TITLE: Where does the proof for commutative rings break down in the non-commutative ring when showing only two ideals implies the ring is a field?
QUESTION [9 upvotes]: We know in a commutative ring, if the only ideals are trivial and the whole ring, then the ring is a field, which is proved by every ideal is contained in a maximal ideal, which is proved by Zorn's lemma.
But in non-commutative case, we can find counter-examples that the only ideals are trivial and the whole ring, but there are elements in the ring that do not have multiplicative inverse. 
My question is: where does the proof for commutative rings break down in the non-commutative ring? 
I think the only possibility is some ideals may not be contained in a maximal ideal, is that true?

REPLY [4 votes]: For a counterexample take the first Weyl algebra $\mathbf C[p,q]$ where $pq-qp=1$, which is a subalgebra of ${\rm End}_{\mathbf C}\mathbf C[X]$. Here $p$ acts on polynomials by $p f(X)=f'(X)$ and $q$ acts by $qf(X)=Xf(X)$. It should be checked $q,p$ are not invertible, yet $pq-qp=1$ means $(p)=(q)=1$. The point is that in noncommutative cases, even principal ideals are not that simple: $(a)$ is equal to all possible sums $\sum\limits_{i=1}^n a_i ab_i$ and a relation of the form $\sum\limits_{i=1}^n a_i ab_i=1$ doesn't imply invertibility.