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Is it possible to construct a flat family $$ \phi:\mathbb{A}_{\mathbb{C}}^8=\operatorname{Spec} \mathbb{C}[x,y,z,w,a,b,c,d]\longrightarrow \operatorname{Spec} \mathbb{C}[t_1, t_2, t_3] =\mathbb{A}_{\mathbb{C}}^3 $$ so that $$ \phi^{-1}((0,0,0)) = Z(xy+zw,ab+cw+d^2,x+a+c) $$ while $$ \phi^{-1}((t_1, t_2, t_3)) = Z(xy+zw,ab+cw,x+a+c), $$ for some $(t_1, t_2, t_3)\not=(0,0,0)$?

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    Does $Z(\alpha, \beta, \gamma)$ mean the locus where $\alpha = \beta = \gamma =0$? If so, then no. The expressions you give are $5$-dimensional, while the fibers of a flat map from an $8$-dimensional space to a $1$-dimensional space are $7$-dimensional. Maybe you wanted the target to be $\mathbb{A}^3$?2012-07-03
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    Thank you for the correction David! I see it now why that needed to be changed. =)2012-07-04
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    I changed Spec\;\mathbb{C} to \operatorname{Spec}\mathbb{C} (with no "\;"). That's standard usage. \operatorname{} does not only prevent the letters in "Spec" from being italicized as if they were variables; it also causes proper spacing to appear before and after it. (The spacing part doesn't work on Wikipedia. But maybe it will if they finish their project to switch from what they're using to mathJax.)2012-07-04
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    Thank you Michael! I will try to remember to use \operatorname{Spec} in the future.2012-07-04
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    Your fiber above $(t_1, t_2, t_3)$ doesn't depend on $(t_1, t_2, t_3)$, this is strange and actually impossible. Do you mean the condition holds for some point $(t_1, t_2, t_3)$^?2012-07-04
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    Yes, thank you QiL. I meant for some point $(t_1,t_2, t_3)$.2012-07-04
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    This is just a hunch, but I suspect you are not asking the question you want to ask. I suspect you want to ask whether there is a closed subset of $\mathbb{A}^8 \times \mathbb{A}^1$, flat over $\mathbb{A}^1$, whose fiber over $0$ is your first set and whose fiber over $1$ is your second set. That would be a flat family linking the two subschemes of $\mathbb{A}^8$. (If that is the right question though, it is more polite to ask it as a new question, rather than un-accept QiL's correct answer to what you asked.)2012-07-05
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    Thank you David! Yes, you are absolutely correct about what I've been thinking for the past an hour or so. I thought about QiL's answer, which is correct for the above problem, and I am about to formulate and post what I really want to ask as another question.2012-07-05

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No. Suppose such a $\phi$ exists. Let $(x_0,y_0,z_0,w_0, a_0, b_0, c_0, d_0)$ be a point of $\phi^{-1}(t_1,t_2,t_3)$. Then $(x_0,y_0,z_0,w_0, a_0, b_0, c_0, 0)$ belongs to $\phi^{-1}(t_1,t_2,t_3)\cap \phi^{-1}(0,0,0)=\emptyset$!

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    Wow, thank you QiL! I'm still thinking about why the preimage of the two points must be empty...2012-07-04
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    @math-visitor: the pre-images of two distinct points are disjointed, this is purely set-theoretical.2012-07-05
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    Thank you QiL! I guessed that this must be one of the properties of a flat map. =)2012-07-05
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    @math-visitor, no, the flatness doesn't count here. If $f: X \to Y$ is a map between two sets and $y_1\ne y_2$ are points of $Y$, then $f^{-1}(y_1)\cap f^{-1}(y_2)=\emptyset$.2012-07-06
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    Of course, thank you QiL!2012-07-06