Image of an ideal in a quotient I am confused about a statement made in Dummit and Foote.
Let $R$ be a PID and let $p$ be a prime in $R$. Let $F$ denote the field $R/(p)$, and let $M=R/(a)$ where $a$ is a nonzero element of $R$. A statement in the book claims that $M/pM\cong F$ if $p$ divides $a$ in $R$. 
For the proof, the authors make the following claim: "This follows from the Isomorphism Theorems: note first that $p(R/(a))$ is the image of the ideal $(p)$ in the quotient $R/(a)$, hence is $(p)+(a)/(a)$.
My question is how does this follow from the Isomorphism Theorems? The elements of $p(R/(a))$ are of the form, I believe, $rp+(a)$. But why would the collection of these as $r$ ranges over $R$ be $(p)+(a)/(a)$ and not simply $(p)/(a)$?

It can't be $(p)/(a)$, as in general the ideal $(p)$ dos not contain $(a)$. 
To see the detailsconsider the restriction of the canonical homomorphism $\varphi: R\longrightarrow R/(a)$ to the ideal $(p)$. Its kernel is $(p)\cap (a)$, hence by the first isomorphism theorem, its image $(p)\cdot R/(a)$ is isomorphic to $(p)\big/(p)\cap(a)$.
By the second isomorphism theorem, we have
$$(p)\big/(p)\cap (a)\simeq (p)+(a)\big/(a).$$
