08.02 Principal ideal domains (P.I.D.s)

Definition P.I.D.

• $R$,
  • integral domain

$R$ is said to be P.I.D. if every ideal $I$ is principal.

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Remark

• Porposition 1 in Seciton 8.2 proves every Euclidean Domain is a P.I.D.

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Proposition 7.

• $R$,
  • P.I.D.
• $(p) \neq 0 \subseteq R$,
  • prime ideal

Then $(p)$ is maimal ideal.

proof.

Let $(p) \supseteq I = (m)$ be ideal containing $(p)$. Since $p \in (m)$, $p = rm$ for some $r \in R$. $p \in (p)$ and $(p)$ is prime ideal so that either $r \in (p)$ or $m \in (p)$ holds.

(i) $r \in (p)$. In this case, $r = ps$ for some $s \in R$. $p = rm = psm$. $R$ is integral domain so that $sm = 1$. Hence $m$ is unit so that $(m) = I$.

(ii) $m \in (p)$. In this case, $(p) = (m) = I$.

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Corollary 8.

• $R$,
  • commutative ring,

If $R[x]$ is a P.I.D. or Eucliean Domain, then $R$ is a field.

proof.

Suppose that $R[x]$ is a P.I.D.. $R[x]$ is of course a integral domain. Since $R$ is a subring of $R[x]$, $R$ is a integral domain. $R[x]/(x)$ is isomorphic to a integral domain $R$. The ideal $(x)$ is a nonzero prime ideal in $R[x]$ by Proposition 13 in Seciton 7.4. Since $R[x]$ is P.I.D., $(x)$ is a maximal ideal by porposition 7. By Proposition 12 in Section 7.4, $R[x]/(x)$ is field.

$\Box$