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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.

Dima
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  • Is $\sqrt{\sqrt 2+1}$ really the root of $x^4-2x-1$? – Тyma Gaidash Nov 08 '22 at 18:11
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    @TymaGaidash Looks like a typo. It should be $x^4-2x^2-1$. – Christian E. Ramirez Nov 08 '22 at 18:13
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    @TymaGaidash sorry I meant $x^4-2x^2-1$ – Dima Nov 08 '22 at 18:15
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    "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 08 '22 at 18:17
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    By far the simplest way to do this: Apply Eisenstein's criterion to $f(x+1)=x^4+4x^3+4x^2-2$. – Christian E. Ramirez Nov 08 '22 at 18:18
  • How do you know the left side isn't in $\mathbb Q$? – mr_e_man Nov 08 '22 at 18:21
  • @mr_e_man I think it would be in $\mathbb{Q}$ only if $b=0$, but if $b=0$, then $0=\alpha^2+b\alpha+c=\alpha^2+c$, and thus $c=-(1+\sqrt{2})$, but $c\in \mathbb{Q}$. Is that right? – Dima Nov 08 '22 at 18:48
  • For comparison, consider $$\sqrt2\big((1)^2+\sqrt{3-2\sqrt2};(1)\big)=2$$ which is in $\mathbb Q$. – mr_e_man Nov 08 '22 at 18:59
  • You need to prove that $\sqrt{1+\sqrt2}$ is not in $\mathbb Q(\sqrt2)$. – mr_e_man Nov 08 '22 at 19:00
  • So, suppose $(x+y\sqrt2)^2=1+\sqrt2$ for some rational $x,y$. That means $$x^2+2y^2=1,\quad2xy=1.$$ Can you show that this is impossible? – mr_e_man Nov 08 '22 at 19:12
  • @mr_e_man I was about to write this :), I tried solving this system but it only has complex solutions. So $\alpha \notin \mathbb{Q} (\sqrt{2})$. – Dima Nov 08 '22 at 19:17
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    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. – Paramanand Singh Nov 09 '22 at 02:07
  • how do you conclude that $f$ is not reducible? why 1 and -1 ? – NotaChoice Nov 10 '22 at 17:26
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    @NotaChoice see proposition 9.4.11 in Dummit and Foote, if you don’t have the book, please let me know so that I write the proposition for you. – Dima Nov 11 '22 at 02:30

2 Answers2

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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 $3$.

TheSilverDoe
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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 $\mathbb Q$.

mathcounterexamples.net
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