Algebraic number theory course of the University of Oxford (B3.4)

Original solutions to all three questions of the University of Oxford Part B algebraic number theory paper, sat on 30 May 2023. Source: Oxford's released examination PDF, bundle pages 241–242; paper pages 1–2. The cover prints the formal code A12171W without a final digit. In 2(b)(iii), the printed assertion about a principal ideal domain is false as written; the solution identifies the error and proves the intended principal ideal ring assertion and its consequence.

For a number field $K$ of degree of a field extension $n$, an integral basis is a basis ${\omega }_1,\dots ,{\omega }_n$ over $\mathbb{Q}$ such thatThus every algebraic integer in $K$ has a unique expression with integer coefficients in this basis. Merely requiring its entries to be algebraic integers would not suffice: for example, $1,2\sqrt{2}$ is a rational basis of $\mathbb{Q}(\sqrt{2})$ but misses the algebraic integer $\sqrt{2}$ in its integer span.

Here is a proof of the existence of an integral basis. Start with any rational basis $v_1,\dots ,v_n$. Each $v_j$ has a nonzero integer multiple ${\alpha }_j$ that is an algebraic integer. Indeed, ifchoose a positive integer $q$ clearing all the denominators. Then $qv$ satisfies the monic polynomialThe ${\alpha }_j$ are still a rational basis. Put $M={\sum }_j\mathbb{Z}{\alpha }_j$ and form the matrixIts entries are integers, because products and sums of algebraic integers are algebraic integers and a rational algebraic integer is an integer. The nondegeneracy of the trace form of a number field gives $D\ne 0$: if $x\ne 0$, then $\mathrm{Tr}(x{x}^{-1})=n\ne 0$.

For $x={\sum }_j{x}_j{\alpha }_j\in {\mathcal{O}}_K$, each $\mathrm{Tr}(x{\alpha }_i)$ is an integer, soThe adjugate matrix formula for $T^{-1}$ gives $D{x}_j\in \mathbb{Z}$. ConsequentlyThe submodule theorem for free modules over a principal ideal domain says that every subgroup of the finite-rank free abelian group $D^{-1}M$ is again a free abelian group of rank at most $n$. Since ${\mathcal{O}}_K$ contains $M$, its rank is exactly $n$. A basis of this free abelian group is the required integral basis.

Write $x=r+s\sqrt{d}$ with $r,s\in \mathbb{Q}$. The field trace and field norm areThe element is an algebraic integer if and only if $T,N\in \mathbb{Z}$: necessity follows from its two conjugates; sufficiency follows from the monic polynomial $X^2-TX+N$.

Suppose $x$ is an algebraic integer. Set $a=2r\in \mathbb{Z}$. ThenWrite $2s=h/k$ in lowest terms, with $k>0$. Then $k^2\mid d$, and the hypothesis that $d$ is a square-free integer forces $k=1$. Thus $b=2s\in \mathbb{Z}$, and the remaining condition isThe only squares modulo four are zero and one. For $d\equiv 2(\mathrm{mod}4)$, an odd $b$ is impossible, after which $a$ is even. For $d\equiv 3(\mathrm{mod}4)$, an odd $b$ would require $a^2\equiv 3(\mathrm{mod}4)$, again impossible. For $d\equiv 1(\mathrm{mod}4)$, the condition is precisely $a\equiv b(\mathrm{mod}2)$.

Conversely, $\sqrt{d}$ satisfies $X^2-d$, and when $d\equiv 1(\mathrm{mod}4)$, the element $\omega =(1+\sqrt{d})/2$ satisfiesTheir integer spans therefore consist of algebraic integers. In the last case,This proves the ring of integers of a quadratic field formula. The required integral basis isTheir rational linear independence follows from $\sqrt{d}\notin \mathbb{Q}$.

Put $\alpha =\sqrt{m_1}$, $\beta =\sqrt{m_2}$ and $F=\mathbb{Q}(\alpha )$. The hypotheses make $L/\mathbb{Q}$ a biquadratic extension. To check the degree of a field extension is four, if $\beta =x+y\alpha$ with rational $x,y$, squaring gives $xy=0$. The case $y=0$ would make $m_2$ a rational square; the case $x=0$ would make $m_2/m_1$ a rational square. Neither can occur for the specified distinct coprime square-free integers. Hence $[L:F]=2$.

