I am quite confused about an elementary verification of something that's part of the Theorem in the title. $E(a,b)$ is an elliptic curve defined by the equation $Y^2Z = X^3+aXZ^2+bZ^3$. The statement I was trying to verify the following:
If $a' = c^4a$ and $b' = c^6b$, for a $c \neq 0$, then $\varphi: (x:y:z) \mapsto (c^2x:c^3y:z)$ is an isomorphism from $E(a',b') \to E(a,b)$.
In Milnes notes, no proof for this is given. It is fairly obvious, that this map has to be an isomorphism (it is linear) if it is well defined - that is, if it really maps on $E(a,b)$. My problem is to proof that it really maps on $E(a,b)$. Question: How do you verify this morphism is well defined?
The following are the thoughts I've had so far: Now, for our map to satisfy $\varphi (E(a',b')) \subset E(a,b)$, we need to verify that for all points $P = (x:y:z) \in E(a',b')$ the point $\varphi(P) = (c^2x: c^3y : z)$ satisfies the equation $Y^2Z = X^3+aXZ^2+bZ^3$. Plugging $\varphi(P)$ into the equation yields $$\tag{1} c^6y^2z = c^6x^3+c^2axz^2+bz^3.$$ On the other hand we know, any point $P \in E(a',b')$ satisfies the equation $y^2z = x^3+a'xz^2+b'z^3$. This, according to our prerequisites, is the same as $$ \tag{2} y^2z = x^3+c^4axz^2+c^6bz^3. $$ So the problem now is, that the point $(x:y:z)$ would satisfy equation (2) and we, somehow, would have to deduce that it also satisfies equation (1). How to do this, is not clear to me. Furthermore, the weird thing at this point is, that if we had mapped from $E(a,b) \to E(a',b')$, everything else being the same, all things would cancel out as equation (1) would be $$\tag{3} c^6y^2z = c^6x^3+c^6axz^2+c^6 bz^3,$$ while equation (2) would be $$\tag{4} y^2z = x^3+axz^2+bz^3. $$ Deducing equation (3) from (4) is obvious.
I don't think the direction of the morphism in Milne's notes is simply a typo, because it shows up consistently in this theorem and the proof. Am I getting something wrong, or is the direction of the morphism a mistake in the lecture notes?
You have things the wrong way around. The isomorphism is from $E(a, b)$ to $E(a',b')$. In particular if we evaluate the defining equations for $E(a', b')$ at $(c^2x, c^3y, z)$ we have $$c^6 y^2 = c^6 x^3 + c^2 a' x + b'$$ and upon dividing by $c^6$ we verify the result.