# 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

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 \(\mathbb{Q}\).

*blynn@cs.stanford.edu*💡