# 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?

• An example would be an irrational rotation algebra, which is a simple non-commutative unital $C^{*}$-algebra (hence a simple non-commutative unital ring). Commented Sep 3, 2017 at 0:38

## 2 Answers

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.

• 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. Commented Oct 25, 2014 at 7:05
• 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. Commented Oct 25, 2014 at 7:07
• Well, what i wrote shows very clearly that there are none! Commented Oct 25, 2014 at 7:09
• 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? Commented Oct 25, 2014 at 7:13
• There is always the zero ideal, which is maximal if there are no other ideals... Commented Oct 25, 2014 at 7:13

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!!!

• Do you know an example of a non-simple non-commutative ring? Commented Oct 1, 2015 at 13:14
• @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. Commented Sep 3, 2017 at 0:29