Alternative proof of $I$ maximal implies $I$ prime

Question

The proof I know to show maximal implies prime for an ideal $I$ in a commutative ring $R$ goes as follows:

$I$ maximal $\iff$ $R/I$ is a field $\implies$ $R/I$ is an integral domain $\iff$ $I$ is prime.

Surely, it has to be possible to show the same as follows:

Let $I$ be maximal and $ab \in I$. Assume $a \notin I$. Then $I$ is properly contained in $I + (a)$ and hence $1 \in I + (a)$ that is, there exists $r \in R$ and $i \in I$ such that $i + ra = 1$. Similarly, one can find $i^\prime$ and $r^\prime$ such that $i^\prime + r^\prime b = 1$. So we get that $ra - r^\prime b \in I$.

Now I'm stuck. How can I deduce that either $a$ or $b$ have to be in $I$?

Answer 1

Your goal is to prove that, when $I$ is maximal, $ab\in I$ implies that either $a\in I$ or $b\in I$. Note that this is equivalent to proving that $ab\in I$ and $a\notin I$ implies $b\in I$.

If $ab\in I$ and $a\notin I$, then $I+(a)=R$ and hence there is some $r\in R$ and $i\in I$ such that $i+ra=1$ (you've already noticed this). Multiplying by $b$, $$ib+rab=b.$$ $I$ is an ideal and $i\in I$, so that $ib\in I$, and we also have $ab\in I$ by assumption. Therefore, we can conclude that $ib+rab=b\in I$.

Answer 2

Let $I\subset R$ be a maximal ideal. Suppose $ab\in I$ yet $a,b\notin I$. Then $I+(a)=I+(b)=R$, so $$R=(I+(a))\cdot (I+(b))=I^2+I\cdot (a)+I\cdot (b)+(a)\cdot (b)\subseteq I+(ab)=I$$ a contradiction, hence either $a$ or $b$ is in $I$ thus $I$ is prime.

Answer 3

The proof that you seek works more generally and it is highly instructive to present it this way. Notice that an ideal $\rm\:P\:$ is prime iff its complement $\rm\:\overline P\:$ is a monoid, i.e. $\rm\:a,b\not\in P\:\Rightarrow\:ab\not\in P.\:$ Krull discovered a very useful generalization of this primality test, namely $(3)$ below, which states that prime ideals are precisely the ideals that are maximal in the complement of some multiplicative submonoid M of the ring, i.e. ideals maximal w.r.t. not meeting M.

Theorem $\ $ TFAE for an ideal $\rm\:P\:$ in a ring $\rm\:R.$

$\rm(1)\ \ \:P\:$ is prime.

$\rm(2)\ \ \:\overline P\:$ is a submonoid $\rm\:M\:$ of $\rm\:R\:$$ $ (where $\rm\:\overline S := R\backslash S = $ set-theoretic complement of $\rm\:S\:$ in $\rm\:R)$

$\rm(3)\ \ \: P\:$ is an ideal maximal in $\rm\:\overline M\:$ for a monoid $\rm\:M\subseteq \left<R,*\right>,\:$ i.e. ideal $\rm\:I\supsetneq P\Rightarrow I\cap M\ne \{\ \}$

Proof $\ (1 \Rightarrow 2)\ $ Since $\rm\:P\:$ is prime, $\rm\:a,b\not\in P\:\Rightarrow\:ab\not\in P,\:$ i.e. $\rm\:a,b\in \overline P\:\Rightarrow\:ab\in \overline P.\:$ Furthermore $\rm\:1\in \overline P,\:$ hence $\rm\:\overline P\:$ is a submonoid of $\rm\:M.$

$(2\Rightarrow 3)\ $ Let $\rm\: M = \overline P.$

$(3\Rightarrow 1)\ $ Assume $\rm\:a,b\not\in P.\:$ Then $\rm\:P + (a)\supsetneq P\:$ so it meets $\rm\:M,\:$ i.e. $\rm\ m = p + a\:\!r\in M,\ p\in P.\:$ Similarly $\rm\:P + (b)\supsetneq P\:$ $\Rightarrow$ $\rm\: m' = p'+b\:\!r'\in M,\ p'\in P.\:$ Therefore

$$\rm\ \ \ ab\in P\ \Rightarrow\ mm' = (p+a\:\!r)(p'+b\:\!r') \in P \cap M = \{\ \}\ \Rightarrow\Leftarrow$$

So, having proved $\rm\:ab\not\in P\:$ when $\rm\:a,b\not\in P,\:$ we deduce that $\rm\:P\:$ is prime. $\ \ $ QED

Your result is simply the special case $\rm\:M = \{1\},\:$ since $\rm\:P\:$ is maximal in $\rm\:\overline{\{1\}}\:$ iff $\rm\:P\:$ is maximal. Notice, in particular, that the proofs in the other answers are essentially the same as the proof above of $(3\Rightarrow 1)$ for the special case $\rm\:M = \{1\}.$ But the proof for general $\rm\:M\:$ is no more difficult. Further, this view is well-worth knowing, since it lends insight into the essence of that matter.

