Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 2 / 20H (Number Fields)

Since $10\equiv 2(\mathrm{mod}4)$, the ring of integers is ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{10}]$, with field norm $N(x+y\sqrt{10})=x^2-10y^2$. Norms $\pm 2$ and $\pm 3$ are impossible: modulo five they would require a square congruent to 2 or 3. An algebraic integer is a unit exactly when its norm is $\pm 1$.

If 2 factored into two nonunits, their absolute norms would both be 2, since $N(2)=4$; this is impossible. The same argument for $N(3)=9$ excludes a factorization of 3. Finally $ϵ+1=4+\sqrt{10}$ has norm 6, so a nonunit factorization would require absolute norms 2 and 3, also impossible. Hence 2, 3 and $ϵ+1$ are irreducible. The conjugate $\overline{ϵ}+1$ is irreducible by the same argument.

The equality $6=2\cdot 3=(4+\sqrt{10})(4-\sqrt{10})$ gives inequivalent factorizations into irreducibles: no norm-six factor is associated to either 2 or 3, whose norms are 4 and 9. A principal ideal domain is a unique factorization domain. Consequently ${\mathcal{O}}_K$ is not a principal ideal domain.

Write $r=\sqrt{10}$. The ideals are ${\mathfrak{p}}_2=(2,r)$, ${\mathfrak{p}}_3=(3,r+1)$ and ${\overline{\mathfrak{p}}}_3=(3,r-1)$, with residue fields of orders 2, 3 and 3 respectively. Thus their ideal norms are 2, 3 and 3.

The generators of ${\mathfrak{p}}_2^2$ are $4,2r,10$. They are multiples of 2, and the integer combination of 4 and 10 giving 2 proves ${\mathfrak{p}}_2^2=(2)$. Similarly ${\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$ is generated by $9,3r+3,3r-3,r^2-1=9$. The difference of the middle generators is 6; together with 9 this generates 3. Therefore the product is $(3)$.

The ideals ${\mathfrak{p}}_2$ and ${\mathfrak{p}}_3$ are coprime, so their product equals their intersection. The element $4+r$ belongs to both, hence $(4+r)\subseteq {\mathfrak{p}}_2{\mathfrak{p}}_3$. Both ideals have norm 6, so equality holds. Likewise $2+r$ belongs to ${\mathfrak{p}}_2$ and ${\overline{\mathfrak{p}}}_3$, and its norm is $-6$. ThusThe discriminant is 40. The Minkowski bound for ideal classes in a real quadratic field is $\sqrt{40}/2=\sqrt{10}<4$, so every ideal class has an integral representative of norm 1, 2 or 3. The only possible prime ideals with those norms are the ones just listed. Their classes satisfy $[{\mathfrak{p}}_2{]}^2=1$ and $[{\mathfrak{p}}_3]=[{\overline{\mathfrak{p}}}_3]=[{\mathfrak{p}}_2]$. Thus there are at most two classes. The ideal ${\mathfrak{p}}_2$ is not principal, since a generator would have norm $\pm 2$, already ruled out. Therefore the class number is exactly two.

The element $ϵ=3+\sqrt{10}>1$ has norm $-1$. To prove that it is a fundamental unit, let $u=x+y\sqrt{10}$ be a positive real unit with $1<u<ϵ$. Its conjugate is $\pm 1/u$, so $y=(u-\overline{u})/(2\sqrt{10})>0$. For norm $-1$, the increasing function $u+1/u$ gives $y<(ϵ+1/ϵ)/(2\sqrt{10})=1$; for norm $+1$, $y<(ϵ-1/ϵ)/(2\sqrt{10})=3/\sqrt{10}<1$. Either contradicts integer $y$. There is no positive unit between one and $ϵ$. For any positive unit, multiply by a suitable integer power of $ϵ$ to place it in $[1,ϵ)$; it must then equal one. HenceTo classify the norm-six elements, a principal ideal generated by such an element has norm 6. Unique prime-ideal factorization and part (b) force it to be either ${\mathfrak{p}}_2{\mathfrak{p}}_3=(ϵ+1)$ or ${\mathfrak{p}}_2{\overline{\mathfrak{p}}}_3=(ϵ-1)$. Therefore its generator is an associate of one of these two elements. Since their norms are respectively 6 and $-6$, the unit multiplying the first must have even exponent and the unit multiplying the second must have odd exponent. Combining the two cases gives all integer solutions:Conversely every displayed element is integral, since $ϵ$ is a unit, and has norm 6. This proves completeness of the norm-six equation in the quadratic field of square root ten.

The PDF's restriction to $n=0,1,2,\dots$ misses conjugate solutions. In particular $(x,y)=(4,-1)$ satisfies the equation but is not in that restricted family; it is the $n=-1$ member. A nonnegative-index version becomes correct by adding the conjugates of every displayed element, with both signs. The restriction as printed cannot describe all signed integer pairs.