[SOLVED] Operations with ideals in a ring.

Feb 2009
189
0
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
May 2008
2,295
1,663
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 \(\displaystyle x=\sum_{i=1}^n r_ia \cdot s_ib=\left(\sum_{i=1}^n {r_is_i} \right)ab \in Rab.\)
 
  • Like
Reactions: ENRIQUESTEFANINI
Feb 2009
189
0
I'm terribly sorry. I considered a general commutative ring so, not necessarily with unit.
 
Last edited:

NonCommAlg

MHF Hall of Honor
May 2008
2,295
1,663
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:

\(\displaystyle (a)(b)=(Ra + \mathbb{Z}a)(Rb + \mathbb{Z}b)=R^2ab+Rab + \mathbb{Z}ab \subseteq Rab + \mathbb{Z}ab= (ab) \subset P.\)
 
  • Like
Reactions: ENRIQUESTEFANINI
Feb 2009
189
0
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:

NonCommAlg

MHF Hall of Honor
May 2008
2,295
1,663
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
May 2008
2,295
1,663
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

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