Let \[ X\subseteq\AA^n_\CC \] be an affine variety of dimension \(d\), and let \(p\in X\). Show that \[ \dim_\CC T_pX=d \] if \(p\) is smooth, while \[ \dim_\CC T_pX>d \] if \(p\) is singular.
Solution
Choose generators \[ I(X)=(f_1,\ldots,f_m) \] and let \[ F=(f_1,\ldots,f_m):\AA^n_\CC\longrightarrow\AA^m_\CC. \] Write \[ r=\rank J_F(p), \] where \(J_F(p)\) is the Jacobian matrix at \(p\).
1. The Zariski tangent space satisfies\[ T_pX=\ker dF(p) \]and therefore\[ \boxed{\dim_\CC T_pX=n-r.} \]
The Zariski tangent space at \(p\) is the kernel of the differentials of generators of \(I(X)\), that is, \(T_pX=\ker dF(p)\). The linear map \[ dF(p):\CC^n\longrightarrow\CC^m \] has matrix \(J_F(p)\) and rank \(r\). Rank-nullity gives \[ \dim_\CC\ker dF(p) = n-r. \]
2. If \(p\) is smooth, then\[ \dim_\CC T_pX=d. \]
By the Jacobian criterion, \[ p\text{ smooth} \quad\Longleftrightarrow\quad r=n-d. \] Substituting this into step 1 gives \[ \dim_\CC T_pX = n-(n-d) = d. \]
3. If \(p\) is singular, then\[ \dim_\CC T_pX>d. \]
At a singular point the Jacobian rank is strictly smaller than the codimension: \[ r<n-d. \] Hence \[ n-r > n-(n-d) = d. \] Step 1 identifies the left-hand side with \(\dim_\CC T_pX\).
4. Therefore\[ \boxed{ p\text{ is smooth} \quad\Longleftrightarrow\quad \dim_\CC T_pX=d. } \]
Step 2 gives equality at smooth points, while step 3 shows that every singular point has strictly larger tangent dimension.
5. Q.E.D.
Steps <1>1–<1>4 prove both assertions.