Tag Archives: finite

Finite extensions of the rationals contain only finitely many roots of unity

Let K be a finite extension of \mathbb{Q}. Prove that K contains only finitely many roots of unity.


Suppose to the contrary that K contains infinitely many roots of unity. Now for each n, there are only finitely many primitive roots of unity (in fact \varphi(n) of them). So for each m, the number of primitive roots of unity of order at most m is finite. In particular, for any m, there exists a primitive nth root of unity for some n > m.

Let m be the degree of K over \mathbb{Q}. If \zeta \in K is a primitive kth root of unity, then [\mathbb{Q}(\zeta):\mathbb{Q}] = \varphi(k) by Corollary 42 on page 555, where \varphi denotes the Euler totient. By this previous exercise, since k is arbitrarily large, \varphi(k) is arbitrarily large. So there exists a primitive kth root \zeta such that \varphi(k) > m, a contradiction since \mathbb{Q}(\zeta) \subseteq K.

There are only finitely many Pythagorean triples (x,y,z) with any one parameter fixed

Suppose (x,y,z) is a solution to x^2 + y^2 = z^2 in \mathbb{N}^+. Show that x^2 \geq 2y+1. Deduce that if we fix any of x, y, or z, then there are only finitely Pythagorean triples (x,y,z).


Suppose x^2 + y^2 = z^2. Now x^2 = z^2 - y^2, and z > y. (We may assume that xyz \neq 0.) Now x^2 = (z+y)(z-y) \geq z+y > 2y. Since x^2 and 2y are integers, x^2 \geq 2y+1.

With x fixed, there are only finitely many possible y, as they must satisfy 2y+1\leq x^2. With x and y fixed, z is determined, so there are only finitely many solutions with fixed x. Symmetrically, there are finitely many solutions with fixed y.

With z fixed, we have z > x,y, so there are certainly only finitely many solutions.

Compute the action of two group elements on a set

Let A be a nonempty set and let k \in \mathbb{Z}^+ such that k \leq |A|. The symmetric group S_A acts on the set A^k by \sigma \cdot ( a_i )_{i = 1}^k = ( \sigma(a_i) )_{i=1}^k.

  1. Prove that this is a group action.
  2. Describe explicitly how the elements (1\ 2) and (1\ 2\ 3) act on the 16 elements of \{ 1, 2, 3, 4 \}^2.

  1. We have \mathsf{id}_A \cdot ( a_i )_{i = 1}^k = ( \mathsf{id}_A(a_i) )_{i = 1}^k = ( a_i )_{i = 1}^k. Moreover, if \sigma, \tau \in S_A, we have \sigma \cdot (\tau \cdot ( a_i )_{i = 1}^k) = \sigma \cdot ( \tau(a_i) )_{i = 1}^k = ( \sigma(\tau(a_i)) )_{i = 1}^k = ( (\sigma \circ \tau)(a_i) )_{i = 1}^k = (\sigma \circ \tau) \cdot ( a_i )_{i = 1}^k. Thus the mapping is a group action.
  2. We have

    (1\ 2) \cdot (1,1) = (2,2)
    (1\ 2) \cdot (1,2) = (2,1)
    (1\ 2) \cdot (1,3) = (2,3)
    (1\ 2) \cdot (1,4) = (2,4)
    (1\ 2) \cdot (2,1) = (1,2)
    (1\ 2) \cdot (2,2) = (1,1)
    (1\ 2) \cdot (2,3) = (1,3)
    (1\ 2) \cdot (2,4) = (1,4)
    (1\ 2) \cdot (3,1) = (3,2)
    (1\ 2) \cdot (3,2) = (3,1)
    (1\ 2) \cdot (3,3) = (3,3)
    (1\ 2) \cdot (3,4) = (3,4)
    (1\ 2) \cdot (4,1) = (4,2)
    (1\ 2) \cdot (4,2) = (4,1)
    (1\ 2) \cdot (4,3) = (4,3)
    (1\ 2) \cdot (4,4) = (4,4)

    Or, in cycle notation, ((1,1)\ (2,2))((1,2)\ (2,1))((1,3)\ (2,3))((1,4)\ (2,4)) ((3,1)\ (3,2))((4,1)\ (4,2)).

    and

    (1\ 2\ 3) \cdot (1,1) = (2,2)
    (1\ 2\ 3) \cdot (1,2) = (2,3)
    (1\ 2\ 3) \cdot (1,3) = (2,1)
    (1\ 2\ 3) \cdot (1,4) = (2,4)
    (1\ 2\ 3) \cdot (2,1) = (3,2)
    (1\ 2\ 3) \cdot (2,2) = (3,3)
    (1\ 2\ 3) \cdot (2,3) = (3,1)
    (1\ 2\ 3) \cdot (2,4) = (3,4)
    (1\ 2\ 3) \cdot (3,1) = (1,2)
    (1\ 2\ 3) \cdot (3,2) = (1,3)
    (1\ 2\ 3) \cdot (3,3) = (1,1)
    (1\ 2\ 3) \cdot (3,4) = (1,4)
    (1\ 2\ 3) \cdot (4,1) = (4,2)
    (1\ 2\ 3) \cdot (4,2) = (4,3)
    (1\ 2\ 3) \cdot (4,3) = (4,1)
    (1\ 2\ 3) \cdot (4,4) = (4,4).

    Or, in cycle notation, ((1,1)\ (2,2)\ (3,3))((1,2)\ (2,3)\ (3,1))((1,3)\ (2,1)\ (3,2)) ((1,4)\ (2,4)\ (3,4))((4,1)\ (4,2)\ (4,3)).