## Prime and Maximal Ideals

An ideal $P$ of $R$ is called prime if $P\ne R$ 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 $\mathbb{Z}$ are $\{0\}$ and $p\mathbb{Z}$ for $p$ prime.

An ideal $M$ of $R$ is maximal if $M\ne R$ 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 $\mathbb{Z}$ but not maximal.

Proposition: Let $f:R\rightarrow S$ 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 $\phi$ be the natural map $S \rightarrow S/Q$. Then

$ker (\phi f) = \{x\in R | f(x) \in Q\} =f^{-1}(Q)$

so $R/f^{-1}(Q) = R/ker(\phi f)$ 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 $f:\mathbb{Z}\rightarrow\mathbb{Q}$. Then $\{0\} = f^{-1}(\{0\})$ is not maximal in $\mathbb{Z}$, though it is maximal in $\mathbb{Q}$.