# Describe the divisors of zero in a ring.

My Problem Is:

Describe the divisors of zero in $$\displaystyle \mathscr{F} (\mathbb{R})$$.

What I have so far:

I tried to adapt the formal definition of a divisors of zero in a ring, to my specific problem:

In the ring $$\displaystyle \mathscr{F} (\mathbb{R})$$, a nonzero function $$f_1(x)$$ is called a divisor of zero if there is a nonzero function $$f_2(x)$$ in the ring such that the product $$f_1(x)f_2(x)$$ or $$f_2(x)f_1(x)$$ is equal to zero.

An example of a pair of divisors of zero in the ring $$\displaystyle \mathscr{F} (\mathbb{R})$$ would be:

$$f_1(x)=x$$, $$f_2(x) = \begin{cases} 1, & x=0 \\ 0, & x \ne 0 \end{cases}$$

I cant figure out how to generalize further to describe all the divisors of zero however...

• since there are no non-zero divisors of $0$ in $\mathbb R$, $f_1(x)$ would have to be $0$ for some $x$ – J. W. Tanner Apr 10 at 22:18

If for some $$x\in\mathbb R$$ we were to have $$f(x)g(x)=0$$, then because $$\mathbb R$$ is a field, either $$f(x)=0$$ or $$g(x)=0$$. Thus, the union of the set of zeroes of $$f$$ and $$g$$ must be the whole of $$\mathbb R$$ (otherwise, we would be able to find a real $$x$$ so that neither of $$f(x),g(x)$$ is $$0$$ but their product is). So, everywhere nonzero functions are not zero divisors, since if $$f(x)$$ is never zero then $$g(x)$$ is always zero and $$g(x)$$ is the zero map. Conversely, if a function $$f$$ vanishes at a nonempty set of points $$S$$, then the function $$g(x)=\begin{cases}1,x\in S\\ 0,x\notin S\end{cases}$$ makes $$f$$ a zero divisor.
I guess that $$\mathscr{F}(\mathbb{R})$$ is the ring of all functions $$\mathbb{R}\to\mathbb{R}$$.
If $$f$$ never vanishes, then…
If $$f$$ vanishes somewhere, then… (generalize the example you have).