Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Radicals and the Ring of Quotients

  1. Dec 21, 2017 #1
    1. The problem statement, all variables and given/known data
    Let ##S## be a multiplicative subset of a commutative ring ##R## with identity. If ##I## is an ideal in ##R##, then ##S^{-1}(\mbox{ Rad } I) = \mbox{Rad}(S^{-1}I)##.

    2. Relevant equations


    3. The attempt at a solution

    If ##x \in S^{-1}(\mbox{ Rad } I)##, then ##x = \frac{r}{s}## for some ##r \in \mbox{ Rad } I## and ##s \in S##. Hence, there exists an ##n \in \Bbb{N}## such that ##r^n \in I##; moreover, since ##S## is multiplicative, ##s^n \in S##. Therefore ##\left( \frac{r}{s} \right)^n = \frac{r^n}{s^n} \in S^{-1}I## which means that ##\frac{r}{s} \in \mbox{Rad}(S^{-1}I)##.

    Now suppose that ##x \in \mbox{Rad}(S^{-1}I)##. Then there exists a natural number ##n## such that ##S^{-1}I \ni \left( \frac{r}{s} \right)^n = \frac{r^n}{s^n}##. Therefore ##r^n \in I## which implies ##r \in \mbox{Rad } I##, and so ##\frac{r}{s} \in S^{-1}(\mbox{ Rad } I)##.

    Is it really that easy, or did I make a simple blunder?
     
  2. jcsd
  3. Dec 21, 2017 #2

    fresh_42

    User Avatar
    2017 Award

    Staff: Mentor

    I don't follow your conclusion ##\operatorname{Rad}(S^{-1}I) \subseteq S^{-1}\operatorname{Rad}(I)##. With ##x \in \operatorname{Rad}(S^{-1}I)## we have ##x^n=rs^{-1}\; , \;r \in I##. How can we assume, that ##x## is of the form ##r^ns^{-n}\,##?
     
  4. Dec 21, 2017 #3

    fresh_42

    User Avatar
    2017 Award

    Staff: Mentor

    I don't follow your conclusion ##\operatorname{Rad}(S^{-1}I) \subseteq S^{-1}\operatorname{Rad}(I)##. With ##x \in \operatorname{Rad}(S^{-1}I)## we have ##x^n=rs^{-1}\; , \;r \in I\,.## How can we assume, that ##x## is of the form ##r^ns^{-n}\,##?
     
  5. Dec 21, 2017 #4
    Let me try again. Suppose that ##x \in \mbox{Rad}(S^{-1}I)##. Then ##x = \frac{r}{s}## with ##s \in S## and ##r \in R## such that ##x^n = \frac{r^n}{s^n} \in S^{-1} I##. Then by definition ##r^n \in I## and ##s^n \in S## and therefore ##r \in \mbox{Rad } I##. Hence ##\frac{r}{s} \in S^{-1} \mbox{Rad}(I)##.
     
  6. Dec 21, 2017 #5

    fresh_42

    User Avatar
    2017 Award

    Staff: Mentor

    For ##x \in \operatorname{Rad}(S^{-1}I) ## we have ##x^n=s^{-1}r## with ##r \in I## by definition. But how do we know that ##x \in RS^{-1}\,?## What have I missed?
     
  7. Dec 21, 2017 #6
    Well, ##I \subseteq R##, so ##S^{-1} I \subseteq S^{-1} R##.
     
  8. Dec 21, 2017 #7

    fresh_42

    User Avatar
    2017 Award

    Staff: Mentor

    But you only have this for ##x^n## and not for ##x## itself. However, you keep making assumptions on ##x## which have to be proven.

    Edit and hint: We want to have ##x=s^{-1}r'## with ##r' \in \operatorname{Rad}(I)## that is ##r'^{\,n} =(sx)^n \in I##.
     
    Last edited: Dec 21, 2017
  9. Dec 23, 2017 #8
    Okay. Here is another try. If ##x \in \mbox{ Rad }(S^{-1})##, then ##x^n \in S^{-1}I## for some ##n \in \Bbb{N}## and therefore ##x^n = \frac{r}{s}## with ##r \in I##. But ##x## is also an element in ##S^{-1}R##, so that ##x = \frac{t}{k}## for ##t \in R## and ##k \in S##. This implies ##\frac{t^n}{k^n} = \frac{r}{s}## which happens if and only if there exists an ##s_1 \in S## such that ##s_1(st^n - rk^n) = 0## or ##s_1s t^n = r s_1 k^n \in I##. Since ##I## is an ideal, ##s_1^n s^n t^n = r s_1^n k^n s^{n-1} \in I##. Hence ##x = \frac{t}{k} = \frac{s_1 s t}{s_1 sk}## with ##(s_1 s t)^n \in I##, which means ##x \in S^{-1} \mbox{ Rad }(I)##.

    How does this sound?
     
  10. Dec 23, 2017 #9

    fresh_42

    User Avatar
    2017 Award

    Staff: Mentor

    Certainly too complicated. And again, why is ##x \in S^{-1}R\;##? Anyway, you don't need this detour.

    ##x \in \operatorname{Rad}(S^{-1}I)## means ##x^n=s^{-1}r\;(r \in I)## and we get
    $$
    (sx)^n=s^nx^n=s^{n-1}r \in I \Longrightarrow sx \in \operatorname{Rad}I \Longrightarrow x \in S^{-1}(\operatorname{Rad}I)
    $$
     
  11. Dec 23, 2017 #10
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted