The product of two parabolas: a block Jacobian and the direct-sum formula

Setup: $k$ is a field and $k[x,y]$, $k[u,v]$, $k[x,y,u,v]=k[x][y][u][v]$ are commutative rings by [F1]; $f=y-x^2$ and $g=v-u^2$ are nonzero polynomials, $I_1=(f)$, $I_2=(g)$ are principal ideals and $J=(f,g)$ is the ideal generated by the two-element set $\{f,g\}$ by [F2]; $A_1=k[x,y]\mathrm{/}{I}_1$, $A_2=k[u,v]\mathrm{/}{I}_2$, $B=k[x,y,u,v]\mathrm{/}J$, $C_1=\mathrm{Spec}{A}_1$, $C_2=\mathrm{Spec}{A}_2$, $X=\mathrm{Spec}B\subseteq {\mathbb{A}}_k^4$; the origin $0=(0,0,0,0)$ satisfies $f(0)=0$ and $g(0)=0$ because $f$ and $g$ have no constant term; and $T_x$ denotes the intrinsic tangent space of [F14].

The origins are $k$-rational points: the evaluation $\mathrm{ev}:k[x,y,u,v]\to k$, $h\mapsto h(0,0,0,0)$ has kernel $(x,y,u,v)$, maximal by [F11]; $f$ and $g$ lie in that kernel because $f(0)=g(0)=0$, so the kernel contains the ideal $J=(f,g)$ generated by them by [F2], and [F10] factors $\mathrm{ev}$ uniquely through the quotient $B=k[x,y,u,v]\mathrm{/}J$ as a $k$-algebra homomorphism $\delta :B\to k$; $\delta$ is surjective because $\mathrm{ev}$ is, and [F12] renders a $k$-algebra homomorphism $B\to k$ as a $k$-rational point of $X$, namely the origin $0$ with $B\mathrm{/}\ker \delta \cong k$; the same argument applied to $k[x,y]$ and $A_1$ gives a $k$-algebra homomorphism ${\delta }_1:A_1\to k$, a $k$-rational point $0_1$ of $C_1$, and applied to $k[u,v]$ and $A_2$ gives a $k$-algebra homomorphism ${\delta }_2:A_2\to k$, a $k$-rational point $0_2$ of $C_2$.

The product structure: by [F9] there is a $k$-algebra homomorphism $k[x,y]\to B$ with $x\mapsto \widehat{x}$, $y\mapsto \widehat{y}$, and it kills $f$ because $\widehat{y}={\widehat{x}}^2$ in $B$, so [F10] factors it as a $k$-algebra homomorphism $\alpha :A_1\to B$ with $\alpha (\bar{x})=\widehat{x}$ and $\alpha (\bar{y})=\widehat{y}$; likewise a $k$-algebra homomorphism $\beta :A_2\to B$ is defined by $\beta (\bar{u})=\widehat{u}$ and $\beta (\bar{v})=\widehat{v}$, using $\widehat{v}={\widehat{u}}^2$; by [F8] there is a unique $k$-algebra homomorphism $\phi :A_1{\otimes }_k{A}_2\to B$ with $\phi (a\otimes 1)=\alpha (a)$ and $\phi (1\otimes b)=\beta (b)$, where $A_1{\otimes }_k{A}_2$ is a commutative $k$-algebra with multiplication rule $(a\otimes b)(a^{\mathrm{\prime }}\otimes b^{\mathrm{\prime }})=a{a}^{\mathrm{\prime }}\otimes b{b}^{\mathrm{\prime }}$ by [F7]; in the other direction [F9] gives a $k$-algebra homomorphism $\psi :k[x,y,u,v]\to A_1{\otimes }_k{A}_2$ with $x\mapsto \bar{x}\otimes 1$, $y\mapsto \bar{y}\otimes 1$, $u\mapsto 1\otimes \bar{u}$, $v\mapsto 1\otimes \bar{v}$, and the multiplication rule of [F7] computes $\psi (f)=\psi (y)-\psi (x{)}^2=(\bar{y}\otimes 1)-(\bar{x}\otimes 1{)}^2=(\bar{y}-{\bar{x}}^2)\otimes 1=0$ and $\psi (g)=\psi (v)-\psi (u{)}^2=1\otimes (\bar{v}-{\bar{u}}^2)=0$, so [F10] factors $\psi$ through the quotient $B=k[x,y,u,v]\mathrm{/}J$ as a $k$-algebra homomorphism $\bar{\psi }:B\to A_1{\otimes }_k{A}_2$ with $\bar{\psi }(\widehat{x})=\bar{x}\otimes 1$, $\bar{\psi }(\widehat{y})=\bar{y}\otimes 1$, $\bar{\psi }(\widehat{u})=1\otimes \bar{u}$ and $\bar{\psi }(\widehat{v})=1\otimes \bar{v}$.

