Do isogenies with rational kernels tend to be surjective?

Question

Dear MO Community,

this is a pretty vague title, so let me tell you the precise observation I have made.

Consider the family of elliptic curves over $\mathbf{Q}$ having a rational $5$-torsion point $P$. They are given by $$E_d: Y^2 + (d+1)XY +dY=X^3+dX^2,$$ for $d \in \mathbf{Q}^*$ and $P=(0,0)$. Let $\eta: E_d \rightarrow E_d'$ the isogeny whose kernel consists exactly of the five rational $5$-torsion points.

Now assume that $E_d$ has rank $1$. After modding out torsion the Mordell-Weil group is isomorphic to $\mathbf{Z}$, and hence, $\eta$ induces an injective group homomorphism $\mathbf{Z} \rightarrow \mathbf{Z}$ which is either an isomorphism or has cokernel of size $5$.

It seems to me that this map tends to be an isomorphism.

To be precise, among all $d$, such that the numerator and denominator is bounded by $100$, there are $3,038$ elliptic curves of analytic rank equal to $1$ (out of $6,087$ total curves), and among those the above map $\mathbf{Z} \rightarrow \mathbf{Z}$ is an isomorphism in $91.2\%$ of the cases.

So I wonder, what you should expect on average? $50\%$? $100\%$?

Maybe, this was just a coincidence in a small database. Maybe someone has seen a similar behaviour somewhere else. Maybe this is nothing new and I just haven't heard about it. I am curious to read what you think about it.

Many thanks.

Answer 1

Here is a rather long comment in which I try to justify why I think that the majority will have a surjective map on the Mordell-Weil group. I would not want to guess what the % is.

Let $E$ be an elliptic curve defined over $\mathbb{Q}$ with a rational point of order $p>2$. Write $\varphi : E \to E'$ for the quotient of $E$ by this rational torsion point. Suppose $E$ has rank $1$. Let us also make some simplifying assumptions first: Suppose that the reduction at $p$ is not additive, that the Tate-Shafarevich groups of $E$ and $E'$ do not have $p$-torsion, and that $E'$ has no $p$-torsion.

This implies that the curves have everywhere semistable reduction (at least if $p>3$). This again implies that the map $E\to E'$ is etale and the real Neron periods change by $\Omega' = \tfrac{1}{p}\Omega$. Let us consider the BSD formula which is known to be invariant under this isogeny. The quotient of the two formulae for the $L$-value gives a relation like $$ 1 = \frac{w \cdot h \cdot c \cdot s}{t^2} $$ where first $w = \Omega/\Omega' = p$, then $t=p$ is the quotient of the order of the torsion group of $E$ by the one of $E'$. Next $h$ is the quotient of the regulator of $E$ by the regulator of $E'$. Hence $h=1/p$ if the map $\varphi$ on the Mordell-Weil group is surjective and $h=p$ otherwise. Then $s=1$ is the quotient of the order of Shas and $c$ is the quotient of the Tamagawa numbers of $E$ by the ones of $E'$. So we find in our case that $$ c = \prod_v \frac{ c_v(E)}{c_v(E')} \qquad \text{ is $p^2$ if our map surjective and $1$ otherwise.} $$ (If $s>1$, this has to be divided by $s$.) Now at all places $v$ the reduction is semistable (a part from when $p=3$ and the type is IV or IV*). For such places the quotient of Tamagawa numbers is easy to compute. If the reduction is non-split, then the quotient is $1$. Otherwise, either $c_v(E)/c_v(E')$ is $p$ or $1/p$. However the second case can only occur if $v\equiv 1 \pmod{p}$. So I would think that the latter is less frequent.

Let $a$ be the number of split places for which the quotient of Tamagawa numbers is $p$ and $b$ the number of split places when it is $1/p$. Then $c = p^{a-b}$. We are asking if $a-b$ is $2$ (surjective case) or $0$. My remark above suggests that $a$ is likely to be larger than $b$.

However $a$ and $b$ are not free: in fact $a-b$ is equal to the difference between the dimensions $\hat d$ of the $\hat\varphi$-Selmer group and the dimension $d$ of the $\varphi$-Selmer group. From considering the descents, one also sees now that $d,\hat{d}\leq 2$ and $d+\hat{d} \geq 2$ (still assuming trivial contribution from Shas). This leaves the possibilities $a-b= \hat{d}-d$ to be either $-2,0,2$ where the first option is excluded by the above.

