Definition 111.35.1. For any ring R we denote R[\epsilon ] the ring of dual numbers. As an R-module it is free with basis 1, \epsilon . The ring structure comes from setting \epsilon ^2 = 0.
111.35 Tangent Spaces
Exercise 111.35.2. Let f : X \to S be a morphism of schemes. Let x \in X be a point, let s = f(x). Consider the solid commutative diagram
\xymatrix{ \mathop{\mathrm{Spec}}(\kappa (x)) \ar[r] \ar[dr] \ar@/^1pc/[rr] & \mathop{\mathrm{Spec}}(\kappa (x)[\epsilon ]) \ar@{.>}[r] \ar[d]& X \ar[d] \\ & \mathop{\mathrm{Spec}}(\kappa (s)) \ar[r] & S }
with the curved arrow being the canonical morphism of \mathop{\mathrm{Spec}}(\kappa (x)) into X. If \kappa (x) = \kappa (s) show that the set of dotted arrows which make the diagram commute are in one to one correspondence with the set of linear maps
\mathop{\mathrm{Hom}}\nolimits _{\kappa (x)}( \frac{\mathfrak m_ x}{\mathfrak m_ x^2 + \mathfrak m_ s\mathcal{O}_{X, x}}, \kappa (x))
In other words: describe such a bijection. (This works more generally if \kappa (x) \supset \kappa (s) is a separable algebraic extension.)
Definition 111.35.3. Let f : X \to S be a morphism of schemes. Let x \in X. We dub the set of dotted arrows of Exercise 111.35.2 the tangent space of X over S and we denote it T_{X/S, x}. An element of this space is called a tangent vector of X/S at x.
Exercise 111.35.4. For any field K prove that the diagram
\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(K[\epsilon _1]) \ar[d] \\ \mathop{\mathrm{Spec}}(K[\epsilon _2]) \ar[r] & \mathop{\mathrm{Spec}}(K[\epsilon _1, \epsilon _2]/(\epsilon _1\epsilon _2)) }
is a pushout diagram in the category of schemes. (Here \epsilon _ i^2 = 0 as before.)
Exercise 111.35.5. Let f : X \to S be a morphism of schemes. Let x \in X. Define addition of tangent vectors, using Exercise 111.35.4 and a suitable morphism
\mathop{\mathrm{Spec}}(K[\epsilon ]) \longrightarrow \mathop{\mathrm{Spec}}(K[\epsilon _1, \epsilon _2]/(\epsilon _1\epsilon _2)).
Similarly, define scalar multiplication of tangent vectors (this is easier). Show that T_{X/S, x} becomes a \kappa (x)-vector space with your constructions.
Exercise 111.35.6. Let k be a field. Consider the structure morphism f : X = \mathbf{A}^1_ k \to \mathop{\mathrm{Spec}}(k) = S.
Let x \in X be a closed point. What is the dimension of T_{X/S, x}?
Let \eta \in X be the generic point. What is the dimension of T_{X/S, \eta }?
Consider now X as a scheme over \mathop{\mathrm{Spec}}(\mathbf{Z}). What are the dimensions of T_{X/\mathbf{Z}, x} and T_{X/\mathbf{Z}, \eta }?
Remark 111.35.7. Exercise 111.35.6 explains why it is necessary to consider the tangent space of X over S to get a good notion.
Exercise 111.35.8. Consider the morphism of schemes
f : X = \mathop{\mathrm{Spec}}(\mathbf{F}_ p(t)) \longrightarrow \mathop{\mathrm{Spec}}(\mathbf{F}_ p(t^ p)) = S
Compute the tangent space of X/S at the unique point of X. Isn't that weird? What do you think happens if you take the morphism of schemes corresponding to \mathbf{F}_ p[t^ p] \to \mathbf{F}_ p[t]?
Exercise 111.35.9. Let k be a field. Compute the tangent space of X/k at the point x = (0, 0) where X = \mathop{\mathrm{Spec}}(k[x, y]/(x^2 - y^3)).
Exercise 111.35.10. Let f : X \to Y be a morphism of schemes over S. Let x \in X be a point. Set y = f(x). Assume that the natural map \kappa (y) \to \kappa (x) is bijective. Show, using the definition, that f induces a natural linear map
\text{d}f : T_{X/S, x} \longrightarrow T_{Y/S, y}.
Match it with what happens on local rings via Exercise 111.35.2 in case \kappa (x) = \kappa (s).
Exercise 111.35.11. Let k be an algebraically closed field. Let
\begin{eqnarray*} f : \mathbf{A}_ k^ n & \longrightarrow & \mathbf{A}^ m_ k \\ (x_1, \ldots , x_ n) & \longmapsto & (f_1(x_ i), \ldots , f_ m(x_ i)) \end{eqnarray*}
be a morphism of schemes over k. This is given by m polynomials f_1, \ldots , f_ m in n variables. Consider the matrix
A = \left( \frac{\partial f_ j}{\partial x_ i} \right)
Let x \in \mathbf{A}^ n_ k be a closed point. Set y = f(x). Show that the map on tangent spaces T_{\mathbf{A}^ n_ k/k, x} \to T_{\mathbf{A}^ m_ k/k, y} is given by the value of the matrix A at the point x.
Comments (3)
Comment #4094 by Benjamin Voulgaris Church on
Comment #4095 by Benjamin Voulgaris Church on
Comment #4152 by Johan on