Tag Archives: quadratic integer ring

Exhibit an algebraic integer in a quadratic integer ring having a given norm and trace

Find an algebraic integer \alpha \in \mathbb{Q}(\sqrt{D}) having norm 31 and trace 17.


Let \alpha = a+b\sqrt{D} and \overline{\alpha} = a-b\sqrt{D}. Now N(\alpha) = \alpha\overline{\alpha} = a^2 - Db^2 and \mathsf{tr}(\alpha) = \alpha + \overline{\alpha} = 2a. We wish to find a, b, and D such that a+b\sqrt{D} is an algebraic integer, a^2-Db^2 = 31, and 2a = 17. If such an integer exists, then a = \frac{17}{2} is a half-integer- in particular, we must have D \equiv 1 mod 4. Substituting, and letting b = \frac{b_0}{2}, we have Db_0^2 = 165 = 3 \cdot 5 \cdot 11. Thus D = 165 and b_0 = \pm 1, so that b = \pm \frac{1}{2}. Indeed, we can verify that N(\frac{17}{2} + \frac{1}{2}\sqrt{165}) = 31 and \mathsf{tr}(\frac{17}{2} + \frac{1}{2}\sqrt{165}) = 17.

Factor the principal ideals in an algebraic integer ring which are generated by ramified rational primes

Let K = \mathbb{Q}(\sqrt{-5}) and let \mathcal{O} be the ring of integers in K. Factor (p) in \mathcal{O}, where p is a ramified rational prime.


By Theorem 9.6, if p does not divide the discriminant d = -20 of K (using Theorem 6.11), then p is not ramified. Now 20 = 2^2 \cdot 5, so that if p is ramified in K, it is either 2 or 5.

We claim that (2) = (2,1+\sqrt{-5})^2. Indeed, the (\supseteq) direction is clear, and we have 2 = -(1+\sqrt{-5})^2 + 2 \cdot (1+\sqrt{-2}) - 2 \cdot 2. We claim also that P = (2,1+\sqrt{-5}) is maximal. To this end, let a+b \sqrt{-5} \in \mathcal{O}, and say a - b \equiv c mod 2 where c \in \{0,1\}. Evidently, a+b\sqrt{-5} \equiv c mod P; if c \equiv 0 mod 2 then a+b\sqrt{-5} \equiv 0 mod P, and if c \equiv 1 mod 2 then a+b\sqrt{-5} \equiv 1 mod P. Now suppose 1 \in P; then 1 = 2(a+b\sqrt{-5}) + (1+\sqrt{-5})(h+k\sqrt{-5}). Comparing coefficients mod 2, we have 0 \equiv h+k \equiv 1 mod 2, a contradiction. So \mathcal{O}/(2,1+\sqrt{-5}) = \{ \overline{0}, \overline{1}\}, and thus \mathcal{O}/(2,1+\sqrt{-5}) \cong \mathbb{Z}/(2) is a field. Hence P is maximal, and (2) = (2,1+\sqrt{-5})^2 is the prime factorization of (2).

Certianly (5) = (\sqrt{-5})^2. We claim that Q = (\sqrt{-5}) is prime. To see this, note that 5 \in (\sqrt{-5}). If a+b\sqrt{-5} \in \mathcal{O}, then a+b\sqrt{-5} \equiv a \equiv a_0 mod Q, where a_0 \in \{0,1,2,3,4\}. Suppose t \in Q \cap \mathbb{Z}. Then t = (a+b\sqrt{-5})\sqrt{-5} for some a,b \in \mathbb{Z}. Comparing coefficients, we have t \equiv 0 mod 5. In particular, r \not\equiv s mod Q, where r,s \in \{0,1,2,3,4\} are distinct. Thus \mathcal{O}/(\sqrt{-5}) \cong \mathbb{Z}/(5) is a field, so that (\sqrt{-5}) is maximal. Thus (5) = (\sqrt{-5})^2 is the prime factorization of (5).

Factor (10) in QQ(sqrt(-6))

Find two distinct factorizations of 10 in \mathbb{Q}(\sqrt{-6}). Factor 10 as a product of four ideals in \mathbb{Q}(\sqrt{-6}).


Evidently 10 = 2 \cdot 5 = (2 + \sqrt{-6})(2 - \sqrt{-6}); moreover, these factorizations are distinct since, as the only units in \mathbb{Q}(\sqrt{-6}) are \pm 1, 2 is not associate to either of 2 \pm \sqrt{-6}.

