| Version 13 |
Version 12 |
| \PMlinkescapeword{chain} |
\PMlinkescapeword{chain} |
| \PMlinkescapeword{equivalent} |
\PMlinkescapeword{equivalent} |
| \PMlinkescapeword{identity} |
\PMlinkescapeword{identity} |
| \PMlinkescapeword{necessary} |
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| \PMlinkescapeword{order} |
\PMlinkescapeword{order} |
| \PMlinkescapeword{property} |
\PMlinkescapeword{property} |
| \PMlinkescapeword{simple} |
\PMlinkescapeword{simple} |
| \PMlinkescapeword{unity} |
\PMlinkescapeword{unity} |
|
|
| \begin{thm} |
\begin{thm} |
| Let $\mathcal{R}$ be a ring with unity. |
Let $\mathcal{R}$ be a ring with unity. |
| Every proper ideal of $\mathcal{R}$ lies in a maximal ideal of $\mathcal{R}$. |
Every proper ideal of $\mathcal{R}$ lies in a maximal ideal of $\mathcal{R}$. |
| \end{thm} |
\end{thm} |
|
|
| Applying this theorem to the zero ideal gives the following corollary: |
Applying this theorem to the zero ideal gives the following corollary: |
|
|
| \begin{cor} |
\begin{cor} |
| Every ring $\mathcal{R}\neq 0$ with unity has a maximal ideal. |
Every ring $\mathcal{R}\neq 0$ with unity has a maximal ideal. |
| \end{cor} |
\end{cor} |
|
|
| {\it Proof of theorem}. |
{\it Proof of theorem}. |
| Let $\mathcal{I}$ be a proper ideal of $\mathcal{R}$, |
Let $\mathcal{I}$ be a proper ideal of $\mathcal{R}$, |
| and let $\Sigma$ be the partially ordered set |
and let $\Sigma$ be the partially ordered set |
| $$\Sigma=\{ \mathcal{A} \mid \mathcal{A} \hbox{ is an ideal of } |
$$\Sigma=\{ \mathcal{A} \mid \mathcal{A} \hbox{ is an ideal of } |
| \mathcal{R},\hbox{ and } |
\mathcal{R},\hbox{ and } |
| \mathcal{I}\subseteq\mathcal{A}\neq \mathcal{R}\}$$ |
\mathcal{I}\subseteq\mathcal{A}\neq \mathcal{R}\}$$ |
| ordered by inclusion. |
ordered by inclusion. |
|
|
| Note that $\mathcal{I}\in\Sigma$, so $\Sigma$ is non-empty. |
Note that $\mathcal{I}\in\Sigma$, so $\Sigma$ is non-empty. |
|
|
| In order to apply Zorn's Lemma |
In order to apply Zorn's Lemma |
| we need to prove that every non-empty \PMlinkname{chain}{TotalOrder} |
we need to prove that every non-empty \PMlinkname{chain}{TotalOrder} |
| in $\Sigma$ has an upper bound in $\Sigma$. |
in $\Sigma$ has an upper bound in $\Sigma$. |
| Let $\{\mathcal{A}_{\alpha}\}$ be a non-empty chain of ideals in $\Sigma$, |
Let $\{\mathcal{A}_{\alpha}\}$ be a non-empty chain of ideals in $\Sigma$, |
| so for all indices $\alpha,\beta$ we have |
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}\subseteq \mathcal{A}_{\beta}\quad \text{ or }\quad \mathcal{A}_{\beta}\subseteq |
| \mathcal{A}_{\alpha}.$$ |
\mathcal{A}_{\alpha}.$$ |
| We claim that $\mathcal{B}$ defined by |
We claim that $\mathcal{B}$ defined by |
| $$\mathcal{B}=\bigcup_{\alpha}\mathcal{A}_{\alpha}$$ |
$$\mathcal{B}=\bigcup_{\alpha}\mathcal{A}_{\alpha}$$ |
| is a suitable upper bound. |
is a suitable upper bound. |
| \begin{itemize} |
\begin{itemize} |
| \item $\mathcal{B}$ is an ideal. Indeed, let $a,b\in\mathcal{B}$, |
\item $\mathcal{B}$ is an ideal. Indeed, let $a,b\in\mathcal{B}$, |
| so there exist $\alpha,\beta$ such that $a\in |
so there exist $\alpha,\beta$ such that $a\in |
| \mathcal{A}_{\alpha}$, $b\in\mathcal{A}_{\beta}$. Since these two |
\mathcal{A}_{\alpha}$, $b\in\mathcal{A}_{\beta}$. Since these two |
| ideals are in a totally ordered chain we have |
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}\subseteq \mathcal{A}_{\beta}\quad \text{ or }\quad \mathcal{A}_{\beta}\subseteq |
| \mathcal{A}_{\alpha}$$ Without loss of generality, we assume |
\mathcal{A}_{\alpha}$$ Without loss of generality, we assume |
| $\mathcal{A}_{\alpha}\subseteq \mathcal{A}_{\beta}$. Then both |
