Fraleigh's proof that $ M $ is a maximal ideal if and only if $ R/M $ is a field I was reading Fraleigh's abstract algebra textbook and he gave a proof about the theorem that if $ R $ is a commutative ring with unity, then $ M $ is a maximal ideal if and only if $ R/M $ is a field. There are several spots in the proof that I don't understand, so any help would be appreciated.
The proof assumed the following lemma, which I have successfully proved.
Lemma: Let $ \phi: R \to R' $ be a ring homomorphism and let $ N $ be an ideal of $ R, $ then $ \phi[N] $ is an ideal of $ \phi[R], $ but not necessary an ideal of $ R' $ and if $ N' $ be an ideal of either $ \phi[R] $ or $ R', $ then $ \phi^{-1}[N'] $ is an ideal of $ R. $
Proof of theorem: Suppose $ M $ is a maximal ideal of $ R, $ then $ R/M $ is a nonzero commutative ring with unity. Let $ (a + M) \in R/M $ with $ a \notin M $ so that $ a + M $ is not the additive identity of $ R/M. $ Suppose that $ a + M $ doesn't have the multiplicative inverse in $ R/M, $ then the set $ (R/M)(a + M) = \{(r + M)(a + M)|(r + M) \in R/M \} $ doesn't contain $ 1 + M $ and $ (R/M)(a + M) $ is an ideal of $ R/M. $ It is nontrivial because $ a \notin M $ and it is proper because it doesn't contain $ 1 + M. $ If we define the map $ \gamma: R \to R/M, $ then $ \gamma $ is a homomorphism and hence using the lemma, $ \gamma^{-1} [(R/M)(a + M)] $ is a proper ideal of $ R $ containing $ M. $
This is a contradiction since $ M $ is the maximal ideal of $ R. $ Thus $ a + M $ must have a multiplicative inverse in $ R/M. $
Question: So $ \gamma^{-1} [(R/M)(a + M)] $ is an ideal of $ R $ according to the lemma, but why does it must contain $ M? $
Conversely, suppose that $ R/M $ is a field. If $ N $ is any ideal of $ R $ such that $ M \subset N \subset R $ and $ \gamma $ is a homomorphism of $ R $ onto $ R/M, $ then $ \gamma[N] $ is an ideal of $ R/M $ with $ \{(0 + M) \} \subset N \subset R. $ This is a contradiction because a field doesn't contain any nontrivial proper ideal. Hence $ M $ must be maximal.
Question: So $ \gamma[N] $ is an ideal of $ R/M $ according to the lemma, but why must $ \gamma[N] $ be a nontrivial proper ideal of $ R/M? $
 A: Viewing $R$ as an abelian group and $M$ as a subgroup, the lattice isomorphism theorem (aka correspondence theorem) for abelian groups (https://en.wikipedia.org/wiki/Correspondence_theorem_(group_theory)) shows that $\gamma^{-1}[(R/M)(a+M)])$ contains $M$ because $(R/M)(a+M)$ is a subgroup of $R/M$. You can trace the proof in any text on abstract algebra to see why this is so.
$\gamma[N]$ is an ideal because $\gamma$ is surjective. It is non-trivial because the lattice isomorphism theorem gives a bijection of subgroups containing $M$ with subgroups of $R/M$. Since $N\neq M$, their images in $R/M$ must be distinct. It is proper because $N\neq R$, and again they must have distinct images in $R/M$.
Note that the lattice isomorphism extends to a theorem about ideals, and further to one about modules, but that extra level of specification is unnecessary to apply it here.
A: You can show it directly-let $J = \{ra + M: r \in R\}$. We have that this is an ideal of $R/M$ since:
$(ra + M) - (r'a + M) = (ra - r'a) + M = (r-r')a + M \in J$, so $J$ is an additive subgroup, and for any $x + M \in R/M$:
$(x + M)(ra + M) = x(ra) + M = (xr)a + M \in J$.
Now if $\gamma: R \to R/M$ is the map $r \mapsto r + M$, we consider $\gamma^{-1}(J)$. We could also show directly this is an ideal of $R$, but you say you understand this.
If $m \in M$, we have $\gamma(m) = m + M = 0 + M$ (since $m - 0 = m \in M$), and
$0 + M = 0a + M \in J$, so $\gamma$ maps every element of $M$ inside $J$.
The answer to your second question is trivial: we have $\gamma(\gamma^{-1}(J)) = J$, as we do for any surjective function (the image of a pre-image set, is the original set).
Note that the surjectivity of $\gamma$ is key: if $\gamma(R) \neq R'$ for a ring-homomorphism $\gamma: R \to R'$, then for an additive subgroup $I' \subseteq \gamma(R)$ we might not be able to show that given $r' \in R'$ and $x' \in I'$, that $r'x' \in \gamma(R)$, much less $I'$. 
