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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    $\begingroup$ I think it is not true for a ring without identity! For example take $2\mathbb Z$ and its maximal ideal $4\mathbb Z$. $\endgroup$
    – Bill
    Commented May 13, 2017 at 7:39
  • $\begingroup$ upvote for the edit. i too just realized that I too had read that stupid theorem $\endgroup$
    – Krishan
    Commented Sep 8, 2022 at 12:52
  • $\begingroup$ If $xy\in M$ and $x,y\notin M,$ $M+(x)=M+(y)=R$ i.e., $m_1+r_1x=m_2+r_2y=1$. Now $1=(m_1+r_1x)(m_2+r_2y)=m_1m_2+r_2ym_1+r_1xm_2+r_1r_2xy\in M.$ $\endgroup$
    – nkh99
    Commented May 2 at 10:18

5 Answers 5


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.

  • $\begingroup$ Thanks for showing me this, good to learn some less standard proofs $\endgroup$
    – Freeman
    Commented Sep 29, 2011 at 11:15
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    $\begingroup$ @Freeman The above proof is simply an ideal-theoretic translation of a well-known proof of Euclid's Lemma for integers - see my answer $\endgroup$ Commented Feb 25, 2014 at 20:51
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    $\begingroup$ 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 $A$. $\endgroup$ Commented Jul 4, 2019 at 20:28
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    $\begingroup$ @user2734153 It is not clear how you are deducing $B\neq R$ but it may well be circular or incorrect. $\endgroup$ Commented Nov 1, 2020 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


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

  • $\begingroup$ 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! :) $\endgroup$
    – Freeman
    Commented Sep 29, 2011 at 11:06
  • $\begingroup$ @BodyDouble Thanks for the edit. I agree that it didn't make sense. $\endgroup$
    – user1729
    Commented Apr 27, 2021 at 8:39

Worth emphasis: the common proof (e.g. 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
$\!{\!\begin{align} \color{#0a0}{Ax\!+\!ay}=&\,\color{#c00}{\bf 1}^{\phantom{|^{|}}}\!\!\!\:\!,\,\ A\ \mid\ ab\ \ \ \Rightarrow\, A\ \mid\ b.\ \ \ {\bf Proof}\!:\, A\ \mid\ Ab,ab\, \Rightarrow\:\! A\,\mid Abx\!\!+\!aby = \smash{(\!\overbrace{\color{#0a0}{Ax\!+\!ay}}^{\textstyle\color{#c00}{\bf 1}}\!)} b = \:\!b\\ \color{#0a0}{(A,\ \ \ a)}=&\,\color{#c00}{\bf 1},\,\ A\ \mid\ ab\ \ \, \Rightarrow\ A\ \mid\ b.\ \ \ {\bf Proof}\!:\, A\ \mid\ Ab,ab\, \Rightarrow\:\! A\,\mid (Ab,\ \ ab) = (\color{#0a0}{A,\ \ \ a})\ \ b =\, b\\ \color{#0a0}{A\!+\!(a)}=&\,\color{#c00}{\bf 1},\,\ A\supseteq\! (ab)\:\! \Rightarrow\, A \supseteq\! (b).\: {\bf Proof}\!:\, A \supseteq Ab,ab \,\Rightarrow A\supseteq Ab\!+\!(ab)\! =(\color{#0a0}{A\!+\!(a)})b =\! (b)\\ \color{#0a0}{A +{\cal A}}\ =&\,\color{#c00}{\bf 1},\,\ A\supseteq {\cal A B}\, \Rightarrow\, A \supseteq\, {\cal B}.\,\ {\bf Proof}\!:\, A \supseteq\! A{\cal B},\!{\cal AB}\!\Rightarrow A\supseteq A{\cal B}\!+\!\!{\cal AB} =(\color{#0a0}{A+{\cal A}}){\cal B} = {\cal B} \end{align}}$

The third ideal form is precisely the common 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 becomes integer arithmetic, where the the gcd distributive law becomes the distributive law in the ring of integers.

  • $\begingroup$ may you elaborate on the step $(Ab,ab) = (A,a)b$? All i see is $(A,a)b | Ab,ab$ so $(A,a)b | (Ab,ab)$... $\endgroup$
    – Bryan Shih
    Commented Feb 8, 2017 at 9:12
  • $\begingroup$ @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). $\endgroup$ Commented Feb 8, 2017 at 16:35
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    $\begingroup$ 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 $R/A$. $\endgroup$ Commented May 26, 2018 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 $ab \notin P$.

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    $\begingroup$ One can see this nice theorem on the first page of Kaplansky's book entitled "Commutative rings". $\endgroup$ Commented Oct 30, 2019 at 7:43
  • $\begingroup$ This well-known theorem has already been proved here in many places, e.g. here a few years prior, from a more conceptual viewpoint. $\endgroup$ Commented Nov 1, 2020 at 4:55

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