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The cusp parametrization $t \mapsto (t^{2}, t^{3})$ is bijective but not an isomorphism

Statement refuted

A bijective morphism of affine varieties need not be an isomorphism.

Let $C : = V (v^{2} - u^{3}) \subseteq A_{k}^{2},$ and define $ν : A_{k}^{1} \to C, t \mapsto (t^{2}, t^{3}) .$ Every point of $C$ has the form $(a^{2}, a^{3})$ for a unique $a \in k$ , so $ν$ is bijection on points.

Its pullback on coordinate rings is $k [C] = k [u, v] / (v^{2} - u^{3}) ⟶ k [t], u \mapsto t^{2}, v \mapsto t^{3} .$ This homomorphism is injective but not surjective, because $t$ is not in its image. Hence Affine morphisms are contravariantly equivalent to coordinate-ring homomorphisms shows that $ν$ is not an isomorphism.

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6 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Algebraic Geometry, Example 3.29(a) (standard reference, not scraped)
Michael Artin, Notes for a Course in Algebraic Geometry, Example 2.5.7 (standard reference, not scraped)

The cusp parametrization t↦(t2,t3)t↦(t2,t3) is bijective but not an isomorphism