The question I am working on defines some algebraic number $\zeta$. It is asking for me to find a polynomial $P(x)\in\mathbb{Q}[x]$ such that $\mathbb{Q}[\zeta]\cong\mathbb{Q}[x]/P(x)\mathbb{Q}[x]$.
Now this looks an awful lot like the First Isomorphism Theorem to me, where $P(x)\mathbb{Q}[x]$ is the kernel of our homomorphism $\mathbb{Q}[x]\rightarrow \mathbb Q[\zeta]$, which is essentially the evaluation of $\mathbb Q[x]$ at $\zeta$.
But the kernel of that map would just be any rational polynomial with one root being $\zeta$, would it not? Am I oversimplifying the scenario here or is this really just a fancy way of asking for the minimal polynomial of $\zeta$?
Here is an example showing that you do indeed need the polynomial to be irreducible. Consider $\mathbb{Q}[i]$. As $i$ is a root of $x^2+1$, and as $x^2+1$ is irreducible over $\mathbb{Q}$, it can be shown that $$ \mathbb{Q}[i]\cong\mathbb{Q}[x]/(x^2+1). $$ What if we tried another polynomial with $i$ as a root, such as $(x^2+1)^2$? Is it true that $$ \mathbb{Q}[i]\cong\mathbb{Q}[x]/((x^2+1)^2)? $$ Note that any polynomial in the ideal $((x^2+1)^2)$ which is not a constant polynomial has degree at least $4$. Thus $x^2+1$ is not in the ideal. However, $(x^2+1)^2$ is in the ideal. Thus in $\mathbb{Q}[x]/((x^2+1)^2)$, we have a nonzero element $a$ such that $a^2=0$. Thus $\mathbb{Q}[x]/((x^2+1)^2)$ is not an integral domain. However, it is easy to see that $\mathbb{Q}[i]$ is an integral domain. Thus $$ \mathbb{Q}[i]\not\cong\mathbb{Q}[x]/((x^2+1)^2), $$ exemplifying what can go wrong when the polynomial is not irreducible.