Download Math 210B: Algebra, Homework 7

Math 210B: Algebra, Homework 7
Ian Coley
February 26, 2014
Problem 1.
Construct a field with 9 elements and give its addition and multiplication tables.
Solution.
We realise this field as the quotient F3 [X]/(X 2 + 1).
+
1 2 X
X+1
1
2 0 X+1 X+2
2
1 X+2 X
X
2X
2X+1
X+1
2X+2
X+2
2X
2X+1
2X+2
X+2
X
X+1
2X+2
2X
2X+1
×
2 X X+1 X+2
2
1 2X 2X+2 2X+1
X
2
X+2 2X+2
X+1
2X
1
X+2
X
2X
2X+1
2X+2
2X
2X+1
2X+1 2X+2
2X+2 2X
0
1
1
2
2
0
X
X+1
X+2
2X
X
1
2X+1
X+1
2
2X+1
X+2
X+1
2
2X
2X+2
X
2X+2
2X
2X+1
2
0
1
X+2
X
X+1
2X+2
X+1
2X+1
X
2
X+2
1
2X
Problem 2.
Let f and g be two irreducible polynomials of the same degree n over a finite field F . Show
that there is a polynomial h ∈ F [X] such that f (h(X)) is divisible by g(X).
Solution.
Note first that both f and g split over the unique finite field K/F where |K| = |F |n . Since
any root of f or g is of degree at most n, and the splitting fields of f and g have degree at
1
most n. Since there is only one (up to isomorphism) degree n extension of F , all roots must
lie in it.
Consider the field structure on K given by f and g via F [X]. Since we have shown that
F [X]/(f ) ∼
= F [X]/(g), let ϕ : F [X]/(f ) → F [X]/(g) be such an isomorphism. We must
have ϕ(f (X)) = 0 since f (X) = 0 ∈ F [X]/(g). But we may consider ϕ to be a polynomial
by examining its image h(X). In this case, we see that h(f (X)) = f (h(X)) = 0 ∈ F [X]/(g),
that is, g(X) | f (h(X)). This completes the proof.
Problem 3.
Show that for any finite field Fq and n ∈ N there is an irreducible polynomial over Fq of
degree n.
Solution.
If we let q = pm , then we know there exists a finite field K of order pmn . Since this field is
unique, we can embed Fq into K, since Fq is isomorphic to the (unique) subfield of K of order
pm . Therefore K/Fq is a degree n extension. We know that we can express K as K = Fq (α)
for some α ∈ K by the primitive element theorem. Therefore we have n = [K : Fq ] = deg mα ,
where mα is the minimal polynomial of α over Fq . Hence mα is an irreducible polynomial of
degree n.
Problem 4.
Let f be an irreducible polynomial of degree n over a finite field Fq . Prove that f divides
n
X q − X.
Solution.
We know by Problem 7 that f splits in Fqn . Therefore all the roots of f , say {α1 , . . . , αn },
n
lie in Fqn . Since every element of Fqn satisfies X q − X = 0, we know that (X − αi ) | Fqn ,
n
hence lcm{X − αi } | X q − X. Since these linear polynomials are coprime, lcm{X − αi } =
Q
n
qn
− X as claimed.
i=1 (X − αi ) = f . Hence f | X
Problem 5.
Describe all q such that −1 is a square in Fq .
Solution.
We use the fact that the multiplicative group of a finite field is cyclic. Let F×
q = hαi. Let
m
2m
m ∈ N be minimal so that α = −1. Therefore α = 1, where 2m = q − 1.
Now, suppose q is a power of 2. Then −1 = 1, so it is a square. Now suppose q is an odd
prime power, so we have q ≡ 1 or 3 mod 4. If q ≡ 1 mod 4, then q − 1 ≡ 0 mod 4. Since
2m ≡ 0 mod 4, m is even, say m = 2n. Then
(αn )2 = αm = −1,
so αn is a square root of −1. However if q ≡ 3 mod 4, we have that 2m ≡ 2 mod 4, so m
is odd. Suppose that αn is a square root of −1. Since our chosen m was minimal, we have
m | 2n, hence m | n. But then 2m | 2n, and since α2m = 1, this implies that α2n = 1, which
is a contradiction since we assumed α2n = −1. Therefore only when q ≡ 1 mod 4 is −1 a
square in Fq .
