j-invariant of an elliptic curve

Remarks

If the characteristic of $K$ is not $2$ or $3$ and $E$ is $y^2 = x^3 + Ax + B$, then $c_4 = -48A$ and $\Delta = -16(4A^3 + 27B^2)$, so $$j(E) = 1728 \, \frac{4A^3}{4A^3 + 27B^2}.$$ The value does not depend on the chosen Weierstrass equation of $E$. Over an algebraically closed field, two elliptic curves are isomorphic if and only if they have the same $j$-invariant. Moreover, every element of the field is the $j$-invariant of some elliptic curve. Example: $y^2 = x^3 - x$ has $A = -1$ and $B = 0$, so $j = 1728 \cdot (-4)/(-4) = 1728$. Non-example: $y^2 = x^3$ has $\Delta = 0$, so it is not an elliptic curve and $c_4^3/\Delta$ is not defined. Over a field that is not algebraically closed, curves with the same $j$-invariant can fail to be isomorphic: $y^2 = x^3 - x$ and $y^2 = x^3 + x$ both have $j = 1728$ but are not isomorphic over $\Q$.