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Let $D$ be a integral domain with characteristic $p>0$. It is easy to prove that $p$ is prime.

Now, if $R$ is a commutative ring with unity and characteristic $p$, does $p$ be a prime number implies $R$ be a integral domain, or there is an counterexample?

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No. Take $x=(0,1)$ and $y=(1,0)$ in $\mathbb{Z}_p\times \mathbb{Z}_p$. You get $xy=0$.

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Let $p$ a prime, and consider the quotient ring $\Bbb{Z}_{p}[x]/(x^2)$. This is a commutative ring of characteristic $p$, but the square of the image of $x$ in it is zero.

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