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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 and QiaochuYuan 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 $0$.

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Good question. Something perhaps related is Lemma 2.2 in arxiv.org/abs/1006.3125v3 . – darij grinberg Jan 18 at 0:23
up vote 13 down vote accepted

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 $p>0$.]

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This argument is a hair higher-powered than @QiaochuYuan's below, but has the virtue of showing that one can't achieve functoriality (subject to the reasonable conditions you mention) even if the only morphisms are isomorphisms. You are making the very slight additional assumption that finite-dimensional vector spaces are not sent to the $0$ module (but the stronger apparent assumption that dimension is preserved is not necessary.) The bracketed result you mention at the end of your post (where I guess $W_2$ is $W_p$?) also sounds interesting; do you know a reference? – L Spice Jan 18 at 19:29
    
Sorry; on further thought, I actually don't see why we must have $s(U_{ij}(k)) = U_{ij}(W(k))$. It's certainly not forced by the argument to that point, which would allow us to replace $s$ by $\operatorname{Int}(n) \circ s$ for any $n$ in the normaliser of the diagonal torus. Of course, in the case I've described, we can just conjugate further to arrange the desired equality, but I'm not sure that I see why that's always possible. – L Spice Jan 18 at 20:46
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@LSpice: I was assuming that your desired hypothetical functor composed with reduction mod p gives the identity functor, so with vector spaces of finite dimension assumed to go to finite free modules it follows that the rank is preserved (so not giving the 0 module if we plug in $k^n$). For the bracketed result, where $W_2$ refers to the functor of length-2 Witt vectors (e.g., $W(k)/(p^2)$ on $k$-points for a perfect field $k$ of characteristic $p$), see A.6 of the book "Pseudo-reductive Groups". – nfdc23 Jan 18 at 22:10
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@LSpice: Note that ${\rm{Int}}(n) \circ s$ isn't a section! We do have that $s(U_{ij}(k)) \subset U_{ij}(W(k))$, because $U_{ij}(W(k))$ is the set of points in ${\rm{GL}}_n(W(k))$ on which the $(q-1)$-torsion in the diagonal acts through conjugation via scaling on matrix entries through the root $a_{ij}:t \mapsto t_i/t_j$ (such roots being pairwise distinct characters on $(q-1)$-torsion since $q > 3$). The relevance of this uses crucially that $s$ is really a section. – nfdc23 Jan 18 at 22:19
    
Oops, you are quite right on all counts. I was forgetting both that $s$ was a section, and that $s(T(k))$ fills out the $(q - 1)$-torsion in $T(W(k))$ (rather than just being some subgroup therein). Thanks! – L Spice Jan 19 at 4:05

Let me restrict my attention to the only case I understand, which is $A = \mathbb{F}_p, W(A) = \mathbb{Z}_p$ (so here I mean $p$-typical Witt vectors). You don't provide any conditions you want your functor to satisfy, but I presume you don't just want e.g. the zero functor, so here is a very simple no-go for getting extra properties: no such functor can both

  • be additive, and
  • send finite free modules to finite free modules.

This is for the very simple reason that $\mathbb{F}_p$ (the endomorphism ring of a finite free $\mathbb{F}_p$-module of rank $1$) admits no ring homomorphism to $M_n(\mathbb{Z}_p)$ for any $n$, since the latter is torsion-free.

More generally this argument shows that the image of any additive functor from $\mathbb{F}_p$-modules to $\mathbb{Z}_p$-modules necessarily lands in modules whose underlying abelian groups are $p$-torsion. Of course there is a natural additive functor given by restriction of scalars along the quotient map $\mathbb{Z}_p \to \mathbb{F}_p$ with this property.

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You are right about the implicit assumptions that I had in mind. This is an elegantly simple argument, and I like it a lot; but I accepted @nfdc23's because it has the added advantage of ruling out functoriality even if the only morphisms are isomorphisms. – L Spice Jan 18 at 19:31

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