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If the quotient ring $R/I$ is an integral domain, then $I\ne R$, because an integral domain has $1\ne 0$. For $ab\in I$,There are no nonzero zero divisors in an integral domain, so $a+I=0+I$ or $b+I=0+I$. Thus $a\in I$ or $b\in I$, proving $I$ is a prime ideal.

If $R/I$ is a field, again $I$ is proper. Let $J$ be an ideal properly containing $I$, and choose $a\in J\setminus I$. The nonzero element $a+I$ has a multiplicative inverse $b+I$, so $ab-1\in I\subset J$. Since $ab\in J$, subtraction gives $1\in J$, hence $J=R$. Therefore $I$ is a maximal ideal. These arguments establish the relevant prime ideal quotient criterion and maximal ideal quotient criterion.

By the prime ideal quotient criterion,For a principal ideal, this is precisely the assertion that $p$ is a prime element. This closes the cycle and proves that (i)--(v) are equivalent.

Now let $\phi :R\to S$ be a surjective ring homomorphism, where $R$ is a principal ideal domain and $S$ is an integral domain. WriteIf $a=0$, the first isomorphism theorem for rings makes $\phi$ an isomorphism. Otherwiseis an integral domain, so the equivalences just proved show that $(a)$ is maximal and $S$ is a field.

Next suppose that the polynomial ring $R[X]$ is a principal ideal domain. It follows first that $R$ is an integral domain. For nonzero $a\in R$, the idealis principal. Since $f$ divides the nonzero constant $a$, it is constant; since it also divides $X$, that constant divides the coefficient $1$, so $f$ is a unit. Therefore $(a,X)=R[X]$, and there are polynomials $g,h$ such thatSetting $X=0$ gives $1=ag(0)$, so every nonzero $a\in R$ is a unit. Hence

Finally, let $R$ be an integral domain in which every two nonzero elements have a greatest common divisor. If an irreducible $p$ divides $ab$, then $\gcd (p,a)$ is either a unit or an associate of $p$. In the second case $p\mid a$. In the first case, the Euclid lemma in a greatest-common-divisor domain gives $p\mid b$. Thus every irreducible element of $R$ is prime.

Suppose $I$ is prime. Ifthen $ab\in I$, so $a\in I$ or $b\in I$. Thus one factor is zero and $R/I$ is an integral domain.

Conversely, if $R/I$ is an integral domain and $ab\in I$, thenso $a+I=0+I$ or $b+I=0+I$. Therefore $a\in I$ or $b\in I$, and $I$ is prime. This proves the prime ideal quotient criterion.