# 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

## Thickenings

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

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

becomes

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

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

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

QED.

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 $\{db : b \in B\}$

——————————————–

QED.

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.