# A question about principal ideals and order

1. Jan 1, 2013

### Artusartos

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

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

2. Jan 1, 2013

### Number Nine

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
3. Jan 1, 2013

### Artusartos

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 $$\tau \in I$$ of smallest order Use Ex. 3.27 on p.130 to prove that $$I=(\tau)$$

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 $$f=x^nu$$. It follows easily that $$f \in I$$ if and only if $$x^n \in I$$, and moreover f divides g if an only if $$ord(f) \leq ord(g)$$. 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 $$x^n \in I$$ and every element of I is a multiple of $$x^n$$, hence $$I=(x^n)$$. Since very ideal has the form $$(x^n)$$ or (0), k[[x]] is a PID.