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I'm looking for an example of a ring R which is not an integral domain but the localization of R at every maximal ideal is an integral domain

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    What are your thoughts? What have you tried so far? – 2017-02-10

2 Answers 2

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Take $A = \Bbb Z/(6)$. You can verify that $A$ has two maximal ideals; namely, $\mathfrak{m}_1 = 2\Bbb Z/(6)$ and $\mathfrak{m}_2 = 3\Bbb Z/(6)$. I leave it to you to check that \begin{align*} A_{\mathfrak{m}_1}&\cong\Bbb F_2,\\ A_{\mathfrak{m}_2}&\cong\Bbb F_3, \end{align*} both of which are integral domains.

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Some geometry can help: disconnected thing cannot be integral, but each component could be integral. For example, two points would suffice.

Concretely, take $A = \mathbb C[x]/(x(x-1))$. Two two maximal ideals are $(x)$ and $(x-1)$. Localizing at any one of them gives $\mathbb C$, but $A$ is clearly not a domain.

In fact, this is the only kind of examples that one can give. If $A$ is Noetherian, then $A$ is integral iff $A$ is stalk-locally integral and $\operatorname{Spec} A$ is connected.

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    It is such a shame how such a well-thought answer which - 1) gives geometric insight, - 2) gives a result, which states what further assumptions one needs to make it a local property - has less upvotes than a simple 'Ok $\mathbb Z/6\mathbb Z$ is an example'-answer. – 2017-02-10