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This is a simple question so I am hoping the answer is quite simple as well.

Suppose $\phi:X\rightarrow Spec\; \mathbb{C}[t]$ is a map such that the algebraic variety $\phi^{-1}(0)$ is a complete intersection. Does that mean $\phi$ is flat?

Note that I am not imposing any conditions on $X$.

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Dear math-visitor, it took me somr time to understand what you meant: you should replace "so that the algebraic variety..." by "such that the algebraic variety..." –  Georges Elencwajg Jul 3 '12 at 7:23
    
Thank you Georges! I was trying to formulate the question and at the same time, type this question. I will change it. –  math-visitor Jul 3 '12 at 7:25
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up vote 3 down vote accepted

No.
If you take for $X$ the subvariety $X\subset \mathbb A^2(\mathbb C)$ defined by $xy=0$, the first projection $\phi: X\to \mathbb A^1_\mathbb C=Spec(\mathbb C[t])$ is not flat because the dimension of the fibers $\phi^{-1}(t)$ jumps at $t=0$, nevertheless $\phi^{-1}(0)\cong \mathbb A^1(\mathbb C)$ is a complete intersection.

Edit
It is easy to see purely algebraically that $\phi$ is not flat.
Indeed, the dual map $\phi^*:\mathbb C[t]\to \mathbb C[x,y]/(xy):t\mapsto \bar x$ is not flat because over a PID (like $\mathbb C[t]$) the flat modules are exactly the torsion-free modules and here the element $\bar y\in \mathbb C[x,y]$ is a torsion element: $t\cdot \bar y=\bar x \bar y=0$

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Thank you Georges! Your answers are always helpful! –  math-visitor Jul 3 '12 at 7:38
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@math-visitor: it may be worth to note a kind of converse: if $X$ is smooth and $\phi$ is flat, then $\phi^{-1}(0)$ is locally complete intersection. –  user18119 Jul 3 '12 at 9:02
    
Of course, and thank you QiL! =) –  math-visitor Jul 3 '12 at 9:03
    
Dear @QiL, thank you for that comment. However I don't find it evident: is it in some algebraic geometry book? –  Georges Elencwajg Jul 3 '12 at 9:25
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@GeorgesElencwajg: I'm not sure if this is what you are looking for but here's it is: emis.de/journals/UIAM/actamath/PDF/35-243-246.pdf –  math-visitor Jul 4 '12 at 4:57
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