I am stuck in a true/false question. It is

In a finite commutative ring, every prime ideal is maximal.

The answer says it's false.

Well what I can say is (Supposing the answer is right)

$(1)$ The ring can't be Integral domain since finite integral domain is a field.

$(2)$ There can't be unity in the ring since in that case the result would be true.(By the Theorem that if $R$ is a commutative ring with unity then an ideal $I$ is prime iff $R/I$ is Integral Domain)

$(3)$ All the elements are zero divisors since if there is at least one non- zero divisor, there will be a unity and so $(2)$ would follow.

So at the end, I am in search of a finite commutative with all elements as zero -divisors, having no unity and obviously a prime ideal in it which is not maximal.

What kind of strange looking ring is this (if possible) ? Any hints??

  • $\begingroup$ Actually, any non zero principal ideal is prime then. The ideal $\{ (0,0), (0,2) \}$ is also prime as $(0,2)$ is not a product of two elements, and the ideal $\{ (0,0), (0,2), (2,2) \}$ contains this, so not maximal either. $\endgroup$
    – Siddhartha
    Commented Aug 29, 2020 at 19:20
  • $\begingroup$ @Siddhartha Sorry ,I can't follow , why any non-zero principal ideal is prime? $\endgroup$ Commented Aug 29, 2020 at 19:31
  • $\begingroup$ The ideal $\{ (0,0), (2,0) \}$ is not prime since $(0,0) = (2,2)(2,2)$. Okay, then this is not an example. Sorry for the boo-boo. Let me think. $\endgroup$
    – Siddhartha
    Commented Aug 29, 2020 at 19:41
  • $\begingroup$ This is not true. Such a domain is called a Dedekind domain (holds for every nonzero ideal since clearly R/{0} will never be a field unless R is a field itself) $\endgroup$
    – uhhhhidk
    Commented Aug 30, 2020 at 2:36

2 Answers 2


There is no counterexample, because even if the ring has no identity, the quotients by primes must have identity.

Every nonzero finite ring without zero divisors has a multiplicative identity, so the quotient would in fact be a finite domain with identity, and hence a field.

  • 1
    $\begingroup$ I can’t see the interpretation which would make your t/f question “false”. $\endgroup$
    – rschwieb
    Commented Aug 30, 2020 at 2:37
  • $\begingroup$ @rschweib If $R$ is without unity , I can't seem to convince myself how would we formally write a unity in $R/I$ which would have been possible easily if $R$ did have the unity. So can you please at least prove the existence of unity in $R/I$ considering $I$ is prime and $R$ is finite commutative without unity. Also with respect to the link you provided, It's been never said (in the question) that the ring is without zero divisors ..so there may be an example with zero divisors. Kindly clarify these questions , sir before I accept your answer. $\endgroup$ Commented Aug 30, 2020 at 7:31
  • $\begingroup$ As for the interpretation you are asking , the statment is false means there should exist a finite commutative ring (possibly without unity) and a prime ideal in it which is not maximal. $\endgroup$ Commented Aug 30, 2020 at 7:34
  • $\begingroup$ @user710290 Firstly, I proved the existence in the link. You should read it. Secondly, I’m talking about the quotient $R/P$ for a prime ideal: by the definition of prime, it would have no zero divisors. Thirdly, I did not mean “interpretation of your question” I meant interpretation of your definitions. $\endgroup$
    – rschwieb
    Commented Aug 30, 2020 at 11:12
  • $\begingroup$ @user-492177: Choose an arbitrary non-zero element $a \in R/I$. Since $I$ is prime, $R/I$ has the property that $xy = 0$ implies $x=0$ or $y=0$. Thus the multiplication by $a$ map on $R/I$ is injective and, since $R$ and thus $R/I$ is finite, also surjective. This means there is an element $x \in R/I$ with $ax = a$. Multiplication by any $b \in R/I$ yields $axb = ab$, hence (by the injectivity) $xb = b$. This means $x$ is the unity. $\endgroup$
    – Thrash
    Commented Mar 9, 2021 at 23:30

The answer is false. $I$ is prime means $R/I$ is a domain. Which implies $R/I$ is a field which implies that $I$ is maximal.

  • $\begingroup$ You mean the statement is true? But if $R$ doesn't have a unity then how do we expect to have a unity in integral domain $R/I$ considering $I$ as prime $\endgroup$ Commented Aug 29, 2020 at 21:35
  • $\begingroup$ Finite domain always is a field hence it has unity. If you do not know that fact prove it as an exercise. $\endgroup$
    – markvs
    Commented Aug 29, 2020 at 21:38
  • 1
    $\begingroup$ Yes I know that but my question is if $R$ is just a commutative ring without unity then $I$ is prime ideal iff $R/I$ is domain...is this true? Isn't the unity required in $R$ for this result? $\endgroup$ Commented Aug 29, 2020 at 21:42
  • $\begingroup$ Yes it is the definition. $\endgroup$
    – markvs
    Commented Aug 29, 2020 at 22:13
  • 5
    $\begingroup$ I think this is just a silly miscommunication — it might be more clear if this answer said "the answer is incorrect". "The answer is false" is a bit ambiguous, because it might mean "the correct answer to the question 'In a finite commutative ring, every prime ideal is maximal.' is 'false'" or it might mean "the answer OP was given to the question 'In a finite commutative ring, every prime ideal is maximal.' is incorrect" (which is the negation of the previous interpretation). $\endgroup$ Commented Aug 30, 2020 at 2:20

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