# Is an irreducible ideal always a prime ideal in a finite commutative ring?

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?

• 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 ... . – Chris Leary May 10 '20 at 14:12
• @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? – Claudia May 10 '20 at 23:36
• 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 – 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.