## 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 $D < 0$.