Past exam of the number theory course of the University of Oxford / 2009 / B (Algebraic Number Theory)

The set ${\mathcal{O}}_K$ consists of the elements of $K$ integral over $\mathbb{Z}$, meaning roots of monic polynomials in $\mathbb{Z}[X]$.

First prove the finite-module criterion for integrality in a slightly stronger form, for any nonzero finitely generated $\mathbb{Z}$-submodule $M\subseteq K$ stable under multiplication by $\alpha$. Choose generators $m_1,\dots ,m_r$ and write $\alpha m_i={\sum }_j{a}_{ij}{m}_j$ with integer coefficients. Multiply the vector equation $(\alpha I-A)m=0$ by its adjugate matrix. This gives $\det (\alpha I-A)m_i=0$ for every generator. Some generator is nonzero and $K$ is a field, so $\det (\alpha I-A)=0$. The polynomial $\det (XI-A)$ is monic with integer coefficients, proving integrality of $\alpha$. This implication needs no prior closure theorem for ${\mathcal{O}}_K$.

Conversely, if $\alpha$ satisfies a monic polynomial of degree $d$, then $M=\mathbb{Z}[\alpha ]$ is generated by $1,\alpha ,\dots ,{\alpha }^{d-1}$ and is stable under $\alpha$. Every $\beta \in M$ also maps $M$ into itself because it is an integer polynomial in $\alpha$; the finite-module criterion for integrality just proved makes every such $\beta$ integral. Thus this $M$ really lies in ${\mathcal{O}}_K$, proving the stated equivalence without circularity.

For two integral elements $\alpha ,\beta$, the module spanned by ${\alpha }^i{\beta }^j$ for $i<d_{\alpha }$, $j<d_{\beta }$ contains one, is finitely generated, and is stable under both elements by their monic equations. It is therefore stable under $\alpha +\beta$ and $\alpha \beta$. The finite-module criterion for integrality proves that both are integral. Negatives follow by replacing a multiplication matrix by its negative, and $0,1$ are integral. Hence $\boxed{{\mathcal{O}}_K\text{ is a ring}}$.

The integral basis of the cubic field of discriminant minus 140 can be obtained as follows. For the specified field, Eisenstein at two makes $X^3+2X+2$ irreducible, and $\theta$ is integral because the polynomial is monic. The discriminant of a monic cubic $X^3+rX+s$ is $-4r^3-27s^2$, soHere ${\Delta }^2$ is the printed notation for the signed discriminant, namely the square of a Vandermonde product of the conjugates; that product need not be real.

Suppose $\eta =(a+b\theta +c{\theta }^2)/2$ is integral with the prescribed binary coefficients. The root traces give $\mathrm{Tr}(1)=3$, $\mathrm{Tr}(\theta )=0$, $\mathrm{Tr}({\theta }^2)=-4$. Thus $\mathrm{Tr}(\eta )=(3a-4c)/2$ must be an integer, forcing $a=0$. If $b=0$, nonzeroness of the coefficient triple forces $c=1$, giving $\eta ={\theta }^2/2$. If $b=1$, multiply by $\theta$ and use ${\theta }^3=-2\theta -2$:Closure of the ring then again gives ${\theta }^2/2\in {\mathcal{O}}_K$.

But $N_{K/\mathbb{Q}}(\theta )=-2$, so the field norm of ${\theta }^{-1}$ is $-1/2$. The field norm of an algebraic integer is an integer, proving ${\theta }^{-1}\notin {\mathcal{O}}_K$. Dividing the defining cubic by $2\theta$ gives ${\theta }^2/2=-1-{\theta }^{-1}$. Thus ${\theta }^2/2$ cannot be integral either.

Finally the discriminant-index formula for an integral lattice says $-140=[{\mathcal{O}}_K:\mathbb{Z}[\theta ]{]}^2\mathrm{disc}(K)$. The square of the integer index divides $140$, so its only possibilities are one and four, giving index one or two. If the index were two, a nonzero coset would supply an integral element $(a+b\theta +c{\theta }^2)/2$ after reducing coefficients modulo two, with a nonzero binary triple. The preceding argument would force the impossible integrality of ${\theta }^2/2$. Hence

ideals are coprime when $A+B={\mathcal{O}}_K$, equivalently $1=a+b$ for some $a\in A$, $b\in B$. Always $AB\subseteq A\cap B$. If $x\in A\cap B$, then $x=xa+xb\in AB$, giving $\boxed{AB=A\cap B}$ in the coprime case. For nonzero integral ideals, the Chinese remainder theorem gives ${\mathcal{O}}_K/(AB)\cong {\mathcal{O}}_K/A\times {\mathcal{O}}_K/B$, and taking finite cardinalities yields $\boxed{N(AB)=N(A)N(B)}$.

