# the existence of a closed subset of $\mathbb{A}^8\times \mathbb{A}^1$ which is flat over $\mathbb{A}^1$

I've been doing some reading on deformation theory and one way it is used is to study singularities on varieties while perturbing the varieties. I would actually like to use deformation theory to construct a flat family linking the following two subschemes of $\mathbb{A}_{\mathbb{C}}^8$.

I would like to thank David Speyer who helped me to formulate this question from my other post.

Does there exist a closed subset of $\mathbb{A}_{\mathbb{C}}^8\times \mathbb{A}_{\mathbb{C}}^1$, which is flat over $\mathbb{A}_{\mathbb{C}}^1$, whose fiber over $0$ is the set $$Z(x_1 y_1 + z_1 y_2, x_2 y_2 +z_3^2, x_3 y_3-z_3 y_2, x_1-x_2+x_3)$$ and whose fiber over $1$ is the set $$Z(x_1 y_1 + z_1 y_2,x_2 y_2, x_3 y_3-z_3 y_2, x_1-x_2+x_3)?$$

Note that the only difference between the two sets is that the second polynomial is a monomial.

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Dear math-visitor, I didn't think very hard about your particular equations, but is the obvious candidate $Z(x_1y_1 + z_1y_2, x_2 y_2 + (1-t)z_3^2, x_3y_3- z_3 y_2, x_1- x_2 + x_3)$ not flat over the $t$-line for some reason? Regards, – Matt E Jul 5 '12 at 4:01
Yes, that was what I was thinking of originally, but are there relatively simple/easy ways to check whether or not the map you wrote is flat? – math-visitor Jul 5 '12 at 4:19
Dear math-visitor, Since over a Dedekind domain flatness is the same as torsion free, you just have to check that the total space of the family doesn't have a component supported over some closed point of $\mathbb A^1$. This could happen if your equations were sufficiently complicated, and conspired in some way to produce such a component for a particular value of $t$. Probably they don't, and even if they did, one could likely modify my suggestion slightly so that the modified family is flat. Regards, – Matt E Jul 5 '12 at 4:44
Thank you Matt! I will try that. – math-visitor Jul 5 '12 at 4:53
@MattE This may be a naive question but suppose I am able to show that the obvious candidate is a flat morphism over the $t$-line. Then if the variety over $t=0$ is a complete intersection in $\mathbb{A}_{\mathbb{C}}^8$, then would the variety over $t=1$ be a complete intersection in $\mathbb{A}_{\mathbb{C}}^8$ as well? We could check this with a computer software for the above example, but there are more complicated examples where using a computer may be impossible or difficult. – math-visitor Jul 5 '12 at 7:00