Let $C:$ $(Y^2=X^3+X^2)\subset \mathbb{A}^2$; the familiar parametrization $$ \varphi\colon \mathbb{A}^1 \to C,$$ given by $$ T \mapsto (T^2-1,T^3 -T)$$ is a polynomial map, but is not an isomorphism (why not?). Find out whether the restriction $$ \varphi'\colon \mathbb{A}^1\setminus\{1\} \to C$$ is an isomorphism.
Miles Reid, Undergraduate Algebraic Geometry, Problem 4.7. (it is assumed that the underlying field $k$ is algebraically closed)
My understanding: We know that $\varphi$ is an isomorphism (of affine varieties) iff $$ \varphi^* \colon k[C] \to k[\mathbb{A}^1] $$ defined by $$ \varphi^* (g) := g \circ \varphi$$ is an isomorphism (of rings). Now $$k[C]=k[X,Y] / (Y^2 - X^3 - X^2),$$ and denote $k [\mathbb{A} ^1 ]$ by $k[T]$. The image of $\varphi^*$ is easily seen to be the $k$ algebra generated by $T^2-1$ and $T^3-T$, which is not the whole of $k[T]$, so $\varphi^*$ is not an isomorphism, hence neither is $\varphi$.
For the second part, asking whether the restriction is an isomorphism, I see that defining $$ \psi (x,y) = y/x$$ where $x\ne 0$, and $\psi(0,0)=-1$, we get an inverse to $\varphi$. But $\psi$ is not a polynomial map.
My question: How to show $\varphi'$ is or isn't an isomorphism ?
Remark that $C$ is a curve with a node $N=(0,0)$, and, informally speaking, the map $\varphi$ realizes $C$ as the "quotient" of $\mathbf A^1$ obtained by identifying the points $\pm 1$ (which both map to $N$ on $C$). One can even make that precise : any morphism of $\mathbf A^1$ into an affine variety which maps $\pm 1$ to the same point factors uniquely through $\varphi$.
The restriction of $\varphi$ to $\mathbf A^1-\{1\}$ is bijective on points, but it is not an isomorphism. The point $-1$ on $\mathbf A^1$ is sent to the node $N$, and we see geometrically that $C$ has two "branches" passing through $N$ whereas $\mathbf A^1-\{1\}$ has only one "branch" passing through $-1$. In other words the two curves do not look similar in small neighborhoods of these two points. This suggests that we might use this to prove that the map is not an isomorphism.
In order to make this formal, one should look at the local rings at the corresponding points, and show that they are not isomorphic (or better yet, that $\varphi$ doesn't induce an isomorphism between them). Every local ring of $\mathbf A^1$ is a DVR, but the maximal ideal $(X,Y)$ of the local ring of $N$ is not principal: both $X$ and $Y$ are needed to generate it. Roughly speaking, since $1+X$ is a unit in this local ring, the equation defining $C$ around $N$ is essentially $Y^2=X^2$, and this shows that both $X$ and $Y$ are needed.
You can also prove it by showing that $\varphi$ doesn't induce an isomorphism between the Zariski tangent spaces of the two points. The tangent space of $-1$ on $\mathbf A^1$ is one-dimensional, but the tangent space of $N$ is two-dimensional.