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The required property isFor a nonzero element $x$ of an integral domain $R$, this follows fromIndeed, $\overline{a}\overline{b}=0$ in the quotient ring exactly when $x\mid ab$. The prime element condition says this forces $x\mid a$ or $x\mid b$, exactly the absence of zero divisors in the quotient ring. The nonunit condition makes the quotient ring nonzero. Being an integral domain is preserved by every ring isomorphism.

With the usual definition of a prime element, $x\ne 0$ is understood. The prime ideal formulation works for every ideal of any commutative ring, including the zero ideal; it avoids this convention about elements.

Every multiple of $x_1\cdots x_r$ is a multiple of each $x_j$, giving one inclusion. For the reverse inclusion, a principal ideal domain is a unique factorization domain, so each irreducible element $x_j$ is a prime element.

Let $a\in {\bigcap }_{j=1}^r(x_j)$. The case $a=0$ is immediate. Otherwise, write $a=x_{1}b$. Since $x_2\mid a$ and the prime element $x_2$ cannot divide the non-associate irreducible element $x_1$, it divides $b$. Write $a=x_1{x}_2{b}_2$, and repeat. At stage $j$, the prime element $x_j$ divides none of $x_1,\dots ,x_{j-1}$, so it divides the remaining factor. Hence $x_1\cdots x_r\mid a$, proving

Let $R$ be a unique factorization domain and let $p$ be a prime element. The only prime ideals contained in the principal ideal $(p)$ are $(0)$ and $(p)$. To prove this, suppose $0\ne P⊊(p)$. Then $p\notin P$. For $0\ne f\in P$, unique factorization allows us to write $f=p^{k}g$ with $g\notin (p)$. Since $P$ is a prime ideal and $p\notin P$, repeated primality gives $g\in P\subseteq (p)$, a contradiction.

In the particular polynomial ring $\mathbb{Z}[t]$, when $p$ is a prime number, the same proof uses only the integer coefficients: factor out the largest common power of $p$ from the coefficients of a nonzero polynomial $f$. This gives $f=p^{k}g$ with $g\notin (p)$ directly, without requiring the general unique factorization domain theorem for polynomial rings.

Let $p\ge 3$ be a prime number dividing $n^2+3$, and set $\omega =(-1+i\sqrt{3})/2$. The case $p=3$ is $N(1-\omega )=3$. For $p>3$, useThe Eisenstein integer $p$ divides this product, but divides neither factor: divisibility by the rational integer $p$ in the basis $1,\omega$ requires both integer coefficients to be divisible by $p$, whereas the second coefficients are $2$ and $-2$. Thus $p$ is not a prime element of $\mathbb{Z}[\omega ]$.

The Euclidean norm on the Eisenstein integers makes this ring a principal ideal domain. Since irreducible elements of a principal ideal domain are prime, $p$ admits a factorization $p=uv$ into two nonunits. The positive integral Eisenstein norm is multiplicative, and its value is one exactly on units: when $N(u)=1$, its complex conjugate is an inverse. Thus $N(u),N(v)>1$ and $N(u)N(v)=p^2$, forcing both norms to be $p$. Writing $u=a+b\omega$ proves the displayed representation.

Every nonzero nonunit in a unique factorization domain is a finite product of irreducible elements. Such a factorization is unique up to permutation and multiplication of the factors by units. Irreducibles are prime elements, so divisibility can be checked by comparing their multiplicities. This applies to a polynomial ring over a field and is useful for intersections of localizations inside its fraction field.