Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
add comment

1 Answer

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.