Algebraic Number Theory, PS 1

Bryden Cais

October 21, 2002

1. Let L/K be an extension of number fields. It is not true that for α ∈ L, Tr(α), N(α) ∈ O_{K} implies α ∈ O_{L}. As a counterexample, set K = ℚ and let α be a root of

f(x) = x^{3} + \frac{1}{3}x − 1

and let L = K(α) Then we have Tr(α) = 0, N(α) = 1. It is clear that f is irreducible: if it were not, it would have a rational root. But such a root would have numerator and denominator dividing 3, which (after a brief search) is impossible. It follows that f is the minimal monic polynomial of α so that α is not integral.

If deg(L/K) = 2 then Tr_{L/K}(α), N_{L/K}(α) ∈ O_{K} if and only if α ∈ O_{L}. One direction is clear. For the reverse, observe that α is a root of the monic polynomial

x^{2} − Tr_{L/K}(α)x + N_{L/K}(α),

and is therefore integral over O_{K} (and hence over ℤ). Since α ∈ L, we have α ∈ O_{L}.

2. Let d be square free and p a prime not dividing 2d. Set K = ℚ(\sqrt{d}). Since Δ_{K} = d or 4d, we see that p does not ramify in K. Now d is not a square modulo p if and only if x^{2} − d has no solutions in ℤ/(p). But in any case, x^{2} − d has a root in O_{K} (obviously) and hence has a root in O_{K}/℘ for any prime ℘ over p. Therefore, d is not a square modulo p if and only if

O_{K}/℘ ≄ ℤ/(p).

Since deg(K/ℚ) = 2, it follows that O_{K}/℘ is a degree two extension of ℤ/(p) and hence that d is not a square modulo p if and only if p remains prime in K.

3. Let D be a dedekind domain with a finite number of prime ideals. We show that D is a PID. Let 𝔭_{1}, 𝔭_{1}, …, 𝔭_{n} be the distinct prime ideals of D. Pick α_{i} ∈ p_{i} \ p_{i}^{2} such that α ∉ 𝔭_{j} for j ≠ i. We know that such a choice is always possible by, for example, the chinese remainder theorem (problem 4). Let us look at the prime decomposition of (α_{i}). Since a_{i} ∉ 𝔭_{j} for i ≠ j, we see that 𝔭_{j} |̸(α_{i}). Similarly, since α_{i} ∉ 𝔭_{i}^{2}, we have 𝔭_{i}^{2} |̸(α_{i}). It follows that (α_{i}) = p_{i}, that is, each prime ideal is principal. Hence, since D is a dedekind domain and we therefore have unique factorization into prime ideals, every ideal of D is principal.

4. Define

f : O ⟶ ⨁_{i=1}^{n} O/𝔞_{i}

by

f(x) = (x mod 𝔞_{1}, x mod 𝔞_{2}, …, x mod 𝔞_{n}).

We claim that f is surjective. It is sufficient to show that we can obtain the element (0, …, 0, 1, 0, …, 0) where the 1 occurs in the i^{th} place, for any i. Since 𝔞_{i} + 𝔞_{j} = O for i ≠ j, we have

𝔞_{i} + ⋂_{j≠i} 𝔞_{j} = O.

To see this, we proceed by induction: let α ∈ 𝔞_{i}, β ∈ 𝔞_{j} satisfy α + β = 1. Let I ⊂ O be any ideal such that I + 𝔞_{i} = O and I + 𝔞_{j} = O, and let γ ∈ I, α′ ∈ 𝔞_{i} satisfy γ + α′ = 1. Then we have

αγ + βγ = γ

and hence that

α′ + αγ + βγ = 1.

It follows that

α + α′β + αγβ + β^{2}γ = 1.

