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

**ynj** · August 18th 2009, 06:00 PM

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

**NonCommAlg** · August 18th 2009, 08:10 PM

Originally Posted by ynj

Let $P$ be a prime ideal, $R$ be a commutative ring with identity, $S=R-P$ . Then $S$ will be a multiplicative subset of $R$
Prove that $S^{-1}R$ has a unique maximal ideal..
140.112.21.20/latex/algebra/hungerford/remember/ch3.pdf ....................
see the definition of $S^{-1}R$ ....
Thank you

let $M=S^{-1}P=\{\frac{x}{s}: \ x \in P, \ s \in S \}.$ see that $M$ is a proper ideal of $S^{-1}R.$ to show that $M$ is the unique maximal ideal of $S^{-1}R,$ you only need to show that if $z \notin M,$ then $z$ is invertible:

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

**ynj** · August 18th 2009, 08:29 PM

Originally Posted by NonCommAlg

you only need to show that if $z \notin M,$ then $z$ is invertible:

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

**NonCommAlg** · August 18th 2009, 08:43 PM

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

**ynj** · August 18th 2009, 08:46 PM

Originally Posted by NonCommAlg

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

oooooooooooo...........it is a concept mistake.....................Sorry..
Thanks!

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