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Let $F$ is a field. I know that $F[x]$ is a principal ideal domain. I know examples of ideals in $F[x]$ like $\{a_0+a_1x+\dots+a_nx^n|a_0=0\}$. I need to know all ideals. Same with $F[[x]]$: I know some examples of ideals but can't say that I know all ideals of the ring. Is there some technique to find all ideals of the rings?

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the ideals of $F[x]$ are all principal, as you say, so every ideal is uniquely determined by the monic polynomial of minimal degree it contains. $F[[x]]$ is local, the unique maximal ideal is $(x)$ so every ideal is contained in $(x)$ (power series with non-zero constant term are invertible) –  yoyo Feb 4 '12 at 16:34
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@yoyo thanks for the nice comment. Please convert it to an answer and I will approve it as the solution. –  Sergey Filkin Feb 4 '12 at 16:44
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The following answer was given by yoyo in the comments:

The ideals of $F[x]$ are all principal, as you say, so every ideal is uniquely determined by the monic polynomial of minimal degree it contains. $F[[x]]$ is local, the unique maximal ideal is $(x)$ so every ideal is contained in $(x)$ (power series with non-zero constant term are invertible).

And as a consequence of this we easily see that any ideal $I$ of $F[[x]]$ is of the form $(x^m)$ for some non-negative integer $m$. This is easily seen as follows. Let $m$ the degree of the lowest degree non-zero term of any power series $a(x)$ in $I$. Then $a(x)=x^mu$, where $u$ is a unit of $F[[x]]$. Thus $(x^m)\subseteq I$. The reverse inclusion is trivial.

Essentially the same argument works for all DVRs (=discrete valuation rings).

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Oops, too many crusaders :) I was just adding a post containing stuff about the power series ring when you edited. Oh well :) –  rschwieb Jun 20 '13 at 19:29
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