The nonidentity field automorphism of $L/F$ fixes $\alpha$ and sends $\beta$ to $-\beta$. Thus, for $\gamma =(r+s\alpha )+(t+u\alpha )\beta$, the field norm and field trace areIf $\gamma$ is an algebraic integer, so is its conjugate, and their sum and product are algebraic integers. Conversely, put $T={\mathrm{Tr}}_{L/F}(\gamma )$ and $N=N_{L/F}(\gamma )$. The identityshows that $\gamma$ is integral over $\mathbb{Z}[T,N]$. If $T,N$ are algebraic integers, the ring generated by algebraic integers $\mathbb{Z}[T,N]$ is integral over $\mathbb{Z}$. Since integral dependence is transitive, $\gamma$ is an algebraic integer. This proves the trace and norm integrality criterion for a quadratic extension, including elements already in $F$.

Again put $\alpha =\sqrt{m_1}$ and $\beta =\sqrt{m_2}$. The ring of integers of each of the three quadratic subfields has integral basis $1,\alpha$, $1,\beta$, or $1,\alpha \beta$, respectively: their square-free radicands are respectively congruent to $3,2,2$ modulo four. The field traces of the algebraic integer $\gamma$ to those subfields areThey are algebraic integers, so comparison with each integral basis givesThis use of all three subfields is needed to control $d$ as well as the other coefficients.

The field norm to $\mathbb{Q}(\alpha )$ isIts two coefficients must be integers. The second condition says $ab$ is even, since $m_2$ is even. Reducing the numerator of the first coefficient modulo four givesReduction modulo two first gives $a\equiv b(\mathrm{mod}2)$. Since $ab$ is even, both $a$ and $b$ are even. The same congruence now gives $c^2+d^2\equiv 0(\mathrm{mod}2)$, or $c\equiv d(\mathrm{mod}2)$. Therefore every algebraic integer has the formwith four integer coefficients.

For the converse, $1,\alpha ,\beta$ are algebraic integers, andHere $4\mid 1+m_1$ and $2\mid m_2$, so both coefficients are integers. The monic polynomial $X^2-w^2$ and integral dependence is transitive show that $w$ is an algebraic integer. The four displayed generators are rationally linearly independent, since their coefficient matrix in $1,\alpha ,\beta ,\alpha \beta$ has determinant $1/2$. Hence the integral basis for coprime biquadratic radicands congruent to three and two isIn this formula $\sqrt{m_1{m}_2}$ denotes the product $\alpha \beta$ of the chosen square roots; this convention also handles negative radicands.

Put $\alpha =\sqrt{m_1}$, $\beta =\sqrt{m_2}$, andThe proposed integral basis for coprime biquadratic radicands congruent to two and one isEach entry is an algebraic integer, by the ring of integers of a quadratic field formula and closure under products. Their coefficient matrix in $1,\alpha ,\beta ,\alpha \beta$ has determinant $1/4$, so they are rationally linearly independent. It remains to show that they span every algebraic integer.

Write $\gamma =(a+b\alpha +c\beta +d\alpha \beta )/2$ initially with rational $a,b,c,d$. The field traces to $\mathbb{Q}(\alpha )$ and $\mathbb{Q}(\alpha \beta )$ are $a+b\alpha$ and $a+d\alpha \beta$. Since $m_1\equiv m_1{m}_2\equiv 2(\mathrm{mod}4)$, the corresponding integral basis in each subfield gives $a,b,d\in \mathbb{Z}$. The field trace to $\mathbb{Q}(\beta )$ is $a+c\beta$. Membership in $\mathbb{Z}[\omega ]$ says that $2a,2c$ are integers of equal parity. As $a$ is already an integer, this forces $c\in \mathbb{Z}$ too.

