Solution to Assignment 2 - The University of Hong Kong

Solution to Assignment 2 - The University of Hong Kong

The University of Hong Kong
DEPARTMENT OF MATHEMATICS
MATH3302/MATH4302 Algebra II
Suggested Solution to Assignment 2
In the following, E, F, K are fields.
1. Let G be an abelian group and a, b ∈ G with ord (a) = m, ord (b) = n.
(a) Prove that:
i.
ii.
iii.
iv.
Let gcd(m, n) = 1. Then ord (ab) = mn.
Let r|m with postive integers r. Then ord (ar ) = m
r.
There exists c ∈ G such that ord (c) = lcm (m, n).
Suppose for all g ∈ G, ord (g) ≤ m. Then for all g ∈ G, ord (g)|m.
(b) Let G be a finite subgroup of the multiplication group F ∗ :=
F − {0} of a field F . Prove that G is cyclic.
(c) Concluding a finite extension of a finite field must contain a primitive elements.
Solution:
(a)
i. Let t = ord (ab). On the one hand, (ab)mn = amn bmn = e ⇒
t|m. On the other hand, (ab)tn = atn btn = atn = e ⇒ m|tn.
Since gcd(m, n) = 1, m|t. Similarly, we also have n|t. Thus
mn|t. Therefore mn = t.
m
ii. Write k = ord (ar ). On the one hand (ar ) r = e ⇒ k| m
r ⇒
r )k = e ⇒ m|rk ⇒ m ≤ rk.
k≤ m
.
One
the
other
hand
(a
r
Thus k = m
r.
iii. By unique factorization of integers, m = pk11 · · · pknn , n =
k
kj
1 ···p jn−m
p
p
si
1 ···psim
jn−m
ps11 · · · psnn . Set a0 = a j1
, b0 = b i1 im where {i1 , . . . , im } =
{i, ki ≥ si }, {j1 , . . . , jn−m } = {j, kj < sj }. Then ord (a0 ) =
ki
sj
k
sj
n−m
pi1 1 · · · pimim , ord (b0 ) = pj11 · · · pjn−m
, so that gcd(ord (a0 ), ord (b0 )) =
0
0
1. Let c = a b . We have ord (c) = ord (a0 )ord (b0 ) = lcm (m, n).
iv. For any g ∈ G, write ord (g) = kg . Since ∃c ∈ G such that
m ≤ lcm (m, kg ) = ord (c) ≤ m. Thus kg |lcm (m, kg ) = m.
(b) Let u ∈ G − {e} be an element of maximal order. We claim
that G =< u >. Write ord (g) = kg and ord (u) = m. For any
g ∈ G, kg |m. Thus g is a root of f (x) := xm − e. Consider
f (x) ∈ A[x] for some algebraically closed field A, we see by the
1
factor theorem that (x−g) is a factor of f (x) for any g ∈ G. Thus
h(x) = Πg∈G (x − g) is a factor of xm − e so that |G| = deg(h) ≤
deg(f ) = m. Hence |G| = m.
(c) If [Fq : Fp ] = n, then |Fq | = pn . Then F∗q =< u > for some
u ∈ F∗q . Thus Fq = Fp (u).
2. (a) Prove that
FLT(F ) =
ax + b a, b, c, d ∈ F, ad − bc 6= 0
cx + d is a group with function composition.
(b) Prove that
ϕ : GL(2,
F) → FLT(F )
a b
ax+b
7→
cx+d
c d
is a surjective group homomorphism.
(c) Find Ker (ϕ).
(d) Define PGL(2, F ) = GL(2, F )/Ker (ϕ). Prove that PGL(2, F ) is
isomorphic to FLT(F ).
(e) Find a subgroup of order 6 of FLT(F ).
Solution:
(a) The closure of operation follows from the condition that ad−bc 6=
0. The identity element is the identity function x and for each
ax+b
−dx+b
cx+d , the inverse is cx−a . The associativity follows from the
associativity of the composition of functions.
(b) Sujectivity follows from the condition ad−bc 6= 0. To show that ϕ
is homomorphism,
let A = (aij ), B = (bij ), C = (cij ) ∈ GL(2, F ),
P2
where cij = k=1 aik bkj . Then
P2
P
a b x+ 2 a1k bk2
P2k=1 1k k1 Pk=1
,
2
k=1 a2k bk1 x+ k=1 a2k bk2
a11 x+a12
P
P2
b11 a x+a +b12
a1k bk1 x+ 2k=1 a1k bk2
21
22
P2
= Pk=1
.
a11 x+a12
2
b21 a x+a +b22
k=1 a2k bk1 x+ k=1 a2k bk2
P
ϕ(AB) = ϕ((cij )) = ϕ( 2k=1 aik bkj ) =
ϕ(A)ϕ(B) =
a11 x+a12
a21 x+a22
◦
b11 x+b12
b21 x+b22
=
21
(c) Ker ϕ = {(aij ) ∈ GL(2, F ) |
a11 x+a12
a21 x+a22
22
= x} = {λI | λ ∈ F \ {0}}.
