So I've been given these two rings:
$R = Map(\mathbb{R, R})$ which I need to prove is NOT an integral domain and
$R/I$ where $R$ is defined as above and $I = \{f \in R | f(1) = 0\}$ which I need to prove IS an integral domain.
So what I've determined so far is that in $R$ the zero element is $f(x)=0$ since its always $0$ so anything added to it will always be unchanged. I was thinking that I could use a piecewise function like
$g(x) = \left\{\begin{array}{rcl} 0 \mbox{ if } x<0\\ 0\text{ if }x >0 \end{array}\right.$
That leaves $g(x) = 0$ undefined but its zero in every other way so despite not being the zero element so any other map $h$ multiplied with it will almost be $0$ at any point. I'm just not sure because $g(x) = 0$ is undefined so $(gh)(x) = 0$ will be undefined, not $0$. I'm not sure if that's enough to show $gh = 0$ where $g\neq{0}$.
Moreover, I'm not sure why the quotient ring $R/I$ is an integral domain even though $R$ isn't.
Help would be much appreciated.
Hints:
For part 1, try to design two non-zero functions $f \neq 0$ and $g \neq 0$ such that $fg = 0$. E.g., what can you do if $f(x) = 0$ when $x = 0$ and $f(x) = 1$ otherwise.
For part 2, what can you say about the mapping $R \to \Bbb{R}$ that maps $f$ to $f(1)$? Is it well-defined, one-to-one, onto, a homomorphism? If it is a homomorphism, what is its kernel?