Commutative Algebra - Prime and Maximal Ideals

An ideal $P$ of $R$ is called prime if $P \neq R$ and for all $x, y \in R$ , if $x y \in P$ then $x \in P$ or $y \in P$ .

It is easily verified that if $P$ is a nonzero ideal, then $P$ is prime if and only if $R / P$ is an integral domain. In particular, ${0}$ is prime if and only if $R$ is an integral domain.

Example: The prime ideals of $ℤ$ are ${0}$ and $p ℤ$ for $p$ prime.

An ideal $M$ of $R$ is maximal if $M \neq R$ and there is no ideal $I$ such that $M \subset I \subset A$ where the inclusions are strict.

It it easily verified that if $M$ is a nonzero ideal then $M$ is maximal if and only if $A / M$ is a field. This implies all maximal ideals are prime. The converse is not true in general, for example ${0}$ is prime in $ℤ$ but not maximal.

Proposition: Let $f : R \to S$ be a ring homomorphism and $Q$ be a prime ideal of $S$ . Then $f^{- 1} (Q) = {x \in R ∣ f (x) \in Q}$ is a prime ideal of $R$ .

Proof: Let $ϕ$ be the natural map $S \to S / Q$ . Then $\ker (ϕ f) = {x \in R ∣ f (x) \in Q} = f^{- 1} (Q)$ so $R / f^{- 1} (Q) = R / \ker (ϕ f)$ and by the fundamental homomorphism theorem this is isomorphic to some subring of $S / Q$ . But $S / Q$ is an integral domain since $Q$ is prime, so $R / f^{- 1} (Q)$ is also an integral domain implying that $f^{- 1} (Q)$ is indeed prime.

Preimages of maximal ideals need not be maximal. For example, consider the identity injection $f : ℤ \to ℚ$ . Then ${0} = f^{- 1} ({0})$ is not maximal in $ℤ$ , though it is maximal in $ℚ$ .