# 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?

• Your question is a bit confusing on first reading. I think you mean "How to find all polynomials ... where $A$ is a set of algebraic numbers". Is that right, if so please fix. – Rob Arthan Mar 24 at 23:44
• Since $\mathbb{Q}(A)$ is a field, you need $(p(x))$ to be maximal in $\mathbb{Q}[x]$, which holds if and only if $p(x)$ is irreducible over $\mathbb{Q}$; in which case, $\mathbb{Q}[x]/(p(x))$ is isomorphic to $\mathbb{Q}(\alpha)$ where $\alpha$ is a root of $p(x)$ (in $\overline{\mathbb{Q}}$). So for a fixed algebraic number $A$, this holds if and only if $p(x)$ is the a nonzero scalar multiple of the irreducible polynomial of $A$ over $\mathbb{Q}$. – Arturo Magidin Mar 24 at 23:46
• “Wouldn’t the polynomial be irreducible over $\mathbb{Q}[x]$”? Techincally, it would be irreducible in $\mathbb{Q}[x]$, but for fields, we typically speak about a polynomial being irreducible over a field $k$ if its coefficients lie in $k$, and the polynomial is irreducible in $k[x]$. This is common nomenclature. – Arturo Magidin Mar 25 at 0:09
• What do you mean, “it happens that other irreducible polynomials can work”? No, for a particular algebraic number $\alpha$, no other polynomials will work except for the irreducible polynomial of $\alpha$ and its associates. – Arturo Magidin Mar 25 at 0:09