# The elements of the extended ideal $I^e\trianglelefteq S^{-1}R$

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.

• $2/1=6/3\in I^e$ and $2\notin I$. Commented May 8, 2019 at 23:27
• @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? Commented May 8, 2019 at 23:32
• Since $6\in I$ and $3\in S$, we have $6/3\in I^e$. But $6/3=2/1$. Commented May 9, 2019 at 14:59

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 $$r'\in I$$.