§ Primitive element theorem
- Let E/k be a finite extension. We will characterize when a primitive element exists, and show that this will always happen for separable extensions.
§ Part 1: Primitive element iff number of intermediate subfields is finite
§ Forward: Finitely many intermediate subfields implies primitive
- If k is a finite field, then E is a finite extension and E× is a cyclic group. The generator of E× is the primitive element.
- So suppose k is an infinite field. Let E/k have many intermediate fields.
- Pick non-zero α,β∈E. As c varies in k, the extension k(α+cβ) varies amongst the extensions of E.
- Since E only has finitely many extensions while k is infinite, pigeonhole tells us that there are two c1=c2 in E such that k(α+c1β)=k(α+c2β).
- Define L≡k(α+c1β). We claim that L=E, which shows that α+c1β is a primitive element for E.
- Since k(α+c2β)=k(α+c1β)=L, this implies that α+c2β∈L.
- Thus, we find that α+c1β∈L and α+c2β∈L. Thus, (c1−c2)βinL. Since c1,c2∈k, we have (c1−c2)−1∈K, and thus β∈L, which implies α∈L.
- Thus L=k(α,β)=k(α+c1β).
- Done. Prove for more generators by recursion.
§ Backward: primitive implies finitely many intermediate subfields
- Let E=k(α) be a simple field (field generated by a primitive element). We need to show that E/k only has finitely many subfields.
- Let ak(x)∈k[x] be the minimal polynomial for α in k. By definition, a is irreducible.
- For any intermediate field k⊆F⊆E, define aF(x)∈F[x] to be the minimal polynomial of α in F.
- Since ak is also a member of F[x] and ak,aF share a common root α and aF is irreducible in F, this means that aF divides ak.
- Proof sketch that irreducible polynomial divides any polynomial it shares a root with (Also written in another blog post): The GCD gcd(aF,ak)∈F[x] must be non constant since aF,ak share a root). But the irreducible polynomial aFcannot have a smaller polynomial ( gcd(aF,ak)) as divisor. Thus the GCD itself is the irreducible polynomial aF. This implies that aF divides aksince GCD must divide ak.
- Since ak is a polynomial, it only has finitely many divisors (upto rescaling, which does not give us new intermediate fields).
- Thus, there are only finitely many intermediate fields if a field is primitive.
§ Interlude: finite extension with infinitely many subfields
- Let F=Fp(t,u) where t,u are independent variables. This has finite degree since it has a vector space basis {tiuj}.
- Let α,β be roots of xp−t and xp−u. Define L≡F(α,β).
- Consider intermediate subfields Fλ≡F(α+λβ) for λ∈F.
- Suppose λ=μ for two elements in F. We want to show that Fλ=Fμ. This gives us infinitely many subfields as F has infinitely many elements. TODO
- Reference
§ Part 2: If E/k is finite and separable then it has a primitive element
- Let K=F(α,β) be separable for α,β∈K. Then we will show that there exists a primitive element θ∈K such that K=F(θ).
- By repeated application, this shows that for any number of generators K=F(α1,…,αn), we can find a primitive element.
- If K is a finite field, then the generator of the cyclic group K× is a primitive element.
- So from now on, suppose K is infinite, and K=F(α,β) for α,β∈F.
- Let g be the minimal polynomial for α, and h the minimal polynomial for β. Since the field is separable, g,h have unique roots.
- Let the unique roots of g be αi such that α=α1, and similarly let the unique roots of h be βi such that β=beta1.
- Now consider the equations α1+fi,jβ1=αi+fi,jβj for i∈[1,deg(g)] and j∈[1,deg(h)].
- Rearranging, we get (α1−αj)=fi,j(βj−β1). Since βj=β1 and α1=αj, this shows that there is a unique fi,j≡(α1−αj)/(βj−β1) that solves the above equation.
- Since the extension F is infinite, we can pick a f∗ which avoids the finite number of fi,j.
- Thus, once we choose such an f∗, let θ≡a1+fb1. Such a θ can never be equal to αi+fβj for any f, since the only choices of fthat make α1+fβ1=αi+fβj true are the fi,j, and f∗ was chosen to be different from these!
- Now let Fθ≡F(θ). Since θ∈K, E is a subfield of K.
- See that K=F(α,β)=F(α,β,α+fβ)=F(β,α+fβ)=F(θ,β)=Fθ(β).
- We will prove that K=Fθ.
- Let p(x) denote the minimal polynomial for β over Fθ. Since K=Fθ(β), if p(x) is trivial, the K=Fθ.
- By definition, β is a root of h(x). Since p(x) is an irreducible over Fθ, we have that p(x) divides h(x)[proof sketch: irreducible polynomial p(x) shares a root with h(x). Thus, gcd(p(x),h(x)) must be linear or higher. Since gcd divides p(x), we must have gcd=p(x) as p(x) is irreducible and cannot have divisors. Thus, p(x), being the GCD, also divides h(x)].
- Thus, the roots of p(x) must be a subset of the roots {βj} of h(x).
- Consider the polynomial k(x)=g(θ−f∗⋅x). β is also a root of the polynomial k(x), since k(β)=g(θ−f∗β), which is equal to g((α+f∗β)−f∗β)=g(α)=0. [since α is a root of g].
- Thus, we must have p(x) divides k(x).
- We will show that βj is not a root of k(x) for j=2. k(βj)=0 implies g(θ−f∗βj)=0, which implies θ−f∗βj=αisince the roots of g are αi. But then we would have θ=αi+f∗βj, a contradiction as θ was chosen precisely to avoid this case!
- Thus, every root of p(x) must come from {βj}. Also, the roots of p(x) must come from the roots of k(x). But k(x) only shares the root β1with the set of roots β2,…,βj. Also, p(x) does not have multiple roots since it is separable. Thus, p(x) is linear, and the degree of the field extension is 1. Therefore, K=E=F(θ).
§ References