Let $R$ be the ring of functions $f:\Bbb{N}\rightarrow \Bbb{Z}/2\Bbb{Z}$. Then $X^2-X$ has infinitely many roots. 
Let $R$ be the ring of functions $f:\Bbb{N}\rightarrow\Bbb{Z}/2\Bbb{Z}$. Prove that the polynomial $X^2-X\in R[X]$ has infinitely many roots.

Am I supposed to show that there are infinitely many $n\in \Bbb{N}$ such that $n^2-n\equiv 0\pmod 2$? Or am I misinterpreting this question?
 A: I assume the operations in $R$ are defined pointwise, based on the standard operations in the field $\Bbb Z/ 2 \Bbb Z$, that is, for two maps $f, g: \Bbb N \to \Bbb Z / 2 \Bbb Z$ we take $(f + g)(n) = f(n) + g(n)$ and $(fg)(n) = f(n)g(n)$ for all $n \in \Bbb N$.  Then for any such $f:\Bbb N \to \Bbb Z / 2 \Bbb Z$, we must have
$(f(n))^2 = f(n), \tag{1}$
since for each $n \in \Bbb N$ we have $f(n) = 0$ or $f(n) = 1$.  This shows that every $f \in R$ is a zero of the polynomial $X^2 - X$.
$X^2 - X \in R[X]$ has a lot of roots.  Indeed, in appears that such functions are in one-to-one correspondence with $2^{\Bbb N}$, the power set of $\Bbb N$.  For any subset $\Bbb M \subset \Bbb N$, we may define $f_{\Bbb M}$ to be the charactersitic function of $\Bbb M$:
$f_{\Bbb M}(n) = 1 \Leftrightarrow n \in \Bbb M; \tag{2}$
likewise, given $f \in R$, taking $\Bbb M_f$ to be the set
$\Bbb M_f = \{ n \in \Bbb N \mid f(n) = 1 \} \tag{3}$
we see that
$f_{\Bbb M_f} = f \tag{4}$
and
$\Bbb M_{f_{\Bbb M}} = M. \tag{5}$
It is as if the value of $f(n)$ can be taken as a "yes" or "no" to the question, "Is $n \in \Bbb M_f \subset \Bbb N$?"
Contrast this to the case in which $R$ is a field.  Then $X^2 - X$ has exactly two zeroes.
