Finding all polynomials $p(x)$ such that $ \mathbb{Q}[x]/p(x) \simeq \mathbb{Q}(A)$ for fixed $A$ algebraic As an example, if we put $F =\mathbb{Q}(\sqrt{2}, \sqrt{3}, ... , \sqrt{n})$, $F$ is the splitting field of $p(x)$ so that we can write
$$
p(x) = (x^2 - 2)(x^2 - 3)\cdots (x^2 - n).
$$
Question: if it is always true the isomorphism above (I'm not sure there is a counter-example for a general case), how can I find all polynomials $p\in\mathbb{Q}[x]$ such that $\mathbb{Q}[x]/p(x) \simeq \mathbb{Q}(A)$ for a fixed algebraic $A$ (or a series of extensions for that matter)?
Are they the [irreducible] polynomials which are multiples of $p(x)$, when $p(x)$ is such that $\mathbb{Q}(A)$ is its splitting field?
 A: Here are some facts:


*

*Not every finite algebraic extension is a splitting field. For instance $\mathbb Q(\sqrt[3]{2})$ is not a splitting field.

*Every finite algebraic extension is of the form $\mathbb Q(\theta)$, for some algebraic number $\theta$, and so is isomorphic to $\mathbb{Q}[x]/p(x)$ for some irreducible polynomial $p$. This is the contents of the primitive element theorem.
For simplicity, I'm talking here about finite algebraic extensions of $\mathbb Q$.
A: By the Primitive Element Theorem, an algebraic extension $F$ of $\mathbb{Q}$ is of the form $\mathbb{Q}(\alpha)$ for some $\alpha\in\mathbb{C}$ if and only if $[F:\mathbb{Q}]$ is finite. Not all such extensions are given by splitting fields, however.
For $\alpha\in\overline{\mathbb{Q}}$, let $n=[\mathbb{Q}(\alpha)\colon\mathbb{Q}]$. Then $\mathbb{Q}[x]/(p(x))$ is isomorphic to $\mathbb{Q}(\alpha)$ if and only if $p(x)$ is irreducible over $\mathbb{Q}$, $\deg(p)=n$, and there is a root $\beta$ of $p(x)$ in $\mathbb{Q}(\alpha)$. There are going to be infinitely many such polynomials (for example, all $\beta=q_1\alpha+q_2$ with $q_1,q_2\in\mathbb{Q}$, $q_1\neq 0$ will do), and I suspect you’ll find it rather difficult to find all such polynomials. Even with something like $\mathbb{Q}(\sqrt{2},\sqrt{3})$, you have infinitely many elements of degree $4$ in that field, generally written as $q$-linear combinations of $1+\sqrt{2}$, $\sqrt{3}$, and $\sqrt{6}$. 
Of course, they can all be written as $q$-linear combinations of $1$, $\alpha,\ldots,\alpha^{n-1}$ (since these elements form a basis for $\mathbb{Q}(\alpha)$ over $\mathbb{Q}$ as a vector space), but not all such combinations work.
