Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 1 / 24G (Algebraic Geometry)

The parametrization $t\mapsto (t^3,t^2)$ maps ${\mathbb{A}}^1$ to the cusp. Off the origin its inverse is $t=x/y$, since $(x/y{)}^2=y$ and $(x/y{)}^3=x$ on $x^2=y^3$ with $y\ne 0$. This is an isomorphism between dense open subsets, so the cusp is birational to ${\mathbb{A}}^1$.

At the origin the differential of the defining polynomial is zero, giving a two-dimensional Zariski tangent space. The tangent space of ${\mathbb{A}}^1$ at every point is one-dimensional. An isomorphism induces tangent-space isomorphisms, so the cusp is not isomorphic to ${\mathbb{A}}^1$. The rational inverse cannot extend regularly through the singular point.

For an irreducible affine variety, its dimension is the minimum of its Zariski tangent space dimensions:The minimum is attained on the smooth locus. Equivalently $T_{x}X$ is the dual of ${\mathfrak{m}}_x/{\mathfrak{m}}_x^2$, or the simultaneous kernel of the differentials of the defining equations at $x$. If “variety” permits reducible sets, apply this definition to each irreducible component and take the maximum; taking the minimum over an arbitrary reducible set with components of unequal dimension would be incorrect.

The ideal $(f)$ is prime, since an irreducible element of the polynomial ring is prime. By the Hilbert Nullstellensatz, it is the full vanishing ideal of $Z(f)$. Its Zariski tangent space at a point $a$ is therefore the kernel of the single linear form $d{f}_a$, and has dimension $n-1$ or $n$.

At least one partial derivative $f_j$ is nonzero. In characteristic zero this is immediate for a nonconstant polynomial. In characteristic $p$, if all partial derivatives vanished, all exponents would be multiples of $p$. Since an algebraically closed field is perfect, $f$ would be a $p$th power of a nonconstant polynomial, contradicting irreducibility.

The nonzero $f_j$ cannot vanish everywhere on $Z(f)$: the Nullstellensatz would give $f_j^r\in (f)$, hence $f\mid f_j$, contradicting $\mathrm{deg}{f}_j<\mathrm{deg}f$. Thus some point has $d{f}_a\ne 0$, and the minimum tangent-space dimension is