Exercise · ZW-UNIQUE-FACTORIZATION-DOMAIN-002 · ★☆☆ Basic
What is the factorization of \(x^4-1\) into irreducibles of \(\mathbb{Q}[x]\)?
Choose one option.
Show answer
- D. \((x-1)(x+1)(x^2+1)\)
Explanation
First \(x^4-1=(x^2-1)(x^2+1)\) and then \(x^2-1=(x-1)(x+1)\). The factor \(x^2+1\) has no rational roots, so it is already irreducible; the factor \(x^2-1\) still splits.
Review the concept: Unique factorization domain
An integral domain \(D\) in which every nonzero nonunit element \(a\) is a product of irreducibles \(p_i\), and this factorization is unique up to order and associates.
\[a=p_1p_2\cdots p_r\]
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