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Let $E/\mathbb C$ be an elliptic curve. It is known that if $E$ has CM, then $j(E)$ is an algebraic integer. My first question is: what about the converse? Is there a way to identify the subset of algebraic integers whose elements correspond to [isomorphism classes of] elliptic curves with CM? The second question is: fixed a number field $K$, does there always exist an elliptic curve over $K$ with CM? And without CM?

Thank you very much!

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For the second question: there are CM elliptic curves over Q. Their base changes to the field are hand are CM as well. E.g., you could take y^2 = x^3 - x over any number field. On the other hand there are non-CM curves as well: just take the j-invariant to lie in the number field you want but to not be an algebraic integer. –  Kestutis Cesnavicius Aug 30 '12 at 13:41
    
However the extra endomorphisms of those curves, in this case $x\to -x$, $y\to iy$, are not defined over $\mathbb Q$. This is because their action on $1$-forms is multiplication by an element of the ring of integers of the imaginary quadratic field, so they cannot be defined over $\mathbb Q$. –  Will Sawin Aug 30 '12 at 16:16
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3 Answers 3

up vote 3 down vote accepted

1)If an elliptic curve has integral $j$-invariant it absolutely DOES NOT NEED to have CM. The class of curves with integral $j$-invariant (let's call that the class of IM Elliptic curves for Integral Modulus) is MUCH MUCH larger than the class of CM Elliptic curves. In fact, one can use Heilbronn's Theorem that class numbers of imaginary quadratic fields tend to infinity to show that over any given number field, there are only finitely many elliptic curves with CM. In particular over $\mathbf{Q}$, there are only 13 CM $j$-invariants. Even for all number fields of a certain degree, there are only a finite number $N(d)$ of elliptic curves with CM over any number field of degree $d$. This gives a way to enumerate all the CM $j$-invariants or "singular moduli," which is about as good as you can hope for in terms of describing the complex numbers which are $j$-invariants of CM elliptic curves. To do this explicitly (say for number fields of degree up to 100) see Mark Watkins' enumeration of imaginary quadratic fields of class number up to 100.

Meanwhile, for any regular integer $n$ (1,2,3, etc) there is at least one elliptic curve over $\mathbf{Q}$ whose $j$ invariant is $n$. Therefore over $\mathbf{Q}$ and therefore over any number field, there are infinitely many non-isomorphic IM elliptic curves.

If you want to say something general about elliptic curves with IM, consider the theorem of Deuring that an elliptic curve has IM if and only if it has potential good reduction. For this and much much more see Serre-Tate's "Good reduction of abelian varieties"

2) Easy proof that the answer is yes, at least as long as you mean "has a CM $j$-invariant" when you say CM: $y^2 = x^3 + 1$ is a CM elliptic curve with $j$-invariant zero defined over any number field. On the other hand, if we take an elliptic curve with $j$-invariant equal to 1/2, say this one: y^2 + x*y = x^3 + 72*x + 13822 , well it doesn't have integral $j$-invariant and therefore can't have CM!

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There is a way to "describe" the j-invariants of elliptic curves with CM over the complex numbers (but I am not sure it is really practical), by using the $\tau$ invariant.

Recall that an elliptic curve over $\mathbb C$ can be descrived as $\mathbb C/\Lambda$, where $\Lambda$ is a lattice in $\mathbb C$; such a lattice can be always put as $\Lambda=\mathbb Z w_1 \oplus \mathbb Z w_2$, with $\tau=w_1/w_2$ in the upper half plane (with imaginary part $>0$). Then $E$ has CM if an only if $\tau$ is algebraic and quadratic imaginary (in our case, the field $\mathbb Q(\tau)$ has degree $2$ over $\mathbb Q$).

But there is a function, the Klein's j function

http://en.wikipedia.org/wiki/J-invariant

that gives the j-invariant of the corresponding elliptic curve. So the description of the j-invariants with CM could be $j(\tau)$ for $\tau$ quadratic imaginary.

Incidentally, the only $\tau$'s algebraic whose $j(\tau)$ is algebraic are the quadratic imaginary (this is a theorem by Theodor Schneider).

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The $j$ invariant must lie on the intersection of the diagonal $x=y$($=j$) with the classical modular curve.

This is because $\Phi_n(j_1,j_2)=0$ if and only if the curve with $j$-invariant $j_1$ and the curve with $j$-invariant $j_2$ have an isogeny between them whose kernel is $\mathbb Z/n$, so $\Phi_n(j,j)=0$ if and only if the curve has an endomorphism of degree $n$ whose kernel is cyclic.

Every CM endomorphism can be factored into a standard, multiplication by $k$, endomorphism and an endomorphism with cyclic kernel. If the CM endomorphism corresponds to a non-rational integer in the imaginary quadratic field, the second component cannot be trivial, so it must be recognized by one of the $\Phi_n$.

Thus it is easy to show that a curve is not a CM curve for a field which contains a non-rational element of norm $\leq n$. Just check all the classical modular polynomials!

If you can compute the degree $d$ of $j$ as an algebraic integer then you know it can only have CM by a field of class number $\leq d$. This is a finite list, so you can find a value of $n$ sufficient to rule out $CM$ entirely by ruling out each field individually. I'm not sure if you can find an explicit bound for $n(d)$ or how good it is.

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