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Let $k$ be an algebraically closed field. Given a morphism between $k$-varieties $\phi: X \to Y$, I want to define the induced map on the tangent spaces as follows. Consider the pullback $\phi^\ast: \mathcal{O}_Y(U) \to \mathcal{O}_X(\phi^{-1}U)$ for any $U \subset Y$ open. This yields a local homomorphism between the stalks $\mathcal{O}_{Y, \phi(p)}$ and $\mathcal{O}_{X, p}$ by sending $$ [U,h]_{\phi(p)} \mapsto [\phi^{-1}U, h \circ \phi]_p, $$ where $[U,f]_q$ denotes a germ at $q$. Now, restricting to a maximal ideal $\mathfrak{m}_{\phi(p),Y}$, we should get a well-defined map $$ \Phi: \mathfrak{m}_{\phi(p), Y}/\mathfrak{m}_{\phi(p), Y}^2 \to \mathfrak{m}_{p, X}/\mathfrak{m}_{p,X}^2. $$ and define the differential of $\phi$ at $p \in X$ as the dual map of $\Phi$, that is $$ d_p \phi: T_pX \to T_pY, \quad \ell \mapsto \ell \circ \Phi, $$ where we use the definition of the tangent spaces as the dual space of the maximal ideals above quotient by their squares.

Question. With this definition, I would like to compute the differential of a polynomial map $\phi: X \to Y$ where $X \subset \mathbb{A}^n$ and $Y \subset \mathbb{A}^m$ are algebraic sets and $$ \phi(x) := (f_1(x), ..., f_m(x)), $$ for polynomial functions $f_i: k^n \to $k. I expect it to be the multiplication by the Jacobian of $f$, but I have trouble to make this argument precise. So, here is my attempt (with all its mistakes): Consider the pullback $k[y_1, ..., y_m] \to k[x_1, ...,x_n], \: y_i \mapsto y_i \circ \phi$ which singles out the $i$th component of the function $\phi$. Since on affine varieties, the stalk is isomorphic to the localization at the respective maximal ideal, we get a map from $$ \Phi_p: k[y_1, ..., y_m]_{\mathfrak{m}_{\phi(p)}} \to k[x_1, ...,x_n]_{\mathfrak{m}_p}, \quad \frac{g}{h} \mapsto \frac{g \circ \phi}{h \circ \phi}. $$ Since $h$ vanishes on $\phi(p)$, it is plain that $h \circ \phi$ vanishes on $p$, so this is well-defined. Now, take a polymomial function $h$ vanishing at $p$, that is $h \in \mathfrak{m}_{\phi(p), Y}$ and consider its linearization $$ h^{(1)}:= \sum_{i=1}^m \frac{\partial h}{\partial y_i}\big\vert_{\phi(p)} d_{\phi(p)} y_i \in \mathfrak{m}_{\phi(p), Y}/\mathfrak{m}_{\phi(p), Y}^2. $$ So, $$ \Phi_p \circ h^{(1)}= \sum_{i=1}^m \frac{\partial (h \circ \phi)}{\partial y_i}\big \vert_{\phi(p)} d_{\phi(p)}y_i = \sum_{i=1}^m \frac{\partial h}{\partial y_i}\big \vert_{\phi(p)} \frac{\partial \phi}{\partial x_i}\big \vert_{p} d_{\phi(p)}y_i = \operatorname{Jac}_f(p) \circ h^{(1)}. $$ using the chain rule on formal deriatives. So, $\Phi_p = \operatorname{Jac}_f(p)$. Is this correct?

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