associate elements কী?

Example 2.20 (A ring without unique factorization)

None of the implications Euclidean $\Rightarrow$ PID $\Rightarrow$ UFD is an equivalence, and the failure of the last is worth seeing once in complete detail. In

the norm is multiplicative and $N(z) = 1$ iff $z \in A^\times = \{\pm1\}$. Consider

All four factors are irreducible: their norms are $4, 9, 6, 6$, and a proper factorization $z = z_1z_2$ would force $N(z_1) \in \{2, 3\}$ — but $a^2 + 5b^2$ never equals $2$ or $3$ ($b = 0$ leaves the non-squares $2, 3$; $\abs b \geq 1$ gives $\geq 5$). Yet $2$ is associate to neither $1 \pm \iu\sqrt5$ (norms $4 \neq 6$): two genuinely different factorizations of $6$ into irreducibles. Equivalently, irreducible $\neq$ prime here: $2$ divides the product $(1 + \iu\sqrt5)(1 - \iu\sqrt5) = 6$ but neither factor (norms again). The ideal-theoretic repair of this failure — factorizing ideals rather than elements — is the birth of algebraic number theory; at our level, the example calibrates how special the Euclidean rings $\Z$, $K[X]$, $\Z[\iu]$ of this chapter really are.

Example 2.26

$X^n - p$ is irreducible over $\Q$ for every prime $p$ and $n \geq 1$ (Eisenstein at $p$): there are irreducible polynomials of every degree over $\Q$ — in stark contrast with $\C$ (degree $1$, d’Alembert–Gauss, proved in Chapter 16) and $\R$ (degrees $1, 2$). The trick of shifting enlarges Eisenstein’s reach: the $p$-th cyclotomic polynomial $\Phi_p = X^{p-1} + \dots + X + 1 = \frac{X^p - 1}{X - 1}$ has

Eisenstein at $p$ ($p \mid \binom pk$ for $0 < k < p$, and $\binom{p}{p-1} = p \not\equiv 0 \bmod p^2$): $\Phi_p(X+1)$, hence $\Phi_p$, is irreducible over $\Q$. This is the algebraic heart of the $17$-gon story told in Chapter 4.