I often see in the literature some arguments like this:
"to show that an elliptic curve $E$ is defined over $\mathbb{F}_p$, we show that the $j$-invariant of the curve is in $\mathbb{F}_p$."
But I thought in general there are elliptic curves not define over $\mathbb{F}_p$, which still have $j$-invariant $\mathbb{F}_p$?
What do they mean by "defined over" when they make a such implication?
- $E$ is really defined over $\mathbb{F}_p$, i.e., the weierstrass equation of $E$ has coefficients in $\mathbb{F}_p$.
- $E$ is not defined over $\mathbb{F}_p$, but it is isomorphic to a curve defined over $\mathbb{F}_p$.
Is it the case that "$E$ is defined over $K$ if and only if $j\in K$" when $K$ is a finite field? I'm sure this is not true when the characteristic of $K$ is $0$, but I haven't thought about explicitly constructing an elliptic curve not defined over $\mathbb{F}_q$ but $j \in \mathbb{F}_q$, $q =p^r$.
The meaning is: $E$ is defined over $\bar K$ and is isomorphic (over $\bar K$) to an elliptic curve defined over $K$.
$E$ with this property will certainly have $j\in K$. For the opposite direction (the one that you're interested in), if $\operatorname{char}K\neq2,3$ and $j\neq0,1728$ then $$y^2=x^3+\frac{3j}{1728-j}\,x+\frac{2j}{1728-j}$$ has $j$-invariant $j$ and coefficients in $K$ (and $y^3=x^3+1$ and $y^3=x^3+x$ have $j=0$ and $j=1728$). In characteristic $2$ and $3$ one needs to do more work.
All of this is taken from Washington's book Elliptic curves.