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Problem 6.
Prove that every element of a finite field can be written as a sum of two squares.
Solution.
Let S be the set of all squares in a finite field F . First, if we assume that char F = 2, then
the the Frobenius automorphism x 7→ x2 is, in particular, onto. Therefore every y ∈ F
has a preimage so that x2 = y. Therefore every element is itself a square, so we can write
02 + x2 = y for all y.
Now if the characteristic of F is odd, let F × = hai. Then x ∈ F × is a square if and only
if x = a2m for some m ∈ N. Therefore there are |F − 1|/2 + 1 squares in F (since we include
0). Let S denote the set of all squares in F , and let b ∈ F be an arbitrary element. Let
T = {b − s : s ∈ S}. Then since
|S ∩ T | = |S| + |T | − |S ∪ T | = 2 · (|F − 1|/2 + 1) − |F |− = |F − 1| + 2 − |F | > 0,
there is always some element in its intersection. Therefore we have t = b − s =⇒ s + t = b
for any b ∈ F . Therefore every element is the sum of two squares.
Problem 7.
Show that any finite extension of a finite field is normal.
Solution.
Suppose that K/F is an extension of finite fields of degree n. Let f ∈ F [X] be an irreducible
polynomial with a root in K. Then since any root of f is of degree at most [K : F ], it lies
in a degree n extension of F . But since all such extensions are isomorphic, every root of f
must lie in K itself. Therefore f splits in K. Since this holds for all irreducible polynomials
f , K is a normal extension.
Problem 8. p
√
Prove that Q( 2 + 2)/Q is a normal extension.
Solution.
p
√
We claim that K = Q( 2 + 2 is the splitting field a polynomial. Consider the polynomial
f (X) = X 4 − 4X 2 + 2.
Then
q
√
√
√
f
2 + 2 = (6 + 4 2) − (8 + 4 2) + 2 = 0.
Further, this polynomial is irreducible by Eisenstein’s p
criterion and p = 2. Therefore it
√
suffices to prove that all roots of f are in K. Clearly − 2 +p 2 is a root of fpwhich is in
√
√
K. It is also easy to verify p
that the other two roots of f are 2 − 2 and − 2 − 2, so
√
we need only to show that 2 − 2 ∈ K. We have
q
q
q
√
√
√
√
2 + 2 · 2 − 2 = 4 − 4 = 2.
3
Therefore we obtain
p
2−
√
2 by
p
√ 2
2+ 2 −2
p
,
√
2+ 2
which is an element of K. Therefore K is a normal extension.
Problem 9.
√
Find a normal closure of the extension Q( 4 2)/Q.
Solution.
As discussed in class, we can construct
a norma closure of this field by adjoining all roots
√
4
of the minimal polynomial of α = 2 over Q. α is a root of X 4 − 2, which is irreducible by
Eisenstein’s criterion, hence this is its minimal polynomial. The roots of this polynomial are
{α, iα, −α, −iα}, i.e. α multiplied by the quartic roots of unity (which we identify for those
in C). These four roots are contained in the field K = Q(α, i), so the normal closure of Q(α
is contained in K. Further, in E = Q(α, iα, −α, −iα), we can find
i=
1 3
· α · iα,
2
so both i, α ∈ E, so Q(α, i) ⊂ K. Therefore K = E, so this field is a normal closure of
Q(α).
Problem 10.
Let F be a field of characteristic p. Show that the rational function field F (X) is a normal
extension of F (X p ).
Solution.
First, F (X) is generated by X over F (X p ). Second, in the polynomial ring F (X p )[Y ], X is
a root of Y p − X p . In F (X), this factors as (Y − X)p , so all the roots of Y p − X p are in
F (X). Therefore F (X) is a normal extension.
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