## Tag Archives: conjugate

### The conjugates of a root of unity are roots of unity

Let $\zeta$ be a root of unity. Show that the conjugates of $\zeta$ are also roots of unity.

If $\zeta$ is a root of 1, then $\zeta$ is a root of $p(x) = x^n - 1$ for some $n$. Thus the minimal polynomial of $\zeta$ over $\mathbb{Q}$ is also a root of $p(x)$, so the conjugates of $\zeta$ are roots of unity.

### 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 $\epsilon = \gamma/\gamma^\prime$.

### A conjugate of a unit is a unit

Let $K$ be an algebraic extension of $\mathbb{Q}$, and let $u \in K$ be an algebraic integer. Prove that if $u$ is a unit, then the conjugates of $u$ are also units.

By Lemma 7.3, the norm $N(u)$ is $\pm 1$, where (by definition) $N(u)$ is the product of the conjugates of $u$. In particular, each conjugate of $u$ divides 1 (in the extended sense, meaning that $1/u^\prime$ is an algebraic integer). Thus each conjugate of $u$ is a unit.

Alternately, if $u$ is a unit then $uv = 1$ for some algebraic integer $v$. Letting $K = \mathbb{Q}(\theta)$, $u = r(\theta)$, and $v = s(\theta)$, and letting $\theta_i$ be the conjugates of $\theta$, then $u_iv_i = r(\theta_i)s(\theta_i) = 1$, where $u_i$ and $v_i$ denote the conjugates of $u$ and $v$, respectively.

### If possible, exhibit an integral basis for a quadratic field whose elements are conjugate

If possible, exhibit and integral basis for $\mathbb{Q}(\sqrt{D})$ whose elements are conjugate for $D \in \{ -3,-1,5\}$.

Let $D$ be a squarefree integer, and let $\omega = \sqrt{D}$ if $D \not\equiv 1$ mod 4 and $(1+\sqrt{D})/2$ if $D \equiv 1$ mod 4. Recall that the integers in $\mathbb{Q}(\sqrt{D})$ have the form $a+b\omega$ where $a,b \in \mathbb{Q}$. Let $\overline{\omega} = -\sqrt{D}$ if $D \not\equiv 1$ mod 4 and $(1-\sqrt{D})/2$ if $D \equiv 1$ mod 4. Note that $\omega\overline{\omega} = -D$ if $D \not\equiv 1$ mod 4 and $(1-D)/4$ if $D \equiv 1$ mod 4 (so that $\omega\overline{\omega} \in \mathbb{Z}$), that $\omega + \overline{\omega} = 0$ if $D \not\equiv 1$ mod 4 and $1$ if $D \equiv 1$ mod 4, and that $\omega - \overline{\omega} = 2\omega$ if $D \not\equiv 1$ mod 4 and $2\omega - 1$ if $D \equiv 1$ mod 4.

Note that $p(x) = (x - (a+b\omega))(x - (a+b \overline{\omega})) = x^2 - 2ax + (a^2 + ab(\omega+\overline{\omega}) + \omega\overline{\omega}b^2 \in \mathbb{Z}[x]$. In particular, if $b \neq 0$ then $p(x)$ is the minimal polynomial of $a+b\omega$, so that the conjugates of $a+b\omega$ are itself and $a+b\overline{\omega}$.

Recall that the discriminant of $\mathbb{Q}(\sqrt{D})$ is $4D$ if $D \not\equiv 1$ mod 4 and $D$ if $D \equiv 1$ mod 4, and that any integral basis will have this discriminant.

Suppose now that $B = \{a+b\omega, a+b\overline{\omega}\}$ is an integral basis of $\mathbb{Q}(\sqrt{D})$, with $b \neq 0$ and $a \neq 0$. (Certainly if $b = 0$ then $B$ cannot be a basis, much less an integral basis; if $a = 0$, then (for example) we cannot generate 1.) Note that $a,b \in \mathbb{Z}$. Then we must have

$\mathsf{det} \left( \left[ \begin{array}{cc} a+b\omega & a + b\overline{\omega} \\ a + b\overline{\omega} & a+b\omega \end{array} \right] \right)^2 = \left\{ \begin{array}{crl} 4D & \mathrm{if} & D \not\equiv 1 \mod 4 \\ D & \mathrm{if} & D \equiv 1 \mod 4. \end{array} \right.$

Note that $(\omega - \overline{\omega})^2 = 4D$ if $D \not\equiv 1$ mod 4 and $D$ if $D \equiv 1$ mod 4. Evidently, this determinant is $(\omega - \overline{\omega})^2 b^2 (2a + b(\omega+\overline{\omega}))^2$, and so we have $(2a + b(\omega + \overline{\omega}))^2 = 1$.

