[proofplan] We compare three local quantities at $p$: the rank of the [Jacobian matrix](/page/Jacobian%20Matrix), the dimension of the Zariski tangent space, and the Krull dimension of the local ring. The generators $f_1,\dots,f_r$ of the full vanishing ideal cut out the tangent space by their first-order parts, so $\dim_k T_pX=n-\operatorname{rank}J(f_1,\dots,f_r)_p$. The hypothesis on irreducible components says that the local dimension of $X$ at $p$ is $d$. Finally, over an [algebraically closed field](/page/Algebraically%20Closed%20Field), smoothness at $p$ is equivalent to regularity of the local ring, i.e. equality between embedding dimension and local dimension. [/proofplan]

[step:Translate the point into the local coordinate ring] Let \begin{align*} S:=k[x_1,\dots,x_n], \end{align*} and let \begin{align*} A:=S/I(X) \end{align*} be the coordinate ring of $X$. Since $p=(a_1,\dots,a_n)\in X$, define the maximal ideal of $S$ corresponding to $p$ by \begin{align*} \mathfrak n_p:=(x_1-a_1,\dots,x_n-a_n)\trianglelefteq S. \end{align*} Because every element of $I(X)$ vanishes at $p$, we have $I(X)\subset \mathfrak n_p$. Hence \begin{align*} \mathfrak m_p:=\mathfrak n_p/I(X)\trianglelefteq A \end{align*} is the maximal ideal corresponding to $p$. Let \begin{align*} R:=A_{\mathfrak m_p} \end{align*} be the local ring of $X$ at $p$, and let \begin{align*} \mathfrak m_R:=\mathfrak m_pR \end{align*} be its maximal ideal. The quotient map $A\to R$ induces a natural $k$-linear isomorphism \begin{align*} \mathfrak m_p/\mathfrak m_p^2 \cong \mathfrak m_R/\mathfrak m_R^2. \end{align*} Indeed, localization at $\mathfrak m_p$ does not change the residue field, which remains $A/\mathfrak m_p\cong k$, and the usual localization isomorphism gives \begin{align*} \mathfrak m_R/\mathfrak m_R^2\cong (\mathfrak m_p/\mathfrak m_p^2)\otimes_A A_{\mathfrak m_p}. \end{align*} Since $\mathfrak m_p$ acts by zero on $\mathfrak m_p/\mathfrak m_p^2$, this [tensor product](/page/Tensor%20Product) is naturally the same $k$-[vector space](/page/Vector%20Space) as $\mathfrak m_p/\mathfrak m_p^2$. [/step]

[step:Compute the tangent space from the Jacobian matrix]Define the Zariski tangent space at $p$ by \begin{align*} T_pX:=\operatorname{Hom}_k(\mathfrak m_p/\mathfrak m_p^2,k). \end{align*} We claim that \begin{align*} \dim_k T_pX=n-\operatorname{rank}J(f_1,\dots,f_r)_p. \end{align*} Let \begin{align*} \theta:k^n\to T_p\mathbb A_k^n \end{align*} be the standard identification sending $v=(v_1,\dots,v_n)$ to the $k$-[linear map](/page/Linear%20Map) \begin{align*} \theta(v):\mathfrak n_p/\mathfrak n_p^2\to k \end{align*} defined by \begin{align*} \theta(v)(x_j-a_j+\mathfrak n_p^2)=v_j \end{align*} for each $1\le j\le n$. Under this identification, a vector $v=(v_1,\dots,v_n)\in k^n$ lies in $T_pX$ precisely when every $f\in I(X)$ has zero first-order part on $v$: \begin{align*} \sum_{j=1}^n \frac{\partial f}{\partial x_j}(p)v_j=0. \end{align*} Since $I(X)=(f_1,\dots,f_r)$, it is enough to impose these equations for $f_1,\dots,f_r$. Thus $T_pX$ identifies with the kernel of the $k$-linear map \begin{align*} L:k^n&\to k^r \end{align*} whose matrix in the standard bases is $J(f_1,\dots,f_r)_p$. Therefore rank-nullity gives \begin{align*} \dim_k T_pX=n-\operatorname{rank}J(f_1,\dots,f_r)_p. \end{align*}[/step]

