Exercise · ZW-UNIQUE-FACTORIZATION-DOMAIN-007 · ★★☆ Intermediate
Which ideal of \(\mathbb{Z}[x]\) shows that \(\mathbb{Z}[x]\) is not a principal ideal domain, even though it is a UFD?
Choose one option.
Show answer
- B. \((2,x)\)
Explanation
A generator of \((2,x)\) would have to divide both 2 and \(x\), and the only common divisors are units; but then the ideal would be the whole ring, and \(1\notin(2,x)\). The other three have a single generator by construction.
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.
\[a=p_1p_2\cdots p_r\]
More practice problems: unique factorization domain
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- ZW-UNIQUE-FACTORIZATION-DOMAIN-006 In ℤ[√-5], the element 6 has the factorizations 2·3 and (1+√-5)(1-√-5). Since the four factors are…
- ZW-UNIQUE-FACTORIZATION-DOMAIN-008 In ℚ[x], how many irreducible factors (with multiplicity, up to associates) does x^4-x^2 have?
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