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Homemaximal ideals of ring of formal power series

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# 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 formal 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 $R[[X]]$.

## Mathematics Subject Classification

13H05*no label found*13J05

*no label found*13C13

*no label found*13F25

*no label found*

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