# Under what conditions is the ideal generated by a zero divisor finite?

Let $R$ be an infinite commutative ring with unity. Suppose $R$ has zero divisors $r,s \in R$, such that $rs = 0$. Then we can consider the ideal $(r)$ generated by $r$. Certainly

$$\{ 0, r, r + r, \ldots, r + r + \cdots + r \} \subseteq (r)$$

where the last element is $(s - 1)r$. Naturally $(r)$ may contain many other elements, but this gets me thinking: under what conditions could the $(r)$ in this example be of finite order?

Thanks,

z.

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Are you assuming that $s$ is a natural number? – Diego Silvera Jan 22 '13 at 1:07
Apparently deep inside my mind, that's what was going on. Ha. I don't want to be assuming that though... – Ziggy Jan 22 '13 at 1:35

Consider the ring $R = (\mathbb{Z}/4\mathbb{Z})[X]$. Then $2 \cdot 2 = 4 = 0_R$, so $2$ is a zero divisor, and yet $$(2) \supseteq \{ 2, 2X, 2X^2, 2X^3, 2X^4, \dots \}$$ so $(2)$ is infinite.

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Ah sorry, I edited my question but forgot to change the name of it! That is a good example of an infinite ideal generated in the way I described. I'm more interested in knowing under what conditions (r) is finite. – Ziggy Jan 22 '13 at 1:32

Let $r \in R$ then we have a natural map $\phi: R \rightarrow (r)$ given by $\phi(s)=rs$. This is a homomorphism of $R$-modules, in particular $R/\ker \phi \cong (r)$ as $R$-modules. The $\ker\phi$ is frequently known as the annihilator of $r$ or even as the ideal quotient $(0:\{r\})$. In particular $(r)$ is finite if and only if the index of $\mathrm{Ann}_R(r)$ is finite.

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Let $R$ be a direct sum of infinitely many copies of $\mathbb{Z}/6\mathbb{Z}.$ Alternatively $R= \oplus~\mathbb{Z}/6\mathbb{Z}$ such that the direct sum is indexed by $\mathbb{Z}$. Obviously $R$ is an infinite commutative ring.

Now let $r = (2,2,\cdots,2,\cdots),$ i.e. the element of $R$ that is $2 (mod 6)$ at each coordinate. The ideal generated by $r$ only contains $3$ elements, and is therefore finite.

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