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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?

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  • $\begingroup$ An example would be an irrational rotation algebra, which is a simple non-commutative unital $ C^{*} $-algebra (hence a simple non-commutative unital ring). $\endgroup$ – Transcendental Sep 3 '17 at 0:38
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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.

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  • $\begingroup$ The problem when you do not have a $1$ is that une union of a totally ordered set of proper ideals need not be a proper ideal. $\endgroup$ – Mariano Suárez-Álvarez Oct 25 '14 at 7:05
  • $\begingroup$ Yes, but I'm looking for an example of a non-commutative ring that DOES have a multiplicative identity element but yet, it doesn't have any two-sided maximal ideals. $\endgroup$ – math.n00b Oct 25 '14 at 7:07
  • $\begingroup$ Well, what i wrote shows very clearly that there are none! $\endgroup$ – Mariano Suárez-Álvarez Oct 25 '14 at 7:09
  • $\begingroup$ Yes. Sorry. What if $\mathscr I=\emptyset$? How are you going to prove that there exists a two-sided ideal which does not contain $1$ in any non-commutative ring? $\endgroup$ – math.n00b Oct 25 '14 at 7:13
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    $\begingroup$ There is always the zero ideal, which is maximal if there are no other ideals... $\endgroup$ – Mariano Suárez-Álvarez Oct 25 '14 at 7:13
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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!!!

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  • $\begingroup$ Do you know an example of a non-simple non-commutative ring? $\endgroup$ – Sigur Oct 1 '15 at 13:14
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    $\begingroup$ @Sigur: Being a functional analyst, I can tell you right away that for any infinite-dimensional Hilbert space $ \mathcal{H} $, the ring $ \mathscr{B}(\mathcal{H}) $ of bounded linear operators on $ \mathcal{H} $ is a non-simple non-commutative ring. It contains the ring $ \mathscr{K}(\mathcal{H}) $ of compact linear operators as a proper two-sided ideal. $\endgroup$ – Transcendental Sep 3 '17 at 0:29

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