explicitly constructing a certain flat family

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 $\mathbb{A}^3$? –  David Speyer Jul 3 '12 at 23:32
Thank you for the correction David! I see it now why that needed to be changed. =) –  math-visitor Jul 4 '12 at 0:08
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.) –  Michael Hardy Jul 4 '12 at 4:22
Thank you Michael! I will try to remember to use \operatorname{Spec} in the future. –  math-visitor Jul 4 '12 at 4:26
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 $(t_1, t_2, t_3)$^? –  user18119 Jul 4 '12 at 21:01

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... –  math-visitor Jul 4 '12 at 22:26
@math-visitor: the pre-images of two distinct points are disjointed, this is purely set-theoretical. –  user18119 Jul 5 '12 at 22:21
Thank you QiL! I guessed that this must be one of the properties of a flat map. =) –  math-visitor Jul 5 '12 at 23:32
@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 $f^{-1}(y_1)\cap f^{-1}(y_2)=\emptyset$. –  user18119 Jul 6 '12 at 10:13
Of course, thank you QiL! –  math-visitor Jul 6 '12 at 19:01