Let \(X\) be a scheme. For any point \(x \in X\), define the Zariski tangent space \(T_x\) to \(X\) at \(x\) to be the dual of the \(k(x)\)-vector space \(\mfm_x / \mfm_x^2\).
Now assume that \(X\) is a scheme over a field \(k\), and let \(k[\eps]/\eps^2\) be the ring of dual numbers over \(k\). Show that to give a \(k\)-morphism \(\Spec k[\eps]/\eps^2 \to X\) is equivalent to giving a point \(x \in X\) rational over \(k\), that is, with \(k(x) = k\), together with an element of \(T_x\).
Solution
Put \[ R=k[\varepsilon]/(\varepsilon^2), \qquad S=\operatorname{Spec}R. \]
1. The scheme \(S\) has a single point, corresponding to the maximal ideal\[ (\varepsilon), \]and its residue field is \(k\).
Every prime ideal contains every nilpotent element, so any prime of \(R\) contains \(\varepsilon\). Prime ideals of \(R\) therefore correspond to prime ideals of \[ R/(\varepsilon)\cong k. \] The field \(k\) has the single prime ideal \((0)\), so \(R\) has the single prime ideal \((\varepsilon)\).
The quotient by this maximal ideal is \[ R/(\varepsilon)\cong k. \]
2. A \(k\)-morphism\[ f:S\longrightarrow X \]determines a point\[ x=f((\varepsilon))\in X \]and a local \(k\)-algebra homomorphism\[ \phi=f^\sharp_x: \mathcal O_{X,x} \longrightarrow R. \]
The source has one point, so the underlying continuous map chooses \(x\). The map on stalks at that point is local because \(f\) is a morphism of schemes. Since \(f\) is a morphism over \(\operatorname{Spec}k\), the stalk map commutes with the structure maps from \(k\), so it is a \(k\)-algebra homomorphism.
3. The point \(x\) in 2 is \(k\)-rational:\[ \boxed{\kappa(x)=k.} \]
Reduce the local homomorphism \[ \phi:\mathcal O_{X,x}\to R \] modulo the maximal ideal \((\varepsilon)\) of \(R\). Since \(\phi\) is local, its inverse image of \((\varepsilon)\) is the maximal ideal \(\mathfrak m_x\). Therefore reduction induces an injective \(k\)-algebra homomorphism \[ \kappa(x) =\mathcal O_{X,x}/\mathfrak m_x \hookrightarrow R/(\varepsilon) =k. \]
The structural map \(k\to\kappa(x)\) followed by this injection is the identity because \(f\) is a \(k\)-morphism. Hence the inclusion \(k\to\kappa(x)\) is an isomorphism: the image of \(\kappa(x)\) in \(k\) contains all of \(k\) and is contained in \(k\). Thus \(\kappa(x)=k\) as a \(k\)-algebra.
4. Under the identification \(\kappa(x)=k\), every local \(k\)-algebra map\[ \phi:\mathcal O_{X,x}\longrightarrow R \]whose reduction is the residue map has a unique form\[ \boxed{ \phi(a)=\bar a+D(a)\varepsilon, } \]where\[ D:\mathcal O_{X,x}\longrightarrow k \]is a \(k\)-derivation and \(\bar a\) is the residue class of \(a\) in \(k\).
Every element of \(R\) has a unique expression \[ c+d\varepsilon, \qquad c,d\in k. \] Since reduction of \(\phi(a)\) modulo \(\varepsilon\) is the residue \(\bar a\), there is a unique scalar \(D(a)\in k\) with \[ \phi(a)=\bar a+D(a)\varepsilon. \]
Additivity of \(\phi\) makes \(D\) additive. Because \(\phi\) is a \(k\)-algebra map, \[ D(c)=0 \qquad(c\in k). \] Finally, \[ \begin{aligned} \phi(ab) &=\phi(a)\phi(b)\\ &=(\bar a+D(a)\varepsilon) (\bar b+D(b)\varepsilon)\\ &=\bar a\bar b +(\bar aD(b)+\bar bD(a))\varepsilon, \end{aligned} \] because \(\varepsilon^2=0\). Comparing the \(\varepsilon\)-coefficients gives \[ D(ab)=\bar aD(b)+\bar bD(a), \] which is the Leibniz rule for a derivation to the residue-field module \(k\).
Conversely, any \(k\)-derivation \(D\) makes the displayed formula additive, multiplicative, and unital, hence defines such a \(k\)-algebra homomorphism.
