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Let $R$ be a commutative ring with $1_R$, $S\subseteq R$ a multiplicative set, $I\trianglelefteq R$ and ideal of $R$ and $\nu:R\longrightarrow S^{-1}R,\ a\longmapsto \nu(a):=\frac{a}{1_R}$ the natural ring homomorphism. There is a well-known result, that then the elements of the extraction $I^e$ of the ideal $I$ under $\nu$ has the following form: $$ I^e=\Big\{ \frac{r}{s}\in S^{-1}R:\ r\in I,\ s\in S \Big\} \trianglelefteq S^{-1}R. $$ Then my textbook writes that due to this lemma, every element $\alpha\in I^e$ has at least one factorisation in the form $\alpha=\frac{r}{s}$, where $r\in R, s\in S$. But, it's not true that if $\alpha =\frac{r}{s}$ with $r\in R,s\in S$ then $r\in I$ and this happens only if $I$ is prime and $I\cap S=\emptyset$.

And as an example it takes $R=\Bbb Z,\ S=\{1,3,3^2,\dots\},\ I=\langle 6 \rangle$.

I can not understand this claim. Could you please explain?

Thank you.

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    $\begingroup$ $2/1=6/3\in I^e$ and $2\notin I$. $\endgroup$
    – user26857
    Commented May 8, 2019 at 23:27
  • $\begingroup$ @user26857 Thank you for your answer. Why $\frac{2}{1}\in I^e$? I can not understand the statement which I wrote in "italics". I.e., we proved that every element of $I^e$ should be in the for $\frac{r}{s}$ with $r\in I,s\in S$. How do we take elements inside $I^e$, but without this form? $\endgroup$
    – Chris
    Commented May 8, 2019 at 23:32
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    $\begingroup$ Since $6\in I$ and $3\in S$, we have $6/3\in I^e$. But $6/3=2/1$. $\endgroup$
    – user26857
    Commented May 9, 2019 at 14:59

1 Answer 1

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Note, that elements of $S^{-1}R$ are not pairs $(r,s)$ (which we may denote by $\frac{r}{s}$), where $r\in R$ and $s\in S$, but equivalence classes of such elements by the following relation: $(r_1,s_1)\sim(r_2,s_2)$ if and only if there exists $s_3\in S$, such that $(r_1s_2-r_2s_1)s_3=0$. If $(r_1,s_1)\sim(r_2,s_2)$, we write $\frac{r_1}{s_1}=\frac{r_2}{s_2}$. That's why $\frac{2}{1}$ and $\frac{6}{3}$ are the same elements of $S^{-1}R$, where $R=\mathbb{Z}$ and $S=\{3^n|n\in\mathbb{N}\}$. So if we take all equivalence classes of pairs of the form $(i,s)$, where $i\in I$ and $s\in S$, then we get $I^e$, but it doesn't mean that there are no representatives $(r',s')$ of such classes with $r'\notin I$.

As for the statement in italics: if $\frac{i}{s}=\frac{r'}{s'}$, then $r'ss_3\in I$ for some $s_3\in S$. Then use definition of prime ideal and the fact that $I\cap S=\varnothing$ to prove that $r'\in I$.

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  • $\begingroup$ Perfect! Thank you for your answer. $\endgroup$
    – Chris
    Commented May 15, 2019 at 18:16

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