Let \(k\) be algebraically closed.
A morphism whose underlying map of topological spaces is a homeomorphism need not be an isomorphism.
Let \(\phi: \AA^1 \to \AA^2\) be defined by \(t \mapsto (t^2, t^3)\). Show that \(\phi\) is a bijective bicontinuous morphism of \(\AA^1\) onto the curve \(y^2 = x^3\), but that \(\phi\) is not an isomorphism.
Let the base field \(k\) have characteristic \(p > 0\), and define \(\rho: \AA^1 \to \AA^1\) by \(t \mapsto t^p\). Show that \(\rho\) is bijective and bicontinuous but not an isomorphism. This is called the Frobenius morphism.
Solution
For part (a), put \(Y=Z(y^2-x^3)\subseteq\AA^2\) and write \(A(Y)=k[x,y]/(y^2-x^3)\).
1. The map \(\phi\) is a morphism from \(\AA^1\) onto \(Y\) and is bijective.
Its coordinate functions \(t^2,t^3\) are polynomial, and \[ (t^3)^2=(t^2)^3, \] so its image lies in \(Y\).
Let \((a,b)\in Y\). If \(a=0\), then \(b^2=a^3=0\), hence \((a,b)=(0,0)=\phi(0)\). If \(a\ne0\), put \(t=b/a\). Using \(b^2=a^3\) gives \[ t^2=\frac{b^2}{a^2}=a, \qquad t^3=t\,t^2=\frac{b}{a}a=b, \] so \((a,b)=\phi(t)\). For a point of the form \(\phi(t)\), this inverse returns \(0\) when \(t=0\) and returns \(t^3/t^2=t\) when \(t\ne0\). It is therefore a two-sided set-theoretic inverse, proving injectivity and surjectivity.
2. The bijection \(\phi:\AA^1\to Y\) is bicontinuous.
Every morphism of varieties is continuous, so only continuity of the inverse remains. The closed subsets of \(\AA^1\) are \(\AA^1\) itself and finite sets: a proper closed subset is the zero set of a nonzero polynomial in one variable, hence is finite. Because \(k\) is algebraically closed, every point of \(Y\) is closed, so every finite subset of \(Y\) is closed. Thus \(\phi\) sends every closed subset of \(\AA^1\) to a closed subset of \(Y\): the whole line maps to \(Y\), and a finite set maps to a finite set. Hence \(\phi\) is a closed continuous bijection, so its inverse is continuous.
3. The morphism \(\phi\) is not an isomorphism, completing part (a).
The induced homomorphism on coordinate rings is \[ \phi^*:A(Y)\longrightarrow k[t], \qquad x\longmapsto t^2, \qquad y\longmapsto t^3. \] Its image is the subring \(k[t^2,t^3]\). Every monomial of positive degree in this subring has exponent \(2i+3j\ge2\), so no element of \(k[t^2,t^3]\) has a nonzero \(t^1\) coefficient. In particular \(t\notin k[t^2,t^3]\), and therefore \(\phi^*\) is not surjective. An isomorphism of affine varieties induces an isomorphism of coordinate rings, so \(\phi\) cannot be an isomorphism.
Assume now that \(\operatorname{char}k=p>0\) for part (b).
4. The Frobenius morphism \(\rho:\AA^1\to\AA^1\), \(t\mapsto t^p\), is bijective and bicontinuous.
The map is a morphism because \(t^p\) is a polynomial. If \(a^p=b^p\), then in characteristic \(p\), \[ (a-b)^p=a^p-b^p=0, \] so \(a=b\); thus \(\rho\) is injective. For every \(c\in k\), the polynomial \(T^p-c\) has a root in the algebraically closed field \(k\), so \(c=t^p\) for some \(t\in k\); hence \(\rho\) is surjective.
As in step 2, every proper closed subset of \(\AA^1\) is finite. The bijection \(\rho\) therefore sends closed sets to closed sets, while every morphism is continuous. Thus \(\rho\) is a closed continuous bijection and hence a homeomorphism.
5. The Frobenius morphism is not an isomorphism.
Its pullback on coordinate rings is \[ \rho^*:k[u]\longrightarrow k[t], \qquad u\longmapsto t^p. \] The image is \(k[t^p]\). Every exponent occurring in a polynomial in \(t^p\) is divisible by \(p\), so \(t\notin k[t^p]\). Thus \(\rho^*\) is not surjective and cannot be an isomorphism of coordinate rings. Consequently \(\rho\) is not an isomorphism of affine varieties.
6. Q.E.D.
Steps <1>1–<1>3 prove part (a), and steps <1>4–<1>5 prove part (b).