2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.