A nondegenerate projective quadric

Set up the objects and charts. All monomials $X_i^2$ have total degree two, so $F$ is homogeneous of degree two [F4] and $F\mathrm{\ne }0$. By [F3] the set $Q=V_{+}(F)\subseteq {\mathbf{P}}_k^n$ is the projective algebraic set of zeros of $F$, and by [F5] the normalization of the $i$th coordinate identifies $D_{+}(X_i)$ with ${\mathbf{A}}_k^n$, so by [F6] the intersection $Q\cap D_{+}(X_i)$ is the affine hypersurface $V(g_i)$ cut out by $g_i=1+{\sum }_{j\mathrm{\ne }i}{x}_j^2$ in the $n$ ratio coordinates; write $Q_i=Q\cap D_{+}(X_i)$ and $A_i=k[x_j:j\mathrm{\ne }i]\mathrm{/}(g_i)$ for its coordinate ring [F6, F29]. The charts $D_{+}(X_i)$ for $i=0,\dots ,n$ cover ${\mathbf{P}}_k^n$ and hence $Q$ [F2]. Since $k$ is a field [F7] with $\mathrm{char}k\mathrm{\ne }2$ [F8], the element $2=2\cdot 1_k$ is nonzero, and by [F9] the nonconstant polynomial $t^2+1$ has a root $\iota \in k$; then $F(1,\iota ,0,\dots ,0)=1+{\iota }^2=0$, so the class $p=[1:\iota :0:\cdots :0]$ is a point of $Q$ [F2, F3] and $Q\mathrm{\ne }\mathrm{\varnothing }$. Each $g_i$ has degree $2$ in every variable $x_j$ with $j\mathrm{\ne }i$ [F10, F11], and $1+{\sum }_{j\mathrm{\ne }i}{x}_j^2=0$ holds in $A_i$.

Squarefreeness of the chart equations. Each $g_i$ is of the shape $g_i=c+{\sum }_{j\in S}{ϵ}_j{x}_j^2$ with $c=1$, $S=\{0,\dots ,n\}\setminus \{i\}$, $\mathrm{\mid }S\mathrm{\mid }=n\ge 2$ and all coefficients ${ϵ}_j=1\mathrm{\ne }0$. Suppose some $g_i$ were not squarefree; since $k[x_j:j\mathrm{\ne }i]$ is a unique factorisation domain [F15], some irreducible element $q$ occurs in $g_i$ at least twice [F16], that is $g_i=q^{2}r$ with $r\mathrm{\ne }0$. Fix a variable $x_j$ and view the ring as a one-variable polynomial ring in $x_j$ over the remaining variables [F10]; that coefficient ring is an integral domain [F12] and degrees in $x_j$ add on products [F13], so $q^2=q\cdot q$ gives ${\mathrm{deg}}_{x_j}(g_i)=2{\mathrm{deg}}_{x_j}(q)+{\mathrm{deg}}_{x_j}(r)$. For $j\in S$ the left side is $2$ and for $j\mathrm{\notin }S$ it is $0$, so ${\mathrm{deg}}_{x_j}(q)={\mathrm{deg}}_{x_j}(r)=0$ for every $j\mathrm{\notin }S$. If also ${\mathrm{deg}}_{x_j}(q)=0$ for every $j\in S$, then $q$ is a nonzero constant, hence a unit [F14], contradicting irreducibility [F16]. 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 $r$ not involving $x_z$, and comparing $x_z$-coefficients in $g_i=q^{2}r$ gives $2q_0{q}_{1}r=0$. Since $2\mathrm{\ne }0$ in $k$ [F7, F8], $q_1\mathrm{\ne }0$, $r\mathrm{\ne }0$ and the coefficient ring is a domain [F12], this forces $q_0=0$, so $q=x_z{q}_1$; as $q$ is irreducible and $x_z$ is a nonunit [F14], the cofactor $q_1$ is a unit [F16] and $q=u{x}_z$ for a unit $u$. Then $g_i=u^2{x}_z^{2}r$ is divisible by $x_z$, hence vanishes after substituting $x_z=0$; but that substitution leaves $c+{\sum }_{j\in S,j\mathrm{\ne }z}{ϵ}_j{x}_j^2$, whose term ${ϵ}_j{x}_j^2$ for some $j\in S\setminus \{z\}$ (here $\mathrm{\mid }S\mathrm{\mid }\ge 2$ is used) is a nonzero monomial, so the substituted polynomial is nonzero — a contradiction. Hence every $g_i$ is nonconstant and squarefree.

