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Ideal

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In ring theory, an ideal is a special kind of subset of a ring. Two-sided ideals in rings are the kernels of ring homomorphisms; in this way, two-sided ideals of rings are similar to normal subgroups of groups.

Specifially, if A is a ring, a subset \mathfrak{a} of A is called a left ideal of A if it is a subgroup under addition, and if xa \in \alpha, for all x\in R and a\in \mathfrak{a}. Symbolically, this can be written as 0\in \mathfrak{a}, \qquad \mathfrak{a+a\subseteq a}, \qquad A \mathfrak{a \subseteq a} . A right ideal is defined similarly, but with the modification \mathfrak{a}A \subseteq \mathfrak{a}. If \mathfrak{a} is both a left ideal and a right ideal, it is called a two-sided ideal. In a commutative ring, all three kinds of ideals are the same; they are simply called ideals. Note that the right ideals of a ring A are exactly the left ideals of the opposite ring A^0.

An ideal has the structure of a pseudo-ring, that is, a structure that satisfies the properties of rings, except possibly for the existence of a multiplicative identity.

Contents

Examples of Ideals; Types of Ideals

In the ring \mathbb{Z}, the ideals are the rings of the form n \mathbb{Z}, for some integer n.

In a field F, the only ideals are the set \{0\} and F itself.

In general, if A is a ring and x is an element of A, the set Ax is a left ideal of A. Ideals of this form are known as principal ideals.

By abuse of language, a (left, right, two-sided) ideal of a ring A is called maximal if it is a maximal element of the set of (left, right, two-sided) ideals distinct from A. A two-sided ideal \mathfrak{m} is maximal if and only if its quotient ring A/\mathfrak{m} is a field.

An ideal \mathfrak{p} is called a prime ideal if ab\in \mathfrak{p} implies a\in \mathfrak{p} or b \in \mathfrak{p}. A two-sided ideal \mathfrak{p} is prime if and only if its quotient ring A/ \mathfrak{p} is a domain. In commutative algebra, the notion of prime ideal is central; it generalizes the notion of prime numbers in \mathbb{Z}.

Generated Ideals

Let A be a ring, and let (x_i)_{i\in I} be a family of elements of A. The left ideal generated by the family (x_i)_{i\in I} is the set of elements of A of the form \sum_{i \in I} a_i x_i, where (a_i)_{i \in I} is a family of elements of A of finite support, as this set is a left ideal of A, thanks to distributivity, and every element of the set must be in every left ideal containing (x_i)_{i\in I}. Similarly, the two-sided ideal generated by (x_i)_{i\in I} is the set of elements of A of the form \sum_{i\in I} a_i x_i b_i, where (a_i)_{i\in I} and (b_i)_{i \in I} are families of finite support.

The two-sided ideal generated by a finite family {a_1, \dotsc, a_n} is often denoted (a_1, \dotsc, a_n).

If (\mathfrak{a}_i)_{i\in I} is a set of (left, right, two-sided) ideals of A, then the (left, two sided) ideal generated by \bigcup_{i\in I} \mathfrak{a}_i is the set of elements of the form \sum_i x_i, where x_i is an element of \mathfrak{a}_i and (x_i)_{i\in I} is a family of finite support. For this reason, the ideal generated by the \mathfrak{a}_i is sometimes denoted \sum_{i\in I} \mathfrak{a}_i.

Multiplication of Ideals

If \mathfrak{a} and \mathfrak{b} are two-sided ideals of a ring A, then the set of elements of the form \sum_{i\in I} a_ib_i, for a_i \in \mathfrak{a} and b_i \in \mathfrak{b}, is also an ideal of A. It is called the product of \mathfrak{a} and \mathfrak{b}, and it is denoted \mathfrak{ab}. It is generated by the elements of the form ab, for a\in \mathfrak{a} and b\in \mathfrak{b}. Since \mathfrak{a} and \mathfrak{b} are two-sided ideals, \mathfrak{ab} is a subset of both \mathfrak{a} and of \mathfrak{b}, so \mathfrak{ab \subseteq a \cap b} .

Proposition 1. Let \mathfrak{a} and \mathfrak{b_1, \dotsc, b}_n be two-sided ideals of a ring A such that \mathfrak{a+b_i} = A, for each index i. Then A = \mathfrak{a + b}_1 \dotsc \mathfrak{b}_n = \mathfrak{ a} + \bigcap_i \mathfrak{b}_i .

Proof. We induct on n. For n=1, the proposition is degenerately true.

Now, suppose the proposition holds for n-1. Then \begin{align*} A = (\mathfrak{a + b}_1 \dotsm \mathfrak{b}_{n-1})(\mathfrak{a} + \mathfrak{b}_n) &= \mathfrak{a \cdot a} ... which proves the proposition. \blacksquare

Proposition 2. Let \mathfrak{b}_1, \dotsc \mathfrak{b}_n be two-sided ideals of a ring A such that \mathfrak{b}_i + \mathfrak{b}_j = A, for any distinct indices i and j. Then \bigcap_{1\le i \le n} \mathfrak{b}_i = \sum_{\sigma \in S_n} \mathfrak{b}_{\sigma(1)} \dotsm \mathfrak{b}_{\sigma(n)} , where S_n is the symmetric group on \{ 1, \dotsc, n\}.

Proof. It is evident that \sum_{\sigma \in S_n} \mathfrak{b}_{\sigma(1)} \dotsm \mathfrak{b}_{\sigma(n)} \subseteq \bigcap_{1\le i \le n} \mathfrak{b}_... We prove the converse by induction on n.

For n=1, the statement is trivial. For n=2, we note that 1 can be expressed as b_1 + b_2, where b_i \in \mathfrak{b}_i. Thus for any b\in \mathfrak{b}_1 \cap \mathfrak{b}_2, b = (b_1+b_2)b = b_1b + b_2b \in \mathfrak{b}_1 \mathfrak{b}_2 + \mathfrak{b}_2 \mathfrak{b}_1 .

Now, suppose that the statement holds for the integer n-1. Then by the previous proposition, \mathfrak{b}_n + \bigcap_{1\le i \le n-1} \mathfrak{b}_i = A, so from the case n=2, \begin{align*}\bigcap_{1\le i \le n} \mathfrak{b}_i &= \mathfrak{b}_n \bigcap_{1\le i \le n-1} \mathfrak{b}_i \\&\sub... as desired. \blacksquare

Problems

Problem 1

See also

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