Questions about evaluation map from $\mathbb Z[[X]]$ to $\mathbb Q$ and $\mathbb Z[[X]]$ to $\mathbb C$. Let $ℤ[[]]$ denote the ring of formal power series with coefficients in $ℤ$. 
Question 1: If $\alpha \in ℚ$ is unit such that $\alpha \notin ℤ$, then can there exist a ring homomorphism $\phi :ℤ[[]]\to ℚ$ such that $\phi $ sends $x\in ℤ[[]]$ to $\alpha \in ℚ$? If the answer is yes, then for what kinds of $\alpha $ with the above property does there exist such ring homomorphism?
Question 2: Can you answer the same question with replacing $ℚ$ by $ℂ$?
 A: The only homomorphism $\mathbb{Z}[[X]]\to\mathbb{Q}$ is the one sending a power series to its constant term (so, $X$ goes to $0$).  First, note that this is the only homomorphism sending $X$ to $0$, since $\mathbb{Z}[[X]]/(X)\cong\mathbb{Z}$ and there is only one homomorphism $\mathbb{Z}\to\mathbb{Q}$.  Now suppose $\varphi:\mathbb{Z}[[X]]\to\mathbb{Q}$ has $\varphi(X)=\alpha$ for some $\alpha\neq 0$.  Note that by Hensel's lemma, $1+4nX$ has a square root in $\mathbb{Z}[[X]]$ for any $n\in\mathbb{Z}$.  Applying $\varphi$, we find that $1+4n\alpha$ must have a square root in $\mathbb{Q}$ for any $n$.  But since $\alpha\neq 0$, we can pick $n$ such that $1+4n\alpha<0$ and reach a contradiction.
On the other hand, for $\alpha\in\mathbb{C}$ there exists a homomorphism $\varphi:\mathbb{Z}[[X]]\to\mathbb{C}$ such that $\varphi(X)=\alpha$ iff $1/\alpha$ is not an algebraic integer (or $\alpha=0$).
To prove this, let $f\in\mathbb{Z}[X]$ be the minimal polynomial of $\alpha$ over $\mathbb{Z}$ (or $0$ if $\alpha$ is transcendental).  If such a $\varphi$ exists, then it induces a homomorphism $\mathbb{Z}[[X]]/(f)\otimes\mathbb{Q}\to\mathbb{C}$, and in particular $\mathbb{Z}[[X]]/(f)\otimes\mathbb{Q}$ is a nonzero ring.  Conversely, if $\mathbb{Z}[[X]]/(f)\otimes\mathbb{Q}$ is a nonzero ring, we can pick a prime ideal $P$ in it (we pick $P=0$ if $f=0$) and consider the fraction field $K$ of the quotient $(\mathbb{Z}[[X]]/(f)\otimes \mathbb{Q})/P$.  Note that $\mathbb{Q}[X]/(f)$ is already a field unless $f=0$, so the subfield $k\subseteq K$ generated by $X$ is isomorphic to the fraction field of $\mathbb{Q}[X]/(f)$ (with $X$ mapping to $X$).  So by definition of $f$, there is a homomorphism from $k\to\mathbb{C}$ which maps $X$ to $\alpha$, and then this homomorphism can be extended to $K$ since $\mathbb{C}$ is algebraically closed and has transcendence degree over $k$ at least that of $K$ since $|K|\leq 2^{\aleph_0}=|\mathbb{C}|$.
Thus such a homomorphism $\varphi$ exists iff the ring $\mathbb{Z}[[X]]/(f)\otimes\mathbb{Q}$ is nonzero, or equivalently iff $\mathbb{Z}[[X]]/(f)$ is a ring of characteristic $0$.  Note moreover that $1/\alpha$ is an algebraic integer iff $f(0)=\pm 1$.  Indeed, observe that $\alpha$ is a root of a polynomial $g$ iff $1/\alpha$ is a root of the polynomial $X^{\deg g}g(1/X)$, so the minimal polynomial of $1/\alpha$ over $\mathbb{Z}$ is $X^{\deg f}f(1/X)$.  In particular, its leading coefficient is the constant term of $f$, so $1/\alpha$ is an algebraic integer iff that constant term is $\pm 1$.
So finally, it suffices to show that $\mathbb{Z}[[X]]/(f)$ is a ring of characteristic $0$ iff $f(0)\neq \pm 1$.  In the case where $\alpha$ is transcendental and $f=0$, this is trivial.  If $f(0)=\pm 1$ then $f$ is a unit in $\mathbb{Z}[[X]]$ so $\mathbb{Z}[[X]]/(f)$ is the zero ring and does not have characteristic $0$.  Now suppose $f(0)\neq \pm 1$ but $\alpha$ is algebraic.    Note that $f$ is the minimal polynomial over $\mathbb{Z}$ of an algebraic number so it has trivial content.  Combined with $f(0)\neq \pm 1$, this means there is a prime $p\in\mathbb{Z}$ which divides the constant term of $f$ but which does not divide some other coefficient of $f$.   Now suppose $\mathbb{Z}[[X]]/(f)$ has nonzero characteristic and let $n$ be its characteristic.  That means $n$ is divisible by $f$ and is the least positive integer which is divisible by $f$.  Say $n=fg$ for some $g\in\mathbb{Z}[[X]]$.  Then in particular $n=f(0)g(0)$ so $n$ is divisible by $p$.  But $p$ does not divide $f$, so since $p$ is prime in $\mathbb{Z}[[X]]$ (since the quotient $\mathbb{Z}[[X]]/(p)\cong\mathbb{F}_p[[X]]$ is a domain) it must divide $g$.  But then we can write $n/p=f(g/p)$ so $n/p$ is divisible by $f$, contradicting the minimality of $n$.  Thus $\mathbb{Z}[[X]]/(f)$ must have characteristic $0$, as desired.
