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.
Best Answer
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.