Let R = {(a,, a,, a, ...)}, where each a; E Z. Let I = {(a,, a, az, ...)}, where only a finite number of terms are nonzero. Prove that I is not a principal ideal of R.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Let R = {(a,, a,, a, ...)}, where each a; E Z. Let I = {(a,, a,
az, ...)}, where only a finite number of terms are nonzero. Prove
that I is not a principal ideal of R.
Transcribed Image Text:Let R = {(a,, a,, a, ...)}, where each a; E Z. Let I = {(a,, a, az, ...)}, where only a finite number of terms are nonzero. Prove that I is not a principal ideal of R.
Expert Solution
Step 1: Given Information:

Here given that R=a1, a2, a3......., where each aiZ.

Consider, I=a1, a2, a3........

It has to be proved that I is not principal ideal of  R.

Step 2: Calculation:

Consider, I is a principal ideal of R that is an element that exists in b such that

I=<b>, here <b>is defined as ideal and it is generated by the b.

Since, I is generated by the b, then each element of I must be the form rb for the some r in R.

Since, I contains finitely -many non-zero terms, hence b also have finitely many non-zero terms.

Consider, b=b1, b2, b3, .......bk, 0, 0...., here bk is last non-zero term.

Now for any r=r1, r2, r3........ in R.

Therefore, we have rb=r1 b1, r2 b2,  ,........,bk rk, 0,0.....

Consider the element s=0,0,0,....,0,1,1,1... in R.

As there are many infinitely non-zero terms in ss cann't be in I.

However, I is a principal ideal, and some element t must be existed in R.

Since, I is generated by t, then s=rt for some r in R.

It follows, there are many infinitely non-zero terms in s, it contradicts that I is generated by b.

And each element of I is the form of rb having many finitely non-zero terms.

Hence, it can be concluded that I is not ideal principal of R.

 

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