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

See similar textbooks

Similar questions

Prove that Z Q has the universal property of inverting non-zero elements.
The approximated value for e using the Maclaurin series expansion considering a true relative error that is conforming to one significant figure is: O a. 4.052 O b. 3,885 Oc3414 O d. 2414
Show that the set of all real numbers of the forma0 + a1π + a2π2 + · · · + anπnwith ai ∈ Z and n ∈ N is a subring of R that contains both Z and π
Prove, without using Theorem 1.12 or copying the proof, that for all a, b ∈ Q with a < b, there exists a c ∈ Q with a < c < b
Do the sets [0,1] and [0,1/2)∪(1/2,1] have the same cardinality? Either give an explicit bijection between them or show that no such bijection exists
Let S = {x | 6x – 1 E Q} and T be the - set of all odd integers. Prove that S and T have the same cardinality (no need to prove bijectivity of functions). Do not use the Schroder-Bernstein Theorem.
= m2 +1. Let A1, A2,..., An be a family of distinct finite sets of positive integers where Show that there exist at least m +1 of these sets, say Ak, Ak2,..., Akm+1, such that Ak, U Ak, ordered set P = ({A1, A2, ... , An}, C).) Ak, does not hold for all distinct kr, ks, kt. (Hint: Consider the partially
4. 225
Show that the set Q + Q √ 3 is a field. Q denotes set of all rational numbers
Use Cardinality and bijection to prove
(1) Prove that if R is an integral domain of character- istic p where p is a prime number. Then for any a, b € R, (a + b)² = a³ + b³.
Show work