1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quotients of Principal Ideals

  1. Jun 16, 2011 #1
    1. The problem statement, all variables and given/known data

    Let R be a commutative ring, [itex] a, b \in R [/itex] disjoint elements of the ring. Let [itex] (a), (b) [/itex] denote the principal ideals of a and b respectively. If [itex] \bar b \in R/(a) [/itex] is the class of b in the quotient ring, show that
    [tex] R/(a)/(\bar b) \cong R/(a,b) [/tex]


    3. The attempt at a solution

    This is just an application of the third (second for some people) isomorphism theorem which states that if [itex] I \subseteq J \subseteq R [/itex] are ideals, then [itex] (R/I)/(J/I) \cong R/J [/itex]
    What I need to show is that
    [tex] (\bar b) = \frac{ (a,b)}{(a)} [/tex]
    and this is where I'm having trouble.

    Indeed, we know that [itex] \bar b = b + (a) [/itex] in R/(a). So then I figure [itex] (\bar b) = (b+(a)) R [/itex]. If I write this out set-wise
    [tex] (b+(a))(R) = \{ (b+ar_1)r_2 : r_1, r_2 \in R \} = \{b r_2 + a r_3 : r_2, r_3 \in R\} = bR + a R = (a,b) [/tex]
    But we know, via correspondence theorem, that ideals of [itex] R/(a) [/itex] must have the form [itex] J/(a) [/itex] for some ideal J of R. It seems like I'm close, but something in my reasoning here is wrong.
     
  2. jcsd
  3. Jun 16, 2011 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Hi Kreizhn! :smile:

    Can't you apply the first isomorphism theorem. I.e. find a surjection

    [tex]f:(a,b)\rightarrow (\overline{b})[/tex]

    with kernel (a).
     
  4. Jun 16, 2011 #3
    Thanks for the reply micromass, you really seem to jump on these questions :smile:

    Are we allowed to do this though? I mean, (a,b) need not be a ring. I guess we could view (a,b) as a group, use the 1st IT then show it's a ring isomorphism, or maybe we are applying a categorical decomposition of the morphism?

    Anyway, assuming that we can do this, I think that my issue is that I'm having trouble seeing what [itex] (\bar b) [/itex] looks like. Let me regurgitate my thinking:

    Case 1

    Well, we know that the canonical projection [itex] \pi: R \to R/(a) [/itex] has kernel (a). My first thought is that maybe we can use the restriction to (a,b) and apply the fundamental theorem of homomorphisms. Unfortunately, it's not clear to me if the restriction of [itex] \pi [/itex] to (a,b) is surjective on on [itex] (\bar b) [/itex], so I'm hesitant to continue in this fashion.

    Case 2

    Instead, I'll try to make the homomorphism from scratch. Let a typical element of (a,b) be [itex] ar_1 + br_2 [/itex] for [itex] r_1,r_2 \in R [/itex]. I want to map this to an element [itex] \bar b r_3 [/itex] for some [itex] r_3 \in R [/itex]. Since a,b are fixed, the only thing we can play with is [itex] r_1,r_2,r_3 [/itex]. If we want [itex] \ker f = (a) [/itex] then whenever [itex] r_2 = 0 [/itex] we want [itex] \bar b r_3 = (a) [/itex].

    Now, [itex] \bar b r_3 = (b + (a)) r_3 = b r_3 + (a) r_3 = br_3 + (a) [/itex] since (a) is an ideal. So I'm thinking we just set
    [itex] f: ar_1 + br_2 \mapsto b r_2 + (a) [/itex]?
    This is surjective, but I've only shown that [itex] (a) \subseteq \ker f [/itex]. In particular, if b is a left-zero divisor, it seems like the kernel could be bigger.

    Alternatively, I could just be greatly over complicating this.
     
  5. Jun 16, 2011 #4

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Very good remark. There are two possible ways to clear this up. Firstly, we can see (a,b) as an R-module and then use the first iso theorem for modules.
    However, (a,b) actually is a ring. It doesn't have an identity, but for the rest it is a ring. So you can apply the first isomorphism theorem for rings without identity. (which is true by thesame proof).

    I like this approach. Basically, we must simply show that there is an element which is being sent to [itex]\overline{b}[/itex]. But isn't b sent to [itex]\overline{b}[/itex].
     
  6. Jun 16, 2011 #5
    Hmm..okay. In that case the canonical projection would map
    [tex] \pi \big|_{(a,b)}: ar_1 + br_2 \mapsto ar_1 + br_2 + (a) [/tex]

    But this brings something else up. Since (a) is a subgroup of the underlying abelian group of the ring, [itex] ar_1 + (a) = (a) [/itex] right? Does this mean in general that

    [tex] aR + bR + (a) = (a) + (b) + (a) = (a) + (b) [/tex]
    which would imply that [itex] (a,b) \cong (a,b)/(a) [/itex].
     
  7. Jun 16, 2011 #6

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    OK, I agree with all of this.

    Why would that imply that? I don't quite see how this follows. This isn't true of course. Maybe you mean

    [tex](\overline{a},\overline{b})=(a,b)/(a)[/tex]

    which is true. But note that [itex]\overline{a}=0[/itex] (in R/(a)).
     
  8. Jun 16, 2011 #7
    I think I might see where I screwed up: I messed up the addition. My original thought process was as follows: if we write out elements of (a,b)/(a), they would look like elements of (a,b) with (a) thrown on the end. That is,
    [tex] ar_1 + br_2 + (a) [/tex]
    so that (a,b)/(a) = aR + bR + (a). Clearly this is where the error happened.

    So what did I did wrong? I think this is it, but maybe you could clear it up for me.

    When I write [itex] aR + bR + (a) [/itex] this is really addition on the quotient ring [itex] [aR+(a)]+[bR+(a)] [/itex] which is what gives us [itex] (\bar a, \bar b) [/itex] right? But I think where I'm messing up is that the addition signs are technically different. To be more precise,

    [tex] [aR\ +_R\ (a)]\ +_{R/(a)}\ [bR\ +_R\ (a)] = (\bar a, \bar b)[/tex]

    which is why I can't just blindly do the addition. Does this make sense?
     
  9. Jun 16, 2011 #8

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    I'm starting to see what you did. The problem is that you can't just write things like bR+(a), without saying what it means.
    You can write b+(a) only because it's a notation. The real element is , the equivalence class of b. So b+(a) is only notation. And Rb+(a) is certainly not standard. What do you mean with this. Do you mean

    [itex]\{rb+(a)~\vert~r\in R\}[/itex]?

    In this case, it is indeed true that

    [tex]Rb+(a)=(\overline{b})=(a,b)/a[/tex]
     
  10. Jun 16, 2011 #9
    I just looked over my first post, and I see the mistake I made there as well. Namely, that

    [tex] (\bar b) = \bar b R/(a). [/tex]

    I forgot the quotient by (a)!
     
  11. Jun 16, 2011 #10
    Yeah, that's ultimately what I meant.

    Edit: I think the lesson today is that I need to be more careful :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Quotients of Principal Ideals
  1. Principal Ideals (Replies: 4)

  2. Principal Ideal Domain (Replies: 3)

Loading...