Operations on Ideals - Hello Experts

  • Context: Graduate 
  • Thread starter Thread starter DukeSteve
  • Start date Start date
  • Tags Tags
    Operations
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 3K views
DukeSteve
Messages
9
Reaction score
0
Hello Experts,

I post this question here because in the homework topics there is no abstract algebra!
Please help me I want to understand it:

I have a ring R with unit. Also I am given n - natural number, I_n is the set {x in R: n*x = 0}

I have to prove or refute: Given n, m natural numbers:

A) Is I_n + I_m is an ideal of the form of I_k?
B) Is I_n intersection with I_m is an ideal of the form of I_k?
C) Is I_n union with I_m is an ideal of the form of I_k?

I just used the Bezout's identity that d = ax+by for any d is a common devisor of a,b, and x,y are integers.
And I get that A is a proof.
For B I don't know how to start...
 
Physics news on Phys.org
DukeSteve said:
Hello Experts,

I post this question here because in the homework topics there is no abstract algebra!
Please help me I want to understand it:

I have a ring R with unit. Also I am given n - natural number, I_n is the set {x in R: n*x = 0}

I have to prove or refute: Given n, m natural numbers:

A) Is I_n + I_m is an ideal of the form of I_k?
B) Is I_n intersection with I_m is an ideal of the form of I_k?
Let's write I_n=nR##.
##nR \cap mR = (l.c.m.)R## which is again an ideal of this form. Bezout in part A) gave you the greatest common divisor, now we need the least common multiple.
C) Is I_n union with I_m is an ideal of the form of I_k?
No. E.g. ##2\mathbb{Z} \cup 3\mathbb{Z} = \{\,\ldots,-6,-4,-3,-2,0,2,3,4,6,\ldots\,\}## but ##2+3=5 \notin 2\mathbb{Z} \cup 3\mathbb{Z}##.
I just used the Bezout's identity that d = ax+by for any d is a common devisor of a,b, and x,y are integers.
And I get that A is a proof.
Yes, right ideals to be exact: ##(nR+mR)\cdot R \subseteq nR+mR## and ##nR+mR = dR = kR## with Bezout's lemma, correct.
For B I don't know how to start...