= Solution
By the <prime ideal quotient criterion>,
$$
R/(p)\text{ is an integral domain}
\quad\Longleftrightarrow\quad
(p)\text{ is a prime ideal}.
$$
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 $\varphi:R\to S$ be a surjective ring homomorphism, where $R$ is a <principal ideal domain> and $S$ is an integral domain. Write
$$
\ker\varphi=(a).
$$
If $a=0$, the <first isomorphism theorem for rings> makes $\varphi$ an isomorphism. Otherwise
$$
S\cong R/(a)
$$
is 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 ideal
$$
(a,X)=(f)
$$
is 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 that
$$
1=ag(X)+Xh(X).
$$
Setting $X=0$ gives $1=ag(0)$, so every nonzero $a\in R$ is a unit. Hence
$$
\boxed{R\text{ is a field}}.
$$
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.
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