Tag Archives: quadratic field

Exhibit a quadratic field as a field of matrices

Let K = \mathbb{Q}(\sqrt{D}), where D is a squarefree integer. Let \alpha = a+b\sqrt{D} be in K, and consider the basis B = \{1,\sqrt{D}\} of K over \mathbb{Q}. Compute the matrix of the \mathbb{Q}-linear transformation ‘multiplication by \alpha‘ (described previously) with respect to B. Give an explicit embedding of \mathbb{Q}(\sqrt{D}) in the ring \mathsf{Mat}_2(\mathbb{Q}).


We have \varphi_\alpha(1) = a+b\sqrt{D} and \varphi(\alpha)(\sqrt{D}) = bD + a\sqrt{D}. Making these the columns of a matrix M_\alpha, we have M_\alpha = \begin{bmatrix} a & bD \\ b & a \end{bmatrix}, and this is the matrix of \varphi_\alpha with respect to B. As we showed in the exercise linked above, \alpha \mapsto M_\alpha is an embedding of K in \mathsf{Mat}_2(\mathbb{Q}).

Compare to this previous exercise about \mathbb{Z}[\sqrt{D}].

Exhibit a pair of algebraic integers in QQ(sqrt(-5)) which are relatively prime but for which Bezout’s identity does not hold

Show that 3 and 1 + \sqrt{-5} are relatively prime as algebraic integers in \mathbb{Q}(\sqrt{-5}), but that there do not exist algebraic integers \lambda,\mu \in \mathbb{Q}(\sqrt{-5}) such that 3\lambda + (1+\sqrt{-5})\mu = 1.


Let a+b\sqrt{-5} be an arbitrary integer in \mathbb{Q}(\sqrt{-5}). Since -5 \equiv 3 \not\equiv 1 mod 4, a and b are integers. Note that N(a+b\sqrt{-5}) = a^2 + 5b^2. If this element has norm 3, then taking this equation mod 5 we have a^2 \equiv 3. Note, however, that the squares mod 5 are 0, 1, and 4. In particular, no algebraic integer in \mathbb{Q}(\sqrt{-5}) has norm 3.

Now N(3) = 9 and N(1+\sqrt{-5}) = 6; if these elements have a factorization, then some factor must be a unit. Thus both 3 and 1 + \sqrt{-5} are irreducible as integers in \mathbb{Q}(\sqrt{-5}). By Theorem 7.7 in TAN, the units in this ring are precisely \pm 1, so that 3 and 1 + \sqrt{-5} are not associates.

If \delta|3 and \delta|(1+\sqrt{-5}), then since N(\delta) \neq 3, N(\delta) = \pm 1, and thus by Lemma 7.3 in TAN \delta is a unit. That is, every common divisor of 3 and 1+\sqrt{-5} is a unit, so that these elements are relatively prime.

Suppose now that there are integers a,b,c,d such that, with \lambda = a+b\sqrt{-5} and \mu = c+d\sqrt{-5}, we have 3\lambda + (1+\sqrt{5})\mu = 1.Comparing coefficients, we have 3a+c-5d = 1 and 3b+c+d = 0. Mod 3, these equations reduce to c+d \equiv 1 and c+d \equiv 0, a contradiction. So no such \lambda and \mu exist.

Describe the associates of an element in a given quadratic integer ring

Describe the associates of \sqrt{-3} in \mathbb{Q}(\sqrt{-3}) and of 2 in \mathbb{Q}(\sqrt{2}).


By Theorem 7.7 in TAN, the units in \mathbb{Q}(\sqrt{-3}) are \pm 1, (1 \pm \sqrt{-3})/2, and (-1 \pm \sqrt{-3})/2. So the associates of \sqrt{-3} are \pm \sqrt{-3} and (\pm 3 \pm \sqrt{-3})/2.

By Theorem 7.9 in TAN, the units in \mathbb{Q}(\sqrt{2}) are (1+\sqrt{2})^k with k \in \mathbb{Z}. So the associates of 2 are 2(1+\sqrt{2})^k with k \in \mathbb{Z}.

Characterize the units in QQ(sqrt(5))

Characterize the units in \mathbb{Q}(\sqrt{5}).


Note that every element of \mathbb{Q}(\sqrt{5}) is real; in particular, every element is either positive or negative. Note also that \lambda = (1+\sqrt{5})/2 is a unit with inverse (-1+\sqrt{5})/2.

We begin with a lemma.

