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



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


