Question

1. If a and b in K are algebraic over F of degree m and n respectively, then ab is
algebraic over F of degree​

Answer 1

Answer:

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Answer 2

Answer:

First we tend to build some easy observations. F(a) is that the smallest field containing F and

a, and conjointly that F[a] = F(a). Thus F(a, b) = F[a](b) = F[a][b] is that the smallest field containing

F and conjointly each a and b. so F[b][a] = F(b, a) = F(a, b) = F[a][b]. Now, we've the chain of

containment F ⊆ F[a] ⊆ F[a][b] = F(a, b) and then we are able to write

                      [F(a, b) : F]=[F[a] : F] · [F[a][b] : F[a]]

but thus m = [F[a] : F] divides [F(a, b) : F]. By symmetry, n conjointly divides [F(a, b) : F] and then

[F(a, b) : F] ≥ mn since m and n area unit comparatively prime.

 On the opposite hand, g(b) = zero for a few g(x) ∈ F[x] of degree n. Note F[x] ⊆ F[a][x] and then

g(x) ∈ F[a][x] similarly. So g(b) = zero. Let’s let h(x) ∈ F[a][x] be the stripped polynomial for b, so

that h|g. It follows that deg h ≤ n and then [F[a][b] : F[a]] = deg h ≤ n. Thus,

        mn ≤ [F(a, b) : F] = [F[a] : F] · [F[a][b] : F[a]] = m · [F[a][b] : F[a]] ≤ mn

and so we've our desired equality.

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