Chapter 15: Problem 4
Compute the coefficients of the Swinnerton-Dyer polynomial $$ f=(x+\sqrt{-1}+\sqrt{2})(x+\sqrt{-1}-\sqrt{2})(x-\sqrt{-1}+\sqrt{2})(x-\sqrt{-1}-\sqrt{2}) \in \mathbb{Z}[x] $$ and its factorizations modulo \(p=2,3,5\). Prove that \(f\) is irreducible.
Short Answer
Expert verified
The polynomial \(f = x^4 - 3x^2 + 6x - 2\) is irreducible over \(\mathbb{Z}\). It factors as \(x(x^3+1)\) modulo 2, \((x^2+1)^2\) modulo 3, and remains irreducible modulo 5.
Step by step solution
01
Multiply Pairwise Conjugates
Start by recognizing that the expression \((x+\sqrt{-1}+\sqrt{2})(x-\sqrt{-1}-\sqrt{2})\) and \((x+\sqrt{-1}-\sqrt{2})(x-\sqrt{-1}+\sqrt{2})\) represent product of conjugates. Simplify these by multiplying them:\[(x+\sqrt{-1}+\sqrt{2})(x-\sqrt{-1}-\sqrt{2}) = (x+\sqrt{-1})^2 - (\sqrt{2})^2 = x^2 + 2x + 1 - 2\]\[(x+\sqrt{-1}-\sqrt{2})(x-\sqrt{-1}+\sqrt{2}) = (x-\sqrt{2})^2 + 1 = x^2 - 2x + 2\]
02
Simplify Intermediate Products
Use the simplified expressions from the previous step:\[= (x^2 + 2x - 1)(x^2 - 2x + 2)\]Next, expand the product using the distributive property (FOIL method):\[(x^2 + 2x - 1)(x^2 - 2x + 2) = x^4 - 2x^3 + 2x^2 + 2x^3 - 4x^2 + 4x - x^2 + 2x - 2\]Combine like terms:\[= x^4 - 3x^2 + 6x - 2\]
03
Compute Coefficients over \(\mathbb{Z}\)
The polynomial is now written in its expanded form as \(x^4 - 3x^2 + 6x - 2\). The coefficients are: 1 (for \(x^4\)), 0 (for \(x^3\)), -3 (for \(x^2\)), 6 (for \(x^1\)), and -2 (constant term).
04
Factorization Modulo \(p=2\)
Consider the polynomial \(x^4 - 3x^2 + 6x - 2\) modulo 2. Reducing the coefficients modulo 2 gives:\[x^4 - 3x^2 + 6x - 2 \equiv x^4 + x \pmod{2}\]Determine factors: \[x(x^3 + 1)\]Further, factor \((x^3+1)\) into \((x+1)(x^2+x+1)\) over \(\mathbb{F}_2\).
05
Factorization Modulo \(p=3\)
Reduce the coefficients of \(x^4 - 3x^2 + 6x - 2\) modulo 3:\[x^4 - 3x^2 + 6x - 2 \equiv x^4 + 0x^2 + 0x + 1 \equiv (x^2+1)^2 \pmod{3}\]The factorization is \((x^2+1)^2\) over \(\mathbb{F}_3\).
06
Factorization Modulo \(p=5\)
Reduce the polynomial \(x^4 - 3x^2 + 6x - 2\) modulo 5:\[x^4 - 3x^2 + 6x - 2 \equiv x^4 + 2x - 2 \pmod{5}\]Check possible quadratic factors:No further factorization can be found, confirming that it remains irreducible over \(\mathbb{F}_5\).
07
Prove Irreducibility over \(\mathbb{Z}\)
Use the factorization results modulo small primes. A polynomial is irreducible over \(\mathbb{Z}\) if it can't be factored into non-trivial polynomials with integer coefficients and remains irreducible modulo these primes. Since \(f\) does not factor into polynomials of lower degree over these small primes except as products of irreducible polynomials, and these products do not lift to nontrivial factorization over \(\mathbb{Z}\), \(f\) is irreducible over \(\mathbb{Z}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Swinnerton-Dyer polynomial
The Swinnerton-Dyer polynomial is a fascinating entity due to its unique structure involving roots composed of sums of different square roots. In this particular problem, the polynomial is constructed using complex numbers and square roots: \( f=(x+\sqrt{-1}+\sqrt{2})(x+\sqrt{-1}-\sqrt{2})(x-\sqrt{-1}+\sqrt{2})(x-\sqrt{-1}-\sqrt{2}) \). This formulation utilizes a clever combination of conjugates to form a polynomial with integer coefficients.
Each term in the product denotes a root of the polynomial in the complex plane. The conjugate pairs, such as \((x+\sqrt{-1}+\sqrt{2})(x-\sqrt{-1}-\sqrt{2})\), suggest that when expanded, they will result in real coefficients, since the imaginary parts cancel out.
This process of converting complex roots into a polynomial of real coefficients is known as constructing a polynomial with conjugate pairs. This offers us a tidy representation of the polynomial in the form of integer coefficients, which is more accessible for further analysis or computation.
Each term in the product denotes a root of the polynomial in the complex plane. The conjugate pairs, such as \((x+\sqrt{-1}+\sqrt{2})(x-\sqrt{-1}-\sqrt{2})\), suggest that when expanded, they will result in real coefficients, since the imaginary parts cancel out.
This process of converting complex roots into a polynomial of real coefficients is known as constructing a polynomial with conjugate pairs. This offers us a tidy representation of the polynomial in the form of integer coefficients, which is more accessible for further analysis or computation.
irreducibility proof
Proving the irreducibility of a polynomial involves showing that it cannot be factored into non-trivial polynomials of lower degree with coefficients in a specified field or ring. For the polynomial \(f\) in this exercise, the proof of irreducibility is carried out over both the integers and small finite fields.
The polynomial \(x^4 - 3x^2 + 6x - 2\) was checked for potential factorization over fields with low characteristics, such as modulo 2, 3, and 5. Each field's factors were examined in order to detect any reducibility:
The polynomial \(x^4 - 3x^2 + 6x - 2\) was checked for potential factorization over fields with low characteristics, such as modulo 2, 3, and 5. Each field's factors were examined in order to detect any reducibility:
- Modulo 2, the polynomial simplifies significantly, but after factoring, it maintains irreducible components such as \(x^2 + x + 1\).
- Modulo 3, it resolves into \((x^2+1)^2\), revealing no simple binomial factors.
- And modulo 5, it does not permit any further factorization, remaining irreducible.
finite field factorization
Factorization over finite fields provides insight into the structure of polynomials in different algebraic contexts, such as fields with a fixed number of elements. In this exercise, the polynomial \(f\) was examined over \(\mathbb{F}_2\), \(\mathbb{F}_3\), and \(\mathbb{F}_5\) to understand its behavior under different modulus conditions.
Working over finite fields with small primes involves reducing the polynomial's coefficients by the prime and examining possible divisions:
Working over finite fields with small primes involves reducing the polynomial's coefficients by the prime and examining possible divisions:
- In \(\mathbb{F}_2\), the polynomial simplifies to \(x^4 + x\), allowing a partial factorization to \(x(x^3 + 1)\), then further to a more complex product.
- Over \(\mathbb{F}_3\), it turns into \((x^2+1)^2\), which shows some symmetry akin to certain classical problems in number theory.
- In \(\mathbb{F}_5\), the polynomial remains unchanged apart from simple reductions, which points to its stubbornness against factorizing in this field.