Subring isomorphic to a quotient ring Let $I$ range over sequences $(i_1,\dots,i_n,\dots)$ of
natural numbers with only finitely many nonzero terms. Call this set of sequences $\mathbb{N}^{(\mathbb{N})}$. Consider formal sums
\begin{equation}
\displaystyle \sum_{I \in \mathbb{N}^{(\mathbb{N})}}c_Ix^{I},
\end{equation}
with the $c_I \in \mathbb{Z}$, and where $x^I$ is a formal element we think of as representing the (finite) monomial
\begin{equation}
\displaystyle \prod_{n=1}^{\infty} x_{n}^{i_n}.
\end{equation}
With no restriction on $c_I$ and equipped with polynomial-like addition and multiplication, the collection of all such formal sums forms a ring
\begin{equation} 
R=\mathbb{Z}[[x_i|1\leq i \leq \infty]].
\end{equation}It is easy to see that the ring $S=\mathbb{Z}[x_i|1 \leq i \leq \infty]$ of polynomials in infinitely many variables (the only difference in the construction of $S$ from that of $R$ is that of imposing restrictions on the coefficients $c_I$) is a subring of $R$. My question is, can we find a subset $E$ of $R$ s.t. $R/(E) \cong S$.
 A: There's a substantial obstruction which is that $R$ has many units and $S$ has very few units. More formally, the only units in $U$ are the constants $\pm 1$ (e.g. because the degree of a product is the sum of the degrees) whereas in $T$ every series with constant term $\pm 1$ is invertible, basically because $(1 - x)^{-1} = 1 + x + x^2 + \dots$.
This means that if $f : R \to S$ is a homomorphism then $f$ sends every element of $T$ which can be expressed as a sum of units to the constant subring $\mathbb{Z}$ of $U$. In fact every element of $R$ is a sum of units: if $f \in R$ has constant term $f(0)$ then write it as a sum of $f(0) - 1$ copies of $1$ and $f - f(0) + 1$ which has constant term $1$.
The conclusion is that every homomorphism $f : R \to S$ lands in $\mathbb{Z}$ and so in particular cannot be surjective. (I don't think I've ever seen this "subring generated by units" construction before, it's pretty cute. I was about to ask a new question about this and was helpfully pointed to this math.SE question about it.)
