Timeline for Flatness and intersection of fibers
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Mar 31, 2015 at 17:57 | comment | added | Allen Knutson | 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.) | |
| Mar 31, 2015 at 16:40 | comment | added | Karl Schwede | 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. | |
| Mar 31, 2015 at 16:06 | history | undeleted | Ron | ||
| Mar 31, 2015 at 16:06 | history | deleted | Ron | via Vote | |
| Mar 31, 2015 at 14:57 | history | edited | Ron |
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| Mar 31, 2015 at 14:06 | history | edited | Ron | CC BY-SA 3.0 |
deleted 6 characters in body; edited tags; edited title
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| Mar 31, 2015 at 13:56 | comment | added | Ron | @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? | |
| Mar 31, 2015 at 13:22 | comment | added | Matthieu Romagny | 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 $\mathbb{A}^2_{\mathbb{C}}$. | |
| Mar 31, 2015 at 13:01 | history | asked | Ron | CC BY-SA 3.0 |