# Formal power series and polynomial ring

Let $R$ be a ring then the set $R^R$, the functions from $R$ to itself, has a natural ring structure (pointwise addition and multiplication).

Some questions:

1. Is $R^R$ ring isomorphic to $R[[x]]$, the ring of formal power series?
2. In general, $R[x]$, the polynomial ring, is not isomorphic to $R[[x]]$. I've seen some examples of this, for example $\mathbb{Z}[x]$ is not ring isomorphic to $\mathbb{Z}[[x]]$ by comparing units. Is it possible to have a ring $R$ such that $R[x]$ and $R[[x]]$ are ring isomorphic? In case yes, what are some examples?
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Note that $\Big|R[[x]]\Big|=|R|^\omega$, while $\left|R^R\right|=|R|^{|R|}$. Thus, if $|R|=2^\omega=\mathfrak c$ (e.g., if $R$ is $\Bbb R$ or $\Bbb C$), then $\left|R^R\right|=2^{\mathfrak c}>2^\omega=\Big|R[[x]]\Big|$, and the two rings have different cardinalities and therefore certainly aren’t isomorphic. The fact that the two sets need not even have the same cardinality seems to me to suggest that isomorphism must be the exception.
$\Bbb Z[[x]]$ is not isomorphic to $\Bbb Z^{\Bbb Z}$. The invertible power series are precisely those with constant term $1$ or $-1$, and the invertible sequences are precisely those whose terms are all $1$ or $-1$. Thus, in $\Bbb Z^{\Bbb Z}$ the square of every invertible element is the multiplicative identity, something that clearly is not true in $\Bbb Z[[x]]$.
In $\Bbb Q[[x]]$ the map $f(x)\mapsto xf(x)$ maps the invertible elements bijectively to the non-invertible elements. Suppose that $\sigma=\langle a_q:q\in\Bbb Q\rangle\in\Bbb Q^{\Bbb Q}$ is such that the map $\tau\mapsto\sigma\tau$ maps the invertible elements of $\Bbb Q^{\Bbb Q}$ to the non-invertible elements. Then $\sigma$ cannot be invertible, so there must be a $q\in S$ such that $a_q=0$. Let $\tau_0,\tau_1\in\Bbb Q^{\Bbb Q}$ be invertible sequences differing only at the index $q$; then $\sigma\tau_0=\sigma\tau_1$, so the map $\tau\mapsto\sigma\tau$ cannot be a bijection. This shows that $\Bbb Q[[x]]$ is not isomorphic to $\Bbb Q^{\Bbb Q}$.
@user10: No, it has cardinality $2^\omega$, like the real line. –  Brian M. Scott Nov 14 '12 at 4:48