Radical chart ideals. Since $g_i$ is squarefree, [F18] says the ideal $(g_i)$ of $k[x_j:j\mathrm{\ne }i]$ is already radical, so the reduced classical chart $V(g_i)$ has vanishing ideal $I(V(g_i))=(g_i)$ and coordinate ring $k[V(g_i)]=A_i$ [F29], a reduced finitely generated $k$-algebra [F30]. Moreover $g_i$ is a nonzero nonunit of $k[x_j:j\mathrm{\ne }i]$: it is nonzero and nonconstant by step 2.1, and a unit would have to be a nonzero constant [F14]. [F14, F18, F29, F30, step 2.1, given, algebra] The associated scheme. To interpret the structure morphism in the Example, glue the reduced affine schemes $\mathrm{Spec}{A}_i$ as follows. Write $x_j^{(i)}=X_j\mathrm{/}{X}_i$ on chart $i$. On the overlap with chart $h$, invert $x_h^{(i)}$ and identify $x_i^{(h)}=1\mathrm{/}{x}_h^{(i)}$ and $x_j^{(h)}=x_j^{(i)}\mathrm{/}{x}_h^{(i)}$ for $j\mathrm{\ne }i,h$. Substitution sends $g_h$ to $g_i\mathrm{/}(x_h^{(i)}{)}^2$, so it induces an isomorphism $(A_h{)}_{x_i^{(h)}}\cong (A_i{)}_{x_h^{(i)}}$ with inverse obtained by interchanging $i,h$. The ratio formulas compose identically on triple overlaps. Thus [F45, F46] glue these affine schemes to a reduced finite-type $k$-scheme $Q^{\mathrm{s}\mathrm{c}\mathrm{h}}$: reducedness holds on each chart by the radical-ideal calculation above, and the cover has $n+1$ charts of finite-type algebras. Its closed-point charts and local rings are the classical $Q_i$ and their local rings by [F20, F32, F33], and their identifications are precisely [F6]. This is the associated scheme of the stated reduced classical variety. In the scheme assertions below, $Q$ denotes this $Q^{\mathrm{s}\mathrm{c}\mathrm{h}}$; every scheme point, including a nonclosed one, lies in some $\mathrm{Spec}{A}_i$.

The chart hypersurfaces have dimension $n-1$. The polynomial ring $k[x_1,\dots ,x_n]$ is an integral domain [F12], so its zero ideal is prime [F19] and the corresponding nonempty irreducible algebraic set is ${\mathbf{A}}_k^n$ itself by the Nullstellensatz correspondence [F20]; by [F21], $\dim {\mathbf{A}}_k^n=n$. The polynomial $g_i$ is a nonzero nonunit of $k[{\mathbf{A}}_k^n]=k[x_1,\dots ,x_n]$ by step 3.1, so [F22] applies with $X={\mathbf{A}}_k^n$: the zero set $V(g_i)$ is nonempty and every irreducible component of $V(g_i)$ has dimension $n-1$.

