Could anyone explain to me why maximal ideals are prime?
I'm approaching it like this, let $R$ be a commutative ring with $1$ and let $A$ be a maximal ideal. Let $a,b\in R:ab\in A$.
I'm trying to construct an ideal $B$ such that $A\subset B \neq A$ As this would be a contradiction. An alternative idea I had was to prove that $R/A$ is an integral domain, but this reduces to the same problem.
EDIT: Ergh.. just realized that I've learnt a theorem that states is $A$ is a maximal ideal then $R/A$ is a field
$\endgroup$ 16 Answers
$\begingroup$Here’s a proof that doesn’t involve the quotient $R/A$.
Suppose that $A$ is not prime; then there are $a,b\in R\setminus A$ such that $ab\in A$. Let $B$ be the ideal generated by $A \cup \{a\}$; $B = \{x+ar: x\in A\text{ and }r\in R\}$. Clearly $A \subsetneq B$, so $B = R$, $1_R \in B$, and hence $1_R = x + ar$ for some $x\in A$ and $r\in R$. Then $$b = b1_R = b(x+ar) = bx + bar.$$ But $bx \in bA \subseteq RA = A$, and $bar \in Ar \subseteq AR = A$, so $b \in A$. This contradiction shows that $A$ is prime.
$\endgroup$ 4 $\begingroup$$A$ is an maximal ideal $\Rightarrow$ $R/A$ is a field $\Rightarrow$ $R/A$ is an integral domain $\Rightarrow$ $A$ is prime
$\endgroup$ 0 $\begingroup$Let $A$ be a maximal ideal. Then $R/A$ contains no proper ideals, by the correspondence theorem. Indeed, $R/A$ is a field (assuming that $R$ contains an identity). Hence, $A$ is a prime ideal.
Theorem. $R/A$ is a field.
Proof. Let $i+A\in R/A$ such that $i+A\neq 0+A$. We want to prove that $i+A$ is a unit. So set $B=A+Ri=\{a+ri: a\in A, r\in R\}$.
Now, you (yourself!) need to prove that $B$ is an ideal, and that $A\subset B$ properly. Since $A$ is maximal this means that $B=R$.
As $B=R$ we have $1 \in B$, hence there exists some $a\in A, r\in R$ such that $a + ri = 1$. Then $1+A=(a+ri)+A=ri+A=(r+A)(i+A)$, and so $i+A$ is a unit, as required. QED
$\endgroup$ 2 $\begingroup$Worth emphasis: the common proof in Brian's answer is simply an ideal-theoretic form of the common Bezout-based proof of $ $ Euclid's Lemma $ $ for integers. To highlight the analogy we successively translate the Bezout-based proof into the language of gcds and (principal) ideals.
Euclid's Lemma in Bezout form, gcd form and ideal form
$\!{\!\begin{align} \color{#0a0}{Ax\!+\!ay}=&\,\color{#c00}{\bf 1},\,\ A\ \mid\ ab\ \ \ \Rightarrow\, A\ \mid\ b.\ \ \ {\bf Proof}\!:\, A\ \mid\ Ab,ab\, \Rightarrow\, A\,\mid Abx\!\!+\!aby = \smash{(\!\overbrace{\color{#0a0}{Ax\!+\!ay}}^{\textstyle\color{#c00}{\bf 1}}\!)} b = b\\ \color{#0a0}{(A,\ \ \ a)}=&\,\color{#c00}{\bf 1},\,\ A\ \mid\ ab\ \ \ \Rightarrow\, A\ \mid\ b.\ \ \ {\bf Proof}\!:\, A\ \mid\ Ab,ab\, \Rightarrow\, A\,\mid (Ab,\ \ ab) = (\color{#0a0}{A,\ \ \ a})\ \ b =\, b\\ \color{#0a0}{A\!+\!(a)}=&\,\color{#c00}{\bf 1},\,\ A\supseteq\! (ab)\, \Rightarrow\, A \supseteq\! (b).\, {\bf Proof}\!:\, A \supseteq Ab,ab \,\Rightarrow A\supseteq Ab\!+\!(ab)\! =(\color{#0a0}{A\!+\!(a)})b =\! (b)\\ \color{#0a0}{A +{\cal A}}\ =&\,\color{#c00}{\bf 1},\,\ A\supseteq {\cal A B}\, \Rightarrow\, A \supseteq\, {\cal B}.\,\ {\bf Proof}\!:\, A\, \supseteq\! A{\cal B},\!{\cal AB}\!\Rightarrow\! A\supseteq A{\cal B}\!+\!\!{\cal AB} =(\color{#0a0}{A+{\cal A}}){\cal B} = {\cal B} \end{align}}$
The third ideal form is precisely the same proof as in Brian's answer. The last form shows that the proof works more generally for coprime (i.e. comaximal) ideals $\,A,\, {\cal A},\, $ i.e. $\, A+{\cal A}= 1.\,$ In the second proof for integers, we can read $\,(A,a)\,$ either as a gcd or an ideal. Read as a gcd the proof employs the universal property of the gcd $\, d\mid m,n\iff d\mid (m,n)\,$ and the gcd distributive law $\,(Ab,ab) = (A,a)b.\,$ In the first proof (by Bezout) the gcd arithmetic is traded off for integer arithmetic, so the the gcd distributive law becomes the distributive law in the ring of integers.
$\endgroup$ 3 $\begingroup$For a completely different approach: An ideal is prime if and only if it is maximal with respect to the exclusion of a nonempty multiplicatively closed subset. (This theorem is extremely useful in commutative ring theory.) By definition, maximal ideals are maximal with respect to the exclusion of {1}.
For the proof of the nontrivial direction of that theorem, let $P$ be an ideal maximal with respect to the exclusion of a nonempty multiplicatively closed subset $S$. Then $P$ is proper. Pick $a,b \notin P$. Since $(P + (a))(P + (b)) \subseteq P + (ab)$ contains an element of $S$, we conclude that $ab \notin P$.
$\endgroup$ 2 $\begingroup$Let $R$ be a commutative ring with identity. Suppose $\mathfrak{m}$ is a maximal ideal in $R$ and $ab \in \mathfrak{m}$. If $a \not\in \mathfrak{m}$, then $I$ be the ideal generated by $a$ and $\mathfrak{m}$. Since $\mathfrak{m}$ is maximal, it follows $I=R$. Thus there exist elements $m \in \mathfrak{m}$ and $t \in A$ such that $m+at=1$. Hence $b=bm+bat \in \mathfrak{m}$ (as $ab \in \mathfrak{m}$).
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