$\mathcal{A}_{\alpha}\subseteq \mathcal{A}_{\beta}$. Then both |
| $a,b\in \mathcal{A}_{\beta}$, and $\mathcal{A}_{\beta}$ is an |
$a,b\in \mathcal{A}_{\beta}$, and $\mathcal{A}_{\beta}$ is an |
| ideal of the ring $\mathcal{R}$. Thus $a+b\in |
ideal of the ring $\mathcal{R}$. Thus $a+b\in |
| \mathcal{A}_{\beta}\subseteq \mathcal{B}$. |
\mathcal{A}_{\beta}\subseteq \mathcal{B}$. |
|
|
| Similarly, let $r\in \mathcal{R}$ and $b\in \mathcal{B}$. As |
Similarly, let $r\in \mathcal{R}$ and $b\in \mathcal{B}$. As |
| above, there exists $\beta$ such that $b\in \mathcal{A}_{\beta}$. |
above, there exists $\beta$ such that $b\in \mathcal{A}_{\beta}$. |
| Since $\mathcal{A}_{\beta}$ is an ideal we have |
Since $\mathcal{A}_{\beta}$ is an ideal we have |
| $$r\cdot b \in \mathcal{A}_{\beta}\subseteq\mathcal{B}$$ |
$$r\cdot b \in \mathcal{A}_{\beta}\subseteq\mathcal{B}$$ |
| and |
and |
| $$b\cdot r \in \mathcal{A}_{\beta}\subseteq\mathcal{B}.$$ |
$$b\cdot r \in \mathcal{A}_{\beta}\subseteq\mathcal{B}.$$ |
| Therefore, $\mathcal{B}$ is an ideal. |
Therefore, $\mathcal{B}$ is an ideal. |
|
|
| \item $\mathcal{B}\neq \mathcal{R}$, otherwise $1$ would belong to |
\item $\mathcal{B}\neq \mathcal{R}$, otherwise $1$ would belong to |
| $\mathcal{B}$, so there would be an $\alpha$ such that $1\in |
$\mathcal{B}$, so there would be an $\alpha$ such that $1\in |
| \mathcal{A}_{\alpha}$ so $\mathcal{A}_{\alpha}=\mathcal{R}$. But |
\mathcal{A}_{\alpha}$ so $\mathcal{A}_{\alpha}=\mathcal{R}$. But |
| this is impossible because we assumed $\mathcal{A}_{\alpha}\in |
this is impossible because we assumed $\mathcal{A}_{\alpha}\in |
| \Sigma$ for all indices $\alpha$. |
\Sigma$ for all indices $\alpha$. |
|
|
| \item $\mathcal{I}\subseteq\mathcal{B}$. |
\item $\mathcal{I}\subseteq\mathcal{B}$. |
| Indeed, the chain is non-empty, |
Indeed, the chain is non-empty, |
| so there is some $\mathcal{A}_\alpha$ in the chain, |
so there is some $\mathcal{A}_\alpha$ in the chain, |
| and we have $\mathcal{I}\subseteq\mathcal{A}_\alpha\subseteq\mathcal{B}$. |
and we have $\mathcal{I}\subseteq\mathcal{A}_\alpha\subseteq\mathcal{B}$. |
| \end{itemize} |
\end{itemize} |
| Therefore $\mathcal{B}\in\Sigma$. Hence every chain in $\Sigma$ |
Therefore $\mathcal{B}\in\Sigma$. Hence every chain in $\Sigma$ |
| has an upper bound in $\Sigma$ and we can apply Zorn's Lemma to |
has an upper bound in $\Sigma$ and we can apply Zorn's Lemma to |
| deduce the existence of $\mathcal{M}$, a maximal element (with |
deduce the existence of $\mathcal{M}$, a maximal element (with |
| respect to inclusion) in $\Sigma$. |
respect to inclusion) in $\Sigma$. |
| By definition of the set $\Sigma$, |
By definition of the set $\Sigma$, |
| this $\mathcal{M}$ must be a maximal ideal of $\mathcal{R}$ |
this $\mathcal{M}$ must be a maximal ideal of $\mathcal{R}$ |
| containing $\mathcal{I}$. |
containing $\mathcal{I}$. |
| \qed |
Q.E.D. |
|
|
| Note that the above proof never makes use of |
Note that the above proof never makes use of |
| the associativity of ring multiplication, |
the associativity of ring multiplication, |
| and the result therefore holds also in non-associative rings. |
and the result therefore holds also in non-associative rings. |
| The result cannot, however, be generalized to rings without unity. |
The result cannot, however, be generalized to rings without unity. |
|
|
| Note also that the use of the Axiom of Choice (in the form of Zorn's Lemma) |
Note also that the use of the Axiom of Choice (in the form of Zorn's Lemma) |
|
is necessary,
|
is necessary:
|
|
as there are models of ZF in which the above theorem and corollary fail.
|
there are models of ZF in which the above theorem and corollary fail. |
