2
$\begingroup$

Find the minimal polynomial for the roots $\sqrt{2}$ and $\sqrt{3}$ over $\mathbb{Q}$.

I know the definition of a minimal polynomial is defined for a single root, but can we talk about minimal polynomials for multiple roots simultaneously?

I have only seen methods for finding the minimal polynomial of a single root.

Would it just be the product of the minimal polynomials for each root separately?

I am trying to better my understanding of finding the degree of field extensions such as $ \left[ \mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q} \right] $

$\endgroup$
6
  • 2
    $\begingroup$ What you are saying has the flavor of finding a primitive element $a$ of the extension $\mathbb{Q}(\sqrt{2},\sqrt{3})$ over $\mathbb{Q}$. This is $\mathbb{Q}(a)=\mathbb{Q}(\sqrt{2},\sqrt{3})$ and then finding the minimal polynomial of $a$ over $\mathbb{Q}$. If the final goal is just to find the degree, it looks to me simpler just to look at the intermediate extension $\mathbb{Q}\subset\mathbb{Q}(\sqrt{2})\subset\mathbb{Q}(\sqrt{2},\sqrt{3})$. And showing that $\sqrt{3}\notin \mathbb{Q}(\sqrt{2})$. $\endgroup$
    – YAlexandrov
    Mar 13, 2018 at 13:37
  • $\begingroup$ I would (first) interpret the question as looking for 2 separate minimal polynomials, one for $\sqrt2$ and one for $\sqrt3$. I doubt a reducible polynomial would be called a 'minimal polynomial'. $\endgroup$
    – Berci
    Mar 13, 2018 at 13:37
  • $\begingroup$ @YAlexandrov, Could you give a suggestion on how to find the primitive element for a given field extension? $\endgroup$
    – Zed1
    Mar 13, 2018 at 13:46
  • $\begingroup$ There's some theorem that in such a case, for $K(\alpha,\beta)$ with infinite $K$, one can always find a primitive element in the form of $\alpha+k\beta$. $\endgroup$
    – Berci
    Mar 13, 2018 at 13:51
  • $\begingroup$ $a=\sqrt{2}+\sqrt{3}\in\mathbb{Q}(\sqrt{2},\sqrt{3})$. Then $\sqrt{3}=\frac{1}{2}a+\frac{1}{2}\frac{1}{a}$ and $\sqrt{2}=\frac{1}{2}a-\frac{1}{2}\frac{1}{a}$. Therefore, $\mathbb{Q}(\sqrt{2}+\sqrt{3})=\mathbb{Q}(\sqrt{2},\sqrt{3})$. $\endgroup$
    – YAlexandrov
    Mar 13, 2018 at 13:58

1 Answer 1

Reset to default
1
$\begingroup$

As previous comments point out, you should find a primitive element of the field extension first, in this case $\sqrt2+\sqrt3$ is the element you want to find. Next, it is a standard approach to find such element's minimal polynomial: $$\begin{align} \text{Let } \theta= \sqrt{2}+\sqrt{3}, (\sqrt{2}+\sqrt{3})^2 = 5+2\sqrt{6} &\Rightarrow \theta^2 - 5=2\sqrt{6} \Rightarrow (\theta^2-5)^2=24 \\ &\Rightarrow \theta^4-10\theta^2 + 1= 0 \end{align} $$ It is easy to verify that $f(x)=x^4-10x^2+1$ is irreducible over $\mathbb{Q}$ by the rational root test and solving for a quadratic factor (which doesn't exist). So $f$ is the minimal polynomial for $\theta$.

$\endgroup$
1

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .