# A question about principal ideals and order

• Artusartos
In summary, the conversation discusses the concept of ideal and its notation in a commutative ring. It also addresses the confusion around the use of "order" in relation to taking powers in rings and the proof that the ring of formal power series k[[x]] is a PID if k is a field.
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 $$r_1(b_1)^n + r_2(b_2)^n + ... + r_k(b_k)^n = 1$$?

This isn't really a calculus question, but...
since all rings have an identity element, there must be an "n" in R such that b^n = 1, right?

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).

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Number Nine said:
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).

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.

What does "order" mean in this context?

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## 1. What is a principal ideal?

A principal ideal is a special type of ideal in a ring that is generated by a single element. In other words, it is the set of all elements that can be obtained by multiplying the generator by any element in the ring.

## 2. How is the order of a principal ideal determined?

The order of a principal ideal is determined by the order of the generator. For example, if the generator has an order of n, then the principal ideal will have n elements.

## 3. Can a principal ideal have more than one generator?

No, a principal ideal can only have one generator. This is because the definition of a principal ideal requires that it is generated by a single element.

## 4. What is the relationship between principal ideals and subrings?

A principal ideal is a type of ideal, which is a subset of a ring. Similarly, a subring is also a subset of a ring. However, a principal ideal is a special type of ideal that is generated by a single element, while a subring can be generated by multiple elements.

## 5. How are principal ideals used in algebraic structures?

Principal ideals are used to study the structure of rings, particularly in abstract algebra. They help in understanding the properties and relationships of elements within a ring, and can be used to prove theorems and solve problems.

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