# [SOLVED] Operations with ideals in a ring.

#### 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.

#### NonCommAlg

MHF Hall of Honor
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 $$\displaystyle x=\sum_{i=1}^n r_ia \cdot s_ib=\left(\sum_{i=1}^n {r_is_i} \right)ab \in Rab.$$

• ENRIQUESTEFANINI

#### ENRIQUESTEFANINI

I'm terribly sorry. I considered a general commutative ring so, not necessarily with unit.

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#### NonCommAlg

MHF Hall of Honor
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:

$$\displaystyle (a)(b)=(Ra + \mathbb{Z}a)(Rb + \mathbb{Z}b)=R^2ab+Rab + \mathbb{Z}ab \subseteq Rab + \mathbb{Z}ab= (ab) \subset P.$$

• ENRIQUESTEFANINI

#### 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.

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#### NonCommAlg

MHF Hall of Honor
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 $$\displaystyle I=(c).$$

#### NonCommAlg

MHF Hall of Honor
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

$$\displaystyle ab+a'c=a(b+0)+a'(0+c) \in A(B+C).$$

Hi: