Reduced and integral rings

Question

Let $R$ be a commutative ring with unit. Are the following true?

1. If $\operatorname{Spec}(R)$ is irreducible i.e, cannot be written as union of two proper closed subsets, and $R$ is reduced, then $R$ is integral domain.

Conversely,

2. If $R$ is integral domain, then $\operatorname{Spec}(R)$ is irreducible.

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The problem is equivalent to showing that $(0)$ is a prime ideal, this seems to follow but I don't see how. I think for 2. we have $\operatorname{Spec}(R) = V(\{0\}) =\overline{\{0 \}}$.

Answer 1

1. If $\operatorname{Spec} R$ is irreducible, it has a single minimal prime ideal, and the intersection of the minimal primes is the nilradical.
2. Your argument is correct.

Answer 2

Part 1 is an if and only if. This question Spec R is irreducible shows that $Spec(R)$ is irreducible if and only if the nilradical $N$ is a prime ideal. So $R/N$ is an integral domain, but $R=R/N$ since $R$ is reduced.