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'existence of maximal ideals'
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| Title of object: |
existence of maximal ideals |
| Canonical Name: |
EveryRingHasAMaximalIdeal |
| Type: |
Theorem |
| Created on: |
2003-09-08 16:00:56 |
| Modified on: |
2006-02-03 16:30:57 |
| Classification: |
msc:13A15, msc:16D25 |
| Keywords: |
maximal ideal, commutative ring, identity, axiom of choice |
| Synonyms: |
existence of maximal ideals=existence of maximal ideals |
Revision comment (for changes between this and next version):
| more general result (correction #8922) |
Preamble:
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amsfonts}
\newtheorem{thm}{Theorem}
\newtheorem{defn}{Definition}
\newtheorem{prop}{Proposition}
\newtheorem{lemma}{Lemma}
\newtheorem{cor}{Corollary}
\newcommand{\Nats}{\mathbb{N}}
\newcommand{\Ints}{\mathbb{Z}}
\newcommand{\Reals}{\mathbb{R}}
\newcommand{\Complex}{\mathbb{C}}
\newcommand{\Rats}{\mathbb{Q}} |
Content:
\PMlinkescapeword{chain}
\PMlinkescapeword{equivalent}
\PMlinkescapeword{identity}
\PMlinkescapeword{order}
\PMlinkescapeword{property}
\PMlinkescapeword{simple}
Let $\mathcal{R}\neq 0$ be a ring with identity $1$. 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 $\mathcal{R}\neq 0$ with identity 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 \PMlinkname{chain}{TotalOrder}
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
by
$$\mathcal{B}=\bigcup_{\alpha}\mathcal{A}_{\alpha}$$
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
$$r\cdot b \in \mathcal{A}_{\beta}\subseteq\mathcal{B}$$
and
$$b\cdot r \in \mathcal{A}_{\beta}\subseteq\mathcal{B}.$$
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. |
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