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?

  • $\begingroup$ 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. $\endgroup$ – Rob Arthan Mar 24 at 23:44
  • $\begingroup$ 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 $\mathbb{Q}$. $\endgroup$ – Arturo Magidin Mar 24 at 23:46
  • $\begingroup$ “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. $\endgroup$ – Arturo Magidin Mar 25 at 0:09
  • $\begingroup$ 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. $\endgroup$ – Arturo Magidin Mar 25 at 0:09
  • $\begingroup$ You seem to be confusing $\mathbb{Q}[x]/(p(x))$ with the splitting field of $p(x)$. Note that $\mathbb{Q}[x]/(p(x))$ is isomorphic to the field obtained by adjoining to $\mathbb{Q}$ a single root of $p(x)$. This may, or may not, be the same as the splitting field of $p(x)$. For example, $\mathbb{Q}(\sqrt{2},\sqrt{5})$ is not isomorphic to $\mathbb{Q}[x]/((x^2-2)(x^2-5))$, as the latter is actually isomorphic to the ring $\mathbb{Q}(\sqrt{2})\times\mathbb{Q}(\sqrt{5})$. However, $\mathbb{Q}(\sqrt{2},\sqrt{5})$ is isomorphic to the splitting field of $(x^2-2)(x^2-5)$ (cont)$ $\endgroup$ – Arturo Magidin Mar 25 at 0:20

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.


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 $\mathbb Q$.

  • $\begingroup$ @RickAlmeida: In general you do not have that $\mathbb{Q}[x]/(p(x))\cong \mathbb{Q}(r_1,\ldots,r_n)$. That will only hold if the extension is a normal extension. Generally, you only get one root of $p(x)$ in the extension. If you started with a polynomial $p(x)$, the splitting field will be $\mathbb{Q}(r_1,\ldots,r_n)=\mathbb{Q}(\beta) \cong \mathbb{Q}[x]/(q(x))$ for some $\beta$ and $q(x)$ the minimal polynomial of $\beta$ (by the Primitive Element Theorem). But $q(x)$ is not necessarily the $p$ you started with. $\endgroup$ – Arturo Magidin Mar 25 at 2:37
  • $\begingroup$ And, no, you don’t get a unique $p(x)$, because in general there are infinitely many $\beta$ that work, and they can’t all come from the same poluynomial. $\endgroup$ – Arturo Magidin Mar 25 at 2:38
  • $\begingroup$ @RickAlmeida: If you start with $p(x)$, then $\mathbb{Q}[x]/(p(x))$ is not, in general, a normal extension; and the splitting field of $p(x)$ is not, in general, equal to $\mathbb{Q}[x]/(p(x))$. For example, take $p(x) = x^3-2$. Then $\mathbb{Q}[x]/(p(x)) \cong \mathbb{Q}(\sqrt[3]{2})$,which is not a normal extension; and the splitting field of $p(x)$ is not $\mathbb{Q}[x]/(p(x))$. However, if $F$ is the splitting field of $x^3-2$, there does exist a $\beta$ in $F$ (in fact, infinitely many such $\beta$s) such that $F=\mathbb{Q}(\beta) \cong\mathbb{Q}[x]/(q(x))$, (cont) $\endgroup$ – Arturo Magidin Mar 25 at 2:51
  • $\begingroup$ @RickAlmeida (cont) where $q(x)$ is the irreducible polynomial of $\beta$. There are infinitely many such $q(x)$ that will “work”, because there are infinitely many $\beta$s that work. But $q(x)$ is not $x^3-2$, the polynomial we started with. $\endgroup$ – Arturo Magidin Mar 25 at 2:52

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