# Some questions regarding computation of the Mordell-Weil group

I was reading the theory relevant to Selmer and Shafarevich-Tate groups from Silverman's AEC. And I have a lot of doubts related to these topics:

1. First, I don't understand the reasoning behind the name 'homogeneous spaces'? Does it have something to do with homogeneous spaces from Lie theory?

Let's talk by 'descent via isogeny' method now. Some of the ideas are still hazy to me and I am having difficulty in tying together all the things discussed in Chapter X of the book.

Here is what I understood: We want to compute $$E(K)$$ for an arbitrary number field, $$K$$, but the whole argument in the Mordell-Weil theorem tells us that it is sufficient to compute the weak Mordell-Weil group, $$E(K)/ mE(K)$$ ($$m$$ is as given by the descent theorem and is not arbitrary). Now, using the following exact sequence: $$0 \longrightarrow \frac{E'(K)[\hat{\phi}]}{\phi(E(K)[m])} \longrightarrow \frac{E'(K)}{\phi(E(K))} \longrightarrow \frac{E(K)}{mE(K)} \longrightarrow \frac{E(K)}{\hat{\phi}(E'(K)} \longrightarrow 0$$ w see that to compute $$\frac{E(K)}{mE(K)}$$, it is sufficient to compute $$\frac{E'(K)}{\phi(E(K))}$$ and $$\frac{E(K)}{\hat{\phi}(E'(K)}$$. Further, using the exact sequence: $$0 \longrightarrow \frac{E'(K)}{\phi(E(K))} \longrightarrow S^{(\phi)}(E/K) \longrightarrow \Sha(E/K)[\phi]$$

in certain cases, like when Sha group is trivial we are able to compute to our group $$\frac{E'(K)}{\phi(E(K))}$$ (and similarly the other one and hence the group $$E(K)/mE(K)$$. Now my questions are:

1. Why do we use an isogeny $$\phi$$ with degree $$m$$ instead of directly using the isogeny 'multiplication-by-$$m$$'?
2. In general, it is not that easy to compute E(K) as the example 4.10 in Chapter X of the book shows, right? I mean in that particular example, we didn't have to go through the whole descent procedure to compute E(K) from E(K)/2E(K) but that need not happen in general, right? Also, in the same example, we picked a 2-isogeny but $$m$$ need not be $$2$$ here, right?

3. Also is it correct that 'descent via isogeny' method, that is computing the abovementioned two groups work when there is at least one $$\phi$$-torsion point that is also $$K$$-rational? (This doesn't really seem to be a very effective way to compute the weak Mordell-Weil group to me though.)

4. And as for its name (2-descent via isogeny), is it because we are reducing the problem of finding $$E(K)/mE(K)$$ to first finding two Selmer groups and then their corresponding quotient groups?

It would be great if someone here could help confirm or deny things I have said above. They will help a great deal in understanding this whole discussion of Mordell-Weil group computation.

And as always, thank you all for reading and responding to my queries!

Edit: I forgot that I was on mathoverflow and not math StackExchange. While I repost this to the math StackExchange, I think it would be okay to let it remain here as well?

• To the question at the top: the definition is exactly the same as that in Lie theory, except we require the maps to be algebraic instead of just continuous/smooth - it's a space with a free and transitive action of our elliptic curve. Mar 2, 2020 at 19:32
• The questions below: 1. The smaller the degree of the isogeny the easier it is to compute. We always have $[m]$ but if $m>10$ there is little hope to be able to compute it directly. So if there is an isogeny of prime degree (e.g. $2$ or $3$) that is defined over $\mathbb{Q}$ then use it. --- 2. Right. --- 3. Descent by isogeny can be done if the isogeny is defined over $K$ that is a weaker assumption than saying that the kernel contains a $k$-rational point. -- 4. Well, yes. Mar 2, 2020 at 19:53
• For answer 1. I don't see why direct computation of $E(K)/mE(K)$ would be difficult in case m>10, is it because then it has more elements ($m^2$) than the group $E'(K)/\phi(E(K))$ (and also the other one)does ($m$) and thus overall it's difficult to compute the bigger group? Mar 3, 2020 at 18:51