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Let $E/\mathbb{Q}$ be an elliptic curve with a $p$-torsion point. Does this imply that $E/pE$ is isomorphic to $E$? If not, are there any conditions I can assume such that this is true?

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I guess you mean a $\mathbf{Q}$-rational $p$-torsion point, i.e., $E[p](\mathbf{Q})\neq 0$. I don't know what you mean by $E/pE$. Do you mean $E(\overline{\mathbf{Q}})/pE(\overline{\mathbf{Q}})$, or $E(\mathbf{Q})/pE(\mathbf{Q})$? In the former case, the quotient is zero because multiplication by $p$ is surjective on geometric points of $E$. If you mean the latter, then the answer is no, because the quotient is finite, but $E(\mathbf{Q})$ could have positive rank. – Keenan Kidwell Dec 18 '12 at 18:06
The question seems incompletely thought-out, but my guess is that you mean that if $X$ is a $\mathbb Q$-rational point of period $p$ on $E$, you’re wondering whether $E/\langle X\rangle$ is isomorphic to $E$. Please clarify! – Lubin Dec 18 '12 at 19:12
up vote 2 down vote accepted

$\def\QQ{\mathbb{Q}}$This question deals specifically with the issue of things being defined over $\mathbb{Q}$. It assumes you are already happy with the $\mathbb{C}$ picture, which I discuss in my other answer. In this question, my hypotheseses are that $E$ and $E/z$ are defined over $\mathbb{Q}$, for $z$ a $p$-torsion point. I will show that $E$ is not isomorphic to $E/z$ over $\mathbb{Q}$. Suppose otherwise.

Let $\pi$ denote the composition $E \to E/z \cong E$. So $E$ has an endomorphism whose kernel is $\mathbb{Z}/p$. Let $R$ be the endomorphism ring of $E$. Note that $R$ acts on $H^0(E, \Omega^1)$. Since $H^0(E, \Omega^1)$ is a one dimensional $\mathbb{Q}$-vector space, this gives a map $R \to \QQ$. This map is injective (that's a special fact about characteristic zero). So, for any elliptic curve $E$ over $\mathbb{Q}$, the endomorphism ring of $E$ OVER $\QQ$ is a subring of $\mathbb{Q}$. Endomorphism rings of elliptic curves are also finitely generated $\mathbb{Z}$-modules, so this shows that the endomorphism ring of $E$ OVER $\QQ$ is always $\mathbb{Z}$. So any endomorphism defined over $\mathbb{Q}$ is of the form $w \mapsto m \cdot w$ and has kernel $(\mathbb{Z}/m)^2$, not isomorphic to the cyclic group $\mathbb{Z}/n$. Contradiction, QED. See Silverman, Arithmetic of Elliptic Curves, Chapter III.9 for more on endomorphism rings of elliptic curves.

It is interesting to see what happens to the $\mathbb{Z}$ examples to make them not work over $\mathbb{Q}$. For example, let $E$ be the elliptic curve $y^2 = x^3-x$. As a complex curve, $E$ is $\mathbb{C}/\mathrm{Span}_{\mathbb{Z}}(1,i)$. The $2$-torsion points are $(\pm 1, 0)$ and $(0,0)$. The first two correspond to $1/2+\mathrm{Span}_{\mathbb{Z}}(1,i)$ and $i/2 +\mathrm{Span}_{\mathbb{Z}}(1,i)$, whicle the last one corresponds to $(1+i)/2 + \mathrm{Span}_{\mathbb{Z}}(1,i)$. Note that the last lattice is a rotation-dilate of $\mathrm{Span}_{\mathbb{Z}}(1,i)$; in other words, multiplication by $(1+i)$ maps the lattice $\mathrm{Span}_{\mathbb{Z}}((1+i)/2, 1)$ to $\mathrm{Span}_{\mathbb{Z}}(1,i)$. So, as complex curves, $E$ should be isomorphic to the quotient of $E$ by the $2$-torsion point $(0,0)$.

If I haven't made any errors, translation by $(0,0)$ is given by $$(x,y) \mapsto \left( \frac{-1}{x}, \frac{y}{x^2} \right).$$ Define $$X := x-x^{-1} \quad Y:= y +y x^{-2}.$$ So $X$ and $Y$ are invariant for translation by $(0,0)$, and I assert without proof that they generate the coordinate ring of the quotient. They obey the relation $$Y^2 = X(X^2+4).$$ (Exercise!) Call this curve $F$. Since $F$ has only one rational $2$-torsion point, and $E$ has three of them, they are not isomorphic over $\mathbb{Q}$.