The Jacobian computation: the ideal $J=(f,g)$ is presented by the finite generating list $(f,g)$ by [F2], and every element of $J$ vanishes at $0$ since $f(0)=g(0)=0$, so the Jacobian matrix of this list at $0$ is defined by [F3]; its rows are the formal partial derivatives of $f=y-x^2$ and $g=v-u^2$, namely $(-2x,1,0,0)$ and $(0,0,-2u,1)$, so that at $0$ the matrix is $\left(\begin{array}{cccc}0& 1& 0& 0\\ 0& 0& 0& 1\end{array}\right)$; a vector $w=(w_1,w_2,w_3,w_4)\in k^4$ lies in the kernel exactly when $w_2=0$ and $w_4=0$, so $\ker J(0)=\{(a,0,b,0):a,b\in k\}$, a two-dimensional subspace with $k$-basis $(1,0,0,0),(0,0,1,0)$ because $1\mathrm{\ne }0$ in $k$ by [F13]; by [F4] the coordinate-velocity map gives a canonical $k$-linear isomorphism $T_{0}X\cong \ker J(0)$, hence $T_{0}X\cong \{(a,0,b,0)\}$ is two-dimensional.

The two factors: for $C_1=\mathrm{Spec}(k[x,y]\mathrm{/}(f))$ the Jacobian matrix of the single equation $f$ at $0_1$ is the $1\times 2$ matrix $(-2x,1)$ evaluated at $(0,0)$, namely $(0,1)$, by [F3]; its kernel is $\{(a,0):a\in k\}\cong k$, one-dimensional, and [F4] gives $T_{0_1}{C}_1\cong \{(a,0)\}$; for $C_2=\mathrm{Spec}(k[u,v]\mathrm{/}(g))$ the Jacobian matrix of $g$ at $0_2$ is $(-2u,1)$ evaluated at $(0,0)$, again $(0,1)$, with kernel $\{(b,0):b\in k\}\cong k$, so $T_{0_2}{C}_2\cong \{(b,0)\}$ and each factor contributes exactly one dimension.

The maps of step 2.2 are inverse: the composites $\bar{\psi }\circ \phi$ and the identity of $A_1{\otimes }_k{A}_2$ are $k$-algebra homomorphisms out of the coproduct that restrict to the same maps on the two factors, since $\bar{\psi }(\phi (\bar{x}\otimes 1))=\bar{\psi }(\widehat{x})=\bar{x}\otimes 1$ and likewise on $\bar{y},\bar{u},\bar{v}$, and every element of $A_1$ is a polynomial in the two classes $\bar{x},\bar{y}$ and every element of $A_2$ a polynomial in $\bar{u},\bar{v}$, so the uniqueness clause of [F8] gives $\bar{\psi }\circ \phi ={\mathrm{id}}_{A_1{\otimes }_k{A}_2}$; conversely $\phi \circ \bar{\psi }$ and ${\mathrm{id}}_B$ are $k$-algebra homomorphisms $B\to B$ whose composites with the quotient map $k[x,y,u,v]\to B$ agree on $x,y,u,v$ by those same displayed identities, so by the uniqueness of factorization in [F10] together with the uniqueness in [F9] they are equal and $\phi \circ \bar{\psi }={\mathrm{id}}_B$; hence $\phi$ is a $k$-algebra isomorphism $A_1{\otimes }_k{A}_2\cong B$ and, taking the base ring $k$ in [F6], $C_1{\times }_k{C}_2=\mathrm{Spec}{A}_1{\times }_{\mathrm{Spec}k}\mathrm{Spec}{A}_2\cong \mathrm{Spec}(A_1{\otimes }_k{A}_2)\cong \mathrm{Spec}B=X$, with the projections corresponding to $\alpha$ and $\beta$, that is, to the coordinate projections; moreover the $k$-rational point $\delta$ of step 2.1 corresponds under this isomorphism to the pair $({\delta }_1,{\delta }_2)$, because $\delta \circ \phi :A_1{\otimes }_k{A}_2\to k$ is by [F8] the unique $k$-algebra homomorphism restricting to ${\delta }_1$ and ${\delta }_2$ on the two factors, which is the product point $(0_1,0_2)$.

