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Find the zero divisors of the ring $R=C[0,1]$ the continuous functions $f:[0,1] \to [0,1]$.

I could thought of a set $S$ that I think is included in the set of zero divisors, but I am not sure if $S$ contains all the zero divisors. $S=\{f:[0,1] \to [0,1]: \text{there is} \space \epsilon>0, (x_0-\epsilon,x_0+\epsilon) \subset [0,1], f((x_0-\epsilon,x_0+\epsilon))=0\}$

I would like some help to find the zero divisors. Is the set $S$ a subset of the set of zero divisors? How can I show that?

user16924
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2 Answers2

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Well, no doubt $S$ is a subset of the set of zero divisors. In fact, it contains all zero divisors in the specified ring.

Let $f\in C[0,1]\setminus S$, and we show that $f$ is not a zero divisor. Since $f\not\in S$, the set of zeros of $f$, $Z:=\{x\in[0,1]|f(x)=0\}$, does not contain any open interval. Assume now $g$ is another function such that $f\cdot g=0$. It follows that $g$ has a zero in every open interval, or in other words the set of zeros of $g$ is dense. Since $g$ is continuous, it follows that $g\equiv0$.

Amitai Yuval
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    +1. Alternatively, if $fg = 0$ and $g \ne 0$, say $g(x_0) \ne 0$, then by continuity $g$ does not vanish in a neighborhood of $x_0$, so $f$ must – zcn Sep 12 '14 at 06:59
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Yes, you're correct. If $f$ is a zero divisor, then there is some nonzero $g \in C[0,1]$ such that $fg = 0$. Suppose $g(x_0) \neq 0$. By continuity, $g(x)$ is nonzero on some an interval $I := (x_0 - \epsilon, x_0 + \epsilon)$ for some $\epsilon > 0$, and thus $f\vert_I = 0$.

Conversely, if $f \in C[0, 1]$ is zero on such an interval $I$, then one can choose $g$ that is nonzero at $x_0$ but zero on $[0, 1] - I$, and by construction $fg = 0$.

Travis Willse
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