A projective cone with a smooth conic base

Set up the objects. By [F1] the Axiom of Choice is available. Since $k$ is a field [F31] and $\mathrm{char}k\mathrm{\ne }2$ [F32], the element $2=2\cdot 1_k$ is nonzero, hence invertible in $k$. Every monomial of $F=X_0^2+X_1^2-X_2^2$ has total degree two, so $F$ is homogeneous of degree two [F4] and does not involve $X_3$. The points of projective space are the classes $[a_0:\cdots :a_n]$ of [F2]; the hyperplane $V_{+}(X_3)$ consists exactly of the classes $[a_0:a_1:a_2:0]$, which dropping the last coordinate identifies with ${\mathbf{P}}_k^2$, and under this identification $X\cap V_{+}(X_3)=V_{+}(X_0^2+X_1^2-X_2^2)=C$ [F3]. Also $F(v)=0$, so $v\in X$ [F3], and $1+0-1=0$, so $[1:0:1]\in C$; both $X$ and $C$ are nonempty.

Chart equations. For each $i\in \{0,1,2,3\}$ the normalization of the $i$th coordinate identifies $D_{+}(X_i)$ with ${\mathbf{A}}_k^3$ [F5], and [F6] identifies $X\cap D_{+}(X_i)$ with the affine hypersurface obtained by setting $X_i=1$ in $F$. Explicitly, in the chart's own coordinates, $X\cap D_{+}(X_3)$ is $V(g_3)$ for $g_3=x_0^2+x_1^2-x_2^2$, while $X\cap D_{+}(X_0)$, $X\cap D_{+}(X_1)$ and $X\cap D_{+}(X_2)$ are $V(g_0)$, $V(g_1)$ and $V(g_2)$ for $g_0=1+x_1^2-x_2^2$, $g_1=x_0^2+1-x_2^2$ and $g_2=x_0^2+x_1^2-1$; in these three charts the ratio $x_3=X_3\mathrm{/}{X}_i$ does not occur. The vertex $v=[0:0:0:1]$ lies in $D_{+}(X_3)$ with coordinates $(0,0,0)=0$. In the plane $V_{+}(X_3)\cong {\mathbf{P}}_k^2$ with coordinates $X_0,X_1,X_2$ the same lemma gives $C\cap D_{+}(X_i)=V(h_i)\subseteq {\mathbf{A}}_k^2$ for $i=0,1,2$, with $h_0=1+x_1^2-x_2^2$, $h_1=x_0^2+1-x_2^2$ and $h_2=x_0^2+x_1^2-1$, the same displayed quadratics read in the two remaining coordinates.

Open-chart locality. Let $W$ be a classical prevariety over $k$ and let $U\subseteq W$ be open with $p\in U$. By [F21] the stalk ${\mathcal{O}}_{W,p}$ is the filtered colimit of the ${\mathcal{O}}_W(V)$ over the open neighbourhoods $V$ of $p$, and the neighbourhoods of $p$ contained in $U$ are cofinal among all of them: for any open $V\ni p$ the intersection $V\cap U$ is an open neighbourhood of $p$ contained in $U$ and in $V$. Hence the stalks of ${\mathcal{O}}_W$ and of the restricted sheaf at $p$ are canonically isomorphic, ${\mathcal{O}}_{W,p}\cong {\mathcal{O}}_{U,p}$, and the maximal ideals correspond; by [F22] these are local rings with residue field $k$. Since the tangent space, the dimension of the local ring and regularity are invariants of this local ring [F23], the point $p$ has the same tangent space, the same local dimension and the same regularity status computed in $W$ and in $U$.

