Past exam of the number theory course of the University of Oxford / 2011 / B (Algebraic number theory)

An integral basis is a $\mathbb{Z}$-basis of the ring of integers ${\mathcal{O}}_K$; it consists of $n$ algebraic integers and is also a $\mathbb{Q}$-basis of $K$.

The lattice $M$ has finite index $h=[{\mathcal{O}}_K:M]>1$. Choose a prime $p\mid h$. The finite abelian group ${\mathcal{O}}_K/M$ has an element of order $p$, represented by $\alpha \in {\mathcal{O}}_K\setminus M$. ThusIf all $c_j$ were divisible by $p$, cancellation in the characteristic-zero field would put $\alpha$ in $M$, a contradiction. The discriminant-index formula for an integral lattice giveswhere ${\Delta }^2(K)$ is an integer. Hence $p^2\mid {\Delta }^2(M)$, and the required integral element is $\alpha =p^{-1}{\sum }_j{c}_j{w}_j$.

The polynomial is irreducible by Eisenstein's criterion at $q$, so $[K:\mathbb{Q}]=n$. Its constant coefficient gives the field normThe defining equation givesIf $q\nmid r$ and $0\le i<n$, thenhas $q$-adic valuation $i-n<0$. It is not an integer. The norm of an algebraic integer is an integer, so $r{\theta }^i/q$ is not integral.

Suppose now that $\gamma$ is integral. Multiply it by ${\theta }^{n-1}$. All terms except $r_0{\theta }^{n-1}/q$ are integral, since their powers of $\theta$ are at least $n$. Thus this remaining term is integral, and the norm obstruction forces $q\mid r_0$. Subtract the integer $r_0/q$ from $\gamma$. Multiply the remainder by ${\theta }^{n-2}$; the only term of exponent below $n$ is $r_1{\theta }^{n-1}/q$, so $q\mid r_1$. Continuing in this way proves $q\mid r_j$ for every $j$. ThereforeFor $n=1$ the same conclusion follows directly from the norm obstruction with $i=0$.

Let $\theta =\sqrt[3]{3}$. The power-basis discriminant isIf $\mathbb{Z}[\theta ]$ were a proper sublattice of ${\mathcal{O}}_K$, part (a) would produce a prime $p$ whose square divides $243$ and a new integral element with denominator $p$. The only possible prime is $3$. But $x^3-3$ is Eisenstein at $3$, and part (b) says that every integral element of $(1/3)\mathbb{Z}[\theta ]$ already belongs to $\mathbb{Z}[\theta ]$. This is a contradiction. Hence

The Kummer-Dedekind theorem states the following. If $K=\mathbb{Q}(\theta )$ with $\theta$ integral and $p\nmid [{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$, factor the minimal polynomial modulo $p$ as $\bar{f}={\prod }_i{\bar{g}}_i^{e_i}$ into distinct monic irreducible factors. ThenThe index restriction is essential.

For the proposed ideal relation, reflexivity uses $\alpha =\beta =1$ and symmetry interchanges $\alpha ,\beta$. If $(\alpha )I_1=(\beta )I_2$ and $(\gamma )I_2=(\delta )I_3$, multiplication gives $(\alpha \gamma )I_1=(\beta \delta )I_3$. All multipliers remain nonzero. Thus it is an equivalence relation, namely equality of ideal classes.

Put $s=\sqrt{-2}$. The minimal polynomial is $x^2+2$ and the given full power basis makes every prime eligible for the theorem. Modulo $3$ its roots are $\pm 1$; modulo $5$ and $7$ it is irreducible; modulo $17$ its roots are $\pm 7$. ConsequentlyThe two inert ideals have norms $25$ and $49$; each split factor has norm equal to its rational prime.

For $p_1\equiv 1(\mathrm{mod}8)$, quadratic reciprocity and its supplementary laws give $(-2/p_1)=1$. Thus $x^2+2$ has two distinct roots modulo $p_1$, giving distinct prime ideals $P,P^{\prime }$ withThe given principal ideal domain property makes $P=(x+ys)$ for some integers $x,y$. Taking norms, which are positive here, gives $p_1=N(P)=x^2+2y^2$.

Let $\omega =(1+\sqrt{-3})/2$, whose minimal polynomial is $x^2-x+1$. For $p_2\equiv 1(\mathrm{mod}3)$, the cyclic group ${\mathbb{F}}_{p_2}^{\times }$ has a nontrivial cube root of unity $t$. Then $-t$ and $-t^2$ are distinct roots of $x^2-x+1$. By the Kummer-Dedekind theorem,Since the given ring of integers is a principal ideal domain, write $Q=(a+b\omega )$. Its norm isMultiplication by $4$ completes the square:Thus $x=2a+b$ and $y=b$ supply the required integers.

