Quotient ring of principal ideal domain is principal ideal.

$R$ : principal ideal domain, $I$ : ideal of $R$.

Since that $R$ is the principal ideal domain, every ideal of $R$ is principal ideal.
Let such ideal be $H$. Also, let $J =\{a \in R \mid a + I \in H\}$ (Could be interpreted as the set of the representatives of cosets in $R/I$).

Let's prove that $J$ is the ideal of $R$.
Since that ring must have the additive identity, (which implies that the ring is non-empty set) we can say that $0 + I \in H, 0 \in J$.
Also, $a, b \in J, \forall a + I, b + I \in H$ which implies $(a + I) - (b + I) = (a - b) + I \in H$. So, $a - b \in J$.
Since that $H$ is the ideal of $R/I$, $ar + I, ra + I \in H, \forall a \in J, \forall r \in R$ which implies $ar, ra \in J$. Therefore, we can conclude that $J$ is the ideal of $R$.

Next, let's prove $H = <j + I>, \exists j \in J$.
$(\Longleftarrow)$
    Obiviously, $<j + I> \subseteq H, \exists j \in J$.

$(\Longrightarrow)$
    If $a + I \in H$, then $a \in J$.
    Then $a = rj, \exists r \in R, \exists j \in J$, because if $R$ is the principal ideal domain, $J =          <j>, \exists j \in R$.
    Hence, $a + I = (r + I)(j + I) \implies a + I \in <j + I> \implies H \subseteq <j + I>$.

$\therefore H$ is principal ideal and $R/I$ is principal ideal domain.