We claim that (10) = (2,2+\sqrt{-6})(2,2-\sqrt{-6})(5,2+\sqrt{-6})(5,2-2\sqrt{-6}). To see the (\supseteq) direction, note that this ideal product is generated by all 2^4 possible selections of one generator from each factor. Name the generators \alpha_1,\alpha_2,\beta_1,\beta_2 such that \alpha_1\alpha_2 = \beta_1\beta_2 = 10. Now any selection which includes \alpha_1 (without loss of generality) and does not include \alpha_2 must include both \beta_1 and \beta_2, and so this ideal product is in (10).

Note also that 5 \cdot 5 \cdot (2 + \sqrt{-6}) \cdot (2 - \sqrt{-6}) - 6 \cdot 2 \cdot 2 \cdot (2 + \sqrt{-6}) \cdot (2 - \sqrt{-6}) = 10.

Show that two ideals are distinct

Let \mathcal{O} be the ring of integers in \mathbb{Q}(\sqrt{-5}). Show that the ideals (3,1+2\sqrt{-5}) and (7,1-2\sqrt{-5}) are distinct.


Suppose to the contrary that these ideals are equal. Then in particular, we have 3 = 7(a+b\sqrt{-5}) + (1-2\sqrt{-5})(c+d\sqrt{-5}). Comparing coefficients yields the two integer equations 7a+c+10d = 0 and 7b+d-2c = 0. Mod 7, we have c+3d\equiv 3 and d+5c \equiv 0. Solving this system by substitution yields 0 \equiv 3 mod 7; hence the system has no solution.

So these ideals are distinct.

Compute the discriminant of a basis for a given ideal in a quadratic integer ring

Let \mathcal{O} be the ring of integers in \mathbb{Q}(\sqrt{D}). Find a basis for the ideal (4) over \mathbb{Z}. Then compute the discriminant of this basis.


Suppose first that D \not\equiv 1 mod 4.

Note that (4) = \{4a + 4b\sqrt{D} \ |\ a,b \in \mathbb{Z} \}. We claim that \{4,4\sqrt{D}\} is a basis for (4). Certainly (4,4\sqrt{D})_\mathbb{Z} = (4). Now if 4a+4b\sqrt{D} = 0 and b \neq 0, then \sqrt{D} = -a/b is rational, a contradiction. So 4a = 0, and a = b = 0. So \{4,4\sqrt{D}\} is a basis for (4) over \mathbb{Z}. Evidently, the discriminant of this basis is \Delta[4,4\sqrt{D}] = 1024D. Recall from Theorem 6.11 in TAN that the discriminant of \mathbb{Q}(\sqrt{D}) is \Delta_\mathcal{O} = 4D. Thus \Delta[4,4\sqrt{D}] = 16^2 \Delta_\mathcal{O}.

Now suppose D \equiv 1 mod 4.

Now (4) = \{ 4a + 2b(1+\sqrt{D}) \ |\ a,b \in \mathbb{Z}\}. We claim that \{4,2+2\sqrt{D}\} is a basis. Certainly (4,2+2\sqrt{D})_\mathbb{Z} = (4). If 4a + 2b + 2b\sqrt{D} = 0 and b \neq 0, then -(2a+b)/b = \sqrt{D}, a contradiction. So b = 0, and thus a = b = 0. Hence \{4,2+2\sqrt{D}\} is a basis for (4). Evidently the discriminant of this basis is \Delta[4,2+2\sqrt{D}] = 256D. Again recall from Theorem 6.11 in TAN that \Delta_\mathcal{O} = D; hence \Delta[4,2+2\sqrt{D}] = 16^2 \Delta_\mathcal{O}.

Find a basis for a given ideal in a quadratic integer ring

Let I = (3+i,7+i) be considered as an ideal in \mathbb{Z}[i]. Find a generator and a basis for I. Draw a diagram to visualize the elements of I on the complex plane.


Note that 7+i = (1+i)(2-i) and 3+i = (1+i)(4-3i), so that I \subseteq (1+i). Conversely, since -2i(3+i) + i(7+i) = 1+i, we have (1+i) \subseteq I. Hence I = (1+i).

We claim that B = \{1+i, -1+i\} is a basis for I over \mathbb{Z}. To see this, note that -1+i = i(1+i), so that (1+i,-1+i)_\mathbb{Z} \subseteq (1+i); certainly (1+i) \subseteq (1+i, -1+i)_\mathbb{Z}, so these sets are equal. Now suppose a(1+i) + b(-1+i) = 0. Comparing coefficients, we have a-b = 0 and a+b = 0, so that a = b = 0. Thus B is free as a generating set for I over \mathbb{Z}.

We can visualize this ideal in the complex plane as in the following diagram.

(1+i) in ZZ[i]

The ideal (1+i) in the Gaussian integers

Show that a given generating set is a basis for an ideal in a quadratic integer ring

Show that B = \{5+i, 2+3i\} is a basis for the ideal (2+3i) in \mathbb{Z}[i].


First we wish to show that (5+i,2+3i)_\mathbb{Z} = (2+3i); the (\supseteq) direction is clear. Now suppose \zeta = a(5+i) + b(2+3i) = (5a+2b) + (a+3b)i; note that \zeta = (2+3i)(a+b-ai) \in (2+3i), as desired.

Now suppose we have integers a,b \in \mathbb{Z} such that (5+1)a + (2+3i)b = 0. Comparing coefficients, we have 5a+2b = 0 and a+3b = 0. Evidently this system has only the solution a = b = 0, so that B is \mathbb{Z}-linearly independent. Hence B is a basis for (2+3i).

Find a generator and a basis for a given ideal in a quadratic integer ring

Show that A = \{4a+2\sqrt{2}b \ |\ a,b \in \mathbb{Z}\} is an ideal in the ring of algebraic integers in \mathbb{Q}(\sqrt{2}). Find a generator and a basis for this ideal.


First, since 2 \not\equiv 1 mod 4, the elements of A are certainly algebraic integers. (Integers in \mathbb{Q}(\sqrt{2}) have the form m+n\sqrt{2} with m,n \in \mathbb{Z}.) We claim that A = (2\sqrt{2}). To see this, note that if 4a+2\sqrt{2}b \in A, then 4a+2\sqrt{2}b = 2\sqrt{2}(b+a\sqrt{2}) \in (2\sqrt{2}). Conversely, if 2\sqrt{2}(a+b\sqrt{2}) \in (2\sqrt{2}), then 2\sqrt{2}(a+b\sqrt{2}) = 4b+2\sqrt{2}a \in A. So A = (2\sqrt{2}), and more generally, A is an ideal in the ring of integers in \mathbb{Q}(\sqrt{2}).

Now we claim that B = \{4, 2\sqrt{2}\} is a basis of A. It is certainly a generating set over \mathbb{Z}. Suppose now that 4a + 2\sqrt{2}b = 0. If b \neq 0, then \sqrt{2} = -4a/2b, so that \sqrt{2} is rational, a contradiction. So b = 0, and we have 4a = 0, so a = 0. Thus B is \mathbb{Z}-linearly independent, and hence a basis for A.

Factor some given algebraic integers in a quadratic field

Factor 5 + \sqrt{3} in \mathbb{Q}(\sqrt{3}) and 7 + \sqrt{-5} in \mathbb{Q}(\sqrt{-5}).


Note that N(5+\sqrt{3}) = 5^2 - 3 = 22 = 2 \cdot 11. If \alpha = \beta\gamma, then N(\beta)N(\gamma) = 2 \cdot 11, so (without loss of generality) N(\beta) = 2. Suppose \beta = h+k\sqrt{3}; then h^2-3k^2 = 2. Mod 3, we have h^2 \equiv 2. This is a contradiction since 2 is not a square mod 3. That is, no algebraic integer in \mathbb{Q}(\sqrt{3}) has norm 2. (Similarly, no integer has norm 11.) Thus 5+\sqrt{3} is irreducible as an integer in \mathbb{Q}(\sqrt{3}).

Evidently, (1+\sqrt{-5})(2-\sqrt{-5}) = 7+\sqrt{-5}. Now N(1+\sqrt{-5}) = 6 and N(2-\sqrt{-5}) = 9. We claim that no integer in \mathbb{Q}(\sqrt{-5}) has norm 3; to that end, suppose h^2+5k^2 = 3. We must have k = 0, so that h^2 = 3, a contradiction. In particular, both 1+\sqrt{-5} and 2-\sqrt{-5} are irreducible in \mathbb{Q}(\sqrt{-5}).

The polynomial x² – √2 is irreducible over ZZ[√2]

Prove that p(x) = x^2 - \sqrt{2} is irreducible over \mathbb{Z}[\sqrt{2}].


In this previous exercise, we showed that \mathbb{Z}[\sqrt{2}] is a Euclidean domain, and thus a unique factorization domain.

If p(x) is reducible, then it has a root; say \alpha = a+b\sqrt{2}. Evidently, then, we have (a^2 + 2b^2) + 2ab \sqrt{2} = \sqrt{2}. Comparing coefficients, 2ab = 1 for some integers a and b, a contradiction. Thus p(x) is irreducible over \mathbb{Z}[\sqrt{2}].