Exercise · ZW-IRREDUCIBLE-ELEMENT-002 · ★☆☆ Basic
Which polynomial is irreducible in \(\mathbb{Q}[x]\)?
Choose one option.
Show answer
- A. \(x^2+1\)
Explanation
The polynomial \(x^2+1\) has no rational (not even real) roots, so it does not split into linear factors. The other three factor as \((x-1)(x+1)\), \((x-2)(x+2)\) and \((x-1)^2\).
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-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)?
- ZW-IRREDUCIBLE-ELEMENT-009 A polynomial f∈ℚ[x] of degree 3 is irreducible if and only if: