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 $$\mathbb{Q}$$.

Ben Lynn blynn@cs.stanford.edu 💡