Minimal polynomial over $\mathbb Q(\sqrt{-2})$

Question

Find the minimal polynomial for $\sqrt[3]{25} - \sqrt[3]{5} $ over $\mathbb Q$ and $\mathbb Q(\sqrt{-2})$.

I have done the first part of this, over $ Q$, and have a polynomial. But I do not know how to do this over $ Q \sqrt{-2}$

Answer 1

Let $\alpha = \sqrt[3]{25}-\sqrt[3]{5}$. As mentioned, the minimal polynomial of $\alpha$ over $\mathbb Q$ has degree $3$ and so $[\mathbb Q(\alpha):\mathbb Q]=3$.

Now

$[\mathbb Q(\alpha,\sqrt{-2}):\mathbb Q]= [\mathbb Q(\alpha,\sqrt{-2}):\mathbb Q(\alpha)][\mathbb Q(\alpha):\mathbb Q]= 3[\mathbb Q(\alpha,\sqrt{-2}):\mathbb Q(\alpha)] $

On the other hand,

$[\mathbb Q(\alpha,\sqrt{-2}):\mathbb Q]= [\mathbb Q(\alpha,\sqrt{-2}):\mathbb Q(\sqrt{-2})][\mathbb Q(\sqrt{-2}):\mathbb Q]= 2[\mathbb Q(\alpha,\sqrt{-2}):\mathbb Q(\sqrt{-2})] $

Therefore, $[\mathbb Q(\alpha,\sqrt{-2}):\mathbb Q(\sqrt{-2})]$ is a multiple of $3$ and so is $3$ because $\alpha$ is a root of polynomial of degree $3$ over $\mathbb Q \subseteq \mathbb Q(\sqrt{-2})$.

Thus, the minimal polynomial of $\alpha$ over $\mathbb Q(\sqrt{-2})$ coincides with its minimal polynomial over $\mathbb Q$.

Answer 2

$f(x) = x^3 + 15x - 20$ is still a polynomial over $\Bbb Q(\sqrt{-2})$ with $\sqrt[3]{25}-\sqrt[3]{5}$ as a root. The minimal polynomial you are looking for therefore has to be a factor of $f$. Can you find those factors (over, say $\Bbb C$)? Are there any of them (apart from $f$ itself) that both is a polynomial over $\Bbb Q(\sqrt{-2})$ and have $\sqrt[3]{25} - \sqrt[3]5$ as a root?

Answer 3

Suppose $f$ is an irreducible polynomial in $\Bbb Q[X]$, and it has a nontrivial irreducible factor $g$ over $K = \Bbb Q(\sqrt{-2})$.

$g$ can't have rational coefficients, so its conjugate $\overline g$ is another factor of $f$ over $K$. Since $g$ and $\overline g$ are coprime, $g \overline g$ is also a factor of $f$, but it has rational coefficients, so you must have $f = \lambda g \overline g$, and the degree of $f$ must be even.

In your case, $f$ has degree $3$ so this is impossible : it has to stay irreducible over $K$.

Answer 4

Put $\theta=\sqrt[3] 5$ so $$x=\theta(\theta-1)\Rightarrow x^3=\theta^3(\theta^3-3\theta^2+3\theta-1)\Rightarrow x^3+15x-20=0$$ This is the minimal polynomial (irreducible by Einsenstein with $p=5$) of $\theta(\theta-1)$ over $\mathbb Q$ and also the m. p. of the same element $\theta(\theta-1)$ over any $K=\mathbb Q(t)\supset \mathbb Q$ providing that it is irreducible in $K[x]$.

We can use multiplication of degrees but (another way), splitting in $\mathbb C$ one has $$x^3+15x-20=(x-x_1)(x-\alpha)(x-\bar\alpha)$$ where $$x_1=\theta(\theta-1)\space\text {and}\space \alpha=\frac 12\left((1-i\sqrt3)\theta-(1+i\sqrt3)\theta^2\right)$$ hence it is directly seen that $x^3+15x-20$ is not decomposable on $\mathbb Q(\sqrt{-2})[x]$