Coderivations of $S(V)$ correspond to linear maps $S(V) \to V.$ Only over characteristic $0$?

Question

Definition. Let $k$ be a commutative ring. Let $V$ be a $k$-module. We turn the symmetric algebra $\mathrm{S}\left(V\right)$ of $V$ into a graded Hopf algebra by defining the comultiplication \begin{align} \Delta : \mathrm{S}(V) \to \mathrm{S}(V) \otimes \mathrm{S}(V) \end{align} by \begin{align} & \Delta(v_1v_2\ldots v_n) \\ = {} & \sum\limits_{i=0}^n \ \ \sum\limits_{\sigma\in\operatorname{Sh}(i,n-i)} \left(v_{\sigma(1)}v_{\sigma(2)}\ldots v_{\sigma(i)}\right) \otimes (v_{\sigma(i+1)}v_{\sigma(i+2)}\ldots v_{\sigma(n)}), \end{align} where $\operatorname{Sh}(i,n-i)$ denotes the set $\left\lbrace \sigma\in S_n \mid \sigma(1) < \sigma(2) < \cdots < \sigma\left(i\right) \text{ and } \sigma(i+1) < \sigma(i+2) < \cdots < \sigma\left(n\right) \right\rbrace$ of all $(i,n-i)$-shuffles. The counit of this Hopf algebra is simply the projection from $S(V)$ onto $k$.

Definition. Let $k$ be a commutative ring, and $C$ be a $k$-coalgebra. A coderivation of $C$ means a $k$-linear map $c:C\to C$ such that $\Delta \circ c = \left(c\otimes\operatorname{id} + \operatorname{id}\otimes c\right)\circ \Delta$.

Remark. Coderivations behave, in some sense, dually to derivations (which is not surprising since the condition $\Delta \circ c = \left(c\otimes\operatorname{id} + \operatorname{id}\otimes c\right)\circ \Delta$ is a kind of dual to the Leibniz identity, when the latter is written in pointfree notation): First of all, if $c:C\to C$ is a coderivation, then $c^{\ast} : C^{\ast}\to C^{\ast}$ is a derivation. The converse holds at least when $C$ is finite-dimensional. As an exercise in reversing arrows, the reader can prove that $\varepsilon\circ c=0$ for every coderivation $c$ of a coalgebra (in analogy to the equality $d\left(1\right)=0$ which holds for every derivation $d$ of an algebra).

Theorem. Let $k$ be a field of characteristic $0$. Let $V$ be a $k$-vector space. Then, the maps \begin{align} \operatorname{Hom}\left(S\left(V\right),V\right) &\to \operatorname{Coder}\left(S\left(V\right)\right), \\ X &\mapsto \mu \circ \left(\operatorname{id}\otimes X\right) \circ \Delta \end{align} (where $\mu$ is the multiplication morphism $S\left(V\right)\otimes S\left(V\right)\to S\left(V\right)$) and \begin{align} \operatorname{Coder}\left(S\left(V\right)\right) &\to \operatorname{Hom}\left(S\left(V\right),V\right), \\ c &\mapsto \pi_1\circ c \end{align} (where $\pi_1$ is the projection from $\mathrm{S}\left(V\right)=\bigoplus\limits_{i\in\mathbb N}\mathrm{S}^i\left(V\right)$ onto the addend $\mathrm{S}^1\left(V\right)=V$) are mutually inverse isomorphisms.

This is how I understand Chapter 5, Theorem 4.20 in Eckhard Meinrenken, Clifford algebras and Lie groups (revision 12/2011). (Unfortunately, the statement of the theorem in Meinrenken's text is obscured by the fact that one direction of the isomorphism - the easy one - is not written out explicitly.) The proof given in this text uses the characteristic-$0$ assumption: first, by assuming "WLOG" that a generic element of $\mathrm{S}\left(V\right)$ has the form $e^v$ for some $v\in V$ (this is made formal by going over to formal power series, but stripped of this formality, this is exactly what is known as umbral calculus for over a century), and second, by using the fact that the primitives of $\mathrm{S}\left(V\right)$ all come from $V$.

Question. Does the above theorem hold in arbitrary characteristic?

I am sure this is intimately related to the question whether $\mathrm{S}\left(V\right)$ is the cofree graded cocommutative coalgebra over $V$ (or at least whether it is "cogenerated" in degree $1$, whatever this means precisely!). Unfortunately, the only case when I am sure of this is the characteristic-$0$ case, so this is of no help to me. Loday-Valette does not seem to care for positive characteristic too much, either, and it is difficult for me to find any other source.

Answer 1

Let us assume $k$ has characteristic $p$. The problem is (to me at least) easier to understand by dualising (assume that we are only looking at homgeneous derivations so that we can take the graded dual and have no problems at least if $V$ is finite-dimensional, things will go wrong even here). Then the statement would say that any linear map $V\to V\subset S(V^\ast)^\ast$ has a unique extension to a derivation of $S(V^\ast)^\ast$. However, $S(V^\ast)^\ast$ is the divided power algebra on $V$ which is not generated as an algebra by its degree $1$ elements so the uniqueness doesn't follow as for the symmetric algebra. In fact for $V=kx$ we have that the divided power algebra is generated as a commutative algebra by $x_n:=\gamma_{p^n}(x)$ and relations $x_n^p=0$. Hence one can arbitrarily choose the value of a derivation on the $x_n$ which makes it very clear that the derivation is not determined by its value on $x_1$.