Let $R$-algebra $A$. If $P⊂A$ is a minimal prime ideal then $p=P \cap R$ consists of zerodivisors for $A$?

Question

We have: Let $R$ be a Noetherian commutative ring. Suppose $P⊂R$ is a minimal prime ideal. Then it is a theorem that $P$ consists of zero-divisors.

But how to prove this?

The $R$-algebra $A$ is assumed to be finitely generated as an $R$-module. For each $P\in\operatorname{Min}A$ the prime ideal $p=P ∩ R$ consists of zerodivisors for $A$.

thank for all