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.