Ma5c HW 1, Spring 2016
Problem 1. Show that p(x) = x3 + 9x + 6 is irreducible in Q[x]. Let θ be a root of p(x). Find the inverse of 1 + θ in
Q(θ).
Proof. The polynomial
p(x) = x3 + 9x + 6
is such that p = 3 divides each of the non-leading coefficients (0,9 and 6), but p2 does not divide 6 and p does
not divide the leading coefficient, 1. It is thus irreducible by Eisenstein’s criterion. Multiplying θ + 1 by an
arbitrary element of Q(θ) and setting the product equal to 1 gives
(1 + θ)(a + bθ + cθ2 ) = a + θ(a + b) + θ2 (b + c) + cθ3 = 1
but θ3 = −9θ − 6, giving
(1 + θ)(a + bθ + cθ2 ) = (a − 6c) + θ(a + b − 9c) + θ2 (b + c)
We solve a − 6c = 1, a + b − 9c = 0 and b + c = 0.

   
1 0 −6
a
1

   
1 1 −9  b  = 0

   
0 1 1
c
0
giving c = 1/4, b = −1/4
(1 + θ)−1 =
5 θ θ2
− + .
2 4
4
Problem 2. Prove that x3 − nx + 2 ∈ Z[x] is irreducible for n 6= −1, 3, 5.
Proof. If p were reducible, then it would have to factor into a linear a quadratic factor (or three linear factors).
In either case, p would have a rational root α. By the rational root theorem, α would be an integer dividing 2,
hence α ∈ {±1, ±2}. Plugging these values into p, gives us n ∈ {−1, 3, 5}, precisely the values n is not allowed
to take.
Problem 3. Suppose F = Q(α1 , . . . , αn ) where αi2 ∈ Q for i = 1, 2 . . . , n. Prove that
√
3
2∈
/ F.
Proof. One has a tower of extensions Q =: F0 ⊆ F1 := Q(α1 ) ⊆ . . . ⊆ Fi := Qi (α1 , . . . , αn ) ⊆ . . . ⊆ Fn := F.
Since αi2 ∈ Q ⊆ Fi−1 for all i, [Fi : Fi−1 ] is either 1 or 2 for every i. Since field degrees are multiplicative in
towers, we have [F : Q] = 2m for some nonnegative integer m ≤ n.
√
√
√
The minimal polynomial of 3 2 is X 3 − 2, hence [Q( 3 2) : Q] = 3. If F contained 3 2, then it would contain
√
Q( 3 2), and we should have 3|2m , which is obviously impossible.
Problem 4. A field F is said to be formally real if −1 is not expressible as a sum of squares in F. Let F be a formally
real field, let f (x) ∈ F[x] be an irreducible polynomial of odd degree and α be a root of f (x). Prove F(α) is also formally
real. [There is a hint in the book]
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Ma5c HW 1, Spring 2016
Proof. As the hint suggests, let f (x) be a minimal degree counterexample. It is of odd degree > 1 (or F(α) = F)
so that
−1 = p1 (α)2 + . . . + pm (α)2 .
Now we use the isomorphism
F(α) = F[x]/(f (x))
to say there exists a polynomial g(x) such that
p1 (x)2 + . . . + pm (x)2 = −1 + f (x)g(x)
We can choose pi to have degree smaller than f . Then the RHS has degree less than 2 deg f , hence deg g < deg f .
We now claim the degree of the left hand side is even. Suppose the maximal degree of the pm is d, then as
polynomials over F we may write
pi (x) =
d
X
ai,j xj ,
j=0
where ai,d may be 0 but there exists at least one coefficient, ak,d that is nonzero, then the coefficient of x2d is


X ai,d 2

a2i,k 1 +
ai,k
i6=k
over F, which is non-zero by the assumption F is formally real. This tells us the degree of g(x) is in fact
odd, hence, it must also have an irreducible factor, q(x) of odd degree and deg q < deg f . We may write
g(x) = q(x)r(x), but now
−1 + q(x) (r(x)f (x)) = p1 (x)2 + . . . + pm (x)2
so that −1 is a square over F[x]/(q(x)) contradicting the degree minimality of f .
Problem 5. Let K1 and K2 be two finite extensions of a field F contained in the field K. Prove the F-algebra K1 ⊗F K2
is a field if and only if [K1 K2 : F] = [K1 : F][K2 : F].
Proof. Suppose that K1 ⊗F K2 is a field. Define a map K1 × K2 → K1 K2 by (k1 , k2 ) → k1 k2 . This map is
F-bilinear, so it induces a field homomorphism φ : K1 ⊗F K2 → K1 K2 . But a nontrivial field homomorphism is
always injective and moreover it is clear that φ is surjective, thus φ is an isomorphism of fields. Considering
them as vector spaces over F we obtain
[K1 K2 : F ] = dimF (K1 ⊗F K2 ) = dimF (K1 )dimF (K2 ) = [K1 : F][K2 : F].
Conversely, suppose that [K1 : F][K2 : F] = [K1 K2 : F ], then let K1 have basis {αi } and K2 have basis {βj },
then {αi βj } is a basis for K1 K2 , so we may construct an isomorphism of vector spaces extending φ : αi βj → αi ⊗βj
linearly. As an isomorphism of F-algebras, this is a field.
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