Zero divisors in ring of real valued functions.

Question

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.

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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 :)

Answer 1

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.

Answer 2

If you're not assuming continuity, then the zero divisors are precisely the functions that vanish somewhere. If you have a function $f$ that vanishes somewhere, say at $x$, then you can define a function $g$ that takes the value 1 at $x$, and 0 everywhere else. Then $fg = 0$, but neither $f,g$ are 0.

Edit: I just want to add that sometimes in math instead of trying to directly prove it, you can try to work intuitively, for example, think about what kinds of zero divisors you can come up with. Ie, try to come up with two nonzero functions that multiply to 0. For example, you could have the function $f(x) = x$, $g(x) = x^2-1$, then $(fg)(x) = x(x^2-1)$, which is obviously not the zero function. So that doesn't work. Maybe you can come up with some other examples, and see which work, which don't...etc.

Answer 3

Does this precisely describe the set of zero divisors? \begin{align*} \{f \in\mathscr{F}(\mathbb{R}): f(x_0) = 0 \text{ for some } x_0\in \mathbb{R} \} \end{align*}

(The reason is that we can always find a non-zero element $g\in\mathscr{F(\mathbb{R})}$ such that $f(x)g(x) = 0$ for all $x\in\mathbb{R}$. This holds as long as $g(x)\ne 0$ whenever $f(x)=0$, and vice-versa.)