Now to my simplifying assumptions. I think the only one restricting to the non-generic case is that $s=1$. I could image that one could push the above argument further.

A related result is the fact that for curves $E$ with a $p$-torsion point with $p>3$, it is almost impossible that that $\prod c_v(E)$ is not divisible by $p$. Lorenzini shows that there are only finitely many such $E$.

Hopefully someone can extend and complete my attempt to answer this.

Answer 2

I recycled my code from the other thread to test this. There are 559 elliptic curves of conductor < 300000 that have rank 1 and a rational 5 torsion point. Of these 559 curves there are 452 for which the map $\mathbb Z \to \mathbb Z$ induced by $\eta$ is surjective (this is about 81%).

I did the same computation for rational 7 torsion point. The problem now becomes that the dataset is very small because there are only 31 elliptic curves of conductor < 300000 of rank 1 with a rational 7 torsion point. But the remarkable thing is that for 30 of these 31 cases the map $\mathbb Z \to \mathbb Z$ is surjective.

Edit: I updated the results after discovering a small bug in my code that caused some curves in the database to be skipped in the test.

Answer 3

I tried the same as Chris, but I don't think it gives you an argument for the surjectivity. As it is too long for a comment, I post an answer.

Assume we are given an elliptic curve $E$ over $\mathbf{Q}$ with rational $5$-torsion and denote by $\eta:E \rightarrow E'$ the isogeny modding out the rational $5$-torsion. As I mentioned in my question this curve is given by a rational non-zero number $d$. Write $d=u/v$, with $u$ and $v$ coprime non-zero integers. Now we apply the equation of Cassels and Tate which encodes the invariance of BSD: $$ \frac{\# sha(E/\mathbf{Q})}{\# sha(E'/\mathbf{Q})} = \frac{R_{E'}}{R_E} \cdot \frac{ \# E(\mathbf{Q})^2_{tor} }{\# E'(\mathbf{Q})^2_{tor} } \cdot \frac{P_{E'}}{P_E} \cdot \prod_{p \leq \infty}\frac{c_{E',p}}{c_{E,p}}.$$

Now assume, that the elliptic curve has rank 1.

As Chris already stated, the regulator quotient equals $5^a$, for $a \in \{\pm 1 \}$, where $a=1$ if and only if the induced map of $\eta$ on the free part of the Mordell-Weil group is surjective. Hence this is the case we are interested in.

For the torsion quotient we have, that it equals $5^b$, for $b \in \{0,2 \}$, where $b=2$ if and only if $d$ is not a fifth power. This is true for $100\%$ of the cases.

For the periods and Tamagawa numbers we have, that they are equal to $5^{c-1}$, for $$c=\# \{p \equiv 1(5),\ p \mid u^2+11uv-v^2\} + \# \{p=5,\ p^3 \mid u^2+11uv-v^2 \}$$ $$-\# \{p,\ p\mid uv \},$$ where $p$ ranges over the finite primes. (The results on the torsion quotient and on the period and Tamagawa quotient (= local quotient) can be found here http://arxiv.org/abs/1206.1822/) (The case $5^3\mid u^2+11uv-v^2$ happens if and only if $u \equiv 7v (25)$.)

Hence, in 100% of the cases we have

$$ \frac{\# sha(E/\mathbf{Q})}{\# sha(E'/\mathbf{Q})} = 5 \cdot 5^{\{\pm 1 \}} \cdot 5^c.$$

I don't see any reason why a negative value of $c$, which you should expect quite often, should force $a$ to be $1$, and not be (completely) subsumed in the Sha quotient.

So, instead of having a pro argument for $a=1$, this is just a pro argument, that the isogeny $\eta$ can alter the $5$-primary part of Sha in an arbitrary way.

(In case you wonder whether $c$ can be positive, look at $u=1$, $v=76971487$. Then $uv$ is prime and $u^2+11uv-v^2$ factors as $-1 \cdot 11 \cdot 31 \cdot 41 \cdot 61 \cdot 131 \cdot 151 \cdot 331 \cdot 1061$, hence $c=7$.)

(In my database of 3,038 curves, there is the following situation: $c=0$ for 247 curves. For all of them $\eta$ is NOT surjective. $c=-2$ for 2258 curves. For 20 of them $\eta$ is NOT surjective. $c=-4$ for 533 curves. For all of them $\eta$ is surjective.)