The direct-sum comparison: by [F5] applied to the $k$-schemes $C_1,C_2$ and their $k$-rational points $0_1,0_2$ of step 2.1, the two projections induce a canonical $k$-linear isomorphism $T_{(0_1,0_2)}(C_1{\times }_k{C}_2)\cong T_{0_1}{C}_1\oplus T_{0_2}{C}_2$, which by step 2.4 is $\cong k\oplus k\cong k^2$; the isomorphism $X\cong C_1{\times }_k{C}_2$ of steps 2.2 and 3.1 carries $0$ to $(0_1,0_2)$ and its projections are the coordinate projections, so a based dual-number point of $X$ with velocity $(a,0,b,0)$, as in step 2.3, projects to based points of the factors with velocities $(a,0)$ and $(b,0)$; hence the direct-sum isomorphism is compatible with the block splitting $k^4=k_{(x,y)}^2\oplus k_{(u,v)}^2$ of the coordinates, and the two computations of $T_{0}X$ agree: both are two-dimensional, the direct-sum side reconstructing exactly the kernel $\{(a,0,b,0)\}$; no characteristic hypothesis enters either computation.

Conclusion: for $X=\mathrm{Spec}(k[x,y,u,v]\mathrm{/}(y-x^2,v-u^2))$ over any field $k$, the origin has tangent space $T_{0}X\cong \{(a,0,b,0):a,b\in k\}$, the kernel of the block Jacobian matrix $\left(\begin{array}{cccc}0& 1& 0& 0\\ 0& 0& 0& 1\end{array}\right)$, a two-dimensional $k$-vector space; the direct-sum formula for the product $C_1{\times }_k{C}_2\cong X$ reproduces the same two dimensions from the one-dimensional factor tangent spaces $T_{0_1}{C}_1\cong k$ and $T_{0_2}{C}_2\cong k$, and no characteristic, perfectness, reducedness or algebraic-closedness hypothesis is used.

Boundary and scope dispositions: $X$, $C_1$, $C_2$ and the field $k$ are nonempty, the origins being $k$-rational points by step 2.1 and $k$ being a field with $0\mathrm{\ne }1$ by [F13], so no object here is empty and the only empty lists would be empty generating lists, which do not occur since $I_1=(f)$, $I_2=(g)$ and $J=(f,g)$ are presented by one or two exhibited elements (steps 1.1 and 2.3); the zero cases are the origin $0$, where all four coordinates vanish, the vanishing $f(0)=g(0)=0$ of step 1.1, the vanishing entries $-2x$ and $-2u$ at the origin, the zero vector of $\ker J(0)$, and the zero tangent map of a constant based dual-number point, while the kernel itself is not zero but two-dimensional because the entries $1$ never vanish (step 2.3); there is one point $0$, one product isomorphism, two equations, one Jacobian matrix with two rows, one kernel description, and each of the two factors contributes exactly one dimension (steps 3.1, 2.3 and 2.4); the degenerate instances of the direct-sum formula, a zero tangent factor, do not occur because both factors are one-dimensional (step 2.4), no equation degenerates to the zero polynomial since $f$ and $g$ have $y$- and $v$-coefficient $1$ (step 1.1), and the characteristic does not enter at all, every evaluated entry of the two Jacobian matrices being $0$ or $1$ (steps 2.3 and 2.4), so the characteristic-two case is not exceptional and no hypothesis is excluded; the dimension endpoints are one for each factor and two for the product, the extreme parameter value $a=b=0$ giving the zero vector, and the entries $-2x$, $-2u$ evaluated at the origin are the endpoint values that vanish there (steps 2.3 and 2.4); both directions of the kernel characterization are proved in step 2.3, the forward one because a vector of the kernel satisfies $w_2=0$ and $w_4=0$, the reverse one because $w_2=w_4=0$ makes both rows annihilate $w$, and the algebra isomorphisms of steps 2.2 and 3.1 are two-sided inverses by construction; no Axiom of Choice or dependent choice is used, since $f$, $g$, the two maps $\alpha ,\beta$, the assignment defining $\psi$ and the Jacobian evaluations are all exhibited explicitly, no basis or neighbourhood is selected, and every cited supplier is choice-free, with no def-axiom-of-choice dependency declared.