Now the field norm to $\mathbb{Q}(\beta )$ isBy the ring of integers of a quadratic field formula, $A/2$ is an integer and has the same parity as $ac-m_{1}bd$. Thus $A$ is even, giving $a\equiv c(\mathrm{mod}2)$, andUsing $m_1\equiv 2$, $m_2\equiv 1$ and $a\equiv c(\mathrm{mod}2)$, this becomesHence $b\equiv d(\mathrm{mod}2)$. Finally,an expression with integer coefficients. This proves the asserted integral basis without assuming a general theorem about products of elements of an integral basis.

Write $R={\mathcal{O}}_K$. We prove unique factorization of ideals in a number field, rather than assuming this conclusion as a property of a Dedekind domain. Three preliminary facts about $R$ will be useful. Its additive group is a finite-rank free abelian group, so every ideal is finitely generated and the ascending chain condition holds. It is an integrally closed domain, by integral dependence is transitive. Finally, every nonzero prime ideal is a maximal ideal. For this last fact, choose $0\ne a\in P$. Multiplication by $a$ is an injective endomorphism of the finite-rank free abelian group $R$ with nonzero determinant, so $R/(a)$ is finite. Consequently $R/P$ is a finite integral domain and hence a field.

First, every nonzero proper ideal contains a product of nonzero prime ideals. Otherwise choose a maximal counterexample $J$ under inclusion, using the ascending chain condition. It cannot be a prime ideal. Choose $x,y\notin J$ with $xy\in J$. The strictly larger ideals $J+(x)$ and $J+(y)$ each contain a product of nonzero prime ideals, allowing the empty product $R$ when necessary. Multiplying those products gives a product contained ina contradiction.

Next fix a nonzero prime ideal $P$, choose $0\ne a\in P$, and choose a product of nonzero prime idealswith $r$ minimal. Because $(a)\subseteq P$, primality and maximality imply $P_j=P$ for some $j$; reorder so that $P_1=P$. Minimality provides $b\in P_2\cdots P_r$ with $b\notin (a)$, taking the remaining product to be $R$ if $r=1$. ThusDefine the fractional idealIt contains $R$ and $x$. The ideal $P{P}^{-1}$ lies between $P$ and $R$, so it is $P$ or $R$. If it were $P$, then $xP\subseteq P$. Applying the determinant trick to a finite generating set of an ideal for $P$ gives a monic polynomial over $R$ satisfied by $x$. Since $R$ is an integrally closed domain, this would put $x$ in $R$, a contradiction. ThereforeThis proves that each nonzero prime ideal is an invertible fractional ideal, independently of the desired factorization theorem.

Now take a nonzero proper ideal $A$ and a maximal ideal $P$ containing it. The ideal $B=P^{-1}A$ satisfiesThe first inclusion is strict: if $A=B$, the same $x\in P^{-1}\setminus R$ would stabilize $A$, and the determinant trick would again force $x\in R$. Repeating gives a strictly ascending chain until it reaches $R$; the ascending chain condition ensures this happens after finitely many steps. The identities $A=PB$ then express $A$ as a product of nonzero prime ideals.

For uniqueness, suppose $P_1\cdots P_r=Q_1\cdots Q_s$. Since the right-hand side is contained in $P_1$, some $Q_j$ is contained in $P_1$. Both are nonzero maximal ideals, so $Q_j=P_1$. Multiplication by $P_1^{-1}$ cancels this factor. Repeat. A nonempty product of proper prime ideals cannot equal $R$, so both products finish together. This proves uniqueness up to reordering, including repeated factors.

By the correspondence theorem for ideals, an ideal of $R/P^n$ has the form $J/P^n$ with $P^n\subseteq J\subseteq R$. If $J\ne R$, apply unique factorization of ideals in a number field to $J$. Every prime ideal $Q$ dividing $J$ contains $J$ and hence $P^n$. Primality gives $P\subseteq Q$, and maximality gives $Q=P$. Thus $J=P^i$ for some $i\ge 1$.

The inclusion $P^n\subseteq P^i$ forces $i\le n$: if $i>n$, cancellation using the invertible fractional ideal $P^n$ would give $R\subseteq P^{i-n}$, which is impossible. The zero ideal corresponds to $i=n$, while the unit ideal corresponds to $i=0$. Consequently the nonzero proper ideals are exactlyWhen $n=1$, there are none, in agreement with $R/P$ being a field.

