We know that $\mathbb{Z} [ x ]$ is not a principal ideal ring, because we can construct the ideal $\langle 2,x \rangle$. Why we couldn't take as an example of non-principal ideal in it all polynomials with the same given root $y$? i.e. $$P(x)=a_nx^n+...+a_1x+a_0=(x-y)\cdot (\dots$$ $$P(x) \cdot Q(x)=(x-y)^2 \cdot \dots$$ $$P(x)+Q(x)=(x-y)(\dots)$$ So, $I$ is the ideal. Am I right?

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    $\begingroup$ But it's principal. $\endgroup$
    – Wojowu
    Sep 14, 2016 at 19:32
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    $\begingroup$ Take $y=0$ and you'll see easily the ideal is principal. $\endgroup$
    – egreg
    Sep 14, 2016 at 19:39

1 Answer 1


The ideal $I = \{ f \in {\mathbb Z}[x] \mid f(a) = 0 \}$ is principal: $I = \langle x - a \rangle$.

  • $\begingroup$ Thank you ;) I'm so inattentive... I'm so sorry for it. $\endgroup$
    – Nicholas S
    Sep 14, 2016 at 19:43

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