Difference between revisions of "Ideal"

Revision as of 20:24, 19 June 2008

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>

@@ Line 26: / Line 26: @@
 If <math>(\mathfrak{a}_i)_{i\in I}</math> is a set of (left, right, two-sided) ideals of <math>A</math>, then the (left, two sided) ideal generated by <math>\bigcup_{i\in I} \mathfrak{a}_i</math> is the set of elements of the form <math>\sum_i x_i</math>, where <math>x_i</math> is an element of <math>\mathfrak{a}_i</math> and <math>(x_i)_{i\in I}</math> is a family of finite support.  For this reason, the ideal generated by the <math>\mathfrak{a}_i</math> is sometimes denoted <math>\sum_{i\in I} \mathfrak{a}_i</math>.
+== Multiplication of Ideals ==
+If <math>\mathfrak{a}</math> and <math>\mathfrak{b}</math> are two-sided ideals of a ring <math>A</math>, then the set of elements of the form <math>\sum_{i\in I} a_ib_i</math>, for <math>a_i \in \mathfrak{a}</math> and <math>b_i \in \mathfrak{b}</math>, is also an ideal of <math>A</math>.  It is called the product of <math>\mathfrak{a}</math> and <math>\mathfrak{b}</math>, and it is denoted <math>\mathfrak{ab}</math>.  It is generated by the elements of the form <math>ab</math>, for <math>a\in \mathfrak{a}</math> and <math>b\in \mathfrak{b}</math>.  Since <math>\mathfrak{a}</math> and <math>\mathfrak{b}</math> are two-sided ideals, <math>\mathfrak{ab}</math> is a subset of both <math>\mathfrak{a}</math> and of <math>\mathfrak{b}</math>, so
+<cmath> \mathfrak{ab \subseteq a \cap b} . </cmath>
+The ideals of <math>A</math> constitute a [[pseudo-ring]].
+'''Proposition 1.''' Let <math>\mathfrak{a}</math> and <math>\mathfrak{b_1, \dotsc, b}_n</math> be two-sided ideals of a ring <math>A</math> such that <math>\mathfrak{a+b_i} = A</math>, for each index <math>i</math>.  Then
+<cmath> A = \mathfrak{a + b}_1 \dotsc \mathfrak{b}_n = \mathfrak{ a} + \bigcap_i \mathfrak{b}_i . </cmath>
+''Proof.''  We induct on <math>n</math>.  For <math>n=1</math>, the proposition is degenerately true.
+Now, suppose the proposition holds for <math>n-1</math>.  Then
+<cmath> 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, </cmath>
+which proves the proposition.  <math>\blacksquare</math>
 ==Problems==
@@ Line 34: / Line 49: @@
 * [[Quotient ring]]
 * [[Subring]]
+* [[Normal subgroup]]
 * [[Krull's Theorem]]
 * [[Pseudo-ring]]
 [[Category:Ring theory]]

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