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In EGA IV, Sec. 16, Grothendieck defines the sheaf of principal parts as follows: Let $f:X\rightarrow S$ be a morphism of schemes and $\Delta:X\rightarrow X\times_S X$ the diagonal morphism associated to $f$. $\Delta$ is an immersion, so the corresponding morphism $\Delta^{-1}\mathcal{O}_{X\times_S X}\rightarrow\mathcal{O}_X$ is surjective. Let $\mathcal{I}$ denote its kernel and define the sheaves of principal parts as

\[\mathcal{P}_{X/S}^n:=\Delta^{-1}( \mathcal{O}_{X\times_S X}) / \mathcal{I}^{n+1}\]

In their book on Crystalline cohomology, Berthelot and Ogus define the sheaf of principal parts $\mathcal{P}^n_{X/S}$ as

\[(\mathcal{O}_X\otimes_{f^{-1}\mathcal{O}_S}\mathcal{O}_X)/\mathbf{I}^{n+1},\] where $\mathbf{I}$ is the kernel of the multiplication map from the tensor product to $\mathcal{O}_X$.

My question is probably simple, but I don't know how to see it: Why are those definitions equivalent if $X$ and $S$ are not affine and $n>0$?

I've not seen the second definition anywhere else, although it seems somewhat nicer than the first one...

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I fixed an error in the statement. I looked it up to make sure. – Harry Gindi Apr 10 '10 at 18:34
up vote 5 down vote accepted

The statement holds in general if $f : X \to S$ is a morphism of locally ringed spaces. The fibred product of locally ringed spaces can be constructed explicitly without gluing constructions, and also restricts to the fibred product of schemes. See this article (german; shall I translate it?) for details. I will make use of the explicit description given there. Also I use stalks all over the place. Probably this is not the most elegant proof, but it works.

First we construct a homomorphism $\mathcal{O}_X \otimes_{f^{-1} \mathcal{O}_S} \mathcal{O}_X \to \Delta^{-1} \mathcal{O}_{X \times_S X}$. For that we compute the stalks at some point $x \in X$ lying over $s \in S$:

$(\mathcal{O}_X \otimes_{f^{-1} \mathcal{O}_S} \mathcal{O}_X)_x = \mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x},$

$(\Delta^{-1} \mathcal{O}_{X \times_S X})_x = \mathcal{O}_{X \times_S X,\Delta(x)} = (\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x})_{\mathfrak{q}}$,

where $\mathfrak{q}$ is the kernel of the canonical homomorphism

$\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x} \to \kappa(x), a \otimes b \mapsto \overline{ab}.$

Thus we get, at least, homomorphisms between the stalks (namely localizations). In order to get sheaf homomorphisms out of them, the following easy lemma is useful:

(*) Let $F,G$ be sheaves on a topological space $X$ and for every $x \in X$ let $s_x : F_x \to G_x$ be a homomorphism. Suppose that they fit together in the sense that for every open $U$, every section $f \in F(U)$ and every $x \in U$ there is some open neighborhood $x \in W \subseteq U$ and some section $g \in G(W)$ such that $s_y$ maps $f_y$ to $g_y$ for all $y \in W$. Then there is a sheaf homomorphism $s : F \to G$ inducing $s$.

This can be applied in the above situation: Every section in a neighborhood of $x$ in $\mathcal{O}_X \otimes_{f^{-1} \mathcal{O}_S} \mathcal{O}_X$ induced by an element in $\mathcal{O}_X(U) \otimes_{\mathcal{O}_S(V)} \mathcal{O}_X(U)$ for some neighborhoods $U$ of $x$ and $V$ of $s$ such that $U \subseteq f^{-1}(V)$. This yields a section in $\mathcal{O}_{X \times_S X}$ on the basic-open subset $\Omega(U,U,V;1)=U \times_V U$ and thus a section of $\Delta^{-1} \mathcal{O}_{X \times_S X}$ on $U$. It is easily seen, that this construction yields the natural map on the stalks.

