Example of a homeomorphic regular morphism of affine algebraic sets that's not an isomorphism of affine algebraic sets? As the title suggests, can anyone give me an example of a homeomorphic regular morphism of affine algebraic sets that is not an isomorphism of affine algebraic sets? Many thanks in advance.
 A: We expand on Takumi Murayama's comment. The standard example is to take $$X = \mathbb{A}_k^1, \text{ }Y = V(x^3 - y^2) \subset \mathbb{A}_k^2,$$ and define$$F: X \to Y,\text{ }F(t) = (t^2, t^3).$$It is not hard to see this is a bijection. Because the Zariski closed subset of $X$, resp. $Y$, are $X$ itself, resp. $Y$ itself, together with all finite subsets, $F$ is a homeomorphism. But it is not an isomorphism, because the map of coordinate rings is not an isomorphism.
A more interesting example is the aforementioned Frobenius morphism (ubiquitous in positive characteristic algebra). Let $k$ be an algebraically closed field of positive characteristic $p$. Let $n \ge 1$ and define$$F: \mathbb{A}_k^n \to \mathbb{A}_k^n,\text{ }F(x_1, \dots, x_n) = (x_1^p, \dots, x_n^p).$$This is a bijection because every element of $k$ has a unique $p$th root. Moreover, for every polynomial $g \in k[x_1, \dots, x_n]$, $g^p = F^*(h)$ for some element $h \in k[x_1, \dots, x_n]$. Therefore,$$V(g) = V(g^p) = F^{-1}(V(h)) \implies F(V(g)) = V(h).$$So $F$ is a closed, continuous bijection, i.e., $F$ is a homeomorphism. However, $F$ is not an isomorphism since there is no $h \in k[x_1, \dots, x_n]$ such that $F^*h = x_1$.