Pure dimension of the quadric. By [F26] every classical variety is Noetherian with finitely many irreducible components. Let $Z$ be an irreducible component of $Q$. Since the charts $Q_i=V(g_i)$ cover $Q$ [F2, F5], the intersection $Z\cap D_{+}(X_i)$ is nonempty for some $i$, so $W=Z\cap V(g_i)$ is a nonempty open subset of the irreducible space $Z$; by [F25], $W$ is irreducible and dense in $Z$, and by [F24] $\dim W=\dim Z$. The irreducible subset $W$ of $V(g_i)$ is contained in an irreducible component $W^{\mathrm{\prime }}$ of $V(g_i)$ [F27], and step 4.1 gives $\dim W^{\mathrm{\prime }}=n-1$, so $\dim Z=\dim W\le \dim W^{\mathrm{\prime }}=n-1$. Conversely $W^{\mathrm{\prime }}$ is irreducible and contains $W$, so its closure $\overline{W^{\mathrm{\prime }}}$ in $Q$ is irreducible [F27] and contains the dense subset $W$ of $Z$; hence $Z\subseteq \overline{W^{\mathrm{\prime }}}$, and since $Z$ is an irreducible component of $Q$ while $\overline{W^{\mathrm{\prime }}}$ is irreducible and closed, $\overline{W^{\mathrm{\prime }}}=Z$. Therefore $W^{\mathrm{\prime }}\subseteq Z$ and $n-1=\dim W^{\mathrm{\prime }}\le \dim Z$, so $\dim Z=n-1$ for every irreducible component $Z$ of $Q$. Thus $Q$ has pure dimension $n-1$ [F23], and $Q$ is nonempty because $V(g_0)$ is nonempty by step 4.1 and contained in $Q$.

Jacobian rank and regularity of the chart local rings. Fix $i$ and a maximal ideal $\mathfrak{m}$ of $A_i$; by the classical correspondence [F20] the ideal $\mathfrak{m}$ is the evaluation ideal of a closed point $x$ of the chart $V(g_i)\subseteq {\mathbf{A}}_k^n$ with residue field $k$, and by [F32] and [F33] the local ring of $Q$ at $x$ agrees with the local ring of the chart, ${\mathcal{O}}_{Q,x}={\mathcal{O}}_{V(g_i),x}\cong (A_i{)}_{\mathfrak{m}}$ (using $k[V(g_i)]=A_i$ from step 3.1). By [F28] applied to the reduced chart and by step 5.1, $\dim (A_i{)}_{\mathfrak{m}}=\dim {\mathcal{O}}_{V(g_i),x}={\max }_{x\in X_j}\dim X_j=n-1$, the maximum running over the components of $V(g_i)$ through $x$. The Jacobian matrix of the one-element generating list $(g_i)$ of the ideal of $A_i$ is the $1\times n$ row with entries $\mathrm{\partial }{g}_i\mathrm{/}\mathrm{\partial }{x}_j=2x_j$ [F34]. If $x_j\in \mathfrak{m}$ for every $j\mathrm{\ne }i$, then ${\sum }_{j\mathrm{\ne }i}{x}_j^2\in \mathfrak{m}$, and since $1+{\sum }_{j\mathrm{\ne }i}{x}_j^2=0$ in $A_i$ by step 1.1 this gives $1\in \mathfrak{m}$, impossible; hence some $x_j\mathrm{\notin }\mathfrak{m}$, and because $2\mathrm{\ne }0$ in $k$ [F7, F8] the image of $2x_j$ in $A_i\mathrm{/}\mathfrak{m}=k$ is nonzero, so ${\mathrm{rank}}_{k}J(\mathfrak{m})=1=n-(n-1)=n-\dim (A_i{)}_{\mathfrak{m}}$. Since $\mathfrak{m}$ is a $k$-rational point, the Jacobian criterion [F35] applies without a perfectness hypothesis and shows that $(A_i{)}_{\mathfrak{m}}$ is a regular local ring. As $\mathfrak{m}$ was an arbitrary maximal ideal of $A_i$, every maximal localization of $A_i$ is regular.

