# the localization of a ring is an integral domain iff the annihilators of zero divisors are comaximal ideals

I would appreciate any help with the following problem. let $R$ be a commutative ring (with $1$). I need to show that the following are equivalent

i) for every prime ideal $P$, the localization $R_P$ is an integral domain

ii) for every maximal ideal $M$, $R_M$ is an integral domain

iii) for $x,y \in R$ such that $xy = 0$ we have $Ann(x) + Ann(y) = R$ where $Ann$ denotes the annihilator ideal.

Clearly, i) implies ii), and I showed that ii) implies iii). If iii) were false, then we can find $x,y \in R$ with $xy = 0$ and $Ann(x) + Ann(y) \subset M$ for some maximal ideal $M$. Then, in $R_M$ we have $\frac{xy}{1} = \frac{0}{1}$. Now, we are assuming that $R_M$ is an integral domain, hence either $\frac{x}{1} = \frac{0}{1}$ or $\frac{y}{1} = \frac{0}{1}$. Therefore, there exist $s,t \in R-M$ such that either $sx = 0$ or $ty = 0$. But $sx = 0$ means that $s \in Ann(x)$, hence $s \in M$, which contradicts the fact that $s \in R-M$. Similarly, $ty = 0$ leads to a contradiction. So ii) implies iii).

I am having problems proving that iii) implies i). Let $P$ be a prime ideal of $R$ and suppose that $\frac{x}{s} \cdot \frac{y}{t} =\frac{0}{1}$. This means that there exists $z \in R-P$ with $zxy = 0$. Now, I want to show that either $\frac{x}{s}$ or $\frac{y}{t}$ is equal to $0$, so I attempted to show that either $zx = 0$ or $zy = 0$. From the given assumption we have $Ann(zx) + Ann(y) = R$, but I do not see how to conclude that $Ann(zx) = R$ (or maybe this approach is not even correct).

I am studying this on my own, so this is not a HW problem. I would appreciate any suggestions. Thank you in advance.

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I suppose your approach is correct, but you need to be a little bit more delicate.

Let us begin with a observation concerning localization in general:

### Lemma

If $\mathfrak{p}$ is a prime ideal, then for any $x \in R$ such that $x/1 \neq 0 \in R_\mathfrak{p}$ then $\mathrm{Ann}(x) \subset \mathfrak{p}$.

### Proof

Indeed, suppose that $t \in \mathrm{Ann}(x)$, or in other words $tx = 0$ and $t \not \in \mathfrak{p}$. Then, $1/t$ is a valid element of $R_\mathfrak{p}$ and we can compute: $$0 = tx/1 \cdot 1/t = tx/t = x/1 \neq 0$$ which forms the desired contradiction. $\square$

Now, suppose as you did, that $x/s \cdot y/t = 0 \in R_\mathfrak{p}$. Then you rightly conclude that $xyz = 0$ for some $z \not \in \mathfrak{p}$. It will suffice to prove that $x/1 = 0$ or $y/1 = 0$. Suppose otherwise, namely that $x/1 \neq 0$ and $y/1 \neq 0$. Then, we argue that also $xz/1 \neq 0$, since $z/1$ is invertible. By the lemma, we know that $$\mathrm{Ann}(xz), \mathrm{Ann}(y) \subset \mathfrak{p}$$ so $$\mathrm{Ann}(xz) + \mathrm{Ann}(y) \subset \mathfrak{p} + \mathfrak{p} = \mathfrak{p}$$ But from (iii) we know that $$\mathrm{Ann}(xz) + \mathrm{Ann}(y) = R$$ which is the desired contradiction.

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Thank you very much for your response! I really appreciate your help. – algebra_fan Jun 3 '11 at 0:29
The pleasure is all mine ;) – Jakub Konieczny Jun 3 '11 at 15:13