5. A \(k\)-derivation\[ D:\mathcal O_{X,x}\longrightarrow k \]is equivalent to a \(k\)-linear functional\[ \lambda: \mathfrak m_x/\mathfrak m_x^2 \longrightarrow k. \]
If \[ a,b\in\mathfrak m_x, \] then \[ \bar a=\bar b=0, \] so the Leibniz rule gives \[ D(ab)=0. \] Hence \(D\) vanishes on \(\mathfrak m_x^2\).
Also \(D\) vanishes on \(k\). Since the residue field is \(k\), every \[ a\in\mathcal O_{X,x} \] has a unique decomposition \[ a=\bar a+m, \qquad \bar a\in k, \quad m\in\mathfrak m_x. \] Thus \(D\) is completely determined by its restriction to \(\mathfrak m_x\), and that restriction factors uniquely through a \(k\)-linear map \[ \lambda:\mathfrak m_x/\mathfrak m_x^2\to k. \]
Conversely, given such a linear functional \(\lambda\), define \[ D(a)=\lambda(a-\bar a\bmod\mathfrak m_x^2). \] For \[ a=\bar a+m, \qquad b=\bar b+n, \] one has modulo \(\mathfrak m_x^2\) \[ ab-\bar a\bar b \equiv \bar a n+\bar b m. \] Applying \(\lambda\) gives \[ D(ab)=\bar aD(b)+\bar bD(a), \] so \(D\) is a \(k\)-derivation.
6. Therefore local \(k\)-algebra maps\[ \mathcal O_{X,x}\longrightarrow k[\varepsilon]/(\varepsilon^2) \]reducing to the residue map are naturally parametrized by the Zariski tangent space\[ \boxed{ T_xX = (\mathfrak m_x/\mathfrak m_x^2)^\vee. } \]
Step 4 identifies such local maps with \(k\)-derivations to \(k\), and 5 identifies those derivations with the dual of \(\mathfrak m_x/\mathfrak m_x^2\).
7. Conversely, let \(x\in X\) be \(k\)-rational and let\[ v\in T_xX. \]The corresponding derivation \(D_v\) and formula\[ \phi_v(a)=\bar a+D_v(a)\varepsilon \]define a local \(k\)-algebra homomorphism\[ \phi_v:\mathcal O_{X,x}\longrightarrow R. \]
By 5, \(v\) corresponds to a \(k\)-derivation \(D_v\). Step 4 shows that \(\phi_v\) is a \(k\)-algebra homomorphism.
Its reduction modulo \((\varepsilon)\) is the residue map \[ \mathcal O_{X,x}\to k, \] so \[ \phi_v^{-1}((\varepsilon))=\mathfrak m_x. \] Thus \(\phi_v\) is local.
8. The local homomorphism \(\phi_v\) defines a unique \(k\)-morphism\[ f_v:\operatorname{Spec}R\longrightarrow X \]with underlying point \(x\).
Send the unique point of \(\operatorname{Spec}R\) to \(x\).
For an open set \(U\subseteq X\) containing \(x\), define the pullback on sections as \[ \mathcal O_X(U) \longrightarrow \mathcal O_{X,x} \xrightarrow{\phi_v} R, \] where the first map takes a germ at \(x\). If \(x\notin U\), the inverse image in the one-point source is empty, so use the unique map to the zero ring.
These maps commute with restrictions and define a morphism of sheaves. The only stalk map is \(\phi_v\), which is local by 7, so this is a morphism of schemes. It is a \(k\)-morphism because \(\phi_v\) is a \(k\)-algebra map.
Uniqueness follows because any morphism with underlying point \(x\) is determined by its local-ring map at that point, exactly as in Hartshorne II.2.7.
9. The two constructions are inverse. Hence\[ \boxed{ \operatorname{Hom}_k \left( \operatorname{Spec}k[\varepsilon]/(\varepsilon^2),X \right) \cong \coprod_{x\in X(k)}T_xX. } \]
Starting with a morphism, steps <1>2–<1>6 extract its point \(x\) and the coefficient derivation \(D\), hence the tangent vector \(v\). Reconstructing from \(v\) gives the same stalk homomorphism \[ a\mapsto\bar a+D(a)\varepsilon, \] and therefore the same morphism by 8.
Starting from \((x,v)\), the constructed morphism has stalk map \(\phi_v\), whose \(\varepsilon\)-coefficient is exactly \(D_v\) and hence recovers the original tangent vector.
10. Q.E.D.
Step 9 is the equivalence requested by the exercise.