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
Canonical name	MaximalIdealsOfRingOfFormalPowerSeries
Date of creation	2013-03-22 19:10:49
Last modified on	2013-03-22 19:10:49
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	7
Author	pahio (2872)
Entry type	Result
Classification	msc 13H05
Classification	msc 13J05
Classification	msc 13C13
Classification	msc 13F25