I'm thinking about these statements and I would like to know if I am right.

  1. A principal ideal ring $R$, by definition, is a ring whose ideals are principals, since $R$ is itself an ideal, $R=(x)$, for some $x\in R$.

  2. If $R=(y)$ for some $y$, every ideal of $R$ is principal.

  3. Using (1) and (2), the principal ideal rings are in this form $(x)$, for some element $x$ and vice and versa, every (x) is a principal ideal ring.

Am I wrong? I need some help.

Thanks a lot

  • 6
    $\begingroup$ Every unital ring satisfies $R = (y)$ for some $y$. $\endgroup$ – Ink Jul 16 '13 at 4:26
  • $\begingroup$ @Ink $R$ is not necessarily unital. $\endgroup$ – user42912 Jul 16 '13 at 4:36
  • $\begingroup$ And your point is? $\endgroup$ – Ink Jul 16 '13 at 4:46
  • $\begingroup$ @Ink I would like to know if what I said above is true, maybe I mistaken in some part. $\endgroup$ – user42912 Jul 16 '13 at 4:59
  • $\begingroup$ There's an answer below that tells you why it's false. $\endgroup$ – Ink Jul 16 '13 at 5:00

No argument for (2) has been supplied, and the assertion is not correct.

Consider for example the ring of all polynomials $P(x,y)$ with real coefficients. Certainly the whole ring is a principal ideal, since it is generated by the polynomial $1$. But the ideal generated by $x$ and $y$ is non-principal.

  • $\begingroup$ why the ideal generated by x and y is not principal? $\endgroup$ – user42912 Jul 16 '13 at 5:41
  • 1
    $\begingroup$ This is the ideal of all polynomials that vanish at $(0,0)$. Can you think of a generator? It would have to be a polynomial that is $0$ at $(0,0)$. Since $x$ is in the ideal, it would have to look like $kx$. But $y$ is not $kx$ times a polynomial. $\endgroup$ – André Nicolas Jul 16 '13 at 6:03

Property (2) is completely wrong because any commutative unital ring $R = (1)$ and so this would imply that any commutative unital $R$ is a principal ideal ring! This is certainly not true for example by considering $R = \Bbb{Z}[x]$ and $I = (2,x)$: If $I$ is principal then since $R$ is Noetherian the Krull Hauptidealsatz implies that $I$ has height one which is a contradiction, for

$$0 \subsetneqq (x) \subsetneqq (2,x).$$


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