Differentials I

Setup: k alg. closed. D = \mbox{Spec} k[t]/t^2 and X is an algebraic k-scheme: meaning X \to k is quasi compact and of finite type.  Note there is a bijection

closed pts of X \leftrightarrow X/k (k) = \hom_k(k,X)

where I’ve abbreviated \mbox{Spec} k as k.  The scheme D is a one pt. space so given m \in \hom_k(D,X), there is a backwards map of local rings \mathcal{O}_{X,x} \to k[t]/t^2.  The map is k-linear and so its clear what the map does to elements outside of m_x , the map being a local homomorphism means it restricts to give m_x \to (t), as t^2 = 0 there is a factorization

\begin{array}{ccc} m_x \: \: \: \: & \xrightarrow{\: \: \: \: \: \: \: \: \: \: \:} & \: \: \: \: (t) \\ \searrow & \mbox{} & \nearrow \\ \mbox{} & m_x/m_x^2 & \mbox{} \end{array}

Also there is a canonical isomorphism of k-vector spaces: (t) \cong k sending at \mapsto a, thus the map m \in \hom_k(D,X) gives us a closed point x\in X and a module homomorphism m_x/m_x^2 \to k.  It can be checked there is a bijection

\hom_k(D,X) \leftrightarrow \bigcup_x \hom_k(m_x/m_x^2, k)

and now thickenings and the sheaf of differentials


First the commutative algebra version.  Let A be a ring an I and ideal such that any element of I is nilpotent.  For example consider the maximal ideal m in k[x_1,x_2, .... ]/(x_1^2,x_2^2, ... ).  Every element of m is nilpotent (because it has some x_i as a factor) but note that m^n is nonzero for every n>0.  In any case, for A,I as above note that any prime of A necessarily contains I, hence the surjection A \to A/I gives a homeomorphism on the associated map of spectra.

Now moving to the category of schemes, a thickening T \to T' is a closed immersion which is bijective.  Equivalently, a thickening is a closed immersion defined a qcoh sheaf of ideals I s.t. every local section of I is nilpotent.  Think of T’ as a slightly fatter version of T (thicker), topologically the same but scheme theoretically a little bigger.   A nilpotent thickening is one such that the defining ideal is nilpotent. If I^{n+1} = 0, then I defines an nth order thickening.  Finally, given a (locally) closed immersion X \to Z, it is defined by a sheaf of ideals I so I^{n+1} defines another (locally) closed immersion X_n \to Z and there is a factorization X \to X_n \to Z.  

Here’s an example.  Let E be a qcoh sheaf on X.  Set \mathcal{A} = \mathcal{O}_X\oplus E and define \mathcal{O}_X algebra structure on \mathcal{A}:

(a,e)\cdot (a',e') = (aa', ae' + a'e)

This makes \mathcal{A} into a qcho sheaf of algebras, so I can construct D(E) := \mbox{Spec}_X \mathcal{A} with a given affines morphism to X.  Using the identity map on X, and sheaf map \mathcal{A} \to \mathcal{O}_X which is just projection to \mathcal{O}_X, and the universal property of relative spec I get a morphism m\colon X \to D(E).  Note in D(E) the algebra structure makes E into a sheaf of ideals.  In fact X \to D(E) gives a map of sheaves \mathcal{O}_{D(E)} \to \mathcal{O}_X, and by construction, the kernel of this sheaf maps is the sheaf of ideals defined by E, and also E^2 = 0, in summary

X \to D(E)

is a closed immersion which is actually a 1st order thickening.  

Sheaf of Differentials

Say f \colon X \to Y is a morphism of schemes.  Then \Delta_{X/Y} \colon X \to X\times_Y X is at least a locally closed immersion.  Let P^n_{X/Y} by its nth order thickening, so there is a factorization of the diagonal

X \to P^n_{X/Y} \to X\times_Y X

without bothering to worry about pullbacks or pushforwards, there is a map of sheaves \mathcal{O}_{P^n} \to \mathcal{O}_X.  When n = 1, the kernel of this sheaf map (pulled back to a sheaf on X) is the sheaf of relative differentials denoted \Omega_{X/Y}^1:

\Omega_{X/Y} := \ker (\mathcal{O}_{P^n} \to \mathcal{O}_X)

This is looks much nicer in the affines case: X = \mbox{Spec} B and Y = \mbox{Spec} A.  First, the diagonal corresponds to a ring homomorphism B\otimes_A B \to B.  It is surjective with kernel I, so we can write B \cong B\otimes_A B/I. So

X \to P^n \to X\times_Y X


B \cong (B\otimes_A B)/I \leftarrow (B \otimes_A B)/I^2 \leftarrow B\otimes_A B

and so \Omega_{X/Y} = \ker (B\otimes_A B/I^2 \to B\otimes_A B/I) = I/I^2.  Now restricting the two projections p_i \colon X\times_Y X \to X, I get the diagram 

\begin{array}{ccc} \mbox{} & \mbox{} & P^n_{X/Y} \\ \mbox{} & \nearrow & \downarrow \downarrow \\ X & = & X \end{array}

So on sheaves there are maps p_i^\# \colon \mathcal{O}_X \to \mathcal{O}_{P^n}.  For n=1 set \delta = p_2^\# - p_1^\#, and let d be the composition

\mathcal{O}_X \xrightarrow{\delta} \ker(\mathcal{O}_{X\times_Y X} \to \mathcal{O}_X) \to \Omega_{X/Y}

It turns out that d is a derivation, in fact a pretty special one.  In the affine case \delta \colon B \to B \otimes_A B is the map b \mapsto 1 \otimes b - b\otimes 1

Here is a proposition for the affines case:

Prop: Let I_{B/A} = \ker(B\otimes_A B \to B), and \delta \colon B \to I_{B/A} be the map above.  Then

  1. \delta is A-linear and annihilates the image of A in B.
  2. \delta(bb') = b\delta(b') + b'\delta(b) + \delta(b)\delta(b')
  3. I_{B/A} is generated as a left B module by \{\delta(b) : b \in B\}


pf: 1 is a simple check, and 2 is a calculation.  Here is the proof of 3.  A general element g \in I_{B/A} can be expressed as \sum_i b_i \otimes b_i'.  That g maps to zero means \sum_i b_ib_i' = 0 so

\begin{array}{cl} g & =\sum_ib_i\otimes b_i' - 0 \\ \mbox{} & =\sum_ib_i\otimes b_i' - \sum_i b_ib_i' \\ \mbox{} & =\sum_ib_i\otimes b_i' - (\sum_ib_ib_i')\otimes 1 \\ \mbox{} & =\sum_i b_i(1 \otimes b_i' - b_i'\otimes 1) \\ \mbox{} & =\sum_i b_i \delta(b_i') \end{array}


Cor. For d the composition B \to I_{B/A} \to I_{B/A}/I_{B/A}^2

  1. d is A linear and annihilates the image of A in B
  2. d(bb') = bd(b') + b'd(b)
  3. \Omega_{B/A} = I/I^2 is generated as a B module by \{db : b \in B\}



Note: In contrast, Hartshorne begins by just defining \Omega_{B/A} to be the module generated by the formal symbols db modulo all the derivation relations.


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