The rank-one 2 by 2 determinantal cone

Let $k$ be an algebraically closed field, put $R=k[x,y,z,t]=k[x][y][z][t]=A[t]$ with $A=k[x,y,z]$ by [F1], and put $f=xt-yz\in R$ and $X=V(f)\subseteq {\mathbb{A}}_k^4$ with its reduced classical structure. In the unique expansion of [F2] the only nonzero coefficients of $f$ are those of the monomials $xt$ and $yz$, both of total degree $2$; the coefficients $1$ and $-1$ are nonzero because $1\mathrm{\ne }0$ in the field $k$ [F9, F10], so $f\mathrm{\ne }0$ and $f$ is homogeneous of degree $2$ by [F25]. The constant coefficient is absent, so $f(0)=0$ and the origin $0=(0,0,0,0)$ is a $k$-rational point of $X$.

For each variable $v\in \{x,y,z,t\}$ the degree of $f$ in $v$ under the expansion of [F2] is $1$: the exponent of $v$ is $1$ in one of the monomials $xt$, $yz$ and $0$ in the other, and both coefficients are nonzero [F9, F10]. By [F3] the same exponent vectors compute these degrees in the presentation of $R$ as a polynomial ring in the other three variables over the field $k$, whose coefficient ring is an integral domain by [F4, F5], and there the degree of a nonzero polynomial is additive on products by [F8]. If $f$ were a unit, say $fg=1$ with $g\in R$, then $g\mathrm{\ne }0$ and $0={\mathrm{deg}}_{x}1={\mathrm{deg}}_{x}f+{\mathrm{deg}}_{x}g=1+{\mathrm{deg}}_{x}g$, impossible since ${\mathrm{deg}}_{x}g\ge 0$; hence $f$ is a nonunit. Therefore $f$ is nonzero, nonconstant and a nonunit of $k[{\mathbb{A}}^4]=R$.

Since $f$ is a nonzero nonunit of the UFD $R$ [F4, F5], $f$ is a finite product of irreducibles [F5]; if some irreducible occurred twice, then $q^2\mid f$ for that irreducible $q$, so it suffices to rule this out. Suppose $q$ is irreducible with $q^2\mid f$, say $f=q^{2}r$: then $q\mathrm{\ne }0$ because $q$ is a nonzero nonunit [F6], and $r\mathrm{\ne }0$ because $f\mathrm{\ne }0$ by step 1.2. Present $R=k[y,z,t][x]$ by [F3]; the degree in $x$ of a nonzero polynomial over the integral domain $k[y,z,t]$ [F4, F5] is additive on products [F8], so step 1.2 gives $1={\mathrm{deg}}_{x}f={\mathrm{deg}}_x(q^2)+{\mathrm{deg}}_{x}r=2{\mathrm{deg}}_{x}q+{\mathrm{deg}}_{x}r$ and hence ${\mathrm{deg}}_{x}q=0$; the same argument with the other three variables gives ${\mathrm{deg}}_{y}q={\mathrm{deg}}_{z}q={\mathrm{deg}}_{t}q=0$. All four degrees of $q$ vanish, so in the expansion of [F2] the only possibly nonzero coefficient of $q$ is the constant one, and $q\mathrm{\ne }0$ makes that constant nonzero: $q$ is a nonzero element of $k^{\times }$ [F10]. By [F7] applied four times through the integral domains $k$, $k[x]$, $k[x,y]$ and $k[x,y,z]$ [F4, F5], the units of $R=k[x,y,z][t]$ are exactly the units of $k$, that is the nonzero elements of $k$ [F10]; hence $q$ is a unit of $R$, contradicting that $q$ is irreducible [F6]. Therefore no irreducible factor of $f$ occurs twice: $f$ is squarefree.

The ring $R$ is an integral domain by [F4, F5], so $I({\mathbb{A}}^4)=I(V(0))=\sqrt{(0)}=(0)$ by [F12] and [F11], and the coordinate ring of ${\mathbb{A}}^4$ is $k[{\mathbb{A}}^4]=R\mathrm{/}(0)=R$ by [F14]; further $\dim {\mathbb{A}}^4=4$ by [F15], and ${\mathbb{A}}^4$ is irreducible because $(0)$ is a proper prime ideal of $R$ and nonempty irreducible algebraic sets correspond precisely to proper prime ideals by [F13]. The element $f\in k[{\mathbb{A}}^4]=R$ is nonzero and a nonunit by step 1.2, so [F16] applies to the irreducible affine algebraic set ${\mathbb{A}}^4$: the zero locus $X=V(f)$ is nonempty and every irreducible component of $X$ has dimension $4-1=3$.

