Bilinear pairing between tangent plane and maximal ideal

Question

I am revisiting some of the exercises from Silverman's AEC and am stuck on 1.3 which asks us to show that \begin{equation} \dim_{\overline{K}}M_P/M_P^2=\dim V \end{equation} if and only if $P$ is nonsingular, where $V$ is a variety given by the single equation $f(X_1,\dots,X_n)=0$. There is a proof of this in Hartshorne and I also found a simpler proof elsewhere but I would like to know how to do this using Silverman's hint by showing that if \begin{equation} T=\{(y_1,\dots,y_n)\in \mathbf{A}^n: \sum_{i=1}^n(\frac{\partial f}{\partial X_i}(P))y_i=0\} \end{equation} then the pairing \begin{equation} M_P/M_P^2 \times T \to \overline{K}, \ \ (g,y) \mapsto \sum_{i=1}^n(\frac{\partial g}{\partial X_i}(P))y_i, \end{equation} is a well-define perfect pairing. It is clear that this pairing is bilinear and by considering the functions $(X_i-a_i)$ where $P=(a_1,\dots,a_n)$, that this pairing is non-degenerate on the right. Also, if $P$ is singular, then $T=\mathbf{A}^n$ and by writing $g(X)=\sum_{i=1}^n(X_i-a_i)g_i(X)$, it follows that $\frac{\partial g}{\partial X_i}(P)=g_i(P)$ so using $e_i$ as a basis vector with a $1$ in the $i$-th position and $0$'s elsewhere, it follows that $g_i(P)=0$ and that $g_i \in m_P$ for all $i$. I'm struggling to see why the pairing is non-degenerate on the left when $P$ is non-singular. In this case, we can assume without loss of generality that $\frac{\partial f}{\partial X_1}(P)\neq 0$. Even in the simple case of an elliptic curve given by Weierstrass equation $y^2-4x^3+g_2x+g_3=0$ at some affine point $P=(a,b)$, I'm not seeing why this is true. In this case, $T$ is one-dimensional. For simplicity, if we assume that $b\neq 0$, then $T$ is spanned by $(1,(12a^2-g_2)/2b)$. If we write $h(x,y)=(x-a)h_1(x,y)+(y-b)h_2(x,y)$, then $(h,T)=0$ is equivalent to saying that $h_1(a,b)+\frac{12a^2-g_2}{2b}h_2(a,b)=0$. I don't see any reason to believe that this implies $h_1(a,b)=h_2(a,b)=0$. Given this relation, why should $h \in m_P^2$? Any help for the general case or this special case would be very much appreciated.

Answer 1

Given a nonzero linear form $g \in M_P/M_P^2$, the gradient $\nabla g$ of $g$ is not a scalar multiple of $\nabla f$ at $P$, because if it were, then $g$ would be congruent to a scalar multiple of $f$ modulo $M_P^2$, and hence we would have $g=0$ in $M_P/M_P^2$, contradicting the hypothesis that $g$ is nonzero in $M_P/M_P^2$. Since $\nabla g(P)$ is not a scalar multiple of $\nabla f(P)$, we have that $\nabla g$ is therefore not perpendicular to $T$. Projecting $\nabla g(P)$ onto $T$, we obtain an element $y \in T$ such that $(g,y) = (\nabla g(P)) \cdot y \neq 0$.