Homework 1-Solutions
Math 582
Spring 2014
1. Let F be a field of
not equal to 2.
√
√ characteristic
(a) Let K = F( D1 , D2 ), where D1 , D2 ∈ F and neither of D1 , D2 , and D1 D2 is a
square in F. Prove that K/F is a Galois extension with Gal(K/F) isomorphic to the
Klein-4 group.
(b) Conversely, suppose that K/F is √
Galois
√ group with Gal(K/F) isomorphic to the
Klein-4 group. Prove that K = F( D1 , D2 ), where D1 , D2 ∈ F and neither of D1 ,
D2 , and D1 D2 is a square in F.
Proof. The proof is straightforward.
2. Determine the Galois group of the splitting field over Q of x8 − 2.
Proof. This is done in the book on page 577.
3. Determine the splitting field of the polynomial x p − x − a over F p , where a 6= 0, a ∈ F p .
Show explicitly that the Galois group is cyclic. [Hint: Show that α 7→ α + 1 is an
automorphism.]
Proof. Let f (x) = x p − x − a. Notice that for all b ∈ F p we have b p = b and thus
f (b) = b p − b − a = −a 6= 0. Therefore none of the roots of f (x) lie in F p . Let α be a
root of f (x). We claim that F p (a) is the splitting field of f (x) over F p . Indeed, notice
that for any b ∈ F p then f (α + b) = (α + b) p − (α + b) − a = α p + b p − α − b − a = 0,
since α is a root of f (x) and b ∈ F p . Thus the roots of f (x) are α, α + 1, . . . , α + p − 1
and thus F p (α) is the splitting field of f (x), since α 6∈ F p .
Now, f (x) is irreducible since if it were reducible any irreducible factor of it will have
to have degree either 1 or p. Hence F p (α) is Galois over F p .
Since f (x) is irreducible then f (x) is the minimal polynomial of α over F p and thus
[F p (α) : F p ] = deg f (x) = p. Hence |Gal(F p (α)/F p )| = p and since p is prime this
means the Galois group is cyclic.
n
4. Prove that x p − x + 1 is irreducible over F p only when n = 1 or n = p = 2. [Hint: Notice
that if α is a root then so is α + a for any a ∈ F pn . Show that this implies F p (α) contains
F pn and that [F p (α) : F pn ] = p.]
n
Proof. Let f (x) = x p − x + 1 and suppose that f (x) is irreducible over F p . Let α be a
n
n−1
n−1
root of f (x). If α ∈ F p then α p = α and if n ≥ 2 then α p = (α p ) p = (α) p = . . . = α.
n
Then f (α) = α p − α + 1 = 1 6= 0. Thus α 6∈ F p .
n
Next notice that for any a ∈ F pn we have a p = a and
n
n
n
f (α + a) = (α + a) p − (α + a) + 1 = α p + a p − α − a + 1
n
n
= (α p − α + 1) + (a p − a) = 0.
Hence α + a is a root of f (x) for all a ∈ F pn and thus f (x) is an irreducible separable
polynomial over F p . Notice that f (x) is the minimal polynomial for α over F p and
F p (α) ' F p [x]/ f (x) ' F p (α + a) for any a ∈ F pn . Hence α + a ∈ F p (α) and thus a ∈
F p (α) for all a ∈ F pn , i.e. F pn ⊂ F p (α).
1
2
Next we observe that F p (α) is the splitting field of f (x) and thus F p (α)/F p is a Galois
extension of degree pn , since f (x) is irreducible. Furthermore, since F(α) is a finite
field extension of F p then G = Gal(F p (α)/F p ) ' Z/pn Z. Then by the Fundamental
Theorem of Galois theory there exists a subgroup H of G that corresponds to F pn . Since
G is cyclic then H is also cyclic. Let σ be a generator of H. Then σ ∈ G and thus σ sends
the root α to another root of f (x). Hence there exists a ∈ F pn such that σ(α) = α + a.
Since < σ >= H then σ fixes all the elements of F pn , since it is the fixed field of H. Thus
σi (α) = σi (α + a) = . . . = α + ia for all i ≥ 0. In particular σ p (α) = α, i.e. σ p = id and
σi 6= id for all 1 ≤ i ≤ p. Therefore, |H| = p and [F p (α) : F pn ] = p.
Combining now the degree extensions for F p ⊂ F pn ⊂ F p (α) we have pn = [F p (α) :
F p ] = [F p (α) : F pn ][F pn : F p ] = pn, since [F pn : F p ] = n. This can only happen if n = 1
or n = p = 2.
q
√
5. Determine the Galois closure of the field Q( 1 + 2) over Q.
q
√
Proof. Let E = Q( 1 + 2). The Galois closure of E over Q is defined as the composite of the splitting fields of the basis
q elements of E over Q. Therefore it is enough to
√
determine the splitting field of α = 1 + 2. We claim that f (x) = x4 − 2x2 − 1 is the
√
minimal polynomial for α over Q. Notice that α2 = 1 + 2 and (α2 − 1)2 = 2. Hence
f (α) = 0. Next notice that f (1) = 1−2−1 = −2 6= 0 and f (−1) = 1−2−1 6= 0. Therefore, by Corollary 11 in Chapter 9 we have that f (x) is irreducible over Q. Therfore f (x)
is the minimal
q polynomial for α over Q.
√
Let β = 1 − 2 and notice that f (β) = 0. Moreover, it is straightforward to check
that f (x) = (x − α)(x + α)(x − β)(x + β). Next observe that β 6∈ R and E ⊂ R. Therefore
β 6∈ E and thus E(β) = Q(α, β) contains the splitting field of f (x) since it contains all
its roots. Since β 6∈ Q(α) then α ∈ Q(β) either. Hence neither Q(α) or Q(β) can be the
splitting field of f (x) and therefore the splitting field of f (x) is Q(α, β). Thus the Galois
closure of E over Q is Q(α, β).