# 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 $R/M?$

• The inverse image of any ideal (of the quotient) contains the kernel of the quotient map (in this case M - since, what is the inverse image of 0, i.e. [M]?). The second question follows the same reasoning as above: there is an inclusion perserving bijection between ideals of the quotient and ideals of $R$ containing $M$. I think this is the third isomorphism theorem? Maybe this will help. – Eoin May 8 '16 at 1:45

## 2 Answers

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 $I'$.

• The image of a pre-image is only equal to the original set in the case that you are working with a surjective function (not any function). – J. David Taylor May 8 '16 at 17:07
• Good point, I have amended my answer. – David Wheeler May 9 '16 at 11:04

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.