(d) By the fundamental theorem of homomorphism, GL(2, F )/Ker ϕ ∼
=
ϕ(GL(2, F )). Since ϕ is surjective, ϕ(GL(2, F )) = FLT(F ).
2
1
x
(e) {x, x1 , 1 − x, 1−x
, 1 − x1 , x−1
}.
3. Let x be non-algebraic over F .
1
(a) Show that F (x) = F ( x1 ) = F (1 − x) = F (1 − x1 ) = F ( 1−x
) =
x
F ( x−1 ).
(b) Let σ1 : g(x) 7→ g(x), σ2 : g(x) 7→ g( x1 ), σ3 : g(x) 7→ g(1 − x), σ4 :
1
1
x
g(x) 7→ g( 1−x
), σ5 : g(x) 7→ g( 1−x
), σ6 : g(x) 7→ g( x−1
) for g(x) ∈
F (x). Show that for i = 1, . . . , 6, σi are automorphisms of F (x)
2 −x+1)3
(ie., σi (I(x)) = I(x)).
over F fixing I(x) := (x
x2 (x−1)2 )
(c) Show that [F (x) : F (I(x))] ≤ 6.
(d) Show that there are only 6 distinct automorphisms of F (x) over
F (I(x)).
(e) Hence show that F (x)/F (I(x)) is separable.
Solution:
(a) It is easy to see that F (x) = F ( x1 ) = F (1 − x) and {x, x1 , 1 − x}
generates the rest.
(b) The check for σi , i = 1, 2, 3 is straight forward. The rest follows
from that fact that they are generated by σi , i = 1, 2, 3.
(c) Consider
f (t) = (t2 − t + 1)3 − I(x)(t2 (t − 1)2 ) ∈ F (I(x))[t].
As x is a root of f , [F (x) : F (I(x))] ≤ deg(f (t)) = 6.
(d) Since 6 = |{σi , i = 1, . . . , 6}| ≤ |Aut (F (x)/F (I(x))| ≤ [F (x) :
F (I(x)] ≤ 6. This forces |Aut (F (x)/F (I(x))| = 6.
(e) Note that σi ’s are actually embeddings of F (x) over F (I(x)).
Thus |Emb (F (x)/F (I(x))| = [F (x) : F (I(x))].
4. Prove that the following two statements are equivalent:
(a) F is a separable field.
(b) Char (F ) = 0; or Char (F ) = p > 0 and for all a ∈ F , there exists
b ∈ F such that bp = a.
Solution: The case that Char F = 0 is clear. We only need to consider
Char F = p > 0.
3
Suppose F is separable. Let a ∈ F and consider f (x) = xp − a ∈ F [x].
Then in some extension field E over F , we can find some α ∈ E such
that a = αp . Then in f (x) = xp −αp = (x−α)p ∈ E[x]. Since minimal
polynomials are irreducible, we have MinPoly F (α)|f (x). Since F is
separable, any irreducible polynomial in F [x] has no multiple roots.
Thus MinPoly F (α) = x − α and α ∈ F .
Conversely, suppose every element in F has a p-th root but F is
not separable, then ∃u ∈ F such that MinPoly F (u) has a multiple root. It follows that MinPoly F (u) is a polynomial in xp . Writting the coefficients of MinPoly F (u) as p-th roots, one notice that
n
MinPoly F (u) = (x − u)p . This is a contradiction to the irreducibility
of MinPoly F (u).
5. (a) Let K/E and E/F be separable. Prove that K/F is separable.
(b) Let K/F be separable, and E/F any extension.
KE/E is separable.
Prove that
(c) Let K/F and E/F are separable. Prove that KE/F is separable.
Solution: Define |K|F to be the number of distinct embeddings of K
over F . Note that for a finite extension K/F , it is separable if and
only if |K|F = [K : F ].
(a) |K|F = |K|E |E|F = [K : E][E : F ] = [K : F ].
(b) Since K/F is finite and separable, we may assume K = F (α) for
some α ∈ K. Then KE = E(α) and |KE|E = |α|E . Note that
K/F is separable implies that α is separable over F and hence
separable over E. Thus |KE|E = |α|E = [E(α) : E] = [KE : E].
(c) [KE : F ] = [KE : E][E : F ] = |KE|E [E : F ] = |KE|E |E|F =
|KE|F .
6. (a) Let K/F be algebraic. Show that any σ ∈ Emb (K/F ) induces
an automorphism of K.
(b) Give an example σ ∈ Emb (K/F ) such that σ : K → K over F
does not induce an automorphism of K (ie., σ(K) 6= K).
Solution:
(a) If u ∈ K and all the distinct roots of MinPoly F (u) are S :=
{α1 , . . . , αr }, then the embedding σ is injective on the finite set
S → S. Thus σ is also surjective on S. This means that σ is
surjective on K.
4
(b) Let σ : Q(π) → Q(π), f (π) 7→ f (π 2 ). Then σ is homomorphism
√
√
but [Q( π) : Q(π)] = 2. Thus π ∈
/ Q(π), which implies that π
has no inverse image, ie., π ∈
/ σ(Q(π)). Hence σ is not surjective.
5