Saturation and Jacobson radical

Question

This question was asked in my assignment on algebraic geometry and I need help in it.

Let A be a commutative ring. A multiplicatively closed subset S in A is called saturated if for all $a,b \in A$, $ab\in S$ implies that $a\in S$ and $b\in S$. Define $\bar{S} =${$ a\in A| \exists b\in A$ with $ab \in S$} to be saturation of S.

(a) Let A' be an ideal in A and let $S_{A'}=1+ A' = ${$1+a' | a\in A'$}. Then prove that S is multiplicatevely closed set in A.

I have done it.

(b) What is the saturation $\bar{S}_{A'}$ of the multiplicatively closed set $S_{A'}$?

Work: I think $S_{A'} = ${$a' \in A' | \exists b \in A' $ such that $a'b\in S_{A'} $}.

So, Saturated set is all $a'\in S_{A'}$ such that $ \exists b\in S_{A'}$ such that a'b =1+a'' $a'' \in S_{A'}$ . I don't think any further simplification is possible.

Am I right?

(c)Prove that $A'S_{A'}^{-1} A\subseteq M_{S_{A'}^{-1} A}$= Jacobson radical of $S_{A'}^{-1} A$.

I have to show that $A'S_{A'}^{-1} A $ is contained in every maximal ideal of $S_{A'}^{-1} A$.

Let there exists a maximal ideal M which doesn't contains $A'S_{A'}^{-1} A $. What would be the contradiction?

Kindly help.

Thanks!

Answer 1

Both of these are garbled:

I think $S_{A'} = ${$a' \in A' | \exists b \in A' $ such that $a'b\in S_{A'} $}.

So, Saturated set is all $a'\in S_{A'}$ such that $ \exists b\in S_{A'}$ such that $a'b =1+a''$[, where] $a'' \in S_{A'}$ .

You miswrote your definition two times in a row. It should read

I think $\bar S_{A'} = \{a \in A | \exists b \in A \text{ such that }ab\in S_{A'} \}$.

and

So, Saturated set is all $a\in A$ such that $ \exists b\in A$ such that $ab =1+a'$[, where] $a' \in A'$ .

Given that latter one, you have $ab-1\in A'$, and I don't see any other better explanation than

$\bar S_{A'}$ is the set of all $a\in A$ such that $a+A'$ is a unit in $A/A'$.

It's funny this should happen but I'm now remembering an anecdote by Arhangelskii where he recounted a mathematician with the unfortunate habit of using $A$'s for everything: $a$, $A$, $\alpha$, $\mathfrak A$, $\mathscr A$ with various diacritics or subscripts.

Next time, use $R$ and $I\lhd R$ please instead of $A'\lhd A$.

For $c$, it suffices ot use the characterization of the Jacobson radical as the set of all elements $x$ such that $xr-1$ is a unit for every $r$ in the ring.

To that end, look at something of that form: $\frac{i}{j+1}\frac{r}{k+1}-1$ where $i, j, k\in I$ and $r\in R$.

If you write $\frac{i}{j+1}\frac{r}{k+1}=\frac{ir}{m+1}$ with $m\in I$, then $\frac{ir}{m+1}-1=\frac{ir-m-1}{m+1}$ is a unit iff $ir-m-1$ is a unit.

You should be able to take it from there...