Ma5c HW 5, Spring 2016
Problem 1. Let p be a prime and let F be a field. Let K be a Galois extension of F whose Galois group is a p-group
(i.e., [K : F] = pk for some k). Such an extension is called a p-extension (note that p-extensions are Galois by definition).
(a) Let L be a p-extension of K. Prove that the Galois closure of L over F is a p-extension of F.
(b) Given an example to show that (a) need not hold if [K : F] is a power of p but K/F is not Galois.
Proof.
(a) Let L be the Galois closure of L over F and L1 , . . . , Ln be the Galois conjugates of L over F (in
some algebraic closure of L), then we claim L = L1 . . . Ln . It should be clear that Li ⊂ L, since σ(L) = L
for all σ ∈ Gal(L/F). So L1 . . . Ln ⊂ L. Given an α with minimal polynomial mα (x) splits in L, as we
did with Theorem 14.13, the roots of mα (x) are contained in L and are given by the Galoi conjugates of
mα (x), hence are contained in L1 . . . Ln , hence, any irreducible with a root in L1 . . . Ln has all roots in
L1 . . . Ln , hence is a Galois extension containing L. This means that L1 . . . Ln contains the Galois closure.
This shows L1 . . . Ln = L.
Secondly we may use corrolary 14.20 to show composites of p-extensions are p-extensions. If Li and Lj are
p-extensions , then corrolary 14.20 tells us
[Li Lj : K] =
[Li : K][Lj : K]
[Li ∩ Lj : K]
is a product of powers of p, hence, must be a p-extension. Inductively, the finite composite L1 . . . Ln is a
composite of fields isomorphic to a p-extension, hence, the composite is also a p-extensions.
√
(b) Let F = Q and L = K = Q( 3 2) then L is of degree 6, which is not a power of 3.
Problem 2. Prove that Fp (x, y)/Fp (xp , y p ) is not a simple extension by explicitly exhibiting an infinite number of
intermediate subfields.
Proof. Note that x is a root of z p − xp ∈ Fp (xp , y p )[z] which is a polynomial of degree p in z. Similarly y is a
root of z p − y p ∈ Fp (xp , y p )[z]. As indeterminants, Fp (x) ∩ Fp (y) = Fp , hence
[Fp (x, y) : Fp (xp , y p )] = p2
We wish to show that for each λ ∈ Fp (xp ), the fields F(xp , y p )(x + λy) = F(xp , y p , x + λy) are distinct. Suppose
F(xp , y p , x + λy) = F(xp , y p , x + µy)
where λ 6= µ, then
(x + λy) − (x + µy) = (λ − µ)y ∈ F(xp , y p , x + λy)
which implies both x and y are in our extension, implying Fp (x, y) = Fp (xp , y p , x + λy) is a degree p2 extension
over Fp (xp , y p ). But, because (x + λy)p = xp + λp y p , this comes as the solution of
z p − (xp + λp y p )
which is degree p, contradicting the statement that this is degree p2 . Since there are an infinite number of
elements of Fp (xp ), there are an infinite number of subfields of Fp (x, y).
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Ma5c HW 5, Spring 2016
Problem 3. Compute the Galois groups of the following polynomials
(a) f (x) = x4 + 3x3 − 3x − 2.
(b) f (x) = x5 + x − 1.
Proof.
(a) The discriminant is −2183 and the group is S4 .
(b) The trick here is that the polynomial factors into
(x2 − x + 1)(x3 + x2 − 1).
The question is what is the Galois group of these two polynomials. By the rational root theorem, both are
irreducible. The discriminant of the first factor is D = −3, while the discriminant of the second polynomial
is −23, we have groups S2 = Z/2Z and S3 respectively. Since it is clear the intersection of the splitting
fields is trivial for each polynomial is trivial, the resuling group is
G∼
= S2 × S3 .
Problem 4. Prove that the discriminant, D, of the polynomial xn + px + q is given by the formula
D = (−1)
n(n−1)
2
nn q n−1 + (−1)
(n−1)(n−2)
2
(n − 1)n−1 pn
Proof. If p = 0, and q 6= 0 then this product is simply
D = (−1)(n−1)(n−2)/2
n
Y
nαjn−1 = (−1)n(n−1)/2 nn q n−1 .
(1)
j=1
as required.
The product of f 0 (αj )’s are almost the discriminant, however, precisely (n − 1)(n − 2)/2 terms in the wrong
order, hence,
D=
Y
(αi − αj )2 =(−1)(n−1)(n−2)/2
i>j
n
Y
(nαjn−1 + p),
(2)
j=1
If p = 0, and q 6= 0 then this product is simply
D = (−1)(n−1)(n−2)/2
n
Y
nαjn−1 = (−1)n(n−1)/2 nn q n−1 .
j=1
as required.
If q = 0 and p 6= 0 f (x) = x(xn−1 − p), which has 0 as a solution, hence, we may split this up into
D = D0
Y
(0 − ζ
√
n−1
a)2 = p2 D0
ζ∈µn−1
where D0 is the discriminant of xn−1 − p, which by (??) is
D = p2 (−1)
(n−1)(n−2)
2
(n − 1)n−1 pn−2 = (−1)
as required.
2
(n−1)(n−2)
2
(n − 1)n−1 pn
(3)
Ma5c HW 5, Spring 2016
If p, q 6= 0, then
n
Y
(nαjn−1 + p) =
j=1
Y nαjn + pαj
j
αj
=
Y n(−pαj − q) + pαj
αj
j
(−1)n Y
(−nq + αj (p(1 − n)))
q
j
−nq
pn (n − 1)n
−nq
pn (n − 1)n Y
− αj =
f
=
q
p(n − 1)
q
p(n − 1)
j
=
=
pn (n − 1)n
q
nq
(−1)n nn q n
−
+q
n
n
p (n − 1)
(n − 1)
= (−1)n nn q n−1 + (n − 1)n−1 pn
Putting this into (??) gives
D=
Y
(αi − αj )2 = (−1)
n(n−1)
2
nn q n−1 + (−1)
(n−1)(n−2)
2
(n − 1)n−1 pn .
i>j
as required.
Problem 5. Prove that the polynomial x4 − px2 + q ∈ Q[x] is irreducible for any distinct odd primes p and q and has
as its Galois group the dihedral group of order 8.
Proof. Firstly, we know the roots, which are obtained from the quadratic formula,
s
p
p ± p2 − 4q
x=±
.
2
None of these roots are in the field, nor could we split these roots into two distinct irreducible factors of degree
2, indicating f is irredicuble. From the book, we have that the resolvant polynomial is
h(x) = x3 + px2 − 4qx − 4pq
which is reducible. This factors into a polynmial of degree 2 and 1, namely x + p is a factor. This means that
G ≤ D8 , which means the group is C4 or D8 ,. As the book also suggests, these two cases, C4 and D8 , are
√
√
distinguished by the intersection with the fixed field for A4 , i.e, if we adjoin the D = (p2 − 4q) q, then the
polynomial remains irreducible, indicating that the group has a transitive intersection with A4 , indicating it is
indeed D8 and not C4 .
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