= Solution
Because $g,h$ are nonconstant, their classes in $\mathbb Q[x]/(gh)$ are nonzero: each has smaller <degree> than $gh$, so neither is a multiple of $gh$. Their product is zero. Thus \b[the <quotient> has <zero divisors>].
If both factors have <degree> one, each <quotient> by one factor is <isomorphic> to $\mathbb Q$, by evaluation at its rational <root>. <Coprimality> ensures the two roots are distinct. Part (c) therefore identifies the ring with $\mathbb Q\times\mathbb Q$. If $(r,s)^2=(0,0)$, then $r^2=s^2=0$ in a <field>, so $r=s=0$.
$$
\boxed{\text{There is no nonzero square-zero element when both factors have degree one.}}
$$
A ring may therefore have <zero divisors> without having nonzero <nilpotent> elements.
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