Affine charts and their schemes. Let $U$ be an affine chart of a classical variety with coordinate ring $k[U]$ [F24] and let $p\in U$ with vanishing ideal ${\mathfrak{m}}_p$. By [F19] there is a canonical isomorphism ${\mathcal{O}}_{U,p}\cong k[U{]}_{{\mathfrak{m}}_p}$; if $p$ is a $k$-rational point, its evaluation ideal is maximal with residue field $k$ [F33], and the affine scheme $\mathrm{Spec}(k[U])$ has the same local ring $k[U{]}_{{\mathfrak{m}}_p}$ at that point [F20]. Therefore the intrinsic tangent space of [F18] and the Jacobian-kernel description of [F16] applied to the chart scheme compute the tangent space of the classical point $p$, and the local dimension and regularity computed on the chart scheme agree with those of $p$ in the variety.

The cone over the conic. Let $p=[a_0:a_1:a_2:0]\in C$ and let $[s:t]\in {\mathbf{P}}^1$. The point $[s{a}_0:s{a}_1:s{a}_2:t]$ lies on the line $\overline{vp}$, and $F(s{a}_0,s{a}_1,s{a}_2,t)=s^2(a_0^2+a_1^2-a_2^2)=0$ by $F$-homogeneity and $X_3$-independence [F3, F4], so the line is contained in $X$. Conversely, if $[b_0:b_1:b_2:b_3]\in X$ has $(b_0,b_1,b_2)\mathrm{\ne }0$, then $[b_0:b_1:b_2:0]\in C$ and the point lies on the line joining it to $v$; the only point of $X$ with $b_0=b_1=b_2=0$ is $v$, by [F2]. Thus $X$ is the union of $v$ with the lines joining $v$ to the points of $C$.

Squarefreeness of the chart equations. Each of the seven chart polynomials displayed in step 1.2 is nonzero, nonconstant, and of the shape $f=c+{\sum }_{j\in S}{ϵ}_j{x}_j^2$ with $c\in k$, $\mathrm{\mid }S\mathrm{\mid }\ge 2$ and nonzero coefficients ${ϵ}_j\in k$: the four three-variable equations are $x_0^2+x_1^2-x_2^2$, $1+x_1^2-x_2^2$, $x_0^2+1-x_2^2$, $x_0^2+x_1^2-1$, and the three two-variable equations are the last three read in two variables. Suppose such an $f$ were not squarefree, so that some irreducible factor $q$ occurs in it at least twice [F25]; then $f=q^{2}r$ with $r\mathrm{\ne }0$. Fix a variable $x_j$, view $k[x_1,\dots ,x_N]$ as the one-variable polynomial ring in $x_j$ over the remaining variables [F29, F30], and use that this coefficient ring is a domain [F13] and that degrees in $x_j$ add [F26]; for $q^2=q\cdot q$ this gives ${\mathrm{deg}}_{x_j}(f)=2{\mathrm{deg}}_{x_j}(q)+{\mathrm{deg}}_{x_j}(r)$. For $j\in S$ this degree is $2$, and for $j\mathrm{\notin }S$ it is $0$, so ${\mathrm{deg}}_{x_j}(q)={\mathrm{deg}}_{x_j}(r)=0$ for $j\mathrm{\notin }S$. If ${\mathrm{deg}}_{x_j}(q)=0$ for every $j\in S$ as well, then $q$ is a nonzero constant, hence a unit [F27], contradicting irreducibility [F28]. Otherwise fix $z\in S$ with ${\mathrm{deg}}_z(q)\ge 1$; then ${\mathrm{deg}}_z(q)=1$ and ${\mathrm{deg}}_z(r)=0$, so $q=q_0+x_z{q}_1$ with $q_1\mathrm{\ne }0$ and $r$ involving no $x_z$, and comparing $x_z$-coefficients in $f=q^{2}r$ gives $2q_0{q}_{1}r=0$. The coefficient $2$ is nonzero in $k$ [F31, F32], $q_1$ and $r$ are nonzero, and the coefficient ring is a domain [F13], so $q_0=0$; then $q=x_z{q}_1$, and since $q$ is irreducible and $x_z$ is a nonunit [F27], the cofactor $q_1$ is a unit [F28], so $q=u{x}_z$ with $u$ a unit [F27]. Then $f=u^2{x}_z^{2}r$ is divisible by $x_z$, but $\mathrm{\mid }S\mathrm{\mid }\ge 2$ provides $j\in S$ with $j\mathrm{\ne }z$, and the monomial ${ϵ}_j{x}_j^2$ has no factor $x_z$, contradicting divisibility. Hence every displayed chart polynomial is nonconstant and squarefree.

