Is the set of hyperreal numbers a quotient ring?

It is easy to see that the set of real sequences $\mathbb{R}^{\mathbb{N}}$ is a ring. It suffices to define, for all $r,s\in\mathbb{R}^{\mathbb{N}}$, the operations $r\oplus s =(r_n+s_n)_{n\in\mathbb{N}}$ and $r\odot s=(r_n\cdot s_n)_{n\in\mathbb{N}}$.

Let $\mathcal{U}$ be a nonprincipal ultrafilter on $\mathbb{N}$. For all $r\in\mathbb{R}^{\mathbb{N}}$, we define the set $r^{(0)}=\{n\in\mathbb{N} \mid r_n=0\}$.

My question: Is the set of the $\textit{almost null sequences}$ $$\mathbb{I} = \{r\in\mathbb{R}^{\mathbb{N}}\mid r^{(0)}\!\in\mathcal{U}\}$$ a two-sided ideal of $\mathbb{R}^{\mathbb{N}}$? I think yes, because if $s\in\mathbb{I}$ and $r\in\mathbb{R}^{\mathbb{N}}$, then $(s\odot r)^{(0)}\in\mathcal{U}$ (i.e. the product of any sequence and an almost null sequence is almost null).

If yes, is the set of the hyperreal numbers the quotient ring $\mathbb{R}^{\mathbb{N}}\diagup \mathbb{I}$? In this case two sequences should belong to the same class if their difference is almost null (namely, they match on an index set which belongs to the ultrafilter $\mathcal{U}$)

• $\mathbb{R}^\mathbb{N}$ is commutative, so there's no need to say "two-sided". – Eric Wofsey Jul 8 '16 at 20:38

Yes, $\mathbb{I}$ is an ideal. One needs to check closure under addition, but this is straightforward: The intersection of two sets in the ultrafilter is in the ultrafilter.
Note that the hyperreals are much more than a field. In particular, the ultrapower construction gives, for each function $f:\mathbb{R}^n\to\mathbb{R}$, and for every relation $A\subset \mathbb{R^n}$, a natural extension to the ultrapower that preserves first-order properties.