I'm working though Pinter's A book of Abstract Algebra and would like a quick verification on a simple problem. Exercise 17.B2 asks
Describe the divisors of zero in $\mathcal{F}(\mathbb{R})$.
Note that $\mathcal{F}(\mathbb{R})$ denotes the ring of all real valued functions endowed with pointwise multiplication and addition.
My attempt
If a non-zero element $f$ is a zero divisor of $\mathcal{F}(\mathbb{R})$, then there exists a non-zero $g$ such that
$$\forall x \in \mathbb{R}: \quad f(x)g(x) = 0.$$
Because $f$ is non-zero, there exists an $x_1 \in \mathbb{R}$ such that $f(x_1) \neq 0$. I claim that if, in addition, there is an $x_2 \in \mathbb{R}$ such that $f(x_2) = 0$, then $f$ is a zero-divisor. To see this, define $g$ by
$$ g(x) = \begin{cases}
0 &\text{if}\;\;\; x \neq x_2 \\
1 &\text{if}\;\;\; x = x_2 \end{cases}$$
Then it is clear that $g$ is non-zero and $fg=0$. Thus, $f$ is a zero-divisor.
I claim, further, that such $f$ are the only zero divisors. To see this, suppose that there does not exist an $x_2 \in \mathbb{R}$ such that $f(x_2) = 0$. Then $f(x) \neq 0$ for every $x \in \mathbb{R}$. But then for any $x \in \mathbb{R}$, we can divide though by $f(x)$ in $f(x)g(x) = 0$ to obtain
$$g(x) = \frac{0}{f(x)} = 0.$$
Which means that all $g$ that satisfy $fg=0$ must have $g(x) = 0$ for all $x \in \mathbb{R}$ contradicting the requirement that $g$ is non-zero.
I'm trying to learn this on my own, and am having some trouble understanding and gaining intuition regarding zero-divisors. So even though this may seem simple, I would like some help and criticism 🙂
Best Answer
Yes, that's a fine argument. For a slightly different approach, note that if $f$ has no zeros, then we can define an inverse for it via
$$\tilde{f}(x) = \frac{1}{f(x)}$$
Since $f(x) \ne 0$ for all $x$, this makes sense. Then it's easy to check that $f \tilde f = 1 = \tilde f f$ is the function which is identically $1$ (which is the multiplicative identity in the ring), so that $f$ is invertible as a ring element. Invertible elements are never zero divisors.