Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 4 / 19H (Number Fields)

For a square-free integer $\delta \equiv 2$ or $3(\mathrm{mod}4)$, the ring of integers of the quadratic field is ${\mathcal{O}}_{\mathbb{Q}(\sqrt{\delta })}=\mathbb{Z}[\sqrt{\delta }]$. Square-freeness is the usual convention underlying the first request: congruence alone would not suffice, for example $\delta =27$ gives the same field as $\delta =3$.

Now take the stipulated square-free $\delta \equiv 3(\mathrm{mod}4)$ and put $r=\sqrt{2}$, $s=\sqrt{\delta }$. For any integral $\alpha =a+br+cs+drs$, relative traces to the three quadratic subfields show $2a,2b,2c,2d\in \mathbb{Z}$. Those subfields all have integral bases $1$ and their square root, including $\mathbb{Q}(\sqrt{2\delta })$. Write $\alpha =(A+Br+Cs+Drs)/2$ with integer coefficients.

Its relative field norm to $\mathbb{Q}(s)$ isThus $AC$ is even, and the constant numerator is divisible by four. Since $\delta \equiv 3(\mathrm{mod}4)$, these conditions force $A,C$ both even: one odd makes the numerator odd, and both odd contradicts $AC$ even. They then force $B,D$ to have the same integer parity. Every algebraic integer consequently belongs to the lattice generated byConversely the last generator $w$ satisfies $w^2=(1+\delta )/2+\sqrt{\delta }$ and the monic integer equationAll four generators are integral and linearly independent over $\mathbb{Q}$, so this lattice is exactly ${\mathcal{O}}_L$, proving this integral basis of an odd biquadratic field containing square root of two without an unproved index assumption.

Here ${\mathcal{O}}_K=\mathbb{Z}[s]$ with $s=\sqrt{-14}$ and field discriminant $-56$. The imaginary-quadratic Minkowski bound for ideal classes is $(2/\pi )\sqrt{56}<5$, so each ideal class has an integral representative of ideal norm at most four. Representatives therefore use prime ideals above two and three. PutThe prime two ramifies, with ${\mathfrak{p}}^2=(2)$, and three splits into $\mathfrak{q}\overline{\mathfrak{q}}=(3)$. No element has field norm two, because $N(a+bs)=a^2+14b^2$; thus $[\mathfrak{p}]$ is nontrivial of order two. Norm-four ideals represent $[\mathfrak{p}{]}^2=1$ and add no generator, so the class group is generated by $[\mathfrak{p}],[\mathfrak{q}]$.

The element $2-s$ has field norm 18. It belongs to $\mathfrak{p}$ and $\mathfrak{q}$, but not $\overline{\mathfrak{q}}$, since reduction there gives $s\equiv 1(\mathrm{mod}3)$. Its ideal therefore factors asThe exponents follow from the ideal norm 18, or from the root $s\equiv 2(\mathrm{mod}9)$ defining ${\mathfrak{q}}^2$. Hence $[\mathfrak{q}{]}^2=[\mathfrak{p}]$, so $[\mathfrak{q}]$ has order four and already generates all possible classes. ThereforeAs a separate check, the reduced forms of discriminant of a binary quadratic form $-56$ are $[1,0,14],[2,0,7],[3,2,5],[3,-2,5]$, giving exactly four classes.