The following is an example that I made up in order to understand a certain concept in one of Eisenbud's books.
Consider $R = k[x_1,x_2,x_3,x_4]$ and let $I = \left< x_1 x_2+x_3 x_4 +x_2 + x_3, x_1 x_4+ x_2 x_3+ x_2 +x_3 \right>$ and $J = \left< x_1 x_2+x_3 x_4, x_1 x_4+ x_2 x_3\right>$ be ideals in $R$. The difference between $I$ and $J$ is that $J$ doesn't have any of the linear terms.
Consider $S = R[t]$ and write $I' = \left< x_1 x_2+x_3 x_4 + tx_2 + tx_3, x_1 x_4+ x_2 x_3+ tx_2 +tx_3 \right>$.
How do you know that $\pi: Spec(R/I')\rightarrow Spec(k[t])$ is flat?
If a morphism is flat and I know that the fiber $\pi^{-1}(1)$ is a complete intersection, then doesn't this mean $\pi^{-1}(0)$ is also a complete intersection?
Also, where can I find more information on invariants of flat families?
Thanks all.
Edit By the way, I checked that both $I$ and $J$ have codim 2 in $R$ but I would like to relate the two ideals/varieties via deformation theory.
Let us compute a Groebner basis of $I'$ using Macaulay2.
First, set up the ring and define the ideal, which will be named $I$ because the prime is a bit annoying:
Let M2 compute a Groebner basis for us (it will use the grevlex monomial ordering, by default)
and let it show us the generators of the initial monomial of I
This means that $\mathbb Q[x_1,x_2,x_3,x_4,t]/I'$ has as $\mathbb Q$-basis the classes of the monomials of $\mathbb Q[x_1,x_2,x_3,x_4,t]$ which are not divisible by $x_2x_3$, $x_1x_2$ nor $x_1^2x_4$. It is more or less clear now that that quotient is free as a $k[t]$-module: as a $k[t]$-module is has as a basis the monomials in $x_1$, $x_2$, $x_3$ and $x_4$ which are not divisible by any of those same three monomials.