I'm working through Silverman's Arithmetic of Elliptic Curves Exercise 1.7 (3 parts) and I have a question about the last part.
(a) In $\mathbb{P}^1$ at least one of $S$ or $T$ is non-zero, so the map is well-defined at all $[S,T] \in \mathbb{P}^1$. Therefore, $\phi$ is a morphism.
(b) You can easily check that $\psi: [X,Y,Z]\mapsto [Y,X]$ works.
(c) The only problematic point on $\psi$ is $[0,0,1]$. I tried to do a bit of the usual manipulation, but was not able to make $\psi$ well-defined on $[0,0,1]$. For instance: $$ [Y,X] = [YX^2,X^3] = [YX^2,Y^2Z] = [X^2,YZ] $$ This leads me to believe that $\psi$ is not actually a morphism. However, for $\phi$ to be an isomorphism there just has to exist an inverse morphism $\phi^{-1}$ to $\phi$, and this does not necessarily have to be $\psi$ as I have defined it.
I'm wondering how can I (formally) show that $\phi$ is not an isomorphism?
EDIT: I realize that $Y^2Z=X^3$ has a cusp at $[0,0,1]$. So I guess can someone show me how to prove that an isomorphism of projective varieties must preserve singularities?
If $i: X\rightarrow \mathbb{P}^n_k$ is a projective variety over a field $k$ and if $x\in X$ is a point, you may define the local ring $\mathcal{O}_{X,x}$ at $x$. By definition
L1. $\mathcal{O}_{X,x} \cong lim_{x \in U}\mathcal{O}_{X}(U)$,
The local ring in L1 is invariant wrto choice of embedding $i$, it only depends on the structure sheaf $\mathcal{O}_X$ (the structure sheaf is an "intrinsic" object that does not depend on $i$). The maximal ideal $\mathfrak{m}(x) \subseteq \mathcal{O}_{X,x}$ is also independent of choice of embedding. Hence the notion "non singular" does not depend on choice of embedding or isomorphism.