[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*}
[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]
[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]
