I was going through Exercises in Dummit&Foote, which does not assume identity in the definition of a ring, and reached the following exercise:

Prove that in a Boolean ring ($a^2 = a$ for all $a$) every prime ideal is a maximal ideal.

This was relatively easy assuming identity, but isn't so clear without identity. My questions,then, are:

  • Can this be solved without assuming identity?
  • The proof that maximal ideals are prime requires identity. It states that if $R$ is a commutative ring with identity and $I$ is a maximal ideal, then $R/I$ is a field, and thus an integral domain. Therefore, $I$ is a prime ideal. Does something similar apply to commutative rings without identity? Are maximal ideals necessarily prime in such rings? If so, what is an example of such a ring?

To address your last question: It is not true in general that maximal ideals are necessarily prime in (commutative) rings without identity.

Consider the ring without identity $R=2\mathbb{Z}$. The ideal $4\mathbb{Z}$ is maximal in $R$ (since it has prime index as a subgroup), but it is not prime: $2\times 2\in 4\mathbb{Z}$, but $2\notin 4\mathbb{Z}$.

On the other hand,

Proposition. Let $R$ be a ring, not necessarily commutative, not necessarily with identity, such that $RR=R$ (in particular, this holds if $R$ has an identity). If $\mathfrak{M}$ is a maximal ideal, then $\mathfrak{M}$ is a prime ideal; that is, if $\mathfrak{AB}\subseteq \mathfrak{M}$ for ideals $\mathfrak{A}$ and $\mathfrak{B}$, then $\mathfrak{A}\subseteq\mathfrak{M}$ or $\mathfrak{B}\subseteq \mathfrak{M}$.

Proof. Let $\mathfrak{M}$ is a maximal ideal of $R$, and $\mathfrak{A},\mathfrak{B}$ are ideals such that neither $\mathfrak{A}$ nor $\mathfrak{B}$ are contained in $\mathfrak{M}$; we will show that $\mathfrak{AB}$ is not contained in $\mathfrak{M}$. Indeed, maximality of $\mathfrak{M}$ gives $\mathfrak{A}+\mathfrak{M}=\mathfrak{B}+\mathfrak{M} = R$, so $$R = RR = (\mathfrak{A}+\mathfrak{M})(\mathfrak{B}+\mathfrak{M}) = \mathfrak{AB}+\mathfrak{AM}+\mathfrak{MB}+\mathfrak{MM}\subseteq \mathfrak{AB}+\mathfrak{M}\subseteq R,$$ so $\mathfrak{AB}+\mathfrak{M}=R$, hence $\mathfrak{AB}$ is not contained in $\mathfrak{M}$. $\Box$

The condition that $RR=R$ is a bit tricky. There are rings in which this does not hold but the implication holds anyway: for example, take an abelian group that has no maximal ideals (e.g., $\mathbb{Q}$), and make it into a ring by defining $ab=0$ for all $a,b$. Then ideals are subgroups, and the absence of maximal ideals means that the implication holds by vacuity. If $RR\neq R$ and there is a maximal ideal that contains $RR$, then that ideal will be a witness to the implication not holding, as occurs for example above with $R=2\mathbb{Z}$.

  • $\begingroup$ Very nice and thorough answer! $\endgroup$ – Alex Becker Jan 2 '12 at 3:31
  • $\begingroup$ Great, that result was exactly the type of thing I was wondering about. Thanks to those who answered the first question as well. $\endgroup$ – Carl Jan 2 '12 at 4:41

You don't need an identity. Let $P$ be a prime ideal and let $I$ be an ideal which strictly contains $P$. I will show that $I$ is the whole ring.

Let $i \in I - P$. For any element $r$ in the ring, $$ i(r-ri) = ri-ri^2 = ri - ri = 0 \in P. $$ Since $P$ is a prime ideal, this implies that $r-ri$ is in $P$. Since $ri \in I$, this shows that $r = (r-ri)+ri \in I$.


Neither do you need to assume that the ring is unital nor the ring is commutative.

Definition: Those rings where for each $x \in R$ there is a positive integer $n(x) >1$ (depending on $x$) s.t. $x^{n(x)}=x,$ are called Jacobson-rings or J-rings.

General result due to Jacobson: J-rings are commutative. For the proof, you may take a look at Non-commutative Rings by Herstein)

In any J-ring, every prime ideal is maximal.

Proof: Let $p$ be a prime ideal which is not maximal. So there is a maximal ideal $m$ of $R$ s.t. $p \subsetneq m \subsetneq R.$ Let $x \in m\setminus p.$ By the given property, there is a natural number $n>1$ s.t. $x^n=x$ or $x^n-x=0.$ Let $y \in R$ be arbitrary. Then $x^ny-xy=0 \to x(x^{n-1}y-y)=0 \in p$, therefore, either $x \in p$ or $x^{n-1}y-y \in p.$ Now $x \in m\setminus p$, we conclude that $x^{n-1}y-y \in p \subsetneq m.$ We know that $x \in m$ therefore, $y \in m.$ Since $y$ was an arbitrary element of $R$ then $m=R$ which is a contradiction and we're done.



Using your answer it would come out to this. Which looks good to me

Let $R$ be a Boolean ring i.e. $x^2=x$

Let $P$ be a prime ideal such that it is not maximal and $M$ be a maximal Ideal such that $P \subsetneq M \subsetneq R$

Further let $y \in M\backslash P$. Now we know $y^2=y \implies y^2 - y = 0$ now let $z$ be an arbitrary element in $R$. Then $y^2z-yz= 0 \implies y(yz-z) = 0 \in P$ thus either $y \in P$ or $yz-z \in P$ But we claimed $y \in M \backslash P$ so we can conclude that $yz-z \in P$ and we know $y \in M$ but further $z \in P \subsetneq M \implies z \in M$ and $z$ was arbitrarily chosen thus $M=R$, we have a contradiction.

  • 1
    $\begingroup$ The last line is a bit confusing. I think you mean that from $P \subseteq M$, then we get $yz - z \in M$, and then since $yz \in M$ this implies $z \in M$. But $z$ was arbitrarily chosen etc etc $\endgroup$ – Carl Apr 8 '15 at 5:07
  • $\begingroup$ I think I have a hole. If $yz-z \in P$ how can I know for sure $z \in P$ for example: $2Z$ is a prime ideal and $2 \cdot 5 -5 \in 2\mathbb{Z}$ but $5 \notin 2\mathbb{Z}$ $\endgroup$ – oliverjones Apr 8 '15 at 23:08

You don't assume identity; instead, recall that a Boolean ring is defined as having identities. The definition given in parentheses in this problem statement is just a simplification of the specification.

  • $\begingroup$ I think there is a misunderstanding. In the question, "identity" refers to a multiplicative identity element, not to the definition of Boolean ring. Not every Boolean ring has a multiplicative identity, e.g. $\mathbb Z_2\oplus \mathbb Z_2\oplus\mathbb Z_2\oplus\cdots$. $\endgroup$ – Jonas Meyer Jan 2 '12 at 4:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.