Tag Archives: algebraic integer ring

Exhibit an extended principal generator for an ideal in an algebraic integer ring

Let A = (3,1+2\sqrt{-5}) be an ideal in the ring of integers of K = \mathbb{Q}(\sqrt{-5}). Find an algebraic integer \kappa such that A = (\kappa) \cap \mathcal{O}_E, where E = K(\kappa).


We saw in the text that the class number of K is 2. Note that A^2 = (3), as indeed 3 = (1+2\sqrt{-5})^2 - 2 \cdot 3^2. Let \kappa = \sqrt{3}. By our proof of Theorem 10.6, A = \{ \tau \in \mathcal{O}_K \ |\ \tau/\kappa \in \mathcal{O}_K \}.

If A² is principal over an algebraic number field, then A² is principal over any extension

Let K_1 and K_2 be algebraic number fields with K_1 \subseteq K_2, and having integer rings \mathcal{O}_1 and \mathcal{O}_2, respectively. Suppose (A)_{\mathcal{O}_1} \subseteq \mathcal{O}_1 is an ideal such that (A)^2 is principal. Show that (A)_{\mathcal{O}_2}^2 is principal in \mathcal{O}_2.


We will let G_1 and G_2 denote the ideal class groups of K_1 and K_2, respectively. For brevity, if S \subseteq \mathcal{O}_1 is a subset, then we will denote by (S)_1 the ideal generated by S in \mathcal{O}_1 and by (S)_2 the ideal generated by S in \mathcal{O}_2.

We claim that if (A)_1 \sim (B)_1, then (A)_2 \sim (B)_2. To see this, suppose we have \alpha and \beta such that (\alpha)_1(A)_1 = (\beta)_1(B)_1. Then (\alpha A)_1 = (\beta B)_1. Now if x = \sum r_i \alpha a_i \in (\alpha A)_2, then each \alpha a_i is in (\beta B)_1, and so has the form \sum s_{i,j} \beta b_j. So x \in (\beta B)_2. Conversely, (\beta B)_2 \subseteq (\alpha A)_2, and so (A)_2 \sim (B)_2.

The mapping A \mapsto (A)_2 then induces a well-defined mapping \Psi : G_1 \rightarrow G_2 given by [A] \mapsto [(A)_2]. Moreover, since \Psi([A][B]) = \Psi([AB]) = [(AB)_2] = [(A)_2][(B)_2] = \Psi(A) \Psi(B), \Psi is a group homomorphism.

In particular, if A^2 \sim (1) in \mathcal{O}_1, then \Psi(A) \sim (1) in \mathcal{O}_2, as desired.

The exponent of the smallest power of an ideal which is principal divides the class number

Let \mathbb{O} be the ring of integers in an algebraic number field K of class number k. Let A be an ideal. Show that if k is minimal such that A^k is principal, then k|h.


This k is precisely the order of [A] in the class group of K. The result then follows by Lagrange’s Theorem.

The class number of QQ(sqrt(-6)) is 2

Compute the class number of \mathbb{Q}(\sqrt{-6}).


Recall that the ring of integers in this field is \mathcal{O} = \mathbb{Z}[\sqrt{-6}], and that \{1,\sqrt{-6}\} is an integral basis. We will now give an upper bound on the constant C which is proven to exist in Theorem 10.2.

Let A be an ideal in \mathcal{O}, and let t = \lfloor \sqrt{N(A)} \rfloor. That is, t is the largest integer such that t^2 \leq N(A). Now consider \{b_0 + b_1\sqrt{-6} \ |\ 0 \leq b_0,b_1 \leq t\}. Since (t+1)^2 < N(A), there exist some \beta_1,\beta_2 in this set such that \beta_2 - \beta_1 \in A. Say \alpha = \beta_2 - \beta_1 = a_0+a_1\sqrt{-6}, where (clearly) |a_0|,|a_1| \leq t. By the triangle inequality, |N(\alpha)| = |a_0^2 + 6a_1^2| \leq |a_1|^2 + 6|a_1|^2 \leq t^2 + 6t^2 = 7t^2 \leq 7 \cdot N(A). By our proof of Theorem 10.3, every class of ideals in \mathcal{O} contains an ideal whose norm is at most 7. We will now construct all of the ideals having norm k where k \in \{1,2,3,4,5,6,7\}.

