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

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    I think it is not true for a ring without identity! For example take $2\mathbb Z$ and its maximal ideal $4\mathbb Z$. – Bill May 13 '17 at 07:39

6 Answers6


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.

Brian M. Scott
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  • Thanks for showing me this, good to learn some less standard proofs – Freeman Sep 29 '11 at 11:15
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    @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
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    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 04 '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 01 '20 at 05:18

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

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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

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  • 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 '21 at 08:39

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.

Bill Dubuque
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  • 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 08 '17 at 09:12
  • @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 08 '17 at 16:35
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    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 $ab \notin P$.

Jason Juett
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    One can see this nice theorem on the first page of Kaplansky's book entitled "Commutative rings". – Tusif Ahmed Oct 30 '19 at 07:43
  • This well-known theorem has already been proved here in many places, e.g. [here](https://math.stackexchange.com/a/146757/242) a few years prior, from a more conceptual viewpoint. – Bill Dubuque Nov 01 '20 at 04:55

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 01 '20 at 06:03