= Solution
<Gauss lemma for polynomials> states that the product of two <primitive polynomials> over a <unique factorization domain> is primitive. In particular, a primitive polynomial in $\mathbb Z[X]$ is reducible over $\mathbb Q$ exactly when it is reducible over $\mathbb Z$.
<Eisenstein criterion> says that a primitive polynomial
$$
F(X)=a_nX^n+\cdots+a_0\in\mathbb Z[X]
$$
is irreducible over $\mathbb Q$ if there is a prime $p$ such that $p\nmid a_n$, $p\mid a_j$ for every $j<n$, and $p^2\nmid a_0$. To prove it, suppose $F=GH$ with both factors of positive degree. After removing contents, Gauss's lemma lets us take $G,H\in\mathbb Z[X]$ primitive. Reduction modulo $p$ gives
$$
\overline F=a_nX^n=\overline G\,\overline H,
$$
so both reduced factors are monomials of positive degree. Their constant terms are therefore divisible by $p$, making $p^2$ divide $a_0=G(0)H(0)$, a contradiction.
An <algebraic integer> is a complex number satisfying a monic polynomial in $\mathbb Z[X]$. Let $\alpha$ be one and let $m_\alpha\in\mathbb Q[X]$ be its monic <minimal polynomial>. The usual proof using polynomial content shows that $m_\alpha\in\mathbb Z[X]$. The set
$$
I_\alpha=\{f\in\mathbb Z[X]:f(\alpha)=0\}
$$
is the kernel of the polynomial-evaluation <ring homomorphism>, hence an ideal. Every $f\in I_\alpha$ is divisible by $m_\alpha$ in $\mathbb Q[X]$. Because $m_\alpha$ is monic, polynomial division of $f$ by $m_\alpha$ takes place in $\mathbb Z[X]$, and the remainder must vanish by minimality. Therefore
$$
\boxed{I_\alpha=(m_\alpha)},
$$
where $m_\alpha$ is monic and irreducible.
For
$$
f(X)=X^4+2X^3-3X^2-4X-11,
$$
translation gives
$$
f(X+1)=X^4+6X^3+9X^2-15.
$$
This is Eisenstein at $p=3$, so $f(X+1)$ and therefore $f(X)$ are irreducible over $\mathbb Q$. Hence
$$
\boxed{\mathbb Q[X]/(f)\text{ is a field}}.
$$
Irreducibility and Gauss's lemma also show that $(f)$ is a <prime ideal> of $\mathbb Z[X]$, so
$$
\boxed{\mathbb Z[X]/(f)\text{ is an integral domain}}.
$$
It is not a field. The class of $2$ is nonzero but not a unit: reducing further modulo $2$ gives a nonzero quotient
$$
\mathbb F_2[X]/(\overline f),
$$
in which the image of $2$ is zero, whereas a unit must remain a unit under every unital quotient map. Thus
$$
\boxed{\mathbb Z[X]/(f)\text{ is not a field}}.
$$
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