Let's investigate the conjecture on elliptic periods.

We want to find a counter-example, i.e., we want to find a curve over $\mathbb{F}_p$ such that a subgroup of order $ℓ$ is defined over an extension of degree $r|(ℓ-1)$, and such that the periods of that subgroup fall in a field smaller than $\mathbb{F}_{p^r}$.

To avoid pitfalls, we prefer having $r>2$, $ℓ≠p$, and maybe even $r≠p$. Since we have to mod-out the automorphisms of the curve, the smallest possible $ℓ$ to exhibit a counter-example is $13$, so that $(ℤ/13ℤ)^*/\{±1\}=\mathcal{C}_2\times\mathcal{C}_3$.

So, let's set $ℓ=13$, $r=3$. The elliptic period is defined as $x(P)+x([5]P)$ for $P$ a point of order 13 in a well chosen subgroup.

We look for a curve that has $13$-torsion points in an extension of degree $3$.

Here's an idea to generate tons of examples: take a curve over $ℚ$ with this property, and reduce modulo many primes. We are lucky: there are infinitely many such curves (http://arxiv.org/pdf/1211.2188.pdf, prop 31), and one such example is curve 147b1 in Cremona's database.

The curve is singular at $3$ and $7$

And certainly supersingular at $2$

It's $13$-division polynomial factors as expected

Let's look at multiplication by $5$. It's action on the abscissa is given by a rational fraction in $x$.

As we expect, multiplication by $5$ sends the abscissas of one factor to the other. We can verify this with a resultant.

By the same resultant trick, we get the minimal polynomial of the elliptic period. Remember: the period is defined as $x(P)+x([5]P)$.

Quite big coefficients, but let's look at the leading one

Ahah! Only the singular prime appears in it! Of course not a coincidence.

Let's simplify

Surprisingly short!

Just to be sure, let's check that this does not depend on the choice of the initial point.

So, now we look for a prime $p≠2,3,7$ such that the $13$-torsion points live in $\mathbb{F}_{p^3}$, but the period lives in $\mathbb{F}_p$.

We can read this directly on the polynomials.

Bad luck! Or maybe not.

One thing is obvious: if the first two split, the third splits as well. The opposite is less obvious. Let's work it out.

Call $P_1$ and $P_2$ the two factors of $φ_{13}$ of degree $3$. We want to prove that if $P_1$ (and $P_2$) is irreducible, then the minimal polynomial of the period is irreducible too. Let's suppose this is not the case and see what happens.

The first component of the multiplication-by-$5$ defines a rational mapping $I(x) = ψ_5(x)/φ_5^2(x)$. Since the only poles of $I(x)$ are in $E[5]$, its restriction to $V(P_1P_2)$ defines a polynomial mapping $i(x) = I(x) \bmod P_1P_2$, sending the roots of $P_1$ onto the roots of $P_2$ and vice-versa. This is obvious from the properties of the resultant, but let's double check.

The mapping $x + i(x)$ sends the roots of $P_1$ and the roots of $P_2$ onto the ellptic periods. If we suppose that $\mathrm{Res}_x(P_1(x), y - x - i(x)) = \mathrm{Res}_x(P_2(x), y - x - i(x))$ splits over $\mathbb{F}_p$, then by Galois invariance $x + i(x) = t \bmod P_1$ for a constant $t∈\mathbb{F}_p$, and since the periods do not depend on the initial choice of $P_1$ or $P_2$, we conclude that $i(x) = (t - x) \bmod P_1P_2$.

Hence, we can just compute $x + i(x) \bmod P_1P_2$ and see for what primes it is a constant.

No suprise: $7$ is the only divisor in the denominators.

And bad news: the only prime that divides all coefficients is $2$, but we had ruled this one out before. So no reduction of $E$ is ever going to yield a counter-example.

Just for completeness, let's do the same check using norms instead of traces

At least now we have a criterion to do a brute-force search relatively fast: look for a curve $E/\mathbb{F}_p$ such that its $13$-division polynomial has two factors of degree $3$, and such that multiplication-by-$5$ modulo these two factors equals $t-x$ for some constant $t∈\mathbb{F}_p$.

This criterion easily extends to all primes $r,ℓ$ such that $4r=ℓ-1$ and where $i^2=-1\bmod ℓ$.

I'm testing $ℓ=13,29,53$. No luck so far (tested some thousands of primes).

The test is quite slow, mainly due to the factorization step in test_p. Using modular polynomials à la SEA could speed it up a bit.

It was suggested that we use sage.schemes.elliptic_curves.isogeny_small_degree.isogenies_prime_degree_genus_0 for $ℓ=13,29$, but a quick profiling shows that this function is (significantly) slower than our naive factorization.

Finally, we can generalize the test we performed on 147b1 to other curves over $ℚ$ that have 13-torsion points in a cubic extension.

Table 1 in http://arxiv.org/pdf/1509.00528v2.pdf parameterizes these curves by

$$j(t) = \frac{(t^4 - t^3 + 5t^2 + t + 1)(t^8 - 5t^7 + 7t^6 - 5t^5 + 5t^3 + 7t^2 + 5t + 1)^3}{t^{13} (t^2 - 3t - 1)}.$$

Below is a very fast test that runs through all this curves, and looks for prime factors common to all coefficients of $x+i(x)\bmod P_1P_2$, except the constant coefficient, and excluding singular primes and those that yield $j$-invariant $0$ and $1728$.

We can test the conjecture for traces and norms

We're intensively testing the conjecture for many more $j$-invariants, and no counter-example has been found yet.

On the other hand, to vaguely support the conjecture, perturbing the period function just slightly quickly yields common factors: