Existence of ring homomorphism from $\mathbb{Z}[[x]] \to \mathbb{Z} $

Question

Let $\mathbb{Z}[[x]]$ denote the ring of formal power series with coefficients in $\mathbb{Z}$.

Can there exist a ring homomorphism $$\Phi : \mathbb{Z}[[x]] \to \mathbb{Z}$$ such that $\Phi $ sends $x\to 2$? i.e. $\Phi(x)=2$.

Please provide me with a proof that such ring homomorphism cannot/can exist.

Answer 1

No such $\Phi$ can exist: $1 + x$ is a unit of $\mathbb{Z}[[x]]$, so $\Phi(1 + x)$ would have to be a unit of $\mathbb{Z}$. On the other hand, if $\Phi(x) = 2$, we would necessarily have $\Phi(1 + x) = \Phi(1) + \Phi(x) = 1 + 2 = 3$, giving a contradiction.

Answer 2

If $f = \sum_{n\geq 0} 2^nx^{n}$ then $f = 1 + 2xf$. So $\Phi(f) = 1 + 2 \Phi(x) \Phi(f)$ or $\Phi(f) = 1 + 4\Phi(f)$, so $\Phi(f) = -1/3$, impossible.