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Suppose that $E/\mathbb{Q}$. Is it the case that for any $r > 0$ there exists a finite field extension $\mathbb{Q} \subset K$ such that the rank of $E/K$ is greater than $r$?

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Yes. It is enough to know that $E/\bar{{\mathbb Q}}$ has infinite rank, then you can take any minimal subfield containing a rank $r+1$ subgroup, it will be a number field (${\mathbb Q}$ extended by finitely many generators).

Here's a sketch of a very stupid, probably overkill proof. There must be something simpler and more elegant, but until someone more knowledgeable comes, here you go.

Over an algebraically closed field, the torsion points form a subgroup $T$ isomorphic to $({\mathbb Q}/{\mathbb Z})^2$. Let $F_\infty$ be the smallest field containing these. Then points with coordinates in $\bar{{\mathbb Q}}$ but not in $F_\infty$ cannot be torsion, therefore it is enough to prove that $\bar{\mathbb Q}$ is not a finite extension of $F_\infty$.

$Gal(F_\infty/{\mathbb Q})$ acts linearly on $T$ therefore it will be isomorphic to some subgroup of $Aut({\mathbb Q}/{\mathbb Z})^2 \cong GL_2(\hat{ \mathbb Z})$ where $\hat{ \mathbb Z}$ is the inverse limit $\lim_n {\mathbb Z}/n{\mathbb Z}$.

But $Gal(\bar{\mathbb Q}/{\mathbb Q})$ is much larger than that, e.g. it has all the symmetric groups $S_n$ as quotients. Therefore $\bar{\mathbb Q}$ is not a finite extension of $F_\infty$.

[EDIT:] Suppose you have a set of linearly independent points $S$. Note that any points linearly dependent of $S$ will be in the extension of $F_\infty$ by the coordinates of points in $S$ denoted $F_\infty(S)$. Take any algebraic $x'$ outside of this field (you can do this since $\bar{\mathbb Q}$ is not a finite extension), and solve for $y'$ so that $(x',y')$ is on the curve. Adjoin this to $S$ to get a new linearly independent set $S'$. Start with $S=\emptyset$ and repeat $r+1$ times.

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Good point, thanks. I confess that I don't know how to prove that $GL_2(\hat{ \mathbb Z})$ doesn't contain symmetric groups of arbitrary order...but I believe it. –  Jonah Sinick Sep 30 '12 at 23:18
    
I would check the last paragraph. The only finite subgroups of $G_{\mathbb{Q}}$ are cyclic groups of order 2. I think what you say in the first paragraph is sufficient to answer the question. –  jspecter Oct 1 '12 at 0:27
    
I don't understand why the fact that $\overline{\mathbf{Q}}$ has infinite degree over $F_\infty$ implies that $E(\overline{\mathbf{Q}})$ has infinite rank. As you say, every element of $E(\overline{\mathbf{Q}})-E(F_\infty)$ is of infinite order...but I don't see how the argument you give shows that the are infinitely many linearly independent points in this set difference. –  Keenan Kidwell Oct 1 '12 at 2:48
    
jspecter: Of course you're right. What I really meant is $S_n$ appears as a quotient. –  Tib Oct 1 '12 at 8:59
    
Keenan Kidwell: I thought that is straightforward. I made an edit to answer that. Jonah Sinick: It is enough to prove that $F_\infty$ has no subfields with Galois group $S_n$. But the point is that the claim $\bar{\mathbb Q}=F_\infty(S)$ for finite $S$ is absurd for many reasons. (It would be Kronecker's Jugendtraum taken to the extreme.) –  Tib Oct 1 '12 at 9:13
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Well, here's an idea that may or may not work. For the curve $y^2 = x^3 + ax + b$, take a large (but finite) set of integer (or even algebraic) values for $x$, and for each $x$, solve for $y$ to produce a point on the curve. All of these points will lie in some finite extension of $\mathbb{Q}$. Now, the question is, can one prove that enough of these points will be linearly independent? It certainly seems like this would be the case, if the $x$'s are chosen generically enough.

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