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.

  • $\begingroup$ Have you looked at any non-smooth example already? $\endgroup$ – Martin Brandenburg Feb 18 at 0:08
  • $\begingroup$ @MartinBrandenburg To be honest, I have not. I was expecting this to work even in the non-smooth case. Do you think that it fails to be flat for non-smooth complete intersections in general? $\endgroup$ – clarkkent Feb 18 at 0:19
  • $\begingroup$ This is really a statement in commutative algebra: a local complete intersection ring is Cohen-Macaulay, but if the ideal of torsion elements were nonzero then there would be an embedded prime (which cannot exist for a Cohen-Macaulay ring). $\endgroup$ – ulrich Feb 19 at 13:19
  • $\begingroup$ @ulrich Thank you for the reply. I am sorry but I am confused. 1. Is it absolutely necessary that the coordinate ring of $X'$, call it $A$, will be a Cohen-Macaulay (CM) ring? 2. Assume that $A$ is CM. That means that it will be a CM module over itself. Does that mean that it will be a CM module over the $p$-adic integers? We require flatness over $\mathbb{Z}_p$. 3. I don’t see why it is necessary for $A$ to be finite as a module over $\mathbb{Z}_p$. Is finiteness not required to conclude that a CM module has no embedded primes? See [link] (stacks.math.columbia.edu/tag/0BUS). $\endgroup$ – clarkkent Feb 19 at 23:22

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.


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 $i > 0$. $\square$


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.


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