(k=1) There is only one ideal of norm 1, namely A_1 = (1).

(k=2) If A is an ideal of norm 2, then A divides (2). Note that (2,\sqrt{-6})^2 = (4,2\sqrt{-6},-6) = (2). Now 4 = N((2)) = N((2,\sqrt{-6}))^2, so that N((2,\sqrt{-6})) = 2. By Corollary 9.15, (2,\sqrt{-6}) is prime. Thus A_2 = (2,\sqrt{-6}) is the only ideal of norm 2.

(k=3) If A is an ideal of norm 3, then A divides (3). Note that (3,\sqrt{-6})^2 = (9,3\sqrt{-6},-6) = (3). Now 9 = N((3)) = N((3,\sqrt{-6}))^2, so that N((3,\sqrt{-6})) = 3. By Corollary 9.15, (3,\sqrt{-6}) is prime. Thus A_3 = (3,\sqrt{-6}) is the only ideal of norm 3.

(k=4) Suppose A is an ideal of norm 4. If A is not prime, then it is a product of two prime ideals of norm 2; the only such ideal is (2). If A is prime, then by Theorem 9.19, A divides (2). But no prime ideals dividing (2) have norm 4. So the only ideal of norm 4 is A_4 = (2).

(k=5) If A is an ideal of norm 5, then A divides (5). Note that (5) = (5,2+\sqrt{-6})(5,2-\sqrt{-6}), since 5 = 5^2 - 2(2+\sqrt{-6})(2-\sqrt{-6}). We claim that both of these factors are proper. Indeed, if 1 \in (5,2+\sqrt{-6}), then 1 = 5(a_b\sqrt{-6}) + (2+\sqrt{-6})(h+k\sqrt{-6}) for some a,b,h,k \in \mathbb{Z}. Comparing coefficients mod 5, we have 1 \equiv 0, a contradiction. Thus (5,2+\sqrt{-6}) is proper. Similarly, (5,2-\sqrt{-6}) is proper. Now 25 = N((5)) = N((5,2+\sqrt{-6}))N((5,2-\sqrt{-5})), and neither factor is 1. So N((5,2+\sqrt{-6})) = N((5,2-\sqrt{-6})) = 5. Thus A_5 = (5,2+\sqrt{-6}) and A_6 = (5,2-\sqrt{-6}) are the only possible ideals of norm 5 in \mathcal{O}. Note that the discriminant of this field is not divisible by 5, so that these factors are distinct by Theorem 9.6. (Though this last step is not necessary.)

(k=6) If A is an ideal of norm 6, then A divides (6) = (2)(3). Thus the prime factors of A come from the set \{(2,\sqrt{-6}), (3,\sqrt{-6})\}. The only such ideal is A_7 = (\sqrt{-6}).

(k=7) If A is an ideal of norm 7, then A divides (7). Note that (7) = (1+\sqrt{-6})(1+\sqrt{-7}). Moreover, by Corollary 9.11, N((1+\sqrt{-6})) = N((1-\sqrt{-6})) = 7, and by Corollary 9.15, these ideals are prime. So the only ideals of norm 7 are A_8 = (1+\sqrt{-6}) and A_9 = (1-\sqrt{-6}).

Certainly we have A_1 \sim A_4 \sim A_7 \sim A_8 \sim A_9 \sim (1).

Note that A_2A_3 = A_7 \sim (1) and that A_2 = A_4 \sim (1). Thus A_2 \sim A_3.

We also have A_5A_6 = (5) and A_5^2 = (1-2\sqrt{-6}), since 1-2\sqrt{-6} = 25 + 2(2+\sqrt{-6})^2 - 2 \cdot 5 \cdot (2+\sqrt{-6}). So A_5A_6 \sim A_5^2, and thus A_5 \sim A_6.

