1.31. Let p be a prime and let q be a prime that divides p - 1. (a) Let a € F and let b = a(P-¹)/4. Prove that either b = 1 or else b has order q. (Recall that the order of b is the smallest k ≥ 1 such that bk = 1 in F. Hint. Use Proposition 1.30.) (b) Suppose that we want to find an element of F of order q. Using (a), we can randomly choose a value of a € F and check whether b = a(p-¹)/9 satisfies b 1. How likely are we to succeed? In other words, compute the value of the ratio 1} (Hint. Use Theorem 1.31.) #{a € F*: a(P-1)/q; Fa #FP

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Chapter2: Second-order Linear Odes
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### Exercise 1.31: Finding an Element of a Given Order in a Finite Field

#### Problem Statement:

Let \( p \) be a prime number, and let \( q \) be another prime number that divides \( p - 1 \).

##### Part (a):
- Let \( \alpha \in \mathbb{F}_p^* \), the multiplicative group of non-zero elements of the finite field \( \mathbb{F}_p \).
- Define \( b = \alpha^{(p-1)/q} \). Prove that either \( b = 1 \) or \( b \) has order \( q \).

**Hint**: Recall that the order of \( b \) is the smallest integer \( k \geq 1 \) such that \( b^k = 1 \) in \( \mathbb{F}_p^* \). Use Proposition 1.30 for guidance.

##### Part (b):
- Suppose that we want to find an element of \( \mathbb{F}_p^* \) with order \( q \). Using part (a), we can randomly choose a value of \( \alpha \in \mathbb{F}_p^* \) and check whether \( b = \alpha^{(p-1)/q} \) satisfies \( b \neq 1 \).
- How likely are we to succeed? In other words, compute the value of the ratio:

\[
\frac{\{ \alpha \in \mathbb{F}_p^* : \alpha^{(p-1)/q} \neq 1 \}}{|\mathbb{F}_p^*|}
\]

**Hint**: Use Theorem 1.31 to aid in your calculation.

#### Explanation for Ratio Calculation:
To understand the likelihood of finding such an element \( b \neq 1 \) with the condition given, the goal is to evaluate the proportion of elements in \( \mathbb{F}_p^* \) that do not become 1 when raised to the power \((p-1)/q\). The formula provided represents this ratio, comparing the set of all such \( \alpha \in \mathbb{F}_p^* \) to the total number of elements in \( \mathbb{F}_p^* \).

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Transcribed Image Text:### Exercise 1.31: Finding an Element of a Given Order in a Finite Field #### Problem Statement: Let \( p \) be a prime number, and let \( q \) be another prime number that divides \( p - 1 \). ##### Part (a): - Let \( \alpha \in \mathbb{F}_p^* \), the multiplicative group of non-zero elements of the finite field \( \mathbb{F}_p \). - Define \( b = \alpha^{(p-1)/q} \). Prove that either \( b = 1 \) or \( b \) has order \( q \). **Hint**: Recall that the order of \( b \) is the smallest integer \( k \geq 1 \) such that \( b^k = 1 \) in \( \mathbb{F}_p^* \). Use Proposition 1.30 for guidance. ##### Part (b): - Suppose that we want to find an element of \( \mathbb{F}_p^* \) with order \( q \). Using part (a), we can randomly choose a value of \( \alpha \in \mathbb{F}_p^* \) and check whether \( b = \alpha^{(p-1)/q} \) satisfies \( b \neq 1 \). - How likely are we to succeed? In other words, compute the value of the ratio: \[ \frac{\{ \alpha \in \mathbb{F}_p^* : \alpha^{(p-1)/q} \neq 1 \}}{|\mathbb{F}_p^*|} \] **Hint**: Use Theorem 1.31 to aid in your calculation. #### Explanation for Ratio Calculation: To understand the likelihood of finding such an element \( b \neq 1 \) with the condition given, the goal is to evaluate the proportion of elements in \( \mathbb{F}_p^* \) that do not become 1 when raised to the power \((p-1)/q\). The formula provided represents this ratio, comparing the set of all such \( \alpha \in \mathbb{F}_p^* \) to the total number of elements in \( \mathbb{F}_p^* \). --- This modified text can be used as educational content to explain a
### Theorem 1.31 (Primitive Root Theorem)
Let \( p \) be a prime number. Then there exists an element \( g \in \mathbb{F}_p^* \) whose powers give every element of \( \mathbb{F}_p^* \), i.e.,
\[ \mathbb{F}_p^* = \{1, g, g^2, g^3, \ldots, g^{p-2}\}. \]

Elements with this property are called **primitive roots** of \( \mathbb{F}_p \) or generators of \( \mathbb{F}_p^* \). They are the elements of \( \mathbb{F}_p^* \) having order \( p - 1 \).

#### Proof
See [126, Chapter 20] or one of the texts [33, 47, 53, 90, 101].

The theorem is highlighting an important concept in finite field theory, particularly the existence of generators within the multiplicative group of a finite field. These generators are crucial for applications such as cryptography and the study of cyclic groups in abstract algebra.
Transcribed Image Text:### Theorem 1.31 (Primitive Root Theorem) Let \( p \) be a prime number. Then there exists an element \( g \in \mathbb{F}_p^* \) whose powers give every element of \( \mathbb{F}_p^* \), i.e., \[ \mathbb{F}_p^* = \{1, g, g^2, g^3, \ldots, g^{p-2}\}. \] Elements with this property are called **primitive roots** of \( \mathbb{F}_p \) or generators of \( \mathbb{F}_p^* \). They are the elements of \( \mathbb{F}_p^* \) having order \( p - 1 \). #### Proof See [126, Chapter 20] or one of the texts [33, 47, 53, 90, 101]. The theorem is highlighting an important concept in finite field theory, particularly the existence of generators within the multiplicative group of a finite field. These generators are crucial for applications such as cryptography and the study of cyclic groups in abstract algebra.
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