Why is it necessary for a ring to have multiplicative identity?

Question

I have read earlier that in a ring $(R,+,.)$ the following needs to hold:

$(R,+)$ is an abelian group
multiplication is associative and closed
left and right distribution laws hold.

However, I recently came across the fact that every ring has to have a multiplicative identity. Can anyone please clarify this? Is it needed for the ring to have a multiplicative identity?

(In fact it was mentioned that it is one of the reasons why $ker(f)$ is not a subring where $f$ is a ring homomorphism as the additive identity and the multiplicative identity are not usually in the same subset.)

Further in 2 different places I have noticed that there is a difference on whether the mapping $f(1) \to 1$ is a necessary condition for $f$ to be a ring homomorphism. I think this is also related to my doubt as to whether the multiplicative identity is in fact a necessary condition for defining a ring.

A ring does not need to have a multiplicative identity. Probably what your source meant was that in this book or perhaps in this chapter or something like that, all rings will have a multiplicative identity. — David, Sep 02 '15 at 06:42
a few authors define a ring to have a multiplicative identity but most do not. The latter use the name "unit ring" or "ring with identity" to distinguish between the 2 — Alessandro Codenotti, Sep 02 '15 at 06:43
Definitions in mathematics (especially in algebra) are usually made to capture some observed notion, so that we may study such things abstractly. The multiplicative identity of a ring is definitely something important that occurs in many systems and so yes it deserves to be a part of the definition. — fretty, Sep 02 '15 at 07:12
Somewhat related post on meta: Does anyone believe that there are rings without unit elements? — Martin Sleziak, Sep 02 '15 at 09:23
Aside: $(R,+)$ is a semigroup, not a group. However, this situation has a weird property that any homomorphism between "semigroups with inverses and identities" actually turns out to preserve inverses and identities, so "semigroup with inverses and identities" turns out to be the same concept as "group". Unfortunately, many introductions have the weird feature that they say "group" but proceed to study "semigroups with inverses and identities", and so people start out their algebra career with a misunderstanding that carries over to situations where the difference is actually relevant. :( — , Sep 02 '15 at 17:55
Bjorn Poonen from MIT has a write-up explaining why having 1 makes sense: http://www-math.mit.edu/~poonen/papers/ring.pdf — Atticus Christensen, Sep 02 '15 at 17:58
@Hurkyl: I don't understand the distinction you're making. What is the difference between a "semigroup with inverses and identity" and a group? — Jim, Sep 02 '15 at 18:19
@Jim: A priori, one would expect a lot. A homomorphism between groups has to satisfy, for example, $f(1) = 1$, because the constant $1$ is part of the structure. The same for homomorphisms between monoids. A homomorphism between semigroups with identity, however, isn't required to map the identity to the identity (and there are examples where it doesn't) -- having an identity isn't part of the algebra structure, it's just a property that some semigroups have. It's a weird coincidence that homomorphisms between semigroups with both an identity and inverses actually preserve them. — , Sep 02 '15 at 18:53
@Hurkyl: Homomorphisms are not a priori part of the definition of the object. The object is distinct from the category of such objects. — Jim, Sep 02 '15 at 23:00
@Jim: A lot of communication is through established convention, and the principles behind the convention are often even a good guide for doing new mathematics. You can ignore it, but it's at your own peril. (also, insert generic criticism of being overly focused on details of set-theoretic constructions) — , Sep 02 '15 at 23:23
@Hurkyl: That's poor form. It is a fact that a ring under its addition operation does indeed form a group and the established convention is that the notation $(R, +)$ stands for this group. I am fully aware of the principal behind focusing on homomorphisms, I just expect you to communicate your point without making false statements that are likely to confuse the type of student who would ask whether rings contain a $1$ or not. :) — Jim, Sep 03 '15 at 03:27

Jim · Accepted Answer · 2022-05-26T12:27:19.280

Many authors take the existence of $1$ as part of the definition of a ring. In fact, I would disagree with Alessandro's comment and claim that most authors take the existence of $1$ to be part of the definition of a ring. There is another object, often called a rng (pronounced "rung"), which is defined by taking all the axioms that define a ring except you don't require there to be a $1$.

Rng's are useful in of themselves, for example functions with compact support over a non-compact space do not form a ring, they form a rng. But there is also a theorem that states that every rng is isomorphic to an ideal in some ring. So studying rings and their ideals is sufficient, and this is why it is so popular to include the existence of $1$ as one of the axioms of a ring.

So to summarize, there isn't really a reason why it's necessary for rings to have a $1$, it certainly does not follow from the other axioms. It's just a choice of terminology: Do you say rings have a $1$ and if they don't have a $1$ call them rngs, or do you say rings don't need a $1$ and when they do have it call them rings with unity?

For completeness, note that "a rng with a multiplicative identity" is still a different concept from "a ring": the difference being what is required of a homomorphism. A homomorphism between rngs with multiplicative identity is not required to map the identity to the identity, but a homomorphism between rings is required to satisfy $f(1) = 1$. As an example that this matters, if $R$ is a ring, the map $R \to R \times R: x \mapsto (x,0)$ is a rng homomorphism, but not a ring homomorphism. — , Sep 02 '15 at 18:03

score 8 · Answer 2 · answered Sep 02 '15 at 07:31

I'm currently teaching out of the 4th edition of Stewart's Galois Theory textbook. Stewart defines a ring to be what other authors might call a commutative ring with unity. The reason is simple: in this book, there is not much call for noncommutative rings, nor for rings without unity, and it gets old writing "commutative ring with unity" over and over, when that's the only kind of ring you need.

Stewart then defines a subring of a ring to be a subset of a ring closed under addition, subtraction, and multiplication. Note that a subring doesn't have to have unity – a subring doesn't have to be a ring, in this book. Well, it's a convention. As long as it's explained to the reader, and the author is consistent with it, I think it's fine.

Then he goes and spoils it by asking, in Exercise 16.2, whether the rings $\bf Z$ and $2\bf Z$ are isomorphic.

Imo requiring rings to have $1$ but subrings not to have $1$ is a really bad convention, because it kind of messes with the whole category theoretic / universal algebraic perspective that sub-$\mathsf{T}$-algebras are subsets on which there exists a $\mathsf{T}$-algebra structure such that the inclusion is a homomorphism. — goblin GONE, Sep 02 '15 at 10:26
@goblin: It depends on whether your definition is "there exists a multiplicative identity" or "there is a constant $1$ that is a multiplicative identity". (with the former, I think the structure so defined isn't even a variety of universal algebras!) — , Sep 02 '15 at 18:07

score 2 · Answer 3 · answered Sep 02 '15 at 21:16

Lang and a few other authors use "Ring" to mean "Ring with unity" and say "Ring without unity" for what I'd call a Ring.

This is because Rings with unity are by far the most interesting. There are few things you can say of/do to a ring (or ring without unity to you) but there are MANY MANY things you can do with rings with unity (rings to you)

score 0 · Answer 4 · answered Sep 02 '15 at 08:04

I just thought I would expand on Jim's answer and provide a source that discusses this very question about whether a ring should assume the existence of an identity or not in a bit more detail. (This is definitely one of the better puns in algebra that I have come across. Ring without the $i$ for no identity.)

There is a chapter by D.D. Anderson at the beginning of the book on multiplicative ideal theory about rngs. You can see the introduction in the look inside which talks a bit about the history of this. http://link.springer.com/chapter/10.1007/978-0-387-36717-0_1#page-1

Why is it necessary for a ring to have multiplicative identity?

4 Answers4