Definition 1

An ideal $p$ of a ring $A$ is called prime if the ring $A/p$ is an integral domain.

By this definition, an ideal $p$ of a ring $A$ is prime if the following two conditions hold: (1) $p \neq A$; (2) if $x, y$ are two elements of $A$ such that $x \notin p$ and $y \notin p$, then $xy \notin p$.

These conditions can also be expressed by saying that the product of any finite family of elements of $\mathfrak{g}p$ belongs to $\mathfrak{g}p$, as applying this condition to the empty set yields $1 \notin p$.

A maximal ideal $m$ of $A$ is prime since $A/m$ is a field; then it follows from Krull's theorem (Algebra, Chapter I, § 8, no. 7, Theorem 2) that every ideal of

() With the exception of the statements placed between two asterisks: \ ... *, the results of this chapter depend only on Books I to VI and Chapter I, §§ 1–3 of this Book.

A other than $\mathbf{A}$ is contained in at least one prime ideal. In particular, for prime ideals to exist in a ring $\mathbf{A}$, it is necessary and sufficient that $\mathbf{A}$ be not reduced to 0.

Let $f : \mathbf{A} \to \mathbf{B}$ be a ring homomorphism and $q$ an ideal of $\mathbf{B}$. Set $p = f^{-1}(q)$; the homomorphism $\overline{f} : \mathbf{A}/p \to \mathbf{B}/q$ derived from $f$ by taking quotients is injective. Suppose that $q$ is prime; as the ring $\mathbf{B}/q$ is an integral domain, so is $\mathbf{A}/p$, being isomorphic to a subring of $\mathbf{B}/q$; consequently the ideal $p = f^{-1}(q)$ is prime. In particular, let $\mathbf{A}$ be a subring of $\mathbf{B}$; for every ideal $q$ of $\mathbf{B}$, $q \cap \mathbf{A}$ is a prime ideal $\mathbf{A}$. If $f$ is surjective, $\overline{f}$ is an isomorphism; the conditions "p is prime" and "$q$ is prime" are then equivalent. Hence, if $p$ and $a$ are ideals of $\mathbf{A}$ such that $a \subset p$, a necessary and sufficient condition for $p$ to be prime is that $p/a$ be prime in $\mathbf{A}/a$.