I'm trying to solve a couple of problems involving ring localization and I'm not sure if my solutions are right or if I understand the idea of localization correctly.

Let $A$ be a commutative ring, $S$ a set closed under multiplication with $1 \in S \subset A - \{0\}$, $J$ an ideal in $A$ and $i : A \rightarrow A_S$ with $i(a)=\frac{a}{1}$. Prove that

$$ \langle i(a) : a \in J\rangle = \{ \tfrac{a}{s} \in A_S : a\in J \}$$

I'll start by proving $\subset$. Let $p \in \langle i(a) : a \in J\rangle$, then $ p = \sum _{j \in I} s_j i(a_j)$ with $s_j \in A_S$ and $a_j \in J$.

Then, $p = \sum _{j \in I} \frac{\alpha_j}{\beta_j} \frac{a_j}{1} $ with $\alpha_j \in A$, $a_j \in J$ and $\beta_j \in S$. Since $J$ is an ideal $\alpha_j a_j = a'_j \in J$. Working through the sum then:

$$p = \frac{\sum_j a'_j ( \prod_{k \not= j} \beta_k)}{\prod_{j\in I} \beta_j} = \frac{a''}{s'} \, \text{for some} \, a'' \in J, \, s' \in S $$

Now I need to prove the other inclusion.

If $p \in \{ \frac{a}{s} \in A_S : a\in J \}$ then obviously $p = \frac{a}{s}$ for some $a\in A$ and $s \in S$ so $p = \frac{1}{s} i(a)$ and it's done.

I understand that the elements of $A_S$ are classes of the form $(a,s)$ with $a\in A$ and $s\in S$ but, is $\frac{1}{s}$ always an element of $A_S$ even if there's no $a \in A$ so that $i(a)$ is in the same class as $\frac{1}{s}$?

I don't think it's necessarily true that there's an $a \in A$ so that there's a $t \in S$ with $t(as - 1) = 0$ for all $s \in S$ so that would lead me to think no.

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    $\begingroup$ Some MathJax advice: < and > mean "less than" and "greater than", and produce spacing correct for that meaning only; to make angle brackets, use \langle and \rangle. $\endgroup$ – Zev Chonoles Jun 16 '15 at 20:36
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    $\begingroup$ The set of all elements of $A_S$ of the form $i(a)$ comprise a representation of $A$ in $A_S$. You have, for instance, $\mathbb Z$ in $\mathbb Q$, and of course $\mathbb Q$ has many non-integers. (The example is misleading in that $i$ is injective in that case, which is not always true). $\endgroup$ – user714630 Jun 16 '15 at 20:38
  • $\begingroup$ @ZevChonoles Thanks, I'll keep that in mind. $\endgroup$ – John Williams Jun 16 '15 at 20:41
  • $\begingroup$ @KarlKronenfeld Right, I think I see what you mean. I do have all the pairs $(a,s)$ in $A_S$ but $i(A)$ is not $A_S$ so there's no problem. Sounds pretty obvious now that I read it aloud :/ Thanks $\endgroup$ – John Williams Jun 16 '15 at 20:47

First prove that $$ J^e=\left\{\frac{a}{s}:a\in J, s\in S\right\} $$ is an ideal of $S^{-1}A$ (I prefer this notation over $A_S$). This poses no problem, but just to sketch the proof:

  1. $0/1\in J^e$, because $0\in J$
  2. if $a,b\in J$ and $s,t\in S$, then $$ \frac{a}{s}+\frac{b}{t}=\frac{at+bs}{st}\in J^e $$ because $at+bs\in J$ and $st\in S$
  3. if $a\in J$, $b\in A$ and $s,t\in S$, then $$ \frac{a}{s}\frac{b}{t}=\frac{ab}{st}\in J^e $$ because $ab\in J$ and $st\in S$

Then the inclusion $\subset$ follows by observing that, for $a\in J$, $i(a)=\frac{a}{1}\in J^e$, because $1\in S$, and from the fact that $J^e$ is an ideal.

For the converse, an element $a/s$ with $a\in J$ can be written as $$ \frac{a}{s}=\frac{a}{1}\frac{1}{s}=i(a)\frac{1}{s} $$ so it belongs to the ideal of $S^{-1}$ generated by $i(a)$, which is obviously contained in $\langle i(a):a\in J\rangle$.


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