Let $R$ a finite commutative unital ring.

Prove that an ideal of $R$ is irreducible if and only if it is prime.

It is clear that prime ideals are always irreducible.

But when irreducibility implies prime. It is known that it is true if $R$ is a PID. However, is it sufficient for $R$ to be a finite commutative unital ring?

Let $P$ an irreducible ideal such that $P=I_1\cap I_2$ for some $I_1,I_2$ ideals of $R$. Since $P$ is irreducible, $P=I_1$ or $P=I_2$.

If $P=I_1$, in particular, $I_1\subseteq P$. And if $P=I_2$, in particular, $I_2\subseteq P$. Therefore $P$ is prime.

But we have supposed that $P$ is the intersection of two ideals. Is this always possible/true in a finite commutative unital ring?

  • $\begingroup$ Let $P$ be irreducible. If it follows that $P$ is prime, then $R/P$ must be an integral domain. You might try showing that if $P$ is not irreducible then it cannot be prime. Use the quotient by $P$ and my opening remark. $R$ is finite, so $R/P$ is, too. If $R/P$ is a domain, then ... . $\endgroup$ – Chris Leary May 10 '20 at 14:12
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    $\begingroup$ @ChrisLeary I have proved that if $P$ is prime, then is irreducible. So we can say that If $P$ is not irreducible, then it is not prime or we cannot say that? How can we suppose that $R/P$ is a domain if we don't know about zero divisors? $\endgroup$ – Claudia May 10 '20 at 23:36
  • $\begingroup$ My comment was a bit garbled. Prime implies irreducible. You need to prove irreducible implies prime. I was thinking that you could try to show $R/P$ is a field, assuming $P$ irreducible and knowing $R$ is finite. At the moment, I'm not sure I had the correct idea how things would go. Anyway, not irreducible implies not prime, but that's not what you want to prove $\endgroup$ – Chris Leary May 11 '20 at 3:04

No, it isn't.

In $\mathbb Z/4\mathbb Z$, the ideal $4\mathbb Z/4\mathbb Z$ is meet-irreducible but not prime.


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