Tangent spaces of products over a field

By [F1], choose affine open neighbourhoods $U=\mathrm{Spec}A$ of $x$ and $V=\mathrm{Spec}B$ of $y$. Their structure maps make $A$ and $B$ $k$-algebras by [F2]. The fibre-product theorem [F5] gives an open affine neighbourhood of $(x,y)$ in $X{\times }_{k}Y$ with coordinate ring $R=A{\otimes }_{k}B$; the two projections correspond to its canonical $k$-algebra maps from $A$ and $B$.

Let $e_x:A\to k$ and $e_y:B\to k$ be the maps of the rational points, and put $D=k[ϵ]\mathrm{/}({ϵ}^2)$. By [F7] and [F4], a tangent vector at $x$ or $y$ is represented in these charts by a $k$-algebra map $\alpha :A\to D$ or $\beta :B\to D$, with reductions $e_x$ and $e_y$. Conversely, any such pair determines by [F6] a unique $k$-algebra map $\phi :R\to D$ satisfying $\phi (a\otimes b)=\alpha (a)\beta (b)$. Its reduction is $a\otimes b\mapsto e_x(a)e_y(b)$, so it is based at $(x,y)$. Restriction along the two projection maps recovers $\alpha$ and $\beta$; therefore post-composition by the projections gives a bijection between the based dual-number maps of the product and pairs of based dual-number maps of the factors.

Write $\alpha (a)=e_x(a)+ϵD_x(a)$ and $\beta (b)=e_y(b)+ϵD_y(b)$; [F8] says $D_x,D_y$ are the derivations representing the two tangent vectors. Since ${ϵ}^2=0$, the map of step 2.1 satisfies $\phi (a\otimes b)=e_x(a)e_y(b)+ϵ(e_y(b)D_x(a)+e_x(a)D_y(b))\mathrm{.}$ Thus its coefficient derivation is linear in $(D_x,D_y)$. Conversely, restriction of the coefficient derivation of $\phi$ along the two projection maps returns $D_x,D_y$. The bijection in step 2.1 and its inverse are therefore $k$-linear, proving the asserted natural vector-space isomorphism. Its construction uses only the projections, so it is independent of the chosen affine neighbourhoods.

If either factor has zero tangent space, its based maps consist only of the constant map at that point, and step 2.1 pairs it with the based maps of the other factor; if both tangent spaces are zero, the product tangent space is zero as well. For a one-dimensional tangent factor with generator derivation $D_x$, step 3.1 sends $(D_x,0)$ to the coefficient derivation $a\otimes b\mapsto e_y(b)D_x(a)$; a generator $D_y$ in the other factor is sent to $a\otimes b\mapsto e_x(a)D_y(b)$. Each scalar multiple is sent to the same scalar multiple; no one-dimensional exception occurs. The formula also covers nonsmooth and nonreduced schemes because it uses only their based dual-number maps. The zero vector is the constant based map, and the zero pair corresponds to the constant map at $(x,y)$. If either scheme is empty, there is no point pair and the assertion has no instance. Only one affine neighbourhood for each of the two fixed points is used, no bases are chosen, and no Axiom of Choice is needed. The statement contains no iff claim.