Ring localization and ideals

Question

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.

Answer 1

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 $\langle i(a):a\in J\rangle$.