Download Homework 8 Solutions - UCSD Math Department

Homework 8 Solutions
Problem 6.5.2. Give addition and multiplication tables for the finite field F9 , and find a generator for the cyclic
group of nonzero elements under multiplication.
Proof. The polynomial x2 + 1 is irreducible over F3 since it has no roots in F3 and thus affords the representation
F3 [x]/hx2 + 1i of F9 . Note that F9× is cyclic of order 8. Let u = x + hx2 + 1i I claim that hu + 1i = F×
9 . Since
u2 + 1 = 0 we calculate (u + 1)2 = 2u, (u + 1)3 = 1 − u and (u + 1)4 = 2 from which Lagrange’s Theorem implies
that u + 1 has order 8.
Problem 6.5.4. Find the splitting fields over F3 for the following polynomials.
(a) x4 + 2
(b) x4 − 2
Proof. (a) Observe that x4 + 2 = x4 − 1 = (x + 1)(x − 1)(x2 + 1). Therefore the splitting field of x4 + 2 is the
splitting field of x2 + 1 which is F3 [x]/hx2 + 1i. Indeed, F3 [x]/hx2 + 1i = F3 (u) where u = x + hx2 + 1i and
x2 + 1 = (x − u)(x + u).
(b) First we have x4 − 2 = (x2 + 2x + 2)(x2 + x + 2). Note that x2 + 2x + 2 is irreducible over F3 [x] so
F = F3 [x]/hx2 + 2x + 2i is a field. I claim that F is the splitting field of x4 − 2. If u = x + hx2 + 2x + 2i then
u2 + 2u + 2 = 0 and one can verify that x2 + 2x + 2 = (x − u)(x + u + 2) and x2 + x + 2 = (x + u)(x + 2u + 1).
Thus f (x) splits over F and moreover F = F3 (u) so F must be the splitting field.
Problem 6.5.5. Show that x3 − x − 1 and x3 − x + 1 are irreducible over F3 . Construct their splitting fields and
explicitly exhibit the isomorphism between these splitting fields.
Proof. Both polynomials are irreducible over F3 since they have no roots in F3 .
Let p(x) = x3 − x − 1 and let E = F3 [x]/hp(x)i. If u = x + hp(x)i then note that u + a is a root of p(x) for
each a ∈ F3 . Therefore E = F3 (u) is the splitting field of p(x).
Let q(x) = x3 − x + 1 and let F = F3 [x]/hq(x)i. If v = x + hq(x)i then note that v + a is a root of q(x) for
each a ∈ F3 . Therefore F = F3 (v) is the splitting field of q(x).
Note that p(−v) = −v 3 +v +1 = −(v 3 −v −1) = −q(v) = 0 and thus by Lemma 6.4.3 the ring homomorphism
φ : E → F3 (−v) = F3 (v) = F determined by φ(u) = −v and φ(a) = a for a ∈ F3 is a well-defined isomorphism.
Problem 6.5.6. Show that x3 − x2 + 1 is irreducible over F3 . Construct its splitting field and explicitly exhibit
the isomorphism between this field and the splitting field of x3 − x + 1 over F3 .
Proof. Let r(x) = x3 − x2 + 1 then r(x) has no roots is F3 so it’s irreducible. Let L = F3 [x]/hr(x)i then L is a
field and if w = x + hr(x)i then w is a root of r(x) so w3 = w2 − 1. Also note that
r(w3 ) = (w3 )3 − w6 + 1 = (w2 − 1)3 + w6 + 1 = w6 − 1 − w6 + 1 = 0.
Therefore w3 is also a root of r(x). Since r(x) has degree 3 with roots w and w3 other root must also be in
L = F3 (w) so L is the splitting field of r(x).
Using the notation from problem 6.5.5 note that v 6= 0 so v −1 ∈ F and r(v −1 ) = v −3 − v −2 + 1 = v −3 (1 −
v + v 3 ) = v −3 q(v) = 0. Thus by Lemma 6.4.3 the ring homomorphism ψ : L → F3 (v −1 ) = F determined by
ψ(w) = v −1 and ψ(a) = a for a ∈ F3 is a well-defined isomorphism.
m
Problem 6.5.7. Show that if g(x) is irreducible over Fp and g(x) | (xp − x), then deg(g(x)) is a divisor of m.
m
m
Proof. Since g(x) | xp − x and Fpm is the splitting field of xp − x there is a root u of g(x) in Fpm . Therefore
m = [Fpm : Fp ] = [Fpm : Fp (u)][Fp (u) : Fp ] where deg(g(x)) = [Fp (u) : Fp ] since g(x) is irreducible. Hence
deg(g(x)) | m.
Problem 6.5.8. Let m, n be positive integers with gcd(m, n) = d. Show that, over any field, the greatest common
divisor of xm − 1 and xn − 1 is xd − 1.
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Proof. Let d(x) = gcd(xm − 1, xn − 1) then d | m and d | n so by Lemma 4.1.8 we have xd − 1 | xm − 1 and
xd − 1 | xn − 1. Therefore xd − 1 | d(x) by definition of d(x).
Since d = gcd(m, n) = xm + ny for some x, y ∈ Z and m and n are positive so one of x and y is positive
and the other is non-positive. Without loss of generality d = am − bn where a, b ≥ 0. By definition of d(x) we
have d(x) | xm − 1 and d(x) | xn − 1 and therefore xm ≡ 1 [d(x)] and xn ≡ 1 [d(x)]. Hence xam ≡ 1 ≡ xbm [d(x)]
but gcd(x, d(x)) = 1 since x - xm − 1 and therefore xd = xam−bn ≡ 1 [d(x)]. Thus d(x) | xd − 1 and hence
d(x) = xd − 1.
Problem 6.5.10. Let p be an odd prime.
(a) Show that the set S of squares in Fpn contains (pn + 1)/2 elements.
(b) Given a ∈ Fpn , let T = {a − x | x ∈ S}. Show that T ∩ S 6= ∅.
(c) Show that every element of Fpn is a sum of two squares.
(d) What can be said about F2n ?
Proof. (a) Recall that F×
pn = Fpn \{0} is a finite cyclic group generated by a non-zero element u ∈ Fpn . Now
n
define a map f : Fp → Fpn by f (uk ) = u2k . Clearly f is a group homomorphism and ker(f ) consists of the
roots of the polynomial x2 − 1, namely ±1. Note that S = im(f ) ∪ {0} and therefore |S| = |im(f )| + 1 =
n
n
|F×
pn / ker(f )| + 1 = (p − 1)/2 + 1 = (p + 1)/2.
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(b) By part (a) there are (p + 1)/2 squares and therefore |T | = |S| since g : S → T defined by g(x) = a − x
is easily verified to be a bijection. If S ∩ T = ∅ then Fpn would contain pn + 1 elements which is a contradiction.
Therefore T ∩ S 6= ∅.
(c) Suppose a ∈ Fpn . By (b) T ∩ S 6= ∅ so there is an x ∈ S such that a − x = y ∈ S. Therefore a = x + y is
a sum of two squares.
(d) Note that 1 = −1 in F2n and thus ker(f ) = {1} where f was defined in (a). Hence the argument in (a)
n
n
shows that |S| = |F×
2n | + 1 = 2 − 1 + 1 = 2 = |F2n |. Therefore every element of F2n is a square.
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