28
$\begingroup$

Reading the basic definition of rings, I wondered if there are samples of rings whose multiplicative identity is not the number 1 or number 1-based (for instance the identity matrix is 1-based).

E.g. for $\Bbb Z$, if the definition of multiplication is modified (creating a non-standard algebra), could the multiplicative identity of the ring be another number, or the definition of multiplication must be "canonical" and must not be modified?

Is there a ring (currently in use for some field of Mathematics) sample of such non-1-based multiplicative identity?

I am learning by myself so I apologize if the question does not make much sense, thank you!

Update 2015/05/11: I will include some links to those wiki pages that were useful to understand the concepts written in the answers.

Idempotent Element

Abelian Group

Homomorphism

Identity Element

Subring

$\endgroup$
  • 3
    $\begingroup$ Well, usually the multiplicative identity in a ring is always called 1. I guess you specifically mean the number 1. In that case, the answer is indeed yes. $\endgroup$ – Qudit May 3 '15 at 4:50
  • 9
    $\begingroup$ To me, a multiplicative identity is a "generalization" (for lack of better term) of $1$, so it is necessarily "$1$-based" to some extent, e.g. the identity matrix. $\endgroup$ – White Shirt May 3 '15 at 4:51
  • 3
    $\begingroup$ One more point: we insist that $1 \neq 0$ or things are not interesting. Otherwise we just call the multiplicative identity "$1$" if you like. $\endgroup$ – MonkeyKing May 3 '15 at 4:59
  • 3
    $\begingroup$ Thank you everybody, I see that my point of view was very basic. I will keep the question open for a while to gather different explanations. All of them are worthy! $\endgroup$ – iadvd May 3 '15 at 5:01
  • 3
    $\begingroup$ @Qudit: I may have misinterpreted, but I think that the OP means that they won't accept an answer yet. This is just as well, because better examples may be coming. The choice of phrase "keep open" may be a bit misleading. This is a bit unfortunate in the sense that "closing a question" on our site has a technical meaning other than that the asker considers their question "(a) closed (case)", and "open" is synonymous to "no longer a mystery" :-) $\endgroup$ – Jyrki Lahtonen May 3 '15 at 17:21
60
$\begingroup$

Consider $S = \{0, 2, 4, 6, 8\}$ with usual addition and multiplication modulo $10$. Then the identity element is $6$.

$\endgroup$
  • 45
    $\begingroup$ It's perhaps worth emphasizing that this is isomorphic to the familiar ring $\Bbb Z_5$ (simply because this is the only ring of that order up to isomorphism). This illustrates that "$1$-based" isn't a rigorous concept. $\endgroup$ – Travis May 3 '15 at 4:56
  • 16
    $\begingroup$ It may also be worth pointing out that this can also be viewed as coming from the isomorphism of rings between $\Bbb{Z}_2\times\Bbb{Z}_5$ and $\Bbb{Z}_{10}$. The obvious idempotents $(1,0)$ and $(0,1)$ of the product ring are mapped to the cosets of $5$ and $6$ respectively. Whenever $R$ is a ring and $e$ an idempotent, the set $eRe$ becomes a ring with $e$ as a neutral element. $\endgroup$ – Jyrki Lahtonen May 3 '15 at 17:16
37
$\begingroup$

Consider the set $R$ of $2\times2$ matrices of the form $$ R=\left\{\left(\begin{array}{cc}x&x\\x&x\end{array}\right)\big\vert\,x\in\Bbb{R}\right\}. $$ The operations are the usual matrix multiplication and addition. I leave it to you to verify that $R$ is a ring, and that the matrix you get by setting $x=1/2$ is the multiplicative neutral element of $R$.

$\endgroup$
  • $\begingroup$ Good one, thank you for the example. $\endgroup$ – iadvd May 3 '15 at 6:48
  • 9
    $\begingroup$ Similar examples abound. Using an idempotent of a bigger ring as the neutral element is the common idea. $\endgroup$ – Jyrki Lahtonen May 3 '15 at 9:11
  • $\begingroup$ I'm getting a bit of deja vu vibe here. I couldn't find it, but I think this ring has shown up on the site earlier. Where? Anyone? $\endgroup$ – Jyrki Lahtonen May 7 '15 at 4:53
  • 1
    $\begingroup$ @Jyrky Lahtonen, sorry I tried to find it but no luck! I looked for "neutral element" and matrices-related questions... I found some other questions closer to the samples of the present topic but just that... i.e. math.stackexchange.com/questions/1222451/… $\endgroup$ – iadvd May 7 '15 at 5:47
19
$\begingroup$

Let $\Omega$ be an arbitrary non-empty set. Let $\mathfrak{P}(\Omega)$ denote the power set (set of all subsets) of $\Omega$.

