## 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.