Krull's criterion is useful because any ideal disjoint from a monoid $\rm\:M\:$ can, by Zorn's Lemma, be expanded to one maximally so. This yields a frequently useful method for contructing prime ideals. The essence of the matter is clarified when one studies local-global methods, in particular the important technique of localization, e.g see here and see here. From this viewpoint, the ideals in $(3)$ are simply ideals that extend to maximal ideals in the localization $\rm\:M^{-1} R,\:$ and their primality follows from general principles.

Answer 4

The two proofs so far are either for commutative rings with unity, or for the notion of "completely prime" as opposed to prime. Here's a proof that holds in more generality.

Recall that in a not-necessarily-commutative ring $R$, the definition of prime ideal is that $I$ is a prime ideal if and only if $I\neq R$, and for any ideals $\mathfrak{A}$ and $\mathfrak{B}$, $\mathfrak{AB}\subseteq I$ implies $\mathfrak{A}\subseteq I$ or $\mathfrak{B}\subseteq I$. This is a weaker condition that "completely prime" (the element-wise version); for example, in the ring of $n\times n$, $n\gt 1$, matrices over a field $F$, the trivial ideal $(0)$ is prime, but not completely prime (since we can have $a,b\in R$ with $ab=0$ but $a\neq 0\neq b$). But every completely prime ideal is necessarily prime. In commutative rings with unity, the two notions coincide.

The proof is very similar to Alex Becker's computation.

Theorem. Let $R$ be a ring, not necessarily commutative, not necessarily with unity, such that $R^2$ is not contained in any maximal ideal of $R$ (in particular, this holds if $R$ has a unity). If $\mathfrak{M}$ is a maximal ideal of $R$, then $\mathfrak{M}$ is a prime ideal of $R$. Conversely, if $R^2$ is contained in a maximal ideal $\mathfrak{N}$, then $\mathfrak{N}$ is not prime.

Proof. Assume $R^2$ is not contained in any maximal ideal, and let $\mathfrak{M}$ be maximal in $R$. Let $\mathfrak{A},\mathfrak{B}$ be two ideals of $R$ that are not contained in $\mathfrak{M}$; we prove that $\mathfrak{AB}\not\subseteq \mathfrak{M}$.

Since $\mathfrak{M}$ is maximal, and $\mathfrak{A},\mathfrak{B}\not\subseteq \mathfrak{M}$, then $\mathfrak{A}+\mathfrak{M}=\mathfrak{B}+\mathfrak{M}=R$. If $\mathfrak{AB}\subseteq \mathfrak{M}$, then we have: $$\begin{align*} R^2 &= (\mathfrak{A}+\mathfrak{M})(\mathfrak{B}+\mathfrak{M}) \\ &=\mathfrak{AB}+\mathfrak{AM} + \mathfrak{MB}+\mathfrak{M}^2\\ &\subseteq \mathfrak{AB}+\mathfrak{M}\\ &\subseteq \mathfrak{M}+\mathfrak{M}\\ &=\mathfrak{M}. \end{align*}$$ But this contradicts our assumption that $R^2$ is not contained in any maximal ideal; therefore, $\mathfrak{AB}\not\subseteq \mathfrak{M}$, as desired.

If $\mathfrak{N}$ is a maximal ideal that contains $R^2$, then $\mathfrak{A}=\mathfrak{B}=R$ are a witness to the non-primality of $\mathfrak{N}$. $\Box$

If we assume $R^2=R$ (instead of simply that $R^2$ is not contained in any maximal ideal of $R$), then the proof can be done directly, instead of as by contradiction, by showing that $\mathfrak{AB}+\mathfrak{M}=R$.

For an example where $R^2\neq R$ but we still have that the implication holds (vacuously), take $R=\mathbb{Q}$ with zero multiplication. Then ideals corresponds to subgroups, and since $\mathbb{Q}$ has no maximal subgroups, it is still true that every maximal ideal is also a prime ideal.