What are the ring morphisms $\mathbb{Q}[[X]]\to R$ for a ring $R$? If $\mathbb{Q}[[X]]$ is the ring of power series over a field $F$, then can we describe ring morphisms $\mathbb{Q}[[X]] \to R$ for rings $R$ in simple terms?
I am guessing that a "substitution principle" wouldn't make sense for series unless $R$ has a notion of convergence of series. 
 A: $\bullet$ If $R$ is not a $\mathbb{Q}$-vector space (i.e $R$ is not of caracteristic $0$ or does not contain the fraction field of $\mathbb{Z}.1_R$), then $Hom(\mathbb{Q}[[X]],R) = \emptyset$ : because a morphism $f : \mathbb{Q}[[X]] \rightarrow R$, will gives an injection $\mathbb{Q} \rightarrow R, \quad x \mapsto f(x)$.
$\bullet$ If $f : \mathbb{Q}[[X]] \rightarrow R$ is not injective, then its kernel is a non zero ideal, hence of the form $(X^n)$ for some $n$. So $f$ factor through $\mathbb{Q}[X]/X^n$, and the set $Hom_{Ring}(\mathbb{Q}[X]/X^n,R) = Hom_{\mathbb{Q}-Alg}(\mathbb{Q}[X]/X^n,R)$ is well known.
$\bullet$ If $f$ is injective, then it is quite nasty (at least if one assume axiom of choice). For example, take $\Omega$ an algebraic closure of $\mathbb{Q}((X))$. If is known there exists an element $f \in Aut(\Omega/\mathbb{Q}(X))$ such that $f(\exp(X)) \neq \exp(X)$. This yields a morphism
$$\mathbb{Q}[[X]] \rightarrow \Omega, \quad y \mapsto f(x)$$
which sends $X$ to $X$ but is different from the natural map.