If $D \not\equiv 1$ mod 4, then $\omega+\overline{\omega} = 0$, and we have $4a^2 = 1$. This equation has no solutions in $\mathbb{Z}$, and thus $\mathbb{Q}(\sqrt{D})$ does not have an integral basis whose elements are conjugate.

If $D \equiv 1$ mod 4, then $\omega + \overline{\omega} = 1$, and we have $(2a+b)^2 = 1$, so that $2a+b = \pm 1$. This gives a necessary condition on $a$ and $b$ if we wish $B$ to be an integral basis.

Recall that $\{1,\omega\}$ is an integral basis of $\mathbb{Q}(\sqrt{D})$. If $\{\alpha, \beta\}$ is another integral basis, then there is a (unique) invertible matrix $P$ such that $P[1\ \omega]^\mathsf{T} = [\alpha\ \beta]^\mathsf{T}$. Say $P = \left[ \begin{array}{cc} x & y \\ z & w \end{array} \right]$, $\alpha = a+b\omega$, and $\beta = a+b\overline{\omega}$; comparing coefficients, and recalling that $\overline{\omega} = 1 - \omega$ (since $D \equiv 1$ mod 4), we have $P = \left[ \begin{array}{cc} a & b \\ a+b & -b \end{array} \right]$. Computing the determinant, we have $-b(2a+b) = \pm 1$. Combined with our necessary condition, $b = \pm 1$. If $b = 1$, then $a = -1$, and if $b = -1$, then $a = 1$. Thus our only candidates for $B$ are $B_1 = \{ -1+\omega, -1+\overline{\omega}\}$ and $B_2 = \{ 1-\omega, 1-\overline{\omega}\}$. Or, in terms of $\omega$, $B_1 = \{-1+\omega,\omega\}$ and $B_2 = \{1-\omega,\omega\}$. It is easy to see that both $B_1$ and $B_2$ are integral bases.

To summarize, if $D \equiv 1$ mod 4, then $\pm B$ is an integral basis whose elements are conjugate, where $B = \{1-\omega,\omega\}$. If $D \not\equiv 1$ mod 4, then no such integral basis exists.

### Find a basis for QQ(sqrt(D)) whose elements are conjugate

Let $D \in \mathbb{Z}$ be squarefree. Find a basis $\alpha, \overline{\alpha}$ for $\mathbb{Q}(\sqrt{D})$ where $\overline{\alpha}$ is the conjugate of $\alpha$.

Let $\alpha = a+b\sqrt{D}$. Evidently, $\alpha$ is a root of $p(x) = x^2 - 2ax + (a^2 - Db^2)$; the other root is $a - b\sqrt{D}$. If $b \neq 0$, then this polynomial has no rational roots, and so is irreducible over $\mathbb{Q}$. In this case, $p(x)$ is the minimal polynomial of $\alpha$, and so $\overline{\alpha} = a - b\sqrt{D}$.

Note that if $b = 0$, then $\{\alpha,\overline{\alpha}\}$ cannot be a basis for $\mathbb{Q}(\sqrt{D})$ as it contains only one element. Thus we may assume that $b \neq 0$.

We claim that $\{\alpha, \overline{\alpha}\}$ is a basis for $\mathbb{Q}(\sqrt{D})$ so long as $a \neq 0$. To that end, suppose first that $\{\alpha,\overline{\alpha}\}$ is a basis; if $a = 0$, then we have $\alpha = b\sqrt{d}$ and $\overline{\alpha} = -b\sqrt{D}$. Now $1 = xb\sqrt{D} - yb\sqrt{D}$, with $x,y \in \mathbb{Q}$. So $\sqrt{D} = 1/b(x-y)$, a contradiction since $\sqrt{D}$ is not rational. Thus $a \neq 0$.

Conversely, suppose $a \neq 0$. To show that $\{\alpha,\overline{\alpha}\}$ is a basis, it suffices to show that it generates $\mathbb{Q}(\sqrt{D})$, as we know that this field has degree 2 over $\mathbb{Q}$. To achieve this, we will show that $1$ and $i$ are linear combinations of $\alpha$ and $\overline{\alpha}$. Indeed, evidently we have $1 = \frac{1}{2a}\alpha + \frac{1}{2a}\overline{\alpha}$ and $i = \frac{1}{2b}\alpha - \frac{1}{2b} \overline{\alpha}$, as desired.