We also that A_5A_2 = (2+\sqrt{-6}) since 5 \cdot 2 + 5 \cdot \sqrt{-6} - 2(2)(2+\sqrt{-6}) = 2+\sqrt{-6}). So A_5A_2 \sim A_5^2, and thus A_5 \sim A_2. So A_2 \sim A_3 \sim A_5 \sim A_6.

Finally, we claim that A_2 is not principal. If (2,\sqrt{-6}) = (\alpha), then N(\alpha) divides both 4 and 6, and so is either 1 or 2. But no element of \mathbb{Z}[\sqrt{-6}] has norm 2, and A_2 is proper; so A_2 is not principal, and A_2 \not\sim (1).

Thus \mathbb{Q}(\sqrt{-6}) has two ideal classes.

Detect equivalences among ideals of an algebraic integer ring

Determine which of the following ideals of \mathbb{Z}[\sqrt{-6}] are equivalent: A = (2,\sqrt{-6}), B = 5-2\sqrt{-6},2+2\sqrt{-6}), C = (2+\sqrt{-6}), and D = (5,1+2\sqrt{-6}).


First we argue that B is principal. Indeed, B = (5-2\sqrt{-6}, 2+2\sqrt{-6}) = (7,2+2\sqrt{-6}) = (1+\sqrt{-6})(1-\sqrt{-6},2). Now (1-\sqrt{-6})(1+\sqrt{-6}) - 3 \cdot 2 = 1 \in (1-\sqrt{-6},2), so that B = (1+\sqrt{-6})(1) = (1+\sqrt{-6}). Indeed, 5-2\sqrt{-6} = (1+\sqrt{-6})(-1-\sqrt{-6}) and 2+2\sqrt{-6} = 2(1+\sqrt{-6}), and 1+\sqrt{-6} = (1-\sqrt{-6})(5-21\sqrt{-6}) + 4(2+2\sqrt{-6}).

In particular, B \sim C \sim (1).

Now we claim that AD and A^2 are both principal. Indeed, AD = (2,\sqrt{-6})(5,1+2\sqrt{-6}) = (10,2+4\sqrt{-6},5\sqrt{-6},-12+\sqrt{-6}) = (10,2+4\sqrt{-6},5\sqrt{-6},-10) = (10,2-\sqrt{-6},5\sqrt{-6}). Since N(2-\sqrt{-6}) = 10 and (2-\sqrt{-6})(-3+\sqrt{-6}) = 5\sqrt{-6}, we have AD = (2-\sqrt{-6}). So AD \sim (1). Similarly, A^2 = (s,\sqrt{-6})^2 = (4,2\sqrt{-6},-6) = (2), so that A^2 \sim (1). Now AD \sim A^2, so that A \sim D.

Finally, we claim that A is not principal. If (2,\sqrt{-6}) = (\alpha), then \alpha|2 and \alpha|\sqrt{-6}, so that N(\alpha) divides both 4 and 6. So N(\alpha) \in \{1,2\}. Note, however, that no element of \mathbb{Z}[\sqrt{-6}] has norm 2 since the equation a^2 + 6b^2 = 2 has no solutions in \mathbb{Z}. If N(\alpha) = 1, then we have 1 \in D. But then 1 = 2(h+k\sqrt{-6}) + \sqrt{-6}(u+v\sqrt{-6}), so that 1=2h-6v, which is impossible mod 2. So D is not principal.

In summary, we have B \sim C \sim (1), A \sim D, and (1) \not\sim D.

The class number of QQ(sqrt(3)) is 1

Compute the class number of K = \mathbb{Q}(\sqrt{3}).


Recall that \mathbb{Z}[\sqrt{3}] = \mathcal{O} is the ring of integers in K, and that \{1,\sqrt{3}\} is an integral basis for K.

