n$. Regarding heights, the canonical height is a quadratic function, and the naive height is approximately quadratic, and so yes, $h([n] P)$ is roughly $n^2 h(P)$ for any $n$ and any $P$. Regards, –  Matt E Jun 19 '12 at 4:54
    
@MattE Thanks for clearing up the confusion! –  Eugene Jun 19 '12 at 4:56

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