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

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.

• @TymaGaidash Looks like a typo. It should be $x^4-2x^2-1$. Nov 8, 2022 at 18:13
• @TymaGaidash sorry I meant $x^4-2x^2-1$
– Dima
Nov 8, 2022 at 18:15
• "It is also irreducible as the only possible roots are $\pm 1$" - that only rules out linear factors, you need to rule out quadratic factors too (or notice $f(x+1)$ satisfies conditions of Eisenstein criterion).
– Sil
Nov 8, 2022 at 18:17
• By far the simplest way to do this: Apply Eisenstein's criterion to $f(x+1)=x^4+4x^3+4x^2-2$. Nov 8, 2022 at 18:18
• You have already proved that the polynomial has no linear factor (and hence no cubic factor as well). You can also observe that if $a$ is a root of polynomial then $-a$ is also a root and hence if it has a quadratic factor then we must have the polynomial expressed as $(x^2+px+q)(x^2-px+q)$. Derive an obvious contradiction now. Nov 9, 2022 at 2:07

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 $$3$$.

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 $$\mathbb Q$$.

• No, they didn't prove that $f$ is irreducible. The degree $2$ case needs more work. Nov 8, 2022 at 18:25
• @mr_e_man The OP wrote and proved that $\alpha$ is not of degree $2$. Nov 8, 2022 at 19:36
• Are you referring to the comments, or the original question? Nov 8, 2022 at 19:42
• @mr_e_man Have you read the original question? Nov 8, 2022 at 19:48
• The question says "The right hand side is in $\mathbb Q$, but the left isn’t, which is a contradiction." This reasoning is wrong, or incomplete. Why isn't the left side in $\mathbb Q$? Nov 8, 2022 at 19:52

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 ...

• This should possibly have been a new question, but I try and explain it anyway. The trouble is that something like the following may happen. Do the same with $\beta=\sqrt{3+2\sqrt2}$. You get $\beta^2=3+2\sqrt2$ and then $$0=(\beta^2-3)^2-8=\beta^4-6\beta^2+1.$$ But $$x^4-6x^2+1=(x^4-2x^2+1)-4x^2=(x^2-1)^2-(2x)^2$$ is the difference of two squares and hence factors as $(x^2-2x-1)(x^2+2x-1)$. I was mean and set this up. The explanation is that $$(1+\sqrt2)^2=3+2\sqrt2,$$ so we really had $\beta=\alpha^2$. Or, phrased differently, it was possible to denest the square root. Nov 10, 2022 at 18:39
• Nov 10, 2022 at 18:42
• Be prepared for this to get deleted. You really shouldn't post follow up questions as answers. Nov 10, 2022 at 18:56
• still the polynomial $x^4-2x^2-1$ is irreducible, only then I can make the conclusion. The question is how I can guarantee that there is no cubic. Nov 11, 2022 at 3:00
• I am confused actually. The original question asks to explain why the minimal polynomial of $\alpha$ has degree $4$ : the person asking found that $\alpha$ is a root of $p(x)$ and that $p(x)$ is irreducible and has degree $4$. Isn't this enough as a justification? What does he want to explain exactly? Nov 11, 2022 at 3:30