They are isomorphic over $\mathbb{C}$. Specifically, let $$X=2i x \ \mbox{and} \ Y = (2-2i) y.$$

So this is an example of a case where $E$, $z$ and (hence) $E/z$ are all defined over $\mathbb{Q}$, where $E$ is isomorphic to $E/z$ over $\mathbb{C}$, but where they are not isomorphic over $\mathbb{Q}$.

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$\def\QQ{\mathbb{Q}}\def\CC{\mathbb{C}}\def\ZZ{\mathbb{Z}}$ I'm not sure which level to answer this question at. This answer explains the situation over $\CC$, which you should understand first before thinking about $\QQ$.

Let $E$ be an elliptic curve over $\CC$, isomorphic to $\CC/\Lambda$ for some complex lattice $\Lambda$. Then $p$-torsion points on $E$ are the cosets $z+\Lambda$ for $z$ in $\frac{1}{p} \Lambda/\Lambda$. The quotient of $E$ by the torsion point $z+\Lambda$ is the quotient of $\CC$ by the lattice generated by $\Lambda$ and $z$.

For example, if $E = \CC/\mathrm{Span}_{\ZZ}(1, i)$, then the $3$-torsion points on $E$ are the images of the eight points $\pm 1/3$ and $\pm i/3$ and $\pm 1/3 \pm i/3$. The quotient of $E$ by $1/3$ (for example) is $\CC/\mathrm{Span}_{\ZZ}(1/3, i)$. As you can see, the lattices $\mathrm{Span}_{\ZZ}(1, i)$ and $\mathrm{Span}_{\ZZ}(1/3, i)$ are NOT rotation-dilates of each other, so the curves are NOT isomorphic, even over $\CC$.

Let's see when this could happen. Let $E=\CC/\Lambda$. There is a $p$-torsion point on $E$ such that $E \cong E/z$ if and only if there is a lattice $\Lambda'$ containing $\Lambda$ with $\Lambda'/\Lambda \cong \ZZ/p$ AND such that $\Lambda = \alpha \Lambda'$ for some complex number $\alpha$. Such an $\alpha$ must be a quadratic algebraic integer, since it maps a rank $2$ lattice into itself. And, since $|\Lambda'/\alpha \Lambda'|=p$, we see that $\alpha$ has norm $p$. So such an elliptic curve must have complex multipication by a ring of quadratic algebraic integers which contains an element of norm $p$. There are only finitely many such rings, and each of them acts by CM on finitely many elliptic curves (over $\CC$).

For example, let $p=5$. The only complex quadratic algebraic integers of norm $5$ are $\pm 1 \pm 2 i$, $\pm 2 \pm i$ and $\pm \sqrt{5} i$. The only possible CM rings corresponding to these are $\ZZ[i]$, $\ZZ[2i]$ and $\ZZ[\sqrt{5} i]$. The elliptic curves which have CM by these are $\CC/\mathrm{Span}_{\ZZ}(1,i)$, $\CC/\mathrm{Span}_{\ZZ}(1,2i)$, $\CC/\mathrm{Span}_{\ZZ}(1,\sqrt{5} i)$ and $\CC/\mathrm{Span}_{\ZZ}(2,1+\sqrt{5}i)$. So there are $4$ elliptic curves $E$ over $\mathbb{C}$ which have $5$-torsion points $z$ such that $E \cong E/z$. For every other elliptic curve $E$ and $5$-torsion point $z$, we will have $E \not \cong E/z$.

I can think of two deeper question you might be asking: (1) If $E$ is defined over $\mathbb{Q}$ and the $p$-torsion point $z$ is defined over $\mathbb{Q}$, is $E \cong E/z$ over $\mathbb{Q}$? Answer: Always NO. When I have more time, I'll try to write up a detailed answer explaining this. By the way, the situation is the same if, rather than requiring $z$ be defined over $\mathbb{Q}$, you only assume that the cyclic subgroup generated by $z$ is so defined. If you don't even assume that much, then $E/z$ needn't be defined over $\mathbb{Q}$.

(2) You could be asking for $E$ defined over $\mathbb{Q}$, for $z$ (or the subgroup it generates) to be defined over $\mathbb{Q}$ but for the isomorphism $E \cong E / z$ to be defined over $\mathbb{C}$. In that case, the answer is that your hypotheses are already very restrictive, by work of Mazur (the case of a rational torsion point) (the case of a rational cyclic subgroup). It should be possible to go through the small list of cases where those hypotheses are satisfied and work out when the answer is "yes" and when it is "no", but I don't want to try that unless I am certain it is what you are looking for.

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@Matt E: This is the sort of question where you usually give a characteristically excellent answer; please feel free to add or improve. – David Speyer Dec 19 '12 at 15:45

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