Why does the ring element get absorbed here?

Question

Suppose $M$ is finitely generated over $R$ that is a PID. Let $x \in M$, $\ann(x) = (a)$, $p\in R$ be irreducible. Then

1. If $p|a$ then $Rx/pRx \approx R/(p)$
2. If $p \not| a$, then $pRx = Rx.$

The proof goes like this in the textbook

1. Let $\phi(r) = rx$, then $\ker \phi = (a)$. Since $p|a$, then $(p)$ under $\phi$ is $pRx$, so $(p)/(a) = pRx$ and $R/(a) = Rx$, so $R/(p) = Rx/pRx$.

Okay so elements in $(p)$ are $pd$, and under $\phi$ this is $pdx \in pRx$ because $d$ is arbitrary.

2. If $p \not|a$ then $gcd(p,a)=1$ so there exists $s,t$ such that $sp + ta = 1$. Therefore $rx = 1rx = psrx + tart = psrx \implies Rx = pRx$.

Now I don't get that conclusion. Since $rx = psrx$ isn't $psrx \in psRx$? $s$ here isn't arbitrary like above.

Answer 1

Since s and r are both in R, we have sr is in R, so psrx is in pRx