By [F23], $f=\det \left(\begin{array}{cc}x& y\\ z& t\end{array}\right)=xt-yz$, so $X$ is the set of $2\times 2$ matrices over $k$ with vanishing determinant; a square matrix over $k$ is invertible exactly when its determinant is a unit of $k$ by [F24], and the units of the field $k$ are its nonzero elements [F10], so $X$ is exactly the source's locus of noninvertible $2\times 2$ matrices. By step 1.1 $f$ is homogeneous of degree $2$, so for $a\in X(k)$ and $\lambda \in k$ the evaluation rule of [F2] and distributivity and commutativity in $k$ [F9] give $f(\lambda a)=(\lambda a_x)(\lambda a_t)-(\lambda a_y)(\lambda a_z)={\lambda }^2(a_x{a}_t-a_y{a}_z)={\lambda }^{2}f(a)=0$, hence $\lambda a\in X(k)$: the variety $X$ is invariant under scaling about the origin, a cone with vertex $0$.

The principal ideal $(f)$ is radical. Indeed, if $g^N\in (f)$ with $N\ge 1$, then $f\mid g^N$; writing the squarefree element $f$ as $u{q}_1\cdots q_m$ with $u$ a unit and the $q_i$ pairwise nonassociate irreducibles [F5, F6], each $q_i$ is prime by [F4] and divides $g^N$, hence divides $g$, and the pairwise nonassociate primes $q_1,\dots ,q_m$ all dividing $g$ have a product dividing $g$, so $f\mid g$ and $g\in (f)$; thus $\sqrt{(f)}=(f)$ by [F11]. Therefore $I(X)=I(V(f))=\sqrt{(f)}=(f)$ by [F12], so $X$ is a reduced classical variety whose coordinate ring is $k[X]=R\mathrm{/}(f)=k[x,y,z,t]\mathrm{/}(f)$ by [F14], presented by the actual ideal $(f)$ that the Jacobian and tangent computations of [F18, F19] use.

Every closed point $a\in X$ lies on at least one irreducible component of $X$, and all components have dimension $3$ by step 2.2; since $X$ is a reduced classical finite-type space over the algebraically closed field $k$ by step 3.1, [F17] gives $\dim {\mathcal{O}}_{X,a}={\max }_{a\in X_i}\dim X_i=3$ for every closed point $a\in X$.

The ideal $(f)$ has the one-element generating list $(f)$. At a point $a\in X(k)$ the Jacobian matrix of [F18] is the one-row matrix $J_{(f)}(a)=({\mathrm{\partial }}_{x}f(a),{\mathrm{\partial }}_{y}f(a),{\mathrm{\partial }}_{z}f(a),{\mathrm{\partial }}_{t}f(a))$; the monomial derivative formula of [F18] gives ${\mathrm{\partial }}_x(xt)=t$, ${\mathrm{\partial }}_t(xt)=x$, ${\mathrm{\partial }}_y(yz)=z$, ${\mathrm{\partial }}_z(yz)=y$, and all other variable derivatives of the two monomials are zero, with the coefficients $1$ and $-1$ read in $k$ [F9, F10]; hence $J_{(f)}(a)=(a_t,-a_z,-a_y,a_x)$. By [F19] the coordinate-velocity map gives a canonical $k$-linear isomorphism $T_{a}X\cong \ker J_{(f)}(a)$, where $T_{a}X$ is the intrinsic tangent space of [F20] at the $k$-rational point $a$; step 3.1 identifies the reduced classical variety with $\mathrm{Spec}(R\mathrm{/}(f))$ at its rational points and the actual ideal is the one used by [F18], so no hypothesis on the characteristic enters.

