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Friday, October 11, 2013

Finite Extensions of Q and Roots of Unity (13.6.5)

Dummit and Foote Abstract Algebra, section 13.6, exercise 5:

MathJax TeX Test Page Prove there are only a finite number of roots of unity in any finite extension K of Q.

Proof: Let ψ(k)=p1p2...pk, where pi is the ith prime.

Lemma 1 (Local Extrema of Totient Ratio): If nψ(k) then φ(n)/n(p11)(p21)...(pk1)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=qα111(q11)qα212(q21)...qαm1m(qm1)qα11qα22...qαmm=(q11)(q21)...(qm1)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=pjnψ(k), and otherwise let qv be this prime and set m=pjn/qv<nψ(k). In the first case we see φ(m)/m=pj1pjφ(n)/n<φ(n)/n and in the second φ(m)/m=qv(pj1)(qv1)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 (p11)(p31)...(pk1)>z (index 2 is missing). We show when n>ψ(k) that φ(n)>z. Let k be such that ψ(k1)<nψ(k) so that k>k3. We observeφ(n)=(φ(n)/n)n(p11)(p21)...(pk1)p1p2...pkp1p2...pk1=(p11)(p21)...(pk1)pk(p11)(p31)...(pk11)(p11)(p31)...(pk1)>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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