The symmetric difference $A\Delta B$ for $A,B\subset\Omega$ is defined as follows:

$\qquad A\Delta B \equiv (A \backslash B) \cup (B \backslash A$).

Now $(\mathfrak{P}(\Omega), \Delta, \emptyset)$ is an abelian group and $(\mathfrak{P}(\Omega), \Delta, \emptyset, \cap, \Omega)$ is a commutative ring (with $\Delta$ as addition, $\emptyset$ as zero, $\cap$ as multiplication, and $\Omega$ as the unit of multiplication).

There are no "number-like" entities involved at all, everything is build using only the most basic set theory. The unit of multiplication is just a set. The set $\Omega$ could be anything. Families of subsets with certain properties that admit the above construction are used in measure theory and are called rings. However, the fact that these families are rings in the algebraic sense is not very useful, it's more like just a curious coincidence.

Moreover, the resulting ring is essentially the same as $(\mathbb{Z}/2\mathbb{Z})^{\Omega}$ with pointwise operations, so that $\Omega$ corresponds to the constant function $\Omega \mapsto 1$, so we get "the number $1$" (mod 2) again. The question by itself is not really meaningful, because one can always take any ring $\mathcal{R}$, call its unit of multiplication "1", and declare that $\mathcal{R}$ is just yet another kind of "numbers", so that your multiplicative unit becomes "number 1-based".

$\endgroup$
  • 1
    $\begingroup$ thank you, I will need some time and make some extra readings to understand your explanation :) $\endgroup$ – iadvd May 3 '15 at 11:18
  • $\begingroup$ Have I messed up some details? I would be thankful for a comment on the down-vote, then I could correct the mistake if there is one. $\endgroup$ – Andrey Tyukin May 3 '15 at 20:26
  • $\begingroup$ I am not the downvoter, but I assume $A \backslash B$ means those elements of $A$ not in $B$. $\endgroup$ – Mark Hurd May 5 '15 at 5:24
  • 1
    $\begingroup$ By ISO 31-11 the notation $A\backslash B$ stands for "The set of elements which belong to A but not to B.", so yes, your assumption is correct... So what? I don't really know what your point is, but I'll just add some parentheses around the relative complements, the operator precedence between $\cap$ and $\backslash$ could be unclear. $\endgroup$ – Andrey Tyukin May 5 '15 at 11:19
  • $\begingroup$ Your ISO reference shows that notation is (meant to be) clear enough without explanation. My only point was almost every other notation was defined/explained within the answer. $\endgroup$ – Mark Hurd May 7 '15 at 6:52
14
$\begingroup$

Let $A$ be a Abelian group and let $R$ be the set of all homomorphisms from $A$ to itself. Then $R$ is a ring under the operations of pointwise addition and function composition and the multiplicative identity is the identity mapping.

$\endgroup$
  • 3
    $\begingroup$ You can't add two ring homomorphisms to get a ring homomorphism. This works for group homomorphisms... $\endgroup$ – Will Sawin May 3 '15 at 15:39
  • $\begingroup$ Sorry, you are right. I edited it. $\endgroup$ – Qudit May 3 '15 at 17:10
11
$\begingroup$

The ring $2\Bbb{Z}/10\Bbb{Z} = \{0, 2, 4, 6, 8\}$ has $6$ as its multiplicative identity. There are many examples like this, but many authors/mathematicians tend to rule these out when talking about rings for a good reason: the inclusion $2\Bbb{Z}/10\Bbb{Z} \hookrightarrow \Bbb{Z}/10\Bbb{Z}$ ought to be a ring homomorphism, mapping $1 \mapsto 1$.


Edit: to address comments and clarify. I tend to to think of ring meaning unital ring and homomorphism to mean unital ring homomorphism. In this way, they form a category, which is a nice way of thinking of rings and the ways that they map to one another in one cohesive context. In many rich areas of algebra (representation theory of groups, Hopf algebras, quantum groups, etc.) the rings not only have units, but counits as well (more structure!)

