Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 25G / Solution

Let $X\subseteq {\mathbb{A}}^n$ have ideal $I(X)$, and let $p\in X$. Its Zariski tangent space isIf $X$ is irreducible, $p$ is smooth whenotherwise it is singular. Equivalently, for generators of $I(X)$, the Jacobian has the maximum rank $n-\dim X$ at $p$.

Now let the irreducible affine plane cubic be $X=V(f)\subseteq {\mathbb{A}}^2$. Suppose distinct points $p,q\in X$ were both singular. Parametrize their joining line byand put $h(t)=f(\ell (t))$. Then $\mathrm{deg}h\le 3$. At $t=0$,so zero is a root of multiplicity at least two. The same argument at $t=1$ gives another root of multiplicity at least two. Therefore $h$ has at least four roots counted with multiplicity and must vanish identically. The line through $p,q$ is then contained in $X$, so its linear equation divides $f$, contradicting irreducibility. This proves singular points of an irreducible affine plane cubic are unique when they exist.

For the density statement, embed an irreducible affine variety $X$ of dimension $d$ in ${\mathbb{A}}^n$ and choose generators $f_1,\dots ,f_r$ of its ideal. By dimension from minimum tangent dimension, some point has tangent dimension $d$, so the Jacobian has rank $n-d$ there. Some $(n-d)\times (n-d)$ minor $\Delta$ is consequently nonzero on $X$. The distinguished open setis nonempty, and at each of its points the Jacobian rank is at least $n-d$. Since every tangent space has dimension at least $d$, the rank is also at most $n-d$. Thus every point of $X_{\Delta }$ is smooth. A nonempty open subset of an irreducible topological space is dense, proving density of the smooth locus.

Finally write the smooth irreducible projective hypersurface asfor an irreducible homogeneous polynomial $F$. The closure of ${\pi }^{-1}(X)$ is its affine cone over a projective hypersurfaceAt a nonzero point $y\in Y$, simultaneous vanishing of all partial derivatives of $F$ would make the projective point $[y]\in X$ singular. Since $X$ is smooth, this cannot happen. Hence every nonzero point of $Y$ is smooth, and the origin is its only possible singular point.

Both possibilities occur. If $X$ is a projective hyperplane, then $F$ is linear and $Y$ is an affine linear subspace, hence smooth even at the origin. For a singular cone, take the smooth conicIts affine cone isAll first derivatives vanish at the origin, so the vertex is singular, while the preceding argument shows that it is the only singular point.

Solved by gpt-5.6-sol high.