Partial derivatives of the chart equations. By the formal monomial rule of [F17], ${\mathrm{\partial }}_{x_0}{g}_3=2x_0$, ${\mathrm{\partial }}_{x_1}{g}_3=2x_1$ and ${\mathrm{\partial }}_{x_2}{g}_3=-2x_2$; for $g_0=1+x_1^2-x_2^2$ the partials are $2x_1$, $-2x_2$ and $0$ with respect to the three chart variables, for $g_1=x_0^2+1-x_2^2$ they are $2x_0$, $-2x_2$ and $0$, and for $g_2=x_0^2+x_1^2-1$ they are $2x_0$, $2x_1$ and $0$; the same three pairs of nonzero partials occur for $h_0,h_1,h_2$ in two variables. Since $2\mathrm{\ne }0$ in $k$ [F31, F32], the three partials of $g_3$ vanish simultaneously exactly at the origin $(0,0,0)$, the two nonzero partials of $g_0$ vanish exactly where $x_1=x_2=0$, and at every such point $g_0=1\mathrm{\ne }0$, so no point of $V(g_0)$ has all its partials zero; the same computation shows that no point of $V(g_1)$, $V(g_2)$, $V(h_0)$, $V(h_1)$ or $V(h_2)$ has all its partials zero, and $V(h_i)=V(g_i)\cap \{x_3=0\}$ inside $V(g_i)$ for $i=0,1,2$.

Tangent space at the vertex. The origin is a $k$-rational point of the affine scheme $\mathrm{Spec}(k[x_0,x_1,x_2]\mathrm{/}(g_3))$: by the squarefreeness of step 2.1 the ideal $(g_3)$ is radical and is the vanishing ideal of $V(g_3)$ [F8], so the coordinate ring of the chart is $k[x_0,x_1,x_2]\mathrm{/}(g_3)$ [F24], and the evaluation ideal $(x_0,x_1,x_2)$ is maximal with residue field $k$ [F33]. By [F16] the tangent space at that point is the kernel of the Jacobian matrix of [F17], whose single row is $J_{g_3}(0)=(0,0,0)$ by step 2.2, so the kernel is all of $k^3$ and ${\dim }_k{T}_0=3$. By the chart comparisons of steps 1.3 and 1.4 this is the tangent space of the vertex, $T_{v}X\cong k^3$, of dimension three.

The base conic is smooth. Every point of $C$ lies in $C\cap D_{+}(X_i)$ for at least one $i\in \{0,1,2\}$ [F2], and on that chart $C\cap D_{+}(X_i)=V(h_i)\subseteq {\mathbf{A}}_k^2$ with $h_i$ nonconstant and squarefree by steps 1.2 and 2.1. By the gradient test [F7], a point $a\in V(h_i)$ is singular precisely when both partial derivatives of $h_i$ vanish at $a$; by step 2.2 none of the three polynomials has such a point, so no point of the conic is a singular point of its chart. By the open-chart locality of step 1.3 (applied to the conic $C$ and its open chart $C\cap D_{+}(X_i)$), no point of $C$ is singular on $C$: the conic has no singular point, that is, it is a smooth conic.

