Let \(X\) be an affine variety over \(\CC\), let \(p\in X\), and let \[ \mco_{X,p} \] be the local ring with maximal ideal \(\mfm_p\). Show canonically that \[ T_pX \cong \dualof{(\mfm_p/\mfm_p^2)}. \]
Solution
Put \[ \mco=\mco_{X,p}, \qquad \mfm=\mfm_p. \] Since \(p\) is a complex point, \[ \mco/\mfm\cong\CC. \] We regard \(\CC\) as an \(\mco\)-module through evaluation at \(p\).
1. Every \(\CC\)-derivation\[ D:\mco\longrightarrow\CC \]annihilates \(\mfm^2\).
Let \[ a,b\in\mfm. \] Their residues at \(p\) are zero. The Leibniz rule for a derivation into the residue-field module gives \[ D(ab) = a(p)D(b)+b(p)D(a) = 0. \] Thus \(D\) vanishes on every product of two elements of \(\mfm\), hence on the ideal \(\mfm^2\).
2. Restriction induces a linear map\[ \Phi: \Der_\CC(\mco,\CC) \longrightarrow \Hom_\CC(\mfm/\mfm^2,\CC). \]
For a derivation \(D\), define \[ \Phi(D)([a])=D(a), \qquad a\in\mfm. \] Step 1 shows that this depends only on the class of \(a\) modulo \(\mfm^2\). Additivity and \(\CC\)-linearity of \(D\) make \(\Phi(D)\) a \(\CC\)-linear functional.
3. Every linear functional\[ \lambda:\mfm/\mfm^2\longrightarrow\CC \]determines a derivation\[ D_\lambda:\mco\longrightarrow\CC \]by\[ D_\lambda(a) = \lambda\bigl([a-a(p)]\bigr). \]
For every \(a\in\mco\), the difference \[ a-a(p) \] lies in \(\mfm\), so the displayed formula is defined.
It is visibly additive and \(\CC\)-linear. For \(a,b\in\mco\), \[ \begin{aligned} ab-a(p)b(p) &= a(p)(b-b(p)) + b(p)(a-a(p))\\ &\qquad + (a-a(p))(b-b(p)). \end{aligned} \] The last term lies in \(\mfm^2\), so applying \(\lambda\) modulo \(\mfm^2\) gives \[ D_\lambda(ab) = a(p)D_\lambda(b) + b(p)D_\lambda(a). \] Thus \(D_\lambda\) satisfies the Leibniz rule and is a derivation into the residue field.
4. The maps of steps <1>2–<1>3 are inverse isomorphisms:\[ \boxed{ \Der_\CC(\mco,\CC) \cong \dualof{(\mfm/\mfm^2)}. } \]
Let \(D\) be a derivation. Since derivations kill constants, \[ D(a)=D(a-a(p)). \] Hence the derivation reconstructed from \(\Phi(D)\) is \(D\) itself.
Conversely, if \(\lambda\) is a functional and \(a\in\mfm\), then \(a(p)=0\), so \[ \Phi(D_\lambda)([a]) = D_\lambda(a) = \lambda([a]). \] Thus both composites are identities.
5. The Zariski tangent space \(T_pX\) is canonically\[ \Der_\CC(\mco_{X,p},\CC). \]
Embed \[ X\subseteq\AA^n_\CC \] and write \[ p=(a_1,\ldots,a_n). \] Choose generators \[ I(X)=(f_1,\ldots,f_m). \]
A derivation \[ D:\mco_{X,p}\longrightarrow\CC \] is determined by the values \[ v_j=D(x_j-a_j). \] For every relation \(f_i=0\) in the coordinate ring, the chain rule for polynomials gives \[ 0 = D(f_i) = \sum_{j=1}^n \frac{\partial f_i}{\partial x_j}(p)v_j. \] Thus \[ v=(v_1,\ldots,v_n) \] lies in the kernel of the Jacobian matrix, which is \(T_pX\) by definition.
Conversely, if \[ v\in\ker J(p), \] the directional derivative \[ h\longmapsto \sum_{j=1}^n v_j\frac{\partial h}{\partial x_j}(p) \] is a derivation on \(\CC[x_1,\ldots,x_n]\) that annihilates \(I(X)\), because the chosen generators and therefore the whole ideal have zero directional derivative at \(p\). It descends to the coordinate ring and extends uniquely to the localization \(\mco_{X,p}\). Hence every tangent vector gives a unique derivation.
These constructions are inverse and independent of the chosen generators, so \[ T_pX \cong \Der_\CC(\mco_{X,p},\CC). \]
6. Therefore\[ \boxed{ T_pX \cong \dualof{(\mfm_p/\mfm_p^2)}. } \]
Step 5 identifies \(T_pX\) with the derivation space, and step 4 identifies that derivation space with the dual of the cotangent space \(\mfm_p/\mfm_p^2\).
7. Q.E.D.
Step 6 is the required canonical isomorphism.