If we have an elliptic curve E over a field k and we pick a nonsquare d in k{0}. Suppose E is isomorphic to E^(d). (E^(d) is the quadratic twist) Why must j(E) = 1728 and why is k(sqrt(d)) = k(sqrt(1))?

$\begingroup$ The way you've phrased this question suggests that it was something you were told, or have read, rather than something you discovered. Is this correct? (just curious) $\endgroup$ – Yemon Choi Feb 18 '10 at 20:03
To make things concrete, suppose that we're working in characteristic $\ne 2,3$ (we can do something similar in those cases, though it gets messier), and the equation of the curve is
$E : y^2 = x^3 + a x + b$
and the equation of the twist is:
$E^{(d)} : d y^2 = x^3 + a x + b$.
Multiplying this by $d^3$:
$(d^2 y)^2 = (d x)^3 + a d^2 (d x) + d^3 b$
for this to be isomorphic to $E$ over $k$ it is necessary and sufficient that there is a $\lambda \in k^*$ such that (see Silverman, or any other book on elliptic curves) such that $a d^2 = a \lambda^4$, $b d^3 = b \lambda^6$. If $a,b \ne 0$ then we must have $\lambda^6 = d^3$ and $\lambda^4 = d^2$, which shows that $\lambda^2 = d$ which you've ruled out. If $b=0$ you have, by taking square roots, $\lambda^2 = \pm d$, and only the minus sign is possible. If $a=0$ (and thus $b \ne 0$) then $\lambda^6 = d^3$. But then $(\lambda^3/d)^2 = d^3/d^2 = d$, which you've also ruled out. Thus, only $b=0$ is possible, which is $j=1728$.
Various forms of $E$ over $k$ are classified by $H^1(k,Aut(E))$ and the quadratic twist corresponds to the image of the class $d\in k^\times/(k^\times)^2=H^1(k,{\mathbb Z}/2)$ under the map $H^1(k,{\mathbb Z}/2)\to H^1(k,Aut(E))$ induced by the map ${\mathbb Z}/2\to Aut(E)$ which sends nontrivial element of ${\mathbb Z}/2$ to the automorphism $x\mapsto x$ of $E$. You are interested in the case when the image of $d$ under this map is zero. At the very least this means that $Aut(E)$ is strictly greater than ${\mathbb Z}/2$; but there are very few elliptic curves with this property..