[SOLVED] Operations with ideals in a ring.

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• May 18th 2010, 03:15 PM
ENRIQUESTEFANINI
[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.
• May 18th 2010, 04:30 PM
NonCommAlg
Quote:

Originally Posted by ENRIQUESTEFANINI
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 $x=\sum_{i=1}^n r_ia \cdot s_ib=\left(\sum_{i=1}^n {r_is_i} \right)ab \in Rab.$
• May 18th 2010, 05:16 PM
ENRIQUESTEFANINI
I'm terribly sorry. I considered a general commutative ring so, not necessarily with unit.
• May 18th 2010, 05:59 PM
NonCommAlg
Quote:

Originally Posted by ENRIQUESTEFANINI
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:

$(a)(b)=(Ra + \mathbb{Z}a)(Rb + \mathbb{Z}b)=R^2ab+Rab + \mathbb{Z}ab \subseteq Rab + \mathbb{Z}ab= (ab) \subset P.$
• May 19th 2010, 07:55 AM
ENRIQUESTEFANINI
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.
• May 19th 2010, 08:50 AM
NonCommAlg
Quote:

Originally Posted by ENRIQUESTEFANINI
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 $I=(c).$
• May 19th 2010, 09:00 AM
NonCommAlg
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

$ab+a'c=a(b+0)+a'(0+c) \in A(B+C).$
• May 19th 2010, 10:16 AM
ENRIQUESTEFANINI
Hi:
Your reply has been most helpful and clarifying. Thanks a lot.

Enrique.