Thread: maximal ideal in ring of quotients of R with respect to S

1. maximal ideal in ring of quotients of R with respect to S

Let $\displaystyle P$be a prime ideal,$\displaystyle R$be a commutative ring with identity,$\displaystyle S=R-P$. Then $\displaystyle S$will be a multiplicative subset of $\displaystyle R$, that is, $\displaystyle 1\in S,a,b\in S\Rightarrow ab\in S$
Define $\displaystyle S^{-1}R$ as the set $\displaystyle R\times S$.
define $\displaystyle (a,s)=(a_1,s_1)\Leftrightarrow\exists s_0\in S,s_0(as_1-a_1s)=0$
we may write$\displaystyle (a,s)$ as $\displaystyle \frac{a}{s}$
$\displaystyle \frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$
define the times operation
$\displaystyle \frac{a}{b}\frac{c}{d}=\frac{ac}{bd}$
Then we can prove that $\displaystyle S^{-1}R$ is a commutative ring of identity.
Prove that $\displaystyle S^{-1}R$has a unique maximal ideal..
Thank you

2. Originally Posted by ynj
Let $\displaystyle P$be a prime ideal,$\displaystyle R$be a commutative ring with identity,$\displaystyle S=R-P$. Then $\displaystyle S$will be a multiplicative subset of $\displaystyle R$
Prove that $\displaystyle S^{-1}R$has a unique maximal ideal..
140.112.21.20/latex/algebra/hungerford/remember/ch3.pdf ....................
see the definition of $\displaystyle S^{-1}R$....
Thank you
let $\displaystyle M=S^{-1}P=\{\frac{x}{s}: \ x \in P, \ s \in S \}.$ see that $\displaystyle M$ is a proper ideal of $\displaystyle S^{-1}R.$ to show that $\displaystyle M$ is the unique maximal ideal of $\displaystyle S^{-1}R,$ you only need to show that if $\displaystyle z \notin M,$ then $\displaystyle z$ is invertible:

let $\displaystyle z=\frac{y}{t},$ where $\displaystyle y \in R, \ t \in S.$ since $\displaystyle z \notin M,$ we have $\displaystyle y \notin P$ and so $\displaystyle y \in S.$ therefore $\displaystyle w=\frac{t}{y} \in S^{-1}R$ and obviously $\displaystyle zw=1.$ Q.E.D.

3. Originally Posted by NonCommAlg
you only need to show that if $\displaystyle z \notin M,$ then $\displaystyle z$ is invertible:
classic!
I have thought of the same $\displaystyle M$..but i do it in another way and it seems that it is impossible...
a theorem says $\displaystyle M$ is a maximal field in $\displaystyle R$iff $\displaystyle R/M$is a field.
but i cannot find the "1" in the $\displaystyle R/M$
if $\displaystyle \frac{a}{b}+M$is the "1", then $\displaystyle (\frac{a}{b}+M)(\frac{c}{d}+M)=\frac{ac}{bd}+M=\fr ac{c}{d}+M$
it requires that $\displaystyle \frac{ac}{bd}-\frac{c}{d}\in M$
$\displaystyle \frac{ac}{bd}-\frac{c}{d}=\frac{cd(a-b)}{bdd}$
so it requires $\displaystyle cd(a-b)$ is alway in $\displaystyle P$, right?
so $\displaystyle a-b\in P$,but $\displaystyle b\in S$, that implies $\displaystyle a\in S$........So I am confused..
Is there any mistake??
Thank you

4. well, like all other quotient rings, the identity element of the quotient ring $\displaystyle \frac{S^{-1}R}{M}$ is $\displaystyle 1 + M,$ where $\displaystyle 1=\frac{1}{1}$ is the identity element of $\displaystyle S^{-1}R.$

5. Originally Posted by NonCommAlg
well, like all other quotient rings, the identity element of the quotient ring $\displaystyle \frac{S^{-1}R}{M}$ is $\displaystyle 1 + M,$ where $\displaystyle 1=\frac{1}{1}$ is the identity element of $\displaystyle S^{-1}R.$
oooooooooooo...........it is a concept mistake.....................Sorry..
Thanks!