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A question about principal ideals and order

  1. Jan 1, 2013 #1
    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?

    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
  2. jcsd
  3. Jan 1, 2013 #2
    This isn't really a calculus question, but...
    Not all rings have a multiplicative identity. If we assume that R does...
    This is strange notation. Are you raising b to a power (in which case n should be an integer) or multiplying it by an element in R? In the former case, there is no guarantee that b has finite order, so this is not true. In the latter case, this is only true if b has an inverse in R.

    I'm not quite sure about the context in which "order" is being used here, but the "order of I" would have the usual meaning: The number of elements in I. Be careful about taking powers in rings: not all elements have multiplicative inverses, and so not all elements have an n such that bn = 1 (otherwise bbn-1 = 1, so b has an inverse; a contradiction).
    Last edited: Jan 1, 2013
  4. Jan 1, 2013 #3
    I'm sorry, I didn't realize that it was the wrong category...

    This was the context that I meant...

    If k is a field, prove that the ring of formal power series k[[x]] is a PID.

    Hint: If I is a nonzero ideal, choose [tex]\tau \in I [/tex] of smallest order Use Ex. 3.27 on p.130 to prove that [tex]I=(\tau)[/tex]

    This is the answer:

    We solve both parts at once. We saw in #3.27 that for any nonzero f in k[[x]] of order n, there is a unit u such that [tex]f=x^nu[/tex]. It follows easily that [tex]f \in I[/tex] if and only if [tex]x^n \in I[/tex], and moreover f divides g if an only if [tex]ord(f) \leq ord(g)[/tex]. Putting this all together: suppose I is a nonzero ideal such that n is the smallest order among nonzero elements of I, and let f be an element of order n in I. Then [tex]x^n \in I[/tex] and every element of I is a multiple of [tex]x^n[/tex], hence [tex]I=(x^n)[/tex]. Since very ideal has the form [tex](x^n)[/tex] or (0), k[[x]] is a PID.

    What does "order" mean in this context?
    Last edited: Jan 1, 2013
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