TITLE: Lifting a complete intersection in $\mathbb{P}^n_{\mathbb{F}_p}$ to $\mathbb{Z}_p$
QUESTION [7 upvotes]: 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.

REPLY [3 votes]: 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.