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.