# The Character Group of Q

## Mathematics 676 with Conrad at University of Michigan - Ann Arbor *

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- University of Michigan - Ann Arbor
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- Mathematics 676
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- The Character Group of Q

__0 is in the kernel of . Replacing by the continuous homomorphism x 7! (Nux), we may assume that (1) = 1. To summarize, we may assume is a continuous homomorphism from R to S1 which contains 1 in its kernel and which has positive imaginary part for small positive numbers. According to what we are trying to prove, we now expect that (x) = e2 ixy for some (positive) integer y, and this is what we shall show. Since the reciprocals of the prime powers generate the dense subgroup Q of R, by continuity it su ces to nd an integer y such that (1=pn) = e2 iy=pn for all primes p and integers n 1. Actually, we’ll show for any integer m > 1 that there is an integer ym such that (1=mn) = e2 iym=mn for all n 1. Since (1=(pq)n)pn = (1=qn) and (1=(pq)n)qn = (1=pn), it follows that ypq yq 2\nqnZ = f0g, so ypq = yq, and similarly that ypq = yp, so yp = yq. Thus for all primes p, the integers yp are the same, and this common integer y solves our problem. Since (1=mn)mn = (1) = 1, (1=mn) = e2 icn=mn for some integer cn such that 0 cn__ 0 and sin(2 cn=mn) is positive and arbitrarily small. Since 0 cn < mn, the cosine condition implies cn=mn2(0;1=4)[(3=4;1) for large n. We can’t have cn=mn in (3=4;1) for large n, since then sin(2 cn=mn) is negative (here is where we use the assumption that (x) is in the rst quadrant for small positive x). Thus for all large n, cn=mn 2 (0;1=4). Since sin(2 cn=mn) is arbitrarily small for all large n, it follows by the nature of the sine function and the location of cn=mn that cn=mn is arbitrarily small for n large. To x ideas, 0 < cn=mn < 1=(m + 1) for all large n. Then 0 < cn+1=mn < 12 KEITH CONRAD m=(m + 1) for all large n, so jcn=mn cn+1=mnj< 1 for all large n. Since the left hand side of this last inequality is an integer, it must be zero, so all cn’s are equal for n su ciently large. Call this common value ym. Thus (1=mn) = e2 iym=mn for all large n, hence for all n 1; for example, if (1=m100) = e2 iym=m100, then raising both sides to the m98-th power we see that (1=m2) = e2 iym=m2. Acknowledgments. I thank Randy Scott, Eric Sommers and Ravi Vakil for looking over a preliminary version of this manuscript. References [1] J.B. Conway, A Course in Functional Analysis, 2nd ed., Springer-Verlag, New York, 1990. [2] I. M. Gel’fand, M. I. Graev, I. I. Pyatetskii-Shapiro, Representation Theory and Automorphic Functions, Academic Press, 1990. [3] K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory, 2nd ed., Springer-Verlag, New York, 1986. [4] J. Tate, Fourier Analysis in Number Fields and Hecke’s Zeta-Functions, in: Algebraic Number Theory, Academic Press, New York, 1967, 305-347. [5] L. Washington, On the Self-Duality of Qp, American Mathematical Monthly 81 (4) 1974, 369-370.

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