# An example of a non-commutative ring with multiplicative identity 1 in which the only (two sided) ideals are 0 and the whole ring

Is there any example of a non-commutative ring with 1 in which the only ideals are (0) and the whole ring, yet the elements do not have multiplicative inverses?

I thought an example of all 2x2 matrices with entries from any fields (like R, the real number), and it is clearly that not all of them are invertible, thus do not have multiplicative inverses, but how do I show the ideals are only (0) and the whole ring?

Thank you so much!

• You are asking for simple rings which are not necessarily division rings. A big and interesting subclass of those are the central simple algebras ( en.m.wikipedia.org/wiki/Central_simple_algebra), of which matrix algebras $M_n(k)$ over fields $k$ are indeed a standard example. Commented Nov 28, 2017 at 6:25
• I added the "ring-theory" tag to your post. Commented Nov 28, 2017 at 6:39
• But I don't know how to show the ideals to be only (0) and the whole ring.. I get basic idea that if for any 2x2 nonzero matrix in an ideal, we may eventually get the identity matrix in the ideal and therefore the ideal is the whole ring, but I don't know how to show it.. @TorstenSchoeneberg Commented Nov 28, 2017 at 6:52
• DaRT search yields seven at present, four of which haven't been mentioned yet, I think. Commented Nov 28, 2017 at 12:24
• The answers say it all now. A last thing to add is that all this relies on the distinction between two-sided and one-sided ideals in the noncommutative setting. A ring which has neither proper left- nor proper right-ideals would indeed be a division ring. In particular, "artinian" in Qiaochu Yuan's answer means "left-artinian", and it is fascinating to see that, e.g., matrix algebras over an infinite field have infinitely many distinct left ideals (although they fall into finitely many isoclasses), and e.g. the Weyl algebra even has an infinite descending chain of left ideals. Commented Nov 28, 2017 at 20:39

These rings are called simple rings. As Torsten says in the comments, a nice class of examples which are not division rings is the matrix algebras $$M_n(k)$$. More generally, by the Artin-Wedderburn theorem the artinian simple rings are precisely the rings of the form $$M_n(D)$$ where $$D$$ is a division algebra. If the center of $$D$$ is $$k$$ then these are central simple algebras over $$k$$ and are classified by the Brauer group of $$k$$.

To prove that $$M_n(D)$$ is simple it's cleaner to prove a more general result:

Claim: The two-sided ideals of $$M_n(R)$$, for any ring $$R$$, are of the form $$M_n(I)$$ where $$I$$ is a two-sided ideal of $$R$$.

Corollary: $$M_n(R)$$ is simple iff $$R$$ is simple.

Proof. Let $$X \in M_n(R)$$ be any element. The ideal generated by $$X$$ consists of linear combinations of elements of the form

$$e_{ij} X e_{kl}$$

where $$1 \le i, j, k, l \le n$$; this matrix has only a single nonzero component, namely the $$il$$ entry, whose value is $$X_{jk}$$. So by picking $$i, j, k, l$$ appropriately we see that we can arrange for any particular component of $$X$$ to end up in any other component; in other words, the ideal generated by $$X$$ is $$M_n(I)$$ where $$I$$ is the ideal of $$R$$ generated by the components of $$X$$. The desired result follows upon taking sums of ideals, which gives more generally that the ideal generated by any collection of matrices is $$M_n(I)$$ where $$I$$ is the ideal of $$R$$ generated by their components. $$\Box$$

I said above that the artinian simple rings are the rings of the form $$M_n(D)$$, so let's close with a non-artinian example. Maybe the most famous non-artinian simple rings are the Weyl algebras, of which the one-variable version can be written

$$k[x, \partial]/(\partial x - x \partial = 1)$$

where $$k$$ is a field of characteristic zero; this can be thought of as the algebra of differential operators on $$k[x]$$, with $$\partial$$ acting by differentiation by $$x$$.

Claim: With the above hypotheses, the Weyl algebra is simple.

Proof. Let $$f = \sum f_{ij} x^i \partial^j$$ be an element of the Weyl algebra; we will show directly that if $$f$$ is nonzero then the ideal it generates is the entire Weyl algebra. First, observe that the monomials $$x^i \partial^j$$ form a basis of the Weyl algebra; there are various ways to prove this, it is a PBW-type result. So $$f = 0$$ iff $$f_{ij} = 0$$ for all $$i, j$$.

The defining relation of the Weyl algebra can be written $$[\partial, x] = 1$$, where $$[a, b] = ab - ba$$ is the commutator bracket. Now, the commutator bracket $$[\partial, -]$$ is always a derivation, so it follows that

$$[\partial, x^i] = \partial x^i - x^i \partial = ix^{i-1}$$

while $$[\partial, \partial] = 0$$ and hence $$[\partial, \partial^j] = 0$$. This gives

$$[\partial, x^i \partial^j] = ix^{i-1} \partial^j$$

and hence

$$[\partial, f] = \sum f_{ij} ix^{i-1} \partial^j.$$

That is, computing the commutator by $$\partial$$ has the effect of differentiating the polynomial part of $$f$$ (and note that $$[\partial, f] = \partial f - f \partial$$ lies in the ideal generated by $$f$$). So we can repeatedly apply $$[\partial, -]$$ to $$f$$ until all of its polynomial parts vanish except the ones of highest degree; hence we can assume WLOG that $$f$$ in fact has the form

$$f = \sum f_j \partial^j$$

where at least one $$f_j$$ is zero (this is where we need both the assumption that the $$x^i \partial^j$$ form a basis and that $$k$$ has characteristic zero). At this point we can now instead apply the derivation $$[-, x]$$, which satisfies $$[\partial, x] = 1$$ and hence by induction

$$[\partial^j, x] = j\partial^{j-1}$$

which gives

$$[f, x] = \sum f_j j \partial^{j-1}.$$

So we can again repeatedly "differentiate" until $$f$$ is a nonzero constant, which clearly generates the entire Weyl algebra. $$\Box$$

• "The ideal generated by X consists of elements of the form ..." -- I think it would be better for understanding if you edit that to "... contains, in particular, the elements of the form ..." Otherwise, great answer. Commented Nov 28, 2017 at 20:17
• What I meant to say was "consists of linear combinations of elements of the form..." I need this stronger statement and not just the containment. Commented Nov 28, 2017 at 20:52

Matrix is a very important tool non-commutative ring. Following theorem can help to us for your question.
Theorem: Let R be a ring, $M_{n}(R)$ be the ring of all $n$ x $n$ matrices over $R$.
$I\unlhd M_{n}(R) \iff$ $I=M_{n}(U)$ for a uniquely determined ideal $U$ of $R$.
In particular, $R$ is simple $\iff$ $M_{n}(R)$ is simple.
Consequently, $M_{n}(D)$ is a simple for any division ring $D$.
Hence, you can find a lot of examples via above statements.

• What does $\unlhd$ mean in this context? Commented Nov 28, 2017 at 7:49
• Two sided ideal. Commented Nov 28, 2017 at 7:49