Past exam of the algebraic number theory course of the University of Oxford / 2023 / 3

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.