Since α +α′β ∈ 𝔞_{i} and αγβ +β^{2}γ ∈ I ∩𝔞_{j}, we are done. Therefore, there exists γ ∈ ⋂_{j≠i} 𝔞_{j} and α ∈ 𝔞_{i} such that α + γ = 1. Then γ ∈ O satisfies γ mod 𝔞_{j} = 0 for j ≠ i and γ mod 𝔞_{i} = 1. It follows that f is surjective. The kernel of f is the ideal of all x ∈ O with x ∈ 𝔞_{i} for all i. That is,

ker f = ⋂_{i=1}^{n} 𝔞_{i} := 𝔞.

We then have the isomorphism

O/𝔞 ≃ ⨁_{i=1}^{n} O/𝔞_{i}.

5. Let K = ℚ(\sqrt{−5}) and L = ℚ(\sqrt{−6}).

(a) We first show that O_{K} is not a UFD. Consider the two factorizations

2 · 3 = (1 + \sqrt{−5})(1 − \sqrt{−5}) = 6.

We claim that 2, 3, (1±\sqrt{−5}) are all irreducible elements of O_{K}. Suppose not. Then, for example, we could write 2 = (a + b\sqrt{−5})(c + d\sqrt{−5}) with a, b, c, d ∈ ℤ since O_{K} = ℤ[\sqrt{−5}]. Taking norms gives

4 = (a^{2} + 5b^{2})(c^{2} + 5d^{2}),

from which it follows (since we are assuming our factorization of 2 to be non trivial) that a^{2}+5b^{2} = c^{2} + 5d^{2} = 2. But 2 is not a square modulo 5 and we have reached a contradiction. Similar considerations show that if 3 is not irreducible, then 3 = a^{2} + 5b^{2} for some a, b ∈ ℤ, again a contradiction. Now if O_{K} were a UFD then we would have 2|(1 ± \sqrt{−5}). In either case, this requires that 4|6, which is a contradiction.

(b) We use the Minkowski bound to determine the class numbers of K, L. We have

Δ_{K} = −20,
Δ_{L} = −24,

and

d_{K} = d_{L} = 2.

Observe that s = 1 in both cases. The Minkowski bound is then

\frac{2!}{2^{2}} (\frac{4}{π} ) \sqrt{20} < 3

for K and

\frac{2!}{2^{2}} (\frac{4}{π} ) \sqrt{24} < 4

for L. Therefore, every ideal class of K contains an O_{K} ideal of norm at most 2. Therefore, we only need to consider those ideals of O_{K} of norm exactly 2. But any ideal of norm 2 is a prime above 2, and we know that 2 totally ramifies in K since it divides the discriminant. Moreover, since x^{2} − 5 ≡ (x + 1)^{2} mod 2, we have (since O_{K} = ℤ[\sqrt{5}]),

(2) = (2, 1 + \sqrt{5})^{2}.

Moreover, it is not difficult to see that (2, 1 + \sqrt{−5}) is not principal: if it were, we could write 2 = a^{2} + 5b^{2} for a, b ∈ ℤ, which is absurd as we have already seen. Thus, Cl_{O_{K}} consists of two elements and is hence isomorphic to ℤ/(2). For L, we proceed as above and find

(2) = (2, \sqrt{−6})^{2}
(3) = (3, \sqrt{−6})^{2}.

As before, it is not hard to see that (2, \sqrt{−6}), (3, \sqrt{−6}) are not principal. However, we have

(2, \sqrt{−6})(3, \sqrt{−6}) = (6, 2\sqrt{−6}, 6\sqrt{−6}, −6) = (\sqrt{−6}),

so that if [α] denotes the ideal class of α ⊂ O_{L}, we have

[(2, \sqrt{−6})] = [(2, \sqrt{−6})]^{−1},

from which it follows that Cl_{O_{L}} ≃ ℤ/(2).

6. Suppose that K is a norm Euclidean imaginary quadratic field. Then given any α, β ∈ O_{K}, there exists

q, r ∈ O_{K} such that

α = βq + r

and

N_{K/ℚ}(r) < N_{K/ℚ}(β).

