Ideal কী?

অন্য নাম: quotient ring

Definition 1.25 University Mathematics — Year 2 · অধ্যায় 1 — Sets and Structures

Let $A$ be a commutative ring. An ideal $I \subseteq A$ is an additive subgroup such that $a x \in I$ for all $a \in A$ , $x \in I$ . Kernels of ring morphisms are ideals; $I = A$ iff $1 \in I$ iff $I$ contains a unit. The ideal generated by $x$ is $xA = \{xa\}$ (a principal ideal).

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Example 1.27 (A polynomial gcd, two ways)

Compute $\gcd(X^3 - 1,\ X^2 - 1)$ in $\Q[X]$ . By Euclid:

X^3 - 1 = X\,(X^2 - 1) + (X - 1), \qquad X^2 - 1 = (X + 1)(X - 1) + 0 ,

so the gcd is $X - 1$ , and back-substitution gives the Bézout relation

X - 1 = 1\cdot(X^3 - 1) - X\cdot(X^2 - 1).

By ideals: the ideal $(X^3 - 1)\Q[X] + (X^2 - 1)\Q[X]$ is principal (Theorem 1.26); it contains $X - 1$ (the display) and is contained in $(X - 1)\Q[X]$ (both generators vanish at $1$ , hence are multiples of $X - 1$ ): the monic generator is $X - 1$ . The closing insight: the ideal viewpoint identifies the gcd without dividing — common roots locate the ideal, and Euclid merely certifies it.

Example 1.34 (An evaluation morphism and its kernel)

Take $A = \begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}$ and the evaluation $\varepsilon_A \colon \R[X] \to \mathcal{M}_2(\R)$ , $P \mapsto P(A)$ . Since $A^2 = 0$ ,

P(A) = P(0)\,I + P'(0)\,A = \begin{pmatrix} P(0) & P'(0)\\ 0 & P(0)\end{pmatrix},

(only the constant and linear terms of $P$ survive). Hence $\ker\varepsilon_A = \{P : P(0) = P'(0) = 0\} = X^2\,\R[X]$ : a principal ideal, exactly as Theorem 1.26 predicts, generated by the monic $X^2$ of least degree in the kernel — the minimal polynomial of $A$ , star of Chapter 3. The image is the two-dimensional commutative algebra $\{aI + bA\}$ : evaluation morphisms shrink the infinite-dimensional $\R[X]$ onto small, computable algebras.

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Definition 2.1 University Mathematics — Year 3 · অধ্যায় 2 — Rings and Arithmetic

An ideal $I$ of a ring $A$ is an additive subgroup such that $AI \subseteq I$ . The quotient ring $A/I$ is the quotient group $(A, +)/I$ with the multiplication $(a + I)(b + I) = ab + I$ : well defined, since changing $a$ to $a + x$ , $b$ to $b + y$ ( $x, y \in I$ ) changes $ab$ by $ay + xb + xy \in I$ . The projection $\pi \colon A \to A/I$ is a surjective ring morphism with kernel $I$ , and kernels of ring morphisms are exactly the ideals.

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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

A = \Z[\iu\sqrt5] = \{a + \iu b\sqrt5 : a, b \in \Z\}, \qquad N(a + \iu b\sqrt5) = a^2 + 5b^2,

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

6 = 2 \cdot 3 = (1 + \iu\sqrt5)(1 - \iu\sqrt5).

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.

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