Local dimension at the vertex and singularity. The affine space ${\mathbf{A}}_k^3$ is irreducible: its coordinate ring $k[x_0,x_1,x_2]$ is an integral domain [F13] and the zero ideal is prime because the quotient by it is that domain [F15], and the corresponding nonempty algebraic set is ${\mathbf{A}}^3$ itself by the Nullstellensatz correspondence [F14]. Moreover $\dim {\mathbf{A}}_k^3=3$ [F12]. The polynomial $g_3$ is a nonzero nonunit of $k[x_0,x_1,x_2]$ [F25, F27], so by the principal-subvariety theorem [F11] every irreducible component of $V(g_3)$ has dimension $3-1=2$. The local dimension at the origin is therefore $\dim {\mathcal{O}}_{V(g_3),0}={\max }_{0\in X_i}\dim X_i=2$ [F10], and by step 1.3 the local ring of the cone at the vertex, hence also its local dimension, is the same: $\dim {\mathcal{O}}_{X,v}=2$. With ${\dim }_k{T}_{v}X=3$ from step 3.1, the numerical criterion of the classical dimension test [F9] shows that the vertex satisfies ${\dim }_k{T}_{v}X\mathrm{\ne }{\dim }_{v}X$ and is therefore a singular point of the cone; equivalently, all partials of $g_3$ vanish at the origin by step 2.2, and the gradient test [F7] makes the origin a singular point of the chart $V(g_3)$. Thus the vertex has tangent dimension three, local dimension two, and is singular.

The cone is smooth away from the vertex. Let $q\in X$ with $q\mathrm{\ne }v$. If $q$ had $X_0=X_1=X_2=0$, then by [F2] it would equal $[0:0:0:1]=v$, so $q\in D_{+}(X_i)$ for some $i\in \{0,1,2\}$. On that chart $X\cap D_{+}(X_i)=V(g_i)$ with $g_i$ nonconstant and squarefree by steps 1.2 and 2.1, and by step 2.2 no point of $V(g_i)$ has all three partial derivatives equal to zero; hence no point of the chart is singular by the gradient test [F7], in particular $q$ is not singular, and step 1.3 transfers this conclusion from the chart to $X$. Together with step 4.1, the singular locus of the cone is exactly the vertex $\{v\}$.

Boundary and scope dispositions. Nonemptiness: $v\in X$ and $[1:0:1]\in C$ by step 1.1, and each of the four cone charts and three conic charts is nonempty because the corresponding equation is satisfied at the origin of the vertex chart and at $x_1=0,x_2=1$ or $x_0=1,x_1=0$ in the other charts; the empty case therefore has no instance. Zero cases: the vertex is the origin, the zero tuple, of the vertex chart; the Jacobian row there is the zero row of step 3.1 with kernel all of $k^3$, and the zero vector lies in every kernel and every tangent space. One: each chart uses one defining equation, one Jacobian row, and the vertex is the one singular point of the cone by steps 4.1 and 5.1. Degenerate case: the excluded characteristic two is genuinely degenerate, since there $X_0^2+X_1^2-X_2^2=(X_0+X_1+X_2{)}^2$ is a square and the partials of step 2.2 become $0$; the hypothesis $\mathrm{char}k\mathrm{\ne }2$ enters through [F31, F32] in steps 1.1, 2.1 and 2.2. Endpoints: the statement concerns a fixed quadric over a field and has no ordered parameter or interval; its numerical values, the tangent dimension three and the local dimension two at the vertex and the dimension one of the conic, are discrete, so no endpoint case arises. Nonempty choice: AC is declared in [F1] and used through the cited suppliers [F7], [F9], [F10], [F11], [F12], [F14] and [F19], each cited at the step that uses it; the squarefreeness argument of step 2.1, the derivative computations of step 2.2, the chart comparisons of steps 1.3–1.4 and the line containment of step 1.5 use no choice. Both directions of the biconditional in [F7] are used: the reverse direction at the vertex in step 4.1 and the forward direction, through its contrapositive, at every other point in steps 3.2 and 5.1; likewise the numerical criterion of [F9] is used in the direction regular implies equal dimensions, through its contrapositive at the vertex.