Exercise · ZW-IRREDUCIBLE-ELEMENT-007 · ★★☆ Intermediate
In \(\mathbb{Z}[\sqrt{-5}]\), the element 2 divides \(6=(1+\sqrt{-5})(1-\sqrt{-5})\) but divides neither factor. What follows?
Choose one option.
Show answer
- B. 2 is irreducible but not prime
Explanation
We already saw that 2 is irreducible. A prime dividing a product must divide some factor, and 2 does not, so it is not prime: in this ring irreducible does not imply prime.
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-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: