# formal power series over field

Theorem. If $K$ is a field, then the ring $K[[X]]$ of formal power series is a discrete valuation ring with
$(X)$ its unique maximal ideal^{}.

*Proof.* We show first that an arbitrary ideal $I$ of $K[[X]]$ is a principal ideal^{}. If
$I=(0)$, the thing is ready. Therefore, let $I\ne (0)$. Take an element

$$f(X):=\sum _{i=0}^{\mathrm{\infty}}{a}_{i}{X}^{i}$$ |

of $I$ such that it has the least possible amount of successive zero coefficients in its beginning; let its first non-zero coefficient be ${a}_{k}$. Then

$$f(X)={X}^{k}({a}_{k}+{a}_{k+1}X+\mathrm{\dots}).$$ |

Here we have in the parentheses an invertible formal power series $g(X)$, whence get the equation

$${X}^{k}=f(X){[g(X)]}^{-1}$$ |

implying ${X}^{k}\in I$ and consequently $({X}^{k})\subseteq I$.

For obtaining the reverse inclusion, suppose that

$$h(X):={b}_{n}{X}^{n}+{b}_{n+1}{X}^{n+1}+\mathrm{\dots}$$ |

is an arbitrary nonzero element of $I$ where ${b}_{n}\ne 0$. Because $n\ge k$, we may write

$$h(X)={X}^{k}({b}_{n}{X}^{n-k}+{b}_{n+1}{X}^{n-k+1}+\mathrm{\dots}).$$ |

This equation says that $h(X)\in ({X}^{k})$, whence $I\subseteq ({X}^{k})$.

Thus we have seen that $I$ is the principal ideal $({X}^{k})$, so that $K[[X]]$ is a principal ideal domain^{}.

Now, all ideals of the ring $K[[X]]$ form apparently the strictly descending chain

$$(X)\supset ({X}^{2})\supset ({X}^{3})\supset \mathrm{\dots}\supset (0),$$ |

whence the ring has the unique maximal ideal $(X)$. A principal ideal domain with only one maximal ideal is a discrete valuation ring.

Title | formal power series over field |
---|---|

Canonical name | FormalPowerSeriesOverField |

Date of creation | 2015-10-19 9:13:35 |

Last modified on | 2015-10-19 9:13:35 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 7 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 13H05 |

Classification | msc 13J05 |

Classification | msc 13F25 |