Take the field $ Q(\phi) $ when $\phi$ is the golden ratio. Some elements do have square roots in $ Q(\phi) $: $$ \sqrt{1+\phi} = \phi $$ since $\phi^2=1+\phi$. Also, $(1+2 \phi)^2=5+8 \phi$ so $$ \sqrt{5+8 \phi}=1+2 \phi. $$

While the diagonal of a golden rectangle ($ 1 \times \phi $) is of length $\sqrt{1+\phi^2}=\sqrt{2+\phi}$. What is the square root of $2+\phi$?

Which elements in $Q(\phi)$ have square roots in $Q(\phi)$?

  • $\begingroup$ Use \times insted of x to denote multiplication: 1 \times \phi renders as $1 \times \phi$. $\endgroup$
    – CiaPan
    Sep 21, 2015 at 13:03
  • 2
    $\begingroup$ The title does not really reflect the question. $\endgroup$
    – lhf
    Sep 21, 2015 at 13:08
  • $\begingroup$ All algebraic numbers have square roots... And these are algebraic numbers. In other words, the set of algebraic numbers is closed under square roots (and cube roots, and all roots). $\endgroup$
    – lhf
    Sep 21, 2015 at 13:09
  • 1
    $\begingroup$ You can give an answer closely analogous to the answer to the corresponding question for $\mathbb{Q}$ except you need to be more careful about units. The point is that $\mathbb{Z}[\phi]$ has unique factorization, so elements of $\mathbb{Q}(\phi)$ also have unique factorizations (with integer exponents), up to units. A necessary condition is that all of these exponents are even, and then you also need to know which units are squares on top of that. $\endgroup$ Sep 21, 2015 at 22:31
  • $\begingroup$ It might help to remember that $$\phi = \frac{1 + \sqrt{5}}{2}.$$ Then $$(a + b \phi)^2 = \left(a + b\left(\frac{1 + \sqrt{5}}{2}\right)\right)^2 = a^2 + ab + \frac{3b^2}{2} + ab\sqrt{5} + \frac{b^2 \sqrt{5}}{2}.$$ Or it might not, I'm not sure where I was going with this. $\endgroup$
    – Lisa
    Sep 24, 2015 at 21:15

2 Answers 2


$\mathbb Q(\phi)$ is the linear span (over the rationals $\mathbb Q$) of $1$ and $\phi$. $$(x + y \phi)^2 = x^2 + 2 x y \phi + y^2 (1+\phi) = (x^2 + y^2) + (2x+y) y \phi$$ Thus for $a,b \in \mathbb Q$, $a + b \phi$ has a square root in $\mathbb Q(\phi)$ iff there is a solution in the rationals of $$ \eqalign{a &= x^2 + y^2\cr b &= (2x+y) y\cr}$$ Eliminating $y$, we get $$ 5 x^4 + (2b - 6 a) x^2 + (a-b)^2 = 0$$ so what we need is that this polynomial has a rational root $x$. The Rational Root Theorem may be useful (Edit :fixed link typo). Also, it must have real solutions, so the discriminant $16 (a^2 + a b - b^2) \ge 0$.


$\sqrt{2+\phi}$ is a solution of irreducible polynomial $X^4-5X^2+5$, so it is not in $\mathbb Q(\phi)$.


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