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Ideals and commutative rings with unity

  1. Jan 28, 2013 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    From contemporary abstract algebra :

    http://gyazo.com/08def13b62b0512a23505811bcc1e37e

    2. Relevant equations

    "A subring A of a ring R is called a (two-sided) ideal of R if for every r in R and every a in A both ra and ar are in A."

    So I know that since A and B are ideals of a ring R, [itex]ar, ra \in A[/itex] and [itex]br, rb \in B[/itex] for all [itex]a \in A, \space b \in B, \space r \in R[/itex]

    3. The attempt at a solution

    So my guess is to argue the double inclusion for this.

    Case : [itex]A \cap B \subseteq AB[/itex]

    Suppose [itex]k \in A \cap B[/itex], then [itex]k \in A[/itex] and [itex]k \in B[/itex]. We want to show [itex]k \in AB[/itex]

    I'm having trouble seeing how the given facts are supposed to steer the argument from here. Help would be much appreciated.
     
    Last edited: Jan 28, 2013
  2. jcsd
  3. Jan 28, 2013 #2

    Zondrina

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    Hmm I thought about this for an hour or so... and I THINK I may be on to something. I'll continue my flow below :

    Case : [itex]A \cap B \subseteq AB[/itex]

    Suppose [itex]k \in A \cap B[/itex], then [itex]k \in A[/itex] and [itex]k \in B[/itex]. We want to show [itex]k \in AB[/itex].

    Since we know R is a commutative ring with unity, let e denote the unity in R.

    Now since [itex]R = A + B[/itex], we know that [itex]e = a + b[/itex] for some [itex]a \in A[/itex] and [itex]b \in B[/itex]. Now :

    e = a + b
    ke = k(a + b)
    k = ka + kb

    Hence [itex]ka \in A[/itex] and [itex]kb \in B[/itex] hence [itex]k \in A + B[/itex]

    We can take [itex]a \in A[/itex] and we can take [itex]k \in B[/itex]. We can also take [itex]b \in B[/itex] and we can take [itex]k \in A[/itex]. Putting these together we have [itex]ka \in AB[/itex] and [itex]kb \in AB[/itex] so that [itex]k \in AB[/itex] since [itex]a \in A[/itex] and [itex]b \in B[/itex] as desired.

    [itex]∴ A \cap B \subseteq AB[/itex]


    Case : [itex]A \cap B \supseteq AB[/itex]

    Let [itex]ab \in AB[/itex]

    Now, [itex]ab \in A[/itex] since [itex]a \in A[/itex], [itex]b \in R[/itex] and A is an ideal of R. Also, [itex]ab \in B[/itex] since [itex]b \in B[/itex], [itex]a \in R[/itex] and B is an ideal of R. Hence [itex]ab \in A \cap B[/itex].

    [itex]∴ A \cap B \supseteq AB[/itex]

    [itex]∴ A \cap B = AB[/itex]

    I think that's it... If someone could look over it for me and tell me maybe if I missed something or if it's completely wrong that would be great :).
     
  4. Jan 28, 2013 #3

    Dick

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    Writing e=a+b is a great idea. I think you could write it read a little better. Like did you really need to say "Hence [itex]ka \in A[/itex] and [itex]kb \in B[/itex] hence [itex]k \in A + B[/itex]"? Did you ever use that k is in A+B?
     
  5. Jan 28, 2013 #4

    Zondrina

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    Yes good point. I never used it at all so I don't need it there. I wrote it down just to have it around incase I MAY have needed it yknow?

    Does anything else not make sense or has early morning thinking saved me again?
     
  6. Jan 28, 2013 #5

    Dick

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    No, it all looks good to me.
     
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