For a nonzero prime ideal, the chain of finite-index inclusions givesby the given hypothesis. Distinct nonzero prime ideals are maximal and coprime, as are their positive powers. Unique factorization of ideals in a number field therefore gives $N(A)={\prod }_{P}N(P{)}^{v_P(A)}$. Exponents add under ideal multiplication, proving multiplicativity for arbitrary nonzero integral ideals. This is the standard domain of the finite ideal norm used here.

For the Dedekind factorization theorem, choose an integral primitive element $\theta$ of $K$ with monic minimal polynomial of an algebraic element $f\in \mathbb{Z}[X]$, and assume the rational prime $p$ does not divide the index $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$. Factorover ${\mathbb{F}}_p$, with distinct monic irreducible factors. For integer polynomial lifts $f_i$, the distinct prime ideals above $p$ are $P_i=(p,f_i(\theta ))$, andThus the residue degree is $\mathrm{deg}{\bar{f}}_i$, the ramification index is $e_i$, and ${\sum }_i{e}_i\mathrm{deg}{\bar{f}}_i=[K:\mathbb{Q}]$. The index restriction is essential to this direct factorization formula.

For $K=\mathbb{Q}(\sqrt{-2})$, the ring of integers is $\mathbb{Z}[\sqrt{-2}]$, so the index is one and the theorem applies to every prime. Modulo two, $X^2+2=X^2$, giving one prime ideal of ideal norm two. For an odd rational prime, a prime ideal has ideal norm $p$ if the polynomial splits, and ideal norm $p^2$ if it is irreducible. In the former case the supplied condition forces $p\equiv 1$ or $3(\mathrm{mod}8)$; in the latter case every odd square is one modulo eight. ConsequentlyThe odd part of any ideal norm is thus one or three modulo eight, because $3^2\equiv 1$. Multiplication by a power of two leaves possible residues $1,3$ for exponent zero, $2,6$ for exponent one, $4$ for exponent two, and $0$ for exponent at least three. Hence $\boxed{N(I)\not\equiv 5,7(\mathrm{mod}8)}$ for every nonzero integral ideal.

For nonzero ideals, unique factorization of ideals in a number field converts $AB=C^3$ to $v_P(A)+v_P(B)=3v_P(C)$. Coprimality says at most one of the first two exponents is positive, so each is divisible by three. Dividing its exponents by three constructs ideals whose cubes are $A,B$. If a zero ideal is allowed, coprimality forces the other ideal to be the whole ring and the conclusion is immediate with the zero ideal as its own cube.

To compute the Ideal class group of Q of square root minus thirty, put $\theta =\sqrt{-30}$. The ring of integers is $\mathbb{Z}[\theta ]$, with discriminant $-120$. The given Minkowski bound for ideal classes is $2\sqrt{120}/\pi =4\sqrt{30}/\pi <7$. Hence the ideal class group is generated by prime ideals with ideal norm below seven. The rational primes two, three and five all ramify, giving unique prime ideals $P_2=(2,\theta )$, $P_3=(3,\theta )$, $P_5=(5,\theta )$ with $P_i^2=(i)$ and ideal norm $i$. There is no additional prime ideal of ideal norm four because two ramifies with residue degree one; six is not a prime power. Thus these three classes generate and each has order at most two.

As ${\theta }^2=-30$, unique factorization of ideals in a number field gives $(\theta )=P_2{P}_3{P}_5$, so their three classes multiply to one. None is principal: a generator of one would be an integral element $a+b\theta$ of field norm $a^2+30b^2$ equal to two, three or five, and none of those field norm equations has integer solutions. The three nontrivial classes are therefore distinct, with the third the product of the first two. Hence

To rule out integer solutions, suppose $x^2+30=y^3$. Necessarily $y>0$. If a prime $p\mid 30$ divided $y$, then reduction modulo $p$ would force $p\mid x$. Because thirty is squarefree, the valuation of $x^2+30$ at that prime would be exactly one, while that of $y^3$ is a positive multiple of three, a contradiction. Thus $\gcd (y,30)=1$.

If a prime ideal divided both $(x+\theta )$ and $(x-\theta )$, it would divide their difference $(2\theta )$ and their product $(y{)}^3$. Its underlying rational prime would divide $N(2\theta )=120$, hence divide thirty, and would also divide $y$. This contradicts the gcd just proved, so the two ideals are coprime. Their product is $(y{)}^3$, so the first part gives $(x+\theta )=A^3$ for an ideal $A$.

In a ideal class group of exponent two, $[A{]}^3=1$ forces $[A]=1$. Thus $A=(a+b\theta )$ with $a,b\in \mathbb{Z}$. Equality of principal ideals gives $x+\theta =\epsilon (a+b\theta {)}^3$ for a unit $\epsilon$. The only units are $\pm 1$, since $a^2+30b^2=1$ forces $b=0$, $a=\pm 1$; both signs are cubes and can be absorbed into the generator. Comparing the coefficient of $\theta$ after cubing now giveswhich is impossible modulo three. Therefore