State the definition of a principal ideal domain.
Let \(\mathbb{Z}\) be the ring of integers. Is \(\mathbb{Z}[x]\) a principal ideal domain? Either prove that it is, or present an example showing that it is not.
Solution
1. A principal ideal domain is a commutative ring \(R\) withidentity \(1\ne0\), no zero divisors, and every ideal of the form\((a)=\{ra:r\in R\}\) for some \(a\in R\) [@DF04].
2. The ideal \(I=(2,x)\) of \(\mathbb Z[x]\) is proper.
Every element of \(I\) has the form \(2g(x)+xh(x)\) with \(g,h\in\mathbb Z[x]\). Its constant coefficient is \(2g(0)\), which is even. The polynomial \(1\) has odd constant coefficient, so \(1\notin I\) and \(I\ne\mathbb Z[x]\).
3. The ideal \(I\) is not principal, so \(\mathbb Z[x]\) is not aprincipal ideal domain.
Suppose \(I=(f)\). Since \(2\in I\), there is \(g\in\mathbb Z[x]\) such that \(fg=2\). Neither factor is zero. Leading coefficients of nonzero integer polynomials have nonzero product, so degrees add; hence \(\deg f+\deg g=0\). Both factors are constant integers, and \(f\) divides \(2\). Thus \(f\in\{1,-1,2,-2\}\).
2. The possibilities \(f=2\) and \(f=-2\) wouldgive \(I=(2)\), but every coefficient of a polynomial in \((2)\)is even, whereas the coefficient of \(x\) in \(x\in I\) is \(1\).Thus no generator exists.
The same leading-coefficient argument shows that \(\mathbb Z[x]\) is an integral domain.