This question already has an answer here:

The question formuled in the exam was exactly:

''Every maximal ideal is a prime ideal''

Maximal and prime ideals are defined for a commutative ring R, but the proof I have for maximal ideal $\Rightarrow$ prime ideal needs that R is a commutative and unitary ring, because it uses that $R/I$ is field iff $I$ is maximal, and that only happens if R is both commutative and unitary.

Then my question is:

Is there a proof for maximal ideal $\Rightarrow$ prime ideal for R a commutative ring (not neccesarily unitary)?

If not, there must be a counterexample with a maximal ideal which isn't prime in a R commutative (not unitary) ring, right?

I know there are more posts like this, but the difference is that those posts suppose R as a commutative and unitary ring (that's the difference with this one, just to be clear).


marked as duplicate by MooS, Dietrich Burde abstract-algebra Sep 6 '17 at 14:01

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  • $\begingroup$ This is not true, see the remark in this question. $\endgroup$ – Dietrich Burde Sep 6 '17 at 13:59
  • $\begingroup$ $R=2\mathbb{Z}, I=4\mathbb{Z}$. Here $I$ is a maximal ideal, but not a prime ideal. If the ring contains unity, then every maximal ideal is a prime ideal. $\endgroup$ – Krish Sep 6 '17 at 14:00
  • $\begingroup$ @MooS Amusing that the question there is the solution to this question, and the answers there are kind of like the statement of the question here... $\endgroup$ – rschwieb Sep 6 '17 at 14:03
  • $\begingroup$ "In a commutative ring $R,$ the ideal $I$ is prime if and only if the quotient ring $R/I$ is an integral domain." From this result you can deduce what you want. $\endgroup$ – Bumblebee Sep 6 '17 at 14:03

No, there is no proof.

For example, in $2\mathbb Z/4\mathbb Z$, the zero ideal is obviously maximal, but the zero ideal is also obviously not prime.


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