Observe that O_{K} is a lattice in ℂ and that the norm of any α ∈ K is just the square of its usual euclidean distance from the origin (α = u + v\sqrt{−D} = u + iv\sqrt{D} with u, v ∈ ℚ, and N_{K/ℚ}(α) = u^{2} + dv^{2} while the distance of (u, v\sqrt{D}) from the origin is u^{2} + Dv^{2}). Thus, K is norm euclidean if and only if given α, β as above we can find q ∈ O_{K}, t ∈ K such that

\frac{α}{β} = q + t

where N_{K/ℚ}(t) < 1. Therefore, K is norm euclidean if and only if given any point x of ℂ (since any point of ℂ can be approximated as closely as we like by an element α/β of K with α, β ∈ O_{K} as above) we can find a lattice point q ∈ O_{K} of distance less than 1 away from x. Equivalently, K is norm euclidean if and only if the minimum distance from any interior point of a fundamental domain to a lattice point on the boundary of that fundamental domain is strictly less than 1. We then have two cases.

(a) −D ≢ 1 mod 4 : In this case, O_{K} = ℤ ⊕ ℤ\sqrt{−D}. A fundamental cell F for the lattice is the rectangle of side lengths 1, \sqrt{D}. Given a point p = x + iy ∈ F, let d(p) = min_{q∈O_{K}} |p − q|^{2}, where | ∗ | denotes the usual complex absolute value. Then it is obvious that

sup_{p∈F} d(p)

occurs when p = 1/2 + \sqrt{−D}/2. In this case, p is equidistant from the four neighboring lattice points and the distance to any of them is

\frac{\sqrt{1 + D}}{2} .

This is less than 1 if and only if D = 1 or 2.

(b) −D ≡ 1 mod 4 : Here, O_{K} = ℤ ⊕ ℤ (\frac{1+\sqrt{−D}}{2} ) . A fundamental cell F for the lattice is the parallelogram with vertices

\{(0, 0), (1, 0), (1/2, \sqrt{D}/2), (3/2, \sqrt{D}/2)\} .

Let d(p) be as above. Again, it is not difficult to see that

sup_{p∈F} d(p)

occurs when p = 3/4 + \sqrt{D}/4. In this case, we have

d(P) = \frac{1 + D}{16} .

It follows that we must have D < 15. Since we are requiring that −D ≡ 1(4) the only possible choices are

D = 3, 7, 11.

Thus, the only norm euclidean imaginary quadratic fields are

ℚ(\sqrt{−1}), ℚ(\sqrt{−2}), ℚ(\sqrt{−3}), ℚ(\sqrt{−7}), ℚ(\sqrt{−11}).

It is not true that every quadratic imaginary field with class number one is norm euclidean. For example, ℚ(\sqrt{−19}) has class number one but is not norm euclidean as the following counterexample shows:

α = 3 + \sqrt{−19}, β = 4.

In this case, the closest lattice points to α/β are 1 and \frac{1+\sqrt{−19}}{2}. But in one case we have

N_{K/ℚ}(α − β) = 20 ≮ N_{K/ℚ}(β) = 16

while in the other,

N_{K/ℚ} (α − β (\frac{1 + \sqrt{−19}}{2})) = N_{K/ℚ}(1 − \sqrt{−D}) = 20 ≮ N_{K/ℚ}(β) = 16.

7. Let O be a dedekind domain and 𝔞 a nonzero ideal of O. Suppose that

𝔞 = 𝔭_{1}^{a_{1}} … 𝔭_{r}^{a_{r}}.

Then by the chinese remainder theorem we have

O/𝔞 ≃ ⨁_{i=1}^{r} O/𝔭_{i}^{a_{i}}.

Suppose that every ideal of O/𝔭_{i}^{α_{i}} is principal and let I ⊂ O/𝔞 be an ideal of O/𝔞. Then I corresponds to an ideal Ĩ ⊂ O containing 𝔞. Clearly, Ĩ/𝔭_{i}^{a_{i}} is an ideal of O/𝔭_{i}^{a_{i}} and is principal by assumption, say

O/𝔭_{i}^{a_{i}} = x_{i}O/𝔭_{i}^{a_{i}}.

