Sheaves of Principal parts - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T02:24:38Z http://mathoverflow.net/feeds/question/20940 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20940/sheaves-of-principal-parts Sheaves of Principal parts Lars 2010-04-10T17:24:08Z 2010-07-26T08:00:50Z <p>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</p> <p><code>$\mathcal{P}_{X/S}^n:=\Delta^{-1}( \mathcal{O}_{X\times_S X}) / \mathcal{I}^{n+1}$</code></p> <p>In their book on Crystalline cohomology, Berthelot and Ogus define the sheaf of principal parts $\mathcal{P}^n_{X/S}$ as </p> <p><code>$(\mathcal{O}_X\otimes_{f^{-1}\mathcal{O}_S}\mathcal{O}_X)/\mathbf{I}^{n+1},$</code> where $\mathbf{I}$ is the kernel of the multiplication map from the tensor product to $\mathcal{O}_X$.</p> <p>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 $n>0$? </p> <p>I've not seen the second definition anywhere else, although it seems somewhat nicer than the first one...</p> http://mathoverflow.net/questions/20940/sheaves-of-principal-parts/33286#33286 Answer by Martin Brandenburg for Sheaves of Principal parts Martin Brandenburg 2010-07-25T15:46:44Z 2010-07-26T08:00:50Z <p>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 <a href="http://maddin.110mb.com/pdf/faserprodukte.pdf" rel="nofollow">this article</a> (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.</p> <p>First we construct a homomorphism <code>$\mathcal{O}_X \otimes_{f^{-1} \mathcal{O}_S} \mathcal{O}_X \to \Delta^{-1} \mathcal{O}_{X \times_S X}$</code>. For that we compute the stalks at some point $x \in X$ lying over $s \in S$:</p> <p><code>$(\mathcal{O}_X \otimes_{f^{-1} \mathcal{O}_S} \mathcal{O}_X)_x = \mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x},$</code></p> <p><code>$(\Delta^{-1} \mathcal{O}_{X \times_S X})_x = \mathcal{O}_{X \times_S X,\Delta(x)} = (\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x})_{\mathfrak{q}}$</code>,</p> <p>where $\mathfrak{q}$ is the kernel of the canonical homomorphism</p> <p><code>$\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x} \to \kappa(x), a \otimes b \mapsto \overline{ab}.$</code></p> <p>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:</p> <p>(*) 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 $s$.</p> <p>This can be applied in the above situation: Every section in a neighborhood of $x$ in <code>$\mathcal{O}_X \otimes_{f^{-1} \mathcal{O}_S} \mathcal{O}_X$</code> induced by an element in <code>$\mathcal{O}_X(U) \otimes_{\mathcal{O}_S(V)} \mathcal{O}_X(U)$</code> 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.</p> <p>Thus we have a homomorphism <code>$\alpha : \mathcal{O}_X \otimes_{f^{-1} \mathcal{O}_S} \mathcal{O}_X \to \Delta^{-1} \mathcal{O}_{X \times_S X}$</code>. Now let $J$ be the kernel of the multiplication map <code>$\mathcal{O}_X \otimes_{f^{-1} \mathcal{O}_S} \mathcal{O}_X \to \mathcal{O}_X$</code> and $I$ be the kernel of the homomorphism <code>$\Delta^\# : \Delta^{-1} \mathcal{O}_{X \times_S X} \to \mathcal{O}_X$</code>. Then for every $n \geq 1$ our $\alpha$ restricts to a homomorphism</p> <p><code>$(\mathcal{O}_X \otimes_{f^{-1} \mathcal{O}_S} \mathcal{O}_X)/J^n \to (\Delta^{-1} \mathcal{O}_{X \times_S X})/I^n,$</code></p> <p>which is given at $x \in X$ by the natural map</p> <p><code>$(\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x}) / \mathfrak{p}^n \to ((\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x}) / \mathfrak{p}^n)_{\mathfrak{q}}$</code>,</p> <p>where $\mathfrak{p} \subseteq \mathfrak{q}$ is the kernel of the multiplication map <code>$\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x} \to \mathcal{O}_{X,x}$</code>.</p> <p>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 <code>$\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x}$</code>, 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</p> <p><code>$(\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x}) / \mathfrak{p}^n \to (\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{X,x}) / \mathfrak{p}^1 \cong \mathcal{O}_{X,x}$</code>.</p> <p>However, this follows from the observation that the kernel $\mathfrak{p}^1 / \mathfrak{p}^n$ is nilpotent; cf. also <a href="http://mathoverflow.net/questions/31495/when-does-a-ring-surjection-imply-a-surjection-of-the-group-of-units/32919#32919" rel="nofollow">this question</a>.</p> <p>I'm sure that there is also a proof which avoids stalks at all.</p> <p>EDIT: So here is a direct construction of the homomorphism <code>$\mathcal{O}_X \otimes_{f^{-1} \mathcal{O}_S} \mathcal{O}_X \to \Delta^{-1} \mathcal{O}_{X \times_S X}$</code>:</p> <p>Let $p_1,p_2$ be the projections $X \times_S X \to X$. Then we have for $i=1,2$ the homomorphism</p> <p><code>$\mathcal{O}_X \to {p_i}_* \mathcal{O}_{X \times_S X} \to {p_i}_* \Delta_* \Delta^{-1} \mathcal{O}_{X \times_S X} = (p_i \Delta)_* \Delta^{-1} \mathcal{O}_{X \times_S X} = \Delta^{-1} \mathcal{O}_{X \times_S X}$</code>,</p> <p>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.</p>