13
$\begingroup$

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

  1. $(R,+)$ is an abelian group
  2. multiplication is associative and closed
  3. 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.

$\endgroup$
  • 7
    $\begingroup$ 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. $\endgroup$ – David Sep 2 '15 at 6:42
  • 9
    $\begingroup$ 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 $\endgroup$ – Alessandro Codenotti Sep 2 '15 at 6:43
  • 1
    $\begingroup$ 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. $\endgroup$ – fretty Sep 2 '15 at 7:12
  • 4
    $\begingroup$ Somewhat related post on meta: Does anyone believe that there are rings without unit elements? $\endgroup$ – Martin Sleziak Sep 2 '15 at 9:23
  • 2
    $\begingroup$ Bjorn Poonen from MIT has a write-up explaining why having 1 makes sense: www-math.mit.edu/~poonen/papers/ring.pdf $\endgroup$ – user45150 Sep 2 '15 at 17:58
29
$\begingroup$

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?

$\endgroup$
  • 8
    $\begingroup$ 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. $\endgroup$ – Hurkyl Sep 2 '15 at 18:03
9
$\begingroup$

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.

$\endgroup$
  • 9
    $\begingroup$ 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. $\endgroup$ – goblin Sep 2 '15 at 10:26
  • 1
    $\begingroup$ @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!) $\endgroup$ – Hurkyl Sep 2 '15 at 18:07
1
$\begingroup$

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)

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.