Elliptic curves over $\mathbb{Q}$ with no rational torsion and $\mu$-invariant equal to 1 at $p=3$

Question

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

Answer 1

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

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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.)

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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.

Answer 2

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.