We will now compute an upper bound on the constant C which is shown to exist in Theorem 10.2. That is, we will find a number C such that, for every ideal A, there exists an element \alpha \in A such that |N(\alpha)| \leq C \cdot N(A).

To this end, let t = \lfloor \sqrt{N(A)} \rfloor– that is, t is the largest integer such that t^2 \leq N(A). Consider the set of integers \{b_0 + b_1\sqrt{3} \ |\ 0 \leq b_0,b_1 \leq t \}. Since (t+1)^2 > N(A) by our choice of t, there must exist two elements \beta_1,\beta_2 in this set which are congruent mod A. Say \alpha = \beta_2 - \beta_1 = a_0 + a_1\sqrt{3}.

Using the triangle inequality, we have |N(\alpha)| = |a_0^2 - 3a_1^2| \leq |a_0|^2 + 3|a_1|^2 \leq t^2 + 3t^2 = 4t^2 \leq 4 \cdot N(A). Thus the constant C is at most 4.

By our proof of Theorem 10.3, every class of ideals in \mathcal{O} contains an ideal of norm at most 4. We will now compute all of the ideals of norm k for k \in \{1,2,3,4\}.

(k=1) There is only one ideal of norm 1; namely D_1 = (1).

(k=2) Suppose A is an ideal of norm 2. By Theorem 9.19, A is a divisor of (2). We claim that (2) = (2,1+\sqrt{3})^2; indeed, (2,1+\sqrt{3})^2 = (4,2+2\sqrt{3},-2) = (2). Now 4 = N((2)) = N((2,1+\sqrt{3}))^2, so that N((2,1+\sqrt{3})) = 2. By Theorem 9.15, the ideal (2,1+\sqrt{3}) is prime, and so (2) = (2,1+\sqrt{3})^2 is the prime factorization of (2). So D_2 = (2,1+\sqrt{3}) is the only ideal of norm 2 in \mathcal{O}.

(k=3) Suppose A is an ideal of norm 3. By Theorem 9.19, A is a divisor of (3). Note that (3) = (\sqrt{3})^2, and that N((\sqrt{3})) = N(\sqrt{3}) = 3. Thus (3) = (\sqrt{3})^2 is the prime factorization of (3), and D_3 = (\sqrt{3}) is the only ideal of norm 3 in \mathcal{O}.

(k=4) Suppose A is an ideal of norm 4. If A is not prime, then it is a product of two prime ideals of norm 2. The only prime ideal of norm 2 is D_2, so that A = D_2^2 = (2). If A is prime, then by Theorem 9.19, A \supseteq (2). Since (2) has norm 4, in fact A = (2), which in this case is a contradiction. So D_4 = (2) is the only ideal of norm 4.

It is clear that D_1 \sim D_3 \sim D_4. Moreover, note that (1-\sqrt{3})D_2 = (2-2\sqrt{3}, -2) = (2) = (1)(2), so that D_2 \sim D_4.

Thus the class number of \mathbb{Q}(\sqrt{3}) is 1.

The set of ideal classes over an algebraic intger ring is an abelian group

Let \mathcal{O} be the set of integers in an algebraic number field K, and let G = \mathcal{O}/\sim denote the set of equivalence classes of ideals in \mathcal{O} under the relation A \sim B if and only if (\alpha)A = (\beta)B for some nonzero \alpha,\beta \in \mathcal{O}. Prove that G is an abelian group under the usual ideal product of class representatives.


We will denote by [A] the equivalence class containing the ideal A. We showed in this previous exercise that [A][B] = [AB] is a well-defined binary operator on G.

For all [A], [B], [C] \in G, we have [A]([B][C]) = [A][BC] = [A(BC)] = [(AB)C] = [AB][C] ([A][B])[C], so that our product is associative.

For all [A], we have [A][(1)] = [A(1)] = [A], and similarly [(1)][A] = [A].

For all [A], there exists (by Theorem 8.13) an ideal B such that AB is principal. Thus [A][B] = [(1)], and so every element of G has an inverse.

Hence G is a group.

Moreover, we have [A][B] = [AB] = [BA] = [B][A], so that G is abelian.

