Tag Archives: norm

Properties of a matrix norm

Given an n \times n matrix A over \mathbb{C}, define ||A|| = \sum_{i,j} |a_{i,j}|, where bars denote the complex modulus. Prove the following for all A,B \in \mathsf{Mat}_n(\mathbb{C}) and all \alpha \in \mathbb{C}.

  1. ||A+B|| \leq ||A|| + ||B||
  2. ||AB|| \leq ||A|| \cdot ||B||
  3. ||\alpha A|| = |\alpha| \cdot ||A||

Say A = [a_{i,j}] and B = [b_{i,j}].

We have ||A+B|| = \sum_{i,j} |a_{i,j} + b_{i,j}| \leq \sum_{i,j} |a_{i,j}| + |b_{i,j}| by the triangle inequality. Rearranging, we have ||A+B|| \leq (\sum_{i,j} |a_{i,j}| + \sum_{i,j} |b_{i,j}| = ||A|| + ||B|| as desired.

Now ||AB|| = \sum_{i,j} |\sum_k a_{i,k}b_{k,j}| \leq latex \sum_{i,j} \sum_k |a_{i,k}|b_{k,j}|$ by the triangle inequality. Now ||AB|| \leq \sum_{i,j} \sum_{k,t} |a_{i,k}| |b_{t,j}| since all the new terms are positive, and rearranging, we have ||AB|| \leq \sum_{i,k} \sum_{j,t} |a_{i,k}||b_{t,j}| = (\sum_{i,k} |a_{i,k}|)(\sum_{j,t} |b_{t,j}|) = ||A|| \cdot ||B||.

Finally, we have ||\alpha A|| = \sum_{i,j} |\alpha a_{i,j}| = |\alpha| \sum_{i,j} |a_{i,j}| = |\alpha| \cdot ||A||.

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.

In an algebraic integer ring, the smallest positive integer contained in a given ideal divides the norm of the ideal

Let K be an algebraic number field with ring of integers \mathcal{O}, and let A \subseteq \mathcal{O} be and ideal. Suppose m is the smallest positive integer contained in A. Show that m|N(A). Suppose now that N(A) is a rational prime; show that \mathcal{O}/A = \{\overline{t}\ |\ 0 \leq t < p\}.


By Theorem 9.16, N(A) \in A. Now A \cap \mathbb{Z} = (m) is an ideal of \mathbb{Z}, which is principal since \mathbb{Z} is a principal ideal domain. Certainly then m is the smallest positive integer in A. Thus m|N(A).

Now suppose N(A) is a rational prime. Suppose there is a (minimal) integer t \in A such that 0 < t < N(A). By part (1), t|N(A), a contradiction. So none of the integers in (0,N(A)) are in A, and thus t \not\equiv 0 mod A for all t \in (0,N(A)). Moreover, we have t \not\equiv s mod A for all t,s \in (0,N(A)) with s \neq t, since otherwise (assuming s > t) we have s-t \equiv 0 mod A. Since N(A) = p, the cosets \{\overline{t}\ |\ t \in [0,N(A))\} exhaust \mathcal{O} and are mutually exclusive.

In an algebraic integer ring, the norm of an ideal principally generated by a rational integer is the absolute value of the norm of its generator

Let K be an algebraic extension of \mathbb{Q} of degree n, and let \mathcal{O} be the ring of integers in K. Let k \in \mathbb{Z}. Show that the norm N((k)) is precisely |N(k)|. (The first norm is the norm of (k) as an ideal, while the second norm is the norm of k as an element.)


We can assume without loss of generality that k is positive.

Recall that the conjugates of k over K are all the same (Theorem 5.10 in TAN), so that N(k) = k^n.

Let \{\omega_i\} be an integral basis for K. Suppose A is an ideal containing k. By Theorem 9.3 in TAN, there exists an element \alpha = \sum c_i\omega_i such that A = (k,\alpha). Note that we may assume that each c_i is in [0,k). In particular, N((k)) \leq k^n. We claim that this \alpha is in fact unique. To that end, suppose \sum c_i\omega_i \equiv \sum d_i\omega_i mod (k) for some 0 \leq c_i, d_i < k. Then we have \sum c_i\omega_i - \sum d_i \omega_i = \sum ke_i\omega_i for some e_i. Since \{\omega_i\} is an integral basis, we have c_i \equiv d_i mod k for each i, and thus c_i = d_i. So in fact N((k)) = k^n = N(k), as desired.

Compute the norm and discriminant of an ideal in a given algebraic integer ring

Let K = \mathbb{Q}(i) and \mathcal{O} = \mathbb{Z}[i]. Let A = (2) and B = (2+i) be ideals in \mathcal{O}. For I \in \{A,B\}, find a basis for I and compute the discriminant and norm of I. Compute \mathcal{O}/I.


