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Suppose $\psi: X\rightarrow \mathbb{A}_{\mathbb{C}}^1$ is a flat morphism, where $X\subseteq \mathbb{A}_{\mathbb{C}}^n$ with $X$ not needing to be smooth, with $\psi^{-1}(0)$ being a complete intersection.

Question 1: Wouldn't that imply that $\psi^{-1}(c)$ is also a complete intersection for any $c\not=0$?

Now consider a subscheme $Y\subseteq \mathbb{A}_{\mathbb{C}}^n$, where $Y$ is a complete intersection.

Question 2: If $Y\cap \psi^{-1}(0)$ is a complete intersection, then does that imply $Y\cap\psi^{-1}(c)$ is also a complete intersection, where $c\not=0$?

$$ $$ Here is the reason why I am thinking along the above lines. Consider $$ \psi: X = \operatorname{Spec}\left( \dfrac{\mathbb{C}[x,y,z,w,a,b,c,d][t]}{ \left<ab+cw+(1-t)d^2 \right> } \right) \longrightarrow \operatorname{Spec}\mathbb{C}[t]. $$ Then not only is $\psi^{-1}(0)$ a complete intersection, $\psi^{-1}(1)$ is also a complete intersection.

Now take $$ Y= \operatorname{Spec}\left( \dfrac{\mathbb{C}[x,y,z,w,a,b,c,d][t]}{ \left< xy+zw, x+a+c \right> } \right). $$ Then isn't it true that $$ Y\cap\psi^{-1}(0) = \operatorname{Spec} \left( \dfrac{\mathbb{C}[x,y,z,w,a,b,c,d][t]}{ \left< xy+zw, x+a+c, ab+cw+d^2 \right> } \right) $$ while $$ Y\cap\psi^{-1}(1) = \operatorname{Spec} \left( \dfrac{\mathbb{C}[x,y,z,w,a,b,c,d][t]}{ \left< xy+zw, x+a+c, ab+cw \right> } \right)? $$ Please correct me if there is a typo anywhere in the example, or if some thought process isn't entirely correct.

$$ $$

Any thoughts, counter-examples, or references would be great since I have limited deformation theory notes on hand.

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Please define "complete intersection". – user18119 Jul 4 '12 at 20:58
What I mean by complete intersection is that if the fiber has dimension $r$, then the ideal defining the fiber can be generated by the the dimension of the ambient space (which is the dimension of $X$) minus $r$ minus the dimension of the base base (which is $1$). I will re-read what I wrote to make sure it is clear. – math-visitor Jul 4 '12 at 21:26
You should say complete intersection in which space. – user18119 Jul 4 '12 at 21:36
I don't think the definition I gave is correct. Here's a simple example to explain what I mean in the above construction: consider $\mathbb{1}:\mathbb{A}_{\mathbb{C}}^1\rightarrow \mathbb{A}_{\mathbb{C}}^1$ where $\mathbb{1}$ is the identity map. Then the fiber over $\mathbb{1}^{-1}(0)$ is a complete intersection of dimension $0$. In fact, over any other point in the base space, it is still a complete intersection. – math-visitor Jul 4 '12 at 21:37
I will repost my notion of complete intersection and in which space soon. What I wanted to do in this problem is come up with a flat map for this question… using a slightly different strategy so that we obtain $\phi^{-1}((0,0,0))$ and $\phi^{-1}(t_1, t_2, t_3)$ (for some fixed $(t_1, t_2, t_3)$) as in that problem. – math-visitor Jul 4 '12 at 21:43

You should make assumptions about the relative dimension of the new intersection over the target.

Anyway, you should also remember the following. (See Hartshorne, Algebraic Geometry) Whenever we have a morphism of schemes, say, of finite type over a field

$$f:X\to C$$ with $C$ a smooth curve and $X$ equidimensional, the morphism is flat if and only if no irreducible component of $X$ is contained within a fibre.

Milne, Etale Cohomology, in Chapter I, has a section on flat morphisms where this result is stated and, if I'm not mistaken, proven (or a reference is given). Another beautiful reference is Mumford's The Red Book of Varieties and Schemes.

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