Exercise 15.2.9

Proof. (a)

The ring $\mathbb{Z}$ is principal, thus is a UFD with field of fractions $\mathbb{Q}$. By Gauss’s Lemma (Theorem A.3.2, or Theorem A.5.8), if $P,Q\in \mathbb{Z}[u]$ are relatively prime in $\mathbb{Z}[u]$, then $P,Q$ are relatively prime in $\mathbb{Q}[u]$.

Since $P(u),Q(u)$ are relatively prime in $\mathbb{Q}[u]$, there are some polynomials $A,B\in \mathbb{Q}[u]$ such that $A(u)P(u)+B(u)Q(u)=1$, thus the substitution $u\to u^4$ gives $A(u^4)P(u^4)+B(u^4)Q(u^4)=1$. Reasoning by contradiction, suppose that $\mathit{uP}(u^4)$ and $Q(u^4)$ have a common root $\alpha$ in some extension of $\mathbb{Q}$. Since $Q(0)=1$, $\alpha \ne 0$, thus $P({\alpha }^4)=0$. Then $P({\alpha }^4)=Q({\alpha }^4)=0$ implies $1=A({\alpha }^4)P({\alpha }^4)+B({\alpha }^4)Q({\alpha }^4)=0$: this is a contradiction.

So $\mathit{uP}(u^4)$ and $Q(u^4)$ have no common roots in any extension of $\mathbb{Q}$. (b) If $m$ is odd, then $\phi (\mathit{mx})Q_m({\phi }^4)(x))=\phi (x)P_m({\phi }^4(x))$. Reasoning by contradiction, suppose that, for some $x\in ℝ$, $Q_m({\phi }^4(x))=0$. Then $\phi (x)P_m({\phi }^4(x)=0$, so that $\alpha =\phi (x)$ is a common root of $Q_m(u^4)$ and $u{P}_m(u^4)$. Since $P_m,Q_m$ are relatively prime, and $Q_m(0)=1$, this is impossible by part (a).

If $m$ is even, then $\phi (\mathit{mx})Q_m({\phi }^4(x))=\phi (x)P_m({\phi }^4(x)){\phi }^{\prime }(x)$. Suppose that, for some $x\in ℝ$, $Q_m({\phi }^4(x))=0$. Then $\phi (x)P_m({\phi }^4(x)){\phi }^{\prime }(x)=0$. If ${\phi }^{\prime }(x)=0$, then $\sqrt{1-{\phi }^4(x)}=0$, thus ${\phi }^4(x)=1$, and $\phi (x)=\pm 1$. If $\phi (x)\notin \{-1,1\}$, then $\phi (x)P_m(\phi (x))=0$, so that $\alpha =\phi (x)$ is a common root of $Q_m(u^4)$ and $u{P}_m(u^4)$, which is impossible by part (a).

We have proved that $Q_m({\phi }^4(x))\ne 0$ when $\phi (x)\notin \{-1,1\}$.

(Misprint in the sentence of part (b) ? If $\phi (x)=0$, then $Q_m({\phi }^4(x))=Q_m(0)=1\ne 0$, so there is no need to suppose $\phi (x)\ne 0$.) (c) For all $x\in ℝ$, $\phi (x)=0$ if and only if $x=\mathit{k\varpi }$ for some $k\in \mathbb{Z}$.

We must verify that $\phi \left(\frac{2\varpi }{n}\right)\notin \{-1,1\}$. If $n>2$, then $0<\frac{2\varpi }{n}<\varpi$. This proves that $0<\phi \left(\frac{2\varpi }{n}\right)<1$, thus $\phi \left(\frac{2\varpi }{n}\right)\notin \{-1,0,1\}$.

By our version of part (b), this implies that

Note: If we read the proof of Theorem 15.2.5, it is obvious that the denominators $Q_n({\phi }^4(x))$ never vanish, for all $x\in ℝ$, because $1+{\phi }^2(\mathit{nx}){\phi }^2(x)\ne 0$. □