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Ideal

Revision as of 21:24, 19 June 2008 by Boy Soprano II (talk | contribs) (added stuff about ideal multiplication)

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 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 existance of a multiplicative identity.

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 $A$.

Examples 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 principle ideals.

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.

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 \[\mathfrak{ab \subseteq a \cap b} .\] The ideals of $A$ constitute a pseudo-ring.

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 \[A = (\mathfrak{a + b}_1 \dotsm \mathfrak{b}_{n-1})(\mathfrak{a} + \mathfrak{b}_n) = \mathfrak{a \cdot a} + \mathfrak{ab}_n + \mathfrak{b}_1 \dotsm \mathfrak{b}_{n-1} \mathfrak{a} + \mathfrak{b}_1 \dotsm \mathfrak{b}_n = \mathfrak{a} + \mathfrak{b}_1 \dotsm \mathfrak{b}_n \subseteq \mathfrak{ a} + \bigcap_i \mathfrak{b}_i,\] which proves the proposition. $\blacksquare$

Problems

<url>viewtopic.php?t=174516 Problem 1</url>

See also

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