Finding minimal polynomial over $\mathbb{Q}$

Question

We note $m(x)=(x^2-2)^2-3$.

$m(x)$ is a polynomial vanishing at $\alpha=\sqrt{2+\sqrt{3}}$.

But how to prove that $m(x)$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$?

Thanks

Answer 1

It is easy to see that the other roots of $m(x)$ are $\sqrt{2-\sqrt3}$ and $-\sqrt{2-\sqrt3}$. If $m(x)$ was not the minimal polynomial of $\sqrt{2+\sqrt3}$, then it would be reducible in$\mathbb{Q}[x]$. Since $\deg m(x)=4$, we would then be able to express it either as a product of two quadratic polynomials is a the product of a first degree polynonial with a cubic one. This second possibility cannot take place, since $m(x)$ has no rational roots. And if $\alpha$ and $\beta$ are any two distinct roots of $m(x)$, it is easy to show that $(x-\alpha)(x-\beta)\notin\mathbb{Q}[x]$.

Answer 2

The polynomial $m$ has no rational roots, because its constant term is $1$ and neither $1$ nor $-1$ is a root.

If $m$ is reducible, it must be a product of degree two factors; hence $\alpha=\sqrt{2+\sqrt{3}}$ has degree $2$ over $\mathbb{Q}$. Since $\sqrt{3}=\alpha^2-2\in\mathbb{Q}(\alpha)$, we conclude that $\mathbb{Q}(\alpha)=\mathbb{Q}(\sqrt{3})$, by the dimension formula. Hence, for some $x,y\in\mathbb{Q}$, $$ \sqrt{2+\sqrt{3}}=x+y\sqrt{3} $$ Square, compare and find a contradiction.