Ma5c HW 1, Spring 2016
3 Problem 1. Show that p(x) = x + 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 5 θ θ2 (1 + θ)−1 = − + . 2 4 4
3 Problem 2. Prove that x − 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. √ 2 3 Problem 3. Suppose F = Q(α1, . . . , αn) where αi ∈ Q for i = 1, 2 . . . , n. Prove that 2 ∈/ F.
Proof. One has a tower of extensions Q =: F0 ⊆ F1 := Q(α1) ⊆ ... ⊆ Fi := Qi(α1, . . . , αn) ⊆ ... ⊆ Fn := F. 2 Since αi ∈ 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. √ √ √ 3 3 3 The minimal polynomial of 2 is X3 − 2, hence [Q( 2): Q] = 3. If F contained 2, then it would contain √ 3 Q( 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]
1 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 2 2 −1 = p1(α) + ... + pm(α) .
Now we use the isomorphism F(α) = F[x]/(f(x))
to say there exists a polynomial g(x) such that
2 2 p1(x) + ... + pm(x) = −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 d X j pi(x) = ai,jx , j=0
2d where ai,d may be 0 but there exists at least one coefficient, ak,d that is nonzero, then the coefficient of x is 2 2 X ai,d ai,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 2 2 −1 + q(x)(r(x)f(x)) = p1(x) + ... + pm(x)
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 [K1K2 : F] = [K1 : F][K2 : F].
Proof. Suppose that K1 ⊗F K2 is a field. Define a map K1 × K2 → K1K2 by (k1, k2) → k1k2. This map is F-bilinear, so it induces a field homomorphism φ : K1 ⊗F K2 → K1K2. 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
[K1K2 : F ] = dimF(K1 ⊗F K2) = dimF(K1)dimF(K2) = [K1 : F][K2 : F].
Conversely, suppose that [K1 : F][K2 : F] = [K1K2 : F ], then let K1 have basis {αi} and K2 have basis {βj},
then {αiβj} is a basis for K1K2, so we may construct an isomorphism of vector spaces extending φ : αiβj → αi⊗βj linearly. As an isomorphism of F-algebras, this is a field.
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