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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.

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The first thing that pops into my head is that there are easier ways to check something is flat when the base is 1-dimensional and smooth. For example, it is equivalent to check that every associated point maps to the generic point of the base. – Matt May 23 '12 at 4:55
up vote 2 down vote accepted

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:

i1 : R := QQ[x_1 .. x_4, t];

i2 : I = ideal (x_1*x_2+x_3*x_4+t*x_2+t*x_3, x_1*x_4+x_2*x_3+t*x_2+t*x_3);

o2 : Ideal of QQ[x , x , x , x , t]
                  1   2   3   4

Let M2 compute a Groebner basis for us (it will use the grevlex monomial ordering, by default)

i3 : gens gb I

o3 = | x_2x_3+x_1x_4+x_2t+x_3t x_1x_2+x_3x_4+x_2t+x_3t x_1^2x_4-x_3^2x_4+x_1x_3t-x_3^2t+x_1x_4t-x_3x_4t

                                   1                             3
o3 : Matrix (QQ[x , x , x , x , t])  <--- (QQ[x , x , x , x , t])
                 1   2   3   4                 1   2   3   4

and let it show us the generators of the initial monomial of I

i4 : leadTerm I

o4 = | x_2x_3 x_1x_2 x_1^2x_4 |

                                   1                             3
o4 : Matrix (QQ[x , x , x , x , t])  <--- (QQ[x , x , x , x , t])
                 1   2   3   4                 1   2   3   4

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.

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Of course, this simple computation could have been done by hand with no trouble, but this is what we invented computers for! – Mariano Suárez-Alvarez May 23 '12 at 5:36
Thanks Mariano! So is the morphism $\pi$ flat since the generators of the initial ideal of your $I$ (my $I'$) do not involve $t$? I did not know that one may think of the initial ideal this way! – math-visitor May 23 '12 at 6:10
Well, you should prove that since the leading monomials do not involve $t$, the quotient is free as a $k[t]$-module. Once you do that, you have a new trick to prove freeness. Once you do that :) – Mariano Suárez-Alvarez May 23 '12 at 6:13
=) Thanks! I've been attempting to apply a Proposition from Eisenbud's book: $\pi$ is flat over $0$ if and only if the fiber $X_0=\pi^{-1}(0)$ is the limit of the fibers $X_b=\pi^{-1}(b)$ as $b\rightarrow 0$ if and only if no irreducible component or embedded component of $S/I'$ is supported on $X_0$... – math-visitor May 23 '12 at 6:14

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