Ring কী?

Example 7.18

$\Z, \Q, \R, \C$ are commutative rings; $\Z^\times = \{1, -1\}$, $\Q^\times = \Q^*$. Later: polynomial rings $K[X]$ (Chapter 8), matrix rings (non-commutative, Chapter 21), and $\Z/n\Z$ below. In every ring, $0 \times a = 0$ (from distributivity: $0a = (0+0)a = 0a + 0a$), and $(-1)a = -a$.

Example 7.19 (Idempotents: new phenomena in new rings)

In $\Z$, the equation $x^2 = x$, i.e. $x(x - 1) = 0$, has only the solutions $0$ and $1$. In $\Z/6\Z$, testing all classes: $\overline0^2 = \overline0$, $\overline1^2 = \overline1$, $\overline3^2 = \overline9 = \overline3$ and $\overline4^2 = \overline{16} = \overline4$ — four idempotents. The two exotic ones come from zero divisors: $\overline3\,(\overline3 - \overline1) = \overline3 \times \overline2 = \overline6 = \overline0$ with neither factor zero. Such computations calibrate one’s instincts: familiar facts about equations survive in integral domains and fields, but a general ring can and does behave differently — see also the Boolean rings of Exercise 7.10, where every element is idempotent.

Example 7.21 (The binomial theorem in an unfamiliar ring)

Two quick payoffs of the generality. In $\Z/p\Z$ ($p$ prime), the middle binomial coefficients vanish (Theorem 6.23’s first step), so the theorem collapses to the freshman’s dream

a genuine identity there, however criminal it looks over $\R$. And in any commutative ring containing an element $\varepsilon$ with $\varepsilon^2 = 0$, the theorem truncates: $(a + \varepsilon)^n = a^n + n\,a^{n-1}\varepsilon$, all higher terms carrying a factor $\varepsilon^2 = 0$. The coefficient $n\,a^{n-1}$ of $\varepsilon$ is the derivative of $x^n$ — no accident, and a first hint that derivatives are algebra as much as analysis (compare the formal derivative of Chapter 8).