0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ But it's principal. $\endgroup$ – Wojowu Sep 14 '16 at 19:32
  • 1
    $\begingroup$ Take $y=0$ and you'll see easily the ideal is principal. $\endgroup$ – egreg Sep 14 '16 at 19:39
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Thank you ;) I'm so inattentive... I'm so sorry for it. $\endgroup$ – Nicholas S Sep 14 '16 at 19:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.