Lemma: No unit \zeta in \mathbb{Q}(\sqrt{5}) satisfies 1 < \zeta < (1+\sqrt{5})/2. Proof: Suppose to the contrary that \zeta = x+y\sqrt{5} is such a unit. Now N(\zeta) = x^2 - 5y^2 = \pm 1, so that x - y\sqrt{5} = \pm 1/(x+y\sqrt{5}). Since x+y\sqrt{5} > 1, we have -1 < x-y\sqrt{5} < 1. Adding these inequalities, we have 0 < 2x < 3/2 + \sqrt{5}/2, and thus 0 < x < 3+\sqrt{5}. So 0 < x < 5.23 + \varepsilon. Since x \in \mathbb{Z}, we have x \in \{1,2,3,4,5\}. Consider again the inequality 1 < x+y\sqrt{5} < (1+\sqrt{5})/2; for x in this range, we have no integer solutions y. \square

Now let \eta \in \mathbb{Q}(\sqrt{5}) be a unit; say \eta > 0. Since \lambda = (1+\sqrt{5})/2 > 1, there exists k \in \mathbb{Z} such that \lambda^k \leq \eta < \lambda^{k+1}. Now 1 \leq \eta\lambda^{-k} < \lambda, and by the lemma (since \eta\lambda^{-k} is a unit), \eta = \lambda^k.

Thus every unit in \mathbb{Q}(\sqrt{5}) is an integer power of (1+\sqrt{5})/2.

If every element of a quadratic field has nonnegative norm, the field is imaginary

Prove that if every element of a quadratic field \mathbb{Q}(\sqrt{D}) has nonnegative norm, then D < 0.


In particular, N(\sqrt{D}) = -D > 0, so that D < 0.

In a quadratic field, the quotient of two algebraic integers with the same norm is a quotient of an algebraic integer by its conjugate

Let K = \mathbb{Q}(\sqrt{D}) be a quadratic field and let \alpha,\beta \in K be algebraic integers such that N(\alpha) = N(\beta). Prove that there exists an algebraic integer \gamma such that \alpha/\beta = \gamma/\gamma^\prime. Give a nontrivial example.


Note that 1 = N(\alpha)/N(\beta) = N(\alpha/\beta). We showed in this previous exercise that every element of K having norm 1 is a quotient of an algebraic integer by its conjugate. That proof was constructive.

We saw here that 3+4i and 5 are elements of \mathbb{Q}(i) which have the same norm (25) but which are not conjugates or associates. Following the constructive proof given previously, we see that \frac{3+4i}{5} = \frac{2+i}{2-i}.

In a quadratic field, every element of norm 1 is a quotient of an algebraic integer by its conjugate

Let K = \mathbb{Q}(\sqrt{D}) be a quadratic field, and let \epsilon \in K be an element with N(\epsilon) = 1. Prove that there exists an algebraic integer \gamma such that \epsilon = \gamma/\gamma^\prime, where \prime denotes conjugate.


Let \epsilon = a+b\sqrt{D}. If b = 0, then a = \pm 1, and either \epsilon = 1/1 or \epsilon = -\sqrt{D}/\sqrt{D}; similarly if a = -1. So we assume that b \neq 0 and a \neq -1. Now 1 = a^2 - Db^2. Rearranging, we have \frac{a-1}{bD} = \frac{b}{a+1}. Choose some integers k and h such that \frac{k}{h} is equal to this common ratio. Let \gamma = h+k\sqrt{D}; certainly \gamma is an algebraic integer in K. We claim that \epsilon = \gamma/\gamma^\prime.

To that end, note that \gamma^\prime \epsilon = (ah-bkD) + (bh - ak)\sqrt{D}. From \frac{k}{h} = \frac{b}{a+1}, we see that bh - ka = k, and likewise since \frac{k}{h} = \frac{a-1}{bD} we have ha - bkD = h. So \gamma^\prime\epsilon = \gamma, and thus \epsilon = \gamma/\gamma^\prime.

Exhibit elements in a quadratic field having the same norm but which are not conjugate or associate

Exhibit elements in a quadratic field which have the same norm but which are not conjugate or associate.


Let K = \mathbb{Q}(i). If (a,b,c) is a Pythagorean triple, then N(a+bi) = N(c), but a+bi and c are clearly neither conjugate nor associate. For instance, \{3+4i,5\} and \{5+12i,13\}.

In a quadratic field, rational primes have at most two irreducible factors

Let K be a quadratic extension of \mathbb{Q} and let p be a rational prime. Prove that, as an algebraic integer in K, p has at most two irreducible factors.


Recall that the conjugates of p for K are p itself with multiplicity 2. So the norm of p over K is p^2. Since (by Lemma 7.1) the norm of an algebraic integer is a rational integer and the norm is multiplicative, p has at most two irreducible integer factors in K.

17+14sqrt(7) is not the fifth power of an algebraic integer in QQ(sqrt(7))

Prove that 17+14\sqrt{7} is not the fifth power of an algebraic integer in \mathbb{Q}(\sqrt{7}).


Note that the conjugates of 17+14\sqrt{7} are 17 \pm 14\sqrt{7}, so that the norm of 17+14\sqrt{7} in \mathbb{Q}(\sqrt{7}) is 17^2 - 7\cdot 14^2 = -1083 = -3 \cdot 19^2.

Suppose there is an algebraic integer \zeta \in \mathbb{Q}(\sqrt{7}) such that \zeta^5 = 17+14\sqrt{7}. Then -3\cdot 19^2 = N(\zeta)^5; but N(\zeta) is a rational integer, a contradiction. So no such \zeta exists.