Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 2 / 18H / Solution

For the first polynomial, put $\alpha =\sqrt{1+\sqrt{26}}$ and $\beta =5i/\alpha$, so that ${\beta }^2=1-\sqrt{26}$. The roots are $\pm \alpha ,\pm \beta$, andTo justify the degree, a rational quadratic factorization would have the form $(x^2+ux+v)(x^2-ux+w)$. The vanishing linear coefficient implies either $u=0$ or $v=w$. The former requires $v,w=-1\pm \sqrt{26}$, not rational, while the latter requires $v^2=-25$, also impossible over $\mathbb{Q}$. There is no rational root, since its square would have to be $1\pm \sqrt{26}$, so the polynomial is irreducible. The real field $\mathbb{Q}(\alpha )$ has degree four and does not contain $i$, giving degree eight after adjoining it. The group acts transitively on the four roots while preserving their two opposite pairs, so is contained in the square's dihedral group of order eight; the degree makes it the full group.

For the second polynomial, choose $\alpha =\sqrt{3}+i\sqrt{2}$ and $\beta =\sqrt{3}-i\sqrt{2}$. Then ${\alpha }^2=1+2i\sqrt{6}$, ${\beta }^2=1-2i\sqrt{6}$ and $\alpha \beta =5$. Its four roots generateIndeed $\alpha +\beta =2\sqrt{3}$ and $\alpha -\beta =2\sqrt{-2}$ recover both generators. The real quadratic field $\mathbb{Q}(\sqrt{3})$ cannot contain $\sqrt{-2}$, so the degree is four. Independent changes of the two square-root signs give all four automorphisms.