# Ideals and Containment.

• Oct 20th 2012, 12:35 AM
anguished
Ideals and Containment.
Let $R$ be a ring with a proper left ideal $M$ such that, for any $r \in R$, either $r \in M$ or $r$ is a unit.

(a) Prove that if $L$ is a left ideal which contains $M$ then either $L = R$ or $L = M$.

(b) Prove that every proper left ideal of $R$ is contained in $M$.

For (a) we have $M \subseteq L$ so we need to show that we can take an arbitrary $x \in L$ and rearrange it so it looks like an element of $M$. So if we take $x \in L$ then $x = rl$ for some $r \in R$ and $l \in L$ (because L is a left ideal). I'm not too sure how to proceed from here... how can we relate $L$ with $M$?

Thanks,
• Oct 20th 2012, 03:19 AM
Bingk
Re: Ideals and Containment.
If L = M, then we're done. So, assume L properly contains M. That means there's an element x that's in L but not in M.
From the given, what can you say about x?
What can you say about about an ideal that contains a unit?
• Oct 20th 2012, 03:03 PM
anguished
Re: Ideals and Containment.
Quote:

Originally Posted by Bingk
If L = M, then we're done. So, assume L properly contains M. That means there's an element x that's in L but not in M.
From the given, what can you say about x?
What can you say about about an ideal that contains a unit?

Suppose $x \in L$ then that means $x = rl$ for some $r \in R$ and $l \in L$.

If $r \in M$ then $rl \in L$ but doesn't that just show that $M \subseteq L$?

I'm not sure what I can infer beyond that...

Now suppose $r$ is a unit in $R$. Then $r^{-1} \in R$ also since $R$ is a ring.

Now $rL \in L$ since $L$ is a left ideal. Therefore $r.r^{-1} = 1 \in L$.

But for any $r \in R$, since we have $1 \in L$, then $r = r.1 \in L$.

And, since $r$ was an arbitrary element in $R$, then $R = L$. Was that sound reasoning?
• Oct 22nd 2012, 05:24 AM
Bingk
Re: Ideals and Containment.
From the given, r is in M or r is a unit, but not both, right?

So, from where I left off, that means x is a unit, since it's not in M.

After, that yes, your reasoning is sound :) ... except, I think $rL \in L$ should be $rR \subset L$ (with some re-wording).

Basic idea:
x is a unit so it has a multiplicative inverse (denoted 1/x, which is in R).
Since x is in L and L is an ideal, then xr is in L for all r in R.
That means x*(1/x) = 1 is in L (since 1/x is in R).
So, 1r = r is in L for all r in R.
Which gives L = R.