An embedded point at the cusp of a plane cubic

On \(u=1\) the twisted cubic has parametrization \[ x=t^3,\qquad y=t^2,\qquad z=t. \] Projection forgets \(z\), so the image in the \((x,y)\)-plane satisfies \[ x^2=t^6=y^3. \] Conversely the map \(t\mapsto(t^3,t^2)\) has dense image in the irreducible plane curve \(x^2-y^3=0\). Thus the projected curve is the cusp \[ C=V(x^2-y^3), \] with cusp at the origin.

The automorphism used in III.9.8.3 multiplies the coordinate in the projection direction by \(a\) and fixes the other coordinates. Hence the affine parametrization of the twisted cubic becomes exactly the displayed one. For \(a\ne0\) this is isomorphic to the original twisted cubic by rescaling \(z\).

Consider the \(k\)-algebra homomorphism \[ \Phi:k[a,x,y,z]\longrightarrow k[a,t] \] defined by \[ a\mapsto a,\qquad x\mapsto t^3,\qquad y\mapsto t^2,\qquad z\mapsto at. \] Every generator of \(J\) maps to zero, so \[ J\subseteq\ker\Phi. \]

Modulo \(J\), use the relations \[ x^2=y^3,\qquad z^2=a^2y,\qquad yz=ax,\qquad xz=ay^2 \] to reduce every monomial to a \(k[a]\)-linear combination of \[ y^j,\qquad xy^j\quad(j\ge0),\qquad z. \] Their images under \(\Phi\) are respectively \[ t^{2j},\qquad t^{2j+3},\qquad at. \] These are \(k[a]\)-linearly independent: they have distinct powers of \(t\), and a relation involving \(at\) would have coefficient \(c(a)a=0\) in the domain \(k[a]\), hence \(c(a)=0\). Thus the induced map \[ k[a,x,y,z]/J\longrightarrow k[a,t] \] is injective. Consequently \[ J=\ker\Phi, \] and the quotient identifies with the subalgebra \[ B=k[a,t^2,t^3,at]\subseteq k[a,t]. \]

The ring \(B\) is a subring of the domain \(k[a,t]\) containing \(k[a]\). Hence \(B\) is torsion-free as a \(k[a]\)-module. The ring \(k[a]\) is a principal ideal domain, and every torsion-free module over a principal ideal domain is flat. Therefore \(B\) is flat over \(k[a]\).

Thus the closed family defined by \(J\) is the flat extension prescribed in III.9.8.3–9.8.4.

The special fibre is obtained by tensoring with \(k[a]/(a)\), equivalently by adding \(a\) to the total ideal and then setting \(a=0\). The four generators of \(J\) become \[ x^2-y^3,\qquad z^2,\qquad yz,\qquad xz, \] which gives the displayed ideal.

The radical of \(J_0\) is \[ \sqrt{J_0}=(z,x^2-y^3). \] Indeed, \(z^2\in J_0\) forces \(z\) into the radical, and after quotienting by \(z\) the remaining reduced equation is the irreducible cusp equation \(x^2-y^3\). Therefore \[ |X_0|=V(z,x^2-y^3), \] which is exactly the projected cuspidal cubic of step 1.

Let \[ A_0=k[x,y,z]/J_0 \] and denote residue classes by the same letters. The element \(z\in A_0\) is nonzero. Indeed, the normal form of step 3 makes the class of \(z\) a \(k[a]\)-basis element independent from the classes \(y^j,xy^j\). If \(z\) belonged to \(aB\), comparison of the coefficient of that basis element would give \(1\in a,k[a]\), which is impossible. Thus its image in \(B/aB=A_0\) is nonzero. The special-fibre relations give \[ xz=yz=z^2=0. \] Thus the maximal ideal \[ \mathfrak m=(x,y,z) \] annihilates \(z\). Every element outside \(\mathfrak m\) has the form \(c+h\) with \(c\in k^\times\) and \(h\in\mathfrak m\). Since \(\mathfrak m z=0\), such an element sends \(z\) to \(cz\ne0\). Therefore \[ \operatorname{Ann}_{A_0}(z)=\mathfrak m. \] Hence \(\mathfrak m\) is an associated prime of \(A_0\).

The unique minimal prime is \[ \mathfrak p=(z,x^2-y^3), \] and \[ \mathfrak p\subsetneq\mathfrak m. \] Therefore \(\mathfrak m\) is an embedded associated prime. Its closed point is precisely the cusp at the origin. So the special fibre is the cuspidal cubic together with an embedded point at its cusp.

Steps <1>1–<1>4 construct the flat degeneration, steps <1>5–<1>6 identify its special-fibre support with the cuspidal cubic, and step 7 proves that the cusp is an embedded associated point.