Is the "sheaf of derivations" locally free? 
*

*Let $k$ be a field. We require all algebras to be associative commutative, and when unital we require morphisms between them to respect the identity element.


*Let $X$ be a topological space, equipped with a sheaf $\mathcal{O}$ of unital $k$-algebras on $X$. (In particular this makes $X$ a ringed space.)


*Following this MSE answer, we define the "sheaf of $k$-linear derivations" $F$ on $X$
as the sub(pre)sheaf of $\mathcal{Hom}(\mathcal{O},\mathcal{O})$ [where this is the internal hom for sheaves of $k$-vector spaces], given by
$$
F(U) = \bigl\{ D \in \mathcal{Hom}_{\text{Vect}_k^{\mathrm{Open}(X)^{\mathrm{op}}}}(\mathcal{O}|_U,\mathcal{O}|_U) \,\big|\,
\forall \text{ open } V\subseteq U
 ,\, D_V \in \mathrm{Hom}_{\text{Vect}_k}(\mathcal{O}(V),\mathcal{O}(V))
\text{ is a derivation}
\bigr\}
$$
where "derivation" is w.r.t. the algebra structure on $\mathcal{O}(V)$.


*We know $F$ is a sheaf of $k$-Lie algebras, and simultaneously an $\mathcal{O}$-module sheaf (though not compatibly with the Lie algebra structure).

In the case where $X$ is a smooth manifold, and $\mathcal{O}$ is its sheaf of smooth real functions, then the resulting $F$ is canonically the sheaf of smooth vector fields on $X$.
My question is: in the general case, is $F$ at least locally free as an $\mathcal{O}$-module? (Perhaps not finite locally free, in case e.g. $X$ is some sort of infinite-dimensional space?) (And, perhaps we need some extra condition, such as $\mathcal{O}$ having local stalks?) I'm not sure how to approach investigating this.
 A: We know that $k$ is a field. So in $k$, either $2$ is a unit or $3$ is a unit.
Let $p$ be a prime number which, in $k$, is a unit. Then consider the ring $O = k[\epsilon]$, where $\epsilon^p = 0$. Formally, we consider the ring $k[x]/(x^p)$. Note that $O$ is a local ring. $O$ is also a free vector space of dimension $p$, with a basis $\epsilon^0, \epsilon^1, \ldots, \epsilon^{p - 1}$.
Now let us consider what derivations $D$ exist on $O$. Note that since $\epsilon$ generates $O = k[\epsilon]$ as a $k$-algebra, we see that $D$ is uniquely determined by $D\epsilon$.
Let us note that $0 = D0 = D(\epsilon^p) = p \epsilon^{p - 1} D\epsilon$. Therefore, $\epsilon^{p - 1} D\epsilon = 0$. Therefore, $\epsilon \mid D\epsilon$.
Conversely, we see that there is a derivation $D$ sending $\epsilon$ to $\epsilon$. This requires the following Lemma:

Suppose $R$ is a $k$-algebra which, as a $k$-vector space, is spanned by $B$. Then a linear map $g : R \to R$ is a derivation iff for all $a, b \in B$, $g(ab) = g(a) b + a g(b)$.

Proof: suppose $g(ab) = g(a) b + a g(b)$ for all $a, b \in B$. Then $\{x \in R \mid g(ax) = g(a) x + a g(x)\}$ is a subspace of $R$ containing $B$, hence is all of $R$. Then $\{y \in R \mid \forall x \in R (g(yx) = g(y)x + x g(y))\}$ is a subspace of $R$ containing $B$, hence is all of $R$. So $g$ is a derivation. The other direction is immediate. $\square$
So in particular, note that $\{\epsilon^n\}_{n = 0, 1, \ldots, p - 1}$ is a basis of $k[\epsilon]$. Consider the unique linear map $g : k[\epsilon] \to k[\epsilon]$ sending $\epsilon^n$ to $n \epsilon^n$ for $n = 0, 1, \ldots, p - 1$. Then we see that $g$ is a derivation using the above Lemma.
So the derivations on $O$ are, as a $O$-module, isomorphic to the ideal $(\epsilon)$. It is straightforward to show this is not a free module, since the only free module satisfying $\epsilon^{p - 1} x = 0$ for all $x$ is the zero module, which $(\epsilon)$ clearly is not.
Clearly, the above demonstration can be generalised to sheaves on any space. Given a field $k$, we can construct the sheaf $\mathcal{O}$ of locally constant functions on $k[\epsilon]$. This will be a local ring object in the category of sheaves. Its sheaf of derivations will be, as a sheaf of $\mathcal{O}$-modules, isomorphic to the sheaf of locally constant functions on $(\epsilon)$, which is locally free only if the space is empty.
