Stalks of structure sheaf of fibre product？ - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T05:57:04Z http://mathoverflow.net/feeds/question/70503 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70503/stalks-of-structure-sheaf-of-fibre-product Stalks of structure sheaf of fibre product？ MZWang 2011-07-16T13:54:44Z 2011-07-18T07:35:24Z <p>What can I say about it？</p> <p>Can I say the stalks equal the tensor products of the corresponding factors stalks？</p> <p>Thanks！</p> http://mathoverflow.net/questions/70503/stalks-of-structure-sheaf-of-fibre-product/70536#70536 Answer by MZWang for Stalks of structure sheaf of fibre product？ MZWang 2011-07-17T02:43:11Z 2011-07-17T02:43:11Z <p>As Hartshorne chapter III.9.2 claim，an Ox-module （need not be quasi coherent) F's flatness is stable under base change. But the stalks is not the tensor products, how can I prove the claim?</p> http://mathoverflow.net/questions/70503/stalks-of-structure-sheaf-of-fibre-product/70598#70598 Answer by Martin Brandenburg for Stalks of structure sheaf of fibre product？ Martin Brandenburg 2011-07-18T07:19:58Z 2011-07-18T07:35:24Z <p>Let $X,Y$ be $S$-schemes. Then a point of $X \times_S Y$ corresponds to a pair of points $x \in X, y \in Y$ lying over the same $s \in S$ together with a prime ideal <code>$\mathfrak{p} \subseteq \mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{Y,y}$</code> which restricts to the maximal ideals in <code>$\mathcal{O}_{X,x}$</code> resp. $\mathcal{O}_{Y,y}$. The stalk of the structure sheaf in this point is the localization of the tensor product:</p> <p><code>$\mathcal{O}_{X \times_S Y,(x,y,\mathfrak{p})} = (\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{Y,y})_{\mathfrak{p}}$</code>.</p> <p>There are at least two ways to prove these statements: a) Use the universal property of $\text{Spec}(K)$ for a field $K$ to get the points and then use the universal property of $\text{Spec}(R)$ for a local ring $R$ to get their stalks. So this assumes, of course, that you already know that the fiber product exists, but you can recover the description of the elements and the stalks just by using the universal property! But actually, b) you can <em>construct</em> the fiber product as above, also more general in the category of locally ringed spaces. I've written this up <a href="http://maddin.110mb.com/pdf/faserprodukte.pdf" rel="nofollow">here</a>.</p> <p>Now your actual question seems to be:</p> <blockquote> <p>As Hartshorne chapter III.9.2 claim，an Ox-module （need not be quasi coherent) F's flatness is stable under base change. But the stalks is not the tensor products, how can I prove the claim?</p> </blockquote> <p>The statement is the following: If $f : X \to Y, Y' \to Y$ are morphisms, and $\mathcal{F}$ is a module over $X$ which is flat over $f$, then the pullback of $\mathcal{F}$ to $X \times_Y Y'$ is flat over $X \times_Y Y' \to Y'$. I am pretty sure that Hartshorne understands $\mathcal{F}$ to be quasi-coherent here. Otherwise the sketch of proof also does not make sense. But it is also true in general:</p> <p>Pick a point in $X \times_Y Y'$, thus a triple $(x,y',\mathfrak{p})$ as described above. Let $y$ be the underlying point in $Y$. Now <code>$\mathcal{F}_{x}$</code> is flat over <code>$\mathcal{O}_{Y,y}$</code>. By commutative algebra (base change of flat modules), it follows that <code>$\mathcal{F}_x \otimes_{\mathcal{O}_{Y,y}} \mathcal{O}_{Y',y'}$</code> is flat over <code>$\mathcal{O}_{Y',y'}$</code>. Again by commutative algebra (localizations are flat) <code>$(\mathcal{F}_x \otimes_{\mathcal{O}_{Y,y}} \mathcal{O}_{Y',y'})_{\mathfrak{p}}$</code> is flat over <code>$\mathcal{O}_{Y',y'}$</code>. But this is exactly the stalk of the pullback of $\mathcal{F}$ in the given point $(x,y',\mathfrak{p})$.</p>