[guided]The point of this step is to make the Jacobian matrix appear from first-order algebra. The tangent space to [affine space](/page/Affine%20Space) at $p$ is represented by linear functionals on \begin{align*} \mathfrak n_p/\mathfrak n_p^2, \end{align*} where \begin{align*} \mathfrak n_p=(x_1-a_1,\dots,x_n-a_n)\trianglelefteq k[x_1,\dots,x_n]. \end{align*} The classes of $x_1-a_1,\dots,x_n-a_n$ form a $k$-basis of $\mathfrak n_p/\mathfrak n_p^2$, so a tangent vector is the same thing as a tuple $v=(v_1,\dots,v_n)\in k^n$, where $v_j$ is the value of the tangent functional on $x_j-a_j$. Now impose the condition that the tangent vector is tangent to $X$. Since $X$ is defined by the full vanishing ideal $I(X)$, a tangent vector to $X$ must annihilate the first-order part of every function in $I(X)$. For any $f\in I(X)$, Taylor expansion modulo $\mathfrak n_p^2$ gives \begin{align*} f-f(p)\equiv \sum_{j=1}^n \frac{\partial f}{\partial x_j}(p)(x_j-a_j)\pmod{\mathfrak n_p^2}. \end{align*} Because $p\in X$ and $f\in I(X)$, we have $f(p)=0$. Evaluating the corresponding tangent functional determined by $v$ gives the linear condition \begin{align*} \sum_{j=1}^n \frac{\partial f}{\partial x_j}(p)v_j=0. \end{align*} It remains to reduce from all $f\in I(X)$ to the chosen generators. Since $I(X)=(f_1,\dots,f_r)$, every $f\in I(X)$ can be written as \begin{align*} f=\sum_{i=1}^r h_if_i \end{align*} with $h_i\in k[x_1,\dots,x_n]$. Differentiating and evaluating at $p$ gives \begin{align*} \frac{\partial f}{\partial x_j}(p)=\sum_{i=1}^r h_i(p)\frac{\partial f_i}{\partial x_j}(p)+\sum_{i=1}^r f_i(p)\frac{\partial h_i}{\partial x_j}(p). \end{align*} The second sum is zero because $f_i(p)=0$ for every $i$. Hence the first-order condition for all elements of $I(X)$ is generated by the first-order conditions for $f_1,\dots,f_r$. Therefore $T_pX$ is exactly the kernel of the linear map \begin{align*} L:k^n&\to k^r \end{align*} whose matrix is \begin{align*} J(f_1,\dots,f_r)_p=\left(\frac{\partial f_i}{\partial x_j}(p)\right)_{1\le i\le r,\ 1\le j\le n}. \end{align*} By rank-nullity, \begin{align*} \dim_k T_pX=\dim_k\ker L=n-\operatorname{rank}L=n-\operatorname{rank}J(f_1,\dots,f_r)_p. \end{align*}[/guided]

[step:Identify the local dimension with the component dimension through $p$] The irreducible components of $X$ containing $p$ correspond to the minimal prime ideals of $A$ contained in $\mathfrak m_p$. The Krull dimension of the local ring \begin{align*} R=A_{\mathfrak m_p} \end{align*} is the maximum of the dimensions of those irreducible components. By hypothesis, that maximum is $d$. Hence \begin{align*} \dim R=d. \end{align*} [/step]

[step:Convert the rank condition into regularity of the local ring] The embedding dimension of the local ring $R$ is \begin{align*} \operatorname{edim}R:=\dim_k \mathfrak m_R/\mathfrak m_R^2. \end{align*} By the natural isomorphism \begin{align*} \mathfrak m_R/\mathfrak m_R^2\cong \mathfrak m_p/\mathfrak m_p^2 \end{align*} and the definition of $T_pX$, we have \begin{align*} \operatorname{edim}R=\dim_kT_pX. \end{align*} From the tangent-space computation, \begin{align*} \operatorname{edim}R=n-\operatorname{rank}J(f_1,\dots,f_r)_p. \end{align*} Since $\dim R=d$, the condition \begin{align*} \operatorname{rank}J(f_1,\dots,f_r)_p=n-d \end{align*} is equivalent to \begin{align*} \operatorname{edim}R=d=\dim R. \end{align*} By the local algebra criterion for regular local rings, a Noetherian local ring is regular precisely when its embedding dimension equals its Krull dimension. Therefore the Jacobian rank condition is equivalent to $R=\mathcal O_{X,p}$ being a regular local ring. [/step]

[step:Translate regularity back into smoothness at $p$] Because $k$ is algebraically closed, it is perfect. For affine varieties over a perfect field, a point is smooth precisely when its local ring is regular. Applying this smoothness-regularity criterion to the local ring \begin{align*} \mathcal O_{X,p}=R \end{align*} shows that $p$ is smooth if and only if $R$ is regular. By the previous step, $R$ is regular if and only if \begin{align*} \operatorname{rank}J(f_1,\dots,f_r)_p=n-d. \end{align*} This proves the claimed equivalence. [/step]