|
|
|
Viewing Version
3
of
'every ring has a maximal ideal'
|
[ view 'every ring has a maximal ideal'
|
back to history
]
| Title of object: |
every ring has a maximal ideal |
| Canonical Name: |
EveryRingHasAMaximalIdeal |
| Type: |
Theorem |
| Created on: |
2003-09-08 16:00:56 |
| Modified on: |
2003-09-08 16:03:11 |
| Classification: |
msc:13A15, msc:16D25 |
| Keywords: |
maximal ideal, commutative ring, identity, axiom of choice |
| Synonyms: |
every ring has a maximal ideal=existence of maximal ideals |
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{amsthm}
\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
\newtheorem{thm}{Theorem}
\newtheorem{defn}{Definition}
\newtheorem{prop}{Proposition}
\newtheorem{lemma}{Lemma}
\newtheorem{cor}{Corollary}
% Some sets
\newcommand{\Nats}{\mathbb{N}}
\newcommand{\Ints}{\mathbb{Z}}
\newcommand{\Reals}{\mathbb{R}}
\newcommand{\Complex}{\mathbb{C}}
\newcommand{\Rats}{\mathbb{Q}} |
Content:
Let $\mathcal{R}\neq 0$ be a commutative ring with identity. Is
there a maximal ideal in $\mathcal{R}$? This simple property turns
out to be dependent on the axiom of choice. Assuming Zorn's Lemma,
which is equivalent to the axiom of choice, we are able to prove
the following:
\begin{prop}
Every ring $\mathcal{R}$ (as above) has a maximal ideal.
\end{prop}
\begin{proof}
Let $\Sigma$ be the partially ordered set
$$\Sigma=\{ \mathcal{A} \mid \mathcal{A} \text{ is an ideal of }
\mathcal{R},\quad \mathcal{A}\neq \mathcal{R}\}$$ ordered by
inclusion.
Since $0\in \mathcal{R}$, the ideal generated by $0$, $(0)\in
\Sigma$, because $(0)\neq \mathcal{R}$. Hence $\Sigma$ is
non-empty.
In order to apply Zorn's lemma we need to prove that every chain
in $\Sigma$ has an upper bound that belongs to $\Sigma$. Let
$\{\mathcal{A}_{\alpha}\}$ be a chain of ideals {\bf in} $\Sigma$,
so for all indices $\alpha,\beta$ we have
$$\mathcal{A}_{\alpha}\subseteq \mathcal{A}_{\beta}\quad \text{ or }\quad \mathcal{A}_{\beta}\subseteq
\mathcal{A}_{\alpha}$$ We claim that $\mathcal{B}$, defined
$$\mathcal{B}=\bigcup_{\alpha}\mathcal{A}_{\alpha}$$
is such an upper bound.
\begin{itemize}
\item $\mathcal{B}$ is an ideal. Indeed, let $a,b\in\mathcal{B}$,
so there exist $\alpha,\beta$ such that $a\in
\mathcal{A}_{\alpha}$, $b\in\mathcal{A}_{\beta}$. Since these two
ideals are in a totally ordered chain we have
$$\mathcal{A}_{\alpha}\subseteq \mathcal{A}_{\beta}\quad \text{ or }\quad \mathcal{A}_{\beta}\subseteq
\mathcal{A}_{\alpha}$$ Without loss of generality, we assume
$\mathcal{A}_{\alpha}\subseteq \mathcal{A}_{\beta}$. Then both
$a,b\in \mathcal{A}_{\beta}$, and $\mathcal{A}_{\beta}$ is an
ideal of the ring $\mathcal{R}$. Thus $a+b\in
\mathcal{A}_{\beta}\subseteq \mathcal{B}$.
Similarly, let $r\in \mathcal{R}$ and $b\in \mathcal{B}$. As
above, there exists $\beta$ such that $b\in \mathcal{A}_{\beta}$.
Since $\mathcal{A}_{\beta}$ is an ideal we have
$$r\cdot b \in \mathcal{A}_{\beta}\subseteq\mathcal{B}$$
Therefore, $\mathcal{B}$ is an ideal.
\item $\mathcal{B}\neq \mathcal{R}$, otherwise $1$ would belong to
$\mathcal{B}$, so there would be an $\alpha$ such that $1\in
\mathcal{A}_{\alpha}$ so $\mathcal{A}_{\alpha}=\mathcal{R}$. But
this is impossible because we assumed $\mathcal{A}_{\alpha}\in
\Sigma$ for all indices $\alpha$.
\end{itemize}
Therefore $\mathcal{B}\in\Sigma$. Hence every chain in $\Sigma$
has an upper bound in $\Sigma$ and we can apply Zorn's lemma to
deduce the existence of $\mathcal{M}$, a maximal element (with
respect to inclusion) in $\Sigma$. By definition of the set
$\Sigma$, $\mathcal{M}$ must be a maximal ideal in $\mathcal{R}$.
\end{proof}
{\bf NOTE:} Assuming that the axiom of choice does NOT hold,
mathematicians have shown the existence of commutative rings (with
1) that have no maximal ideals. |
|
|
|
|
|
