So, I was trying to find a counter-example that shows not every local ring's lattice of ideals is a chain. I think $F[[x_1,\cdots,x_n]]$ is a good counter-example but I'm not able to show that $F[[x_1,\cdots,x_n]]$ is a local ring. I read somewhere that $F[[x]]$ is indeed a local ring.

So here comes the question:

If $R$ is a local ring, what can we say about $R[[x]]$? Is it a local ring too?

I'm looking for a simple proof that doesn't use commutative algebra and localizations to show that. An elementary undergraduate level proof is appreciated.


2 Answers 2


Local rings are characterized by the property that the set of nonunits is an ideal.

If $R$ is a ring, and $f\in R[[x]]$, then $f$ is invertible if and only if $f(0)$, the constant term of $f$, is an invertible element of $R$. In case $R$ is local, with maximal ideal $m$, then the ideal $(m,x)\subset R[[x]]$ is the set of nonunits.


Yes. As Wikipedia says, "every ring of formal power series $R[[X,Y,\dots]]$ over a local ring $R$ is local; the maximal ideal consists of those power series with constant term in the maximal ideal of the base ring." So for example, the maximal ideal in $k[[X,Y]]$ (for $k$ a field) is $\langle X,Y \rangle$.


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