Question

Let \(m_{0}=x^{2}+1, m_{1}=x^{2}-1, m_{2}=x^{3}+x-1, v_{0}=-x, v_{1}=x+1\), and \(v_{2}=x^{5}-x\) in \(\mathbb{F}_{3}[x]\). (i) How many polynomials \(f \in \mathbb{F}_{3}[x]\) are there with \(f \equiv v_{i}\) mod \(m_{i}\) for \(i=0,1,2\), and \(\operatorname{deg} f \leq 8\) ? Answer this without solving (ii). (ii) Give a list of all \(f\) as in (i).

Short answer

48 polynomials exist satisfying the conditions as the CRT results in \(3^7\) possible solutions.

Step-by-step solution

Check the pairwise coprimality of the modulus terms

We first need to verify that \(m_0 = x^2 + 1\), \(m_1 = x^2 - 1\), and \(m_2 = x^3 + x - 1\) are pairwise coprime in \(\mathbb{F}_3[x]\). To do this, we check that their greatest common divisors (GCD) across pairs are 1.

Calculate GCDs

The GCD of \(x^2 + 1\) and \(x^2 - 1\) is 1 since \((x^2 + 1) - (x^2 - 1) = 2\). Similarly, check the other pairings: \(x^2 + 1\) and \(x^3 + x - 1\), as well as \(x^2 - 1\) and \(x^3 + x - 1\). All GCDs should be 1, confirming that the polynomials are pairwise coprime.

Apply the Chinese Remainder Theorem (CRT)

Because the polynomials are pairwise coprime, the CRT states that there is a unique solution modulo the product \(m_0m_1m_2\) for any given set of congruences. Hence each combination of congruences corresponds uniquely to a polynomial \(f\) mod \(m_0m_1m_2\).

Calculate degree of the common modulus

Determine the degree of \(m_0m_1m_2\). Since \(\deg(m_0) = 2\), \(\deg(m_1) = 2\), and \(\deg(m_2) = 3\), the degree of the product is \(2 + 2 + 3 = 7\).

Count possible functions for the given degree constraint

Since \(\deg(f) \leq 8\) and any polynomial \(f\) satisfying these congruences has degree up to \(7\) (from Step 4), each \(f\) is determined modulo \(m_0m_1m_2\). There are \(3^7\) possible coefficients for \(f\) since each coefficient is in \(\mathbb{F}_3\) (3 choices per coefficient in a polynomial of degree 7 or less).

Verify polynomial solutions (for a separate question)

Provide a list of polynomials that satisfy the congruences: determine polynomials which, when reduced modulo the respective \(m_i\), match \(v_i\). Use the CRT to compute one such polynomial for each combination, but we're not listing them exhaustively since it wasn't needed for this part.

Key concepts

Pairwise Coprime Polynomials

In the context of the Chinese Remainder Theorem (CRT) for polynomials, the concept of pairwise coprime polynomials plays a crucial role. Pairwise coprime means that, for any pair of polynomials among the given set, their greatest common divisor (GCD) is 1. This property allows for the application of CRT, ensuring a unique solution exists for a system of polynomial congruences.

To establish pairwise coprimality, consider the polynomials given in the exercise:

• \(m_0 = x^2 + 1\)
• \(m_1 = x^2 - 1\)
• \(m_2 = x^3 + x - 1\)

These polynomials are from the field \( \mathbb{F}_3[x] \), which consists of polynomials with coefficients in the finite field \( \mathbb{F}_3 \). Checking the pairwise GCDs confirms that each pair \((m_0, m_1), (m_0, m_2), \text{ and } (m_1, m_2)\)\ has a GCD of 1.

This condition means that these polynomials do not share common roots (factors other than a constant factor), which justifies using CRT. This theorem then allows us to combine solutions to individual polynomial congruences into a single solution modulo the product of the moduli.

Polynomial Degree Constraint

The degree of a polynomial is an important factor when applying the Chinese Remainder Theorem in polynomials. In this exercise, it's crucial to find out the constraints on the degree of the resulting polynomial that satisfies the system of congruences.

The sum of degrees of the polynomials moduli \( m_0, m_1, \text{ and } m_2 \) gives us the degree of the combined modulus \( m_0m_1m_2 \).

• \(\deg(m_0) = 2\)
• \(\deg(m_1) = 2\)
• \(\deg(m_2) = 3\)

Thus, \( \deg(m_0m_1m_2) = 2 + 2 + 3 = 7 \).

This means any polynomial solution \( f \), satisfying all the given congruences, must have a degree lower than or equal to \( 7 \). However, the exercise asks for polynomials with a degree constraint up to \( 8 \).

The degree constraint ensures there are enough degrees of freedom to choose coefficients for the polynomial solution. Given the degree constraint, the total number of polynomials that satisfy \( \deg(f) \leq 8 \) in \( \mathbb{F}_3 \) is determined by the degrees of freedom provided by the constraint \( \deg(m_0m_1m_2) \).

Polynomial Congruences

Polynomial congruences are similar to number congruences but applied to polynomial expressions. When working with polynomial congruences, the goal is to find a polynomial \( f \) that satisfies specific conditions with regard to different moduli polynomials.

For example, we want \( f \equiv v_i \mod m_i \) where:

• \(v_0 = -x \) with modulus \( m_0 = x^2 + 1\)
• \(v_1 = x + 1 \) with modulus \( m_1 = x^2 - 1\)
• \(v_2 = x^5 - x \) with modulus \( m_2 = x^3 + x - 1\)

These conditions mean that, when \( f \) is divided by each polynomial \( m_i \), the remainder must be equivalent to \( v_i \). It involves verifying that after substituting these values, the equation holds for each polynomial modulus.

By confirming these congruences, and utilizing the Chinese Remainder Theorem, multiple congruences can be combined into a single polynomial expression with a unique solution for modulo \( m_0m_1m_2 \). Hence, CRT ensures that if \( f \) satisfies the congruences with each modulus, it also satisfies a congruence with the combined modulus polynomial.