At the origin all four entries of $J_{(f)}(0)$ vanish, so $J_{(f)}(0)$ is the zero row and $\ker J_{(f)}(0)=k^4$, of dimension $4$ by [F26, F27]. At a point $a\mathrm{\ne }0$ at least one of $a_x,a_y,a_z,a_t$ is nonzero; if $a_x\mathrm{\ne }0$, the single equation $a_t{v}_1-a_z{v}_2-a_y{v}_3+a_x{v}_4=0$ of the row is solved for $v_4$ as $v_4=(a_z{v}_2+a_y{v}_3-a_t{v}_1)\mathrm{/}{a}_x$, so the kernel is exactly the image of the $k$-linear map $\phi (u_1,u_2,u_3)=(u_1,u_2,u_3,(a_z{u}_2+a_y{u}_3-a_t{u}_1)\mathrm{/}{a}_x)$, which is injective because its first three coordinates are $u_1,u_2,u_3$; the images of the standard basis vectors $e_1,e_2,e_3$ of $k^3$ [F27] are a basis of $\ker J_{(f)}(a)$, since they span by the previous sentence and are linearly independent as $\phi$ is injective, so ${\dim }_k\ker J_{(f)}(a)=3$ by [F26]; the cases $a_y\mathrm{\ne }0$, $a_z\mathrm{\ne }0$, $a_t\mathrm{\ne }0$ are identical with the corresponding coordinate solved for, the divisions being by the nonzero element $a_x\in k^{\times }$ or its analogue, available in every characteristic [F10].

The four partial derivatives of $f$ vanish simultaneously at a $k$-point $a$ exactly when $a_t=a_z=a_y=a_x=0$, that is exactly at the origin. By step 2.1 the polynomial $f$ is nonconstant and squarefree, so the gradient test [F22] applies to the reduced classical variety $X=V(f)$ over the algebraically closed field $k$ and shows for every closed point $a\in X$: the point $a$ is singular if and only if all formal first partials of $f$ vanish at $a$, if and only if $a=0$. This does not yet address nonclosed points. Cover the complement of the origin in the underlying scheme by $D(x),D(y),D(z),D(t)$. On $D(x)$ the equation $xt-yz=0$ eliminates $t$, giving a coordinate-ring isomorphism $(R\mathrm{/}(f){)}_x\cong k[x,x^{-1},y,z]$. Similarly, elimination of $z$ on $D(y)$, of $y$ on $D(z)$, and of $x$ on $D(t)$ identifies their rings with localizations of polynomial rings in three variables over $k$. By [F29], each of these rings is regular at every prime. Thus every point outside the origin, including every nonclosed point, is regular. At the origin ${\dim }_k{T}_{0}X=4\mathrm{\ne }3={\dim }_{0}X$ by steps 5.1 and 4.1, so it is singular by [F21]. Therefore $X_{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}}=\{0\}$ and $X_{\mathrm{r}\mathrm{e}\mathrm{g}}=X\setminus \{0\}$ as subsets of the full scheme.

Boundary and scope dispositions. Nonemptiness: $X$ contains the origin, which is a $k$-rational point by step 1.1, and step 2.2 proves $V(f)$ nonempty again through [F16], so the empty case has no instance. Zero cases: the origin has the zero Jacobian row of step 5.1 with kernel all of $k^4$, the zero vector lies in every kernel and every tangent space, $f$ has zero constant term so $f(0)=0$, and the origin corresponds to the zero $2\times 2$ matrix of vanishing determinant in step 2.3; $f$ itself is nonzero by step 1.2. One: there is one defining equation, one Jacobian row of length four, and one singular point, the origin, in steps 4.2 and 6.1. Degenerate case: no division by $2$ occurs anywhere; the kernel computation of step 5.1 divides only by a nonzero field element, and the gradient test [F22] is characteristic-free, so the conclusions hold in characteristic $2$ as well, where the source's nondegeneracy conventions for quadrics would otherwise require care. Endpoints: the variable degrees of $f$ are all $1$ by step 1.2, the ambient dimension $4$ and the component and local dimensions $3$ are the two ends of the dimension comparison in steps 2.2 and 4.1, and the tangent dimensions $4$ at the origin and $3$ elsewhere of step 5.1 are the two values, the maximum $4$ occurring only at the vertex. Nonempty choice: AC is declared in [F28] and is used through [F12], [F13], [F15], [F16], [F17], [F22] and [F29], each cited at the step that uses it; the polynomial, degree, unit and Jacobian arguments of steps 1.1, 1.2, 2.1, 4.2, 5.1 and 2.3, the basis arguments built on [F26, F27], and the determinant items [F23, F24] are choice-free. Both directions of the gradient biconditional are used at closed points in step 6.1, and the four principal charts cover every nonclosed point outside the origin.