In a commutative ring with 1 if every non zero ideal is prime then is every prime ideal maximal? I make no other assumptions. Please help if further assumptions like noetherianness make the problem easier. However I would like an answer without any assumptions.

  • $\begingroup$ If there are two different maximal ideals $I,J$ then $(I,J) = (1)$ $\endgroup$ – reuns Jun 6 '17 at 19:29
  • 3
    $\begingroup$ Z/9Z isnt a field...but it follows my restrictions $\endgroup$ – CoffeeCCD Jun 6 '17 at 19:34
  • $\begingroup$ @RonitDebnath Strictly speaking you should say something like "every nontrivial ideal". Saying "every nonzero ideal" does not eliminate $R\lhd R$, and a ring is never a prime ideal in itself. $\endgroup$ – rschwieb Jun 7 '17 at 13:49
  • $\begingroup$ @RonitDebnath I think I managed a rather simple classification below. Let me know what you think. $\endgroup$ – rschwieb Jun 7 '17 at 14:11
  • $\begingroup$ @rschwieb oops I completely overlooked the "nonzero ideal" part of the question :-p $\endgroup$ – Alex Macedo Jun 7 '17 at 14:28

It is true without further assumptions: If $P \subsetneq M$ are two prime ideals, we have that $M/P$ is non-zero prime ideal of the integral domain $R/P$. In particular the square of any element generates a non-prime ideal in $R/P$, which gives rise to non-prime ideal in $R$.


Another way to look at this:

Suppose $R$ is such a ring and let $N \subseteq R$ be the nilradical. Let $x \notin N$ be a non-unit. Then $(x^2)$ and $(x)$ are both prime ideals, and $x \cdot x \in (x^2)$ then implies there is some $a \in R$ such that $ax^2=x$. It is clear that this property still holds if $x$ is a unit, so $R/N$ is absolutely flat (or von Neumann regular). This is equivalent to every prime ideal of $R$ being maximal (see Atiyah-MacDonald, Chapter 3, Exercise 11).


Lemma: If every proper ideal of a commutative ring $R$ is prime, $R$ is a field.


Proposition: If every nontrivial ideal of a commutative ring $R$ is prime, the ring is either field, a product of two fields, or a uniserial ring with exactly three ideals.

Proof of proposition: In case $\{0\}$ is prime, the Lemma immediately yields that it is a field. Suppose hereafter that nontrivial ideals exist.

Let $I$ be any nontrivial ideal of $R$. Then $R/I$ satisfies the lemma, and $I$ is a maximal ideal. It should also be apparent that $I$ is a minimal ideal: if there is another nontrivial ideal $I'\subseteq I$, then $R/I'$ being a field implies $I'=I$.

Thus every nontrivial ideal of $R$ is maximal and minimal. Obviously we can conclude now that $R$ is Artinian.

If there is only one maximal ideal $I$, then obviously we are in the three-ideal uniserial ring case.

If there are at least two maximal ideals, $I, I'$, then by minimality and maximality of $I$, $I+I'=R$ and $I\cap I'=\{0\}$. The Chinese remainder theorem yields that $R\cong I\oplus I'$, where each of $I$ and $I'$ are fields.

N.B. Thanks to Moos for suggesting a simplification. My original take on it was to consider the Jacobson radical. Either it was maximal (and then the ring was uniserial) or it was $\{0\}$, and the ring was semisimple, decomposing into only two fields. I like that too but Moos' suggestion is a little more elementary.

Corollary: In the context of the proposition, all prime ideals are maximal, and moreover there are only $0$, $1$, or $2$ nontrivial ideals.

  • $\begingroup$ Seems correct. After you pointed out that any non-trivial ideal is maximal and minimal, you could have finished the job as follows: If there is only one non-trivial ideal, you have the chain ring case. If there are at least two, say $I,J$, we clearly have $I+J=R$ and $I \cap J =0$, i.e. the Chinese remainder theorem yields $R=R/I \times R/J$ is a product of two fields. $\endgroup$ – MooS Jun 7 '17 at 14:55
  • $\begingroup$ @MooS I adapted to your suggestion, if you don't mind. Thanks! $\endgroup$ – rschwieb Jun 7 '17 at 15:03
  • $\begingroup$ Of course, I do not mind. Glad I could simplify argument slightly, though it was already smooth and nice before that :) $\endgroup$ – MooS Jun 7 '17 at 15:11

Lemma If every proper ideal of $R$ is prime then $R$ is a field.

Since the zero ideal is prime, $R$ is a domain. If $x \ne 0, \in R$ then either $x$ is invertible or $(x)$ and $(x)^2$ are both prime. Thus $(x) \subseteq (x)^2$. It follows that $x = yx^2$ for some $y \in R$ and hence $1 = yx$ since $R$ is a domain.

Corollary If every nonzero, proper ideal of $R$ is prime then $R$ is Artinian (every prime ideal is maximal)

If $I$ is a nonzero, proper ideal of $R$ then every proper ideal of $R/I$ is prime by the correspondence theorem. Thus $R/I$ is a field by the Lemma. So $R$ is Artinian.


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.