For every $i\ge 0$, the inclusion $P^{i+1}\subseteq P^i$ is strict: equality, followed by multiplication by the inverse of $P^i$, would give $P=R$. Hence one can choosePut $J=(\pi )+P^{i+1}$. ThenThe classification just proved, applied modulo $P^{i+1}$, leaves no ideal strictly between consecutive powers. It follows thatFor $i=0$, the same argument says $(\pi )+P=R$, using $\pi \notin P$ and maximality.

The extension to every $r\ge 1$ follows by mathematical induction. Suppose $P^i=(\pi )+P^{i+r}$. Multiplication of the initial equality by $P^r$ givesSubstitution yieldsThusTaking $r=n-i$ for $0\le i<n$ givesThis is a principal ideal of the quotient ring $R/P^n$. Its zero ideal is generated by zero, so every ideal of this quotient ring is a principal ideal.

The printed domain assertion is false. For example, take $K=\mathbb{Q}$ and $I=(4)$. Then ${\mathcal{O}}_K/I=\mathbb{Z}/4\mathbb{Z}$ has the nonzero zero divisor $2+(4)$, so it is not an integral domain and cannot be a principal ideal domain. The valid conclusion is that ${\mathcal{O}}_K/I$ is a principal ideal ring. This error is present in the original PDF, not just its transcription.

To prove the corrected conclusion, factorinto powers of distinct nonzero prime ideals. These powers are pairwise comaximal ideals. Indeed, a maximal ideal containing two such powers would contain both underlying prime ideals, forcing them to be equal. The Chinese remainder theorem givesEvery factor is a principal ideal ring by the preceding part. Every ideal $H$ in a finite direct product of rings is a product $H_1\times \cdots \times H_s$ of ideals in the factors: multiplication by the coordinate idempotent elements separates the coordinates of every element of $H$. Write $H_j=(h_j)$. Thenso $R/I$ is a principal ideal ring. This proves the principal ideal property of quotients of a Dedekind domain in the present setting. In fact, this finite quotient ring is an integral domain precisely when $I$ is a prime ideal, in which case it is a field.

For the final deduction, choose any $0\ne a\in I$ and apply the result to the nonzero proper ideal $(a)$. The ideal $I/(a)$ of $R/(a)$ is generated by one residue class $b+(a)$, which can be represented by $b\in I$. For each $x\in I$, some $r\in R$ satisfies $x-rb\in (a)$. HenceThis proves two generators for an ideal in a Dedekind domain; moreover, its first generator can be any prescribed nonzero element of $I$.

Let $d_K$ denote the field discriminant, and put $n=[K:\mathbb{Q}]\ge 2$. Let $r_1$ be the number of real field embeddings and $r_2$ the number of conjugate pairs of nonreal field embeddings, so $n=r_1+2r_2$. The Minkowski bound for ideal classes provides a nonzero integral ideal $A$ satisfyingRearranging gives the Minkowski lower bound for a number-field discriminantBecause $2r_2\le n$ and $\pi /4<1$, this is at leastApply the arithmetic-geometric mean inequality to $1,2,\dots ,n$:It follows thatwhere $n/(n+1)\ge 2/3$ for $n\ge 2$ was used. ThusThe source writes $|\Delta (K{)}^2|>1$. If $\Delta (K)$ denotes the embedding determinant, its square is $d_K$; if it denotes the field discriminant, the displayed result also implies the stated inequality. The conclusion therefore does not depend on that notational convention.

To compute the Ideal class group of Q of square root thirty-five, put $\theta =\sqrt{35}$. The ring of integers of a quadratic field formula gives $R=\mathbb{Z}[\theta ]$, and the field discriminant is $4\cdot 35=140$. Both field embeddings are real, so the Minkowski bound for ideal classes isEvery ideal class therefore has an integral ideal representative of ideal norm at most five. By prime ideal factorization and multiplicativity of the ideal norm, the ideal class group is generated by prime ideals of ideal norm at most five.