However, there are situations in analysis where units are not available. For example, $L^2(\Bbb{R})$, the space of functions $f: \Bbb{R} \to \Bbb{C}$ such that $\int_\Bbb{R} \lvert f(x) \rvert^2 \, dx < \infty$ does not have a multiplicative identity. The constant function $1$ is not square-integrable. And since the arabic numeral $1$ resembles the roman letter $i$ (especially uppercase $I$), these rings without multiplicative identity are sometimes called rngs. :-)

$\endgroup$
  • 2
    $\begingroup$ What do you mean they rule these out? $2\mathbb{Z}/10\mathbb{Z}$ is a ring, but the inclusion you state is not a ring homomorphism, that's perfectly reasonable. $\endgroup$ – Michael Albanese May 3 '15 at 4:59
  • 3
    $\begingroup$ To expand on @MichaelAlbanese's comment: In fact, $2 \Bbb Z / 10 \Bbb Z \hookrightarrow \Bbb Z / 10 \Bbb Z$ is a homomorphism of (not-necessarily-unital) rings, just not (as you say) a homomorphism of unital rings. But this is true for the more primitive reason that the former ring doesn't contain the unit of the latter. In this way, the fact that the ring operations inherited from the latter admit a unit is irrelevant, as unital and non-necessarily-until rings are simply different classes of objects. $\endgroup$ – Travis May 3 '15 at 5:04
  • $\begingroup$ This is pretty much the same as Michael Albanese's earlier answer, no? $\endgroup$ – Gerry Myerson May 3 '15 at 6:54
  • 7
    $\begingroup$ @GerryMyerson: The example is the same, but we answered only minutes apart. $\endgroup$ – Michael Albanese May 3 '15 at 17:14
6
$\begingroup$

It depends on what exactly you mean by $1$. If you mean the real number $1$, then no an arbitrary ring has nothing to do with real numbers and hence nothing to do with the real $1$. However, $1$ is often understood to be defined as an object in a larger structure (here a ring) such that $1x = x1 = x$ for any $x$ in that structure, and this is exactly the definition of the multiplicative identity.

More important is the question of why we define the multiplicative identity to be so. The answer is that then multiplying by $1$ (whatever multiplying may mean) does not change anything. The next question is why we should have two identities, one for addition and one for multiplication, with distributivity of multiplication over addition. The answer is that addition represents some sort of combining while multiplication represents some sort of transforming (in general at least), where often $xy$ actually means "transform $y$ in a way specified by $x$", and so is not always commutative. It is very nice when the transformation commutes with combination, which is exactly the meaning of distributivity. Associativity is a natural part of the idea of combining and transforming. Additive inverse is nice because it corresponds to being able to undo a combination. Combinations of various kinds are also commutative. Thus we get all our ring axioms.

$\endgroup$
4
$\begingroup$

Although "multiply a ring element by another element of the same ring" and "multiply a ring element by an integer" are both called "multiplication", they are actually two distinct operations and should be viewed as such.

Multiplication of ring elements by integers may be defined according to the following:

  • Multiplying a ring element by the integer zero yields the ring's additive identity.
  • Multiplying a ring element by the integer one yields the same ring element.
  • Multiplying a ring element by the whole number one yields the same ring element.
  • For any integers x and y, multiplying a ring element by (x+y) is equivalent to multiplying by x, multiplying by y, and adding the results (using the ring's addition operator).

Some kinds of ring have a basis element such that any element of the ring will be an integer multiple of the multiplicative identity; in such cases, it may make sense to map integers to rings by saying that the integer N maps to N times the additive identity. For such rings, the integer 1 will map to the additive identity, by definition. For rings which do not have such a basis, ring elements may "look like" whole numbers, but will generally have no relation to them.

$\endgroup$
2
$\begingroup$

According to the kind explanation of @user21820, specifically this: "whatever multiplying may mean" I want also to make my own tryout of answer with a sample of algebra whose multiplicative identity is number i-based:

For instance the family of algebras $\Bbb A_i$ whose multiplication function $*_i$ is declared as $a*_ib=\frac{a*b}{i}$. In this case the number $i$ will be the multiplicative identity of each $\Bbb A_i$ because $a*_ii= \frac{a*i}{i}=a$

$\endgroup$
2
$\begingroup$

Let $R$ be a commutative ring with unity. Let $a$ be an idempotent element of $R$ (i.e., $a^2=a$), then the subring $aR=\{ar : r\in R \}$ has $a$ as its unity.

So for instance in $\mathbb{Z}_{30}$, $21$ is an idempotent element. And so $21$ is the identity of $21\mathbb{Z}_{30}=\{0,3,6,9,12,15,18,21,24,27\}$.

$\endgroup$
  • $\begingroup$ thank you, I add a link to the definition of idempotent elements, I needed to read it to follow your kind explanation :) en.m.wikipedia.org/wiki/Idempotent_element $\endgroup$ – iadvd May 7 '15 at 3:27
  • $\begingroup$ Bundrick and Leeson, in their ESSENTIALS OF ABSTRACT ALGEBRA (1972), state, on p. 119, “We are now in a position to construct easily a subring of a ring with identity where the subring has an identity different from the ring identity.” $\endgroup$ – Mike Jones Nov 2 '15 at 13:27

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.