Search the corpus

    $\langle 2,x\rangle$ is not principal in $\ZZ[x]$

    P-YKJVQ

    Area
    Algebra
    Topics
    Ideals, Polynomials, Principal Ideal Domains
    Status
    draft
    • Show that \(\gens{ 2, x }\normal \ZZ[x]\) is not a principal ideal.
    Solution

    Let \[ I=(2,x)\subset\mathbb Z[x]. \] Suppose \(I=(f)\) were principal. Since \(2\in(f)\), the polynomial \(f\) divides the constant polynomial \(2\). Because \(\mathbb Z[x]\) is a domain, \[ \deg f=0, \] so \(f\) is an integer divisor of \(2\). Up to a unit, \[ f=1\quad\text{or}\quad f=2. \]

    If \(f=2\), then \((f)=(2)\) does not contain \(x\). If \(f=1\), then \((f)=\mathbb Z[x]\), but \(I\) is proper: reduction modulo \(2\) followed by evaluation at \(x=0\) gives a homomorphism \[ \mathbb Z[x]\to\mathbb F_2 \] whose kernel contains \((2,x)\) but not \(1\).

    Both possibilities are impossible. Therefore \[ \boxed{(2,x)\text{ is not principal in }\mathbb Z[x]}. \]

    Source Collections