# ideal behind Maximal ideals are Prime ideals

In a set of old lecture notes, I came across this corollary:

Corollary: Any maximal ideal is a prime ideal.

While I do not have the proof for this, what immediately comes to mind is whether this corollary is predicated on the fact that...

Theorem: For any commutative ring $$R$$ and ideal $$A$$, $$R/A$$ is an integral domain iff $$A$$ is a prime ideal

Theorem: For any commutative ring $$R$$ and ideal $$A$$, $$R/A$$ is an field iff $$A$$ is a maximal ideal

Recall: Any finite integral domain is a field.

So, from the two theorem above, we can reconcile the fact that a maximal ideal is a prime ideal (and vice versa) on the supposition that the integral domain in question is finite.

Or is there another proof for the claims in my notes?

• I don't know about whether there are other proofs in your notes, but this is one very standard way to prove that maximal ideals are prime. There is no need to suppose that the integral domain in question is finite. – Arthur Jul 24 '17 at 13:04
• A field is always an integral domain. Why do you mention finiteness? – Cave Johnson Jul 24 '17 at 13:04
• Maximal ideals are always prime but prime ideals are not always maximal since integral domains are not always fields. – walkar Jul 24 '17 at 13:05
• @CaveJohnson Because a finite integral domain is a field. Not all integral domains are necessarily finite. – Mathematicing Jul 24 '17 at 13:05
• But yes, if $p\subseteq R$ is a prime ideal, and $R/p$ is finite, then $p$ is indeed maximal by your argument above. – Arthur Jul 24 '17 at 13:06

We want to show: If $R$ is a commutative ring and an ideal $A \subseteq R$ is maximal then $A$ is prime.
If $A$ is maximal then $R/A$ is a field. Since $R/A$ is a field, then $R/A$ is an integral domain. Since $R/A$ is an integral domain, then $A$ is prime.