If we have an element b in R, then (b) = {rb : r is in R} is a principal ideal generated by b, right? Since R is a commutative ring, and since all rings have an identity element, there must be an "n" in R such that b^n = 1, right?(adsbygoogle = window.adsbygoogle || []).push({});

My question:

Our textbook says: If [tex]b_1, b_2, ... , b_k[/tex] lie in R, then the set of all linear combinations [tex]={r_1b_1 + r_2b_2 + ... + r_kb_k : r_i \in R for all i}[/tex] is an ideal in R. We write [tex]I=(b_1, b_2, ... , b_k)[/tex] in this case, and we call I the ideal generated by [tex]b_1, b_2, ... ,b_k[/tex].

I am a bit confused about what "order" means with regards to [tex](b_1, b_2, ... , b_k)[/tex]. Does it mean a number n where [tex](r_1b_1 + ... + r_kb_k)^n = 1[/tex]...or does it mean [tex]r_1(b_1)^n + r_2(b_2)^n + ... + r_k(b_k)^n = 1[/tex]?

Thanks in advance

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# A question about principal ideals and order

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