Lifting a complete intersection in $\mathbb{P}^n_{\mathbb{F}_p}$ to $\mathbb{Z}_p$

Question

Suppose that you are given a (not necessarily smooth) projective variety $X \subseteq \mathbb{P}^n_{\mathbb{F}_p}$ of codimension $d$ that is a complete intersection. In other words, it can be defined by exactly $d$ homogeneous polynomials and no set of polynomials of cardinality less than $d$ has $X$ as its zero set. Let $f_1$, ..., $f_d$ be a set of polynomials defining $X$. If I lift these polynomials arbitrarily to $\mathbb{Z}_p$ (call the lifts $F_1$, ..., $F_d$), I will get a projective algebraic set $X' \subseteq \mathbb{P}^n_{\mathbb{Z}_p}$.

My question is this: will this $X'$ be flat over $\mathbb{Z}_p$? Equivalently, is the module $\mathbb{Z}_p[x_0,...x_n]/(F_1, ..., F_d)$ flat over $\mathbb{Z}_p$? I know that $\mathbb{Z}_p$ is a DVR, so flatness is equivalent to being torsion-free but I just cannot see how to prove that it is either. If $X$ is smooth, then $X'$ is definitely flat: $X'$ will also be smooth by the Jacobian criterion and hence flat.

Answer 1

You can get a quick proof by using Hartshorne, Theorem III.9.9, which says that it's sufficient to show that the two fibres $X$ and $X' \times_{\mathbb{Z}_p} \mathbb{Q}_p$ have the same Hilbert polynomial. But this is true, since they're both complete intersections defined by polynomials of the same degree.

Answer 2

This is true, and here is one possible proof. There might be easier ones; however I suspect that it will always involve some algebra (not just geometry).

Write $S = \operatorname{Spec} \mathbf Z_p$ and $\mathbf P = \mathbf P^n_{\mathbf Z_p}$ with structure map $\pi \colon \mathbf P \to S$, and let $X \subseteq \mathbf P$ be a complete intersection with generic fibre $X_\eta \subseteq \mathbf P_\eta$ and special fibre $X_s \subseteq \mathbf P_s$. Assume $X$ is cut out by sections $f_1,\ldots,f_r$ of $\mathcal O_{\mathbf P}(d_1), \ldots, \mathcal O_{\mathbf P}(d_r)$ repesctively, such that the $f_i$ remain a regular sequence modulo $p$ (this is equivalent to the setting you're starting with).

Set $\mathscr E = \bigoplus_i \mathcal O_{\mathbf P}(-d_i)$, and consider the Koszul resolution $$0 \to \wedge^r \mathscr E \to \wedge^{r-1} \mathscr E \to \ldots \to \mathscr E \to \mathcal O_{\mathbf P} \to \mathcal O_X \to 0,$$ which is exact since $X$ is a complete intersection (Tag 062F). Since the $\wedge^i\mathscr E$ are locally free, the sheaves $\mathcal Tor_i^{\mathcal O_{\mathbf P}}(\mathcal O_X, \mathcal O_{\mathbf P}/p)$ are computed by the cohomology of $$0 \to \wedge^r \mathscr E \underset{\mathcal O_{\mathbf P}}\otimes \mathcal O_{\mathbf P}/p \to \wedge^{r-1} \mathscr E \underset{\mathcal O_{\mathbf P}}\otimes \mathcal O_{\mathbf P}/p \to \ldots \to \mathscr E \underset{\mathcal O_{\mathbf P}}\otimes \mathcal O_{\mathbf P}/p \to \mathcal O_{\mathbf P}/p \to 0.$$ This sequence is still exact since the $f_i$ modulo $p$ are still a regular sequence, so we conclude that $\mathcal Tor_i^{\mathcal O_{\mathbf P}}(\mathcal O_X, \mathcal O_{\mathbf P}/p) = 0$ for all $i > 0$. $\square$

Answer 3

This is an expanded version of my comment:

The main claim is that the question can be answered from basic facts in (local) commutative algebra. In particular, the same statement holds for a complete intersection in affine space, which I explain below.

We work over any dvr $R$ with residue field $k$. Let $X$ be of codimension $d$ in $\mathbb{A}^n_k$ defined by equations $f_1,f_2,\dots,f_d$. Let $F_1,F_2,\dots,F_d$ be elements of $R[x_1,x_2,\dots,x_n]$ such that $F_i$ is a lift of $f_i$ and let $X'$ be the subscheme of $\mathbb{A}^n_R$ defined by the ideal $(F_1,F_2,\dots,F_d)$. Then $X'$ is a local complete intersection scheme since the uniformizer of $R$ is a nonzero divisor in the polynomial ring and $X$ is a local complete intersection. In particular, $X'$ is Cohen-Macaulay, so it has no embedded points.

Now suppose the coordinate ring $A$ of $X'$ has non-zero $R$-torsion elements. The set of all such elements is a non-zero ideal in $A$ and the support of this ideal (as an $A$-module) is contained in $X'$ (viewed as a subset of $X$). This implies that $A$ has an embedded prime, so we get a contradiction.