Algebraic number theory

Answer key — Algebraic number theory

1. 1.

   $-7 \equiv 1 \pmod 4$ (since $-7 = 4 \cdot (-2) + 1$), so by Theorem 2 $\mathcal{O}_K = \mathbb{Z}\big[\frac{1 + \sqrt{-7}}{2}\big]$ and $d_K = (\omega - \omega')^2 = (\sqrt{-7})^2 = -7$. The element $\omega$ is a root of $x^2 - x + 2$.

2. 2.

   Since $13 \equiv 1 \pmod 4$, Theorem 2 gives $\mathcal{O}_K = \mathbb{Z}\big[\frac{1 + \sqrt{13}}{2}\big]$: the element $\omega = \frac{1 + \sqrt{13}}{2}$ has trace $1$ and norm $\frac{1 - 13}{4} = -3$, so it is a root of $x^2 - x - 3$. The discriminant is $13$.

3. 3.

   $x^2 + 1$ is irreducible mod $p$ iff $-1$ is not a square mod $p$, i.e. iff $p \equiv 3 \pmod 4$. So the inert primes below $25$ are $\{3, 7, 11, 19, 23\}$; $2$ ramifies and $5, 13, 17$ split. This is Fermat's two-squares theorem in the language of ideals.

4. 4.

   In a UFD the irreducible $\pi$ is prime, so $(\pi)$ is a non-zero prime ideal. In a Dedekind domain every non-zero prime ideal is maximal, so from $(\pi) \subseteq \mathfrak{p} \neq R$ we get $(\pi) = \mathfrak{p}$. Thus all non-zero primes are principal, and by unique factorisation of ideals so is every ideal.

5. 5.

   Ideals of norm $1$: only $(1)$. Ideals of norm $2$: they contain $2$ and divide $(2) = \mathfrak{p}^2$, so the only one is $\mathfrak{p} = (2, 1 + \sqrt{-5})$. It is not principal, since $a^2 + 5b^2 = 2$ has no integer solution. So there are exactly two classes, $[(1)]$ and $[\mathfrak{p}]$, and $h_K = 2$; the class group is $\mathbb{Z}/2$, generated by $[\mathfrak{p}]$ with $\mathfrak{p}^2 = (2)$.

6. 6.

   $R$ is Noetherian of dimension one but not integrally closed: $\omega = \frac{1 + \sqrt{-3}}{2}$ is a root of $x^2 - x + 1$ and lies in the fraction field but not in $R$. Moreover $\omega\mathfrak{p} \subseteq \mathfrak{p}$: on generators, $\omega \cdot 2 = 1 + \sqrt{-3} \in \mathfrak{p}$ and $\omega(1 + \sqrt{-3}) = -1 + \sqrt{-3} = (1 + \sqrt{-3}) - 2 \in \mathfrak{p}$. In Lemma B the conclusion '$x$ integral, hence $x \in R$' fails, $\mathfrak{p}$ is not invertible, and indeed $\mathfrak{p}^2 = 2\mathfrak{p}$ with $\mathfrak{p} \ne (2)$. In $\mathcal{O}_K = \mathbb{Z}[\omega]$ the prime $2$ is inert and everything is repaired.

7. 7.

   The three embeddings send $\sqrt[3]{2}$ to $\sqrt[3]{2}$, $\sqrt[3]{2}\,\omega$ and $\sqrt[3]{2}\,\omega^2$ ($\omega = e^{2\pi i/3}$): one real and one complex pair, so $r_1 = 1$, $r_2 = 1$ and the rank is $1 + 1 - 1 = 1$. Indeed $\mathcal{O}_K^\times = \{\pm \varepsilon^n\}$ with fundamental unit $\varepsilon = 1 + \sqrt[3]{2} + \sqrt[3]{4}$ (its inverse is $\sqrt[3]{2} - 1$).

8. 8.

   Mathlib's class UniqueFactorizationMonoid includes the property Irreducible a ↔ Prime a. The lesson proves Irreducible (2 : ℤ√(-5)) (no element has norm $2$, so any factorisation $2 = xy$ has a factor of norm $1$, a unit) and ¬ Prime (2 : ℤ√(-5)) ($2 \mid (1 + \sqrt{-5})(1 - \sqrt{-5})$ but divides neither factor). Together they give not_ufd.

9. 9.

   def inert_primes(d, bound):
       if d % 4 == 1:
           b, c = -1, (1 - d) // 4      # x^2 - x - (d - 1)/4
       else:
           b, c = 0, -d                 # x^2 - d
       out = []
       for p in range(2, bound):
           if any(p % q == 0 for q in range(2, p)):
               continue
           if all((x * x + b * x + c) % p for x in range(p)):
               out.append(p)
       return out
   

   For $d = -1$ the inert primes are those $\equiv 3 \pmod 4$; for $d = 5$ those $\equiv \pm 2 \pmod 5$, including $2$ (because $x^2 + x + 1$ is irreducible mod $2$, not because of $x^2 - 5$); for $d = -5$ those $\equiv 11, 13, 17, 19 \pmod{20}$; for $d = 2$ those $\equiv \pm 3 \pmod 8$. In each case the answer depends only on $p$ modulo $|d_K|$ — quadratic reciprocity at work.