What can one say about the order of the torsion group of an elliptic curve defined over the compositum of all quadratic extensions of $\mathbb{Q}$ ?
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3 Answers
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I think the papers you should have a look at are those two by Fujita, great material and very well-written (in my humble opinion):
1) Y. Fujita, Torsion subgroups of elliptic curves with non-cyclic torsion over ${\mathbb Q}$ in elementary abelian 2-extensions of ${\mathbb Q}$, Acta Arith. 115 (2004) 29–45. MR2102804 (2005j:11041)
2) Fujita, Y.: Torsion subgroups of elliptic curves in elementary abelian $2$-extensions of ${\mathbb Q}$. J. Number Theory, 114, 124–134 (2005) MR2163908 (2006h:11055)
Hope this helps.
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1$\begingroup$ Is there an easier method to prove that order of the group will be finite ?? $\endgroup$– SumanCommented Apr 29, 2013 at 10:25
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1$\begingroup$ The first proof I know of this fact goes back to Laska & Lorenz (J. Reine Angew. Math. 355 (1985) 163-172). They proved that there were at most 31 possible torsion subgroups, and then Fujita showed that there were in fact only 20. Unfortunately, I don't have the paper around right now and I can't recall if their proof of finiteness is easy or not (in fact they might even refer to a previous paper, you know how this goes...). $\endgroup$ Commented May 1, 2013 at 7:24
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$\begingroup$ @Suman First of all, I assume since you accepted this answer that you meant in your original question that the elliptic curves should be defined over $\mathbb{Q}$, and you want to know about $E(K)_tors$ for such curves, where $K$ is the compositum of all quadratic fields. The fact that the curves are defined over $\mathbb{Q}$ is of course crucial in the Fujita & Laska-Lorenz results mentioned above. Secondly, I have an easy way to see that the order of the group is finite. It's maybe too long for a comment here, so see my "answer" below. $\endgroup$ Commented Apr 30, 2015 at 18:46
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Mazur's theorem is not needed here, and besides, it applies only to elliptic curves themselves defined over $\mathbb{Q}$ rather than the said compositum $K^{(2)}$ of all quadratic extensions of $\mathbb{Q}$ (which is how I understand the question).
A simple proof is available by purely local methods. First, if $F/\mathbb{Q}_p$ is a finite extension, then any abelian variety $A/F$ has $|A(F)_{\mathrm{tors}}| < \infty$. This follows from the theory of the formal group; for the elliptic curve case, see Chapter IV, Theorem 6.4 in Silverman's book (supplemented by Ch. VII Ex. 7.6 as noted by René).
Given this, fix any prime. Since $\mathbb{Q}_p$ has just finitely many quadratic extensions, it is possible to embed $K^{(2)}$ in a finite extension $F/\mathbb{Q}_p$. Then the stated finiteness follows from the above quoted result.
answered Apr 30, 2015 at 21:42
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1$\begingroup$ One also needs the fact that the kernel of reduction $A_1(F)$ is of finite index in $A(F)$. (Additionally, for small $p$, one may need to consider kernels of reduction modulo higher powers of $p$ in order to be in a position to deploy the theory of formal groups, but these are readily seen to be of finite index in $A_1(F)$.) This follows essentially from local compactness of $F$; this is the approach suggested in Chapter VII, Exercise 7.6 in Silverman's book. $\endgroup$– R.P.Commented Apr 30, 2015 at 22:22
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$\begingroup$ @René: Thank you, indeed, I negleced to give this reference in addition to Theorem II 6.4. Another proof is also possible using height functions, and the same observation that $K^{(2)}$ has bounded $p$-adic degrees. I outlined it in my initial answer to this question, but I then replaced it by this simpler local argument. $\endgroup$ Commented Apr 30, 2015 at 22:43
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1$\begingroup$ @BobbyGrizzard: Yes, the torsion group is finite for every elliptic curve (or abelian variety) over $\mathbb{Q}^{(d)}$. However, its size is not uniformly bounded unless the elliptic curve is defined over $\mathbb{Q}$ (or over a number field of a bounded global degree), in which case, as you note, Mazur's theorem applies to give a uniform bound. $\endgroup$ Commented May 1, 2015 at 4:07
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1$\begingroup$ @BobbyGrizzard: Indeed, it has to be the chapter on the formal group. Will edit to correct. $\endgroup$ Commented May 1, 2015 at 4:27
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1$\begingroup$ @BobbyGrizzard: What you say about $\mathbb{G}_m$ remains true for the Neron-Tate height on an elliptic curve. This in particular implies the finiteness statement here. But what I'd first written here (using heights) was something simpler and much cruder. You can look at my initial answer by clicking on the edits; I believe this is also what René had in mind. $\endgroup$ Commented May 1, 2015 at 4:46
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Extended comment. OP Asked if there was an easy way to see that the group is finite. Indeed, if $E$ is an elliptic curve over $\mathbb{Q}$ and $K$ is a Galois extension of $\mathbb{Q}$ with only finitely many roots of unity, then $E(K)_\text{tors}$ is finite. You can't have full $n$-torsion if there is not a primitive $n^{\text{th}}$ root of unity in $K$, thanks to the Weil pairing. If you have "half" $n$-torsion, then $\operatorname{Gal}(K/\mathbb{Q})$ acts on this cyclic group of order $n$, which means there is an $n$-isogeny which is defined over $\mathbb{Q}$. For large enough $n$, this is impossible, by Mazur. (I learned this argument from Filip Najman.)
(And it's easy to see that the compositum of all quadratic fields has only finitely many roots of unity. In fact, you could count them on your fingers.)
EDIT: You may need to get creative, as comments suggest, to accomplish this feat with your fingers.
answered Apr 30, 2015 at 18:55
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2$\begingroup$ You can count to 24 on your fingers? $\endgroup$ Commented Apr 30, 2015 at 21:43
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5$\begingroup$ @WillSawin: I can count to $1023$ on my fingers -- can't you? $\endgroup$– LuciaCommented Apr 30, 2015 at 21:52
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4$\begingroup$ Depending on if you can bend your knuckles reliably you can reach $59048 = 3^{10} - 1$. $\endgroup$– AsvinCommented Apr 30, 2015 at 22:28
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1$\begingroup$ @WillSawin thanks for keeping me on my toes ;). $\endgroup$ Commented May 1, 2015 at 4:08
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1$\begingroup$ Is this mathematician's or "athlete's feat"? $\endgroup$ Commented May 23, 2015 at 14:17