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'maximal ideal is prime'
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| Title of object: |
maximal ideal is prime |
| Canonical Name: |
MaximalIdealIsPrime |
| Type: |
Theorem |
| Created on: |
2007-11-23 10:00:05 |
| Modified on: |
2007-11-23 16:35:03 |
| Classification: |
msc:13A15, msc:16D25 |
Revision comment (for changes between this and next version):
Preamble:
% this is the default PlanetMath preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.
% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
%\usepackage{graphicx}
% for neatly defining theorems and propositions
\usepackage{amsthm}
% making logically defined graphics
%\usepackage{xypic}
% there are many more packages, add them here as you need them
% define commands here
\theoremstyle{definition}
\newtheorem*{thmplain}{Theorem}
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Content:
\textbf{Theorem.} In a commutative ring with non-zero unity, any maximal ideal is a prime ideal.
{\em Proof.}\, Let $\mathfrak{m}$ be a maximal ideal of such a ring $R$ and let the ring product $rs$ belong to $\mathfrak{m}$ but e.g. \,$r \notin \mathfrak{m}$. The maximality of $\mathfrak{m}$ implies that\,
$\mathfrak{m}\!+\!(r) = R = (1)$.\, Thus there exists an element \,$m \in \mathfrak{m}$\, and an element\, $x \in R$\, such that\, $m\!+\!xr = 1$.\, Now $m$ and $rs$ belong to $\mathfrak{m}$, whence
$$s = 1s = (m\!+\!xr)s = sm\!+\!x(rs) \in \mathfrak{m}.$$
So we can say that along with $rs$, at least one of its \PMlinkname{factors}{Product} belongs to $\mathfrak{m}$, and therefore $\mathfrak{m}$ is a prime ideal of $R$. |
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