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 $n$th root of unity for some $n > m$.

Let $m$ be the degree of $K$ over $\mathbb{Q}$. If $\zeta \in K$ is a primitive $k$th 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 $k$th 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, $((1,1)\ (2,2)\ (3,3))((1,2)\ (2,3)\ (3,1))((1,3)\ (2,1)\ (3,2))$ $((1,4)\ (2,4)\ (3,4))((4,1)\ (4,2)\ (4,3))$.