The quadric is regular and smooth over $k$. Each $A_i$ is a finitely generated $k$-algebra [F30], and $k$ is a field, hence a Noetherian ring [F43], so $A_i$ is a Noetherian ring [F44]. By step 6.1 every maximal localization of $A_i$ is regular, so by [F37] regularity of the Noetherian ring $A_i$ can be tested at maximal ideals: $A_i$ is regular, that is, every prime localization $(A_i{)}_{\mathfrak{q}}$ is a regular local ring [F36]. Every point of $Q$ lies in a chart $Q_i=D_{+}(X_i)\cap Q$ [F2, F5], and ${\mathcal{O}}_{Q,x}\cong (A_i{)}_{{\mathfrak{q}}_x}$ for the prime ${\mathfrak{q}}_x$ of $A_i$ corresponding to $x$ [F32, F33], so every local ring of $Q$ is regular; the scheme charts constructed in step 3.1 form a finite cover by spectra of finitely generated $k$-algebras, so $Q\to \mathrm{Spec}k$ is a finite-type morphism [F41]. The field $k$ is algebraically closed, hence perfect [F40], so the equivalence of [F39] applies to the finite-type $k$-scheme $Q$: $Q$ is regular if and only if $Q\to \mathrm{Spec}k$ is smooth. Therefore $Q\to \mathrm{Spec}k$ is smooth.

Tangent dimensions and absence of singular points. Let $x$ be a closed point of $Q$. By step 7.1 the local ring ${\mathcal{O}}_{Q,x}$ is regular; by [F38] this means ${\dim }_k{T}_{x}Q=\dim {\mathcal{O}}_{Q,x}$, and by step 5.1 together with [F28] the right side is $n-1$, the maximum of $\dim Z$ over the components $Z$ of $Q$ containing $x$. Hence ${\dim }_k{T}_{x}Q=n-1$, so $x$ is a regular point of the reduced classical finite-type space $Q$ and $x\mathrm{\notin }{Q}_{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}}$ [F42]. Independently, step 6.1 exhibits a nonzero partial $\mathrm{\partial }{g}_i\mathrm{/}\mathrm{\partial }{x}_j=2x_j$ at the point, so the affine gradient test [F17] also makes the corresponding chart point nonsingular. Since every closed point is regular and every local ring of the scheme $Q$ is regular by step 7.1, the quadric has no singular point in either sense.

Boundary and scope dispositions. Nonemptiness: $p=[1:\iota :0:\cdots :0]\in Q$ by step 1.1 and each chart equation $g_i$ has a nonempty zero set by step 4.1, so the empty case has no instance. Zero cases: the origin of each chart is not on the quadric because $g_i(0)=1\mathrm{\ne }0$; this origin is exactly the point where all partials $2x_j$ vanish, so the gradient of each chart equation vanishes only off the quadric, and the zero vector lies in every tangent space. One: each chart is cut by one equation, the Jacobian of the one-element generating list has one row, and the relative dimension of the quadric is $n-1$. Degenerate case: the excluded characteristic two is genuinely degenerate, since there $F=({\sum }_i{X}_i{)}^2$ is a square, the partials of the chart equations vanish identically, and $V_{+}(F)$ is the nonreduced hyperplane ${\sum }_i{X}_i=0$, whose reduced variety is a hyperplane rather than a quadric of dimension $n-1$; the hypothesis $\mathrm{char}k\mathrm{\ne }2$ enters through [F7, F8] in steps 1.1, 2.1 and 6.1. Endpoints: the discrete parameter is $n\ge 2$, the ambient dimension is $n$, the quadric and its tangent spaces have the extreme value $n-1$, and the proof's squarefreeness step uses $\mathrm{\mid }S\mathrm{\mid }=n\ge 2$ exactly at its final divisibility contradiction, so the stated range is the endpoint of the argument. Nonempty choice: AC is declared in [F1] and used only through the AC-assuming suppliers [F18], [F20], [F21], [F22], [F23], [F24], [F26], [F28], [F32], [F37], [F39] and [F42], each cited at the step that uses it; the degree-counting argument of step 2.1, the chart identifications of steps 1.1 and 3.1 and the Jacobian rank computation of step 6.1 make no choice. Biconditional directions: step 6.1 uses the direction rank $=n-\dim A_{\mathfrak{m}}$ implies regular of the Jacobian criterion [F35], step 7.1 uses the direction regular implies smooth of [F39], and step 8.1 uses the direction regular implies $\dim T=\dim \mathcal{O}$ of [F38]; no reverse direction of these three equivalences is used anywhere, so the reverse cases are not applicable.