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?

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

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

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

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

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

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

And, since $\displaystyle r$ was an arbitrary element in $\displaystyle R$, then $\displaystyle R = L$. Was that sound reasoning?

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 $\displaystyle rL \in L$ should be $\displaystyle rR \subset L$ (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.