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Suppose $R=\mathbb{C}[x_1,\ldots, x_n]$ is a polynomial ring with $I$ being an ideal of $R$. Let $I'$ be an ideal of $R[t]$.

If $R[t]/I'$ is flat as a $\mathbb{C}[t]$-module and over $0$, $\dfrac{R[t]}{I'}\otimes_{\mathbb{C}[t]}\dfrac{\mathbb{C}[t]}{(t)}$ is reduced, then over any other point, is $$ \dfrac{R[t]}{I'}\otimes_{\mathbb{C}[t]}\dfrac{\mathbb{C}[t]}{\langle t-c\rangle} \cong \dfrac{R}{I} $$ for $c\not=0$ reduced with $$ dim \left( \dfrac{R[t]}{I'}\otimes_{\mathbb{C}[t]}\dfrac{\mathbb{C}[t]}{(t)}\right) = dim\left(\dfrac{R[t]}{I'}\otimes_{\mathbb{C}[t]} \dfrac{\mathbb{C}[t]}{\langle t-c\rangle} \right)? $$

$$ $$ Here is one example (which may possibly satisfy the above two conditions) that I would like to calculate as a hands-on concrete exercise.

$I' = \langle x_1-x_2+x_3, x_1 y_1 + z_1 y_2, x_3 y_3-z_2 y_2, x_2 y_2 +(1-t)z_2^2\rangle $ in $R[t]=\mathbb{C}[x_1, x_2, x_3, y_1, y_2, y_3,z_1, z_2][t]$.

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Your description of fiber over $0$ is correct, but no for the other fibers. They should be tensor product with $\mathbb C[t]/(t-c)$ for $c\in \mathbb C$. Why the ideal $I$ disappears in your question ? – user18119 Jul 7 '12 at 11:14
Thank you QiL. So $$\dfrac{R[t]}{I'}\otimes_{\mathbb{C}[t]}\mathbb{C}[t,t^{-1}] \not\cong \dfrac{R}{I} \cong \dfrac{R[t]}{I'}\otimes_{\mathbb{C}[t]} \dfrac{\mathbb{C}[t]}{(t-c)}$$ for $c\not= 0$? – math-visitor Jul 7 '12 at 17:47
I meant $I' = \langle x_1-x_2+x_3, x_1 y_1 + z_1 y_2, x_3 y_3-z_2 y_2, x_2 y_2 +(1-t)z_2^2\rangle $ in $R[t]=\mathbb{C}[x_1, x_2, x_3, y_1, y_2, y_3,z_1, z_2][t]$ with $I =\langle x_1-x_2+x_3, x_1 y_1 + z_1 y_2, x_3 y_3-z_2 y_2, x_2 y_2 +z_2^2\rangle$ in $R=\mathbb{C}[x_1, x_2, x_3, y_1, y_2, y_3,z_1, z_2]$. Maybe for this particular example, the special fiber is a complete intersection as well as the fiber over $t=c$ for $c\not=0$ (this is a speculation, not really a fact yet...). For the general case, I haven't yet thought carefully about how to relate the two ideals $I$ and $I'$. – math-visitor Jul 7 '12 at 21:06
up vote 2 down vote accepted

I still don't understand your notation. Neverthless the answer to the question in the title is no.

Example: Let $R=\mathbb C[x_1,x_2]$, let $I'$ be the ideal of $R[t]$ generated by $x_1^2+t-1$. Then $R[t]/I'=\mathbb C[x_1, x_2]$ is flat over $\mathbb C[t]=\mathbb C[x_1^2]$ (it has no torsion). The fiber over $0$ is two copies of the affine line and is reduced, but the fiber over $1$ is given by $\mathbb C[x_1,x_2]/(x_1^2)$ thus non-reduced.

The general principle is the following: let $f : X\to S$ be a morphism of finite type over a noetherian scheme (e.g. $X, S$ are algebraic varieties over a field), then the set of $s\in S$ such that the fiber $X_s$ is geometrically reduced is constructible in $S$, i.e. finite union of locally closed subsets of $S$, see EGA, IV.9.9.5. This holds for many other local properties of algebraic varieties.

Under some conditions (e.g. $f$ is moreover flat and proper), the above set is open in $S$ (EGA, IV.12.2.4(v)). But one can't guarantee that it is equal to $S$.

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Thank you for the counter-example (certainly helps!) and additional references QiL! – math-visitor Jul 7 '12 at 21:09

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