Then we see that I is a principal ideal of O/𝔞 generated by

(x_{1}, x_{2}, …, x_{r}) ∈ ⨁_{i=1}^{r} O/𝔭_{i}^{a_{i}} ≃ O/𝔞.

It therefore suffices to show that for any prime ideal 𝔭, every ideal of O/𝔭^{m} is principal. Let I be any ideal of O/𝔭^{m}. Then I corresponds to an ideal Ĩ of O containing 𝔭^{m}. But since 𝔭^{m} ⊂ Ĩ we have Ĩ|𝔭^{m} and hence that Ĩ = 𝔭^{n} for some n ≤ m. If n = m then I is the zero ideal and obviously principal. Otherwise we may assume that n < m. Let α ∈ 𝔭^{n} \ 𝔭^{n+1}. Then we have the containments

𝔭^{n} ⊇ 𝔭^{m} + (α) ⊋ 𝔭^{m},

since α ∉ 𝔭^{n+1} ⊆ 𝔭^{m}. We therefore must have

𝔭^{n}|𝔭^{m} + (α)|𝔭^{m},

so that

𝔭^{m} + (α) = 𝔭^{i}

for some n ≤ i < m. Suppose n < i. Then α ∈ 𝔭^{m} + (α) = 𝔭^{i}. But α ∉ 𝔭^{n+1} and hence α ∉ 𝔭^{i}. It follows that n = i and

𝔭^{n} = 𝔭^{m} + (α).

Therefore,

I = αO/𝔭^{m}

is principal. We now show that every ideal of O is generated by two elements. Let I ⊂ O be an ideal, and let α ∈ O. We have seen that every ideal of O/(α) is principal. Since (α) ⊂ I, it follows that I/(α) is an ideal of O/(α) and is hence principal, say I/(α) = β̄O/(α). We thus have I = (α, β), where β is any lift of b̄.

8. Let K be a number field, p a prime number and 𝔭 a non zero prime ideal of O_{K}. Set E_{r} = ℚ(ζ_{p^{r}}) and L_{r} = K(ζ_{p^{r}}). We will show that there exists a constant c(𝔭) depending only on 𝔭 (once p is fixed) such that 𝔭 splits into at most c(𝔭) prime factors in L_{r} for all r. Consider the following diagram of field extensions:

[figure: L_{r}; K; E_{r}; ℚ]

Since L_{r} = E_{r}K, it follows that deg(L_{r}/E_{r}) ≤ deg(K/ℚ). Let q = 𝔭 ∩ ℤ. We claim that it suffices to show that there exists a constant c(q) such that q splits into at most c(q) primes in E_{r} for all r. Indeed, let 𝔮 be any prime of E_{r} and write 𝔮 = ℘_{1}^{e_{1}} ⋯ ℘_{s}^{e_{s}} with ℘_{i} primes of O_{L_{r}}. Then we have e_{1}f_{1} + ⋯ + e_{s}f_{s} = deg(L_{r}/E_{r}) ≤ deg(K/ℚ). In particular, since e_{i}, f_{i} ≥ 1, we have s ≤ deg(K/ℚ), so that there exists some constant C independent of r such that every prime of E_{r} splits into at most C factors in L_{r}. We now show that there exists some constant c(q) such that q splits into at most c(q) prime factors in E_{r} for all r > 0. We distinguish two cases:

(a) p = q. Since p totally ramifies in E_{r} for all r, there is exactly one prime above p. In this case, we see that 𝔭 splits into at most deg(K/Q) primes in L_{r} for any r.

