Minimal polynomial of $\sqrt{1+\sqrt{2}}$ over $\mathbb{Q}$

Question

I am trying to determine the minimal polynomial of $\sqrt{1+\sqrt{2}}$ over $\mathbb{Q}$ and explain why does it have degree $4$.

I found that the minimal polynomial is $f(x)=x^4-2x^2-1$. It is monic, and $f(\alpha)=0$. It is also irreducible as the only possible roots are $\pm 1$, however, $f(\pm 1)\ne 0$. Now, I am trying to explain why does it have degree $4$, and here is what I got:

• The degree is not $1$ as $\alpha \notin \mathbb{Q}$.

• The degree is not $2$ as if it was then $\alpha ^2+b \alpha +c=0$ for some $b,c \in \mathbb{Q}$. Then

$1+\sqrt{2}+(\sqrt{1+\sqrt{2}})b+c=0\implies \sqrt{2} (b^2+\sqrt{1+\sqrt{2}} b)=c^2+2c-b^2-1$.

The right hand side is in $\mathbb{Q}$, but the left isn’t, which is a contradiction.

I am stuck at showing that the degree is not $3$ though. If it was three then I think that $[\mathbb{Q}(\alpha):\mathbb{Q}]=3$, and this should lead us at a contraction. I am not sure how to show that.

Answer 1

You have $\mathbb{Q} \hookrightarrow \mathbb{Q}(\sqrt{2}) \hookrightarrow\mathbb{Q}(\sqrt{1+\sqrt{2}})$ : since $[\mathbb{Q}(\sqrt{2}) : \mathbb{Q}]=2$, then $[\mathbb{Q}(\sqrt{1+\sqrt{2}}) : \mathbb{Q}]$ is dividible by $2$, so the minimal polynomial of $\sqrt{1+\sqrt{2}}$ cannot have degree $3$.

Answer 2

If $[\mathbb{Q}(\alpha):\mathbb{Q}]=3$, then $\alpha$ would be the zero of a polynomial $p$ of degree $3$. $p$ would divide $f$. A contradiction as you proved that $f$ is irreducible over $\mathbb Q$.

Answer 3

This is not an answer but I always wanted to understand something :

Let $\alpha= \sqrt{1+\sqrt{2}}$ , the only way to get rid of the square root is to square ; $\alpha^2= 1+\sqrt{2}\ \ $ then $\sqrt{2}= \alpha^2 -1$ so, again squaring ; $2= (\alpha^2 -1)^2= \alpha^4- 2\alpha^2+ 1$ , meaning that $\alpha$ is a root of $$x^4-2x^2-1$$ Now here is the argument that I want you to confirm whether it is valid or not : Since we arrived to that polynomial just by 'getting rid of the square roots' in a 'minimal' way, i.e. this is the polynomial with the smallest degree we can have if we try to find a polynomial with rational coefficients. ( why? because we were just getting rid of the radicals ), and this polynomial is irreducible, then $[\mathbb{Q}(\alpha):\mathbb{Q}]= 4$ .

It seems that this argument is not quite valid because of cubics, or polynomials of degree $3$. I wonder if there is a way, a constructive way, for any root of a cubic to find immediately that polynomial, so that to make this argument valid every time?

I mean starting with $\alpha$ , and just getting rid of the radicals to arrive to the minimal polynomial, which will be of degree $3$, $4$, $5$, $6$, or more ...