Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 1 / 16H / b

The cubic has no rational integer root among the divisors of four, so is irreducible. Its polynomial discriminant is $4-27\cdot 16=-428=-4\cdot 107$. Thus the index $[{\mathcal{O}}_K:\mathbb{Z}[\alpha ]]$ is one or two, since its square divides the discriminant.

Set $\beta =({\alpha }^2+\alpha )/2$. Using ${\alpha }^3=\alpha +4$ givesand hence ${\beta }^3-{\beta }^2-3\beta -2=0$. Therefore $\beta$ is integral. The lattice spanned by $1,\alpha ,\beta$ contains $\mathbb{Z}[\alpha ]$ with index two. Changing basis divides the discriminant by four, giving $-107$, which is squarefree. An integral lattice of full rank with squarefree discriminant cannot have any further index in the full ring of integers. Henceis the required integral basis. The relations also give $\alpha ={\beta }^2-\beta -2$, so this ring can equally be generated by $\beta$.