(b) p ≠ q. Let 𝔮 be a prime over q in E_{r}. Recall from lecture that E_{r}/ℚ is Galois and that

e(𝔮/q) = 1

and f(𝔮/q) = f_{r} is the order of q modulo p^{r}. Since e_{r}f_{r}d_{r} = φ(p^{r}), where d_{r} is the number of primes lying over q in E_{r}, we see that

d_{r} = \frac{p^{r−1}(p − 1)}{f_{r}}

for any r. Now let r_{0} be such that

q^{f_{1}} = 1 + cp^{r_{0}},

where p |̸c. Then f_{1} = f_{r} for 1 ≤ r ≤ r_{0}. Observe that

q^{f_{1}p^{j−1}} ≡ 1 + cp^{r_{0}+j−1} mod p^{r_{0}+j} ≢ 1 mod p^{r_{0}+j}

but

q^{f_{1}p^{j}} ≡ 1 + cp^{r_{0}+j} mod p^{r_{0}+j} ≡ 1 mod p^{r_{0}+j}

for any j ≥ 1. It follows that f_{r_{0}+j}|f_{1}p^{j} but f_{r_{0}+j} |̸f_{1}p^{j−1}, and hence (since f_{1}|f_{r} for any r ≥ 1) that

f_{r_{0}+j} = f_{1}p^{j}.

Therefore,

d_{r_{0}+j} = \frac{p^{r_{0}+j−1}(p − 1)}{f_{1}p^{j}} = \frac{p^{r_{0}−1}(p − 1)}{f_{1}}

for all j ≥ 1. This proves our claim. Putting this analysis together, it is not difficult to see that

c(𝔭) = \frac{p^{r_{0}−1}(p − 1)}{f_{1}} deg(K/Q),

where f_{1} is the order of l = 𝔭 ∩ ℤ modulo p and r_{0} is such that

l^{f_{1}} ≡ 1 mod p^{r_{0}}

but

l^{f_{1}} ≢ 1 mod p^{r_{0}+1}.

9. Let L = K[α] be a finite, separable extension and let f be the minimal monic polynomial of α over K. Set n = deg(f) and let σ_{i} for i = 1, …, n be the n embeddings of L in ℂ. We have

Δ(1, α, α^{2}, …, α^{n−1}) = |\begin{matrix}1 & α & α^{2} & ⋯ & α^{n−1} \\ 1 & σ_{1}(α) & σ_{1}(α)^{2} & ⋯ & σ_{1}(α)^{n−1} \\ ⋮ & & & ⋯ & ⋮ \\ 1 & σ_{n}(α) & σ_{n}(α)^{2} & ⋯ & σ_{n}(α)^{n−1} \end{matrix}|^{2}

= ∏_{1≤i<j≤n}(σ_{i}(α) − σ_{j}(α))^{2}

= (−1)^{n(n−1)/2} ∏_{i≠j}(σ_{i}(α) − σ_{j}(α)),

where we have used the Vandermonde determinant identity in the second line. Now we have

f′(σ_{i}(α)) = ∏_{j≠i}(σ_{i}(α) − σ_{j}(α)),

so that

Δ(1, α, α^{2}, …, α^{n−1}) = (−1)^{n(n−1)/2} ∏_{i=1}^{n} f′(σ_{i}(α)) = (−1)^{n(n−1)/2} N_{L/K}(f′(α)).

10. Let K_{1} = ℚ(α), K_{2} = ℚ(β), where α, β are roots of p = x^{3} − 8x + 15, q = x^{3} + 10x + 1 respectively. We show that K_{1}, K_{2} have the same discriminant but that they are not isomorphic. Let us first compute Δ(1, α, α^{2}) and Δ(1, β, β^{2}). It is clear that p, q are irreducible since any rational root of p would be an integer dividing 15, while a rational root of q would be an integer dividing 1. Neither p nor q have such roots. Thus, by Exercise 9 we have

Δ(1, α, α^{2}) = − N_{K_{1}/ℚ}(3α^{2} − 8)
= −(3a_{0}^{2} − 8)(3a_{1}^{2} − 8)(3a_{2}^{2} − 8)
= −(27a_{0}^{2}a_{1}^{2}a_{2}^{2} − 72(a_{0}^{2}a_{1}^{2} + a_{0}^{2}a_{2}^{2} + a_{1}^{2}a_{2}^{2}) + 192(a_{0}^{2} + a_{1}^{2} + a_{2}^{2}) − 512)

