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Math Help - [SOLVED] Operations with ideals in a ring.

  1. #1
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    [SOLVED] Operations with ideals in a ring.

    Hi:
    Let a, b be elements of a commutative ring R, P ideal in R. I am frustrated by the fact that I cannot prove the following: if (ab) \subset P then (a)(b) \subset P. I think this should be true. If someone could give me a hand I would greatly appreciated it.
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  2. #2
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    Quote Originally Posted by ENRIQUESTEFANINI View Post
    Hi:
    Let a, b be elements of a commutative ring R, P ideal in R. I am frustrated by the fact that I cannot prove the following: if (ab) \subset P then (a)(b) \subset P. I think this should be true. If someone could give me a hand I would greatly appreciated it.
    well, we actually have (ab)=(a)(b). to see this first note that in a commutative ring R (with 1) we have (c)=Rc = \{rc: \ r \in R \}. so you need to show that Rab=Ra \cdot Rb:

    rab=(ra) \cdot b \in Ra \cdot Rb and so Rab \subseteq Ra \cdot Rb. also if x \in Ra . Rb, then x=\sum_{i=1}^n r_ia \cdot s_ib=\left(\sum_{i=1}^n {r_is_i} \right)ab \in Rab.
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    I'm terribly sorry. I considered a general commutative ring so, not necessarily with unit.
    Last edited by ENRIQUESTEFANINI; May 18th 2010 at 04:52 PM.
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    Quote Originally Posted by ENRIQUESTEFANINI View Post
    I'm terribly sorry. I considered a general commutative ring so, not necessarily with unit. Also, I marked this thread as solved. I do not know how to unmark it.
    well, that wouldn't make a lot of difference in the proof. in general in a commutative ring R we have (c)=Rc + \mathbb{Z}c = \{rc + nc: \ r \in R, \ n \in \mathbb{Z} \}. thus:

    (a)(b)=(Ra + \mathbb{Z}a)(Rb + \mathbb{Z}b)=R^2ab+Rab + \mathbb{Z}ab \subseteq Rab + \mathbb{Z}ab= (ab) \subset P.
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    Thank you. It seems as if Zc =def {nc: n an integer}. But then Zc is not a subring of R. Can I still make sums and products involving Zc? May be I could define sums and products of subsets A, B of R by A+B = {a+b: a belongs to A, b belongs to B}, AB = set of all finite sums of the form a_1b_1+...+a_nb_n. But then A(B+C) = AB+AC not necessarily valid.

    Please understand all I want now is to know about Zc and the way to operate with it. As for the proof, I found a longer one, that is in essence that given by you if I could operate freely with Zc.
    Last edited by ENRIQUESTEFANINI; May 19th 2010 at 07:57 AM.
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    Quote Originally Posted by ENRIQUESTEFANINI View Post
    Thank you. It seems as if Zc =def {nc: n an integer}. But then Zc is not a subring of R. Can I still make sums and products involving Zc?
    \mathbb{Z}c is a subset (not subring) of R because nc \in R for all n \in \mathbb{Z}. so rc+nc \in R, for all r \in R, \ n \in \mathbb{Z}, i.e. I=Rc + \mathbb{Z}c \subseteq R. (note that we cannot write rc+nc=(r+n)c because R might

    not have 1 and so n might not be in R.) so you need to show that I is the smallest ideal of R which contains c. that is proved very easily: I is clearly an additive group of R and if s \in R, then

    s(rc+nc)=(sr + ns)c \in Rc \subseteq I. also if J is any ideal of R which contains c, then Rc \subseteq J and \mathbb{Z}c \subseteq J and thus I=Rc + \mathbb{Z}c \subseteq J. that means I=(c).
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    the identity A(B+C)=AB+AC holds for any subsets of R provided that 0 \in B \cap C: to see this, suppose that a,a' \in A, \ b \in B, \ c \in C. then a(b+c)=ab+ac \in AB+AC and

    ab+a'c=a(b+0)+a'(0+c) \in A(B+C).
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    Hi:
    Your reply has been most helpful and clarifying. Thanks a lot.

    Enrique.
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