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I found this solution Ideals of formal power series ring.

I can follow the proof, but I'm a bit confused on what $a$ is. It seems as if in the solution, to define $a$, we need to know what $I$ is first, but then we eventually used $a$ to define $I = (t^a)$. Isn't that a bit circular?

So for a concrete example, suppose our field is $\mathbb{R}$ and we want to find the ideals in $\mathbb{R}[[t]]$. How do we find our $a$ in this case?

sedrick
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    Given $I$, $a$ as defined exists. What's the problem? – Randall Mar 28 '20 at 13:22
  • isn't it that whole point of the problem is that we want to find the ideal $I$? We want to determine the ideals of the ring. – sedrick Mar 28 '20 at 13:22
  • No, $I$ is already handed to you. You want to prove that $I$ has an alternate expression as $(t^a)$. – Randall Mar 28 '20 at 13:23
  • So for example, if the field is $\mathbb{R}$, then what's our $I$? – sedrick Mar 28 '20 at 13:25
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    There is no $I$. There are MANY ideals. The problem is trying to classify them. "Let $I$ be any ideal in this ring. Then $I=(t^a)$ for some $a$." – Randall Mar 28 '20 at 13:26
  • Did you see this in the link you gave? "Let $I$ be an ideal and $p\in I$..." The "let" tells you they are assuming you have an $I$ in your hands, but this $I$ is arbitrary. – Randall Mar 28 '20 at 13:27
  • Ok that just made a lot of sense. I think I finally get what it's saying now. thanks @Randall – sedrick Mar 28 '20 at 13:28
  • The proofs obscure the simple idea: every element has form $,ut^a,$ for a unit $u$ so if $,a\le $ all $a_i$ then $\large (u t^a, u_1 t^{a_1}, u_2 t^{a_2},\ldots) = (t^a, t^{a_1}, t^{a_2},\ldots) = t^a (\color{#c00}1, t^{a_1-a}, t^{a_2-a},\ldots) = t^a(\color{#c00}1) = (t^a)$ $\ \ \ $ – Bill Dubuque Mar 28 '20 at 15:26
  • More generally see DVRs = discrete valuation rings. – Bill Dubuque Mar 28 '20 at 15:38

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