Intuition behind the first exact sequence for Kahler differentials
问题内容
I'm studying algebraic geometry and have a question on Kahler differentials. Let $k$ be a commutative ring with unit and $A$ a $k$-algebra. I already have the intuition if we picture $A$ as some ring of functions on $\operatorname{Spec}{A}$, the the module of relative Kahler differentials $\Omega_{A/k}$ can be thought of as the differentials of those functions, $da$ for $a\in A$, with $k$ taken as constants, so that $dk =0$.
We then have the first exact sequence in Hartshorne: given ring homomorphisms $A\to B\to C$ we have
$$\Omega_{B/A}\otimes_B C\to \Omega_{C/A}\to \Omega_{C/B}\to 0$$
For the proof the reader is referred to Matsumura's Commutative Algebra. Roughly if the homomorphisms are $\phi:A\to B$ and $\psi:B\to C$ one defines $v : \Omega_{B/A}\otimes_B C\to\Omega_{C/A}$ by $v(d_{B/A}(b)\otimes c) = c d_{C/A}(\psi(b))$ and $u:\Omega_{C/A}\to \Omega_{C/B}$ by $u(d_{C/A}(c)) = d_{C/B}(c)$. Then Matsumura shows the exactness of the sequence.
My question here is on the intuition behind this result. I know that $\Omega_{A/k}$ is like a module of globally-defined differentials on $\operatorname{Spec}{A}$, so it is like "global sections of a cotangent bundle". I would hope that this sequence has some important interpretation. In fact, it even gets the name "First Exact Sequence", so surely it is important.
But what is the intuition behind it? Thinking of the rings as rings of functions and of the Kahler differentials modules as modules of globally-defined differentials (or global sections of cotangent bundles), what is the intuition? What the three conditions for exactness really mean here?
回答 (1)
I will answer in the case where everything is of finite type over an algebraically closed field $k$, since it tends to be the most (geometrically) intuitive case.
First, if we take $A = k$ and take $B \to C$ to be $k$-algebras, the exact sequence gives us an interpretation of $\Omega_{C/B}$. Namely, we can think of them as relative $1$-forms. Indeed, the morphism $B \to C$ can be interpreted as coming from pulling back regular functions along the morphism $\mathrm{Spec\ } C \to \mathrm{Spec\ } B$. We can also pull back $1$-forms on $\mathrm{Spec\ } B$ to $1$-forms on $\mathrm{Spec\ } C$, like in the case of manifolds, and the differential of a function is sent to the differential of the pulled back function. We can look at the $C$-span of these forms, that is, the space of all differential forms which are obtained by taking $C$-linear combinations of these pulled back $1$-forms. This is what the first map is doing. Then, $\Omega_{C/B}$ is exactly the quotient of the space of all $1$-forms on $\mathrm{Spec\ } C$ by the $C$-submodule of those coming from $\mathrm{Spec\ } B$.
We can think of the morphism $\mathrm{Spec\ } C \to \mathrm{Spec\ } B$ as parametrizing a family of, say, varieties, given by the different fibers of this map at closed points. Relative differentials are defined by saying that those functions constant on the fibers of $\mathrm{Spec\ } C \to \mathrm{Spec\ } B$ which trivially come from $\mathrm{Spec\ } B$ have differential zero, as if they were constant.
As an example, the relative differentials for the projection on the first $r$ factors $\mathbb{A}_k^n \to \mathbb{A}_k^r$ are precisely those of the form $\sum_{i= r+ 1}^n f_i(x_1, \ldots, x_n) dx_i$.
Now, for the general case, the statement can then be interpreted as follows: If I have morphisms $\mathrm{Spec\ } C \to \mathrm{Spec\ } B \to \mathrm{Spec\ } A$ over $\mathrm{Spec\ } k$, I can first think of $\mathrm{Spec\ } C$ as a family over $\mathrm{Spec\ } A$, and compute the relative one forms $\Omega_{C/A}$. Every relative $1$-form for the family $\mathrm{Spec\ } B \to \mathrm{Spec\ } A$ pulls back to a relative $1$-form in $\Omega_{C/A}$. Intuition tells us that we should be able to compute the space of relative $1$-forms $\Omega_{C/B}$, now thinking of $\mathrm{Spec\ } C$ as a family over $\mathrm{Spec\ } B$, by quotienting by (the $C$-module generated by) these relative $1$-forms which we have pulled back from $\mathrm{Spec\ } B$. The exact sequence then tells us that this is correct!