# Determining the minimal polynomial over $\Bbb{Q}$

I was working on a homework assignment from Hungerford:

Find the minimal polynomial of the element $\sqrt{1+\sqrt{5}}$ over $\Bbb{Q}$.

Naturally the solution would be the polynomial with roots

$$\pm \sqrt{1 \pm \sqrt{5}}$$

Which is found as

$$x = \pm \sqrt{1 \pm \sqrt{5}} \rightarrow x^2 -1 = \pm \sqrt{5} \rightarrow (x^2-1)^2 = 5 \rightarrow$$

$$\text{Minimal Polynomial = } (x^2-1)^2-5$$

Problem is, I quite frankly don't know how to prove this. Hypothetically what if, in the same vein that

$$\sqrt{5 + 2\sqrt{6}} = \sqrt{2} + \sqrt{3}$$

There exists some subtle factorization for $\sqrt{1 + \sqrt{5}}$ with the additionaly property that it is the root of a cubic or quadratic. How do I definitely rule out any of those cases?

• Note that $\sqrt 3 + \sqrt 2$ has degree $4$ over $\Bbb Q$. (But I see your general point.) – coldnumber Aug 10 '15 at 3:42

Maybe there's a nicer way, but you could consider the intermediate extension:

$$\Bbb Q \subseteq \Bbb Q(1+\sqrt 5)\subseteq \Bbb Q(\sqrt{1+\sqrt 5})$$

$1+\sqrt 5$ has degree $2$ over $\Bbb Q$ because it's not rational and it's a root of $x^2-2x-6 \in \Bbb Q[x]$.

Because of the multiplicative property, $[\Bbb Q(\sqrt{1+\sqrt 5}): \Bbb Q]$ must be divisible by $2$, and it is less than $4$ because you showed $\sqrt{1+\sqrt 5}$ is a root of a degree-four polynomial. Hence, it cannot be $3$.

Now you only have to show that $\sqrt{1+\sqrt 5}$ is not an element of $\Bbb Q(1+\sqrt 5)$, and you can show this by contradiction.

Since $\Bbb Q(1+\sqrt 5)$ has degree $2$ over $\Bbb Q$, it has $1, 1+\sqrt 5$ as a basis, so you can try to write

$$\sqrt{1+\sqrt 5} = a+b(1+\sqrt 5) \text{ for some } a,b \in \Bbb Q$$

and derive a contradiction (which might be somewhat tedious).

• Somewhat easier to work with $\mathbb Q [\sqrt 5]$ instead (obviously equivalent). The calculation isn't so terrible. – lulu Aug 10 '15 at 3:54
• @lulu yes, good point! – coldnumber Aug 10 '15 at 3:55
• Can you just clarify the notation $[\Bbb{Q}(\sqrt{1+\sqrt{5}} : \Bbb{Q}]$? – frogeyedpeas Aug 10 '15 at 4:04
• @frogeyedpeas yes, that means the degree of the extension $\Bbb Q\subseteq \Bbb Q(\sqrt{1+\sqrt 5})$, or the degree of $\sqrt{1+\sqrt 5}$ over $\Bbb Q$. – coldnumber Aug 10 '15 at 4:09
• okay this makes sense, and Im pretty sure this is the technique our professor was intending – frogeyedpeas Aug 10 '15 at 4:12

It is enough to show that $(x^2-1)^2-5$, that is, $x^4-2x^2-4$ is irreducible over the rationals. There are no rational roots, so we only need to rule out factorization as a product of quadratics with integer coefficients.

Since there is no $x^3$ term. we can confine attention to factorizations $(x^2+ax+b)(x^2-ax+c)$. If $a=0$, we get $b+c=-2$, $bc=-4$, impossible in integers. If $a\ne 0$, then because there is no $x$ term we need $b=c$, but then $bc$ cannot be $-4$.

• I liked this, its a bit ad hoc, and specific to this case, but to its advantage it is a lot more elementary. – frogeyedpeas Aug 10 '15 at 4:13