Exercise · ZW-IRREDUCIBLE-ELEMENT-009 · ★★★ Advanced
A polynomial \(f\in\mathbb{Q}[x]\) of degree 3 is irreducible if and only if:
Choose one option.
Show answer
- B. it has no roots in \(\mathbb{Q}\)
Explanation
A reducible cubic has a degree-1 factor (the degrees add up to 3), that is, a rational root; and conversely. Over \(\mathbb{C}\) it always has roots, and neither the discriminant nor the coefficients decide reducibility.
Review the concept: Irreducible element
A nonzero nonunit element \(p\) of an integral domain \(D\) with no proper factorization: in every product of two factors equal to \(p\), one of them is a unit.
\[p=ab\ \Rightarrow\ a\ \text{ or }\ b\ \text{is a unit}\]
More practice problems: irreducible element
- ZW-IRREDUCIBLE-ELEMENT-001 Which of these elements is irreducible in ℤ?
- ZW-IRREDUCIBLE-ELEMENT-002 Which polynomial is irreducible in ℚ[x]?
- ZW-IRREDUCIBLE-ELEMENT-003 Which polynomial is reducible in ℚ[x]?
- ZW-IRREDUCIBLE-ELEMENT-004 Which polynomial is irreducible in ℝ[x]?
- ZW-IRREDUCIBLE-ELEMENT-005 In ℤ[i] with norm N(a+bi)=a^2+b^2, which of these elements is irreducible?
- ZW-IRREDUCIBLE-ELEMENT-006 In ℤ[√-5], with N(a+b√-5)=a^2+5b^2, which element is irreducible?
- ZW-IRREDUCIBLE-ELEMENT-007 In ℤ[√-5], the element 2 divides 6=(1+√-5)(1-√-5) but divides neither factor. What follows?
- ZW-IRREDUCIBLE-ELEMENT-008 In ℤ, how many elements are associates of 5 (including 5 itself)?