A maximal ideal is always a prime ideal, and the quotient ring is always a field. In general, not all prime ideals are maximal. 1

In $2\mathbb{Z}$, $4 \mathbb{Z} $ is a maximal ideal. Nevertheless it is not prime because $2 \cdot 2 \in 4\mathbb{Z}$ but $2 \notin 4\mathbb{Z}$. What is that is misunderstand?

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    $\begingroup$ Is $2\mathbb Z$ a ring? $\endgroup$ – Thomas Andrews Jul 10 '12 at 19:58
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    $\begingroup$ @ThomasAndrews It is a ring without unit. $\endgroup$ – azarel Jul 10 '12 at 20:02
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    $\begingroup$ The problem is that depending the book you read, the "ring" term's meaning can go from non-neccesarily-commutative ring without identity (e.g. Hungerford) to commutative ring with identity (e.g. McDonal-Atiyah), passing through ring with identity (e.g. Lang). You only have to be careful, which convecton it is used in each book you read. $\endgroup$ – Josué Tonelli-Cueto Jul 10 '12 at 20:37

As Thomas points out, $2\mathbb Z$ is not a "ring", since it does not contain any identity element $1.$ It is true that every maximal ideal of a commutative ring with identity is prime.


Let $R$ be a ring, not necessarily with identity, not necessarily commutative.

An ideal $\mathfrak{P}$ of $R$ is said to be a prime ideal if and only if $\mathfrak{P}\neq R$, and whenever $\mathfrak{A}$ and $\mathfrak{B}$ are ideals of $R$, then $\mathfrak{AB}\subseteq \mathfrak{P}$ implies $\mathfrak{A}\subseteq \mathfrak{P}$ or $\mathfrak{B}\subseteq \mathfrak{P}$.

(The condition given by elements, $ab\in P$ implies $a\in P$ or $b\in P$, is stronger in the case of noncommutative rings, as evidence by the zero ideal in the ring $M_2(F)$, with $F$ a field, but is equivalent to the ideal-wise definition in the case of commutative rings; this condition is called "strongly prime" or "totally prime". Generally, with noncommutative rings, "ideal-wise" versions of multiplicative ideal properties are weaker than "element-wise" versions, and the two versions are equivalent in commutative rings).

When the ring does not have an identity, you may not even have maximal ideals. But here is what you can rescue; recall that if $R$ is a ring, then $R^2$ is the ideal of $R$ given by all finite sums of elements of the form $ab$ with $a,b\in R$ (that is, it is the usual ideal-theoretic product of $R$ with itself, viewed as ideals). When $R$ has an identity, $R^2=R$; but even when $R$ does not have an identity, it is possible for $R^2$ to equal $R$.

Theorem. Let $R$ be a ring, not necessarily with identity, not necessarily commutative. If $R^2=R$, then every maximal ideal of $R$ is also a prime ideal. If $R^2\neq R$, then any ideal that contains $R^2$ is not a prime ideal. In particular, if $R^2\neq R$ and there is a maximal ideal containing $R^2$, this ideal is maximal but not prime.

Proof. Suppose that $R^2=R$. Let $\mathfrak{M}$ be a maximal ideal of $R$; by assumption, we know that $\mathfrak{M}\neq R$. Now assume that $\mathfrak{A},\mathfrak{B}$ are two ideals such that $\mathfrak{A}\not\subseteq \mathfrak{M}$ and $\mathfrak{B}\not\subseteq\mathfrak{M}$. We will prove that $\mathfrak{AB}$ is not contained in $\mathfrak{M}$ (we are proving $\mathfrak{M}$ is prime by contrapositive). Then by the maximality of $\mathfrak{M}$, it follows that $\mathfrak{M}+\mathfrak{A}=\mathfrak{M}+\mathfrak{B}=R$.

Then we have: $$\begin{align*} R &= R^2\\ &= (\mathfrak{M}+\mathfrak{A})(\mathfrak{M}+\mathfrak{B})\\ &= \mathfrak{M}^2 + \mathfrak{AM}+\mathfrak{MB}+\mathfrak{AB}\\ &\subseteq \mathfrak{M}+\mathfrak{M}+\mathfrak{M}+\mathfrak{AB}\\ &=\mathfrak{M}+\mathfrak{AB}\\ &\subseteq R, \end{align*}$$ hence $\mathfrak{M}\subsetneq\mathfrak{M}+\mathfrak{AB}=R$. Therefore, $\mathfrak{AB}\not\subseteq\mathfrak{M}$. Thus, $\mathfrak{M}$ is a prime ideal, as claimed.

Now suppose that $R^2\neq R$ and $\mathfrak{I}$ is an ideal of $R$ that contains $R^2$. If $\mathfrak{I}=R$, then $\mathfrak{I}$ is not prime. If $\mathfrak{I}\neq R$, then $RR\subseteq \mathfrak{I}$, but $R\not\subseteq \mathfrak{I}$, so $\mathfrak{I}$ is not prime. In particular, if $\mathfrak{M}$ is a maximal ideal containing $R^2$, then $\mathfrak{M}$ is not prime. $\Box$

In your example, we have $R=2\mathbb{Z}$, $R^2=4\mathbb{Z}\neq R$, so any ideal that contains $R^2$ (in particular, the ideal $R^2$ itself) is not prime. And since $4\mathbb{Z}$ is a maximal ideal containing $R^2$, exhibiting a maximal ideal that is not prime. (In fact, $2\mathbb{Z}$ has maximal ideals containing any given ideals; this can be proven directly, or invoking the fact that it is noetherian)

  • $\begingroup$ This is delightful! $\endgroup$ – Manny Reyes Jul 26 '13 at 14:39
  • $\begingroup$ Minor note, much later: the term for "strongly prime" or "totally prime" which I thought is most common is not mentioned: "completely prime ideal." $\endgroup$ – rschwieb Feb 21 '18 at 14:46

If $R$ is a commutative ring such that $R^2=R$ (in particular if R has an identity), then every maximal ideal is prime.


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