# Why are maximal ideals prime?

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

• I think it is not true for a ring without identity! For example take $2\mathbb Z$ and its maximal ideal $4\mathbb Z$. – B.K-Theory May 13 '17 at 7:39

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

• Thanks! That's a much neater proof than the one I have here.. will be making a note of your answer for writing up in my notes! :) – Freeman Sep 29 '11 at 11:06
• @BodyDouble Thanks for the edit. I agree that it didn't make sense. – user1729 Apr 27 at 8:39

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.

• Thanks for showing me this, good to learn some less standard proofs – Freeman Sep 29 '11 at 11:15
• @Freeman The above proof is simply an ideal-theoretic translation of a well-known proof of Euclid's Lemma for integers - see my answer – Bill Dubuque Feb 25 '14 at 20:51
• The proof can be simplified by choosing $a$ as follows. Since $A$ is maximal, $ab$ is not a unit then either $a$ or $b$ is not a unit. Let $a$ be non-unit then $B$ not only properly contains $A$ but is also properly contained in $R$. Which is a contradiction to maximality of $A$. – user2734153 Jul 4 '19 at 20:28
• @user2734153 It is not clear how you are deducing $B\neq R$ but it may well be circular or incorrect. – Bill Dubuque Nov 1 '20 at 5:18

$A$ is an maximal ideal $\Rightarrow$ $R/A$ is a field $\Rightarrow$ $R/A$ is an integral domain $\Rightarrow$ $A$ is prime

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} Ax\!+\!ay=&\,\color{#c00}1,\,\ A\ \mid\ ab\ \ \ \Rightarrow\, A\ \mid\ b.\ \ \ {\bf Proof}\!:\, A\ \mid\ Ab,ab\, \Rightarrow\, A\,\mid Abx\!\!+\!aby = (\!\overbrace{Ax\!+\!ay}^{\large\color{#c00} 1}\!) b = b\\ (A,\ \ \ a)=&\,\color{#c00}1,\,\ A\ \mid\ ab\ \ \ \Rightarrow\, A\ \mid\ b.\ \ \ {\bf Proof}\!:\, A\ \mid\ Ab,ab\, \Rightarrow\, A\,\mid (Ab,\ \ ab) = (A,\ \ \ a)\ \ b =\, b\\ A\!+\!(a)=&\,\color{#c00}1,\,\ A\supseteq\! (ab)\, \Rightarrow\, A \supseteq\! (b).\, {\bf Proof}\!:\, A \supseteq Ab,ab \,\Rightarrow A\supseteq Ab\!+\!(ab)\! =(A\!+\!(a))b =\! (b)\\ A +{\cal A}\ =&\,\color{#c00}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} =(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.

• may you elaborate on the step $(Ab,ab) = (A,a)b$? All i see is $(A,a)b | Ab,ab$ so $(A,a)b | (Ab,ab)$... – Bryan Shih Feb 8 '17 at 9:12
• @CWL It uses the  GCD Distributive Law $\, \gcd(Ab,ab) = \gcd(A,a)b,\,$ as I explained above (the link I gave for that law leads to proofs). – Bill Dubuque Feb 8 '17 at 16:35
• To me this is really the 'best' proof, because while being the most elementary it is also the proof that trivially generalizes to ideals maximal with respect to a closure ($\star$) operation. And for someone with a special affinity for quotients, you could analogously that $a \notin A$ implies $a$ is a fortiori a regular element of $R/A$ (in fact a unit), and $ab \in A$ implies $ab$ is zero in $R/A$. Hence $b$ is zero in $R/A$. But no need at all to mention field or domain structure of $R/A$. – Badam Baplan May 26 '18 at 18:08

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 $ab \notin P$.

• One can see this nice theorem on the first page of Kaplansky's book entitled "Commutative rings". – Tusif Ahmed Oct 30 '19 at 7:43
• This well-known theorem has already been proved here in many places, e.g. here a few years prior, from a more conceptual viewpoint. – Bill Dubuque Nov 1 '20 at 4:55

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 $$ab \in \mathfrak{m}$$).

• This is precisely the same as the proof posted $>9\,$ years prior by Brian, except with less detail. Said in ideal language it is: $\,a\not\in {\frak m}\,$ max $\Rightarrow\,aR+{\frak m}=(1)\Rightarrow b\in b(aR+{\frak m})\subseteq {\frak m},\,$ by $\,ba\in {\frak m},\,$ as explained at length in my answer. Please strive to avoid posting duplicate answers. – Bill Dubuque Nov 1 '20 at 6:03