In particular, $\{1+\sqrt{D}, 1-\sqrt{D}\}$ is a basis for $\mathbb{Q}(\sqrt{D})$ over $\mathbb{Q}$.

### In any extension of QQ, an algebraic element and its complex conjugate are conjugates

Let $E$ be an extension of $\mathbb{Q}$ contained in $\mathbb{C}$. Suppose $\alpha \in E$ is algebraic over $\mathbb{Q}$ with minimal polynomial $p(x)$ and such that the complex conjugate $\overline{\alpha}$ of $\alpha$ is also in $E$. Prove that $\alpha$ and $\overline{\alpha}$ are conjugates in $E$.

Note that complex conjugation is an automorphism of $\mathbb{C}$. Thus $p(\overline{\alpha}) = \overline{p(\alpha)} = \overline{0} = 0$. Since $\overline{\alpha}$ is a root of the minimal polynomial $p(x)$, $\alpha$ and $\overline{\alpha}$ are conjugate in $E$.

### ‘Is conjugate to over F’ is an equivalence relation

Let $F$ be a field and let $E$ be an algebraic extension of $F$. Prove that the relation $\sim$ defined by $\alpha \sim \beta$ if and only if $\alpha$ and $\beta$ are conjugate over $F$ is an equivalence.

Recall that $\alpha$ and $\beta$ are called conjugate precisely when they have the same minimal polynomial over $F$. We can map $E$ to the set $S$ of all minimal polynomials of elements of $E$ by sending each element to its minimal polynomial; call this map $f$. (Note that $f$ is well-defined since minimal polynomials are unique.) Now $\alpha \sim \beta$ precisely when $f(\alpha) = f(\beta)$. Thus it is clear that $\sim$ is an equivalence relation.

### If a is a nonzero algebraic element over F, then all of its conjugates are nonzero

Let $F$ be a field and let $\alpha$ be a nonzero algebraic element over $F$. Prove that the conjugates of $\alpha$ are nonzero.

Let $p(x)$ be the minimal polynomial of $\alpha$ over $F$. Recall that the conjugates of $\alpha$ are precisely the roots of $p(x)$. If $p(x)$ is linear, then the only conjugate of $\alpha$ is $\alpha$ itself. Suppose $p(x)$ has degree at least 2. If one of the roots of $p$ is 0, then in fact $p(x) = xp^\prime(x)$; note that $\alpha$ is a root of $p^\prime$ and that $p^\prime$ has degree strictly smaller than that of $p$, violating the minimalness of $p(x)$. So 0 is not a conjugate of $\alpha$.

### A fact about irreducibles in ZZ[i]

Let $\pi = a+bi$ be an irreducible Gaussian integer with $a,b \neq 0$. Show that if $\pi$ is a factor of its conjugate $\overline{\pi} = a-bi$, then $\pi$ is an associate of $1+i$.

Let $\tau = c+di$ and suppose $\pi\tau = \overline{\pi}$. This equality yields the two equations $ac-bd=a$ and $ad+bc=-b$, which can be rearranged as $a(c-1) = bd$ and $ad = -b(c+1)$. Now $ad^2 = -bd(c+1) = -a(c+1)(c-1)$, so that \$latex $d^2 + c^2 = 1$. Hence $\tau \in \{ 1,-1,i,-i \}$.

If $\tau = 1$, then $a+bi = a-bi$, so that $b = 0$, a contradiction. Similarly, if $\tau = -1$ then $-a-bi = a-bi$ and we have $a = 0$. If $\tau = i$, then $-b+ai = a-bi$, so that $a = -b$. If $\tau = -i$, then $b-ai = a-bi$, and we have $a = b$. In either case, $|a| = |b|$. If $|a| > 1$, then $\pi = a+bi$ has a nontrivial factorization (namely $a(1+i)$) and so is not irreducible. Thus $|a| = 1$, and so $\pi \in \{1+i, 1-i, -1+i, -1-i\}$. These are precisely the associates of $1+i$.

### In the Gaussian integers, the conjugate of a prime is prime

Prove that if $a+bi$ is irreducible in $\mathbb{Z}[i]$, then so is $b+ai$.

Note that conjugation preserves multiplication in $\mathbb{Z}[i]$; $\overline{\alpha\beta} = \overline{\alpha} \overline{\beta}$ for all $\alpha$ and $\beta$. As a consequence, if $\alpha$ is irreducible then $\overline{\alpha}$ is as well.

Note that $b+ai = \overline{(-i)(a+bi)}$. Thus, if $a+bi$ is irreducible, then so is $b+ai$.