Complex multiplication

Remarks

In characteristic $0$, $\End(E)$ is either $\Z$ or an order $\mathcal{O}$ in an imaginary quadratic field $F$, and one then says that $E$ has CM by $\mathcal{O}$. Over $\C$, write $E \cong \C/\Lambda$ for a lattice $\Lambda$; then $\End(E) \cong \set{\alpha \in \C : \alpha \Lambda \subseteq \Lambda}$, which explains the name.

Example: $E \colon y^2 = x^3 - x$ over $\Q$. The map $[i](x, y) = (-x, iy)$ sends $E$ to itself, since $(iy)^2 = -(x^3 - x) = (-x)^3 - (-x)$. It fixes $O$, and $[i] \circ [i] = [-1]$, so $[i] \ne [n]$ for every integer $n$; hence $E$ has CM, by $\Z[i]$.

Non-example: $y^2 = x^3 + x + 1$ over $\Q$ has $j$-invariant $1728 \cdot \frac{4}{31} = \frac{6912}{31}$, which is not an algebraic integer. The $j$-invariant of a CM curve is always an algebraic integer, so this curve has endomorphism ring $\Z$.

Over a finite field the endomorphism ring of every elliptic curve is larger than $\Z$, so the notion is mainly used in characteristic $0$.