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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...

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

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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).

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