Find the minimal polynomial of $\sqrt{i+\sqrt{2}}$ over $\mathbb R, \mathbb Q(i), \mathbb Q(\sqrt{2})$.

I found the minimal polynomial over just $\mathbb Q$ is $x^8-2x^4+9$. Is there a strategy for finding the minimal polynomial in the reals? I know it $x-\sqrt{i+\sqrt{2}}$ but this doesn't help me much. Additionally, what's the difference between the polynomial over $\mathbb Q$ and $\mathbb Q(i), \mathbb Q(\sqrt{2})$?

  • 1
    $\begingroup$ To illustrate what the difference is between the polynomial over $\mathbb Q$ and $\mathbb Q(i), \mathbb Q(\sqrt{2})$, over $\Bbb Q(i)$, you get $(x^2 - i)^2 - 2$ as the polynomial. $\endgroup$
    – Arthur
    Nov 29, 2017 at 13:45
  • $\begingroup$ write $\sqrt{i+\sqrt{2}}=3^{1/4}(\cos \alpha + i \sin \alpha )$. $\endgroup$
    – Lozenges
    Nov 29, 2017 at 13:55
  • $\begingroup$ @Lozenges where did you get that? $\endgroup$
    – K Math
    Nov 29, 2017 at 14:03
  • $\begingroup$ @KMath Because it's a complex number. Complex numbers always have a form like that. $\endgroup$
    – Arthur
    Nov 29, 2017 at 14:05

3 Answers 3


In reals, you can use the rule that in real polynomials, you always get roots in conjugate pairs, so a quadratic real polynomial is enough to get any complex number. That means that as long as you can rewrite $\sqrt{i+\sqrt{2}}$ in proper cartesian form or polar form (into a form which you know how to conjugate), you can write down a second order polynomial with that root. I'll use the polar form:

$$z=\sqrt{i+\sqrt{2}}=\sqrt{\sqrt{{1+\sqrt{2}^2}}\exp(i \arctan{(1/\sqrt{2})})}=\sqrt[4]{3}\exp(i \arctan{(1/\sqrt{2})}/2)$$ Now you just have to write down the polynomial $$(x-z)(x-z^\ast)=0$$ which can be then simplified to $$x^2-2x \operatorname{Re}(z)+|z|^2=0$$ $$x^2-2x \sqrt[4]{3}\cos\left(\frac{\arctan{2^{-1/2}}}{2}\right)+\sqrt{3}=0$$

The ugly terms are the reason that this doesn't work in rational and extended rational field.

When you have different field, you allow different coefficients, so the polynomials will of course be different unless they are all pure rational (in which case the field extension wasn't needed).

Let's compare $\mathbb{Q}(i)$ and $\mathbb{Q}(\sqrt{2})$ solutions.

For both, you first square it:

$$x^2=i+\sqrt{2}$$ For $\mathbb{Q}(i)$ you need to remove the square root but $i$ is ok: $$(x^2-i)^2=2$$ $$x^4-2ix^2-3=0$$ For $\mathbb{Q}(\sqrt{2})$ you do the opposite: you remove the $i$: $$(x^2-\sqrt{2})^2=i^2$$ $$x^4-2\sqrt{2}x^2+3=0$$

These are evidently different. To remove the $i$ or $\sqrt{2}$ to get to pure rational, you will need to square again, getting something of order $8$.


If $r=\sqrt{i+\sqrt{2}}$ (one determination of it, actually) and we look for the minimal polynomial over $\mathbb{Q}(\sqrt{2})$ we find that $i=r^2-\sqrt{2}$ and so $-1=r^4-2\sqrt{2}r^2+2$, so the candidate is $$ X^4-2\sqrt{2}X^2+3=X^4+2\sqrt{3}X^2+3-(2\sqrt{3}+2\sqrt{2})X^2\tag{*} $$ which factors as $$ (X^2+\sqrt{3}-\sqrt{2\sqrt{3}+2\sqrt{2}}X) (X^2+\sqrt{3}-\sqrt{2\sqrt{3}-2\sqrt{2}}X) $$ over $\mathbb{R}$ and is easily seen to have no real roots. If this were a factorization over $\mathbb{Q}(\sqrt{2})$, we'd get $\sqrt{3}\in\mathbb{Q}(\sqrt{2})$, which is false. Therefore the polynomial (*) is irreducible over $\mathbb{Q}(\sqrt{2})$.

In particular $[\mathbb{Q}(r):\mathbb{Q}(\sqrt{2})]=4$.

Can you finish?


The field over which you are working determines the scalars that can be coefficients in the minimal polynomial. So over $\mathbb Q$, the coefficients can be rational numbers (or equivalently, after clearing denominators, integers), and over $\mathbb R$ they can be arbitrary real numbers.

Thus in $\mathbb Q[i]$, the coefficients can be complex numbers whose real and imaginary parts are both rational. So in expanding $x = \sqrt{i+\sqrt{2}}$, you can treat $i$ as a valid coefficient, but you have to remove $\sqrt{2}$. @orion's answer tells you in detail how to do this.


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