Basic Ring Theory - 2

[Tsakanikas Nickos]

Let \( \displaystyle A \) be a non-zero associative and commutative ring with identity \( \displaystyle 1_{A} \). Prove the following:

1. Every maximal ideal of \( \displaystyle A \) is a prime ideal of \( \displaystyle A \).
2. If \( \displaystyle A \) is a principal ideal domain, then every non-zero prime ideal of \( \displaystyle A \) is, in fact, a maximal ideal of \( \displaystyle A \).

[Papapetros Vaggelis]

1. If \(\displaystyle{M\neq A}\) is a maximal ideal of \(\displaystyle{A}\) , then the ring

\(\displaystyle{A/M}\) is a field, so it is integral domain, which means that \(\displaystyle{M}\) is prime.

[Tsakanikas Nickos]

Let \( I \) be a non-zero prime ideal of \( A \). By hypothesis, \[ \exists \ x \in A \ : I = (x) \]Let \( J \) be an ideal of \( A \) such that \( I \subsetneq J \subseteq A \). Again by hypothesis, \[ \exists \ y \in A \ : J = (y) \]Since \( x \in (x) \subset (y) \), \[ \exists \ z \in A \ : x = yz \]So \( yz \in (x) \), and since \( y \notin (x) \) and \( (x) \) is a prime ideal, \( z \in (x) \). So \[ \exists \ w \in A \ : z = xw \]and therefore \[ x = yz = yxw = ywx \implies yw = 1 \implies 1 \in (y) \implies (y)=(1)=A \](the first implication, because \( A \) is an integral domain, so the cancellation law holds). This implies that \( I = (x) \) is a maximal ideal of \( A \).