Is it true that, in the category of $\mathbb{Z}$-modules, $Hom_{\mathbb{Z}}(\mathbb{Z}[x],\mathbb{Z})\cong\mathbb{Z}[[x]]$ and $Hom_{\mathbb{Z}}(\mathbb{Z}[[x]],\mathbb{Z})\cong\mathbb{Z}[x]$?

The first isomorphism is easy since any such homomorphism assigns an integer to $x^i, \forall\ {i>0}$ which defines a power series. For the second one might think similarly that if $S$ is the set of all power series with non-zero constant term then $\mathbb{Z}[[x]]=0\oplus{S}\oplus{x}S\oplus{x^2}S\dots$, but it doesn't quite work since it is not clear how to map $S$.

Is it true that, in the category of $\mathbb{Z}$-modules, $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}[x],\mathbb{Z})\cong\mathbb{Z}[[x]]$ and $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}[[x]],\mathbb{Z})\cong\mathbb{Z}[x]$?

The first isomorphism is easy since any such homomorphism assigns an integer to $x^i, \forall\ {i>0}$ which defines a power series. For the second one might think similarly that if $S$ is the set of all power series with non-zero constant term then $\mathbb{Z}[[x]]=0\oplus{S}\oplus{x}S\oplus{x^2}S\dots$, but it doesn't quite work since it is not clear how to map $S$.

Is it true that $Hom_{\mathbb{Z}}(\mathbb{Z}[x],\mathbb{Z})\cong\mathbb{Z}[[x]]$ and $Hom_{\mathbb{Z}}(\mathbb{Z}[[x]],\mathbb{Z})\cong\mathbb{Z}[x]$  ?

The first isomorphism is easy since any such homomorphism assigns an integer to $x^i, \forall\ {i>0}$ which defines a power series. For the second one might think similarly that if $S$ is the set of all power series with non-zero constant term then $\mathbb{Z}[[x]]=0\oplus{S}\oplus{x}S\oplus{x^2}S\dots$, but it doesn't quite work since it is not clear how to map $S$.

Is it true that, in the category of $\mathbb{Z}$-modules, $Hom_{\mathbb{Z}}(\mathbb{Z}[x],\mathbb{Z})\cong\mathbb{Z}[[x]]$ and $Hom_{\mathbb{Z}}(\mathbb{Z}[[x]],\mathbb{Z})\cong\mathbb{Z}[x]$?

The first isomorphism is easy since any such homomorphism assigns an integer to $x^i, \forall\ {i>0}$ which defines a power series. For the second one might think similarly that if $S$ is the set of all power series with non-zero constant term then $\mathbb{Z}[[x]]=0\oplus{S}\oplus{x}S\oplus{x^2}S\dots$, but it doesn't quite work since it is not clear how to map $S$.

Is it true that $Hom_{\mathbb{Z}}(\mathbb{Z}[x],\mathbb{Z})\cong\mathbb{Z}[[x]]$ and $Hom_{\mathbb{Z}}(\mathbb{Z}[[x]],\mathbb{Z})\cong\mathbb{Z}[x]$ ?

The first isomorphism is easy since any such homomorphism assigns an integer to $x^i, \forall\ {i>0}$ which defines a power series. For the converse one might think similarly that if $S$ is the set of all power series with non-zero constant term then $\mathbb{Z}[[x]]=0\oplus{S}\oplus{x}S\oplus{x^2}S\dots$, but it doesn't quite work since it is not clear how to map $S$.

Is it true that $Hom_{\mathbb{Z}}(\mathbb{Z}[x],\mathbb{Z})\cong\mathbb{Z}[[x]]$ and $Hom_{\mathbb{Z}}(\mathbb{Z}[[x]],\mathbb{Z})\cong\mathbb{Z}[x]$ ?

The first isomorphism is easy since any such homomorphism assigns an integer to $x^i, \forall\ {i>0}$ which defines a power series. For the second one might think similarly that if $S$ is the set of all power series with non-zero constant term then $\mathbb{Z}[[x]]=0\oplus{S}\oplus{x}S\oplus{x^2}S\dots$, but it doesn't quite work since it is not clear how to map $S$.