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What can I say about it?

Can I say the stalks equal the tensor products of the corresponding factors stalks?

Thanks!

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You don't say what your context is (ringed spaces, schemes,...). If it is schemes then no, as the stalk of the product is a local ring and the tensor product of local may not be local. –  Torsten Ekedahl Jul 16 '11 at 14:26
    
Thank you very much! –  MZWang Jul 17 '11 at 1:13
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2 Answers 2

up vote 4 down vote accepted

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 $\mathfrak{p} \subseteq \mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{Y,y}$ which restricts to the maximal ideals in $\mathcal{O}_{X,x}$ resp. $\mathcal{O}_{Y,y}$. The stalk of the structure sheaf in this point is the localization of the tensor product:

$\mathcal{O}_{X \times_S Y,(x,y,\mathfrak{p})} = (\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{Y,y})_{\mathfrak{p}}$.

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 construct the fiber product as above, also more general in the category of locally ringed spaces. I've written this up here.

Now your actual question seems to be:

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?

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:

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 $\mathcal{F}_{x}$ is flat over $\mathcal{O}_{Y,y}$. By commutative algebra (base change of flat modules), it follows that $\mathcal{F}_x \otimes_{\mathcal{O}_{Y,y}} \mathcal{O}_{Y',y'}$ is flat over $\mathcal{O}_{Y',y'}$. Again by commutative algebra (localizations are flat) $(\mathcal{F}_x \otimes_{\mathcal{O}_{Y,y}} \mathcal{O}_{Y',y'})_{\mathfrak{p}}$ is flat over $\mathcal{O}_{Y',y'}$. But this is exactly the stalk of the pullback of $\mathcal{F}$ in the given point $(x,y',\mathfrak{p})$.

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Thanks! This answer is exactly what I want. –  MZWang Jul 20 '11 at 8:42
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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?

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You should have added this to your question above (via the Edit function). Also check out the FAQ. But never mind :) –  Martin Brandenburg Jul 18 '11 at 9:11
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