# [SOLVED] Operations with ideals in a ring.

• May 18th 2010, 02:15 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) $\displaystyle \subset$ P then (a)(b) $\displaystyle \subset$ P. I think this should be true. If someone could give me a hand I would greatly appreciated it.
• May 18th 2010, 03:30 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) $\displaystyle \subset$ P then (a)(b) $\displaystyle \subset$ P. I think this should be true. If someone could give me a hand I would greatly appreciated it.

well, we actually have $\displaystyle (ab)=(a)(b).$ to see this first note that in a commutative ring $\displaystyle R$ (with 1) we have $\displaystyle (c)=Rc = \{rc: \ r \in R \}.$ so you need to show that $\displaystyle Rab=Ra \cdot Rb$:

$\displaystyle rab=(ra) \cdot b \in Ra \cdot Rb$ and so $\displaystyle Rab \subseteq Ra \cdot Rb.$ also if $\displaystyle x \in Ra . Rb,$ then $\displaystyle x=\sum_{i=1}^n r_ia \cdot s_ib=\left(\sum_{i=1}^n {r_is_i} \right)ab \in Rab.$
• May 18th 2010, 04:16 PM
ENRIQUESTEFANINI
I'm terribly sorry. I considered a general commutative ring so, not necessarily with unit.
• May 18th 2010, 04:59 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 $\displaystyle R$ we have $\displaystyle (c)=Rc + \mathbb{Z}c = \{rc + nc: \ r \in R, \ n \in \mathbb{Z} \}.$ thus:

$\displaystyle (a)(b)=(Ra + \mathbb{Z}a)(Rb + \mathbb{Z}b)=R^2ab+Rab + \mathbb{Z}ab \subseteq Rab + \mathbb{Z}ab= (ab) \subset P.$
• May 19th 2010, 06:55 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, 07:50 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?

$\displaystyle \mathbb{Z}c$ is a subset (not subring) of $\displaystyle R$ because $\displaystyle nc \in R$ for all $\displaystyle n \in \mathbb{Z}.$ so $\displaystyle rc+nc \in R$, for all $\displaystyle r \in R, \ n \in \mathbb{Z},$ i.e. $\displaystyle I=Rc + \mathbb{Z}c \subseteq R.$ (note that we cannot write $\displaystyle rc+nc=(r+n)c$ because $\displaystyle R$ might

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

$\displaystyle s(rc+nc)=(sr + ns)c \in Rc \subseteq I.$ also if $\displaystyle J$ is any ideal of $\displaystyle R$ which contains $\displaystyle c,$ then $\displaystyle Rc \subseteq J$ and $\displaystyle \mathbb{Z}c \subseteq J$ and thus $\displaystyle I=Rc + \mathbb{Z}c \subseteq J.$ that means $\displaystyle I=(c).$
• May 19th 2010, 08:00 AM
NonCommAlg
the identity $\displaystyle A(B+C)=AB+AC$ holds for any subsets of $\displaystyle R$ provided that $\displaystyle 0 \in B \cap C$: to see this, suppose that $\displaystyle a,a' \in A, \ b \in B, \ c \in C.$ then $\displaystyle a(b+c)=ab+ac \in AB+AC$ and

$\displaystyle ab+a'c=a(b+0)+a'(0+c) \in A(B+C).$
• May 19th 2010, 09:16 AM
ENRIQUESTEFANINI
Hi: