MHB Is the Analysis of Principal Ideals Correct?

[Math Amateur]

Fraleigh (A First Course in Abstract Algebra) defines principal ideals in section 27 on page 250. His definition is as follows:

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"27.21 Definition

If R is a commutative ring with unity and a \in R , the ideal \{ ra | r \in R \} of all multiples of a is the principal ideal generated by a and is denoted <a>.

An ideal N of R is a principal ideal if N = <a> for some a \in R

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Consider N =\{ ra | r \in R \} ......(1)

If we take r = a in (1) then we have ra = aa = a^2 \in N

If we take r = a and a^2 \in N the we have using (1) again that ra = a^2 a = a^3 \in N

Continuing this, then we have a, a^2, a^3, a^4, a^5 , ... all belonging to N along with the other elements where r \ne a

Is the above analysis correct regarding the nature of principal ideals?

Would really appreciate this issue being clarified.

Peter

 

[I like Serena]

Fraleigh (A First Course in Abstract Algebra) defines principal ideals in section 27 on page 250. His definition is as follows:

===============================================================================================

"27.21 Definition

If R is a commutative ring with unity and a \in R , the ideal \{ ra | r \in R \} of all multiples of a is the principal ideal generated by a and is denoted <a>.

An ideal N of R is a principal ideal if N = <a> for some a \in R

=================================================================================================

Consider N =\{ ra | r \in R \} ......(1)

If we take r = a in (1) then we have ra = aa = a^2 \in N

If we take r = a and a^2 \in N the we have using (1) again that ra = a^2 a = a^3 \in N

Continuing this, then we have a, a^2, a^3, a^4, a^5 , ... all belonging to N along with the other elements where r \ne a

Is the above analysis correct regarding the nature of principal ideals?

Would really appreciate this issue being clarified.

Peter

Hi Peter!

Yes, that is correct.

You might also say that $r=a^n \in R$ for $n \in \mathbb N$, so $a^n \cdot a = a^{n+1} \in N$.
And since you also have unity in R, it follows that $1a \in N$, and therefore $a^n \in N$.

 

[Math Amateur]

Thanks so much for that clarification - can now proceed on with some confidence :D