242: Math Homework 5 John Peca-Medlin
May 1, 2006 2. Let , Q and f (x) = i=0 ai xi Q[x] its accompanying (monic) minimal polynomial, with a0 = 0 and an = 1 (so f () = 0). Let k Z be such that kai Z for all i (we can do this if we let k = lcm{q0 , q1 , . . . , qn }, with qi being the denominator in the irreducible fraction n n n-i i ai (kx)i . Then define ai = pi for all i). Now look at k n f (x) = k n i=0 ai xi = i=0 k q n g(x) = i=0 k n-i ai xi . We see g(x) is monic in Z[x] since each coefficient is an integer, with its lead coefficient being 1, while also g(k) = k n f () = 0. Thus k , as desired. 3. For our specific case, let f (x) = x2 + x + , and let be a zero of f (x) (so f () = 2 + + = 0). Suppose m and n represent the degrees of and . Let S be the Zmodule generated by i j k , with 1 i < n, 1 j < m, and k = 0 or 1. Let S, then r s t = q=0 aq i j k+1 = u=0 au i j + v=0 av i j 2 with au , av Z for each u, v. Since s t f () = 0, we have 2 = - - , and thus = u=0 au i j - v=0 av i j ( + ) S. Thus, by Proposition 6.1.4, . I believe this could be summarized thusly: Let f (x) and be as they were above, and let R = Z[, ], a Z-module, and so we have is integral over R (which just means is a zero of some monic polynomial in R[x]). Thus we have that R[] is also a Z-module, and so . And now for the generalization: let f (x) = xn + 1 xn-1 + + n-1 + n such that i for each i, and suppose is such that f () = 0. Let R = Z[1 , . . . , n ] (so f (x) R[x]), thus R is a finitely generated Z-module. Since is a root of the monic f (x) R[x], then is integral over R and R[] is a finitely generated R-module. Since we had that R is a finitely generated Z-module, then we have that R[] must also be a finitely generated Z-module, and thus . 4. Let the content of a polynomial in Z, written c(g) for a polynomial g(x), be the greatest common divisor of its coefficients (as it is often defined). So we see f (x) is primitive if its content is 1. Suppose f (x), g(x) are primitive, but f (x)g(x) = h(x) is not. Then the content of h(x) > 1. Let p be a prime such that p | c(h), thus p divides every coefficient of h(x) and so h(x) = f (x)g(x) 0 (p). Since f (x) is primitive, then p cannot divide all of its coefficients, so f (x) 0 (p). But by the same argument, we must have g(x) 0 (p). Thus, we must have the product of two non-zero polynomials must also be non-zero modulo p, i.e. f (x)g(x) = h(x) 0 (p), a contradiction. So h(x) must also primitive. be 12. First note 1 + (t | p) is the number of solutions to x2 t (p) and that t t = 0.
n
1
^ 13. Plug f (a) into the desired equality:
p-1 p-1 p-1
^ f (a)a (t) =
s=1 a=1 p-1
p-1
s=1 p-1 -1 s=1 p-1
f (s)-a (s) a (t)
=
a=1 p-1
p
f (s)-a (s)a (t) f (s)a (t - s)t)
s=1 p-1
=
a=1 p-1
p-1
(by 12a)
=
s=1 p-1
p-1 f (s)
a=1
-a (s)a (t)
=
s=1 p-1
p-1 f (s)(t, s) (p-1 f (t)) = (1/p)(pf (t)) = f (t).
s=1
(by 12b)
=
p-1 ^ 14. Let f from the previous problem be the Legendre symbol. Then f (a) = p-1 s=1 f (s)-a (s) = p-1 (-a)s -1 p = p g-a , by simple definitions. s=1 (s | p) 15. Note (t | p) g = gt and the inequality sin x (2/)x for any acute angle x. Using the n n first fact I mentioned, we have | t=m (t | p)| = (1/ p)| t=m gt |. Thus, it is sufficient to show n to show | t=m gt | has a bound by p log p. Notice: -1 n n p-1
gt =
t=m t=m p-1 a=1
a p
p-1
at
n-m
=
a=1
a p
n
at
t=m
=
a=1 p-1
a p a p
am
t=0
at a(n-m+1) - 1 a - a
=
a=1
am
Now use the identiy: 1 - x = 1 - cos x - i sin u = 2 sin2 (x/2) - 2i sin(x/2) cos(x/2) = 2 sin(x/2) (sin(x/2) - i cos(x/2)) = 2 sin(x/2)(-i x/2 )
2
we plug this into what we just had:
n p-1
gt =
t=m a=1 p-1
a p a p a p 1 sin
=
a=1 p-1
a=1 p-1
a(n-m+1) - 1 a - a sin a(n-m+1) p am sin a p a(n-m+1) sin p am sin a p am
(a/2)(n-m+1) a/2 (a/2)(n-m+1) a/2
a=1 p-1
a p
a=1
a (2/)
a p
p-1
=p
a=1
1 2a
p log p The last few steps used the fact that sin x and the Legendre symbol are ...