Equivalent conditions involving (co)primary module (Proposition 3.9 from Eisenbud) Let $R$ be a Noetherian ring and $M$ a f.g. $R$-module. Let $P$ be a prime ideal of $R$. I'm trying to understand why TFAE:
(a) $M$ is $P$-coprimary (i.e. $Ass(M)=\{P\}$)
(b) $P$ is minimal over $Ann(M)$ and every element not in $P$ is a nonzerodivisor on $M$
(c) A power of $P$ annihilates $M$ and every element not in $P$ is a nonzerodivisor on $M$
There is a proof in Eisenbud, but some points are unclear.
$(a)\implies (b)$ Here he refers to Theorem 3.1a that says that $Ass M$ is finite and nonempty set of primes, each containing $Ann(M)$, and $Ass(M)$ includes all primes minimal among primes containing $Ann(M)$. This should imply that $P$ is minimal over $Ann(M)$. But how? If I knew that primes minimal over $Ann(M)$ exist, then I could conclude that they lie in the set $Ass M=\{P\}$, so $P$ would be itself minimal over $Ann(M)$. Do they always exist?
$(b)\implies (c)$ Here he says that it suffices to prove the statement after localizing at $P$ since the elements in $R-P$ are nonzerodivisors. Why is it sufficient? He then says we can assume $R$ is a local ring with maximal ideal $P$; then $P=rad(Ann(M))$ since $P$ is minimal over $Ann(M)$. How is locality of $R$ and maximality of $P$ used here?
$(c)\implies (a)$ This is clear, I think.
 A: 
$(a)\implies (b)$ Here he refers to Theorem 3.1a that says that $Ass M$ is finite and nonempty set of primes, each containing $Ann(M)$, and $Ass(M)$ includes all primes minimal among primes containing $Ann(M)$. This should imply that $P$ is minimal over $Ann(M)$. But how? If I knew that primes minimal over $Ann(M)$ exist, then I could conclude that they lie in the set $Ass M=\{P\}$, so $P$ would be itself minimal over $Ann(M)$. Do they always exist?

In a paragraph after the statement of Theorem $3.1$, Eisenbud shows that primes minimal over a given ideal exist in any ring.
Furthermore, Proposition $3.4$ shows that $\operatorname{Ass} M$ is nonempty.


$(b)\implies (c)$ Here he says that it suffices to prove the statement after localizing at $P$ since the elements in $R-P$ are nonzerodivisors. Why is it sufficient? He then says we can assume $R$ is a local ring with maximal ideal $P$; then $P=rad(Ann(M))$ since $P$ is minimal over $Ann(M)$. How is locality of $R$ and maximality of $P$ used here?

Since every element not in $P$ is a nonzero divisor on $M$, then the canonical map $M \to M_P$ is injective (see Prop $2.1$).
Now what Eisenbud actually proves is that $PR_P = \operatorname{rad}(\operatorname{ann}M_P)$ which implies that $(PR_P)^n \subset \operatorname{ann}M_P$. But since $P \subset PR_P$ and $\operatorname{ann}M_P \subset \operatorname{ann}M$, then we have
$$P^n \subset (PR_P)^n \subset \operatorname{ann}M_P \subset \operatorname{ann}M.$$
