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I'm trying to find all prime ideals in the formal power series ring $k[[x]]$, where $k$ is a field.

I think I've managed to show that all ideals are of the form $\langle x^n \rangle$, $n>0$, i.e. generated by a single element.

I now want to use the condition that an ideal, $I \subset R$ is prime iff $R/I$ is an integral domain. So am I right in thinking that $k[[x]] / \langle x^n \rangle$ is the set of cosets $f + \langle x^n \rangle$ where f is a polynomial of degree less than n? It would be great if I could have some help on showing that this is then an integral domain?

Thanks

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  • $\begingroup$ Yes, every coset has one unique element that is a polynomial of degree less than $n$. In general, $k[[x]]/\langle x^n\rangle$ is not an integral domain, only for specific $n$. $\endgroup$ Commented Feb 9, 2014 at 14:37
  • $\begingroup$ Okay thanks! Why does the polynomial have to be unique? Is it correct that this is then isomorphic to the integers modulo n? $\endgroup$ Commented Feb 9, 2014 at 14:39
  • $\begingroup$ If $\deg f < n$ and $\deg g < n$, then also $\deg (f-g) < n$. If further $x^n \mid (f-g)$, from degree/order considerations it follows that $f-g = 0$. $k[[x]]/\langle x^n\rangle$ is not isomorphic to $\mathbb{Z}/(n)$. It is an $n$-dimensional $k$-vector space (and $k$-algebra). If it is an integral domain, it does in particular contain no nilpotent elements (except $0$). What does that say about $n$? $\endgroup$ Commented Feb 9, 2014 at 14:45
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    $\begingroup$ And in general $(x+\langle x^n\rangle)^n = \langle x^n\rangle$, so the only prime ideal is $\langle x\rangle$. $\endgroup$ Commented Feb 9, 2014 at 15:06
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    $\begingroup$ @DanielFischer: and $0$ (since $k[[x]]$) is a domain). But these are all of them $\endgroup$ Commented Feb 9, 2014 at 19:13

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You've correctly found that the nontrivial ideals of $k[[x]]$ are of the form $(x^n)$ for positive integers $n$.

Since we know $k[[x]]$ is a domain and that its ideals form a chain, we are guaranteed that $(0)$ and the maximal ideal $(x)$ are prime.

The only question then is whether or not $(x^n)$ is prime for powers $n>1$. But they are clearly not prime since for every such $n$, $x,x^{n-1}\notin (x^n)$, but their product is in $(x^n)$.

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