In an algebraic integer ring, multiplication of ideal classes is well defined

Let \mathcal{O} be the ring of integers in an algebraic number field K. Recall that we defined a relation \sim on the set of ideals in \mathcal{O} as follows: A \sim B if and only if there exist nonzero \alpha,\beta \in \mathcal{O} such that (\alpha)A = (\beta)B. Show that if A_1 \sim B_1 and A_2 \sim B_2, then A_1A_2 \sim B_1B_2.


We have \alpha_1,\alpha_2,\beta_1,\beta_2 such that (\alpha_1)A_1 = (\beta_1)B_1 and (\alpha_2)A_2 = (\beta_2)B_2. Then (\alpha_1\alpha_2)A_1A_2 = (\beta_1\beta_2)B_1B_2, so that A_1A_2 \sim B_1B_2.

Characterize the irreducibles in ZZ[sqrt(2)]

Characterize the irreducible elements in \mathbb{Z}[\sqrt{2}].


Suppose \zeta \in \mathbb{Z}[\sqrt{2}] is irreducible. By our proof of Theorem 8.5, it follows that (\zeta) \supseteq (p) for some unique positive rational prime p. That is, p = \zeta\eta for some \eta \in \mathbb{Z}[\sqrt{2}]. Now p^2 = N(\zeta)N(\eta), so that either N(\zeta) = \pm p or N(\zeta) = \pm p^2. (Recall that N(\zeta) is a rational integer and not 1.) Let \zeta = a+b\sqrt{2}.

Suppose N(\zeta) = \pm p. Then \pm p = a^2 - 2b^2, so that a^2 - 2b^2 is a rational prime.

Suppose N(\zeta) = \pm p^2. Now N(\eta) = \pm 1, so that (by Theorem 7.3) \eta is a unit. (In Theorem 7.9 we saw that the units in \mathbb{Z}[\sqrt{2}] have the form \pm(1+\sqrt{2})^k where k \in \mathbb{Z}.) Thus \zeta is an associate of p in \mathbb{Z}[\sqrt{2}]. We claim that in this case, there does not exist a solution (h,k) of the equation h^2 - 2k^2 = \pm p. If so, then \pm p = (h+k\sqrt{2})(h-k\sqrt{2}). But since \pm p is irreducible, either h+k\sqrt{2} or h-k\sqrt{2} is a unit- so p = \pm 1, a contradiction.

In summary, if \zeta = a+b\sqrt{2} is irreducible in \mathbb{Z}[\sqrt{2}], then either (1) N(\zeta) = \pm p is a rational prime or (2) N(\zeta) = \pm p^2 is the square of a rational prime and the equations h^2-2k^2 = \pm p have no solutions (h,k) in \mathbb{Z}.

Conversely, if N(\zeta) = p is prime, then \zeta is irreducible. If N(\zeta = \pm p^2 where p is a rational prime and if h^2-2k^2 = \pm p has no solution, then no element of \mathbb{Z}[\sqrt{2}] has norm \pm p, so that \zeta is irreducible.

No prime ideal in an algebraic integer ring has norm 12

Can there exist an algebraic integer ring \mathcal{O} with a prime ideal A \subseteq \mathcal{O} such that N(A) = 12?


Recall that if A is prime, then it is maximal, and that N(A) is precisely |\mathcal{O}/A|. Since A is maximal, \mathcal{O} is a field. We claim that no field of order 12 exists.

Suppose to the contrary that F is a field of order 12. The characteristic of F is then either 2 or 3, since it must be a prime dividing 12. Moreover, by Cauchy’s Theorem, F contains nonzero elements \alpha and \beta of (additive) order 2 and 3, respectively. That is, 2\alpha = 0 and 3\beta = 0, but no smaller multiple of \alpha or \beta is 0. If F has characteristic 2, then \beta = 0, and if F has characteristic 3, then 4\alpha = \alpha = 0, both contradictions.

Since no field of order 12 exists, no prime ideal can have norm 12.