[proofplan] We use the dual-number definition of the Zariski tangent space: a tangent vector at $p$ is a first-order infinitesimal deformation $p+\varepsilon v$ that still satisfies every equation in $I(X)$ modulo $\varepsilon^2$. Expanding each generator $f_i$ to first order gives exactly the linear equations encoded by the [Jacobian matrix](/page/Jacobian%20Matrix) $J_p$. Because the $f_i$ generate the full vanishing ideal $I(X)$, checking these generators is equivalent to checking every polynomial vanishing on $X$. Finally, the dimension criterion follows from the standard local algebra fact that $T_pX$ is dual to $\mathfrak m_p/\mathfrak m_p^2$ and that a local ring is regular precisely when its embedding dimension equals its Krull dimension. [/proofplan]

[step:Translate tangent vectors into dual-number solutions]Let \begin{align*} D:=k[\varepsilon]/(\varepsilon^2) \end{align*} be the ring of dual numbers over $k$. We write $\varepsilon\in D$ for the image of the polynomial variable. For a vector \begin{align*} v=(v_1,\dots,v_n)\in k^n, \end{align*} define the $k$-algebra homomorphism \begin{align*} \varphi_v:R\to D \end{align*} by \begin{align*} \varphi_v(x_j):=p_j+\varepsilon v_j \end{align*} for every $1\le j\le n$. By the dual-number definition of the Zariski tangent space of the affine algebraic set $X$ at $p$, the vector $v$ belongs to $T_pX$ if and only if \begin{align*} \varphi_v(g)=0 \end{align*} for every $g\in I(X)$, after using that $g(p)=0$ because $p\in X$. Equivalently, \begin{align*} T_pX=\{v\in k^n:\varphi_v(g)=0 \text{ in } D \text{ for every } g\in I(X)\}. \end{align*}[/step]

[guided]The ring $D=k[\varepsilon]/(\varepsilon^2)$ records first-order information because every element has the form $a+\varepsilon b$ with $a,b\in k$, and all second-order terms in $\varepsilon$ vanish. A tangent vector should therefore be interpreted as a first-order motion away from the point $p$. For a chosen vector \begin{align*} v=(v_1,\dots,v_n)\in k^n, \end{align*} we encode the infinitesimal point $p+\varepsilon v$ by the $k$-algebra homomorphism \begin{align*} \varphi_v:R\to D \end{align*} defined on coordinate functions by \begin{align*} \varphi_v(x_j):=p_j+\varepsilon v_j. \end{align*} This is a well-defined $k$-algebra homomorphism because a homomorphism from the [polynomial ring](/page/Polynomial%20Ring) $k[x_1,\dots,x_n]$ is uniquely determined by the images of the coordinate variables. The defining equations of $X$ are precisely the polynomials in $I(X)$. Since $p\in X$, every $g\in I(X)$ satisfies $g(p)=0$. The infinitesimal point $p+\varepsilon v$ lies in $X$ to first order exactly when every such equation still vanishes in $D$. Thus the definition gives \begin{align*} v\in T_pX \end{align*} if and only if \begin{align*} \varphi_v(g)=0 \end{align*} for every $g\in I(X)$. This is the precise algebraic meaning of being tangent to $X$ at $p$.[/guided]

[step:Expand each polynomial to first order at $p$] For every polynomial $g\in R$, the first-order Taylor expansion in the dual-number ring $D$ is \begin{align*} \varphi_v(g)=g(p)+\varepsilon\sum_{j=1}^n \frac{\partial g}{\partial x_j}(p)v_j. \end{align*} Indeed, it is enough to verify this for a monomial \begin{align*} g=x_1^{a_1}\cdots x_n^{a_n} \end{align*} with $a_1,\dots,a_n\in \mathbb N_0$, and then extend by $k$-linearity. Since $\varepsilon^2=0$ in $D$, multiplying the factors \begin{align*} (p_j+\varepsilon v_j)^{a_j} \end{align*} keeps only the terms with at most one factor of $\varepsilon$. Hence the coefficient of $\varepsilon$ is the usual [directional derivative](/page/Directional%20Derivative) \begin{align*} \sum_{j=1}^n \frac{\partial g}{\partial x_j}(p)v_j. \end{align*} Thus the displayed formula holds for every $g\in R$. [/step]

