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How to find out examples over elliptic curves over $\mathbb{Q}$ with no rational torsion and $\mu$-invariant equal to 1 at $p=3$ $?$

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2 Answers 2

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EDIT: There was a problem with my original answer. Details below the bottom line.


If you have any elliptic curve $E/{\mathbb Q}$ with a point of order 3, then we have an exact sequence of Galois modules: $$ 0 \to {\mathbb Z}/3{\mathbb Z} \to E[3] \to \mu_3 \to 0. $$ Here the ${\mathbb Z}/3{\mathbb Z}$ term arises from the point of order 3 defined over ${\mathbb Q}$. The $\mu_3$ term is then forced upon us since the determinant of this Galois representation is the mod 3 cyclotomic character.

Now here is the trick. There is an isogenous curve $E'/{\mathbb Q}$ such that $$ 0 \to \mu_3 \to E'[3] \to {\mathbb Z}/3{\mathbb Z} \to 0 $$ (i.e. the two outer terms have switched!), and moreover this sequence is not split.

I'll say in a moment how to find this curve. But let me first point out that this new curve should have $\mu$-invariant equal to 1 -- this follows from Greenberg's conjecture on $\mu$-invariants. The $\mu$-invariant should be the smallest $n$ such that $E[p^n]$ has a cyclic sub which is odd and ramified. In this case, the sub of $\mu_3$ (which is odd and ramified) gives that $\mu(E')$ is at least 1. Since the sequence is not split, there is no cyclic sub of size 9 (and so conjecturally the curve has $\mu$-invariant 1). (EDIT: Nope. This isn't true. See below. I should be assuming that 3-isogeny class for $E$ has only two curves in it.)

Now, how to find this curve? Just take $E$ and mod out by that ${\mathbb Z}/3{\mathbb Z}$ sub of $E[3]$. The resulting curve then has $\mu_3$ as a sub and ${\mathbb Z}/3{\mathbb Z}$ as quotient. If this extension is non-split, we are done. If not, mod out by ${\mathbb Z}/3{\mathbb Z}$ again. And keep repeating until the extension is non-split. (If this continued indefinitely, then the Tate module of $E$ at 3 would be reducible which it is not.)


EDIT: OK. The problem is that just because the sequence $$ 0 \to \mu_3 \to E'[3] \to {\mathbb Z}/3{\mathbb Z} \to 0 $$ is not split does not mean that $E'[9]$ doesn't contain a cyclic sub of size 9. In fact, if the process described above of modding out by ${\mathbf Z}/3{\mathbf Z}$ takes two steps, then it will contain such a sub and have $\mu$-invariant strictly greater than 1.

In fact, this problem comes up in the example of conductor 19 in Jeff H's answer. In this case, there are 3 isogenous curves of conductor 19 -- 19a1, 19a2, and 19a3 as in Cremona's tables. I'll call them $E_1$, $E_2$ and $E_3$ respectively. For $E_1$ we have $E_1[3] \cong \mu_3 \times {\mathbb Z}/3{\mathbb Z}$. Modding out $E_1$ by the $\mu_3$ gives $E_3$, and for $E_3$ we have a non-split extension. $$ 0 \to {\mathbb Z}/3{\mathbb Z} \to E_3[3] \to \mu_3 \to 0. $$ Modding out $E_1$ by the ${\mathbb Z}/3{\mathbb Z}$ gives $E_2$, and for $E_2$ we have a non-split extension. $$ 0 \to \mu_3 \to E_2[3] \to {\mathbb Z}/3{\mathbb Z} \to 0. $$

So if we had started my solution above with $E = E_3$ (which is a curve with a point of order 3), we would have to first mod out by ${\mathbb Z}/3{\mathbb Z}$ and we arrive at $E_1$ which again has a point of order 3. So we again mod out by ${\mathbb Z}/3{\mathbb Z}$ and we arrive at $E_2$ which has no point of order 3. However, $E_2$ has a cyclic isogeny of degree 9 to $E_3$ whose kernel is odd and ramified. Thus $\mu(E_2)$ is at least 2 (and exactly 2 by Greenberg's conjecture). And every other curve in this isogeny class has a point of order 3.

OK. How to fix this? Well the problem is that third curve in the isogeny class. So instead, we need a curve with a point of order 3 whose 3-isogeny class has size 2. This way we are guaranteed from the start that there is no cyclic sub of size 9. After a quick peek at Cremona's tables, I see that 44a2 works. It has $\mu$-invariant equal to 1 and no 3-torsion.

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You can find many examples like this using Rob Pollack's tables of elliptic curve Iwasawa invariants: http://math.bu.edu/people/rpollack/Data/curves1-5000

Just search the table for a curve with $\mu=1$ at $p=3$ (this is relatively rare), and then check whether it has rational torsion at LMFDB.

Here's what appears to be the example with smallest conductor:

http://www.lmfdb.org/EllipticCurve/Q/19.a1

Edit: This particular example is incorrect, but this method is still probably the easiest way to find such examples.

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  • $\begingroup$ I am basically following what you have suggested for last few days but could not find many examples. 19a1 has $\mu$-invariant equal to 2 not 1 at p = 3. $\endgroup$
    – Suman
    Feb 21, 2014 at 19:02
  • $\begingroup$ Pollack's table claims that $\mu=1$ at $p=3$. $\endgroup$
    – Jeff H
    Feb 21, 2014 at 19:08
  • $\begingroup$ LMFDB follows different nomenclature than Pollack. Pollack is using Cremona numbers which are given in brackets in LMFDB. $\endgroup$
    – Suman
    Feb 21, 2014 at 19:11
  • $\begingroup$ I see now...my mistake! Anyway, this would still be my suggested approach... $\endgroup$
    – Jeff H
    Feb 21, 2014 at 19:34
  • $\begingroup$ Those tables are really old, but I was trying to compute analytic $\mu$-invariants there. These only depend on the corresponding modular form and thus only on the isogeny class. Basically, one normalizes the associated modular symbol to take on values which are integral at $p$ but not always divisible by $p$. This $\mu$-invariant should match the algebraic $\mu$-invariant of one of the curves in the isogeny class, and this might be well understood which curve in the class is being picked out, just not by me. $\endgroup$
    – sibilant
    Jul 15, 2014 at 21:37

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