# Maximal ideal in a ring with unity.

• Jul 15th 2010, 08:11 AM
ENRIQUESTEFANINI
Maximal ideal in a ring with unity.
Hi:
Suppose that $\displaystyle R$ is an arbitrary simple ring such that $\displaystyle R^2 notequal (0)$ and such that $\displaystyle R$ contains a maximal right ideal $\displaystyle A$ (which is certainly the case if $\displaystyle R$ has a unity). Up to here the statement. Now, why does $\displaystyle R$ contain a maximal right ideal if it has a unity? I don't get this. Thanks.
• Jul 15th 2010, 10:56 AM
NonCommAlg
Quote:

Originally Posted by ENRIQUESTEFANINI
Hi:
Suppose that $\displaystyle R$ is an arbitrary simple ring such that $\displaystyle R^2 notequal (0)$ and such that $\displaystyle R$ contains a maximal right ideal $\displaystyle A$ (which is certainly the case if $\displaystyle R$ has a unity). Up to here the statement. Now, why does $\displaystyle R$ contain a maximal right ideal if it has a unity? I don't get this. Thanks.

by Zorn's lemma:

let $\displaystyle (A, \subseteq)$ be the set of all "proper" right ideals of $\displaystyle R$. (proper means $\displaystyle \neq R$.) this set is not empty because it contains $\displaystyle \{0\}.$ now if $\displaystyle \{I_k \}$ is a totally ordered collection of elements of $\displaystyle A,$ then

$\displaystyle J=\bigcup I_k$ is a right ideal of $\displaystyle R$ and $\displaystyle J \neq R,$ because if $\displaystyle J = R,$ then, since $\displaystyle 1 \in R,$ we must have $\displaystyle 1 \in I_k,$ for some $\displaystyle k,$ which is false. so, by Zorn's lemma, $\displaystyle A$ has a "maximal" element, say

$\displaystyle M.$ it is obvious that $\displaystyle M$ is a maximal right ideal of $\displaystyle R.$
• Jul 15th 2010, 10:58 AM
Chandru1
Solution
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}
• Jul 15th 2010, 01:09 PM
ENRIQUESTEFANINI
@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.
• Jul 15th 2010, 02:00 PM
chiph588@
Quote:

Originally Posted by ENRIQUESTEFANINI
@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.

Here's his post: Attachment 18209
• Jul 15th 2010, 02:00 PM
NonCommAlg
Quote:

Originally Posted by ENRIQUESTEFANINI
I tried Zorn's lemma but wasn't wise enough.

who wasn't wise enough? you, me, Zorn or Zorn's lemma? (Wink)

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:

let $\displaystyle A \neq \emptyset$ be a set with a partial ordering $\displaystyle \leq$. suppose that any chain $\displaystyle \mathcal{I} \subseteq A$ has an upper bound in $\displaystyle A$, i.e. there exists $\displaystyle a \in A$ such that $\displaystyle x \leq a,$ for all $\displaystyle x \in \mathcal{I}.$ then $\displaystyle A$ has a maximal

element, i.e. there exists $\displaystyle b \in A$ such that if $\displaystyle y \in A$ and $\displaystyle b \leq y,$ then $\displaystyle b = y.$

by the way, a subset $\displaystyle \mathcal{I}$ of $\displaystyle A$ is called a chain or totally ordered if for every $\displaystyle u,v \in \mathcal{I}$ we either have $\displaystyle u \leq v$ or $\displaystyle v \leq u.$

what i did in my solution was to take $\displaystyle A$ to be the set of all proper right ideals of $\displaystyle R$ and $\displaystyle \leq$ to be the inclusion $\displaystyle \subseteq.$ now if $\displaystyle \mathcal{I}=\{I_k\}$ is any chain in $\displaystyle A$, then $\displaystyle J=\bigcup I_k$ is also in $\displaystyle A.$ (why?)

it is obvious that for any $\displaystyle I_k \in \mathcal{I}$ we have $\displaystyle I_k \subseteq J.$ that means $\displaystyle J$ is an upper bound for the elements of $\displaystyle \mathcal{I}.$ so we can apply Zorn's lemma to get a maximal element for elements of $\displaystyle A.$

that maximal element is clearly a maximal right ideal of $\displaystyle A.$