# 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.

• This is not an answer to your question, but personally I think using $[\![$ (Code: [\![) would look better then $[[$. – Sangchul Lee May 6 '13 at 15:22
• If $\mathbb{Q}$ can imbed into $R$, then at least, there exists a ring-map sending $X$ to zero. – wxu May 6 '13 at 15:42
• Note that all series with nonzero constant must end up in $R^\times$. – Hagen von Eitzen May 6 '13 at 15:52
• My 2c: $\hat{S}=\Bbb Q[\![X]\!]$ is the $(X)$-adic completion of $S=\Bbb Q[X]$. As such, in general it is easy to map into it, but not necessarily out of it (it is an inverse limit). Moreover, any map of rings $S\to R$ induces a map on the completions $\hat{S}\to\hat{R}$, where $\hat{R}$ is an $I$-adic completion of $R$ for some (proper) ideal $I$ in $R$. – A.P. May 6 '13 at 16:16
• This is scary. No doubt you noticed that knowing $f(X)$ only determines the morphism $f$ on the subring $\mathbb{Q}(X)\cap\mathbb{Q}[[X]]$. I foresee axiom of choice and other nastiness coming up. – Jyrki Lahtonen May 6 '13 at 17:23

$\bullet$ 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.