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Prove that if $R$ is a principal ideal ring and $S$ is a multiplicatively closed subset of $R$ then $S^{-1}R$ is also a principal ideal ring.

Thanks for any insight.

user110834
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  • If $(a)$ is a principal ideal of $R$, does $S^{-1}(a)$ form a principal ideal of $S^{-1}R$? – user110834 Nov 24 '13 at 15:06
  • $S$ must be a multiplicatively closed subset of $R$ not additively – An Khuong Doan Nov 24 '13 at 15:19
  • possible duplicate of Is the localization of a PID a PID?. Also http://math.stackexchange.com/questions/137876/a-subring-of-the-field-of-fractions-of-a-pid-is-a-pid-as-well is s slightly broader version with good answers. These are about domains, but really the same arguments apply to commutative rings. I'll leave it up to everyone else to see if it's worth closing or not. – rschwieb Nov 25 '13 at 14:18
  • And I hope commenters can restrain themselves from comments like "your question is wrong because my definition of multiplicatively closed set allows $0\in S$ and my definition of principal ideal ring requires identity." Let's just do the poster a favor and interpret $S$ as not containing $0$ divisors, or else accepting ${0}$ as a principal ideal ring. – rschwieb Nov 25 '13 at 14:24

2 Answers2

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More generally, prove the following: Let $\phi : R \to R'$ be a ring homomorphism such that every element of $R'$ is associated to an element in $\phi(R)$. Note that localizations satisfy this property by construction, but also quotients. If $R$ is a principal ideal ring, then the same is true for $R'$. The proof is straight forward, since there is only one thing you can do: Use the assumption. Every ideal of $R'$ has some generating set, and by the assumption it may be choosen from $\phi(R)$ (since multiplication of units doesn't change an ideal). Take preimages in $R$, use that $R$ is a principal ideal ring, etc.

Martin Brandenburg
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$S^{-1}R$ is a subring of a field (the field of fractions of $R$), and thus is an integral domain.

Let $I \subseteq S^{-1}R$ be an ideal. Let $D = \{a \in R \ |\ \frac{a}{b} \in I \}$.

claim that $D$ is an ideal of $R$.

note that $\frac{0}{s} \in I$, so that $0 \in D$. Now let $\frac{a}{d}$, $\frac{b}{e} \in I$; then $\frac{a}{d} - \frac{e}{d}\frac{b}{e} = \frac{a-b}{d}$, so that $a-b \in D$.

let $r \in R$ and $a \in D$, with $d \in S$. Then $\frac{r}{d} \frac{a}{d} = \frac{ra}{d^2}$, so that $ra \in D$. Thus $D$ is an ideal of $R$, and thus we have $D = (\alpha)$. Let $d \in S$ such that $\frac{\alpha}{d} \in I$; then $(\frac{\alpha}{d}) \subseteq I$. Now let $\frac{a}{b} \in I $with $a = c\alpha$; we have $\frac{a}{b} = \frac{cd}{b} \frac{\alpha}{d}$, so that $I = (\frac{\alpha}{d})$ is principal.

Thus,$S^{-1}R$ is a principal ideal domain.

An Khuong Doan
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  • $R$ is P.I.D then $R$ is obviously an integral domain. (just definition) – An Khuong Doan Nov 24 '13 at 16:19
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    $R$ is not assumed to be a PID, only a principal ideal ring. The claim is also not correct for PID (consider $S={0,1}$, this is multiplicative, but $S^{-1} R = 0$ is not an integral domain by definition). – Martin Brandenburg Nov 24 '13 at 17:30
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    The definition of $D$ is not correct (what is $b$?). – Martin Brandenburg Nov 24 '13 at 17:31
  • @SnowAngel6147 Maybe I've been staring at it for too long but can you explain this step $\frac{a}{d} - \frac{e}{d}\frac{b}{e} = \frac{a-b}{d}$ why is it done in this fashion? like, why $\frac{e}{d}\frac{b}{e}$ ? – oliverjones Apr 17 '15 at 08:46