By Surowski D.

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**Extra info for Workbook in higher algebra**

**Sample text**

Let K2 = F2 (α2 ) ⊇ F2 , where α2 is a root of ˆ 1 (x)). Then there exists an isomorphism f2 (x) = ψ(f ∼ = ψ¯ : K1 −→ K2 , ¯ 1 ) = α2 , and ψ| ¯ F = ψ. 2 Let F1 be a field, let f1 (x) ∈ F1 [x], and let K1 be a splitting field over F1 for f1 (x). Let ∼ = ψ : F1 −→ F2 , ˆ 1 (x)) ∈ F2 [x], and let K2 be a splitting field over F2 for f2 (x). let f2 (x) = ψ(f Then there is a commutative diagram K1 ✻ F1 ψ¯ ✲ K2 ✻ ψ ✲ K2 where the vertical maps are inclusions, and where ψ¯ is an isomorphism. Let F be a field and let F ⊆ F[x].

If F and K are fields with F ⊆ K, we say that K is an extension of F. Of fundamental importance here is the observation that if F ⊆ K is an extension of fields, then K can be regarded as a vector space over F. It is customary to call the F-dimension of K the degree of K over F, and to denote this degree by [K : F]. The following simple result is fundamental. 1 Let F ⊆ E ⊆ K be an extension of fields. Then [K : F] < ∞ if and only if each of [K : E], [E : F] < ∞, in which case [K : F] = [K : E] · [E : F].

Indeed, let q = pn , and let f (x) = xq − x ∈ Fp [x]. 4, part (i) f (x) is separable. Thus if F ⊇ Fp is a splitting field, then it’s easy to see that F consists wholly of the q roots of f (x). 8 For any prime p, and any integer n, there exists a field of order pn . Thus, for any prime power q = pn there exists a unique (up to isomorphism) field of order q. We denote such a field simply by Fq . Finally, we’ll say a few words about Galois groups in this setting. Let F = Fq be the finite field of order q, and let K = Fqn be an extension of degree n.

### Workbook in higher algebra by Surowski D.

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