Sunday, May 12, 2013

Characterization of PIDs (8.3.11)

Dummit and Foote Abstract Algebra, section 8.3, exercise 11:

MathJax TeX Test Page Prove$$R~\text{a Principle Ideal Domain}~⇔R~\text{a UFD}~\land R~\text{a Bezout Domain}$$Proof: The $⇒$ direction is evident, so it suffices to demonstrate the converse. Observe a typical ideal $I \subseteq R$, and choose arbitrary $x∈I$ with $x=u\pi_1^{\alpha_1}...\pi_n^{\alpha_n}$ being its unique decomposition up to units. For any $e∈\mathbb{Z}^+ \cup \{0\}$ and $i$ with $1≤i≤n$ let $A_{i,e}=\{r∈I ~ | ~ \pi_i^e \mid r\}$. For each $i$ let $\beta_i$ be the largest integer such that $A_{i,\beta_i}=I$, which is guaranteed to exist for arbitrary $i$ as the exponent $\alpha_i$ of $\pi_i$ is finite so that as a bound $x∉A_{i,\alpha_i+1}$. For each $i$, choose $a_i$ such that $a_i ∈ I \setminus A_{i,\beta_i+1}$, so that $a_i=\pi_i^{\beta_i}n_i$ with $\pi_i \not \mid n_i$. Letting $A=\bigcup_{1≤i≤n} a_i$ and taking successive greatest common divisors of the $a_i$ and then taking a greatest common divisor of this and $x$, we have a final common divisor $d=\pi_1^{\beta_1}...\pi_n^{\beta_n}$ (since these are the only irreducibles dividing $x$, and for each $i$ there is an element $a_i$ with its minimal $\pi_i$ exponent $\beta_i$, and all the exponents of the $a_j$ and $x$ of $\pi_i$ is at least $\beta_i$ as $a_j,x∈I=A_{i,\beta_i}$), writable as an $R$-linear combination of elements from $A \cup \{x\} \subseteq I$ by 7.2.7. We claim $(d)=I$: Evidently $(d) \subseteq I$, and for the reverse containment, for any $r∈I$ we have $\pi_i^{\beta_i} \mid r$ for applicable $i$ since $r∈I=A_{i,\beta_i}$, implying $I \subseteq (d)$ and now $I=(d)$ is a principle ideal.$~\square$

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