Apply the Dedekind factorization theorem to $X^2-35$. At two and five it has a repeated linear factor, givingAt three, $X^2-35\equiv X^2+1$ has no root, so $(3)$ itself is a prime ideal of ideal norm nine, outside the bound. These are all the relevant prime ideals: a prime ideal of ideal norm four would have to lie above two with residue degree two, but the factorization above two has only the single factor of degree one. Consequently $[P_2]$ and $[P_5]$ generate, and each has order of a group element dividing two.

The algebraic integer $5+\theta$ has field norm $25-35=-10$. It belongs to both $P_2$ and $P_5$. Hence its principal ideal is contained in $P_2\cap P_5=P_2{P}_5$. Both have ideal norm ten, soIt follows that $[P_5]=[P_2{]}^{-1}=[P_2]$, and the ideal class group is generated by one element of order of a group element at most two.

Finally, $P_2$ is not a principal ideal. If $P_2=(a+b\theta )$ with $a,b\in \mathbb{Z}$, the equality of ideal norm and absolute field norm would implyModulo five this requires $a^2\equiv 2$ or $3$, whereas squares modulo five are $0,1,4$. Thus $[P_2]\ne 1$, provingThis computes the ordinary ideal class group, with no positivity restriction on generators.

Let $\alpha =\sqrt[3]{3}$. The polynomial $X^3-3$ is Eisenstein at three, so the number field has degree of a field extension three. Use the permitted ring of integers $R=\mathbb{Z}[\alpha ]$, whose integral basis is $1,\alpha ,{\alpha }^2$. The field traces $\mathrm{Tr}(1)=3$, $\mathrm{Tr}(\alpha )=\mathrm{Tr}({\alpha }^2)=0$, $\mathrm{Tr}({\alpha }^3)=9$, and $\mathrm{Tr}({\alpha }^4)=0$ giveThere is one real field embedding and one conjugate pair of nonreal field embeddings. The Minkowski bound for ideal classes is thereforeEvery ideal class has an integral ideal representative of ideal norm at most four, so it is enough to show that every prime ideal of ideal norm at most four is a principal ideal.

Modulo two,where the quadratic factor is an irreducible polynomial over the finite field ${\mathbb{F}}_2$. The Dedekind factorization theorem givesNowThus $(\alpha -1)=P_2$, by containment and equality of ideal norms, and cancellation in the second identity gives $({\alpha }^2+\alpha +1)=Q_2$. This explicitly handles the prime ideal of ideal norm four, as well as the one of ideal norm two.

Modulo three the polynomial is $X^3$, soThese exhaust the prime ideals of ideal norm at most four. Every bounded representative factors into the listed principal ideals, and hence is a principal ideal. Therefore the Class group of Q of cube root of three is

For $p\equiv 1(\mathrm{mod}20)$, put $\theta =\sqrt{-p}$. Since $-p\equiv 3(\mathrm{mod}4)$, the ring of integers of a quadratic field formula gives $R=\mathbb{Z}[\theta ]$. Modulo five,The two roots are distinct, so the Dedekind factorization theorem givesIn particular $P\ne \overline{P}$.

We prove that $[P]$ has order of a group element greater than $n$. Suppose instead that $P^k$ is a principal ideal for some $1\le k\le n$. Since $P^k\subseteq R$, its generator is in $R$: write $P^k=(a+b\theta )$ with $a,b\in \mathbb{Z}$. Taking ideal norms gives the positive field normBut $5^k\le 5^n<p$, so $b=0$. Thus $a^2=5^k$, forcing $k$ to be even and $a=\pm 5^{k/2}$. This would givecontrary to unique factorization of ideals in a number field and $P\ne \overline{P}$. Consequently no such $k$ exists. The finiteness of the ideal class group ensures that $[P]$ has finite order of a group element, soFor every positive integer $n$, the allowed infinitude of prime numbers in the progression $1(\mathrm{mod}20)$ supplies one larger than $5^n$. Applying the construction proves unbounded element orders in imaginary quadratic ideal class groups. In particular, this establishes elements of arbitrarily large order of a group element, rather than merely unbounded class numbers.