= Solution
{c}
Set $X=0$ to define a surjective <ring homomorphism>
$$
E:\mathbb Q[X,Y]\longrightarrow\mathbb Q[Y],\qquad E(P)=P(0,Y).
$$
Its <kernel of a ring homomorphism> is exactly $(X)$: the constant coefficient in $X$ must vanish, so every remaining monomial is divisible by $X$. The <first isomorphism theorem for rings> therefore gives
$$
\boxed{\mathbb Q[X,Y]/(X)\cong\mathbb Q[Y].}
$$
The <polynomial ring> $\mathbb Q[Y]$ is an <integral domain>, because the product of two nonzero <polynomials> has nonzero leading coefficient. It is not a <field>: $Y$ has no polynomial <multiplicative inverse>, since a nonzero product $YQ(Y)$ has positive <degree of a polynomial>. Consequently \b[$(X)$ is prime and is not maximal]. Concretely, $(X)\subsetneq(X,Y)\subsetneq\mathbb Q[X,Y]$ is a strict chain of <ideals>.
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