I've never really made my way in any detail through the Witt-vector construction. I did read all the articles that a quick Google and MSN search turned up, and none *seemed* to address it, but I could just be unfamiliar with the language; so please pardon me if this is a question with a well known answer.


If it makes a difference whether I'm asking my question about the 'big' Witt vectors or the $p$-typical ones, then please interpret it in the way that makes it more interesting.


With that prelude, the question itself is simple: is there a functor $\text{$A$-Mod} \to \text{$W(A)$-Mod}$, either just for specific algebras $A$ (I have in mind the $p$-typical Witt vectors of a finite field of characteristic $p$), or, possibly, for all commutative algebras? In my motivating case, one can do something awkward like pick a basis for an $A$-module (i.e., a vector space) and simply construct the free $W(A)$-module on that set; but, aside from the ugliness of this approach, it's not even clear to me that it's functorial.


EDIT: As both [nfdc23](https://mathoverflow.net/a/228675/2383) and [QiaochuYuan](https://mathoverflow.net/a/228688/2383) point out, I have additional assumptions in mind; namely, in both cases, that free, finite-rank $A$-modules are taken to non-$0$, free, finite-rank $W(A)$-modules, for nfdc's post that the identity morphism is taken to the identity morphism, and for Qiaochu's post that not every morphism is sent to $0$.



Witt-vector vectors
It is impossible if you want reasonable functoriality, and for finite-dimensional vector spaces to be carried to finite free modules for $A$ a field.


Presumably for $A = k$ a finite field of characteristic $p$, you want the composition of such a functor with "reduction modulo $p$" to be (canonically isomorphic to) the identity functor. (If not, then ignore the rest of what follows.) Then functoriality is impossible to achieve. Indeed, assume it is possible, so we get a homomorphic section $s$ to the reduction map ${\rm{GL}}\_n(W(k)) \rightarrow {\rm{GL}}\_n(k)$ for every $n>0$. 


Let us see that there is no such section when $n \ge 2$ and $q := |k|>3$. (There must be a zillion ways to prove this; I just give the first argument that came to mind.) Let $T \subset {\rm{GL}}\_n$ be a split maximal $k$-torus, so $s(T(k))$ consists of pairwise commuting elements of finite order dividing $q-1$. Since $W(k)[x]/(x^{q-1}-1)$ as a $W(k)$-algebra is a direct product of $q-1$ copies of $W(k)$, it follows that each element of $s(T(k))$ can be diagonalized over $W(k)$ (not just over $W(k)[1/p]$!). Since elements of $s(T(k))$ pairwise commute, we can achieve this diagonalization over $W(k)$ simultaneously, which is to say that we can conjugate to make $s(T(k))$ diagonal. In other words, for some $g \in {\rm{GL}}\_n(W(k))$ (with reduction $g\_0 \in {\rm{GL}}\_n(k)$) the composition $c\_g \circ s \circ c\_{g\_0}^{-1}$ is a section that identifies $c\_{g\_0}(T(k))$ with the $(q-1)$-torsion in the diagonal over $W(k)$. This forces $c\_{g\_0}(T(k))$ to be the diagonal of ${\rm{GL}}\_n(k)$.


By replacing $s$ with $c\_g \circ s \circ c\_{g\_0}^{-1}$ we may assume $s$ carries the diagonal into the diagonal. Now using that $q>3$, so the roots of ${\rm{GL}}\_n$ are pairwise distinct on the group of $k$-points of the diagonal, it follows that $s$ must preserve "root groups"; i.e., for each $i \ne j$ with $1 \le i, j \le n$ (such $i, j$ exist since $n \ge 2$) and $U\_{ij}$ the corresponding copy of $\mathbf{G}\_a$ as a $W(k)$-subgroup of ${\rm{GL}}\_n$ via the $ij$-entry of matrices, we see that $s(U\_{ij}(k)) \subset U\_{ij}(W(k))$. In particular, we get a homomorphic section $s\_{ij}$ to the reduction map $W(k) \rightarrow k$. But that is absurd since $W(k)$ is torsion-free and $k$ is killed by $p$.


[This argument is inspired by the proof that for any semisimple Chevalley group $G$, $G(W\_2(k))$ viewed as a $k$-group in the natural way has no Levi $k$-subgroup for $k$ algebraically closed of characteristic $p>0$.]


