Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose that I am given a subscheme $Y$ of $\mathbf{P}^n_{\mathbf{Z}}$, flat over $\operatorname{Spec}\mathbf{Z}$ and with smooth generic fiber $Y_{\mathbf{Q}}$, defined by the vanishing of some homogeneous polynomials $$ F_1, \ldots, F_k \in \mathbf{Z}[X_0,\ldots,X_n]. $$ How does one determine the set $S$ consisting of primes $p$ such that the fiber $Y_{\mathbf{F}_p}$ is non-smooth?

The way I'm going about the problem seems a bit crude to me, and has some problems. In particular, I can't prove that my algorithm solves the problem.

For simplicity suppose that $k=1$, and write $F=F_1$. In a nutshell, I compute the derivatives $f_i := \frac{\partial F}{\partial X_i}$, and determine an integer $N$ such that $N$ is contained in the ideal $I = (F, f_1, \ldots, f_n) \subset \mathbf{Z}[X_0,\ldots,X_n]$ by repeatedly taking resultants of polynomials in $I$. Then at least for all $p$ not dividing $N$, the fiber $Y_{\mathbf{F}_p}$ is smooth, since for such $p$ the ideal $I_p \subset \mathbf{F}_p[X_0,\ldots,X_n]$ generated by the reduced polynomials $\widetilde{F}$, $\widetilde{f}_1$,$\ldots$,$\widetilde{f}_n$ contains $1$.

One problem is that it's quite expensive computationally. I'm using resultants to eliminate the variables $X_i$ one by one, say in the order $i=0,1,\ldots,n$. Then every time I've eliminated one of the $X_i$s, I get a lot of polynomials that don't contain $X_0$ up to $X_i$, and I take them all into the next elimination round. (You have to be careful when the resultant comes out $0$, and perhaps there are some other subtleties that I'm forgetting, but this is basically it.) Another problem with this is that the final outcome seems to depend on the order in which I take the $i$s. (This may have something to do with singularities being located at infinity with respect to one of the $X_i$s [edit: on second thought, this doesn't make any sense; see comment by François], but I really can't see the geometry of what I'm doing clearly enough to be confident about this.) And I don't see why my method necessarily gives me the whole $S$ - instead of some upper bound on it - even when all orderings of the $X_i$ are taken into account.

My questions:

  1. Does my method give the whole set $S$?
  2. If yes, why? If no, what method does work?
  3. Regardless of the answers to 1. and 2., is there a faster or more natural way of finding the primes of bad reduction?
share|improve this question
1  
Geometrically, taking the resultant of two polynomials corresponds to compute the projection of an intersection. This approach only gives an upper bound for S, because two projections might intersect while the original objects don't (but I don't have a counter-example where all orderings are taking into account). Anyway, a possibly better method would be to use Gröbner bases over the integers, see magma.maths.usyd.edu.au/magma/handbook/text/1112#12186 At least for hypersurfaces you will see the integer N you're interested in as the first member of your Gröbner basis. –  François Brunault Dec 13 '12 at 15:55
    
In the case of a subscheme Y of arbitrary pure dimension d, then I think you should consider the ideal generated by the F_i's together with all (n-d)-minors of the jacobian matrix. –  François Brunault Dec 13 '12 at 16:18
    
That makes a lot of sense, thanks! Your second remark agrees with what I thought myself, I just thought it made the notation simpler if I restricted to $k=1$. –  René Dec 13 '12 at 16:49

1 Answer 1

up vote 3 down vote accepted

Assume for simplicity that $Y$ has pure relative dimension $d$. By considering the standard affine cover of $\mathbf{P}^n_{\mathbf{Z}}$, one easily reduces to the case where $Y=V(F_1,\ldots,F_k)$ is a closed subscheme of $\mathbf{A}^n_{\mathbf{Z}}$. Then the special fiber $Y_p = V(F_{1,p},\ldots,F_{k,p}) \subset \mathbf{A}^n_{\mathbf{F}_p}$ is smooth if and only if for every $x \in Y_p(\overline{\mathbf{F}}_p)$, the Jacobian matrix $(\frac{\partial F_{i,p}}{\partial X_j}(x))$ has rank $n-d$. Note that $Y_p$ is $d$-dimensional by the flatness assumption, so all geometric tangent spaces have dimension $\geq d$, which implies that the rank of the Jacobian matrix is everywhere $\leq n-d$. Thus $Y_p$ is smooth if and only if the ideal generated by $F_{1,p},\ldots,F_{k,p}$ together with all $(n-d)$-minors of the Jacobian matrix is equal to $\mathbf{F}_p[X_1,\ldots,X_n]$.

Now consider the ideal $I$ of $R=\mathbf{Z}[X_1,\ldots,X_n]$ generated by $F_1,\ldots,F_k$ together with all $(n-d)$-minors of the Jacobian matrix. Since the generic fiber of $Y$ is smooth, we have $I \cap \mathbf{Z}=N\mathbf{Z}$ for some integer $N \geq 1$, and the prime factors of $N$ are precisely the primes of bad reduction of $Y$.

Proof. If $p$ doesn't divide $N$ then $(I,p)$ contains $N$ and $p$, so $(I,p)=R$ and $Y_p$ is smooth. If $p$ divides $N$, write $N=p^k M$ with $p$ not dividing $M$. If we had $(I,p)=R$, then we would also have $(I,p^k)=R$. Then $M \in (MI,p^k M)=(MI,N) \subset I$, a contradiction. Thus $Y_p$ is not smooth.

You can compute the integer $N$ using Gröbner bases for polynomials over $\mathbf{Z}$, see e.g. http://magma.maths.usyd.edu.au/magma/handbook/text/1112#12189 for the Magma implementation. I'm not expert in Gröbner bases, so I would appreciate if someone could confirm whether it is always possible to find $I \cap \mathbf{Z}$ when a set of generators of $I$ is given, and whether the Magma implementation works in all cases.

share|improve this answer
1  
Dear François: Thank you very much. I must say though I wonder: in terms of complexity, one might be worse off with the Gröbner basis calculation than if you would upper-bound $S$ using my method and then for each $p \in S$ would check if $X_{\mathbf{F}_p}$ is smooth. I can't prove this, of course, after all you do have to factor an integer $N$ over whose size you don't really seem to have much control –  René Dec 14 '12 at 3:11
    
Dear René: With the Gröbner basis computation, there are a lot of minors to compute, which is indeed an annoying thing. But I think we can combine both approaches: put some minors in $I$ until $I \cap \mathbf{Z} \neq 0$ (probably a generic choice of $d+1$ such minors will do). Once you get an integer $N$ which gives a reasonable upper bound for $S$ (and assuming you can factor $N$!), you can check whether $X_{\mathbf{F}_p}$ is smooth as you suggest. But for this the Gröbner basis computation is useful, because it will give you, I think, univariate polynomials in each variable (...) –  François Brunault Dec 14 '12 at 8:46
    
(...) which vanish on the potential singular points. Then you have a finite number of matrices over the algebraic closure of $\mathbf{F}_p$ whose rank you need to check, which is probably much easier from a computational point of view. –  François Brunault Dec 14 '12 at 8:48

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.