Past exam of the algebraic Geometry course of the University of Oxford / 2011 / 1 / e / Solution

The parametrization lands on the curve because $(t^3{)}^4=(t^4{)}^3$. Its pullback has imageEvery monomial in this subalgebra has exponent $3a+4b$ for nonnegative integers $a,b$. No such exponent is one, so no linear combination of these monomials is $t$. Thus $F^{\#}$ is not surjective and $F$ is not an isomorphism.

It is nevertheless bijective on points. Away from the origin, $x\ne 0$ and $t=y/x$ gives $t^3=x$ and $t^4=y$ using $x^4=y^3$; the origin comes only from $t=0$. The rational inverse cannot extend regularly at the origin: if $t=P(t^3,t^4)/Q(t^3,t^4)$ with $Q(0)\ne 0$, then $tQ(t^3,t^4)$ has a nonzero degree-one term, whereas $P(t^3,t^4)$ has none.This distinguishes a set-theoretic bijection from an isomorphism of varieties in every characteristic.

This is a bijective monomial cusp parametrization with nonregular inverse.