In affine \(3\dash\)space with coordinates \(x, y, z\), let \(C\) be the conic \((x-1)^2 + y^2 = 1\) in the \(xy\dash\)plane, and let \(P\) be the point \((0,0,t)\) on the \(z\dash\)axis. Let \(Y_t\) be the closure in \(\PP^3\) of the cone over \(C\) with vertex \(P\).
Show that as \(t\) varies, the surfaces \(\ts{Y_t}\) form a linear system of dimension \(1\), with a moving singularity at \(P\). The base locus of this linear system is the conic \(C\) plus the \(z\dash\)axis.
This happens in any characteristic.
(Erratum to the printed all-characteristic statement)
There are two literal defects in the printed model.
First, in characteristic \(2\), \[ (x-1)^2+y^2=1 \] reduces to \[ (x+y)^2=0, \] so the displayed \(C\) is a doubled line, not a nonsingular conic. The pencil equation below also loses its second generator because its coefficient is \(2\).
Second, the origin \(O=(0,0,0)\) lies on \(C\). Thus the literal union of lines joining \(O\) to \(C\) is just the plane \(z=0\). It is not the \(t=0\) member \(XZ=0\) of the pencil obtained as the flat/projective limit of the cones with \(t\ne0\).
Accordingly, the printed equations prove the intended moving-singularity phenomenon for \(\operatorname{char}k\ne2\) and \(t\ne0\), with the complete pencil obtained by projective closure in the parameter. The phenomenon itself does occur in every characteristic: one may replace \(C\) by the smooth conic \[ xy+x+y=0, \] which gives the characteristic-free pencil constructed in step <1>6 below.
Solution
Use homogeneous coordinates \[ [X:Y:Z:W] \] on \(\PP^3\), with affine chart \(W=1\) carrying the coordinates \(x,y,z\). Let \[ L=V(X,Y) \] be the projective \(z\)-axis.
1. Assume first that \(\operatorname{char}k\ne2\). The projective closure of the printed conic is\[ C=V(Z,Q), \qquad Q=X^2+Y^2-2XW. \]
Homogenizing \[ (x-1)^2+y^2=1 \] gives \[ X^2+Y^2-2XW=0 \] in the plane \(Z=0\).
Its gradient in that plane is \[ (2X-2W,\ 2Y,\ -2X). \] Because \(2\ne0\), these derivatives have no common projective zero on \(Q=0\). Thus \(C\) is a nonsingular conic.
2. For \(t\ne0\), the cone over \(C\) with vertex\[ P_t=[0:0:t:1] \]is the quadric surface\[ Y_t=V(F_t), \qquad F_t=tQ+2XZ. \]
Take a point \[ Q_0=(u,v,0)\in C \] on the affine chart. A point on the line joining \(P_t\) to \(Q_0\) has the form \[ (x,y,z)=(\lambda u,\lambda v,(1-\lambda)t). \] Since \(t\ne0\), \[ \lambda=1-\frac zt. \] The conic equation \[ u^2+v^2-2u=0 \] therefore becomes \[ \begin{aligned} 0 &=\lambda^2(u^2+v^2)-2\lambda^2u\\ &=x^2+y^2-2x\left(1-\frac zt\right). \end{aligned} \] Multiplying by \(t\) gives \[ t(x^2+y^2-2x)+2xz=0, \] whose homogenization is precisely \[ F_t=t(X^2+Y^2-2XW)+2XZ. \]
Conversely, \(V(F_t)\) contains \(C\) and \(P_t\), and the displayed calculation shows every line \(\overline{P_tQ_0}\) is contained in it. Both the union of these lines and \(V(F_t)\) are irreducible surfaces of degree two for \(t\ne0\), so they are equal.
3. The nonzero-vertex cones lie in the pencil\[ \Lambda=\PP\langle Q,\,XZ\rangle. \]
Step 2 gives \[ F_t=tQ+2XZ. \] Thus every \(Y_t\) with \(t\ne0\) is a member of the projective line of quadrics spanned by \(Q\) and \(XZ\). The two quadrics are linearly independent, so \[ \boxed{\dim\Lambda=1}. \]
As \(t\) ranges through \(k^\times\), the coefficient ratio \([t:2]\) ranges through the open subset of \(\Lambda\) where both coordinates are nonzero. Its closure is the whole pencil. The missing boundary members are the expected projective limits \(Q=0\) and \(XZ=0\).
