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Math Help - maximal ideal in ring of quotients of R with respect to S

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

    Let Pbe a prime ideal, Rbe a commutative ring with identity, S=R-P. Then Swill 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}Rhas a unique maximal ideal..
    Thank you
    Last edited by ynj; August 18th 2009 at 08:08 PM.
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  2. #2
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    Quote Originally Posted by ynj View Post
    Let Pbe a prime ideal, Rbe a commutative ring with identity, S=R-P. Then Swill be a multiplicative subset of R
    Prove that S^{-1}Rhas 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.
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  3. #3
    ynj
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    Quote Originally Posted by NonCommAlg View Post
    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 Riff R/Mis a field.
    but i cannot find the "1" in the R/M
    if \frac{a}{b}+Mis 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
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    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.
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  5. #5
    ynj
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    Quote Originally Posted by NonCommAlg View Post
    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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