Let $f:X \to Y$ be a flat, proper, surjective morphism between noetherian schemes. Assume $Y$ is irreducible and smooth over $\mathbb{C}$. Suppose that $X$ is the union of two schemes $X_1$ and $X_2$ both flat, proper and surjective over $Y$. Is there any known condition under which the scheme theoretic intersection, $X_1.X_2$ is flat over $Y$ (for example, if the generic fibers of the morphisms $X_1 \to Y$ and $X_2 \to Y$ do not intersect)?
4
-
$\begingroup$ Certainly not. Geometrically $Y_1=Spec(B_1)$ and $Y_2=Spec(B_2)$ are closed subvarieties in some affine space $\mathbb{A}^{n+1}_X$ with $X=Spec(A)$ and you are asking: is it true that $Y_1,Y_2$ flat over $X$ imply $Y_1\cap Y_2$ flat over $X$? A counterexample is e.g. $X=\mathbb{A}^1_{\mathbb{C}}$ and $Y_1,Y_2=$ the coordinate axes in $\mathbb{A}^2_{\mathbb{C}}$. $\endgroup$– Matthieu RomagnyCommented Mar 31, 2015 at 13:22
-
$\begingroup$ @Romagny: I wanted to add the condition that $Y_1 \cup Y_2$ is also flat over $X$. I will edit the question. Is there a counterexample in that case as well? $\endgroup$– RonCommented Mar 31, 2015 at 13:56
-
$\begingroup$ You always have a short exact sequence$$ 0 \to O_{X_1 \cup X_2} \to O_{X_1} \oplus O_{X_2} \to O_{X_1 \cap X_2} \to 0. $$ This should tell you exactly what the obstruction is. Indeed, it immediately follows that if $X_1, X_2$ and $X_1 \cap X_2$ are flat, so is $X_1 \cup X_2$, but as you can see from the long exact sequence of Tor, you don't get to deduce that $\text{Tor}^1_{f^{-1} O_Y}(O_{X_1 \cap X_2}, N)$ is zero. $\endgroup$– Karl SchwedeCommented Mar 31, 2015 at 16:40
-
$\begingroup$ Let $X_1 = \{(x,0)\}$, $X_2 = \{(x,x)\}$, $f = $projection to the first factor. Then $X_1,X_2,X_1\cup X_2$ are all flat over $Y$, and their generic fibers do not intersect. But the intersection is supported over $0\in Y$. (Maybe this was Matthieu's example.) $\endgroup$– Allen KnutsonCommented Mar 31, 2015 at 17:57
Add a comment
|