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We know that every intgral domain has prime (or 0) characteristic. Is there en example that the converse isn't true? Does there exist a ring, which is not an integral domain, but has a prime characteristic?

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$\frac{\mathbb Z}{\mathbb Zp}\times\frac{\mathbb Z}{\mathbb Zp}\,.$ For the characteristic $0$ case, consider $\mathbb R\times\mathbb R\,.$

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Yes, of course. Take $\mathbb{F}_p[t]/(t^2)$.

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    $\begingroup$ $\mathbb{F}_p[t]$ are polynomials over given field $\mathbb{F}$ that has p elements, like $\mathbb{Z}_p$? $\endgroup$ – user19502 Oct 22 '13 at 1:03
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    $\begingroup$ @user19502: Yes, exactly. $\mathbb{F}_p$ is another notation for the field $\mathbb{Z}/p\mathbb{Z}$. $\endgroup$ – Najib Idrissi Oct 22 '13 at 13:13

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