[step:Reduce the tangent equations to the chosen generators] Because $I(X)=(f_1,\dots,f_r)$, every $g\in I(X)$ can be written as \begin{align*} g=\sum_{i=1}^r a_i f_i \end{align*} for some polynomials $a_1,\dots,a_r\in R$. Assume first that $v\in k^n$ satisfies \begin{align*} \sum_{j=1}^n \frac{\partial f_i}{\partial x_j}(p)v_j=0 \end{align*} for every $1\le i\le r$. Since $f_i(p)=0$ for all $i$, the first-order expansion gives \begin{align*} \varphi_v(f_i)=0 \end{align*} in $D$ for every $i$. Therefore \begin{align*} \varphi_v(g)=\sum_{i=1}^r \varphi_v(a_i)\varphi_v(f_i)=0. \end{align*} By the dual-number description of $T_pX$, this implies $v\in T_pX$. Conversely, if $v\in T_pX$, then $\varphi_v(f_i)=0$ for every $i$ because each $f_i\in I(X)$. Expanding $f_i$ to first order and using $f_i(p)=0$ gives \begin{align*} \varepsilon\sum_{j=1}^n \frac{\partial f_i}{\partial x_j}(p)v_j=0 \end{align*} in $D$. Since $\varepsilon c=0$ in $D$ holds exactly when $c=0$ in $k$, we obtain \begin{align*} \sum_{j=1}^n \frac{\partial f_i}{\partial x_j}(p)v_j=0 \end{align*} for every $i$. [/step]

[step:Identify the linear equations with the kernel of the Jacobian matrix] The matrix-vector product $J_pv\in k^r$ has $i$-th coordinate \begin{align*} (J_pv)_i=\sum_{j=1}^n (J_p)_{ij}v_j=\sum_{j=1}^n \frac{\partial f_i}{\partial x_j}(p)v_j. \end{align*} The previous step therefore proves \begin{align*} v\in T_pX \end{align*} if and only if \begin{align*} J_pv=0. \end{align*} Hence \begin{align*} T_pX=\{v\in k^n:J_pv=0\}=\ker J_p. \end{align*} [/step]

[step:Apply the local algebra criterion for nonsingularity] Let \begin{align*} A:=k[X]=R/I(X) \end{align*} be the coordinate ring of $X$. Let $\mathfrak m_p\trianglelefteq A$ be the maximal ideal of functions vanishing at $p$, and let \begin{align*} \mathcal O_{X,p}:=A_{\mathfrak m_p} \end{align*} be the local ring of $X$ at $p$. The standard cotangent-space description of the Zariski tangent space gives a natural $k$-linear isomorphism \begin{align*} T_pX\cong \operatorname{Hom}_k(\mathfrak m_p/\mathfrak m_p^2,k). \end{align*} Consequently, \begin{align*} \dim_k T_pX=\dim_k \mathfrak m_p/\mathfrak m_p^2. \end{align*} The number on the right is the embedding dimension of the local ring $\mathcal O_{X,p}$. We now use the standard regular local ring tangent-dimension criterion: the point $p$ is nonsingular precisely when the local ring $\mathcal O_{X,p}$ is regular, and this is equivalent to \begin{align*} \dim_k \mathfrak m_p/\mathfrak m_p^2=\dim \mathcal O_{X,p}. \end{align*} If $X$ is irreducible of dimension $d$, then every closed point $p\in X(k)$ has \begin{align*} \dim \mathcal O_{X,p}=d. \end{align*} Therefore $p$ is nonsingular if and only if \begin{align*} \dim_k T_pX=d. \end{align*} Finally, for a Noetherian local ring essentially of finite type over $k$, the embedding dimension is always at least the Krull dimension. Hence \begin{align*} \dim_k T_pX\ge d. \end{align*} Thus failure of equality is equivalent to \begin{align*} \dim_k T_pX>d, \end{align*} which is exactly the asserted singularity criterion. [/step]