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?
Let be a ring with a proper left ideal such that, for any , either or is a unit.
(a) Prove that if is a left ideal which contains then either or .
(b) Prove that every proper left ideal of is contained in .
For (a) we have so we need to show that we can take an arbitrary and rearrange it so it looks like an element of . So if we take then for some and (because L is a left ideal). I'm not too sure how to proceed from here... how can we relate with ?
Suppose then that means for some and .
If then but doesn't that just show that ?
I'm not sure what I can infer beyond that...
Now suppose is a unit in . Then also since is a ring.
Now since is a left ideal. Therefore .
But for any , since we have , then .
And, since was an arbitrary element in , then . Was that sound reasoning?
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 should be (with some re-wording).
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.