For a point \(P\) on a variety \(X\), let \(\mfm\) be the maximal ideal of the local ring \(\mco_P\). Define the Zariski tangent space \(T_P(X)\) of \(X\) at \(P\) to be the dual \(k\da\)vector space of \(\mfm/\mfm^2\).
For any point \(P \in X\), show that \(\dim T_P(X) \geq \dim X\), with equality if and only if \(P\) is nonsingular.
For any morphism \(\varphi: X \to Y\), show that there is a natural induced \(k\da\)linear map \(T_P(\varphi): T_P(X) \to T_{\varphi(P)}(Y)\).
If \(\varphi\) is the vertical projection of the parabola \(x = y^2\) onto the \(x\da\)axis, show that the induced map \(T_0(\varphi)\) of tangent spaces at the origin is the zero map.
Solution
Let \[ A=\OO_{X,P},\qquad \mathfrak m=\mathfrak m_P. \] Since \(X\) is a variety over the algebraically closed field \(k\), the residue field at the closed point \(P\) is \(k\).
1. The tangent-space dimension is the embedding dimension of the local ring:\[ \dim_kT_P(X)=\dim_k\mathfrak m/\mathfrak m^2. \]
This is immediate from the definition \[ T_P(X)=(\mathfrak m/\mathfrak m^2)^\vee. \] The vector space \(\mathfrak m/\mathfrak m^2\) is finite-dimensional because the local ring is noetherian and its maximal ideal is finitely generated. Its dimension is the minimum number of generators of \(\mathfrak m\), by Nakayama’s lemma, and is called the embedding dimension of \(A\). Dual finite-dimensional vector spaces have the same dimension.
2. For every point \(P\) of the variety,\[ \boxed{\dim T_P(X)\ge\dim X,} \]with equality exactly when \(P\) is nonsingular.
The noetherian local ring \(A\) satisfies \[ \dim A\le\dim_k\mathfrak m/\mathfrak m^2 \] [@Har10a, Chapter I, §5]. One way to see the inequality is to choose a minimal set of \(e=\dim_k\mathfrak m/\mathfrak m^2\) generators of \(\mathfrak m\). Krull’s height theorem gives \[ \dim A=\operatorname{height}\mathfrak m\le e. \] For a closed point of a variety, \(\dim A=\dim X\) [@Har10a, Exercise I.3.12]. Step 1 therefore gives the displayed inequality.
By definition, a noetherian local ring is regular precisely when \[ \dim A=\dim_k\mathfrak m/\mathfrak m^2. \] For varieties, the local ring \(\OO_{X,P}\) is regular exactly when \(P\) is nonsingular [@Har10a, Theorem I.5.1]. Thus equality in the tangent-space inequality is equivalent to nonsingularity, proving part (1).
3. A morphism \(\varphi:X\to Y\) naturally induces a linear tangent map\[ \boxed{T_P(\varphi):T_P(X)\longrightarrow T_{\varphi(P)}(Y).} \]
Put \(Q=\varphi(P)\) and let \[ \mathfrak n\subseteq\OO_{Y,Q} \] be the maximal ideal. The morphism induces the local homomorphism \[ \varphi_P^*:\OO_{Y,Q}\longrightarrow\OO_{X,P}. \] Because it is a local homomorphism, \[ \varphi_P^*(\mathfrak n)\subseteq\mathfrak m. \] Multiplicativity then gives \[ \varphi_P^*(\mathfrak n^2)\subseteq\mathfrak m^2. \] Hence there is a well-defined \(k\)-linear map of cotangent spaces \[ d_P^*\varphi:\mathfrak n/\mathfrak n^2 \longrightarrow \mathfrak m/\mathfrak m^2, \qquad [a]\longmapsto[\varphi_P^*(a)]. \] Dualizing reverses its direction and gives \[ T_P(\varphi) =(d_P^*\varphi)^\vee: (\mathfrak m/\mathfrak m^2)^\vee \longrightarrow (\mathfrak n/\mathfrak n^2)^\vee. \] These are exactly \(T_P(X)\) and \(T_Q(Y)\). The construction uses only the induced local-ring map, so it is natural and respects identities and composition. This proves part (2).
4. For vertical projection of the parabola \(X=V(x-y^2)\) to the \(x\)-axis,\[ \boxed{T_0(\varphi)=0.} \]
The parabola has coordinate ring \[ k[x,y]/(x-y^2)\cong k[y], \] and vertical projection corresponds contravariantly to \[ k[x]\longrightarrow k[y], \qquad x\longmapsto y^2. \] At the origins, the induced map on cotangent spaces is \[ (x)/(x^2) \longrightarrow (y)/(y^2). \] The class of its generator \(x\) maps to the class of \(y^2\), which is zero modulo \((y^2)\). Thus the cotangent map is zero. Its dual is therefore the zero map \[ T_0(X)\longrightarrow T_0(\AA_x^1), \] as required. Geometrically, the parabola has vertical tangent at the origin, so its differential under vertical projection collapses that tangent direction.
5. Q.E.D.
Steps <1>1–<1>2 prove part (1), step 3 proves part (2), and step 4 proves part (3).