For elliptic curve $E/{\overline k}, ~j(E)\in k\iff E$ can be defined over $k$?

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Sorry for my bad English.

Let $k$ be a field of any characteristic, and $E$ be an elliptic curve over $\overline{k}$ where $\overline{k}$ is an algebraic clouser of $k$.

Then we have the $j$-invariant $j(E)\in \overline{k}$.

Now I want to know relationship between a condition $j(E)\in k$ and if $E$ can be defined over $k$; there is an elliptic curve $E'/k$ s.t. $E'\times_k \operatorname{Spec}\overline{k}\cong E$ over $\overline{k}.$

Especially, I want to know the case $k=\mathbb{F}_q$ for a prime power $q$.

Please give me hint, fact, paper, proof, or link.

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Yes this is true. See Prop III.1.4 in Silverman's Arithmetic of Elliptic Curves for a detailed reference.

The proof is simple, if you already know that an elliptic curve is determined up to isomorphism over $\overline k$ by its $j$-invariant. Indeed, given some $j_0\in k$, it's sufficient just to produce a single elliptic curve over $k$ with $j$-invariant $j_0$.

The cases $j_0 = 0, 1728$ are easy. Otherwise, just take $$E\colon y^2 + xy = x^3-\frac{36}{j_0-1728}x - \frac1{j_0-1728}.$$