We claim that \{2,2i\} is a basis for A over \mathbb{Z}. Indeed it is clear that (2,2i)_\mathbb{Z} = (2), and if 2a+2bi = 0, then a = b = 0. Then the discriminant of A is \mathsf{det}^2 \left[ \begin{array}{cc} 2 & 2i \\ 2 & -2i \end{array} \right] = -64 and (using Theorem 9.10 in TAN) N(A) = -4.

Let a + bi \in \mathcal{O} be arbitrary, and say a \equiv a_0 and b \equiv b_0 mod 2, where a_0,b_0 \in \{0,1\}. Certainly a+bi \equiv a_0+b_0i mod A, so that \mathcal{O}/A = \{\overline{0}, \overline{1}, \overline{i}, \overline{1+i}\}. Since N(A) = 4, we know that these cosets are distinct.

We claim now that \{2+i,1-2i\} is a basis for B over \mathbb{Z}. Indeed, note that 1-2i = -i(2+i) \in B, and if \zeta = (2+i)(a+bi) \in B, then zeta = a(2+i) - b(1-2i). Moreover, if a(2+i) + b(1-2i) = 0, then 2a+b = a-2b = 0, so that 5b = 0, and thus b = a = 0. Then the discriminant of B is \mathsf{det}^2 \left[ \begin{array}{cc} 2+i & 1-2i \\ 2-i & 1+2i \end{array} \right] = -100 and the norm of B is N(B) = 5.

Note that N(2+i) = 5 \in B. If a+bi \in \mathcal{O} is arbitrary, then a+bi = a-2b + b(2-i) \equiv a-2b mod B. If a-2b \equiv k mod 5, with k \in \{0,1,2,3,4\}, then a+bi \equiv k mod B. Thus \mathcal{O}/B = \{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4}\}. Since N(B) = 5, these cosets are distinct.

A fact about norms and traces of algebraic integers

Let K be an algebraic number field with ring of integers \mathcal{O}. Let \alpha,\beta \in \mathcal{O} be nonzero. Prove that N(\alpha)\mathsf{tr}(\beta/\alpha) \in \mathbb{Z}.


Recall from Lemma 7.1 that N(\alpha) is a rational integer, so that N(\alpha)\mathsf{tr}(\beta/\alpha) = \mathsf{tr}(N(\alpha)\beta/\alpha). Now N(\alpha)/\alpha is an algebraic integer, so that N(\alpha)\beta/\alpha is an algebraic integer. Thus \mathsf{tr}(N(\alpha)\beta/\alpha) is a rational integer.

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.

Every nonzero ideal in a ring of algebraic integers contains infinitely many rational integers

Let \mathcal{O} be the ring of integers in a fixed finite extension K = \mathbb{Q}(\theta), and let A be a nonzero ideal in \mathcal{O}. Prove that A contains infinitely many rational integers.


We begin with a lemma.

Lemma: Let A \subseteq \mathcal{O} be an ideal. If \alpha \in A, then N(\alpha) \in A, where N denotes the field norm. Proof: Note that N(\alpha) \in K since N(\alpha) is a rational integer. So N(\alpha)/\alpha \in K. Moreover, by definition, N(\alpha)/\alpha is an algebraic integer, being the product of the (certainly integral) conjugates of \alpha for K. So N(\alpha)/\alpha \in \mathcal{O}. Since A is an ideal, \alpha(N(\alpha)/\alpha) = N(\alpha) \in A. \square

Now if A \neq 0, then there exists a nonzero element \alpha \in A. We have N(\alpha) \in A nonzero, and thus kN(\alpha) \in A for all k \in \mathbb{Z}. In particular, A contains all of the \mathbb{Z}-multiples of N(\alpha).

In an algebraic integer ring, two elements whose norms are relatively prime generate the whole ring

Let \mathcal{O} be the ring of algebraic integers in a number field K. Suppose \alpha,\beta \in \mathcal{O} such that N(\alpha) and N(\beta) are relatively prime. Prove that (\alpha,\beta) = (1).


Recall that N(\alpha) and N(\beta) are rational integers. By Bezout’s identity, there exist rational integers h and k such that hN(\alpha) + kN(\beta) = 1. Note that N(\alpha)/\alpha is an algebraic integer (being a product of algebraic integers) and is in K; hence N(\alpha)/\alpha \in \mathcal{O}. Likewise, N(\beta)/\beta \in \mathcal{O}. So h(N(\alpha)/\alpha) \alpha + k(N(\beta)/\beta) \beta = 1, and we have (1) \subseteq (\alpha,\beta). Hence (1) = (\alpha,\beta).

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.