User:Marty/UCSC Math 203 Fall 2008/Lecture 5

As an excellent and important example of our techniques, we will discuss modules of derivations.

=Derivations=

Suppose that $A$ is a ring, and $M$ is an $A$-module.

Definition: A derivation from $A$ to $M$ is a function $d \colon A \rightarrow M$, satisfying the following identities: We write $Der(A,M)$ for the set of derivations from $A$ to $M$.
 * For all $a,b \in A$, $d(ab) = a \cdot d(b) + b \cdot d(a)$. (The Liebniz rule).
 * For all $a,b \in A$, $d(a \pm b) = d(a) \pm d(b)$.

The set $Der(A,M)$ is naturally an $A$-module in the following way:
 * For $\alpha \in A$, and $d \in Der(A,M)$, define $[\alpha d](a) = \alpha \cdot d(a)$.
 * For $d_1, d_2 \in Der(A,M)$, define $[d_1 \pm d_2](a) = d_1(a) \pm d_2(a)$.

A particularly important example is $Der(A,A)$, also sometimes written as $Der(A)$, the ring of (absolute) derivations on $A$.

There is a functor, denoted $Der(A, \bullet)$ from the category of $A$-modules to the category of $A$-modules; for example, if $M \rightarrow N$ is a morphism of $A$-modules, then there is an induced morphism of $A$-modules: $$Der(A,M) \rightarrow Der(A,N).$$

We are interested here, in studying the module $Der(A)$, and related modules of derivations.

=At a point=

Suppose that $p$ is a maximal ideal of $A$, so that $A/p$ is a field. Equivalently, $p$ is a closed point of $Spec(A)$. Let $k = A/p$ be the resulting field. Then $k$ is also an $A$-module in the natural way (it's an $A$-algebra, in fact). It follows that we may consider the module of derivations at $p$: $$Der_p(A) = Der(A,k).$$

In order to think about this, a geometric intuition is extremely helpful:
 * Elements $f \in A$ should be thought of as "functions" on the space $X = Spec(A)$.
 * If $f \in A$, and $\bar f \in A/p$ is its image in the field $k = A/p$, then $\bar f$ should be thought of as the evaluation of $f$ at $p$: $f(p) = \bar f$.
 * A derivation $d \in Der(A,k)$ gives us a way of "differentiating" a "function" $f \in A$, in the sense that the Liebniz (product) rule and the addition rule are satisfied.

To bolster this intuition, observe the following: $$d(f/g) = \frac{\bar f \cdot d(g) - \bar g \cdot d(f) }{\bar g^2},$$ for all $g \in A - p$. Observe that $\bar g^2$ is invertible, in $A/p$.
 * If $d \in Der(A,k)$, then $d(1) = 0$. Indeed, the Liebniz rule implies that $d(1) = d(1 \cdot 1) = 1 \cdot d(1) + 1 \cdot d(1) = d(1) + d(1)$.
 * If $d \in Der(A,k)$, and $f \in A$ "vanishes to order at least 2 at $p$", i.e., $f \in p^2$, then $d(f) = 0$. Indeed, $d(rs) = \bar r d(s) + \bar s d(r) = 0$ if $r,s \in p$.
 * If $d \in Der(A,k)$, then $d$ extends canonically to an element of $Der(A_p,k)$, where $A_p$ is the localization of $A$ at $p$, via the rule:

In other words, the derivative of the "constant function" vanishes, and more generally, the derivative of a function which vanishes to quadratic order. Moreover, the derivative at $p$ is a "local property", it extends to $A_p$, and $d(f)$ can be computed from $d(f/1)$ where $f/1$ is the image of $f$ in $A_p$.

=Relative=

Often, it is the case that one works with rings $A$ in which some elements are thought of as "constants". For example, suppose one is given a ring homomorphism $C \rightarrow A$. One might think of the elements of $C$ being the "constant functions", when viewed via the homomorphism, in $A$. For example, in the ring $\CC[X,Y]$, one often thinks of the subring $\CC$ as the "constants".

