# Concerning the dimension of a complex variety modulo a prime

Let V be a complex affine variety given as the vanishing set of a set of polynomials with integral coefficients. I have 3 questions.

1)

Under what assumption will the dimension of V over C remain the same as the dimension of V computed over some finite field like Z_P, assuming that the prime P does not divide any of the coefficients of the polynomials defining V?

2)

How can such a prime P be found if we have a set of polynomials whose vanishing set is V?

3)

Suppose that V has a singularity over Z_P. Can we conclude that V is singular variety? In other words. Does having a singularity over a finite field immediately imply the variety is not smooth? If not, what guarantees that having a singularity modulo p implies a singularity over C?

• Which definition of dimension are you using? – Qiaochu Yuan Feb 18 '10 at 22:36
• V can easily be singular mod p without being singular. For example consider y^2=x^2+p with p prime. On the other hand if V is non-singular then it will be non-singular mod p for all but finitely many primes p, and the dimension of V mod p will be the same as the dimension of V for all but finitely many primes p. – Kevin Buzzard Feb 19 '10 at 0:57

To make more precise the answer of Felip. You have a scheme $X=Spec(A)$ over $\mathbb Z$, where $A$ is a finitely generate $\mathbb Z$-algebra such that its generic fiber $X_{\mathbb Q}$ (just consider your polynomials as polynomials with rational coefficients) gives $V$ by field extension $\mathbb{C}/\mathbb{Q}$. Of course, $X_{\mathbb Q}$ has the same dimension as $V$, and $X_{\mathbb Q}$ is smooth if and only $V$ is smooth.

Questions 1-2. You want to compare $\dim X_p$ with $\dim X_{\mathbb Q}$. In general $\dim X_p\ge \dim X_{\mathbb Q}$. The equality holds under some flatness condition at $p$. Namely, there is rather general result: if $f: Y\to Z$ is a flat morphism of finite type between noetherian schemes and suppose that $Z$ is integral and universally catenary (e.g.any scheme of finite type over a noetherian regular scheme, so any open subset of $Spec(\mathbb Z)$ is OK), then all [EDIT: non-empty] fibers $Y_z$ of $f$ have the same dimension [EDIT: if the generic fiber of $f$ is equidimensional (i.e. all irreducibles components have the same dimension)].

Problem: your $X$ is not necessarily flat over $\mathbb Z$. But the flatness over $\mathbb Z$ (or any Dedekind domain) is easy to detect: it is equivalent to be torsion free. So consider the ideal $I$ equal to $${ a\in A \mid ka=0 \text{ for some non zero } k \in \mathbb Z \}$$ (don't know how to type "{" and "}"). Then $A/I$ is flat over $\mathbb Z$ and defines a closed scheme of $X$ which coincides with $X$ over an open subset of $\mathbb Z$. Actually, as $I$ is finitely generated, there exists a positive integer $N$ such that $NI=0$. Then $I=0$ over $Spec(\mathbb Z[1/N])$. So for any prime number $p$ prime to $N$, $X_p$ will have dimension $\dim V$. Now you have to compute such a $N$... For hypersurface, $N$ is just the gcd of the coefficients (see comments in Felip's answer). I don't know whether efficient methodes exist in general. Of course if a prime $p$ does not divid any polynomial in the definning ideal of $X$, then this $p$ is OK. But you have to test this property for all polynomials and not just a set of generators (Example: the ideal generated by $T_1+pT_2, T_1$, then $p$ is bad).

Question 3. Suppose $V$ is smooth (and connected for simplicity), and you are looking for the $p$ such that $X_p$ is also smooth. You first proceed as in Question 1 to find an open subset $Spec(\mathbb Z[1/N]$ over which $X$ is flat. Let $d=\dim X_{\mathbb Q}=\dim X_p$ for all $p$ prime to $N$. Write $$X=Spec\big(\mathbb Z[T_1, \ldots T_n]/(F_1,\ldots, F_m)\big)$$ Then $X_p$ is smooth is and only if the Jacobian matrix of the $F_i$'s mod $p$ has rank $n-d$ at all points of $X_p$. Equivalently, the ideal $J\subseteq \mathbb Z[T_1,\ldots, T_n]$ generated by $F_1,...,F_m$ and the rank $n-d$ minors of the Jacobian matrix is the unit ideal mod $p$. Therefore, the computation consists in determining the ideal $J$. As $X_{\mathbb Q}$ is supposed to be smooth, $J$ is generated by a positive integer $M$. Now for all $p$ prime to $MN$, $X_p$ is smooth of dimension $\dim V$.

[EDIT]: The assertion on the dimensions of the fibers needs some hypothesis on the generic fiber of $Y\to Z$. We must assume it is equidimensional. Otherwise $Y_z$ has dimension equal to that of the generic fiber for $z\in Z$ belonging to the image of all irreducible components of $X$. For the initial question, it is better to assum the original variety $V$ is irreducible.

There are some complicated polynomial expressions on the coefficients of the polynomials (obtained e.g. through resultants) such that if P does not divide any of these expressions then the dimension of V mod P is the same as the dimension of V. There are other such expressions such that if V is smooth and P does not divide them, then V mod P is also smooth. This answers all your questions, except for writing down the expressions which is not easy to do in general.

• For example, if V is a hypersurface defined by a single polynomial f, the relevant complicated polynomial expressions are 1) the discriminant of f, and 2) the partial derivatives of f (I should say the ideal generated by these). – Qiaochu Yuan Feb 18 '10 at 23:36
• @Qiaochu. Not quite. A hypersurface will always have codimension 1, so a hypersurface will change dimension mod P if and only if P divides all coefficients of the polynomial. The discriminant (which is the resultant of the polynomial and its partial derivatives) will detect if the hypersurface is smooth or not. – Felipe Voloch Feb 19 '10 at 1:33
• Of course; my mistake! – Qiaochu Yuan Feb 19 '10 at 3:10