Solutions to Homework 8
33. This is Problem 2. of Homework 1 of last semester. One shows that f is continuous and is the
identity on Q ⊂ R; then by the density of the rationals, we have that f = id.
34. (Dummit-Foote, 13.6, #13) (a) Let Z ⊂ D be the center of D. It is a clearly a division ring,
and it is commutative by the definition of the center. Hence it is a field. As it is finite, its prime
subfield must be Fp . Hence Z = Fq where q = pN for some N .
Multiplication turns D into a Z-vector space. Hence the additive group of D is isomorphic to
n
Fq for some n, which proves that |D| = q n .
(b) Let C(x) be the set of all elements of D which commute with x. It is clear that Z ⊂ C(x)
and easy to prove that C(x) is a division ring. Thus C(x) is a Z-vector space. If x ∈
6 Z then
C(x) ( D, hence |C(x)| = q m for m < n.
(c) The class equation for D× is
×
×
|D | = Z(D )| +
r
X
|D× : CD× (xi )|
i=1
where the xi are distinct conjugacy class representatives. By what we’ve already shown, we have
|D× | = q n − 1, |Z(D× )| = q − 1, and |CD× (xi )| = q mi − 1. The result now follows by plugging these
values into the equation.
(d) Since (q n − 1)/(q mi − 1) ∈ Z, we have mi |n by Exercise 4 of 13.5. Since mi < n, no
primitive nth root of 1 is a root of X mi − 1. Since Φn (X) divides X n − 1, it therefore divides
(X n − 1)/(X mi − 1). Plugging in q = X gives the result.
(e) If ζ 6= 1 is a primitive nth root of 1, it is clear that 1 is closer (in the complex plane) to q
than ζ. Therefore
Q |q − ζ| > |q − 1| = q − 1. However, by parts (c) and (d), we have Φn (q)|q − 1,
and Φn (q) = ζ prim (q − ζ). This leads to a contradiction, unless n = 1. Therefore |D| = |Z| and
hence D = Z is a field.
35. (Dummit-Foote,
13.6, #14) Let p1 , . . . , pk be the only primes dividing the values P (1), P (2), . . ..
P
Write P (x) = i bi xi , and suppose N is such that P (N ) = a 6= 0. Using the binomial theorem, we
have
Q(x) = a−1 P (N + ap1 · · · pk x)
= a−1 [b0 + b1 (N + ap1 · · · pk ) + b2 (N + ap1 · · · pk )2 + · · · + bn (N + ap1 · · · pk )n ]
= a−1 [b0 + b1 N + · · · + bn N n + R(x)]
= a−1 [P (N ) + R(x)]
= 1 + a−1 R(x)
where R(x) is divisible by ap1 · · · pk . Thus Q(x) ∈ Z[x], and plugging in any integer M for x shows
that Q(M ) ≡ 1 (mod p1 · · · pk ), because a−1 R(x) is divisible by p1 · · · pk . Picking M such that
|Q(M )| > 1, we see that Q(M ) ≡ 1 (mod pi ) for all i, thus pi 6 | Q(M ) for any i.
1
Therefore there exists a prime q 6= pi for all i, such that q|Q(M ). Clearly then, q|aQ(M ) =
P (N + ap1 · · · pk M ) which contradicts our assumption that only finitely many primes divide the
values P (1), P (2), . . ..
(Dummit-Foote, 13.6, #15) Let a ∈ Z satisfy Φm (a) ≡ 0 (mod p). As in the hint, this means
that am ≡ 1 (mod p), so a is a unit modulo p and thus (a, p) = 1.
Now suppose that there exists d|m such that ad ≡ 1(mod p). Then a satisfies Φd (a) = 0, which
implies that a is a multiple root of xm − 1 modulo p. Therefore a is a root of the derivative of
xm − 1, which is mxm−1 . Thus mam−1 ≡ 0 (mod p), but p divides neither m or a, so this is a
contradiction.
(Dummit-Foote, 13.6, #16) Let p be an odd prime such that Φm (a) ≡ 0 (mod p) and suppose
that p does not divide m. By the previous problem, a is a unit modulo p and the order of a in
(Z/pZ)× is m. But the order of (Z/pZ)× is p − 1, so m|p − 1 and therefore p ≡ 1 (mod m).
(Dummit-Foote, 13.6, #17) By Exercise 14, infinitely many primes divide the integers
Φm (1), Φm (2), . . .
and by Exercise 16, all of these primes either divide m or are congruent to 1 modulo m. Since
only finitely many primes divide m, we conclude that there are infinitely many primes p such that
p ≡ 1 (mod m).
36. Straightforward calculation shows that all of the fractional linear transformations are automorphisms (the condition ad − bc 6= 0 is needed to insure that the map has an inverse). To see
that every automorphism is given by a fractional linear transformation, let φ : k(t) → k(t) be
an automorphism of k(t) over k. Since φ(f (t)/g(t)) = f (φ(t))/g(φ(t)), the map φ is completely
determined by φ(t).
Let φ(t) = f (t)/g(t). We may assume that f (t) and g(t) are relatively prime. Following the
referenced exercise, let h(X) = f (X) − φ(t)g(X) ∈ k(φ(t))[X]. This polynomial is irreducible and
has t as a root. Thus k(t) ' k(φ(t))[X]/(h(X)), and hence
[k(t) : k(φ(t))] = deg(h(X)) = max{deg(f (t)), deg(g(t))} = deg(φ(t)).
Since φ is an automorphism, we must have k(t) = k(φ(t)), which implies that deg(φ(t)) = 1 and
therefore φ is given by a fractional linear transformation. The fact that ad − bc 6= 0 follows from
the fact that the map is invertible (write down a matrix and invert it).
37. Let ζ be a primitive 3rd root of unity, and let α =
√
3
2. The polynomial
(X − (1 + α + α2 ))(X − (1 + ζα + ζ 2 α2 ))(X − (1 + ζ 2 α + ζα2 )) = X 3 − 3X 2 − 3X − 1
is monic of degree three and in Q[X]. Clearly it is satisfied by the element 1 + α + α2 , and therefore
it is the minimal polynomial of this element.
2