Theorem 22.3 Subfields of a Finite Field For each divisor m of n, GF(p") has a unique subfield of order p". Moreover, these are the only subfields of GF(p"). PROOF To show the existence portion of the theorem, suppose that m di- vides n. Then, since p" – 1 = (pm – 1)(pr-m + p-2m + . . . + pm + 1), 1. For simplicity, write p" – 1 = we see that pm - 1 divides p" (pm – 1)t. Let K = {xE GF(p") | x"™ = x}. We leave it as an easy exer- cise for the reader to show that K is a subfield of GF(p") (Exercise 27). Since the polynomial xP – x has at most p™ zeros in GF(p"), we have IKI < p". Let (a) = GF(p")*. Then la' = pm – 1, and since (a')p-I = 1, it follows that a' E K. So, K is a subfield of GF(p") of order p". The uniqueness portion of the theorem follows from the observation that if GF(p") had two distinct subfields of order p", then the polynomial xP" – x would have more than p" zeros in GF(p"). This contradicts Cor- ollary 3 of Theorem 16.2. Finally, suppose that F is a subfield of GF(p"). Then F is isomorphic to GF(p") for some m and, by Theorem 21.5, n = [GF(p"):GF(p)] = [GF(p"):GF( p")][GF(p"):GF(p)] = [GF( p"):GF( p")]m. Thus, m divides n.

Theorem 22.3 Subfields of a Finite Field For each divisor m of n, GF(p") has a unique subfield of order p". Moreover, these are the only subfields of GF(p"). PROOF To show the existence portion of the theorem, suppose that m di- vides n. Then, since p" – 1 = (pm – 1)(pr-m + p-2m + . . . + pm + 1), 1. For simplicity, write p" – 1 = we see that pm - 1 divides p" (pm – 1)t. Let K = {xE GF(p") | x"™ = x}. We leave it as an easy exer- cise for the reader to show that K is a subfield of GF(p") (Exercise 27). Since the polynomial xP – x has at most p™ zeros in GF(p"), we have IKI < p". Let (a) = GF(p")*. Then la' = pm – 1, and since (a')p-I = 1, it follows that a' E K. So, K is a subfield of GF(p") of order p". The uniqueness portion of the theorem follows from the observation that if GF(p") had two distinct subfields of order p", then the polynomial xP" – x would have more than p" zeros in GF(p"). This contradicts Cor- ollary 3 of Theorem 16.2. Finally, suppose that F is a subfield of GF(p"). Then F is isomorphic to GF(p") for some m and, by Theorem 21.5, n = [GF(p"):GF(p)] = [GF(p"):GF( p")][GF(p"):GF(p)] = [GF( p"):GF( p")]m. Thus, m divides n.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Show that the set K in the proof of Theorem 22.3 is a subfield

Theorem 22.3 Subfields of a Finite Field
For each divisor m of n, GF(p") has a unique subfield of order p".
Moreover, these are the only subfields of GF(p").
PROOF To show the existence portion of the theorem, suppose that m di-
vides n. Then, since
p" – 1 = (pm – 1)(pr-m + p-2m + . . . + pm + 1),
1. For simplicity, write p" – 1 =
we see that pm - 1 divides p"
(pm – 1)t. Let K = {xE GF(p") | x"™ = x}. We leave it as an easy exer-
cise for the reader to show that K is a subfield of GF(p") (Exercise 27).
Since the polynomial xP – x has at most p™ zeros in GF(p"), we have
IKI < p". Let (a) = GF(p")*. Then la' = pm – 1, and since (a')p-I = 1,
it follows that a' E K. So, K is a subfield of GF(p") of order p".
The uniqueness portion of the theorem follows from the observation
that if GF(p") had two distinct subfields of order p", then the polynomial
xP" – x would have more than p" zeros in GF(p"). This contradicts Cor-
ollary 3 of Theorem 16.2.

Finally, suppose that F is a subfield of GF(p"). Then F is isomorphic
to GF(p") for some m and, by Theorem 21.5,
n = [GF(p"):GF(p)]
= [GF(p"):GF( p")][GF(p"):GF(p)]
= [GF( p"):GF( p")]m.
Thus, m divides n.

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