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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.

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Similar phenomena are discussed in the highest-ranking answer to this question: mathoverflow.net/questions/113968/…. –  René Feb 27 '13 at 22:03
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But 2 isogenies will not give new information since the kernel of a 2 isogeny is always rational. Let $phi$ be an isogeny of prime degree between two rank 1 curves. Then either $phi$ or it's dual will be surjective modulo the torsion. So if you look at all isogenies of a fixed prime degree and order them by conductor then the ratio surjective - not surjective will always be 50-50. –  Maarten Derickx Feb 28 '13 at 16:58
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@René: I agree with you that the formula (if it is this or anything similar) is hardly ever a $5$-th power in $\mathbb{Q}^{\times}$ if the coordinates $x_0,y_0$ of the point are random rationals. However, they are far from random. For instance, you know from descent that the formula will be a $5$-th power in $\mathbb{Q}_v$ at all good places $v$ (and many more), which translates into stringent conditions on $x_0$ and $y_0$. So I am not sure one can say anything through this approach... –  Chris Wuthrich Mar 1 '13 at 20:54
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The j-invariant of $E$ is: $$\frac{-(d^4 + 12d^3 + 14d^2 - 12d + 1)^3}{d^5(d^2 + 11d - 1)}$$ The j-invariant of $E'$ is: $$\frac{-(d^4 - 228d^3 + 494d^2 + 228d + 1)^3}{d(d^2 + 11d - 1)^5}$$ So, in a sense, $j(E)$ is just arithmetically simpler. For example, computation shows that the modular degree (using $X_0$) of $E$ is usually the smaller one of the two, so its Heegner points should have smaller height, making $\eta$ surjective. –  Dror Speiser Mar 3 '13 at 1:39
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I did some computations regarding the étaleness of the isogeny $\phi : E_d \to E'_d$, with $d$ running through all rational numbers of height $\leq 1000$. There are approximately 96.6% values of $d$ such that the Faltings height of $E_d$ is smaller that that of $E'_d$, which (conjecturally) amounts to say that $\phi$ is étale. Of course it's hard to tell but this ratio seems to approach $29/30$. I did the same for the family of curves with rational $7$-torsion point and now the ratio seems to approach $7/8$. –  François Brunault Mar 5 '13 at 9:20

3 Answers 3

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.

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Dear Chris, thanks for your answer. Unfortunately, I don't think your argument works. See my answer below for my explanation. –  Stefan Keil Mar 1 '13 at 13:47
    
Thanks, Stefan, for your answer. You have thought about it more than I have. See my comments below your answer. –  Chris Wuthrich Mar 1 '13 at 20:56

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.

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Dear Maarten, thanks for your answer. I also checked all elliptic curves of conductor < 1,000,000 that have rank 1 and a rational 5 torsion point. There are 1109 such curves and $84.8\%$ of them are `surjective'. –  Stefan Keil Mar 1 '13 at 13:53
    
Dear Maarten, I just realized that your database misses a few curves. There are 559 elliptic curves of conductor < 300000 that have rank 1 and a rational 5 torsion point. And 452, i.e. 80.9%, give surjectivity. –  Stefan Keil Mar 1 '13 at 15:11
    
Oh, and I should add that I a haven't proved that my database contains all curves up to conductor 1,000,000. It is just very likely, because I searched through all $d\in \mathbf{Q}^*$, such that the numerator and denominator of $d$ is bounded by 50,000. And the last curve of conductor smaller than 1,000,000 in this range has $d=650/4617$. –  Stefan Keil Mar 1 '13 at 15:21
    
Apparently there was a bug in my code that explains why my results where different. –  Maarten Derickx Mar 2 '13 at 16:39

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.)

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Of course I agree with you that the local information translates your question into one about the quotient of Shas above. My assumption was that this quotient $s$ is 1 and I was not following the curves in the order they appear in your family. So I - probably wrongly - would have guessed that this quotient is very often 1 for curves of small conductor. If you restrict your attention to those with $s=1$, do you get the same phenomenon as I ? How many have $s\neq 1$ in your list ? –  Chris Wuthrich Mar 1 '13 at 20:48
    
Your formula shows well that $c$ which is my $-(a-b)$ has a tendancy to be negative. Do you also have conditions from the descent on $c$ implied by the fact that your curve has rank $1$ ? –  Chris Wuthrich Mar 1 '13 at 20:49
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Dear Chris, among the 3038 curves of rank 1 in my database, 2485 (81.8%) curves have $s=1$. (These are the 247 curves with $c=0$ and the 2238 'surjective' curves with $c=-2$. Hence, among the $s=1$-curves 89.0% of them are surjective.) The remaining 20 curves with $c=-2$ and the 533 curves with $c=-4$ have $s = 1/5^2$, and among them the $c=-4$-curves, i.e. 96.4%, are surjective. So, assuming that $s=1$, I agrre that your argument above indicates that $\eta$ is likely to be surjective, but I don't know how often we will have $s=1$ on average. –  Stefan Keil Mar 4 '13 at 14:33

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