In general, Grothendieck suggested that all of algebraic geometry should be relative: one should always consider a pair of rings connected by a homomorphism, such as $C \rightarrow A$, and think of working with the ring $A$ "over the base" $C$. Ordinary ring theory can be thought of as working "over the base" $\ZZ$, since $\ZZ$ maps uniquely into any ring.

With this philosophy, we can describe the module of "relative derivations", as follows: Suppose that $C \rightarrow A$ is a ring homomorphism, and $M$ is an $A$-module. An $M$-valued derivation of $A$, over the base $C$, is a function $d \colon A \rightarrow M$, satisfying the following identities: We write $Der_C(A,M)$ for the resulting $A$-module of $C$-linear derivations.
 * $d(ab) = a \cdot d(b) + b \cdot d(a)$, for all $a,b \in A$. (The Liebniz rule).
 * $d(a \pm b) = d(a) \pm d(b)$ for all $a,b \in A$. (Additivity of derivation)
 * $d(ca) = c \cdot d(a)$, for all $c \in C$ (in particular, $d(c) = 0$, where $c$ denotes the image $c \cdot 1$ of $c$ in $A$).

=Examples=

Consider the ring $A = \CC[X,Y] / \langle Y^2 - X^3 + X^2 \rangle$, over the base ring $C = \CC$. Thus, we limit our attention to $\CC$-linear derivations!

Consider the following two maximal ideals of $A$:
 * $p = \langle X - 0, Y - 0 \rangle$.
 * $q = \langle X - 1, Y - 0 \rangle$.

Geometrically, there is a significant difference when "looking" at $Spec(A)$ near $p$ and $q$. We can "see" this difference by considering derivations. First, observe the following:
 * $A / p \cong \CC$, and $A / q \cong \CC$.
 * The composite morphisms $\CC \hookrightarrow A \twoheadrightarrow A/p$ and $\CC \hookrightarrow A \twoheadrightarrow A/q$ are isomorphisms.

Suppose that $x$ is any maximal ideal of $A$. Then the composite morphism $\CC \hookrightarrow A \twoheadrightarrow A/x$ is an isomorphism, in fact. Suppose that $d \in Der_\CC(A, k_x)$ is a derivation of $A$ "at $x$". Then, we find that if $d(f) = 0$ for all $f \in x$, then $d = 0$. Indeed, since $d(\CC) = 0$, and every element of $A$ can be expressed as $c + f$ for a function $f \in x$ and $c \in \CC$, we find that $d(c + f) = 0$. For this reason, we find that:
 * Derivations in $Der_\CC(A, k_x)$ are completely determined by their values on $\bar f \in x / x^2$.

In this way, the following two $\CC$-vector spaces can be identified: $$Der_\CC(A, k_x) \leftrightarrow Hom_\CC(x / x^2, \CC).$$

This allows us to distinguish the points $p$ and $q$ in a geometric way. We find that: $$Der_\CC(A_p, k_p) \leftrightarrow Hom_\CC(p A_p / (p A_p)^2, \CC).$$ Note that $p A_p / (p A_p)^2$ is spanned by $X, Y$, as a complex vector space. The relation $Y^2 = X^3 - X^2$ has no effect, and we find that: $$dim_\CC(p A_p/ (p A_p)^2) = 2.$$

However, consider $q A_q / (q A_q)^2$. This space is spanned by $X-1$ and $Y$, as a complex vector space. But $X-1 = Y^2/X^2$ in $A_q$, and $Y^2 / X^2 \in (q A_q)^2$. It follows that $X-1 \in (q A_q)^2$. Hence $X-1 = 0$, when viewed in $q A_q / (q A_q)^2$. One can deduce that: $$dim_\CC(q A_q / (q A_q)^2) = 1.$$

One should interpret this geometrically as follows:
 * Near the point $(0,0)$, there are two possible "tangent directions", along which one can take a directional derivative.
 * Near the point $(1,0)$, there is only one possible "tangent direction", along which one can take a directional derivative.