# maximal ideals of ring of formal power series

Suppose that $R$ is a commutative ring with non-zero unity.

If $\mathfrak{m}$ is a maximal ideal of $R$, then  $\mathfrak{M}\,:=\,\mathfrak{m}\!+\!(X)$  is a maximal ideal of the ring $R[[X]]$ of formal power series.

Also the converse is true, i.e. if $\mathfrak{M}$ is a maximal ideal of $R[[X]]$, then there is a maximal ideal $\mathfrak{m}$ of $R$ such that  $\mathfrak{M}\,=\,\mathfrak{m}\!+\!(X)$.

Note.  In the special case that $R$ is a field, the only maximal ideal of which is the zero ideal $(0)$, this corresponds to the only maximal ideal $(X)$ of $R[[X]]$ (see http://planetmath.org/node/12087formal power series over field).

We here prove the first assertion.  So, $\mathfrak{m}$ is assumed to be maximal.  Let

 $f(x)\;:=\;a_{0}\!+\!a_{1}X\!+\!a_{2}X^{2}\!+\ldots$

be any formal power series in $R[[X]]\!\smallsetminus\!\mathfrak{M}$.  Hence, the constant term $a_{0}$ cannot lie in $\mathfrak{m}$.  According to the criterion for maximal ideal, there is an element $r$ of $R$ such that  $1\!+\!ra_{0}\in\mathfrak{m}$.  Therefore

 $1\!+\!rf(X)\;=\;(1\!+\!ra_{0})\!+\!r(a_{1}\!+\!a_{2}X\!+\!a_{3}X^{2}\!+\ldots)% X\;\in\;\mathfrak{m}\!+\!(X)\;=\;\mathfrak{M},$

whence the same criterion says that $\mathfrak{M}$ is a maximal ideal of $R[[X]]$.

Title maximal ideals of ring of formal power series MaximalIdealsOfRingOfFormalPowerSeries 2013-03-22 19:10:49 2013-03-22 19:10:49 pahio (2872) pahio (2872) 7 pahio (2872) Result msc 13H05 msc 13J05 msc 13C13 msc 13F25