Image of an ideal in a quotient

Question

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 $(p)/(a)$?

Answer 1

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 $$(p)\big/(p)\cap (a)\simeq (p)+(a)\big/(a).$$

Answer 2

The isomorphism theorem states that if $A\subseteq B\subseteq R$ and $A$ and $B$ are ideals of $R$, then $R/B\cong\dfrac{R/A}{B/A}$.

If $p$ divides $a$, take $B=(p)$ and $A=(a)$, so we indeed have $A\subseteq B\subseteq R$. We have $M=R/A$ and $F=R/B$, in your notation, so the problem really is to show that $pM=B/A$.

The elements of $B$ have the form $px$ for $x\in B$, so an element of $B/A$ has precisely the form $px+(a)=p(x+(a))$, for some $x\in R$. These are precisely the elements of the form $pm$ for some $m\in M=R/A$. Thus $B/A=pM$, just as we wanted.