Are there interesting rings without unity?

Question

There are several introductory textbooks which define a ring without any reference to a unity. However, nearly all of the rings one encounters in various branches of mathematics are endowed with a $1$. Thus, I'm wondering if some of you could prove me wrong and show me some examples of rings without unity arising naturally in a mathematical theory.

(Two-sided ideals don't count as an example, because normally, they aren't considered as rings in their own right, but as modules or equivalence classes, allowing you to pass to a quotient ring (with $1$).)

Lots in function rings, for example ring of continuous functions with compact support. — André Nicolas, May 05 '13 at 17:41
@AndréNicolas +1, though that is of course a two-sided ideal of the ring of continuous functions. — Hagen von Eitzen, May 05 '13 at 17:51
@HagenvonEitzen: I interpreted the question as asking about rings without unity that are studied for their own sake, whatever that may mean. Then there are the familiar function rings, and relatives, of beginning functional analysis. — André Nicolas, May 05 '13 at 17:57
The ring of compact operators on the infinite dimensional normed space. — Norbert, May 05 '13 at 18:32
So, basic question - can a "ring without unity" - controversial idea since some would define a ring as including unity - always be embedded in a larger ring with unity of which it is an ideal? And are there cases where the ideal is evidently more interesting than the larger ring? — Mark Bennet, May 05 '13 at 18:47
@Norbert: this is, of course, a two-sided ideal in the ring of bounded operators. But as in my answer below I don't see this as a strike against the example. — Qiaochu Yuan, May 05 '13 at 22:47
Various examples are discussed in Non-unital rings: a few examples. See also Does anyone believe that there are rings without unit elements? — Martin, May 05 '13 at 22:53
The ring of integer multiples of any given integer $n > 1$ which is interesting because its quotient with the integers is the ring of integers modulo $n$. — Prime Mover, Apr 06 '21 at 20:49

Qiaochu Yuan · Answer 1 · 2024-07-04T04:50:52.120

Edit, 7/3/24: Apparently my use of the notation $R \oplus \mathbb{Z}$ has been causing confusion, so to be clear: this notation only names the underlying abelian group. The multiplication is the unique multiplication restricting to the original multiplication on $R$ and making $1 \in \mathbb{Z}$ the new unit; this is the Dorroh extension, which is not a name I knew in 2013.

Two-sided ideals don't count as an example

Every nonunital ring $R$ is a two-sided ideal in its "unitization" $R \oplus \mathbb{Z}$ (with an appropriately defined multiplication). Unitization is the left adjoint to the forgetful functor from rings to nonunital rings. More generally, if $R$ is a $k$-algebra, there is a unitization $R \oplus k$ as a $k$-algebra. This is a common construction in, for example, the study of C-algebras (where many naturally-occurring C-algebras such as the algebra of compact operators or group C*-algebras of some locally compact groups are nonunital); it is an algebraic analogue of passing to the one-point compactification (the analogy is via Gelfand-Naimark).

The theory of C*-algebras in particular shows that it's profitable to prove theorems about nonunital rings because you can then apply those theorems to two-sided ideals.

I wrote down some examples and comments in this blog post, which in particular shows that the category of non-unital rings is equivalent to the category of augmented rings (rings together with a morphism $R \to \mathbb{Z}$), and this is a natural category to study; on the geometric side, augmented commutative rings are the rings of functions on "pointed affine schemes."

score 5 · Answer 2 · answered May 05 '13 at 20:15

The path algebra (over, say, $\mathbb{Z}$) of a quiver with infinitely many vertices is a ring without unit. A more "advanced" example is Lusztig's modified enveloping algebra of a Lie algebra. This algebra is important in representation theory and categorification.

score 5 · Accepted Answer · answered May 05 '13 at 23:10

I once thought it was the case that in number theory basically all the relevant rings were commutative with identity. But this is not so! In the theory of automorphic forms and representations, the Hecke algebras which act on representations of various groups (the groups of points of reductive groups over local and global fields, valued in the field, or in the global case, in its completions or its adele ring) are generally non-commutative and do not have multiplicative identities (although there are interesting subalgebras, e.g. the spherical Hecke algebras, which are commutative and do have identities). In the local case, the rings are convolution algebras of locally profinite (meaning totally disconnected and locally compact) groups. Specifically, for $G$ locally profinite, the space $C_c^\infty(G)$ of smooth (meaning locally constant) and compactly supported complex-valued functions is a ring under convolution (for a choice of Haar measure on $G$) and acts on smooth representations of $G$. In fact, smooth representations of $G$ are literally the same as smooth representations of $C_c^\infty(G)$, so it plays the role of the group algebra $\mathbf{C}[G]$ for $G$ finite. But unlike the group algebra, this ring won't usually have a multiplicative identity.

Are there interesting rings without unity?

3 Answers3

Linked