Past exam of the algebra and analysis II for mathematics and philosophy course of the University of Oxford / 2013 / 3 / d

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 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 $r=s=0$.A ring may therefore have zero divisors without having nonzero nilpotent elements.