Are there rings whose multiplicative identity is not the number 1 or number 1-based?

Solution 1:

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

Solution 2:

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 $R$.

Solution 3:

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".

Solution 4:

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.

Solution 5:

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. :-)