= −(27(a_{0}a_{1}a_{2})^{2} − 72((a_{0}a_{1} + a_{1}a_{2} + a_{0}a_{2})^{2} − 2(a_{0} + a_{1} + a_{2})a_{0}a_{1}a_{2}) + 192((a_{0} + a_{1} + a_{2})^{2}

− 2(a_{0}a_{1} + a_{0}a_{2} + a_{1}a_{2})))
= −(27(−15)^{2} − 72((−8)^{2}) + 192(−2(−8)) − 512)
= −4027,

where α = a_{0}, a_{1}, a_{2} are the roots of p. Similarly, we compute

Δ(1, β, β^{2}) = − N_{K_{2}/ℚ}(3β^{2} + 10)
= −(3b_{0}^{2} + 10)(3b_{1}^{2} + 10)(3b_{2}^{2} + 10)
= −(27b_{2}^{2}b_{1}^{2}b_{0}^{2} + 90(b_{0}^{2}b_{1}^{2} + b_{0}^{2}b_{2}^{2} + b_{1}^{2}b_{2}^{2}) + 300(b_{0}^{2} + b_{1}^{2} + b_{2}^{2}) + 1000)

= −(27(b_{0}b_{1}b_{2})^{2} + 90((b_{0}b_{1} + b_{1}b_{2} + b_{0}b_{2})^{2} − 2(b_{0} + b_{1} + b_{2})b_{0}b_{1}b_{2}) + 300((b_{0} + b_{1} + b_{2})^{2}

− 2(b_{0}b_{1} + b_{0}b_{2} + b_{1}b_{2})) + 1000)
= −(27(1) + 90(10^{2}) + 300(−20) + 1000)
= −4027.

Now observe that 4027 is prime so that, in particular, it is square free. It follows that

O_{K_{1}} = ℤ[α]

and

O_{K_{2}} = ℤ[β],

and hence that K_{1}, K_{2} have the same discriminant (-4027). However, K_{1}, K_{2} are not isomorphic. For consider the factorization of 17O_{K_{i}} in O_{K_{i}} for i = 1, 2. We have (as is easily verified)

x^{3} − 8x + 15 ≡ (x + 4)(x + 6)(x + 7) mod 17,

while x^{3} + 10x + 1 is prime modulo 17. It follows that 17 is totally split in O_{K_{1}} but inert in O_{K_{2}}; it follows that K_{1} cannot be isomorphic to K_{2}.

11. Let K = ℚ(ζ_{p^{r}}). We have seen that O_{K} = ℤ[ζ_{p^{r}}] so that

Δ_{K} = Δ(1, ζ_{p^{r}}, ζ_{p^{r}}^{2}, …, ζ_{p^{r}}^{φ(p^{r})−1}) = N_{K/ℚ}(f′(ζ_{p^{r}})),

where f is the minimal monic polynomial of ζ_{p^{r}} over ℚ. We have also seen that

f(x) = \frac{x^{p^{r}} − 1}{x^{p^{r−1}} − 1}.

We therefore have

p^{r}x^{p^{r}−1} = (x^{p^{r−1}} − 1)f′(x) + (p^{r−1}x^{p^{r−1}−1})f(x),

so that

f′(ζ_{p^{r}}) = \frac{p^{r}ζ_{p^{r}}^{p^{r}−1}}{ζ_{p^{r}}^{p^{r−1}} − 1} = \frac{p^{r}}{ζ_{p^{r}}(ζ_{p^{r}}^{p^{r−1}} − 1)}.

Therefore, we have

N_{K/ℚ}(f′(ζ_{p^{r}})) = \frac{N_{K/Q}(p^{r})}{N_{K/Q}(ζ_{p^{r}}) N_{K/Q}(ζ_{p^{r}}^{p^{r−1}} − 1)}.

