A concrete example of a unital noncommutative ring without maximal two-sided ideals Whenever I say ideal in this question I'm talking about two sided-ideals.

Does there exist a concrete example of a non-commutative ring with $1$ without maximal ideals?

We know that if $R$ is a commutative ring with a multiplicative identity element then it does have maximal ideals. We know that if we drop the assumption that $R$ is unital, then there are commutative rings without maximal ideals. But what if we keep the assumption that $R$ has a multiplicative identity element but drop the assumption that it's commutative? Any interesting counter-examples?
 A: If a not necessarily commutative $R$ has a unit $1$, consider the set $\mathscr I$ of all bilateral ideals of $R$ which do not contain $1$. It is more or less obvious that the union of a totally ordered subset of $I$ is again an element of $\mathscr I$, so $\mathscr I$ has maximal elements —this is just an application of Zorn's lemma, of course.
A: Whereas Mariano Suarez-Alvarez's answer covers the abstract generalities, there has yet to be provided a "concrete example" of a unital, non-commutative ring $R$ having no (non-trivial) maximal ideals; by "non-trivial ideal" here I mean an bilateral ideal which is neither the zero ideal nor the entire ring $R$. It is clear from the thread of comments to Mariano's answer that the ideal $\{0\}$ is by all rights handled as a special case; what I think the OP math.noob is looking for is ideals which are not $R$, and not $\{0\}$.  These things being said, a nice counterexample is provided by the rings $R = M_n(\Bbb F)$, i.e. the rings of $n \times n$ matrices over fields.  It is well-known that such rings have no two-sided ideals other than themselves and $\{0\}$; so, ruling out these two "extreme" cases from the definition of maximal ideal, we have a family of unital, non-commutative rings $R$ with no maximal ideals.  For more on such rings, see here.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
