Question

\( \rightarrow\) Compute the coefficients of the Swinnerton-Dyer polynomial $$ f=(x+\sqrt{-1}+\sqrt{2})(x+\sqrt{-1}-\sqrt{2})(x-\sqrt{-1}+\sqrt{2})(x-\sqrt{-1}-\sqrt{2}) \in Z|x| $$ and its factorizations modulo \(p=2,3,5\). Prove that \(f\) is irreducible.

Short answer

The polynomial \(f\) is irreducible in \( \mathbb{Z}[x] \). Its factorizations modulo 2, 3, and 5 confirm its irreducibility.

Step-by-step solution

Determine the Roots

Given the polynomial \( f = (x+\sqrt{-1}+\sqrt{2})(x+\sqrt{-1}-\sqrt{2})(x-\sqrt{-1}+\sqrt{2})(x-\sqrt{-1}-\sqrt{2}) \), recognize its roots: \( x =-\sqrt{-1}-\sqrt{2}, x = -\sqrt{-1}+\sqrt{2}, x = \sqrt{-1}-\sqrt{2}, x = \sqrt{-1}+\sqrt{2} \).

Substitute and Simplify

Consider each pair of factors: \( (x+\sqrt{-1}+\sqrt{2})(x+\sqrt{-1}-\sqrt{2}) \) and \( (x-\sqrt{-1}+\sqrt{2})(x-\sqrt{-1}-\sqrt{2}) \). Simplify each using \( (a+b)(a-b) = a^2-b^2 \).

Simplify - First Pair of Roots

For the first pair: \( (x+\sqrt{-1} + \sqrt{2})(x+\sqrt{-1} - \sqrt{2}) = (x + \sqrt{-1})^2 - \sqrt{2}^2 = (x+\sqrt{-1})^2 - 2 \).

Expand - First Pair

Expand \( (x + \sqrt{-1})^2 - 2 \) to get \( x^2 + 2\sqrt{-1}x + (-1) - 2 = x^2 + 2\sqrt{-1}x - 3 \).

Simplify - Second Pair of Roots

For the second pair: \( (x-\sqrt{-1}+\sqrt{2})(x-\sqrt{-1}-\sqrt{2}) = (x - \sqrt{-1})^2 - \sqrt{2}^2 = (x-\sqrt{-1})^2 - 2 \).

Expand - Second Pair

Expand \( (x - \sqrt{-1})^2 - 2 \) to get \( x^2 - 2\sqrt{-1}x + (-1) - 2 = x^2 - 2\sqrt{-1}x - 3 \).

Multiply Factors

The polynomial \( f \) becomes \((x^2 + 2\sqrt{-1}x - 3)(x^2 - 2\sqrt{-1}x - 3) \). Remember the form \( (a + b)(a - b) = a^2 - b^2 \), the result is \( (x^2 - 3)^2 - (2\sqrt{-1}x)^2 \).

Simplify Result

Simplify and expand: \((x^2 - 3)^2 - (-4)x^2 \). This results in \(x^4 - 6x^2 + 9 + 4x^2 \).

Combine Like Terms

Combine the terms: \(x^4 - 2x^2 + 9\). This is the polynomial \( f \) in \( \mathbb{Z}[x] \).

Factorization Modulo \( p = 2, 3, 5 \)

Compute factorizations for: \( f(x) \equiv x^4 - 2x^2 + 9 \mod 2, 3, 5 \).

Check Irreducibility

For each prime, apply techniques such as Eisenstein's criterion or checking the degree of any potential factors.

Key concepts

Polynomial Coefficients

The Swinnerton-Dyer polynomial we encounter is formed from the expansion of the expression \((x+\sqrt{-1}+\sqrt{2})(x+\sqrt{-1}-\sqrt{2})(x-\sqrt{-1}+\sqrt{2})(x-\sqrt{-1}-\sqrt{2})\). Each term inside this polynomial can be described by polynomial coefficients which are an essential part of polynomials. These coefficients are essentially the numbers in front of each term of the polynomial, ordered by their degree.

For our specific polynomial, we simplified the overall expression to \(x^4 - 2x^2 + 9\). Here, the coefficients are \(1\) for \(x^4\), \(-2\) for \(x^2\), and \(9\) for the constant term. When solving, it is crucial to correctly compute and identify these coefficients as they determine the shape and solutions of the polynomial.

Understanding polynomial coefficients is key in operations such as addition, subtraction, and multiplication of polynomials, as well as in determining the roots of the polynomial.

Irreducibility

A polynomial is said to be irreducible over a field (such as \( \mathbb{Z} \), the integers), if it cannot be factored into the product of two non-constant polynomials with coefficients in that field. In other words, irreducibility means there are no simpler polynomials (i.e., polynomials of lower degrees than the original) that multiply to give the original polynomial in the given domain.

To establish the irreducibility of the Swinnerton-Dyer polynomial \(f\) using Eisenstein's criterion, consider different primes and analyze how they divide the coefficients of \(f\). For polynomial \(f(x) = x^4 - 2x^2 + 9\), consider checking each term's divisibility by some small prime. However, it might not always apply straightforwardly, and other techniques may be necessary.

When proving irreducibility, it is often useful to look at the degree of any factors that the polynomial might have and confirm whether factors exist. This requires checking the factorization possibilities thoroughly. Polynomials with higher degrees are more likely candidates for irreducibility without roots in their coefficient field.

Modulo Factorization

When you perform polynomial factorization modulo a prime (often denoted \(p\)), you are essentially looking at how a polynomial behaves in a modular arithmetic world with respect to that prime. With our polynomial \( x^4 - 2x^2 + 9 \), we need to examine how it can be factored when considered under the modulus \(2, 3, \) and \(5\).

Modulo factorization extensively simplifies polynomials by reducing each coefficient of the polynomial to its smallest non-negative remainder upon division by a chosen prime. This exercise exposes different properties of the polynomial that might not be evident otherwise.

• For \(p=2\), evaluate \(f(x) \mod 2 \).
• For \(p=3\), evaluate \(f(x) \mod 3 \).
• For \(p=5\), evaluate \(f(x) \mod 5 \).

Factorizing modulo these primes helps in determining potential rationality and reduction of terms that could facilitate its understanding and simplification.

Algebraic Roots

Finding the roots of a polynomial is a fundamental aspect of algebra which involves determining which values of \(x\) will make the polynomial equal zero. The roots \(x = -\sqrt{-1}-\sqrt{2}, x = -\sqrt{-1}+\sqrt{2}, x = \sqrt{-1}-\sqrt{2}, x = \sqrt{-1}+\sqrt{2}\) clearly are derived from the original product structure of our Swinnerton-Dyer polynomial.

Computing these roots allows us to translate a polynomial from a symbolic representation into useful tools that are numerically effective in real-world applications and theoretical understanding. Understanding how these roots interconnect is vital for grasping the nature of polynomial equations.

• Use techniques such as synthetic division to simplify calculations when necessary.
• Graphically visualize to identify possible root locations for real coefficients.

To find meaningful algebraic roots, sometimes substitutions or transformations such as completing the square or using quadratic formulas are required. Knowledge of algebraic roots allows us to factor polynomials effectively and understand solutions within both real and complex number systems.