Thus we have a homomorphism $\alpha : \mathcal{O}_X \otimes_{f^{-1} \mathcal{O}_S} \mathcal{O}_X \to \Delta^{-1} \mathcal{O}_{X \times_S X}$. Now let $J$ be the kernel of the multiplication map $\mathcal{O}_X \otimes_{f^{-1} \mathcal{O}_S} \mathcal{O}_X \to \mathcal{O}_X$ and $I$ be the kernel of the homomorphism $\Delta^\# : \Delta^{-1} \mathcal{O}_{X \times_S X} \to \mathcal{O}_X$. Then for every $n \geq 1$ our $\alpha$ restricts to a homomorphism

$(\mathcal{O}_X \otimes_{f^{-1} \mathcal{O}_S} \mathcal{O}_X)/J^n \to (\Delta^{-1} \mathcal{O}_{X \times_S X})/I^n,$

which is given at $x \in X$ by the natural map

$(\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x}) / \mathfrak{p}^n \to ((\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x}) / \mathfrak{p}^n)_{\mathfrak{q}}$,

where $\mathfrak{p} \subseteq \mathfrak{q}$ is the kernel of the multiplication map $\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x} \to \mathcal{O}_{X,x}$.

We want to show that this map is an isomorphism, i.e. that the localization at $\mathfrak{q}$ is not needed. For that it is enough to show that every element in $\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x}$, whose image in $\mathcal{O}_{X,x}$ is invertible, is invertible modulo $\mathfrak{p}^n$. Or in other words: Preimages of units are units with respect to the projection

$(\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x}) / \mathfrak{p}^n \to (\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x}) / \mathfrak{p}^1 \cong \mathcal{O}_{X,x}$.

However, this follows from the observation that the kernel $\mathfrak{p}^1 / \mathfrak{p}^n$ is nilpotent; cf. also this question.

I'm sure that there is also a proof which avoids stalks at all.

EDIT: So here is a direct construction of the homomorphism $\mathcal{O}_X \otimes_{f^{-1} \mathcal{O}_S} \mathcal{O}_X \to \Delta^{-1} \mathcal{O}_{X \times_S X}$:

Let $p_1,p_2$ be the projections $X \times_S X \to X$. Then we have for $i=1,2$ the homomorphism

$\mathcal{O}_X \to {p_i}_* \mathcal{O}_{X \times_S X} \to {p_i}_* \Delta_* \Delta^{-1} \mathcal{O}_{X \times_S X} = (p_i \Delta)_* \Delta^{-1} \mathcal{O}_{X \times_S X} = \Delta^{-1} \mathcal{O}_{X \times_S X}$,

and they commute over $f^{-1} \mathcal{O}_S$. Thus we get the desired homomorphism. But I think stalks are convenient when we want to show that this is an isomorphism when modding out the ideals.

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Hi Martin, thanks, I forgot about this question. If I remember correctly, I came to the same conclusion as you, but I constructed the global map differently: Let $p_1,p_2$ denote the projections $X\times_S X\rightarrow X$, then $p_i\Delta=id$, for both $i$. So if we pull back the morphisms of sheaves $p_i^{-1}\mathcal{O}_X\rightarrow \mathcal{O}_{X\times X}$ via $\Delta$, we obtain two $f^{-1}(\mathcal{O}_S)$-linear morphisms $\mathcal{O}_X\rightarrow\Delta^{-1}\mathcal{O}_{X\times X}$. This should give the same morphism as your construction. – Lars Jul 26 '10 at 8:06
Oops, apparently you edited and I commented simultaneously. – Lars Jul 26 '10 at 8:09
@Martin Brandenburg Ok, I suggest to translate the paper. I'd be interested. – Leo Alonso Dec 12 '12 at 11:10
I had already completed it, but then realized that W. D. Gillam has already done something similar: – Martin Brandenburg Dec 12 '12 at 17:57

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