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Let $R \neq \{0 \}$ be a commutative ring with identity. Suppose that $R$ has only finitely many minimal prime ideals $p_1,\dots , p_s.$ Then $$\sqrt {0} = \bigcap\limits_{i=1}^{s} p_i.$$

I know that $\sqrt 0 = \bigcap\limits_{p\text { prime}} p.$ But from here how do I prove the above theorem? Would anybody please help me regarding this.

Thank you very much.

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Every prime ideal contains a minimal prime ideal, so the intersection of all prime ideals is the same as the intersection of all minimal prime ideals. Given that these are $p_1$, $p_2$, ..., $p_s$, it follows that $$\sqrt 0 = \bigcap\limits_{\text {p prime}} p = \bigcap\limits_{i=1}^{s} p_i.$$

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