Let $k$ be an algebraically closed field and $I = \left<\varphi_1,\ldots,\varphi_m\right>$ an ideal in $k[x_1,\ldots,x_n]$. Let $C$ be the simultaneous zero set of $I$ and suppose $C$ is nonsingular, i.e., the matrix $[D_j\varphi_i(P)]$ has rank $n - 1$ for all $P \in C$. Let $$k[C] = k[x_1,\ldots,x_n]/I$$ be the coordinate ring of $C$. Fix a point $P = (P_1,\ldots,P_n) \in C$ and let $$m_P = \{f \in k[C] : f(P) = 0\} = \left<\{x_i - P_i) : 1 \leq i \leq n\} \right> .$$ Let $$T_P(C) = \{v \in k^n : [D_j\varphi_i(P)]v = 0\}$$ be the tangent line to $C$ at $P$. I'm trying to show that $$m_P/m_P^2 \times T_P(C) \to k, \hspace{.5cm} (f,v) \to \nabla f(P) \cdot v$$ is a perfect pairing.
Suppose $f = \sum_{i=1}^nf_i(x_i - P_i)$ is such that $\nabla f(P) \cdot v = 0$ for all $v \in T_P(C)$. We have $$\nabla f(P) = \sum_{i=1}^n f_i(P) \cdot e_i,$$ where $e_i$ is the $i$-th elementary unit vector in $k^n$. Letting $v = (v_1,\ldots,v_n) \in T_P(C)$ be nonzero, we have $$\sum_{i=1}^n f_i(P)v_i = 0.$$ If $f_i(P) = 0$ for all $i$ then $\sum f_i(x_i - P_i) \in m_P^2$, so we would be done. But the above is not strong enough to conclude this. It seems clear the problem is I haven't used the definition of $T_P(C)$, but I'm not sure how to do this.