Integral domain, field কী?

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 7.26 (How many square roots of $1$?)

Solve $x^2 = \overline 1$ in $\Z/8\Z$ and in $\Z/7\Z$. Testing the eight classes mod $8$: $1^2 = 1$, $3^2 = 9 \equiv 1$, $5^2 = 25 \equiv 1$, $7^2 = 49 \equiv 1$ — four solutions $\{\overline1, \overline3, \overline5, \overline7\}$, even though the polynomial $X^2 - 1$ has degree $2$. In the field $\Z/7\Z$, by contrast, $x^2 = \overline1$ means $(x - \overline1)(x + \overline1) = \overline0$, and a field has no zero divisors: $x = \pm\overline1$, two solutions only. The failure mod $8$ is traceable: $(3-1)(3+1) = 2 \times 4 = 8 \equiv 0$ without either factor vanishing. Moral: the familiar rule “a degree-$d$ equation has at most $d$ roots” is a theorem about integral domains (Corollary 8.8 proves it over fields); in rings with zero divisors it silently fails — which is exactly why the pairing proof of Wilson’s theorem (Exercise 6.11) needed $p$ prime.