Download MATH 601: Abstract Algebra II 5th Homework Partial Solutions

MATH 601: Abstract Algebra II
5th Homework
Partial Solutions
Jonathan Rosenberg
assignment due Wednesday, March 5, 2014
√
Section 14.9, #6. Show that if t is transcendental over Q, then Q(t, t3 − t) is not purely transcendental over Q.
√
3
Solution. We’ll
√ prove something stronger, which is that C(t, t − t) is not purely√transcendental over C.
3
Suppose C(t, t − t) = C(s) for some s transcendental over C. Then t = f (s) and t3 − t = g(s), for some
rational functions f and g with coefficients in C. The rational functions f and g can’t be constant, since t
is transcendental over Q, and are (up to complex scalar factors, which we can ignore) exactly products of
(s − α)± for various α’s (the zeros and poles). (We allow α = ∞ as a possibility, and note that zeros or poles
can have multiplicities. We’ll say α has negative multiplicity −n if it’s a pole of order n, and say that it has
multiplicity zero if α is neither a zero nor a pole.)
Expanding, we have
g(s)2 = t3 − t = f (s) (f (s) + 1) (f (s) − 1) .
(1)
So g has a zero or pole α of order n exactly when f (s)3 − f (s) has a zero or pole at the same point α of
order 2n, or when the orders of f , f + 1, and f − 1 at α add
√ up to 2n.
But now s has to be a rational function of t = f (s) and t3 − t = g(s), so f (s) and g(s) can’t have more
than one common zero (or pole), and can’t have a double common zero (or pole). If f has more than one
zero or a double zero, and these are not cancelled by poles of f + 1 or f − 1, then these are also zeros of g,
and we get a contradiction. Similarly with zeros of f + 1 or f − 1, or with poles instead of zeros. So the only
possibility is that f has a single simple zero or a single simple pole. Looking at (1), we see this is impossible.
Extra problem. Let c, s be transcendental over Q with c2 + s2 = 1 and work in the field L = Q(c, s)
(think of c and s as cos t and sin t respectively). Show that if n is a positive integer, then (c + is)n + (c − is)n
can be written as a polynomial in c with rational coefficients. Deduce that there is a trig identity for cos nt
in the form of a polynomial in cos t. Similarly show that there is a similar identity for sin nt as a polynomial
in sin t, but only if n is odd.
1
Solution. The first part is trivial, since if c = cos t and s = sin t, then
n
1
1 X
cos nt = [(c + is)n + (c − is)n ] =
(is)j + (−is)j cn−j
2
2 j=0
bn/2c
=
X
(−1)k s2k cn−2k
(take j = 2k)
k=0
bn/2c
=
X
(−1)k (1 − c2 )k cn−2k ,
k=0
which is a polynomial in c with rational coefficients.
When it comes to sin nt, we need to consider instead
n
sin nt =
1
1 X
[(c + is)n − (c − is)n ] =
(is)j − (−is)j cn−j
2i
2i j=0
bn/2c
=
X
(−1)k s2k+1 cn−2k−1
(take j = 2k + 1).
k=0
If n is odd, then each n − 2k − 1 is even, and the even powers of c can be converted to polynomials in s,
so we get a formula for sin nt as a polynomial in s with rational coefficients. But when n is even, we need
to show there cannot be such a formula. Recall that L = Q(c, s) and let K = Q(s).√Then K is purely
transcendental over Q and L is a quadratic extension of K generated over K by c = 1 − s2 . Note that
Gal(L/K) is generated by an element σ which fixes s and sends c to −c. When n is even, each n − 2k − 1 is
odd, so σ reverses the sign of every term on the RHS. So sin nt is not fixed by σ and can’t be a polynomial
in s. 2