4. The set-theoretic base locus of the pencil is exactly\[ C\cup L. \]
The base locus is \[ V(Q,XZ). \] If \(Z=0\), the equation \(Q=0\) cuts out precisely \(C\). If \(X=0\), then \[ Q=Y^2, \] so set-theoretically \(Y=0\), giving the \(z\)-axis \[ L=V(X,Y). \] Thus \[ \boxed{|\operatorname{Bs}\Lambda|=C\cup L}. \] Every nonzero-vertex cone contains both: it contains \(C\) by construction, and it contains \(L\) because \(O=(0,0,0)\in C\) and the line joining \(O\) to \(P_t\) is \(L\).
5. For \(t\ne0\), the point \(P_t\) is the unique singular point of \(Y_t\).
The four partial derivatives of \[ F_t=t(X^2+Y^2-2XW)+2XZ \] are \[ (F_t)_X=2tX-2tW+2Z, \] \[ (F_t)_Y=2tY, \qquad (F_t)_Z=2X, \qquad (F_t)_W=-2tX. \] Since \(2t\ne0\), simultaneous vanishing forces \[ X=Y=0, \qquad Z=tW. \] Hence the only singular point is \[ [0:0:t:1]=P_t. \] As \(t\) varies through \(k^\times\), this singular point moves along the punctured affine part of the base line \(L\).
6. The intended phenomenon has an all-characteristic model.
Replace the printed conic by \[ C'=V(Z,Q'), \qquad Q'=XY+XW+YW. \] On the affine chart \(W=1\) this is \[ xy+x+y=0. \]
This conic is smooth in every characteristic. Indeed, \[ (Q'_X,Q'_Y,Q'_W)=(Y+W,\ X+W,\ X+Y). \] If these three derivatives vanish, then \(X=Y=-W\); substituting into \(Q'\) gives \(Q'=-W^2\) (which is \(W^2\) in characteristic \(2\)), so no projective point of \(C'\) has all derivatives zero.
For \(t\ne0\), the same line-parameter elimination as in step 2 gives the cone with vertex \(P_t\): \[ Y'_t =V\bigl(tQ'-Z(X+Y)\bigr). \] Thus the cones lie in the pencil \[ \Lambda'=\PP\langle Q',\,Z(X+Y)\rangle, \] which has dimension one in every characteristic.
Its base locus is \[ V(Q',Z(X+Y)). \] The component \(Z=0\) gives \(C'\). If \(X+Y=0\), then \[ Q'=XY=-X^2, \] so set-theoretically \(X=Y=0\), giving again the axis \(L\). Hence \[ |\operatorname{Bs}\Lambda'|=C'\cup L. \]
Finally put \[ F'_t=tQ'-Z(X+Y). \] Its derivatives are \[ (F'_t)_X=t(Y+W)-Z, \] \[ (F'_t)_Y=t(X+W)-Z, \] \[ (F'_t)_Z=-(X+Y), \qquad (F'_t)_W=t(X+Y). \] At \(P_t=[0:0:t:1]\) they all vanish. Conversely, if all derivatives vanish, then \(X+Y=0\) and \[ Z=t(X+W)=t(Y+W). \] If \(\operatorname{char}k\ne2\), these equations force \(X=Y=0\). If \(\operatorname{char}k=2\), substituting \(Y=X\) into \(F'_t=0\) gives \[ tX^2=0, \] so again \(X=Y=0\) because \(t\ne0\). Then \(Z=tW\), and the unique singular point is \(P_t\).
Thus in every characteristic the pencil \(\Lambda'\) has a singular point that moves along the fixed base line \(L\).
7. The \(t=0\) discrepancy in the printed model is exactly the pencil limit.
The origin \[ O=[0:0:0:1] \] lies on the printed conic \(C\). The literal geometric cone obtained by joining \(O\) to all points of \(C\) lies entirely in the plane \(Z=0\); since the secant lines through a point of a nonsingular plane conic sweep the whole plane, that cone is simply \[ V(Z). \]
By contrast, the \(t\to0\) member of the pencil from step 3 is \[ V(XZ)=V(X)\cup V(Z). \] Thus the pencil is the projective closure of the family of cones with \(t\ne0\), but its special member at parameter \(t=0\) is not the literal cone with vertex \(O\). This is precisely the second defect stated in the erratum.
8. Q.E.D. for the corrected statement.
Steps <1>1–<1>5 prove the intended pencil, base-locus, and moving-singularity claims for the printed model where that model is valid (\(\operatorname{char}k\ne2\), \(t\ne0\)). Step 6 gives an explicit version valid in every characteristic, and step 7 identifies the special \(t=0\) degeneration that prevents the literal printed family from being the whole pencil.