Proof: Let ψ(k)=p1p2...pk, where pi is the ith prime.
Lemma 1 (Local Extrema of Totient Ratio): If n≤ψ(k) then φ(n)/n≥(p1−1)(p2−1)...(pk−1)p1p2...pk. Proof: Collect n such that φ(n)/n is minimal, and then choose n=qα11qα22...qαmm minimal from this collection. Assume αi>1 for some i; thenφ(n)/n=qα1−11(q1−1)qα2−12(q2−1)...qαm−1m(qm−1)qα11qα22...qαmm=(q1−1)(q2−1)...(qm−1)q1q2...qm=φ(n/qi)/(n/qi)and n/qi<n, violating minimality. So n is squarefree. Let pj be the smallest prime not dividing n, which must be ≤pk else n=ψ(k) and φ(n)/n equals the bound given above. If there is no prime larger than pk dividing n then set m=pjn≤ψ(k), and otherwise let qv be this prime and set m=pjn/qv<n≤ψ(k). In the first case we see φ(m)/m=pj−1pjφ(n)/n<φ(n)/n and in the second φ(m)/m=qv(pj−1)(qv−1)pjφ(n)/n<φ(n)/n, invariably violating minimality. Thus the bound holds.◻
Lemma 2: φ(n)→∞. Proof: Choose finite positive z, and let k be such that (p1−1)(p3−1)...(pk−1)>z (index 2 is missing). We show when n>ψ(k) that φ(n)>z. Let k′ be such that ψ(k′−1)<n≤ψ(k′) so that k′>k≥3. We observeφ(n)=(φ(n)/n)n≥(p1−1)(p2−1)...(pk′−1)p1p2...pk′p1p2...pk′−1=(p1−1)(p2−1)...(pk′−1)pk′≥(p1−1)(p3−1)...(pk′−1−1)≥(p1−1)(p3−1)...(pk−1)>z ◻Now, since there are only a finite number of primitive roots for any n, K must contain nth primitive roots for n arbitrarily large. Since the degree φ(n) of the cyclotomic minimal polynomial for these primitive roots also becomes arbitrarily large, we must have K/Q is not finite. ◻
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