The Jacobian kernel computes the tangent space

By [F8], fix a finite list $f_1,\dots ,f_r$ generating $I$. Since $a$ is a $k$-rational point, the affine anti-equivalence [F5] and the base condition [F4] give a $k$-algebra map $A\to k$. Its composite with the quotient map is evaluation at $a$: the polynomial universal property [F6] identifies the composite as the unique map sending $t_j$ to $a_j$. It kills $I$, so $f_i(a)=0$ for each $i$.

For any $v=(v_1,\dots ,v_n)\in k^n$, [F6] gives a unique $k$-algebra map ${\varphi }_v:k[t_1,\dots ,t_n]\to D$ with $t_j\mapsto a_j+ϵv_j$. For a monomial $t^e={\prod }_j{t}_j^{e_j}$, expansion and ${ϵ}^2=0$ give $t^e(a+ϵv)=a^e+ϵ{\sum }_{j:e_j>0}{e}_j{a}_1^{e_1}\cdots a_j^{e_j-1}\cdots a_n^{e_n}{v}_j$. Extending over its finitely many monomials yields $p(a+ϵv)=p(a)+ϵ{\sum }_{j=1}^n(\mathrm{\partial }p\mathrm{/}\mathrm{\partial }{t}_j)(a)v_j$; the integer $e_j$ is read in $k$, including in positive characteristic, and the empty sum and product conventions cover $n=0$.

The map ${\varphi }_v$ kills $I$ exactly when it kills every generator $f_i$. By steps 1.1 and 1.2, ${\varphi }_v(f_i)=ϵ{\sum }_{j=1}^n(\mathrm{\partial }{f}_i\mathrm{/}\mathrm{\partial }{t}_j)(a)v_j$, which is zero exactly when row $i$ of $J_{(f_1,\dots ,f_r)}(a)$ annihilates $v$. Thus ${\varphi }_v$ factors uniquely through $A$ by [F7] exactly when $J(a)v=0$, and its reduction modulo $ϵ$ is $a$. Conversely, any based $k$-morphism $\mathrm{Spec}D\to X$ corresponds by [F4, F5] to a $k$-algebra map $A\to D$ reducing to evaluation at $a$; the images of the coordinates have unique form $a_j+ϵv_j$. By [F6] its composite from the polynomial ring is ${\varphi }_v$, and the same calculation forces $J(a)v=0$. The two constructions are inverse. Their coefficient derivations depend $k$-linearly on $v$, and [F2] identifies them with $T_{a}X$, proving the canonical linear isomorphism in the statement and both membership implications.

Let $g_1,\dots ,g_s$ be another finite generating list of $I$, and write each $g_{\mathrm{\ell }}={\sum }_i{h}_{\mathrm{\ell }i}{f}_i$. The coefficient-of-$ϵ$ formula in step 1.2 is a derivation because each ${\varphi }_v$ is a ring homomorphism. Its product rule in each coordinate direction, together with $f_i(a)=0$, gives $d{g}_{\mathrm{\ell }}(a)={\sum }_i{h}_{\mathrm{\ell }i}(a)d{f}_i(a)$, so each row of $J_g(a)$ lies in the row span of $J_f(a)$. Reversing the lists gives equality of row spans and hence equality of their annihilators, which are the kernels in $k^n$. For $n=0$ all rows are empty and both row spans are zero.

The boundary cases are explicit. If $X$ is empty there is no rational point, so the pointwise statement has no instance. If $r=0$, then $I=(0)$, the matrix has no rows, and the result says $T_a{\mathbf{A}}_k^n=k^n$. If $n=0$ and a rational point exists, its $k$-algebra map $k\mathrm{/}I\to k$ composed with $k\to k\mathrm{/}I$ is the identity, so $I=(0)$; hence $T_{a}X=k^0=0$. In one coordinate, $X=\mathrm{Spec}k[t]\mathrm{/}(t^2)$ at $0$ has Jacobian row $2t{\mathrm{\mid }}_0=0$ (also in characteristic $2$), so its tangent space is all of $k$, as the scheme-theoretic nilpotent structure requires. The zero vector corresponds to the constant based map $t_j\mapsto a_j$. No AC or DC is used: the finite generating tuple is chosen for this single ideal, and no basis or family of choices is made. Steps 2.1 and 2.2 prove both directions of the kernel characterization and generator independence.