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By Mordell-Weil, for any number field $K$ we have $$C(K)=\mathbb{Z}^r \times E(K)_{\mathrm{tors}}$$ As you mention, Mazur showed all the possible options for $E(\mathbb{Q})_{\mathrm{tors}}$ in his f …

You might want to also consider the geometric ("symplectic") version of the conjecture, since Mochizuki alredy has a paper outlining the relationship between Bogomolov's proof and his own IUT theory. …

The way to reconstruct the differential sheaf from the fundamental group is, of course, $$\pi_1(X)\curvearrowright X\curvearrowright \omega_X$$ But I don't think you can get to the differential shea …

The bound in Merel's solution to the Uniform Boundedness conjecture is not explicit, as it relies on Falting's work on the Mordell conjecture. I think this still is the case. But there are known expl …

No, the conjecture is still wide open for rank $r\geq 2$. The closest thing to progress is the work of Bhargava and Shankar that quantifies the rank $0$ case and shows that BSD holds for a positive p …

This was alredy answered in the comments, it is the $p$-adic completion of $\mathbb{Q}_p(p^{1/p^\infty})$ that is a perfectoid field. But here is a reference for completeness Matthias Wulkau, Revie …

The Shafarevich conjecture belongs to the broader program of Inverse Galois theory, and in that context it is just another step in that particular approach to understanding $\mathrm{Gal}(\overline{\ma …

MO is not the place to discuss the validity of preprints, but I think it is safe to say that the finitiness of the Tate-Shafarevich group for elliptic curves over $\mathbb{Q}$ is considered an open pr …

The proof of this result is worked out in detail in page 250 (theorem 7.5) of the book Eberhard Freitag, Reinhardt Kiehl "Etale Cohomology and the Weil Conjecture" You can preview that page in par …

Vojta's general statement (the "conjectured refinement of Roth's theorem", analogous to the second main theorem in Nevanlinna theory) is: Let $X$ be a smooth complete variety over $k$, let $D$ be a n …

If the rank of elliptic curves over $\mathbb{Q}$ is unbounded, there is no known reason for this to be the case. It could very well be that the set of ranks has density $0$ on the naturals, for all we …

You first statement is correct, both ranks are expected to be equal. In particular we have: $$\mathrm{rank}\,\,\mathrm{Sel}_p(E/K)=\mathrm{rank}(E/K)+\mathrm{rank}\,\,Ш(E/K)[p^\infty]$$ So if either …

This is covered at the end of Serre's book "Abelian l-adic Representations and Elliptic Curves". Basically, since $\mathrm{Gal}(\bar{K}/K)$ acts semi-simply on $V_p(E)$ (you might have to go to Lang' …

I accidentally found the answer to this question in the book "The Abel Prize. 2003-2007 The First Five Years", page 74: "A first lecture on zeta and L-functions in the setting of the theory of s …

As a concrete exampe of how much the answer depends on what particular results you are interested, take the problem of understanding the rational points on a non-singular algebraic curve over $\mathbb …