Finite extension of a Galois field

Let $𝑝$ be a prime and $1 \leq 𝑑 \leq 𝑒$ be integers. Then there exists an extension

G F (𝑝 𝑑) ↪ G F (𝑝 𝑒)

iff $𝑑 ∣ 𝑒$ . Moreover, when such an extension exists it is unique and simple.¹ field

Proof

Suppose such an extension exists, whence we have a tower of field extensions $G F (𝑝 𝑒) : G F (𝑝 𝑑) : G F (𝑝)$ , so in particular $[G F (𝑝 𝑑) : G F (𝑝)]$ divides $[G F (𝑝 𝑒) : G F (𝑝)]$ , i.e. $𝑑 ∣ 𝑒$ .

Conversely, assume $𝑑 ∣ 𝑒$ , whence $(𝑝 𝑑 - 1) ∣ (𝑝 𝑒 - 1)$ , and by the same token $(𝑥 𝑝 𝑑 - 1 - 1) ∣ (𝑥 𝑝 𝑒 - 1 - 1)$ . Therefore $(𝑥 𝑝 𝑑 - 𝑥) ∣ (𝑥 𝑝 𝑒 - 𝑥)$ . By Construction and uniqueness, $G F (𝑝 𝑒)$ is the splitting field of the second polynomial. It follows from ^P1 that $G F (𝑝)$ .

For the last statement, note that Finite subgroup of the group of units of a field is cyclic, so in particular $G F (𝑝 𝑒) \times$ has a generator $𝛼$ , which will necessarily generate $G F (𝑝 𝑒)$ over any subfield. If $𝑑 ∣ 𝑒$ this means $G F (𝑝 𝑒) = G F (𝑝 𝑑) (𝛼)$ , so the extension is simple.

tidy | en | SemBr

2009. Algebra: Chapter 0, §VII.5.1, p. 442. ↩

Finite extension of a Galois field

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Finite extension of a Galois field

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