Clearly, N_{K/ℚ}(ζ_{p^{r}}) = 1 and

N_{K/Q}(ζ_{p^{r}}^{p^{r−1}} − 1) = N_{K/Q}(ζ_{p} − 1) = (N_{ℚ(ζ_{p})/ℚ}(ζ_{p} − 1))^{deg(K/ℚ(ζ_{p}))} = p^{p^{r−1}},

since

N_{ℚ(ζ_{p})/ℚ}(ζ_{p} − 1) = ∏_{(p−1,j)=1}(1 − ζ_{p}^{j}) = 1 + 1^{2} + ⋯ + 1^{p−1} = 1.

Therefore,

N_{K/ℚ}(f′(ζ_{p^{r}})) = \frac{p^{rφ(p^{r})}}{p^{p^{r−1}}}
= p^{rp^{r−1}(p−1)−p^{r−1}}
= p^{p^{r−1}(pr−r−1)}.

It follows that

Δ_{K} = ±p^{p^{r−1}(pr−r−1)}.

12. Using Minkowski, we find that every O ideal is equivalent to some ideal J with N(J) ≤ 3. In this case, O = ℤ[\frac{1+\sqrt{−23}}{2}] := ℤ[α], and we look at the irreducible polynomial

X^{2} − X + \frac{1 + 23}{4} = X^{2} − X + 6

modulo 2, 3. This gives the factorizations

(2) = (2, α)(2, α − 1) = P_{2}P̄_{2}
(3) = (3, α)(3, α − 1) = P_{3}P̄_{3}.

Since \frac{1+\sqrt{−23}}{2} \frac{1−\sqrt{−23}}{2} = 6 and the two factors are conjugate, we must have

(\frac{1 + \sqrt{−23}}{2}) = P_{2}P_{3}

up to relabeling. Therefore, Cl_{O} is generated by [P_{2}]. Now, the factorization

\frac{3 + \sqrt{−23}}{2} \frac{3 − \sqrt{−23}}{2} = 2^{3}

tells us that

(\frac{3 + \sqrt{−23}}{2}) = P_{2}^{3}

up to relabeling and units (it cannot be that only a single or double power of P_{2} divides \frac{3+\sqrt{−23}}{2}, for

then we would have (\frac{3+\sqrt{−23}}{2}) = P_{2}^{2}P̄_{2}, up to relabeling, which is impossible since [P_{2}]^{2}[P̄_{2}] = [P_{2}] is

not principal). Therefore, [P_{2}]^{3} = 1 so that [P_{2}] must have order 3. It follows that

Cl_{O} ≃ ℤ/3ℤ

and that the three ideal classes are represented by

1, P_{2}, P_{2}^{2}.

13. Let K be a number field and 𝔪 ⊂ O_{K} an ideal. We show that every ideal class of Cl_{O_{K}} contains an integral ideal 𝔞 such that (𝔪, 𝔞) = 1. For any ideal class 𝔠, let J ∈ 𝔠 be an integral ideal representing 𝔠 (recall that we showed in class that such a J exists). Let

℘_{1}, …, ℘_{n}

be the primes (if any) dividing 𝔪 that do not divide J. Further, write

J = 𝔮_{1}^{e_{1}} ⋯ 𝔮_{s}^{e_{s}}.

By the chinese remainder theorem, we can find α ∈ O_{K} such that

α ∈ 𝔮_{i}^{e_{i}} \ 𝔮_{i}^{e_{i}+1}

and

α ≡ 1 mod ℘_{j}

for all i, j. Then we clearly have

(α) = ∏_{i=1}^{s} 𝔮_{i}^{e_{i}} ∏_{j=s+1}^{s+t} 𝔭_{j}^{e_{j}} := J𝔞

with 𝔮_{i} |̸𝔞 for any i. By construction, 𝔭_{j} |̸𝔪 for any j since if 𝔭_{j}|𝔪 for some j then we have a prime not dividing J that divides 𝔪 and (α), but we chose α to be 1 modulo all of these primes. Therefore, 𝔞 is an integral ideal in the class of J^{−1} (since the product J𝔞 is principal) with (𝔪, 𝔞) = 1. It follows that every ideal class contains an integral ideal 𝔞 with (𝔪, 𝔞) = 1 since every class is the inverse of some other class.