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Hi:
Suppose thatis an arbitrary simple ring such that
and such that
(which is certainly the case if
by Zorn's lemma:Quote:
Originally Posted by ENRIQUESTEFANINI
Hi:
Suppose that
letbe the set of all "proper" right ideals of
.) this set is not empty because it contains
now if
is a totally ordered collection of elements of
then
is a right ideal of
because if
then, since
we must have
for some
which is false. so, by Zorn's lemma,
it is obvious that
is a maximal right ideal of
HI--
Its one of the famous theorems .
for a proof please run this .tex commands
\documentclass[10pt,amssymb]{revtex4}
\usepackage{amsmath}
\begin{document}
\title{ \huge\mdseries Maximal Ideals in Rings with Unity}
\maketitle
\large
\begin{itemize}
\item[\textbf{Theorem}:]{ If $R$ is a ring with $1$ and $I$ is a left ideal of $R$ such that $I \neq R$, then there is a maximal ideal $M$ of the same kind as $I$ such that $I \subseteq M$.}
\item[\textbf{Proof:}]{ Let $I$ be a left ideal of $R$. Consider the family $\mathcal{F}=\mathcal{F}_{1}$ of all ideals in $R$ containing $I$ except the unit ideal $R$, i.e $$\mathcal{F}=\mathcal{F}_{1}=\{ \mathcal{J} \ | \ \text{left ideal in} \ R, J \subseteq I, \ J \neq R\}$$
The theorem is equivalent to showing that $\mathcal{F}$ has a maximal element with set inclusion as the partial order. To apply \textbf{ Zorn's Lemma} to the family $\mathcal{F}$, we have to verify that totally ordered subset $\mathcal{T}$ of $\mathcal{F}$ has an upper bound in $\mathcal{F}$. Given such a $\mathcal{T}$, let $\displaystyle T_{0}= \bigcup\limits_{T \in \mathcal{T}} T$. We will show that $T_{0} \in \mathcal{F}$ (so that $T_{0}$ is obviously an upper bound in for $\mathcal{T}$.) We have $T_{0} \supseteq I$.
\item{ $T_{0}$ is a left ideal of $R$}
\item{ $T_{0} \neq R$}
For if $T_{0}=R$ then $1 \in T_{0}$. Hence $1 \in T$ for some $T \in \mathcal{T}$. But then this will force $T=R$, a contradiction.
Now by Zorn's lemma, $\mathcal{F}$ has a maximal element, say $M$. Since $M \in \mathcal{F}$, we have $M \neq R$, $M \supseteq I$ and $M$ is a left ideal.
\item{ $M$ is a maximal left ideal in $R$.}
For suppose $J$ is a left ideal such that $M \subseteq J \subseteq R$, if $J \neq R$, then $J \in \mathcal{F}$ which implies $M=J$.}
\end{itemize}
\end{document}
@chandru1: I utterly regret not being able to read your proof for the time being, but I'll ask for help on how to do it.
@NonCommAlg: I tried Zorn's lemma but wasn't wise enough. Once again, thanks and good luck with your Ph.D thesis.
Here's his post: Attachment 18209Quote:
Originally Posted by ENRIQUESTEFANINI
who wasn't wise enough? you, me, Zorn or Zorn's lemma? (Wink)Quote:
Originally Posted by ENRIQUESTEFANINI
anywho, i don't know which part of the solution you didn't understand. if it was Zorn's lemma, here is what exatly Zorn's lemma says:
letbe a set with a partial ordering
. suppose that any chain
has an upper bound in
such that
for all
then
element, i.e. there existssuch that if
and
then
by the way, a subsetof
we either have
or
what i did in my solution was to taketo be the inclusion
now if
is any chain in
(why?)
it is obvious that for anywe have
that means
is an upper bound for the elements of
so we can apply Zorn's lemma to get a maximal element for elements of
that maximal element is clearly a maximal right ideal of