Put $s=\sqrt{-5}$ and $P=(2,1+s)$. The polynomial $x^2+5$ reduces to $(x+1{)}^2$ modulo $2$, so the Kummer-Dedekind theorem givesIf $P=(a+bs)$, its norm would imply $a^2+5b^2=2$, impossible for integers $a,b$. Thus $P$ represents the nontrivial ideal class. Since the class number is $2$, every nonprincipal integral ideal $I$ has $[I]=[P]$, and $[IP]=1$. Write $IP=(\gamma )$ with $\gamma \in {\mathcal{O}}_M$; integrality follows from $IP\subseteq {\mathcal{O}}_M$. Multiplying by $P$ givesNorms give $4N(I)=2N(\gamma )$, hence $N(\gamma )=2N(I)$. If $N(I)=p_3\equiv 1(\mathrm{mod}20)$, writing $\gamma =a+bs$ yieldscontradicting the list $0,1,4$ of squares modulo $5$. Therefore no nonprincipal ideal has norm $p_3$.

As $p_3\equiv 1(\mathrm{mod}4)$ and $p_3\equiv 1(\mathrm{mod}5)$, quadratic reciprocity gives $(-5/p_3)=1$. Thus $x^2+5$ has two distinct roots modulo $p_3$ andThe norm obstruction shows that $R$ must be principal, say $R=(x+ys)$. Taking its norm proves

Choose a nonzero integral ideal $B$ whose class is $[A{]}^{-1}$; one obtains such an ideal by multiplying the fractional ideal $A^{-1}$ by a suitable nonzero integer. The assumed Minkowski bound for ideal classes gives $0\ne \beta \in B$ withBecause $(\beta )\subseteq B$ and nonzero ideals of the ring of integers are invertible,It is an integral ideal in class $[B{]}^{-1}=[A]$. By multiplicativity of the ideal norm,The same argument applies if the starting $A$ is a fractional ideal, since its class is unchanged by clearing denominators.

Write $s=\sqrt{-33}$. The discriminant is $-132$, and the bound of part (a) isEvery ideal class therefore has an integral representative of norm at most $7$. The Kummer-Dedekind theorem givesModulo $5$, $x^2+33=x^2+3$ is irreducible, so there is no ideal of norm $5$. Modulo $7$ it has roots $\pm 3$, giving $(7)=R\bar{R}$, where $R=(7,s+3)$. The ideal of $3+s$ has norm $42$, and reduction in the three residue fields shows that it lies in $P,Q,R$. Equality of norms therefore givesIt follows that $[R]=[P][Q]$, because $[P{]}^2=[Q{]}^2=1$. The other prime over $7$ has the inverse class, which is the same class here.

By unique factorization of nonzero ideals, all norms at most $7$ now give only the classes $1,[P],[Q],[PQ]$: norm $4$ gives $P^2=(2)$, norm $6$ gives $PQ$, and norm $7$ was handled above. None of $P,Q,PQ$ is principal, since their norms are $2,3,6$ andAlso $[P]\ne [Q]$, since equality would make $PQ$ principal. Thus there are exactly four classes, all of exponent two:

For degree $3$ there are either three real embeddings or one real embedding and one complex pair. The respective Minkowski bound for ideal classes areUnder the given discriminant bound they are at most $\frac{2}{9}\sqrt{45}<2$ and $\frac{8}{9\pi }\sqrt{45}<2$. Hence every ideal class contains an integral ideal with positive integer norm strictly below $2$, so with norm $1$. The only such ideal is ${\mathcal{O}}_L$. Therefore

Part (a), applied to any ideal class, produces an integral ideal of positive integer norm at most $c_K$, so $c_K\ge 1$. Writing $n=[K:\mathbb{Q}]$ and using $s\le n/2$ givesFor the last inequality, write $n^n/n!={\prod }_{j=1}^n(n/j)$; the first $\lfloor n/2\rfloor$ factors are at least $2$, and the remaining factors are at least $1$. Since $\lfloor n/2\rfloor \ge (n-1)/2$, squaring yieldsAs $\pi /2>1$, the assumed upper bound $D$ forcesFor $D\ge 1$, the ceiling of the right-hand side is a valid integer $M$. If $D=0$, no number field satisfies the premise, so any positive $M$ works.